![INTRODUCTION TO
Signal Processing
Solutions Manual
Sophocles J. Orfanidis
Department of Electrical & Computer Engineering
Rutgers University, Piscataway, NJ 08855
orfanidi@ece.rutgers.edu
Copyright © 2010 by Sophocles J. Orfanidis
Web page: www.ece.rutgers.edu/~orfanidi/i2sp
Chapter 1 Problems
Problem 1.1
The Nyquist interval is [−fs/2, fs/2]= [−4, 4] Hz. The 6 Hz frequency of the wheel lies outside
it, therefore, it will be aliased with f − fs = 6 − 8 = −2 Hz. Thus, the wheel will appear to be
turning at 2 Hz in the opposite direction.
If fs = 12, the Nyquist interval is [−6, 6]. Points on the wheel will appear to be moving up and
down, with either positive or negative sense of rotation.
For the other two sampling frequencies, the Nyquist interval is [−8, 8] or [−12, 12] Hz, and
therefore, the original 6 Hz frequency lies in it and no aliasing will be perceived.
Problem 1.2
The three terms of the signal correspond to frequencies f1 = 1, f2 = 4, and f3 = 6 Hz. Of
these, f2 and f3 lie outside the Nyquist interval [−2.5, 2.5]. Therefore, they will be aliased with
f2 − fs = 4 − 5 = −1 and f3 − fs = 6 − 5 = 1 and the aliased signal will be:
xa(t)= 10 sin(2πt)+10 sin(2π(−1)t)+5 sin(2πt)= 5 sin(2πt)
To show that they have the same sample values, we set t = nT, with T = 1/fs = 1/5 sec. Then,
x(nT)= 10 sin(2πn/5)+10 sin(8πn/5)+5 sin(12πn/5)
But,
sin(8πn/5)= sin(2πn − 2πn/5)= − sin(2πn/5)
and
sin(12πn/5)= sin(2πn + 2πn/5)= sin(2πn/5)
Thus,
x(nT) = 10 sin(2πn/5)−10 sin(2πn/5)+5 sin(2πn/5)
= 5 sin(2πn/5)= xa(nT).
If fs = 10 Hz, then the Nyquist interval is [−5, 5] Hz and only f3 lies outside it. It will be aliased
with f3 − fs = 6 − 10 = −4 resulting in the aliased signal:
xa(t)= 10 sin(2πt)+10 sin(8πt)+5 sin(2π(−4)t)= 10 sin(2πt)+5 sin(8πt)
Problem 1.3
Using the trig identity 2 sin α sin β = cos(α − β)− cos(α + β), we find:
x(t) = cos(5πt)+4 sin(2πt)sin(3πt)= cos(5πt)+2[cos(πt)− cos(5πt)]
= 2 cos(πt)− cos(5πt)
The frequencies in the signal are f1 = 0.5 and f2 = 2.5 kHz. The Nyquist interval is [−1.5, 1.5]
kHz, and f2 lies outside it. Thus, it will be aliased with f2a = 2.5 − 3 = −0.5 giving rise to the
signal:
1](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-1-320.jpg)
![xa(t)= 2 cos(2πf1t)− cos(2πf2at)= 2 cos(πt)− cos(−πt)= cos(πt)
A class of signals aliased with x(t) and xa(t) is obtained by replacing f1 and f2 by their shifted
versions: f1 + mfs, f2 + nfs resulting in:
xmn(t)= 2 cos(πt + 6πmt)− cos(πt − 6πnt)
Problem 1.4
Using a trig identity, we write x(t)= cos(8πt)+ cos(10πt)+ cos(2πt). The frequencies con-
tained in x(t) are thus, 1 Hz, 4 Hz, and 5 Hz. If the sampling rate is 5 Hz, then the Nyquist
interval is [−2.5, 2.5] Hz, and therefore, the 4 Hz and 5 Hz components lie outside it and will be
aliased with the frequencies 4 − 5 = −1 Hz and 5 − 5 = 0 Hz, etc.
Problem 1.5
Using trig identities we find:
x(t)= sin(6πt) 1 + 2 cos(4πt) = sin(6πt)+ sin(10πt)+ sin(2πt)
with frequency content: f1 = 3, f2 = 5, f3 = 1 kHz. The Nyquist interval is [−2, 2] kHz, and the
aliased frequencies are:
f1a = f1 − fs = 3 − 4 = −1, f2a = f2 − fs = 5 − 4 = 1, f3a = f3 = 1
Thus,
xa(t)= sin(−2πt)+ sin(2πt)+ sin(2πt)= sin(2πt)
Problem 1.6
Use the trigonometric identity 2 cos a cos b = cos(a + b)+ cos(a − b) three times to get
x(t) = 2[cos(10πt)+ cos(6πt)]cos(12πt)
= cos(22πt)+ cos(2πt)+ cos(18πt)+ cos(6πt)
The frequencies present in this signal are f1 = 11, f2 = 1, f3 = 9, and f4 = 3 Hz. With a sampling
rate of 10 Hz, only f1 and f3 lie outside the Nyquist interval [−5, 5] Hz, and they will be aliased
with f1 − fs = 11 − 10 = 1 Hz and f3 − fs = 9 − 10 = −1 Hz. The aliased signal will be:
xa(t) = cos(2π(1)t)+ cos(2πt)+ cos(2π(−1)t)+ cos(6πt)
= 3 cos(2πt)+ cos(6πt)
To prove the equality of the samples, replace t = nT = n/10, because T = 1/fs = 1/10. Then,
x(nT)= cos(22πn/10)+ cos(2πn/10)+ cos(18πn/10)+ cos(6πn/10)
But, cos(22πn/10)= cos(2πn+2πn/10)= cos(2πn/10) and similarly, cos(18πn/10)= cos(2πn−
2πn/10)= cos(2πn/10). Therefore, the sample values become
2
x(nT) = cos(2πn/10)+ cos(2πn/10)+ cos(2πn/10)+ cos(6πn/10)
= 3 cos(2πn/10)+ cos(6πn/10)= xa(nT)
If fs = 12 Hz, then f1 and f3 will lie outside of [−6, 6] and will be aliased with 11 − 12 = −1 and
9 − 12 = −3. The aliased signal will be:
xa(t) = cos(2π(−1)t)+ cos(2πt)+ cos(2π(−3)t)+ cos(6πt)
= 2 cos(2πt)+2 cos(6πt)
Problem 1.7
We use the same technique as in the square-wave Example 1.4.6. At a sampling rate of 8 Hz, the
signal frequencies of
{1, 3, 5, 7, 9, 11, 13, 15, . . . }
will be aliased with:
{1, 3, −3, −1, 1, 3, −3, −1, . . . }
Therefore only sin(2πt) and sin(6πt) terms will appear in the aliased signal. Thus, we write it
in the form:
xa(t)= B sin(2πt)+C sin(6πt)
To determine B and C, we demand that xa(t) and x(t) agree at the sampling instants t = nT =
n/8, because T = 1/fs = 1/8 sec. Therefore, we demand
B sin(2πn/8)+C sin(6πn/8)= x(n/8)
Setting n = 1, 2, we get two equations
B sin(2π/8)+C sin(6π/8)= x(1/8)= 0.5
B sin(4π/8)+C sin(12π/8)= x(2/8)= 1
⇒
B
1
√
2
+ C
1
√
2
= 0.5
B − C = 1
The values for x(1/8) and x(2/8) were obtained by inspecting the triangular waveform. Solving
for B and C, we find:
B =
√
2 + 2
4
, C =
√
2 − 2
4
Problem 1.8
For fs = 5 kHz, we have:
xa(t)= sin(2πf1t)
For fs = 10 kHz, we have:
xa(t)= 2 sin(2πf1t)+ sin(2πf2t)
3](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-2-320.jpg)
![Fig. P1.1 Parts (a,b) of Problem 1.8.
Problem 1.9
The audible and inaudible parts are:
x(t)= sin(10πt)+ sin(20πt)
audible
+ sin(60πt)+ sin(90πt)
inaudible
The frequencies of the four terms and their aliased versions are:
fA = 5
fB = 10
fC = 30
fD = 45
⇒
fAa = 5
fBa = 10
fCa = 30 − 40 = −10
fDa = 45 − 40 = 5
a. When there is no prefilter the aliased signal at the output of the reconstructor will be:
ya(t)= sin(10πt)+ sin(20πt)+ sin(−20πt)+ sin(10πt)= 2 sin(10πt)
b. When there is a perfect prefilter, the fC, fD components are removed prior to sampling
thus, the sampled and reconstructed signal will remain unchanged, that is, the audible
part of x(t):
ya(t)= sin(10πt)+ sin(20πt)
c. The attenuations introduced by the practical prefilter shown in Fig. P1.2 are obtained by
determining the number of octaves to the frequencies fC, fD and multiplying by the filter’s
attenuation of 48 dB/octave:
log2
fC
fs/2
= log2
30
20
= 0.585 ⇒ AC = 48 × 0.585 = 28.08 dB
log2
fD
fs/2
= log2
45
20
= 1.170 ⇒ AD = 48 × 1.170 = 56.16 dB
and in absolute units:
4
|HC| = 10−AC/20
= 10−28.08/20
= 0.0394
|HD| = 10−AD/20
= 10−56.16/20
= 0.0016
Thus, the resulting reconstructed signal will be:
ya(t) = sin(10πt)+ sin(20πt)+|HC| sin(−20πt)+|HD| sin(10πt)
= 1.0016 sin(10πt)+0.9606 sin(20πt)
which agrees closely with the audible part of x(t).
10 3020 45
48 dB/octave
Analog Prefilter
kHz
f
H(f)
0
A B
C
D
Fig. P1.2 Prefilter specifications of Problem 1.9.
Problem 1.10
The Fourier series expansion of the periodic function s(t) is:
s(t)=
∞
m=−∞
cme2πjmt/T
, cm =
1
T
T/2
−T/2
s(t)e−2πjmt/T
dt
But within the basic period [−T/2, T/2], s(t) equals δ(t). Thus,
cm =
1
T
T/2
−T/2
δ(t)e−2πjmt/T
dt =
1
T
Using the delta function integral representation:
∞
−∞
e−2πjft
dt = δ(f)
we find:
S(f) =
∞
−∞
s(t)e−2πjft
dt =
1
T
∞
m=−∞
∞
−∞
s(t)e−2πj(f−mfs)t
dt
=
1
T
∞
m=−∞
δ(f − mfs)
Problem 1.11
This is simply the Fourier series expansion of the periodic function X(f) in the variable f, and
x(nT) are the Fourier series expansion coefficients calculated by integrating over one period,
that is, over [−fs/2, fs/2].
5](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-3-320.jpg)
![B ≥ log2
10
√
12 · 10−3
= 11.495
Therefore, B = 12. The actual rms error will be
erms =
R
2B
√
12
=
10
212
√
12
= 0.705 mvolt.
The dynamic range will be 6B = 6 · 12 = 72 dB.
Problem 2.4
At 44.1 kHz sampling rate, one minute of music has 60 sec containing 60 × 44100 samples. And,
for two stereo channels, 2 × 60 × 44100 samples. If each sample is quantized with 16 bits, the
total number of bits will be
2 × 60 × 44100 × 16 bits
or, dividing by 8 we get the total number of bytes:
2 × 60 × 44100 × 16
8
bytes
Dividing by 1024, we get kbytes and by another 1024, Mbytes:
2 × 60 × 44100 × 16
8 × 1024 × 1024
= 10.09 Mbytes
Problem 2.5
Following similar steps as in the previous problem, we have for part (a):
(3 × 60 sec)(16 channels)(48000 samples/sec)(20 bits/sample)
(8 bits/byte)(10242 bytes/Mbyte)
= 330 Mbytes
that is, practically the entire hard disk of a typical home PC. For part (b), if in each channel we
must perform NMAC = 35 operations on each sample, then these operations must be finished
during the sampling interval T = 1/fs between samples:
NMACTMAC ≤ T ⇒ TMAC ≤
T
NMAC
=
1
NMACfs
which gives for the time per MAC:
TMAC =
1
35 · 48 kHz
= 0.595 · 10−3
msec = 595 nsec
This is plenty of time for today’s DSP chips, which can perform a MAC operation within 30–80
nsec. If a single DSP chip handles all 16 channels, then the total number of MACs that must be
performed within the sampling interval T are 16NMAC. This gives:
TMAC =
1
16 · 35 · 48 kHz
=
595
16
= 37 nsec
which is just within the capability of the fastest DSP chips.
In practice, a typical professional digital recording system would require the processing and
mixing of at least 32 channels. The above results suggest that one use at least two DSPs to
multiplex the required operations.
16
Problem 2.6
The mean is
me =
0
−Q
ep(e) de =
1
Q
0
−Q
e de = −
Q
2
and the variance:
σ2
e = E[(e − me)2
]=
0
−Q
(e − me)2
p(e) de =
1
Q
0
−Q
e +
Q
2
2
de =
Q2
12
Problem 2.7
Solving Eq. (2.2.10) for L in terms of ΔB, we find:
L =
22ΔB
π2p
2p + 1
1
2p+1
(P2.1)
With ΔB = 15 and p = 1, 2, 3, we find:
L = 1523, 116, 40
The oversampled audio rates will be, with fs = 44.1 kHz:
Lfs = 67.164, 5.114, 1.764 MHz
Problem 2.8
Using Eq. (P2.1) with ΔB = 7 bits and p = 1, 2, 3, we find:
L = 38, 13, 8
The oversampled speech rates will be, with fs = 8 kHz:
Lfs = 304, 104, 48 kHz
Problem 2.9
Replacing b1 = 1 − b1, we have:
xQ = R(b12−1
+ b22−2
+ · · · + bB2−B
− 0.5)
= R (1 − b1)2−1
+ b22−2
+ · · · + bB2−B
− 0.5
= R −b12−1
+ b22−2
+ · · · + bB2−B
where the first 2−1
canceled the last −0.5.
Problem 2.10
As an example, consider the polynomial of degree M = 3: B(z)= b0 +b1z+b2z2
+b3z3
. Iterating
the algorithm for i = 3, 2, 1, 0, we find:
p = 0
p = ap + b3 = b3
p = ap + b2 = ab3 + b2
p = ap + b1 = a(ab3 + b2)+b1 = a2
b3 + ab2 + b1
p = ap + b0 = a(a2
b3 + ab2 + b1)+b0 = a3
b3 + a2
b2 + ab1 + b0 = B(a)
17](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-9-320.jpg)
![Problem 2.11
/* pol.c - polynomial evaluator */
double pol(M, b, a)
int M;
double *b, a;
int i;
double p = 0;
for (i=M; i>=0; i--)
p = a * p + b[i];
return p;
Problem 2.12
Consider the division of B(z) by (z − a):
B(z)= (z − a)Q(z)+R
where the quotient polynomial Q(z) must have degree M − 1, and the remainder R must be a
constant. Indeed, setting z = a, gives R = b(a). So the objective of the algorithm is to compute
R iteratively. Inserting the polynomial expressions we have:
bMzM
+ bM−1zM−1
+ · · · + b1z + b0 = (z − a)(qM−1zM−1
+ qM−2zM−2
+ · · · + q1z + q0)+R
Equating like powers of z, we find:
bM = qM−1
bM−1 = qM−2 − aqM−1
bM−2 = qM−3 − aqM−2
· · ·
b1 = q0 − aq1
b0 = R − aq0
⇒
qM−1 = bM
qM−2 = aqM−1 + bM−1
qM−3 = aqM−2 + bM−2
· · ·
q0 = aq1 + b1
R = aq0 + b0
Problem 2.13
For a polynomial of degree M = 3, B(z)= b1z + b2z2
+ b3z3
, we iterate the algorithm for i = 3,
2, 1:
p = 0
p = a(p + b3)= ab3
p = a(p + b2)= a(ab3 + b2)= a2
b3 + ab2
p = a(p + b1)= a(a2
b3 + ab2 + b1)= a3
b3 + a2
b2 + ab1 = B(a)
18
natural binary offset binary 2’s C
b1b2b3b4 m xQ = Qm m xQ = Qm b1b2b3b4
— 16 8.0 8 4.0 —
1 1 1 1 15 7.5 7 3.5 0 1 1 1
1 1 1 0 14 7.0 6 3.0 0 1 1 0
1 1 0 1 13 6.5 5 2.5 0 1 0 1
1 1 0 0 12 6.0 4 2.0 0 1 0 0
1 0 1 1 11 5.5 3 1.5 0 0 1 1
1 0 1 0 10 5.0 2 1.0 0 0 1 0
1 0 0 1 9 4.5 1 0.5 0 0 0 1
1 0 0 0 8 4.0 0 0.0 0 0 0 0
0 1 1 1 7 3.5 −1 −0.5 1 1 1 1
0 1 1 0 6 3.0 −2 −1.0 1 1 1 0
0 1 0 1 5 2.5 −3 −1.5 1 1 0 1
0 1 0 0 4 2.0 −4 −2.0 1 1 0 0
0 0 1 1 3 1.5 −5 −2.5 1 0 1 1
0 0 1 0 2 1.0 −6 −3.0 1 0 1 0
0 0 0 1 1 0.5 −7 −3.5 1 0 0 1
0 0 0 0 0 0.0 −8 −4.0 1 0 0 0
Table P2.1 Converter codes for B = 4 bits, R = 8 volts.
Problem 2.14
The quantization width is Q = R/2B
= 8/24
= 0.5 volts. Table P2.1 shows the quantization
levels and their binary codes.
To convert x = 1.2 by rounding, we shift it by half the quantization spacing y = x + Q/2 =
1.2 + 0.25 = 1.45. The following tables show the successive approximation tests for the natural
and offset binary cases. The 2’s complement case is obtained by complementing the MSB of the
offset case.
test b1b2b3b4 xQ C = u(y − xQ)
b1 1 0 0 0 4.0 0
b2 0 1 0 0 2.0 0
b3 0 0 1 0 1.0 1
b4 0 0 1 1 1.5 0
0 0 1 0 1.0
test b1b2b3b4 xQ C = u(y − xQ)
b1 1 0 0 0 0.0 1
b2 1 1 0 0 2.0 0
b3 1 0 1 0 1.0 1
b4 1 0 1 1 1.5 0
1 0 1 0 1.0
The natural binary and 2’C cases agree because x lies in the positive range of both quantizers.
For x = 5.2, we have y = x + Q/2 = 5.45. The tests are shown below:
test b1b2b3b4 xQ C = u(y − xQ)
b1 1 0 0 0 4.0 1
b2 1 1 0 0 6.0 0
b3 1 0 1 0 5.0 1
b4 1 0 1 1 5.5 0
1 0 1 0 5.0
test b1b2b3b4 xQ C = u(y − xQ)
b1 1 0 0 0 0.0 1
b2 1 1 0 0 2.0 1
b3 1 1 1 0 3.0 1
b4 1 1 1 1 3.5 1
1 1 1 1 3.5
For the 2’C case, the quantizer saturates to its maximum positive value. Finally, for x = −3.2, we
have y = −2.95, and the following testing tables:
19](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-10-320.jpg)
![test b1b2b3b4 xQ C = u(y − xQ)
b1 1 0 0 0 4.0 0
b2 0 1 0 0 2.0 0
b3 0 0 1 0 1.0 0
b4 0 0 0 1 0.5 0
0 0 0 0 0.0
test b1b2b3b4 xQ C = u(y − xQ)
b1 1 0 0 0 0.0 0
b2 0 1 0 0 −2.0 0
b3 0 0 1 0 −3.0 1
b4 0 0 1 1 −2.5 0
0 0 1 0 −3.0
In this case, the natural binary scale saturates to its minimum positive value.
Problem 2.15
Similar to the testing tables of Examples 2.4.1 and 2.4.2.
Problem 2.16
These versions are straight forward. To implement truncation instead of rounding, replace the
statement within adc:
y = x + Q/2;
by
y = x;
Problem 2.17
This problem is basically the same as Example 2.4.5. The same program segment of that example
applies here for generating the codes and quantized values. Only the full-scale range is different,
R = 32 volts. In part (c), the amplitude of the sinusoid A = 20 exceeds the quantizer range
R/2 = 16, and therefore both the positive and negative peaks of the sinusoid will saturate to the
positive and negative maxima of the quantizer.
Problem 2.18
The following C program carries out all the required operations. It uses the routine corr of
Appendix A.1 to compute all the sample autocorrelations and cross-correlations.
/* exmpl246.c - quantization noise model
*
* default values:
* N=10 bits
* R=1024
* L=1000, iseed=10, M=50
*/
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
void adc(), corr();
double dac();
20
double ran();
void main()
{
int n, i, L, N, M, *b;
long iseed;
double *x, xq, *e, R, Q, *Ree, *Rex, *Rxx;
double normee, normex;
double v, pi = 4 * atan(1.0);
FILE *fpe, *fpx;
FILE *fpRee, *fpRex, *fpRxx;
printf("enter N, R = ");
scanf("%d %lf", &N, &R);
printf("enter L, iseed, M = ");
scanf("%d %ld %d", &L, &iseed, &M);
fpe = fopen("e", "w");
fpx = fopen("x", "w");
fpRee = fopen("Ree", "w");
fpRex = fopen("Rex", "w");
fpRxx = fopen("Rxx", "w");
b = (int *) calloc(N, sizeof(int));
x = (double *) calloc(L, sizeof(double));
e = (double *) calloc(L, sizeof(double));
Ree = (double *) calloc(M+1, sizeof(double));
Rex = (double *) calloc(M+1, sizeof(double));
Rxx = (double *) calloc(M+1, sizeof(double));
Q = R / (1 << N);
for (n=0; n<L; n++) {
v = 0.25 * R * (2 * ran(&iseed) - 1);
x[n] = 0.25 * R * cos(2 * pi * n / sqrt(131.)) + v;
adc(x[n], b, N, R);
xq = dac(b, N, R);
e[n] = x[n] - xq;
fprintf(fpx, "% lfn", x[n]);
fprintf(fpe, "% lfn", e[n]);
}
corr(L, e, e, M, Ree);
corr(L, e, x, M, Rex);
corr(L, x, x, M, Rxx);
normee = Ree[0];
normex = sqrt(Ree[0] * Rxx[0]);
for (i=0; i<=M; i++) {
fprintf(fpRee, "%lfn", Ree[i]/normee);
fprintf(fpRex, "%lfn", Rex[i]/normex);
fprintf(fpRxx, "%lfn", Rxx[i]/Rxx[0]);
}
21](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-11-320.jpg)
![printf("erms = Q/sqrt(12) = %lfn", Q/sqrt(12.0));
printf("Ree(0) = %lfn", sqrt(normee));
}
Problem 2.19
The mean of v is zero because E[u]= 0.5:
E[v]= 0.5R(E[u]−0.5)= 0
The variance of v is related to the variance of u as follows:
σ2
v = E[v2
]= (0.5R)2
E[(u − 0.5)2
]= (0.5R)2
σ2
u = (0.5R)2 1
12
=
R2
48
Problem 2.20
Inserting x(n)= A cos(2πf0n + φ)+v(n) into the definition of the autocorrelation function, we
find:
Rxx(k) = E[x(n + k)x(n)]
= A2
E[cos 2πf0(n + k)+φ cos(2πf0n + φ)]
+ AE[cos 2πf0(n + k)+φ v(n)]+AE[v(n + k)cos(2πf0n + φ)]
+ E[v(n + k)v(n)]
assuming φ and v(n) are independent random variables, the cross terms will vanish. Indeed, by
the independence assumption the expectation value of the product becomes the product of the
expectation values:
E[v(n + k)cos(2πf0n + φ)]= E[v(n + k)]·E[cos(2πf0n + φ)]= 0
where the last zero follows either from the zero-mean property of v(n) and/or from the zero
mean property of E[cos(2πf0n + φ)]= 0. The latter follows from uniformity of φ whose prob-
ability density will be: p(φ)= 1/(2π). We have:
E[cos(a + φ)]=
2π
0
cos(a + φ)p(φ) dφ =
1
2π
2π
0
cos(a + φ) dφ = 0
Similarly, we have the property:
E[cos(a + φ)cos(b + φ)] =
2π
0
cos(a + φ)cos(b + φ)p(φ) dφ
=
1
2π
2π
0
cos(a + φ)cos(b + φ) dφ
=
1
2
cos(a − b)
Thus, we find
E[cos 2πf0(n + k)+φ cos(2πf0n + φ)]=
1
2
cos(2πf0(n + k)−2πf0n)=
1
2
cos(2πf0k)
Finally, because v(n) is assumed to be zero-mean white noise, we will have:
22
E[v(n + k)v(n)]= σ2
vδ(k)
It follows that the autocorrelation of the noisy sinusoid will be:
Rxx(k)=
1
2
A2
cos(2πf0k)+σ2
vδ(k)
At lag k = 0, we have:
Rxx(0)=
1
2
A2
+ σ2
v
Therefore, the normalized autocorrelation function becomes:
ρxx(k)=
Rxx(k)
Rxx(0)
=
A2
/2
A2/2 + σ2
v
cos(2πf0k)+
σ2
v
A2/2 + σ2
v
δ(k)
Defining the parameter:
a =
A2
/2
A2/2 + σ2
v
⇒ 1 − a =
σ2
v
A2/2 + σ2
v
we finally have:
ρxx(k)= a cos(2πf0k)+(1 − a)δ(k)
Defining the signal to noise ratio as the relative strength of the sinusoid versus the noise in the
autocorrelation function, that is,
SNR =
A2
/2
σ2
v
we may express the parameter a as:
a =
SNR
SNR + 1
⇒ 1 − a =
1
SNR + 1
Problem 2.21
The quantization width of the first ADC will be Q1 = R1/2B1 . The residual signal formed at the
output of the first DAC, x2 = x−x1, is the quantization error, and therefore, it varies of the limits
−
Q1
2
≤ x2 ≤
Q1
2
These limits must serve as the full scale range of the second ADC. Thus, we must choose:
R2 = Q1 =
R1
2B1
, Q2 =
R2
2B2
=
R1
2B1+B2
The following program segment illustrates the sequence of operations:
23](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-12-320.jpg)
![R2 = R1 / (1 << B1);
b1 = (int *) calloc(B1, sizeof(int));
b2 = (int *) calloc(B2, sizeof(int));
for (n=0; n<L; n++) {
x = A * cos(2 * pi * f0 * n);
adc(x, b1, B1, R1);
x1 = dac(b1, B1, R1);
x2 = x - x1;
adc(x2, b2, B2, R2);
}
When there is a third B3-bit stage, the total number of bits will be
B = B1 + B2 + B3
The B2-bit output of the second ADC must be passed into a second DAC to get the corresponding
analog value and subtracted from the input x − x1 to get the third error output, which is then
quantized to B3 bits with a full scale range and quantization width:
R3 = Q2 =
R2
2B2
, Q3 =
R3
2B3
=
R1
2B1+B2+B3
The following loop shows these operations:
R2 = R1 / (1 << B1);
R3 = R2 / (1 << B2);
b1 = (int *) calloc(B1, sizeof(int));
b2 = (int *) calloc(B2, sizeof(int));
b3 = (int *) calloc(B3, sizeof(int));
for (n=0; n<L; n++) {
x = A * cos(2 * pi * f0 * n);
adc(x, b1, B1, R1);
x1 = dac(b1, B1, R1);
adc(x-x1, b2, B2, R2);
x2 = dac(b2, B2, R2);
x3 = x - x1 - x2;
adc(x3, b3, B3, R3);
}
Problem 2.22
The quantized triangularly-dithered signals were generated by the for-loop given in Example 2.5.1.
24
Chapter 3 Problems
Problem 3.1
(a) If the input is doubled the output is not doubled. Thus, the system is not linear. Indeed,
the output due to x1(n)= 2x(n) will be y1(n)= 3x1(n)+5 = 6x(n)+5 = 2(3x(n)+5).
The system is, however, time-invariant since it has constant-in-time coefficients.
(b) The quadratic term x2
(n − 1) breaks linearity because it quadruples whenever the input
doubles. The term x(2n) will break time-invariance. Indeed, let xD(n)= x(n − D) be
a delayed input. The corresponding output yD(n) can be computed from the given I/O
equation:
yD(n)= x2
D(n − 1)+xD(2n)= x2
(n − 1 − D)+x(2n − D)
but, this is not the same as the delayed (shifted) y(n) obtained by replacing n by n − D:
y(n − D)= x2
(n − D − 1)+x(2(n − D))= x2
(n − D − 1)+x(2n − 2D)
Thus, the system is neither linear nor time-invariant.
(c) System is time-invariant, but not linear since the output does not double whenever the
input doubles.
(d) The system is linear, but the term nx(n − 3) breaks time-invariance.
(e) The time-dependent term n breaks both time invariance and linearity.
Problem 3.2
Picking out the coefficients of the x-terms, we find:
(a) h = [3, −2, 0, 4]
(b) h = [4, 1, 0, −3]
(c) h = [1, 0, 0, −1]
Problem 3.3
(a,b) These problems are of the general form:
y(n)= ay(n − 1)+x(n)
with a = −0.9 for (a), and a = 0.9 for (b). Setting x(n)= δ(n) we find the difference
equation for the impulse response:
h(n)= ah(n − 1)+δ(n)
Iterating forward in time with zero initial condition h(−1)= 0, we obtain the causal solu-
tion:
h(0)= ah(−1)+δ(0)= a · 0 + 1 = 1
h(1)= ah(0)+δ(1)= a · 1 + 0 = a
h(2)= ah(1)+δ(2)= a · a + 0 = a2
h(3)= ah(2)+δ(3)= a · a2
+ 0 = a3
· · ·
h(n)= ah(n − 1)+δ(n)= a · an−1
= an
, for n ≥ 1
25](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-13-320.jpg)
![Thus, h(n) will be the geometric series:
h = [1, a, a2
, a3
, . . . ]
(c,d) These problems are similar to
y(n)= ay(n − 2)+x(n)
with a = 0.64 = (0.8)2
for (c), and a = −0.81 = −(0.9)2
for (d). The impulse response
will satisfy the difference equation:
h(n)= ah(n − 2)+δ(n)
A few iterations (with zero initial conditions) give:
h(0)= 1
h(1)= 0
h(2)= ah(0)= a
h(3)= ah(1)= 0
h(4)= ah(2)= a2
h(5)= ah(3)= 0, etc.
Thus, all the odd-indexed h(n) are zero:
h = [1, 0, a, 0, a2
, 0, a3
, 0, a4
, 0, . . . ]
(e) The impulse response satisfies:
h(n)= 0.5h(n − 1)+4δ(n)+δ(n − 1)
A few iterations give:
h(0)= 4
h(1)= 0.5h(0)+1 = 3
h(2)= 0.5h(1)= 3(0.5)
h(3)= 0.5h(2)= 3(0.5)2
, etc.
Thus,
h(n)=
⎧
⎨
⎩
4 if n = 0
3(0.5)n−1
if n ≥ 1
Problem 3.4
We have seen that if h(n)= an
u(n), then the I/O difference equation is
y(n)= ay(n − 1)+x(n)
In cases (a) and (b), we apply this result with a = 0.9 and a = −0.9. In cases (c) and (d), we apply
this result twice and verify that h(n)= an
u(n)+(−a)n
u(n) leads to the difference equation
y(n)= a2
y(n − 2)+2x(n)
26
Thus, for (c) we have a = 0.9, a2
= 0.81, and
y(n)= 0.81y(n − 2)+2x(n)
And, for (d) we have a = 0.9j, a2
= −0.81, and
y(n)= −0.81y(n − 2)+2x(n)
A systematic and simpler way of solving this type of problem will be presented after we cover
z-transforms.
Problem 3.5
The impulse response sequence is explicitly:
h = [4, 3, 3(0.5), 3(0.5)2
, 3(0.5)3
, . . . ]
Replacing these values into the convolutional equation, we get:
yn = h0xn + h1xn−1 + h2xn−2 + h3xn−3 + h4xn−4 + · · ·
= 4xn + 3 xn−1 + 0.5xn−2 + 0.52
xn−3 + 0.53
xn−4 + · · ·
It follows that:
0.5yn−1 = 2xn−1 + 3 0.5xn−2 + 0.52
xn−3 + 0.53
xn−4 · · ·
Subtracting, we have:
yn − 0.5yn−1 = 4xn + 3xn−1 − 2xn−1 or,
yn = 0.5yn−1 = 4xn + xn−1
Problem 3.6
We may write the given impulse response in the form:
h(n)= 5δ(n)+6(0.8)n−1
u(n − 1)= [5, 6, 6(0.8), 6(0.8)2
, . . . ]
Proceeding as in the previous problem we have:
yn = 5xn + 6 xn−1 + 0.8xn−2 + 0.82
xn−3 + · · ·
0.8yn−1 = 4xn−1 + 6 0.8xn−2 + 0.82
xn−3 + · · ·
which gives
yn − 0.8yn−1 = 5xn + 2xn−1
27](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-14-320.jpg)
![Chapter 4 Problems
Problem 4.1
The convolution table is:
hx 1 2 1 1 2 1 1 1
1 1 2 1 1 2 1 1 1
1 1 2 1 1 2 1 1 1
2 2 4 2 2 4 2 2 2
1 1 2 1 1 2 1 1 1
Folding the table, we get
y = [1, 3, 5, 7, 7, 6, 7, 6, 4, 3, 1]
The first and last three samples are the input and output transients, and the middle 5 samples
are the steady-state outputs. The LTI table is:
n 0 1 2 3 4 5 6 7 8 9 10
xh 1 1 2 1 partial output
1 1 1 2 1 x(0)h(n − 0)
2 2 2 4 2 x(1)h(n − 1)
1 1 1 2 1 x(2)h(n − 2)
1 1 1 2 1 x(3)h(n − 3)
2 2 2 4 2 x(4)h(n − 4)
1 1 1 2 1 x(5)h(n − 5)
1 1 1 2 1 x(6)h(n − 6)
1 1 1 2 1 x(7)h(n − 7)
y(n) 1 3 5 7 7 6 7 6 4 3 1
m
x(m)h(n − m)
For the overlap-add method, the input is divided into the following three contiguous blocks:
x = [ 1, 2, 1
x0
, 1, 2, 1
x1
, 1, 1, 0
x2
]
where we padded an extra zero at the end to get a length-3 block. Convolving each block separately
with h gives:
y0 = h ∗ x0 = [1, 3, 5, 6, 4, 1]
y1 = h ∗ x1 = [1, 3, 5, 6, 4, 1]
y2 = h ∗ x2 = [1, 2, 3, 3, 1, 0]
These convolutions can be done by separately folding the three convolution subtables:
block 0 block 1 block 2
hx 1 2 1 1 2 1 1 1 0
1 1 2 1 1 2 1 1 1 0
1 1 2 1 1 2 1 1 1 0
2 2 4 2 2 4 2 2 2 0
1 1 2 1 1 2 1 1 1 0
34
The three subblocks begin at the absolute times n = 0, 3, 6, respectively. It follows from time-
invariance that the corresponding output blocks will also begin at the same absolute times. Thus,
aligning the output blocks according to their absolute timings and adding them up gives the final
result:
n 0 1 2 3 4 5 6 7 8 9 10
y0 1 3 5 6 4 1
y1 1 3 5 6 4 1
y2 1 2 3 3 1
y 1 3 5 7 7 6 7 6 4 3 1
In practice this method is implemented efficiently by computing the individual block convolutions
using the FFT instead of time-domain convolution. For an FIR filter of order M and an FFT of
length N (which is a power of two), the length of each x-block is chosen to be N1 = N − M.
The computational gain of this “fast” convolution method versus the conventional time-domain
“slow” method is approximately
fast
slow
=
log2 N
M
If the input is divided into length-5 subblocks, the last subblock will have length 3:
x = [ 1, 2, 1, 1, 2
x0
, 1, 1, 1
x1
]
The convolutions of the subblocks with the filter are:
y0 = h ∗ x0 = [1, 3, 5, 7, 7, 5, 5, 2]
y1 = h ∗ x1 = [1, 2, 4, 4, 3, 1]
Aligning them at multiples of n = 5, we have:
n 0 1 2 3 4 5 6 7 8 9 10
y0 1 3 5 7 7 5 5 2
y1 1 2 4 4 3 1
y 1 3 5 7 7 6 7 6 4 3 1
Problem 4.2
The convolution table is:
hx 2 2 0 1 -1 0 1 2
2 4 4 0 2 -2 0 2 4
-2 -4 -4 0 -2 2 0 -2 -4
-1 -2 -2 0 -1 1 0 -1 -2
1 2 2 0 1 -1 0 1 2
Folding the table, we get
y = [4, 0, −6, 2, −2, 1, 4, 1, −5, −1, 2]
The first and last three samples are the input and output transients, and the middle 5 samples
are the steady-state outputs. The LTI table is:
35](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-18-320.jpg)
![n 0 1 2 3 4 5 6 7 8 9 10
xh 2 -2 -1 1 partial output
2 4 -4 -2 2 x(0)h(n − 0)
2 4 -4 -2 2 x(1)h(n − 1)
0 0 0 0 0 x(2)h(n − 2)
1 2 -2 -1 1 x(3)h(n − 3)
-1 -2 2 1 -1 x(4)h(n − 4)
0 0 0 0 0 x(5)h(n − 5)
1 2 -2 -1 1 x(6)h(n − 6)
2 4 -4 -2 2 x(7)h(n − 7)
y(n) 4 0 -6 2 -2 1 4 1 -5 -1 2
m
x(m)h(n − m)
For the overlap-add method, the input is divided into the following three contiguous blocks:
x = [ 2, 2, 0
x0
, 1, −1, 0
x1
, 1, 2, 0
x2
]
where we padded an extra zero at the end to get a length-3 block. Convolving each block separately
with h gives:
y0 = h ∗ x0 = [4, 0, −6, 0, 2]
y1 = h ∗ x1 = [2, −4, 1, 2, −1, 0]
y2 = h ∗ x2 = [2, −2, −5, −1, 2]
These convolutions can be done by separately folding the three convolution subtables:
block 0 block 1 block 2
hx 2 2 0 1 -1 0 1 2
2 4 4 0 2 -2 0 2 4
-2 -4 -4 0 -2 2 0 -2 -4
-1 -2 -2 0 -1 1 0 -1 -2
1 2 2 0 1 -1 0 1 2
Aligning the output subblocks at multiples of n = 3, we get:
n 0 1 2 3 4 5 6 7 8 9 10
y0 4 0 -6 0 2 1
y1 2 -4 1 2 -1 0
y2 2 2 -5 -1 2
y 4 0 -6 2 -2 1 4 1 -5 -1 2
If the input is divided into length-5 subblocks, the last subblock will have length 3:
x = [ 2, 2, 0, 1, −1
x0
, 0, 1, 2
x1
]
The convolutions of the subblocks with the filter are:
y0 = h ∗ x0 = [4, 0, −6, 2, −2, 1, 2, −1]
y1 = h ∗ x1 = [0, 2, 2, −5, −1, 2]
Aligning them at multiples of n = 5, we have:
36
n 0 1 2 3 4 5 6 7 8 9 10
y0 4 0 -6 2 -2 1 2 -1
y1 0 2 2 -5 -1 2
y 4 0 -6 2 -2 1 4 1 -5 -1 2
Problem 4.3
Figure P4.1 shows the filter, input, and computed output signals.
0 3 6 10 13 16 20 23 26
0
1
1
2
3
4
3 6 10 13 16 20 23 26
n
n
h(n) x(n)
y(n)
on steady off
Fig. P4.1 h(n), x(n), and y(n) signals of Problem 4.3.
For the direct form, the indices of hm and xn−m are restricted to lie within their respective ranges:
3 ≤ m ≤ 6
10 ≤ n − m ≤ 20
(P4.1)
Solving the second with respect to n, we have:
10 + m ≤ n ≤ 20 + m
Using the first of Eq. (P4.1), we extend the range to:
10 + 3 ≤ 10 + m ≤ n ≤ 20 + m ≤ 20 + 6 ⇒ 13 ≤ n ≤ 26
This is the range of index n of the output yn. To find the limits of summation over m, we change
the sign of the second of Eq. (P4.1) and solve for m:
−20 ≤ m − n ≤ −10 ⇒ n − 20 ≤ m ≤ n − 10
Thus, the inequalities in Eq. (P4.1) are equivalent to
3 ≤ m ≤ 6
n − 20 ≤ m ≤ n − 10
⇒ max(3, n − 20)≤ m ≤ min(6, n − 10)
The final expression for the direct form will be:
37](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-19-320.jpg)
![y(n)=
amax(0,n−L+1)
− an+1
1 − a
In particular, for 0 ≤ n ≤ L − 1, we have max(0, n − L + 1)= 0, and
y(n)=
1 − an+1
1 − a
(input-on transients and steady state)
whereas for n ≥ L, we have max(0, n − L + 1)= n − L + 1, and
y(n)=
an−L+1
− an+1
1 − a
= an−L+1 1 − aL
1 − a
= an−L+1
y(L − 1), (input-off)
Thus, after the input is turned off, the output decays exponentially to zero.
Problem 4.6
From the previous problem, we have at n = 0,
y(0)−ay(−1)=
1 − a0+1
1 − a
− a · 0 = 1 = x(0)
Then, for 1 ≤ n ≤ L − 1, we have x(n)= 1 and
y(n)−ay(n − 1)=
1 − an+1
1 − a
− a
1 − an
1 − a
=
1 − an+1
− a + an+1
1 − a
= 1 = x(n)
Finally, for n ≥ L, we have x(n)= 0 and
y(n)−ay(n − 1)= an−L+1
yL−1 − aan−1−L+1
yL−1 = 0 = x(n)
Problem 4.7
For example, the LTI form is obtained by interchanging the roles played by x and h and their
lengths L and M + 1. The following routine is an implementation of Eq. (4.1.19):
/* convlti.c - convolution in LTI form */
#include <stdlib.h> defines max( ) and min( )
void convlti(M, h, L, x, y)
double *h, *x, *y; h=filter, x=input block, y=output block
int M, L; M=filter order, L=input length
{
int n, m;
for (n = 0; n < L+M; n++)
for (y[n] = 0, m = max(0, n-M); m <= min(n, L-1); m++)
y[n] += x[m] * h[n-m];
}
40
Problem 4.8
The following C program computes the output signals of the two filters (a) and (b) of Example
4.1.8:
/* convex.c - convolution example 4.1.8 */
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define Ma 14
#define Mb 5
void conv();
void main()
{
int K = 50, L = 200, n;
double *x, *ya, *yb;
double ha[Ma+1], hb[Mb+1] = {0.2, -1, 2, -2, 1, -0.2};
x = (double *) calloc(L, sizeof(double));
ya = (double *) calloc(L+Ma, sizeof(double)); Ex. 4.1.8 (a)
yb = (double *) calloc(L+Mb, sizeof(double)); Ex. 4.1.8.(b)
for (n=0; n<=Ma; n++) define filter (a)
ha[n] = 0.25 * pow(0.75, n);
for (n=0; n<L; n++) define input
if (n%K < K/2)
x[n] = 1;
else
x[n] = 0;
conv(Ma, ha, L, x, ya); compute outputs
conv(Mb, hb, L, x, yb);
for (n=0; n<L+Ma; n++)
printf("%lfn", ya[n]);
printf("nn");
for (n=0; n<L+Mb; n++)
printf("%lfn", yb[n]);
}
Problem 4.9
A complete such program is included below.
/* blkfilt.c - FIR filtering by block convolution */
#include <stdio.h>
#include <stdlib.h>
#define MAX 64 memory allocation size
41](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-21-320.jpg)
![void blockcon();
void main(int argc, char **argv)
{
FILE *fph; filter file
double *h, *x, *y, *ytemp;
int M, N, L, i; M = filter order, L = blocksize
int max = MAX, dmax = MAX; initial allocation & increment
if (argc != 3) {
fprintf(stderr, "usage: blkfilt hfile L <xfile >yfilen");
exit(0);
}
if ((fph = fopen(argv[1], "r")) == NULL) {
fprintf(stderr, "can’t open filter file: %sn", argv[1]);
exit(0);
}
L = atoi(argv[2]); block length
h = (double *) calloc(max + 1, sizeof(double));
for (M=0;; M++) { read h
if (M == max) { if necessary, reallocate h
max += dmax;
h = (double *) realloc((char *) h, (max + 1) * sizeof(double));
}
if (fscanf(fph, "%lf", h + M) == EOF) break;
}
M--; M is filter order
if (L < M) {
fprintf(stderr, "blocksize L must be at least %dn", M);
exit(0);
}
h = (double *) realloc((char *) h, (M + 1) * sizeof(double)); final allocation
x = (double *) calloc(L, sizeof(double)); allocate input block
y = (double *) calloc(L + M, sizeof(double)); allocate output block
ytemp = (double *) calloc(M, sizeof(double)); initialized to zero
start filtering:
for (;;) { read input block
for (N=0; N<L; N++) N=L except for last
if (scanf("%lf", x+N) == EOF) goto last;
blockcon(M, h, L, x, y, ytemp); process block
for (i=0; i<L; i++) write output block
printf("%lfn", y[i]);
}
last: process last block
blockcon(M, h, N, x, y, ytemp); last block has N<=L
42
for (i=0; i<N+M; i++) write entire y block
printf("%lfn", y[i]);
}
The dimension of the filter array h is determined on the fly by reading the file of filter coeffi-
cients. Because the length of this file is not known in advance, the program keeps reallocating
the length of h in increments of MAX coefficients until the final length is determined and the final
reallocation is made. The filtering portion of the program is identical to the program segment
discussed at the end of Section 4.1.10.
The program can receive its input from stdin with the user entering the input samples one at a
time. However, only after a group of L samples has been entered followed by a <CR> (carriage
return), will the program output the first group of L outputs. Then, the next group of L input
samples is entered and the program outputs the corresponding group of L outputs, and so on. At
the end, one must enter a <CTRL-Z> (for MSDOS) in order for the program to process the input-off
transients.
Problem 4.10
A complete such program is included below.
/* firfilt.c - FIR filtering by sample-by-sample processing */
#include <stdlib.h>
#include <stdio.h>
#define MAX 64 initial allocation size
double fir();
void main(int argc, char **argv)
{
FILE *fph; filter file
double *h, *w, x, y; filter, states, input, output samples
int M, i;
int max = MAX, dmax = MAX; allocation for h and increment
if (argc != 2) {
fprintf(stderr, "usage: firfilt hfile <xfile >yfilen");
exit(0);
}
if ((fph = fopen(argv[1], "r")) == NULL) {
fprintf(stderr, "can’t open filter file: %sn", argv[1]);
exit(0);
}
h = (double *) calloc(max + 1, sizeof(double)); preliminary allocation
for (M=0;; M++) { read h
if (M == max) { reallocate h, if necessary
max += dmax;
h = (double *) realloc((char *) h, (max + 1) * sizeof(double));
}
if (fscanf(fph, "%lf", h + M) == EOF) break;
43](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-22-320.jpg)
![}
M--; M is filter order
h = (double *) realloc((char *) h, (M + 1) * sizeof(double)); final allocation
w = (double *) calloc(M + 1, sizeof(double));
start filtering:
while(scanf("%lf", &x) != EOF) { keep reading input samples
y = fir(M, h, w, x); compute output sample
printf("%lfn", y); write output sample
}
for (i=0; i<M; i++) { input-off transients
y = fir(M, h, w, 0.0);
printf("%lfn", y);
}
}
As in the case of blkfilt.c, the dimension of the filter array h is determined on the fly by reading
the file of filter coefficients. The program keeps reallocating the length of h in increments of MAX
coefficients until the final length is determined and the final reallocation is made. The filtering
portion of the program is identical to the program segment discussed in Section 4.2.3.
The program can receive its input from stdin with the user entering the input one sample at a
time. After each input is entered followed by a <CR>, the program computes and outputs the
corresponding output sample. (One can also enter a few input samples at a time, followed by a
<CR>, and the program will produce the corresponding outputs.) At the end, one must enter a
<CTRL-Z> (for MSDOS) in order for the program to process the input-off transients.
Problem 4.11
A complete such program is included below.
/* cfirfilt.c - FIR filtering using circular delay-line buffer */
#include <stdlib.h>
#include <stdio.h>
#define MAX 64 initial allocation size
double cfir();
void main(int argc, char **argv)
{
FILE *fph; filter file
double *h, *w, *p; filter, states, circular pointer
double x, y; input and output samples
int M, i;
int max = MAX, dmax = MAX; allocation for h and increment
if (argc != 2) {
fprintf(stderr, "usage: cfirfilt hfile <xfile >yfilen");
exit(0);
}
44
if ((fph = fopen(argv[1], "r")) == NULL) {
fprintf(stderr, "can’t open filter file: %sn", argv[1]);
exit(0);
}
h = (double *) calloc(max + 1, sizeof(double)); preliminary allocation
for (M=0;; M++) { read h
if (M == max) { reallocate h, if necessary
max += dmax;
h = (double *) realloc((char *) h, (max + 1) * sizeof(double));
}
if (fscanf(fph, "%lf", h + M) == EOF) break;
}
M--; M is filter order
h = (double *) realloc((char *) h, (M + 1) * sizeof(double));
w = (double *) calloc(M + 1, sizeof(double));
p = w; initialize p
start filtering:
while(scanf("%lf", &x) != EOF) { keep reading input samples
y = cfir(M, h, w, &p, x); compute output sample
printf("%lfn", y); write output sample
}
for (i=0; i<M; i++) { input-off transients
y = cfir(M, h, w, &p, 0);
printf("%lfn", y);
}
}
It is identical to firfilt.c, except it uses the circular version cfir instead of fir to do the fil-
tering. In addition, the circular pointer p is defined and initialized by the program. This program
must be compiled and linked with cfir.c and wrap.c. The version using the routine cfir1 is
identical.
The following version is based on the routine cfir2 and uses the circular pointer index q instead
of the pointer p. It must be linked with the routines cfir2 and wrap2.
/* cfirfil2.c - FIR filtering using circular delay-line buffer */
#include <stdlib.h>
#include <stdio.h>
#define MAX 64 initial allocation size
double cfir2();
void main(int argc, char **argv)
{
FILE *fph; filter file
double *h, *w; filter, states, circular pointer
double x, y; input and output samples
int M, i, q; circular pointer index
int max = MAX, dmax = MAX; allocation for h and increment
45](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-23-320.jpg)
![if (argc != 2) {
fprintf(stderr, "usage: cfirfil2 hfile <xfile >yfilen");
exit(0);
}
if ((fph = fopen(argv[1], "r")) == NULL) {
fprintf(stderr, "can’t open filter file: %sn", argv[1]);
exit(0);
}
h = (double *) calloc(max + 1, sizeof(double)); preliminary allocation
for (M=0;; M++) { read h
if (M == max) { reallocate h, if necessary
max += dmax;
h = (double *) realloc((char *) h, (max + 1) * sizeof(double));
}
if (fscanf(fph, "%lf", h + M) == EOF) break;
}
M--; M is filter order
h = (double *) realloc((char *) h, (M + 1) * sizeof(double));
w = (double *) calloc(M + 1, sizeof(double));
q = 0; initialize q
start filtering:
while(scanf("%lf", &x) != EOF) { keep reading input samples
y = cfir2(M, h, w, &q, x); compute output sample
printf("%lfn", y); write output sample
}
for (i=0; i<M; i++) { input-off transients
y = cfir2(M, h, w, &q, 0);
printf("%lfn", y);
}
}
Problem 4.12
The following C program is an implementation:
/* cdelfilt.c - Implementation of delay using circular buffer */
#include <stdlib.h>
#include <stdio.h>
double cdelay();
double tap();
void main(int argc, char **argv)
{
double *w, *p; linear buffer & circular pointer
double x, y; input and output samples
int D, i, j;
46
if (argc != 3) {
fprintf(stderr, "usage: cdelfilt i D <xfile >yfilen");
exit(0);
}
i = atoi(argv[1]);
D = atoi(argv[2]);
if (i>D) {
fprintf(stderr, "i must be in the range {0, 1, ..., D}n");
exit(0);
}
w = (double *) calloc(D + 1, sizeof(double));
p = w; initialize p
while(scanf("%lf", &x) != EOF) { keep reading input samples
*p = x; store in circular buffer
y = tap(D, w, p, i); y(n) = x(n-i) = i-th delay
cdelay(D, w, &p); update delay line
printf("%lfn", y); write output sample
}
for (j=0; j<i; j++) { input-off transients
*p = 0; zero input
y = tap(D, w, p, i);
cdelay(D, w, &p);
printf("%lfn", y);
}
}
Problem 4.13
Consider a 3d order filter with internal states wi(n)= x(n − i), i = 0, 1, 2, 3. They are delayed
replicas of each other:
w3(n) = w2(n − 1)
w2(n) = w1(n − 1)
w1(n) = w0(n − 1)
w0(n) = x(n)
Thus, the current states are obtained from the previous ones by shifting the previous contents
of the delay line and reading x(n) into w0(n). Once this shift is done, the filter’s output is
computed by the dot product:
y(n)= h0w0(n)+h1w1(n)+h2w2(n)+h3w3(n)
The following routine firalt.c implements this alternative procedure:
/* firalt.c - Alternative version of FIR filter in direct form */
double firalt(M, h, w, x) Usage: y=firalt(M, h, w, x);
double *h, *w, x; h=filter, w=state, x=input
int M; M=filter order
{
47](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-24-320.jpg)
![int i;
double y; y=output sample
for (i=M; i>=1; i--) update states from previous call
w[i] = w[i-1]; done in reverse order
w[0] = x; read current input sample x
for (y=0, i=0; i<=M; i++)
y += h[i] * w[i]; process current output sample
return y;
}
Note that the previous state vector is updated first into its current value, the current input is read
into w0, and only then is the output y computed. Upon exit, the states are left at their current
values and not updated. They will be updated by the next call. The vector w must be initialized
to zero prior to the first call, just like the fir.c case.
Problem 4.14
Introduce the internal states:
w0(n) = x(n)
w1(n) = x(n − 1)= w0(n − 1)
w2(n) = x(n − 2)= w1(n − 1)
w3(n) = x(n − 3)= w2(n − 1)
Then, the I/O equation together with the state-updating equations will read:
w0(n)= x(n)
y(n)= w0(n)+w1(n)+2w2(n)+w3(n)
and
w3(n + 1) = w2(n)
w2(n + 1) = w1(n)
w1(n + 1) = w0(n)
This leads to the following sample processing algorithm and block diagram:
x y
w0
z-1
z-1
z-1
w1
w2
w3
2
for each input sample x do:
w0 = x
y = w0 + w1 + 2w2 + w3
w3 = w2
w2 = w1
w1 = w0
The sample processing algorithm generates the following output samples:
48
n x w0 w1 w2 w3 y = w0 + w1 + 2w2 + w3
0 1 1 0 0 0 1
1 2 2 1 0 0 3
2 1 1 2 1 0 5
3 1 1 1 2 1 7
4 2 2 1 1 2 7
5 1 1 2 1 1 6
6 1 1 1 2 1 7
7 1 1 1 1 2 6
8 0 0 1 1 1 4
9 0 0 0 1 1 3
10 0 0 0 0 1 1
The first three outputs correspond to the input-on transients (the internal delay registers are
still filling up). The period 3 ≤ n ≤ 7 corresponds to steady state (all delays are filled). The
last three outputs, in general the last M outputs for an M-th order FIR filter, are the input-off
(x = 0) transients (the delays gradually empty out their contents). A C routine and main program
implementing this algorithm are given below:
#include <stdio.h>
#include <malloc.h> declares calloc()
double x[8] = {1,2,1,1,2,1,1,1};
double filter2();
void main()
{
int n;
double y, *w;
w = (double *) calloc(4, sizeof(double)); allocates & initializes w to zero
for (n=0; n<8; n++) { input-on transients & steady-state
y = filter2(x[n], w);
printf("%lfn", y);
}
for (n=8; n<11; n++) { input-off transients
y = filter2(0.0, w);
printf("%lfn", y);
}
}
double filter2(x, w) usage: y = filter2(x, w);
double x, *w;
{
double y;
w[0] = x; read input sample
y = w[0] + w[1] + 2 * w[2] + w[3]; compute output sample
w[3] = w[2]; update internal states
w[2] = w[1];
w[1] = w[0];
49](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-25-320.jpg)
![Notice how the computed y becomes the next w1 and w1 shifts into w2.
Problem 4.18
The impulse response is read off from the coefficients of the I/O equation:
h = [1, 0, −1, 2]
The LTI form is implemented by the following table:
n 0 1 2 3 4 5 6 7 8 9 10
xh 1 0 -1 2 partial output
1 1 0 -1 2 x(0)h(n − 0)
1 1 0 -1 2 x(1)h(n − 1)
2 2 0 -2 4 x(2)h(n − 2)
2 2 0 -2 4 x(3)h(n − 3)
2 2 0 -2 4 x(4)h(n − 4)
2 2 0 -2 4 x(5)h(n − 5)
1 1 0 -1 2 x(6)h(n − 6)
1 1 0 -1 2 x(7)h(n − 7)
y(n) 1 1 1 3 2 4 3 3 3 1 2
m
x(m)h(n − m)
For the overlap-add method, the input is divided into the following length-4 contiguous blocks:
x = [ 1, 1, 2, 2
x0
, 2, 2, 1, 1
x1
, ]
Convolving each block separately with h gives:
y0 = h ∗ x0 = [1, 1, 1, 3, 0, 2, 4]
y1 = h ∗ x1 = [2, 2, −1, 3, 3, 1, 2]
These convolutions can be done by separately folding the two convolution subtables:
block 0 block 1
hx 1 1 2 2 2 2 1 1
1 1 2 2 2 2 2 1 1
0 0 0 0 0 0 0 0 0
-1 -1 -1 -2 -2 -2 -2 -1 -1
2 2 2 4 4 4 4 2 2
The two subblocks begin at the absolute times n = 0, 4, respectively. It follows from time-
invariance that the corresponding output blocks will also begin at the same absolute times. Thus,
aligning the output blocks according to their absolute timings and adding them up gives the final
result:
n 0 1 2 3 4 5 6 7 8 9 10
y0 1 1 1 3 0 2 4
y1 2 2 -1 3 3 1 2
y 1 1 1 3 2 4 3 3 3 1 2
52
The block diagram and sample processing algorithm are:
x y
w0
z-1
z-1
z-1
w1
w2
w3 2
-1
for each input sample x do:
w0 = x
y = w0 − w2 + 2w3
w3 = w2
w2 = w1
w1 = w0
The corresponding circular-buffer version will be:
for each input sample x do:
s0 = ∗p = x
s2 = tap(3, w, p, 2)
s3 = tap(3, w, p, 3)
y = s0 − s2 + 2s3
cdelay(3, w, &p)
where tap-1 is not needed, but tap-2 and tap-3 are. Tap-0 s0 is the content of the current location
pointed to by p, which receives the input sample x. After y is computed, the call to cdelay
circularly decrements the pointer p.
Problem 4.19
Figure P4.2 shows the contents of the registers w = [w0, w1, w1] at successive time instants. Each
time the circular pointer p = pin points to the w-register that receives the current input, whereas
the pointer pout points to the w-register containing the current output, that is, y(n)= x(n − 2).
Therefore, pout is shifted clockwise by 2 with respect to pin:
pout = pin + 2 (modulo 3)
As pin gets decremented circularly, so does pout. If pin points to w[qin] and pout to w[qout], then
pin = w + qin
pout = w + qout
where
qout = (qin + 2)% 3
Thus, the input and output circular indices take on the sequence of values:
qin = 0, 2, 1, 0, 2, 1, 0, 2, 1, 0, 2
qout = 2, 1, 0, 2, 1, 0, 2, 1, 0, 2, 1
The following table lists the contents of w for both the linear and circular buffer implementa-
tions. In the linear case, w0, w1 and w2 are simply the input signal delayed by i = 0, 1, 2 units.
In the circular case, the down-arrow indicates the w-register receiving the current input, that is,
w[qin], whereas the up-arrow indicates the corresponding output sample, that is, w[qout]. In
the circular case, only one w-register changes value at each time instant.
53](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-27-320.jpg)
![for each input sample x do:
y = v1 + h0x
v1 = v2 + h1x
v2 = v3 + h2x
v3 = h3x
For Example 4.2.1, Table P4.1 shows the contents of the vi registers at each time instant. The vi
are initialized to zero. The v1 and v2 entries in each row are obtained by forming v2 + h1x and
v3 + h2x of the previous row. The v3 column is the delayed (down-shifted) version of the h3x
column. The final result is the same as in Example 4.2.1.
n x h0x h1x h2x h3x v1 v2 v3 y = v1 + h0x
0 1 1 2 −1 1 0 0 0 1
1 1 1 2 −1 1 2 −1 1 3
2 2 2 4 −2 2 1 0 1 3
3 1 1 2 −1 1 4 −1 2 5
4 2 2 4 −2 2 1 1 1 3
5 2 2 4 −2 2 5 −1 2 7
6 1 1 2 −1 1 3 0 2 4
7 1 1 2 −1 1 2 1 1 3
8 0 0 0 0 0 3 0 1 3
9 0 0 0 0 0 0 1 0 0
10 0 0 0 0 0 1 0 0 1
Table P4.1 Computations in Problem 4.20.
Problem 4.21
The generalization of the transposed algorithm to order M is straightforward:
for each input sample x do:
for i = 0, 1, . . . , M − 1 do:
vi = vi+1 + hix
vM = hMx
y = v0
where for indexing convenience we introduced v0(n)= y(n)= v1(n)+h0x(n). The following C
routine firtr.c is an implementation:
/* firtr.c - FIR filter in transposed direct form */
double firtr(M, h, v, x) Usage: y=firtr(M, h, v, x);
double *h, *v, x; h=filter, v=state, x=input
int M; M=filter order
{
int i;
56
for (i=0; i<=M-1; i++)
v[i] = v[i+1] + h[i] * x; output & update states
v[M] = h[M] * x; last state
return v[0]; y = v0 = v1 + h0x
}
The following stand-alone C program firfiltr.c is the corresponding version of firfilt.c.
The only difference is that it uses the routine firtr.c instead of fir.c:
/* firfiltr.c - FIR filtering in transposed form */
#include <stdlib.h>
#include <stdio.h>
#define MAX 64 initial allocation size
double firtr();
void main(int argc, char **argv)
{
FILE *fph; filter file
double *h, *v, x, y; filter array, input, output samples
int M, i;
int max = MAX, dmax = MAX; allocation for h and increment
if (argc != 2) {
fprintf(stderr, "usage: firfilt hfile <xfile >yfilen");
exit(0);
}
if ((fph = fopen(argv[1], "r")) == NULL) {
fprintf(stderr, "can’t open filter file: %sn", argv[1]);
exit(0);
}
h = (double *) calloc(max + 1, sizeof(double)); preliminary allocation
for (M=0;; M++) { read h
if (M == max) { reallocate h, if necessary
max += dmax;
h = (double *) realloc((char *) h, (max + 1) * sizeof(double));
}
if (fscanf(fph, "%lf", h + M) == EOF) break;
}
M--; M is filter order
h = (double *) realloc((char *) h, (M + 1) * sizeof(double)); final allocation
v = (double *) calloc((M + 1), sizeof(double));
while(scanf("%lf", &x) != EOF) { keep reading input samples
y = firtr(M, h, v, x); compute output sample
printf("%lfn", y); write output sample
}
for (i=0; i<M; i++) { input-off transients
57](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-29-320.jpg)
![y = firtr(M, h, v, 0.0); compute output sample
printf("%lfn", y);
}
}
Transposed forms are convenient in some applications, such as in the implementation of FIR
decimators and interpolators in multirate applications. MATLAB uses this form to implement its
built-in filtering routines.
58
Chapter 5 Problems
Problem 5.1
The linearity property follows from:
∞
n=−∞
[a1x1(n)+a2x2(n)]z−n
= a1
∞
n=−∞
x1(n)z−n
+ a2
∞
n=−∞
x2(n)z−n
The delay property follows from:
XD(z) =
∞
n=−∞
xD(n)z−n
=
∞
n=−∞
x(n − D)z−n
=
=
∞
k=−∞
x(k)z−(k+D)
= z−D
∞
k=−∞
x(k)z−k
= z−D
X(z)
where we changed summation index from n to k = n − D, so that n = k + D. The convolution
property follows from the linearity and delay properties. Writing y(n) in terms of the delayed
replicas of x(n), we have:
y(n)=
∞
m=−∞
h(m)x(n − m)=
∞
m=−∞
h(m)xm(n)
Taking z-transforms of both sides and using the linearity and delay properties, we get:
Y(z)=
∞
m=−∞
h(m)Xm(z)=
∞
m=−∞
h(m)z−m
X(z)=
∞
m=−∞
h(m)z−m
X(z)
or,
Y(z)= H(z)X(z)
Problem 5.2
a. From the delay property of z-transforms, X(z)= z−5
Δ(z), where Δ(z)= 1 is the z-
transform of δ(n). Thus, X(z)= z−5
. The ROC is the entire z-plane with the exception of
the point z = 0.
b. Similarly, X(z)= z5
Δ(z)= z5
, but now the ROC must only exclude the point at infinity
z = ∞.
c. The unit-step has z-transform: U(z)= 1/(1 − z−1
). Thus,
X(z)= z−5
U(z)=
z−5
1 − z−1
with ROC |z| > 1.
d. We may work with the definition:
X(z) =
∞
n=−∞
u(−n + 5)z−n
=
5
n=−∞
1 · z−n
= z−5
+ z−4
+ z−3
+ z−2
+ z−1
+ 1 + z + z2
+ z3
+ · · ·
= z−5
1 + z + z2
+ z3
+ · · · =
z−5
1 − z
= z−5
U(z−1
)
59](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-30-320.jpg)
![The convergence of the series requires the ROC |z| < 1.
Alternatively, we recognize that x(n) is the delayed version of u(−n), that is, x(n)=
u(−n + 5)= u(−(n − 5)). Using the general property that
g(n)
Z
−→ G(z) ⇒ g(−n)
Z
−→ G(z−1
)
we find X(z)= z−5
U(z−1
).
Problem 5.3
a. Using the general result: an
u(n)−→ 1/(1 − az−1
), ROC |z| > |a|, we obtain: X(z)=
1/(1 − (−0.5)z−1
)= 1/(1 + 0.5z−1
), with ROC |z| > | − 0.5| = 0.5.
b. Method 1: Let w(n)= u(n)−u(n − 10) with z-transform W(z). Using the linearity and
delay properties, we have
W(z)= U(z)−z−10
U(z)= (1 − z−10
)U(z)=
1 − z−10
1 − z−1
Using the modulation property of z-transforms, we have
an
w(n)−→ W(z/a)=
1 − (z/a)−10
1 − (z/a)−1
=
1 − a10
z−10
1 − az−1
With a = −0.5, we have X(z)= (1 − (−0.5)10
z−10
)/(1 + 0.5z−1
).
ROC is the whole z-plane minus the point z = 0.
Method 2: We recognize that x(n) is the finite sequence
x(n)= [1, a, a2
, a3
, a4
, a5
, a6
, a7
, a8
, a9
, 0, 0, 0, · · · ]
and use the definition of z-transforms:
X(z)= 1 + az−1
+ a2
z−2
+ · · · + a9
z−9
=
1 − a10
z−10
1 − az−1
where in the last step we used the finite geometric series:
1 + x + x2
+ · · · + xm
=
1 − xm+1
1 − x
, with x = az−1
and m = 9
c. X(z)=
1
1 − 0.5z−1
+
1
1 + 0.5z−1
=
2
(1 − 0.5z−1)(1 + 0.5z−1)
=
2
1 − 0.25z−2
, ROC |z| >
0.5.
Problem 5.4
Using Euler, 2 cos θ = ejθ
+e−jθ
, with θ = πn/2 and also the fact that ejπ/2
= j and e−jπ/2
= −j,
we rewrite
2(0.8)n
cos(πn/2) = (0.8)n
[ejπn/2
+ e−jπn/2
]
= (0.8)n
[jn
+ (−j)n
]= (0.8j)n
+(−0.8j)n
Thus, the signals in questions (a) and (b) are the same. Their z-transform is
60
X(z)=
1
1 − 0.8jz−1
+
1
1 + 0.8jz−1
=
2
(1 − 0.8jz−1)(1 + 0.8jz−1)
=
2
1 + 0.64z−2
ROC is |z| > |0.8j| = 0.8. An alternative method is to list a few of the signal values and then use
the infinite geometric series to sum them up, that is, with a = 0.8:
x(n)= 2[1, 0, −a2
, 0, a4
, 0, −a6
, 0, a8
, 0, −a10
, 0, · · · ]
we obtain
X(z)= 2[1 − a2
z−2
+ a4
z−4
− a6
z−6
+ a8
z−8
− a10
z−10
+ · · · ]=
2
1 + a2z−2
where we applied the infinite geometric series
1 + x + x2
+ x3
+ · · · =
1
1 − x
with x = −a2
z−2
. Convergence of the geometric series requires that |x| < 1 or | − a2
z−2
| < 1 or
|z| > |a| or in our case |z| > 0.8.
Problem 5.5
The three signals in parts (a,b,c) all have the same z-transform, namely,
X(z)=
1
1 − 0.25z−1
+
1
1 − 4z−1
=
2 − 4.25z−1
1 − 4.25z−1 + z−2
They differ only in their region of convergence: In case (a), both terms are causal, therefore
|z| > 0.25 and |z| > 4. Thus, ROC is |z| > 4. In case (b), the first term requires |z| > 0.25
while the second |z| < 4. Combining, we find ROC 0.25 < |z| < 4. In case (c), both terms are
anticausal, thus, |z| < 0.25 and |z| < 4. Thus, ROC is |z| < 0.25. The signal in (d) is unstable
from both sides of the time axis. The first term in the z-transform expansion
X(z)= −
−1
n=−∞
(0.25)n
z−n
+
∞
n=0
4n
z−n
would require ROC |z| < 0.25 and the second |z| > 4. The intersection of the two ROC sets is
empty, that is, there is no set of z’s for which the z-transform would converge.
Problem 5.6
Method 1: List the values of x(n) and sum the geometric series:
x(n) = [1, 0, −1, 0, 1, 0, −1, 0, 1, 0, −1, 0, . . . ]
X(z) = 1 − z−2
+ z−4
− z−6
+ z−8
− · · ·
= 1 + x + x2
+ x3
+ x4
+ · · · x=−z−2 =
1
1 − x x=−z−2
=
1
1 + z−2
where convergence of the geometric series requires:
61](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-31-320.jpg)
![|x| < 1 ⇒ | − z−2
| < 1 ⇒ |z| > 1
Method 2: Use Euler’s formula to split x(n) as the sum of two terms of the form an
u(n):
x(n)= cos(πn/2)u(n)=
1
2
eπjn/2
+ e−πjn/2
u(n)=
1
2
[jn
+ (−j)n
]u(n)
where we used eπj/2
= j. Taking z-transforms, we find:
X(z)=
1/2
1 − jz−1
+
1/2
1 + jz−1
=
1
(1 − jz−1)(1 + jz−1)
=
1
1 + z−2
where we applied the result
an
u(n)
Z
−→
1
1 − az−1
for a = j and a = −j.
Method 3: Recognize that x(n) is periodic with period 4. Defining the length-4 sequence
g = [1, 0, −1, 0], we see that x(n) will be the periodic replication of g(n):
x = [g, g, g, . . . ] or,
x(n)= g(n)+g(n − 4)+g(n − 8)+g(n − 12)+ · · ·
Taking z-transforms of both sides gives
X(z)= G(z)+z−4
G(z)+z8
G(z)+ · · · = 1 + z−4
+ z−8
+ z−12
+ · · · G(z)
Summing up the series and using G(z)= 1 − z−2
, we get:
X(z)=
G(z)
1 − z−4
=
1 − z−2
1 − z−4
=
1 − z−2
(1 − z−2)(1 + z−2)
=
1
1 + z−2
Problem 5.7
Whenever X(z) is a polynomial, the best method to invert is to just pick out the coefficients of
the polynomial. For (a), we have:
X(z) = (1 − 4z−2
)(1 + 3z−1
)= 1 + 3z−1
− 4z−2
− 12z−3
⇒ x(n) = [1, 3, −4, −12]
For (b), the signal is non-zero only at n = −3, 0, 2:
x(n)= 5δ(n)+3δ(n + 3)+2δ(n − 2)
62
Problem 5.8
a. Carry out a partial fraction expansion:
X(z)=
3(1 + 0.3z−1
)
1 − 0.81z−2
=
3(1 + 0.3z−1
)
(1 − 0.9z−1)(1 + 0.9z−1)
=
A
1 − 0.9z−1
+
B
1 + 0.9z−1
where
A =
3(1 + 0.3z−1
)
1 + 0.9z−1
z=0.9
= 2, B =
3(1 + 0.3z−1
)
1 − 0.9z−1
z=−0.9
= 1
The first ROC is |z| > |0.9| = | − 0.9|, thus both terms will be inverted causally:
x(n)= A(0.9)n
u(n)+B(−0.9)n
u(n)
As expected, the answer is stable because the ROC contains the unit circle. The second
ROC is |z| < |0.9| = | − 0.9|, thus both terms will be inverted anticausally:
x(n)= −A(0.9)n
u(−n − 1)−B(−0.9)n
u(−n − 1)
The answer is unstable, because the ROC does not contain the unit circle.
b. Ordinary partial fraction expansion is not valid in this case because the degree of the
numerator is the same as the degree of the denominator. However, we may still have an
expansion of the form:
X(z) =
6 − 3z−1
− 2z−2
1 − 0.25z−2
=
6 − 3z−1
− 2z−2
(1 − 0.5z−1)(1 + 0.5z−1)
= A +
B
1 − 0.5z−1
+
C
1 + 0.5z−1
where B and C are determined in the usual manner and A is determined by evaluating
X(z) at z = 0:
A =
6 − 3z−1
− 2z−2
1 − 0.25z−2
z=0
=
6z2
− 3z − 2
z2 − 0.25 z=0
=
−2
−0.25
= 8
B =
6 − 3z−1
− 2z−2
1 + 0.5z−1
z=0.5
= −4, C =
6 − 3z−1
− 2z−2
1 − 0.5z−1
z=−0.5
= 2
For the first ROC, |z| > 0.5, the last two terms will be inverted causally:
x(n)= Aδ(n)+B(0.5)n
u(n)+C(−0.5)n
u(n)
For the second ROC, |z| < 0.5, the two terms will be inverted anticausally:
x(n)= Aδ(n)−B(0.5)n
u(−n − 1)−C(−0.5)n
u(−n − 1)
As expected, only the first inverse is stable because its ROC contains the unit circle.
63](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-32-320.jpg)
![c. The degree of the numerator is strictly greater than the degree of the denominator. The
simplest approach in such cases is to use the “remove/restore” method, that is, ignore
the numerator completely, do a partial fraction on the denominator, get its inverse z-
transform, and finally restore the effect of the numerator. To show these steps, write
X(z) =
6 + z−5
1 − 0.64z−2
= (6 + z−5
)·
1
1 − 0.64z−2
≡ (6 + z−5
)W(z)= 6W(z)+z−5
W(z)
In the time-domain, we have then
x(n)= 6w(n)+w(n − 5)
Thus, the problem is reduced to the problem of finding w(n). Doing a partial fraction
expansion on W(z), we find
W(z)=
1
1 − 0.64z−2
=
1
(1 − 0.8z−1)(1 + 0.8z−1)
=
A
1 − 0.8z−1
+
B
1 + 0.8z−1
where A = B = 0.5. For the ROC |z| > 0.8, the two terms are inverted causally:
w(n)= A(0.8)n
u(n)+B(−0.8)n
u(n)
Inserting in x(n)= 6w(n)+w(n − 5), we find
x(n) = 6A(0.8)n
u(n)+6B(−0.8)n
u(n)+A(0.8)n−5
u(n − 5)
+ B(−0.8)n−5
u(n − 5)
Note that the last two terms are active only for n ≥ 5. For the ROC |z| < 0.8, we have the
anticausal/unstable answer:
w(n)= −A(0.8)n
u(−n − 1)−B(−0.8)n
u(−n − 1)
which gives for x(n):
x(n) = −6A(0.8)n
u(−n − 1)−6B(−0.8)n
u(−n − 1)
− A(0.8)n−5
u(−n + 4)−B(−0.8)n−5
u(−n + 4)
The u(−n+4) was obtained as u(−(n−5)−1). Note also that the last two terms are now
active for −n + 4 ≥ 0 or n ≤ 4, that is, x(n) has a slightly causal part extending to the
right up to n = 4. This happened because the strictly anticausal signal w(n) was delayed
(shifted to the right) by 5 time units by the term w(n − 5).
d. The minor new feature of this problem is that the poles are complex-valued. When the
poles are complex, they come in conjugate pairs. In this case, the corresponding residues
are complex conjugates, too. Thus, only half of the residues need be computed:
X(z) =
10 + z−2
1 + 0.25z−2
=
10 + z−2
(1 − 0.5jz−1)(1 + 0.5jz−1)
= A +
B
1 − 0.5jz−1
+
B∗
1 + 0.5jz−1
64
Again, the A-term is needed because the degrees of numerator and denominator polyno-
mials are equal. We find
A =
10 + z−2
10.25z−2
z=0
=
10z2
+ 1
z2 + 0.25 z=0
=
1
0.25
= 4
B =
10 + z−2
1 + 0.5jz−1
z=0.5j
= 3
We only needed to calculate B, and use B∗
for the conjugate pole. For the causal case, we
have
x(n)= Aδ(n)+B(0.5j)n
u(n)+B∗
(−0.5j)n
u(n)
Now, because the last two terms are complex conjugates of each other, we may use the
general identity 2Re(z)= z + z∗
to write
B(0.5j)n
u(n)+B∗
(−0.5j)n
u(n)= 2Re[B(0.5j)n
u(n)]= 6(0.5)n
Re[jn
]u(n)
But, Re[jn
]= Re[ejπn/2
]= cos(πn/2) Thus,
B(0.5j)n
u(n)+B∗
(−0.5j)n
u(n)= 6(0.5)n
cos(πn/2)u(n)
and the final result is
x(n)= 4δ(n)+6(0.5)n
cos(πn/2)u(n)
For the anticausal case, we obtain the anticausal version of the second term, namely,
x(n)= 4δ(n)−6(0.5)n
cos(πn/2)u(−n − 1)
e. The partial fraction expansion is:
X(z)=
6 − 2z−1
− z−2
(1 − z−1)(1 − 0.25z−2)
=
4
1 − z−1
+
1
1 − 0.5z−1
+
1
1 + 0.5z−1
Because ROC is |z| > 1 > 0.5, all terms are inverted causally to give:
x(n)= 4u(n)+(0.5)n
u(n)+(−0.5)n
u(n)
The answer is marginally stable because the pole z = 1 is on the unit circle.
f. The PF expansion gives:
X(z) = −4 +
1
1 + 4z−2
= −4 +
1
(1 − 2jz−1)(1 + 2jz−1)
= −4 +
1/2
1 − 2jz−1
+
1/2
1 + 2jz−1
The two regions of convergence are:
|z| > 2 and |z| < 2
65](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-33-320.jpg)
![They correspond to the causal and anticausal inverses:
x1(n) = −4δ(n)+0.5(2j)n
u(n)+0.5(−2j)n
u(n)
x2(n) = −4δ(n)−0.5(2j)n
u(−n − 1)+0.5(−2j)n
u(−n − 1)
Using Euler’s formula, we may write:
0.5 (2j)n
+(−2j)n
] = 2n 1
2
[jn
+ (−j)n
]= 2n 1
2
[ejπn/2
+ e−jπn/2
]
= 2n
cos(πn/2)
Thus, we can rewrite:
x1(n) = −4δ(n)+2n
cos(πn/2)u(n)
x2(n) = −4δ(n)−2n
cos(πn/2)u(−n − 1)
Only x2(n) is stable because its ROC contains the unit circle, or, because in the second
term n is effectively negative and causes it to decay exponentially for large negative n, that
is, writing n = −|n|, we have:
2n
cos(πn/2)u(−n − 1)= 2−|n|
cos(πn/2)u(−n − 1)→ 0 as n → −∞
g. The PF expansion is in this case:
X(z)=
4 − 0.6z−1
+ 0.2z−2
(1 − 0.5z−1)(1 + 0.4z−1)
= A +
B
1 − 0.5z−1
+
C
1 + 0.4z−1
where
A =
4 − 0.6z−1
+ 0.2z−2
(1 − 0.5z−1)(1 + 0.4z−1) z=0
=
0.2
−0.5 · 0.4
= −1
B =
4 − 0.6z−1
+ 0.2z−2
1 + 0.4z−1
z=0.5
= 2
C =
4 − 0.6z−1
+ 0.2z−2
1 − 0.5z−1
z=−0.4
= 3
The three ROC’s are:
|z| > 0.5, 0.5 > |z| > 0.4, 0.4 > |z|
The corresponding inverses are:
x1(n) = Aδ(n)+B(0.5)n
u(n)+C(−0.4)n
u(n)
x2(n) = Aδ(n)−B(0.5)n
u(−n − 1)+C(−0.4)n
u(n)
x3(n) = Aδ(n)−B(0.5)n
u(−n − 1)−C(−0.4)n
u(−n − 1)
Only x1(n) is stable. Its ROC contains the unit circle. The B-term in x2(n) and both the B
and C terms of x3(n) diverge exponentially for large negative n.
66
Problem 5.9
Consider the z-transform pair:
xa(n)= an
u(n) Xa(z)=
1
1 − az−1
Applying the derivative operator ∂/∂a to the pair, derive the z-transform of the sequence x(n)=
nan
u(n).
Noting that
∂
∂a
xa(n)= nan−1
u(n)
we get
∂
∂a
Xa(z)=
∂
∂a
1
1 − az−1
=
z−1
(1 − az−1)2
Thus,
nan−1
u(n)
Z
−→
z−1
(1 − az−1)2
Multiplying by a, we get:
nan
u(n)
Z
−→
az−1
(1 − az−1)2
Problem 5.10
We note that each application of D brings down a factor of n:
D(an
) = a
∂
∂a
(an
)= anan−1
= nan
D2
(an
) = D(D(an
))= D(nan
)= nD(an
)= n(nan
)= n2
an
D3
(an
) = D(D2
(an
))= D(n2
an
)= n2
D(an
)= n2
(nan
)= n3
an
· · ·
Dk
(an
) = nk
an
Thus, in the z-domain:
nk
an
u(n)
Z
−→ a
∂
∂a
k
1
1 − az−1
In particular, for k = 1, we have:
a
∂
∂a
1
1 − az−1
=
az−1
(1 − az−1)2
Therefore,
nan
u(n)
Z
−→
az−1
(1 − az−1)2
67](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-34-320.jpg)
![Problem 5.16
In all cases, we expand in the appropriate geometric series and pick out the coefficients of the
expansion.
a. The z-transform and its inverse are:
X(z) =
1
1 + z−4
= 1 − z−4
+ z−8
− z−12
+ z−16
− · · ·
x(n) = [1, 0, 0, 0, −1, 0, 0, 0, 1, 0, 0, 0, −1, 0, 0, 0, 1, . . . ]
b. The z-transform and its inverse are:
X(z) =
1
1 − z−4
= 1 + z−4
+ z−8
+ z−12
+ z−16
+ · · ·
x(n) = [1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, . . . ]
c. The z-transform and its inverse are:
X(z) =
1
1 + z−8
= 1 − z−8
+ z−16
− z−24
+ · · ·
x(n) = [1, 0, 0, 0, 0, 0, 0, 0, −1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, −1, . . . ]
d. The z-transform and its inverse are:
X(z) =
1
1 − z−8
= 1 + z−8
+ z−16
+ z−24
+ · · ·
x(n) = [1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, . . . ]
Problem 5.17
For case (a), the roots of the denominator are the solutions of 1 + z−4
= 0 or, z4
= −1. They are
obtained as follows:
z4
= −1 = ejπ
= ejπ+2jπk
⇒ z = e(jπ+2jπk)/4
, k = 0, 1, 2, 3
that is, the four numbers
z = ejπ/4
, e3jπ/4
, e5jπ/4
, e7jπ/4
which are the conjugate pairs:
z = e±jπ/4
, e±3jπ/4
Thus, the z-transform factors as follows and has the PF expansion:
X(z) =
1
1 + z−3
=
1
(1 − ejπ/4z−1)(1 − e−jπ/4z−1)(1 − e3jπ/4z−1)(1 − e−3jπ/4z−1)
=
A
1 − ejπ/4z−1
+
A∗
1 − e−jπ/4z−1
+
B
1 − e3jπ/4z−1
+
B∗
1 − e−3jπ/4z−1
where the coefficients are:
72
A =
1
(1 − e−jπ/4z−1)(1 − e3jπ/4z−1)(1 − e−3jπ/4z−1) z=eπj/4
=
1
4
B =
1
(1 − ejπ/4z−1)(1 − e−jπ/4z−1)(1 − e−3jπ/4z−1) z=e3πj/4
=
1
4
Thus, we find the causal inverse for n ≥ 0:
x(n) =
1
4
ejπn/4
+
1
4
e−jπn/4
+
1
4
e3jπn/4
+
1
4
e−jπn/4
=
1
2
cos(πn/4)+
1
2
cos(3πn/4)
Cases (b,c,d) are done in a similar fashion. See the solution of Problem 5.19 for the denominator
zeros.
Problem 5.18
Let Q(z) and R(z) be the quotient and remainder polynomials of the division of N(z) by D(z).
Then, Q(z) will have degree L − M, R(z) degree M − 1. Thus,
N(z)= Q(z)D(z)+R(z)
It follows that:
H(z)=
N(z)
D(z)
=
Q(z)D(z)+R(z)
D(z)
= Q(z)+
R(z)
D(z)
Because the degree of R(z) is strictly less than the degree of D(z), the term R(z)/D(z) will
admit a PF expansion of the form:
R(z)
D(z)
=
M
i=1
Ai
1 − piz−1
where pi are the M zeros of D(z) or the poles of H(z). Any truly complex poles will come in
conjugate pairs, that is, of the type:
Ai
1 − piz−1
+
A∗
i
1 − p∗
i z−1
which can be assembled into the 2nd order section:
Ai
1 − piz−1
+
A∗
i
1 − p∗
i z−1
=
(Ai + A∗
i )−(Aip∗
i + A∗
i pi)z−1
1 − (pi + p∗
i )z−1 + pip∗
i z−2
where the numerator and denominator coefficients are all real.
73](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-37-320.jpg)
![x
0.8
0.25
y
z-1
z-1
z-1
v1
v2
w1
for each input sample x do:
y = x + 0.25v2 + 0.8w1
v2 = v1
v1 = x
w1 = y
f. Y(z)= 0.9z−1
Y(z)−0.2z−2
Y(z)+X(z)+z−1
X(z)−6z−2
X(z), which implies
H(z)=
1 + z−1
− 6z−2
1 − 0.9z−1 + 0.2z−2
Factor numerator and denominator and expand in partial fractions:
H(z)=
(1 + 3z−1
)(1 − 2z−1
)
(1 − 0.4z−1)(1 − 0.5z−1)
= A +
B
1 − 0.4z−1
+
C
1 − 0.5z−1
where
A =
(1 + 3z−1
)(1 − 2z−1
)
(1 − 0.4z−1)(1 − 0.5z−1) z=0
= −30
B =
(1 + 3z−1
)(1 − 2z−1
)
1 − 0.5z−1
z=0.4
= 136
C =
(1 + 3z−1
)(1 − 2z−1
)
1 − 0.4z−1
z=0.5
= −105
Thus, h(n)= Aδ(n)+B(0.4)n
u(n)+C(0.5)n
u(n). The two zeros at z = −3 and z = 2 are
too far from the unit circle to have any significant effect on the magnitude response. The
two poles at z = 0.4, 0.5 are both in the low frequency range. Thus, the filter will be a low
pass filter. The value of the magnitude response at ω = 0 or z = 1 is |1+1−6|/|1−0.5||1−
0.4| = 13.33, whereas its value at ω = π or z = −1 is |1 −1 −6|/|1 +0.5||1 +0.4| = 2.86.
z-plane
0.5
0.4
-3 2
= poles
= zeros
π
ω
0
13.33
2.86
|H(ω)|
The canonical form realization and its sample processing algorithm are:
80
x
0.9
-6-0.2
y
w0
z-1
z-1
w1
w2
for each input sample x do:
w0 = x + 0.9w1 − 0.2w2
y = w0 + w1 − 6w2
w2 = w1
w1 = w0
The direct form realization and its sample processing algorithm are:
x
0.9
-6 -0.2
y
z-1
z-1
z-1
z-1
v1
v2
w1
w2
for each input sample x do:
y = x + v1 − 6v2 + 0.9w1 − 0.2w2
v2 = v1
v1 = x
w2 = w1
w1 = y
Problem 6.2
a. Setting x(n)= u(n) in the I/O difference equation, we find
y(n)= u(n)+6u(n − 1)+11u(n − 2)+6u(n − 3)
The 4th term is active only for n ≥ 3, the 3d term is active only n ≥ 2, and the 2nd term
only for n ≥ 2. Thus, evaluating at a few n’s we find:
y(n)= [1, 7, 18, 24, 24, 24, 24, . . . ]
The first 3 outputs are the initial transients, the remaining constant values are the steady
part. Note that when you send in a unit step, the output always settles to a constant value
(for a stable filter). That constant value can be easily precalculated as H(1). In the present
case, H(1)= 1 + 6 + 11 + 6 = 24. For the alternating step, we have:
Y(z) =
1 + 6z−1
+ 11z−2
+ 6z−3
1 + z−1
= 1 + 5z−1
+ 6z−2
y(n) = [1, 5, 6, 0, 0, 0, · · · ]
There is a pole zero cancellation: the filter has a zero at z = −1, cutting off the high-
frequency input.
b. Noting that the input z-transform is X(z)= 1/(1 − z−1
), we find
Y(z)= H(z)X(z)=
1 − z−4
1 − z−1
= 1 + z−1
+ z−2
+ z−3
where we used the finite geometric series. Inverting, we get
81](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-41-320.jpg)
![y(n)= [1, 1, 1, 1, 0, 0, 0, 0, . . . ]
The first four samples are the transients, the remaining zeros are the steady state. In the
steady-state, the filter cuts off the unit step from going through because the filter has a
zero at z = 1. The pole of the input canceled the zero of the filter. For the alternating
step:
Y(z)=
1 − z−4
1 + z−1
= 1 − z−1
+ z−2
− z−3
⇒ y(n)= [1, −1, 1, −1, 0, 0, 0, · · · ]
Again, the filter has a high frequency zero at z = −1, cutting off the high frequency input.
Problem 6.3
For filter (a), we have:
Y(z) =
1
(1 − z−1)(1 − 0.25z−2)
=
1
(1 − z−1)(1 − 0.5z−1)(1 + 0.5z−1)
=
A
1 − z−1
+
B
1 − 0.5z−1
+
C
1 + 0.5z−1
y(n) = Au(n)+B(0.5)n
u(n)+C(−0.5)n
u(n)
where A = H(1)= 4/3, B = −0.5, C = 1/6. The A-term is the steady part. For the alternating
step, we have:
Y(z) =
1
(1 + z−1)(1 − 0.25z−2)
=
1
(1 + z−1)(1 − 0.5z−1)(1 + 0.5z−1)
=
A
1 + z−1
+
B
1 − 0.5z−1
+
C
1 + 0.5z−1
y(n) = A(−1)n
u(n)+B(0.5)n
u(n)+C(−0.5)n
u(n)
where A = H(−1)= 4/3, B = 1/6, C = −0.5. The A-term is the steady part. For (b), we have:
Y(z) =
1
(1 − z−1)(1 + 0.25z−2)
=
1
(1 − z−1)(1 − 0.5jz−1)(1 + 0.5jz−1)
=
A
1 − z−1
+
B
1 − 0.5jz−1
+
B∗
1 + 0.5jz−1
y(n) = Au(n)+B(0.5j)n
u(n)+B∗
(−0.5j)n
u(n)
where A = H(1)= 4/5 and
B =
1
(1 − z−1)(1 + 0.5jz−1) z=0.5j
= 0.1 − 0.2j
And, for the alternating step:
Y(z) =
1
(1 + z−1)(1 + 0.25z−2)
=
1
(1 + z−1)(1 − 0.5jz−1)(1 + 0.5jz−1)
=
A
1 + z−1
+
B
1 − 0.5jz−1
+
B∗
1 + 0.5jz−1
y(n) = A(−1)n
u(n)+B(0.5j)n
u(n)+B∗
(−0.5j)n
u(n)
82
where A = H(−1)= 4/5 and
B =
1
(1 + z−1)(1 + 0.5jz−1) z=0.5j
= 0.1 + 0.2j
Problem 6.4
For a unit-step input and the filter (a) of Problem 6.1, we have:
Y(z) = H(z)X(z)=
1
(1 + 0.8z−1)(1 − z−1)
=
A
1 + 0.8z−1
+
B
1 − z−1
y(n) = A(−0.8)n
u(n)+Bu(n)
where A = 1/2.25, B = H(1)= 1/1.8. The B-term represents the steady part. For filter (b):
Y(z) = H(z)X(z)=
1
(1 − 0.8z−1)(1 − z−1)
=
A
1 − 0.8z−1
+
B
1 − z−1
y(n) = A(0.8)n
u(n)+Bu(n)
where A = −4, B = H(1)= 5. The B-term represents the steady part. For filter (c):
Y(z) =
1 + z−1
(1 − z−1)(1 − 0.8z−1)
=
A
1 − z−1
+
B
1 − 0.8z−1
y(n) = Au(n)+B(0.8)n
u(n)
with A = H(1)= 10, B = −9. For filter (d):
Y(z) =
1 − 0.5z−1
(1 − z−1)(1 − 0.8z−1)
=
A
1 − z−1
+
B
1 − 0.8z−1
y(n) = Au(n)+B(0.8)n
u(n)
with A = H(1)= 2.5, B = −1.5. For filter (e):
Y(z) =
1 + 0.25z−2
(1 − z−1)(1 − 0.8z−1)
= A +
B
1 − z−1
+
C
1 − 0.8z−1
y(n) = Aδ(n)+Bu(n)+C(0.8)n
u(n)
where A = Y(0)= 0.3125, B = H(1)= 6.25, C = −5.5625. The B-term is the steady part. For
filter (f):
Y(z) =
1 + z−1
− 6z−2
(1 − z−1)(1 − 0.4z−1)(1 − 0.5z−1)
=
A
1 − z−1
+
B
1 − 0.4z−1
+
C
1 − 0.5z−1
y(n) = Au(n)+B(0.4)n
u(n)+C(0.5)n
u(n)
For the alternating step and filter (a) of Problem 6.1, we have:
Y(z) =
1
(1 + 0.8z−1)(1 + z−1)
=
A
1 + 0.8z−1
+
B
1 + z−1
y(n) = A(−0.8)n
u(n)+B(−1)n
u(n)
83](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-42-320.jpg)
![z-plane
1
= poles
π
ω
0
0.5
|H(ω)|
The canonical form realization and its sample processing algorithm are:
x
y
w0
z-1
z-1
z-1
z-1
z-1
w1
w2
w5
w3
w4
for each input sample x do:
w0 = x + w1
y = w5
w5 = w4
w4 = w3
w3 = w2
w2 = w1
w1 = w0
The direct form realization and its sample processing algorithm are:
x
y
z-1
z-1
z-1
z-1
z-1
z-1
v1
w1
v2
v5
v3
v4
for each input sample x do:
y = w1 + v5
w1 = y
v5 = v4
v4 = v3
v3 = v2
v2 = v1
v1 = x
Filter (c): H(z)= 1/(1 − 0.8z−1
) and the I/O equation becomes:
Y(z)= H(z)X(z)=
X(z)
1 − 0.8z−1
⇒ (1 − 0.8z−1
)Y(z)= X(z) ⇒
Y(z)= 0.8z−1
Y(z)+X(z)
86
and in the time domain
y(n)= 0.8y(n − 1)+x(n)
The rest is as in problem (3d). Filter (d): H(z)= 1/(1 + 0.8z−1
) and the I/O equation becomes:
Y(z)= H(z)X(z)=
X(z)
1 + 0.8z−1
⇒ (1 + 0.8z−1
)Y(z)= X(z) ⇒
Y(z)= −0.8z−1
Y(z)+X(z) ⇒ y(n)= −0.8y(n − 1)+x(n)
The rest is as in problem (3c). Filter (e): With a = −0.8, the impulse response is finite:
h(n)= [1, a, a2
, a3
, a4
, a5
, a6
, a7
, 0, 0, 0, · · · ]
Its z-transform is
H(z)= 1 + az−1
+ a2
z−2
+ a3
z−3
+ a4
z−4
+ a5
z−5
+ a6
z−6
+ a7
z−7
=
1 − a8
z−8
1 − az−1
The point z = a is not really a pole of H(z) because it is canceled by a numerator zero — H(z)
has seven zeros. To find them, solve the equation z8
= a8
and exclude the solution z = a, that
is,
z = ae2πjk/8
= −0.8e2πjk/8
= 0.8ejπ
e2πjk/8
, k = 1, 2, 3, 4, 5, 6, 7
They are shown below
z-plane
= zeros
z=a=-0.8
1
The I/O equation in the time domain is
y(n)= x(n)+ax(n − 1)+a2
x(n − 2)+ · · · + a7
x(n − 7)
Sample-by-sample processing algorithm and block diagram realization:
for each input sample x do:
y = x + aw1 + a2
w2 + a3
w3 + a4
w4 + a5
w5 + a6
w6 + a7
w7
w7 = w6
w6 = w5
w5 = w4
w4 = w3
w3 = w2
w2 = w1
w1 = x
87](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-44-320.jpg)
![x
y
a0
a1
w1
a2
w2
a3
w3
a4
w4
a5
w5
a6
w6
a7
w7
z-1 z-1 z-1 z-1 z-1 z-1
z-1
Filter (f): H(z)=
1
1 − 0.8z−1
+
1
1 + 0.8z−1
=
2
1 − 0.64z−2
Y(z) = H(z)X(z)=
2X(z)
1 − 0.64z−2
⇒ Y(z)= 0.64z−2
Y(z)+2X(z)
y(n) = 0.64y(n − 2)+2x(n)
There is a low frequency pole at z = 0.8 and a high frequency pole at z = −0.8. Thus, the filter
will emphasize both the low and high frequencies:
z-plane
0.8-0.8
= poles
ππ/2
ω
0
|H(ω)|
The direct form realization and its sample processing algorithm are:
x
0.64
y
y
z-1
z-1
w2
w1
2
for each input sample x do:
y = 2x + 0.64w2
w2 = w1
w1 = y
Filters (g,h):
h(n) = 2(0.8)n
cos(πn/2)u(n)= (0.8)n
2Re[jn
]u(n)
= (0.8j)n
u(n)+(−0.8j)n
u(n)
H(z) =
1
1 − 0.8jz−1
+
1
1 + 0.8jz−1
=
2
1 + 0.64z−2
Y(z) = H(z)X(z)=
2X(z)
1 + 0.64z−2
⇒ Y(z)= −0.64z−2
Y(z)+2X(z)
y(n) = −0.64y(n − 2)+2x(n)
88
There is a conjugate pair of poles at midfrequency, z = ±0.8j = 0.8e±π/2
. Thus, the filter will
emphasize mid frequencies:
z-plane
0.8j
-0.8j
= poles
ππ/2
ω
0
|H(ω)|
The direct form realization and its sample processing algorithm are:
x
-0.64
y
y
z-1
z-1
w2
w1
2
for each input sample x do:
y = 2x − 0.64w2
w2 = w1
w1 = y
Problem 6.6
Taking z-transforms, we find:
H(z)= 1 + 0.5z−8
+ 0.52
z−16
+ 0.53
z−24
+ · · · =
1
1 − 0.5z−8
where convergence requires |z|8
> 0.5, or, |z| > (0.5)1/8
. The block diagram and sample pro-
cessing algorithm are:
x
0.5
y
z-8
w8
w0
for each input sample x do:
y = w0 = x + 0.5w8
delay(8, w)
where the call to delay updates the 9-dimensional state vector of the eightfold delay w = [w0,
w1, . . . , w8].
Problem 6.7
Because z = ejω
, we can deduce the transfer function from H(ω):
H(z)=
−0.5 + z−8
1 − 0.5z−8
It can be written in a form that can be expanded in powers of z−8
:
H(z)= −0.5 +
0.75z−8
1 − 0.5z−8
= −0.5 + 0.75z−8
1 + 0.5z−8
+ 0.52
z−16
+ 0.53
z−24
+ · · ·
89](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-45-320.jpg)
![Thus, the impulse response will be
h = [−0.5, 0, 0, 0, 0, 0, 0, 0, 0.75, 0, 0, 0, 0, 0, 0, 0, 0.75(0.5), 0, 0, 0, 0, 0, 0, 0, 0.75(0.5)2
, . . . ]
It is depicted below:
-0.5
0
0.75
0.75/2
0.75/4
n
h(n)
...
Problem 6.8
a. The impulse response h(n) satisfies the difference equation:
h(n)= h(n − 1)+h(n − 2)+δ(n − 1) (P6.1)
Indeed, for n ≥ 2 the δ(n − 1) term is absent and each h(n) is the sum of the previous
two. The sequence is properly initialized at n = 0, 1:
h(0) = h(−1)+h(−2)+δ(−1)= 0 + 0 + 0 = 0
h(1) = h(0)+h(−1)+δ(0)= 0 + 0 + 1 = 1
Next, we take z-transforms of both sides of Eq. (P6.1) to get H(z):
H(z)= z−1
H(z)+z−2
H(z)+z−1
because the z-transform of δ(n − 1) is z−1
. Solving for H(z):
H(z)=
z−1
1 − z−1 − z−2
b. The poles of H(z) are the roots of 1 − z−1
− z−2
= 0 or
z2
− z − 1 = 0 ⇒ z =
1 ±
√
5
2
Denoting the first one by φ, the other one will be:
φ =
1 +
√
5
2
, −
1
φ
=
1 −
√
5
2
Indeed, we have
−
1
φ
= −
2
1 +
√
5
= −
2(1 −
√
5)
(1 +
√
5)(1 −
√
5)
= −
2(1 −
√
5)
1 − 5
=
1 −
√
5
2
Performing a partial fraction expansion on H(z), we get:
H(z)=
z−1
1 − z−1 − z−2
=
z−1
(1 − φz−1)(1 + φ−1z−1)
=
A
1 − φz−1
+
B
1 + φ−1z−1
90
where
A =
z−1
1 + φ−1z−1
z=φ
=
φ−1
1 + φ−2
=
1
√
5
B =
z−1
1 − φz−1
z=−φ−1
= −
φ
1 + φ2
= −
1
√
5
Taking the causal inverse z-transform, we obtain a closed-form solution for the nth Fi-
bonacci number:
h(n)= Aφn
+ B(−φ)−n
, n ≥ 0
It is unstable because φ is greater than 1 (note, φ 1.618).
c. For large n, the exponentially increasing term φn
dominates the expression. Therefore,
the ratio of two successive Fibonacci numbers will tend to:
h(n + 1)
h(n)
→
Aφn+1
Aφn
= φ
This can also be seen from the difference equation of h(n). For n ≥ 1, we have
h(n + 1)= h(n)+h(n − 1)
Dividing by h(n − 1), we have:
h(n + 1)
h(n − 1)
=
h(n)
h(n − 1)
+ 1 or,
h(n + 1)
h(n)
·
h(n)
h(n − 1)
=
h(n)
h(n − 1)
+ 1
If we denote the ratio r(n)= h(n + 1)/h(n), we may write this as:
r(n)r(n − 1)= r(n − 1)+1
The limiting value r(n)→ r will satisfy the limiting equation:
r2
= r + 1
which is the same as that satisfied by φ. Thus, r = φ because they are the positive
solutions.
d. The sequence y(n) satisfies the recursion y(n)= y(n − 1)+y(n − 2) for n ≥ 3. Indeed,
using the property φ2
= φ + 1 and successively multiplying it by powers of φ, we get
y(3) = φ2
= φ + 1 = y(2)+y(1)
y(4) = φ3
= φ2
+ φ = y(3)+y(2)
y(5) = φ4
= φ3
+ φ2
= y(4)+y(3)
· · ·
91](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-46-320.jpg)
![The z-transform of y(n) will be the once-delayed z-transform of the geometric series, that
is,
Y(z)=
z−1
1 − φz−1
Thinking of Y(z) as the output of H(z) for a particular input X(z), we have Y(z)=
H(z)X(z), which may be solved for X(z):
X(z)=
Y(z)
H(z)
=
z−1
1 − φz−1
z−1
(1 − φz−1)(1 + φ−1z−1)
= 1 + φ−1
z−1
which gives the sequence:
x(n)= δ(n)+φ−1
δ(n − 1) or x = [1, φ−1
, 0, 0, 0, 0 . . . ]
Problem 6.9
As in the previous problem, the impulse response satisfies the difference equation:
h(n)= 2h(n − 1)+h(n − 2)+δ(n − 1)
with zero initial conditions. That is, the two start-up values are:
h(0) = 2h(−1)+h(−2)+δ(−1)= 0 + 0 + 0 = 0
h(1) = 2h(0)+h(−1)+δ(0)= 0 + 0 + 1 = 1
Taking z-transforms, we find H(z):
H(z)= 2z−1
H(z)+z−2
H(z)+z−1
⇒ H(z)=
z−1
1 − 2z−1 − z−2
The roots of the denominator polynomial are the solutions of
z2
= 2z + 1 ⇒ z = 1 ±
√
2 = {θ, −θ−1
}
The PF expansion is then
H(z)=
z−1
1 − 2z−1 − z−2
=
z−1
(1 − θz−1)(1 + θ−1z−1)
=
A
1 − θz−1
+
B
1 + θ−1z−1
where
A = −B =
θ
θ2 + 1
=
1
2
√
2
Thus, the nth Pell number will be:
h(n)= Aθn
+ B(−θ)−n
, n ≥ 0
Because the exponentially diverging term θn
dominates the sum, the ratio of two successive
numbers will converge to
92
h(n + 1)
h(n)
→
Aθn+1
Aθn
= θ
The y(n) sequence has z-transform
Y(z)=
z−1
1 − θz−1
=
z−1
(1 − θz−1)(1 + θ−1z−1)
· (1 + θ−1
z−1
)= H(z)(1 + θ−1
z−1
)
and therefore, it may be thought of as the output when the input is:
X(z)= 1 + θ−1
z−1
or,
x(n)= δ(n)+θ−1
δ(n − 1) ⇒ x = [1, θ−1
, 0, 0, 0, 0 . . . ]
Problem 6.10
The input and output z-transforms are:
X1(z)=
1
1 − 0.5z−1
, Y1(z)=
1
1 − 0.5z−1
+
1
1 − 0.4z−1
=
2(1 − 0.45z−1
)
(1 − 0.5z−1)(1 − 0.4z−1)
It follows that transfer function is:
H(z)=
Y1(z)
X1(z)
=
2(1 − 0.45z−1
)
1 − 0.4z−1
Therefore, the second output, having z-transform
Y2(z)=
1
1 − 0.4z−1
will be produced by
X2(z)=
Y2(z)
H(z)
=
0.5
1 − 0.45z−1
⇒ x2(n)= 0.5(0.45)n
u(n)
Problem 6.11
The input and output z-transforms are:
X1(z) =
1
1 − az−1
Y1(z) =
1
1 − az−1
+
1
1 − bz−1
=
2 − (a + b)z−1
)
(1 − az−1)(1 − bz−1)
=
2(1 − cz−1
)
(1 − az−1)(1 − bz−1)
where c = (a + b)/2. Therefore, the transfer function is
H(z)=
Y1(z)
X1(z)
=
2(1 − cz−1
)
1 − bz−1
If the input x2(n)= cn
u(n) is applied it will have z-transform X2(z)= 1/(1 − cz−1
) and will
cause the output:
Y2(z)= H(z)X2(z)=
2(1 − cz−1
)
1 − bz−1
·
1
1 − cz−1
=
2
1 − bz−1
and in the time domain:
y2(n)= 2bn
u(n)
93](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-47-320.jpg)
![Problem 6.12
The input and output z-transforms are:
X(z) =
1
1 − 0.7z−1
Y(z) =
1
1 − 0.7z−1
+
1
1 − 0.5z−1
=
2(1 − 0.6z−1
)
(1 − 0.7z−1)(1 − 0.5z−1)
Therefore the transfer function will be:
H(z)=
Y(z)
X(z)
=
2(1 − 0.6z−1
)
1 − 0.5z−1
Expanding in partial fractions, we have:
H(z)=
2(1 − 0.6z−1
)
1 − 0.5z−1
= 2.4 −
0.4
1 − 0.5z−1
which has the causal inverse:
h(n)= 2.4δ(n)−0.4(0.5)n
u(n)
Problem 6.13
The normalized peak and width frequencies are:
ω0 =
2πf0
fs
=
2π250
5000
= 0.1π [rads/sample]
Δω =
2πΔf
fs
=
2π20
5000
= 0.008π [rads/sample]
Therefore, the Q-factor of the filter will be:
Q =
ω0
Δω
=
0.1π
0.008π
= 12.5
We take the filter poles to be at locations:
p = Rejω0
, p∗
= Re−jω0
The pole radius R may be determined from the approximate relationship
Δω = 2(1 − R)
which gives:
R = 1 −
1
2
Δω = 1 −
1
2
0.008π = 0.9874
The denominator polynomial will be:
1 + a1z−1
+ a2z−2
= (1 − pz−1
)(1 − p∗
z−1
)= 1 − (p + p∗
)z−1
+ p∗
pz−2
which gives:
a1 = −(p + p∗
)= −2R cos ω0 = −2 · 0.9874 · cos(0.1π)= −1.8781
a2 = p∗
p = R2
= 0.9750
94
Therefore, the designed filter will be:
H(z)=
1
1 − 1.8781z−1 + 0.9750z−2
The time constant of the resonator is given in terms of the maximum pole radius (there is only
one radius here):
neff =
ln
ln R
=
ln(0.01)
ln(0.9874)
= 363.18
The approximate expression is obtained by replacing R in terms of Δω:
neff =
ln
ln R
=
ln
ln(1 − Δω/2)
ln
−Δω/2
= −
2 ln
Δω
where we used the small-x approximation ln(1 − x) −x. The approximate numerical value of
the time constant will be:
neff = −
2 ln
Δω
= −
2 ln(0.01)
0.008π
= 366.47
which compares well with the exact value.
Problem 6.14
The -level time constant of a filter is (in sampling instants):
neff =
ln
ln ρ
where ρ = maxi |pi| is the maximum pole radius. In seconds, the time constant is:
τ = neffT =
ln
ln ρ
T
where T = 1/fs is the interval between samples. It follows that the 1 and 2 level time constants
will be
τ1 =
ln 1
ln ρ
T , τ2 =
ln 2
ln ρ
T
And therefore,
τ1 =
ln 1
ln 2
τ2
In particular, for the 60 versus 40 dB time constants, the scale factor is:
ln 1
ln 2
=
ln(0.001)
ln(0.01)
=
3
2
= 1.5
Problem 6.15
Using the approximate expression of Problem 6.13 and = 0.001, we have:
τ = neffT = −
2 ln
Δω
T = −
2 ln
2πΔfT
T = −
ln
πΔf
which gives:
τ = −
ln
πΔf
−
ln(0.001)
πΔf
=
2.2
Δf
95](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-48-320.jpg)
![Problem 6.16
The periodic impulse response may be thought of as the sum of the delayed replicas of one period,
that is,
h(n)= g(n)+g(n − 8)+g(n − 16)+ · · ·
where g(n)= [1, 2, 3, 4, 0, 0, 0, 0] is the basic period of length 8. Taking z-transforms we obtain:
H(z)= G(z)+z−8
G(z)+z−16
G(z)+ · · · = (1 + z−8
+ z−16
+ · · · )G(z)
where G(z)= 1 + 2z−1
+ 3z−2
+ 4z−3
. Using the geometric series on the first factor we obtain:
H(z)=
G(z)
1 − z−8
=
1 + 2z−1
+ 3z−2
+ 4z−3
1 − z−8
The direct and canonical realizations are shown in Fig. P6.1. Note that in the canonical realization
the total number of delays is 8, but the last 5 are lumped together into a fivefold delay z−5
. The
sample processing algorithm for the canonical case is:
x y
2
3
4
z-1
z-1
z-1
x y
2
3
4
w0
w1
w2
w3
w4
z-1
z-1
z-1
z-1
w5
w6
w7
w8
z-1
z-1
z-1
z-1
z-1
z-1
z-1
z-1
z-1
z-1
z-1
z-1
Fig. P6.1 Filter realizations of Problem 6.16.
for each input x do:
w0 = x + w8
y = w0 + 2w1 + 3w2 + 4w3
delay(8, w)
where the call to delay(8, w) implements the updating of the delay line, that is, w8 = w7, w7 = w6,
. . . , w1 = w0.
Problem 6.17
The solution is similar to that of Problem 6.16. Here, the repetition period is 4 and we have:
h(n)= g(n)+g(n − 4)+g(n − 8)+g(n − 12)+ · · ·
96
where g(n)= [0, 1, 2, 3]. In the z-domain, we have:
H(z)= G(z)+z−4
G(z)+z−8
G(z)+ · · · = (1 + z−4
+ z−8
+ · · · )G(z)=
G(z)
1 − z−4
where G(z)= z−1
+ 2z−2
+ 3z−3
. Thus,
H(z)=
z−1
+ 2z−2
+ 3z−3
1 − z−4
The direct and canonical realizations are shown in Fig. P6.2.
x y
2
3
z-1
z-1
z-1
x y
2
3
w0
w1
w2
w3
w4
z-1
z-1
z-1
z-1
z-1
z-1
z-1
z-1
Fig. P6.2 Direct and canonical realizations of Problem 6.17.
The sample processing algorithm for the canonical case is:
for each input x do:
w0 = x + w4
y = w1 + 2w2 + 3w3
delay(4, w)
where the call to delay(4, w) implements the updating of the delay line, that is, w4 = w3, w3 = w2,
. . . , w1 = w0. The cascade realization is obtained by factoring the numerator and denominator
into (up to) 2nd order polynomials:
H(z)=
z−1
+ 2z−2
+ 3z−3
1 − z−4
=
z−1
1 − z−2
1 + 2z−1
+ 3z−2
1 + z−2
≡ H0(z)H1(z)
The block diagram is shown in Fig. P6.3. The corresponding sample processing algorithm is:
for each input x do:
w0 = x + w2
x1 = w1
w2 = w1
w1 = w0
v0 = x1 − v2
y = v0 + 2v1 + 3v3
v2 = v1
v1 = v0
Using partial fractions, we may write:
97](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-49-320.jpg)
![x w0
w1
w2
z-1
z-1
x1 y
3
-1
2
v0
v1
v2
z-1
z-1
H0(z) H1(z)
Fig. P6.3 Cascade realization of Problem 6.17.
H(z) =
z−1
+ 2z−2
+ 3z−3
1 − z−4
=
z−1
+ 2z−2
+ 3z−3
(1 − z−1)(1 + z−1)(1 − jz−1)(1 + jz−1)
=
A
1 − z−1
+
B
1 + z−1
+
C − jD
1 − jz−1
+
C + jD
1 + jz−1
where the PFE coefficients C ± jD are conjugates. The numerical values are:
A =
z−1
+ 2z−2
+ 3z−3
(1 + z−1)(1 − jz−1)(1 + jz−1) z=1
= 1.5
B =
z−1
+ 2z−2
+ 3z−3
(1 − z−1)(1 − jz−1)(1 + jz−1) z=−1
= −0.5
C − jD =
z−1
+ 2z−2
+ 3z−3
(1 − z−1)(1 + z−1)(1 + jz−1) z=j
= −0.5 + j0.5
Taking (causal) inverse z-transforms, we find for n ≥ 0:
h(n) = A + B(−1)n
+(C − jD)jn
+ (C + jD)(−j)n
= A + B(−1)n
+2Re (C − jD)ejπn/2
= A + B(−1)n
+2C cos(πn/2)+2D sin(πn/2)
= 1.5 − 0.5(−1)n
− cos(πn/2)− sin(πn/2)
Evaluating at the first couple of periods, we find:
h(n)= [0, 1, 2, 3, 0, 1, 2, 3, . . . ]
Problem 6.18
Expand the denominator in powers of z−7
and pick out the coefficients of the powers of z−1
. We
have:
H(z)=
1 + z−1
+ z−2
+ z−3
1 − z−7
= (1 + z−1
+ z−2
+ z−3
)(1 + z−7
+ z−14
+ · · · )
The powers of z−7
cause the period-7 replication of the basic period G(z)= 1 + z−1
+ z−2
+ z−3
:
h(n)= [1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, . . . ]
98
x y
w0
w1
w2
w3
w4
z-1
z-1
z-1
z-1
w5
w6
w7
z-1
z-1
z-1
Fig. P6.4 Canonical realizations of Problem 6.18.
The canonical realization is shown in Fig. P6.4.
The sample processing algorithm is:
for each input x do:
w0 = x + w7
y = w0 + w1 + w2 + w3
delay(7, w)
where the call to delay implements the delay-line updates w7 = w6, w6 = w5, . . . , w1 = w0. The
output due to the input x = [3, 2, 1] may be obtained by using the LTI form of convolution, that
is, summing up the delayed replicas of the impulse response, scaled by the input samples:
x0hn = [3, 3, 3, 3, 0, 0, 0, 3, 3, 3, 3, 0, 0, 0, 3, 3, 3, 3, 0, 0, 0, . . . ]
x0hn−1 = [0, 2, 2, 2, 2, 0, 0, 0, 2, 2, 2, 2, 0, 0, 0, 2, 2, 2, 2, 0, 0, . . . ]
x0hn−2 = [0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, . . . ]
yn = [3, 5, 6, 6, 3, 1, 0, 3, 5, 6, 6, 3, 1, 0, 3, 5, 6, 6, 3, 1, 0, . . . ]
Thus, the output also has period 7. The same conclusion can be reached working with z-transforms.
We have for the output z-transform:
Y(z)= H(z)X(z) =
(1 + z−1
+ z−2
+ z−3
)(3 + 2z−1
+ z−2
)
1 − z−7
=
3 + 5z−1
+ 6z−2
+ 6z−3
+ 3z−4
+ z−5
1 − z−7
The expansion of the denominator in powers of z−7
will cause the period-7 replication of the
numerator.
Problem 6.19
The expansion of the denominator, will cause the period-2 replication of the numerator. Because
the numerator has length 4, its period-2 replicas will overlap with each other and must be added
together:
99](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-50-320.jpg)
![H(z)=
1 + z−1
+ z−2
+ z−3
1 − z−2
= (1 + z−1
+ z−2
+ z−3
)(1 + z−2
+ z−4
+ · · · )
or, in the time domain:
h =[1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, . . . ]
+[0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, . . . ]
+[0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, . . . ]+ · · ·
=[1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, . . . ]
The canonical realization and sample processing algorithm are:
x y
w0
w1
w2
w3
z-1
z-1
z-1
for each input x do:
w0 = x + w2
y = w0 + w1 + w2 + w3
w3 = w2
w2 = w1
w1 = w0
Problem 6.20
Writing the sample processing algorithm in terms of the time index n, we have
w0(n) = x(n)+w1(n)
y(n) = w0(n)+w2(n)
w2(n + 1) = w1(n)
w1(n + 1) = w0(n)
or, in the z-domain
W0(z) = X(z)+W1(z)
Y(z) = W0(z)+W2(z)
zW2(z) = W1(z)
zW1(z) = W0(z)
Eliminating W0(z), W1(z), W2(z) in favor of Y(z) and X(z), we obtain:
W1(z)= z−1
W0(z), W2(z)= z−1
W1(z)= z−2
W0(z)
W0(z)= X(z)+z−1
W0(z) ⇒ W0(z)=
X(z)
1 − z−1
Y(z)= W0(z)+z−2
W0(z)= (1 + z−2
)W0(z)= (1 + z−2
)
X(z)
1 − z−1
100
which gives:
H(z)=
Y(z)
X(z)
=
1 + z−2
1 − z−1
Multiplying numerator and denominator by 1 + z−1
, we get:
H(z)=
(1 + z−2
)(1 + z−1
)
(1 − z−1)(1 + z−1)
=
1 + z−1
+ z−2
+ z−3
1 − z−2
which is the same as that of Problem 6.19.
Problem 6.21
The number of harmonics fitting within the Nyquist interval are:
fs
f1
=
240
60
= 4
Therefore, the harmonics will lie at the fourth roots of unity around the unit circle:
ωk = kω1 =
2πk
4
, k = 0, 1, 2, 3
The numerator polynomial 1 − z−4
will have ejωk as roots. Putting the poles just behind the
zeros, we get the desired multinotch filter:
H(z)=
1 − z−4
1 − az−4
where 0 < a < 1 must be chosen to be very near 1.
Problem 6.22
The zeros are at the 16th roots of unity, that is,
zk = ejωk = e2πjk/16
, k = 0, 1, . . . , 15
The poles are just behind the zeros at the same angles:
pk = a1/16
ejωk = a1/16
e2πjk/16
, k = 0, 1, . . . , 15
The zeros and poles are shown in Fig. P6.5. If a is very near 1, the magnitude response will vanish
at the harmonics ωk = kω1 = 2πk/16, and be essentially flat between zeros. Fig. P6.6 shows a
sketch of |H(ω)|.
The impulse response is obtained by expanding H(z) into powers of z−16
. We have with A = 1/a
and B = 1 − A = 1 − 1/a:
H(z)=
1 − z−16
1 − az−16
= A +
B
1 − az−16
= A + B(1 + az−16
+ a2
z−32
+ · · · )
Thus, the causal impulse response will be h(n):
h(n)=
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
A + B = 1, if n = 0
Ban/16
, if n is a non-zero multiple of 16
0, otherwise
101](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-51-320.jpg)
![Problem 6.26
The transfer function is obtained from the I/O difference equation by taking z-transforms of both
sides and solving for the ratio Y(z)/X(z):
Y(z)= 2.5z−1
Y(z)−z−2
Y(z)+3X(z)+3z−2
X(z) ⇒ H(z)=
Y(z)
X(z)
=
3(1 + z−2
)
1 − 2.5z−1 + z−2
Factoring the denominator into its poles and expanding in partial fractions, we get:
H(z)=
3(1 + z−2
)
1 − 2.5z−1 + z−2
=
3(1 + z−2
)
(1 − 2z−1)(1 − 0.5z−1)
= A +
B
1 − 2z−1
C
1 − 0.5z−1
where
A =
3(1 + z−2
)
1 − 2.5z−1 + z−2
z=0
= 3
B =
3(1 + z−2
)
1 − 0.5z−1
z=2
= 5
C =
3(1 + z−2
)
1 − 2z−1
z=0.5
= −5
Thus, the ROCs and corresponding inverses are:
|z| > 2 , h(n)= Aδ(n)+B2n
u(n)+C0.5n
u(n)
2 > |z| > 0.5, h(n)= Aδ(n)−B2n
u(−n − 1)+C0.5n
u(n)
|z| < 0.5, h(n)= Aδ(n)−B2n
u(−n − 1)−C0.5n
u(−n − 1)
The z-transform of the signal g(n) is:
g(n)= cos(πn/2)u(n)
Z
−→ G(z)=
1
1 + z−2
Therefore, the z-transform of the input x(n) will be:
x(n)= g(n)−2g(n − 1)
Z
−→ X(z)= G(z)−2z−1
G(z)= (1 − 2z−1
)G(z)=
1 − 2z−1
1 + z−2
Thus, the output z-transform will be:
Y(z)= H(z)X(z)=
3(1 + z−2
)
(1 − 2z−1)(1 − 0.5z−1)
·
1 − 2z−1
1 + z−2
=
3
1 − 0.5z−1
Inverting causally, we find:
y(n)= 3(0.5)n
u(n)
Problem 6.27
Using the inverse DTFT formula of Eq. (6.3.5), we have:
y(n)=
1
2π
π
−π
H(ω)X(ω)ejωn
dω
Similarly, for the input signal, assuming its spectrum is restricted to be only over the range
[−ωc, ωc]:
104
x(n)=
1
2π
ωc
−ωc
X(ω)ejωn
dω
Using d(ω)= D, the definition of H(ω) over one Nyquist interval will be:
H(ω)=
⎧
⎨
⎩
Ge−jDω
, for 0 ≤ |ω| ≤ ωc
0, for ωc < |ω| ≤ π
where G is its passband gain. Thus, the integration range for y(n) also collapses to [−ωc, ωc],
giving:
y(n)=
1
2π
ωc
−ωc
Ge−jDω
X(ω)ejωn
dω = G
1
2π
ωc
−ωc
X(ω)ejω(n−D)
dω
Comparing this with the expression of x(n), we recognize that
y(n)= Gx(n − D)
Problem 6.28
The PF expansion of H(z) is of the form:
H(z)=
N(z)
M
i=1
(1 − piz−1
)
=
M
i=1
Ai
1 − piz−1
where the i-th coefficient is obtained by deleting the factor (1 − piz−1
) from the left-hand side
and evaluating the remainder at z = pi, that is,
Ai =
⎡
⎢
⎢
⎢
⎣
N(z)
j=i
(1 − pjz−1
)
⎤
⎥
⎥
⎥
⎦
z=pi
=
N(pi)
j=i
(1 − pjp−1
i )
Taking causal inverse z-transforms, we find for h(n):
h(n)=
M
i=1
Aipn
i u(n)
Problem 6.29
The z-transforms of the causal and anticausal sinusoidal inputs are the same up to a negative
sign:
ejω0n
u(n)
Z
−→
1
1 − ejω0 z−1
ejω0n
u(−n − 1)
Z
−→ −
1
1 − ejω0 z−1
with ROCs |z| > 1 and |z| < 1, respectively. The output z-transform will be Y(z)= H(z)X(z).
Thus, in the two cases:
105](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-53-320.jpg)
![Y(z)= ±
N(z)
(1 − ejω0
z−1
)
M
i=1
(1 − piz−1
)
The partial fraction expansion leads to
Y(z)= ±
C
1 − ejω0 z−1
±
M
i=1
Bi
1 − piz−1
where C = H(ω0), as we saw in Eq. (6.3.10), and Bi are obtained by deleting the factor (1−piz−1
)
and evaluating the rest at z = pi:
Bi =
⎡
⎢
⎢
⎢
⎣
N(z)
(1 − ejω0
z−1
)
j=i
(1 − pjz−1
)
⎤
⎥
⎥
⎥
⎦
z=pi
=
N(pi)
(1 − ejω0 p−1
i )
M
j=i
(1 − pjp−1
i )
Comparing with the result of the previous problem, we have:
Bi =
Ai
1 − ejω0 p−1
i
To get the inverse z-transform y(n), we must assume a particular ROC. For the causal case, we
take the ROC to be |z| > 1, so that all the PFE terms will be inverted causally. For the anticausal
case, we take the filter part to be causal and the input anticausal, that is, the ROC will be the
annular region between the maximum pole and the unit circle:
max
i
|pi| < |z| < 1
In this case the C term will be inverted anticausally and the Bi terms causally. These choices for
the ROCs, give:
y(n) = H(ω0)ejω0n
u(n)+
M
i=1
Bipn
i u(n)
y(n) = H(ω0)ejω0n
u(−n − 1)−
M
i=1
Bipn
i u(n)
Adding up the right-sided and left-sided sinusoidal inputs gives rise to the double-sided input:
ejω0n
u(n)+ejω0n
u(−n − 1)= ejω0n
u(n)+u(−n − 1) = ejω0n
, −∞ < n < ∞
where, we used the property u(n)+u(−n − 1)= 1 for all n. It follows from the linearity of the
filter that the corresponding output will be the sum of the two outputs obtained above. This
gives:
y(n) = H(ω0)ejω0n
u(n)+
M
i=1
Bipn
i u(n)+H(ω0)ejω0n
u(−n − 1)−
M
i=1
Bipn
i u(n)
= H(ω0)ejω0n
u(n)+H(ω0)ejω0n
u(−n − 1)
= H(ω0)ejω0n
which is recognized as the standard steady-state sinusoidal response of the filter.
106
Problem 6.30
The PF expansion of H(z) gives:
H(z)=
3 − 5z−1
+ z−2
(1 − 0.5z−1)(1 − 2z−1)
= 1 +
1
1 − 0.5z−1
+
1
1 − 2z−1
The stable ROC containing the unit circle will be 0.5 < |z| < 2. Therefore, the 0.5-pole term will
be inverted causally and the 2-pole anticausally:
h(n)= δ(n)+0.5n
u(n)−2n
u(−n − 1)
The truncated impulse response agrees with h(n) for n ≥ −D and is zero otherwise, that is:
h(n)=
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
δ(n)+0.5n
, n ≥ 0
−2n
, −D ≤ n ≤ −1
0, n ≤ −D − 1
The corresponding z-transform is by definition:
H(z)=
∞
n=0
[δ(n)+0.5n
]z−n
−
−1
n=−D
2n
z−n
The last term can be summed by changing summation variables from n to m = −n:
−1
n=−D
2n
z−n
=
D
m=1
2−m
zm
=
2−1
z(1 − 2−D
zD
)
1 − 2−1z
= −
1 − 2−D
zD
1 − 2z−1
where we used the finite geometric series:
D
m=1
xm
= x + x2
+ · · · xD
= x(1 + x + x2
+ · · · xD−1
)=
x(1 − xD
)
1 − x
Thus,
H(z)= 1 +
1
1 − 0.5z−1
+
1 − 2−D
zD
1 − 2z−1
Comparing with the exact H(z) we have:
H(z)= H(z)−
2−D
zD
1 − 2z−1
or, the error transfer function:
H(z)−H(z)=
2−D
zD
1 − 2z−1
The given input has z-transform:
x(n)= δ(n)−2δ(n − 1)
Z
−→ X(z)= 1 − 2z−1
The error output will be
E(z)= Y(z)−Y(z)= H(z)X(z)−H(z)X(z)= H(z)−H(z) X(z)
or,
E(z)=
2−D
zD
1 − 2z−1
· (1 − 2z−1
)= 2−D
zD
and in the time domain:
e(n)= 2−D
δ(n + D)
By choosing D large enough, the factor 2−D
can be made as small as desired.
107](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-54-320.jpg)
![Chapter 7 Problems
Problem 7.1
Expanding the denominator in powers of z−5
will replicate the numerator periodically with period
5. But because the numerator itself has length 5, the replicas will not overlap, and we will get the
repetition of the numerator coefficients:
h = [0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, . . . ]
The direct and canonical realizations are shown in Fig. P7.1.
x y x y
z-1
z-1
z-1
z-1
z-1
z-1
2
3
4
z-1
z-1
z-1
w5
w0
w1
w2
w3
w4
w5
w0
w1
w2
w3
w4
v0
v1
v2
v3
v4
z-1
2
3
4
z-1
z-1
z-1
z-1
Fig. P7.1 Direct and canonical realizations of Problem 7.1.
The sample processing algorithms are expressed with the help of the routine delay, which updates
the filter’s internal state vectors.
for each input x do:
v0 = x
w0 = w5 + v1 + 2v2 + 3v3 + 4v4
y = w0
delay(4, v)
delay(5, w)
for each input x do:
w0 = x + w5
y = w1 + 2w2 + 3w3 + 4w4
delay(5, w)
where v = [v0, v1, v2, v4], w = [w0, w1, w2, w3, w4.w5] are the 5- and 6-dimensional state vec-
tors needed for the numerator and denominator polynomials. The calls to the delay routines
update the delay line, e.g., the call delay(5, w) is equivalent to the shifts: w5 = w4, w4 = w3,
w3 = w2, w2 = w1, w1 = w0.
Factoring the denominator as 1−z−5
= (1−z−1
)(1+z−1
+z−2
+z−3
+z−4
), we get the cascaded
transfer functions:
H(z)=
z−1
+ 2z−2
+ 3z−3
+ 4z−4
1 − z−5
=
z−1
1 − z−1
·
1 + 2z−1
+ 3z−2
+ 4z−3
1 + z−1 + z−2 + z−3 + z−4
Figure P7.2 shows the cascade of these two filters, with each filter realized in its canonical form.
The corresponding sample processing algorithm is as follows:
108
x1 y
2-1
-1
-1
-1
3
4
z-1
z-1
w0
w1
w2
z-1
w3
z-1
w4
x
z-1
v0
v1
Fig. P7.2 Cascade realization of Problem 7.1.
for each input x do:
v0 = x + v1
x1 = v1
delay(1, v)
w0 = x1 − w1 − w2 − w3 − w4
y = w0 + 2w1 + 3w2 + 4w3
delay(4, w)
where v = [v0, v1], w = [w0, w1, w2, w3, w4] are the internal state vectors for the two filter
sections.
Problem 7.2
In the z-domain, the input to H1(z) is X(z)−Y(z). Therefore, its output will be H1(z) Y(z)
−X(z) . The input and output of H2(z) will be then :
H1(z) Y(z)−X(z) − Y(z)−→ H2(z) H1(z) Y(z)−X(z) − Y(z)
Adding the noise component E(z) to the output of H2(z) will generate Y(z), resulting in:
Y(z)= H2(z) H1(z) Y(z)−X(z) − Y(z) + E(z)
Moving the Y-dependent terms to the left, we get:
1 + H2(z)+H1(z)H2(z) Y(z)= H1(z)H2(z)X(z)+E(z)
and solving for Y(z):
Y(z)=
H1(z)H2(z)
1 + H2(z)+H1(z)H2(z)
X(z)+
1
1 + H2(z)+H1(z)H2(z)
E(z)
Thus, the desired transfer functions are identified to be:
Hx(z) =
H1(z)H2(z)
1 + H2(z)+H1(z)H2(z)
He(z) =
1
1 + H2(z)+H1(z)H2(z)
109](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-55-320.jpg)
![The conditions that Hx(z) be a plain delay and He(z) a double differentiator, gives the two
equations:
H1(z)H2(z)
1 + H2(z)+H1(z)H2(z)
= z−1
1
1 + H2(z)+H1(z)H2(z)
= (1 − z−1
)2
which can be solved for H1(z) and H2(z) giving:
H1(z) =
1
1 − z−1
H2(z) =
z−1
1 − z−1
They are both integrators. The presence of z−1
in the numerator of H2(z) is necessary to make
the overall closed loop computable. See also Problems 12.24 and 12.25.
Problem 7.3
Multiplying the numerator factors, we get the transfer function:
H(z)=
z−1
(1 + 2z−2
)(1 + 3z−2
)
1 − z−6
=
z−1
+ 5z−3
+ 6z−5
1 − z−6
The direct and canonical realizations are shown in Fig. P7.3.
x y x y
z-1
z-1
z-1
z-1
z-1
w5
w6
w0
w1
w2
w3
w4
z-1
z-1
z-1
z-1
z-1
z-1
w5
w6
w0
w1
w2
w3
w4
z-1
z-1
z-1
z-1
z-1
v5
v0
v1
v2
v3
v4
z-1
5
6
5
6
Fig. P7.3 Direct and canonical realizations of Problem 7.3.
The sample processing algorithms are expressed with the help of the routine delay, which updates
the filter’s internal state vectors.
for each input x do:
v0 = x
w0 = v1 + 5v3 + 6v5 + w6
y = w0
delay(5, v)
delay(6, w)
for each input x do:
w0 = x + w6
y = w1 + 5w3 + 6w5
delay(6, w)
110
where v = [v0, v1, v2, v4, v5], w = [w0, w1, w2, w3.w4, w5, w6] are the 6- and 7-dimensional state
vectors needed for the numerator and denominator polynomials. The calls to the delay routines
update the delay line, e.g., the call delay(6, w) is equivalent to the shifts: w6 = w5, w5 = w4,
w4 = w3, w3 = w2, w2 = w1, w1 = w0. Factoring the denominator as
1 − z−6
= (1 − z−2
)(1 + z−2
+ z−4
)= (1 − z−2
)(1 + z−1
+ z−2
)(1 − z−1
+ z−2
)
we get the cascade transfer function:
H(z)=
z−1
1 − z−2
·
1 + 2z−2
1 + z−1 + z−2
·
1 + 3z−2
1 − z−1 + z−2
Figure P7.4 shows the cascade of these three filters, with each filter realized in its canonical form.
z-1
z-1
u0
u1
u2
x x1 x2 y
z-1
z-1
v0
v1
v2 2-1
-1 z-1
z-1
w0
w1
w2 3-1
Fig. P7.4 Cascade realization of Problem 7.3.
The corresponding sample processing algorithm is as follows:
for each input x do:
u0 = x + u2
x1 = u1
delay(2, u)
v0 = x1 − v1 − v2
x2 = v0 + 2v2
delay(2, v)
w0 = x2 + w1 − w2
y = w0 + 3w2
delay(2, w)
where u = [u0, u1, u2], v = [v0, v1, v2], w = [w0, w1, w2], are the internal state vectors for the
three filter sections.
To determine the impulse response, we note that the numerator corresponds to the length-6 signal
b = [0, 1, 0, 5, 0, 6]. Expanding the denominator in powers of z−6
will replicate the numerator
periodically with period 6. But because the numerator itself has length 6, the replicas will not
overlap, and we will get the repetition of the numerator coefficients:
h = [0, 1, 0, 3, 0, 5, 0, 1, 0, 3, 0, 5, 0, 1, 0, 3, 0, 5, . . . ]
Problem 7.4
The z-domain versions of the time domain difference equations are:
v(n) = x(n)+v(n − 1)
y(n) = v(n)+v(n − 2)+v(n − 4)
⇒
V(z) = X(z)+z−1
V(z)
Y(z) = V(z)+z−2
V(z)+z−4
V(z)
111](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-56-320.jpg)
![Note that the numerator could also have been factored as:
1 − 2z−2
+ z−4
= (1 − z−2
)2
= (1 − z−1
)2
(1 + z−1
)2
= (1 − 2z−1
+ z−2
)(1 + 2z−1
+ z−2
)
But the chosen one has the simplest coefficients. Its sample processing algorithm is:
for each input x do:
v0 = x + 0.64v2
x1 = v0 − v2
delay(2, v)
w0 = x1 − 0.64w2
y = w0 − w2
delay(2, w)
x
z-1
z-1
v0
v1
x1
v2 -10.64
z-1
z-1
w0
w1
w2 -1-0.64
y
Fig. P7.11 Cascade realization of Problem 7.7.
Problem 7.8
The given difference equations describe the canonical realization form of the filter:
H(z)=
−0.75 + z−10
1 − 0.75z−10
Its block diagram realization and sample processing algorithm are given below:
x
z-10
w0
w10
0.75
-0.75 y
for each input sample x do:
w0 = x + 0.75w10
y = −0.75w0 + w10
delay(10, w)
The C version of the algorithm is as follows:
double reverb(x, w)
double x, w[11];
{
int i;
double y;
w[0] = x + 0.75 * w[10];
y = - 0.75 * w[0] + w[10];
116
for (i=10; i>=1; i--)
w[i] = w[i-1];
return y;
}
From the transfer function, we obtain the frequency response:
H(ω)=
−0.75 + e−10jω
1 − 0.75e−10jω
=
1 − 0.75e10jω
1 − 0.75e−10jω
e−10jω
=
(1 − 0.75e−10jω
)∗
(1 − 0.75e−10jω)
e−10jω
Noting that the ratio of two conjugate complex numbers has unit magnitude, it follows that H(ω)
will have unit magnitude.
Problem 7.9
The given sample processing algorithm corresponds to the canonical realization of the filter:
H(z)=
1 + z−3
1 − 0.64z−4
Factoring the numerator in the form:
1 + z−4
= (1 + z−1
)(1 − z−1
+ z−2
)
we obtain the cascade factors:
H(z)=
1 + z−3
1 − 0.64z−4
=
1 + z−1
1 − 0.8z−2
·
1 − z−1
+ z−2
1 + 0.8z−2
The cascade realization is shown in Fig. P7.12. Its sample processing algorithm is:
for each input x do:
v0 = x + 0.8v2
x1 = v0 + v1
delay(2, v)
w0 = x1 − 0.8w2
y = w0 − w1 + w2
delay(2, w)
x
z-1
z-1
v0
v1
x1
v2
0.8
z-1
z-1
w0
w1
w2-0.8
-1
y
Fig. P7.12 Cascade realization of Problem 7.9.
117](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-59-320.jpg)
![Problem 7.10
The first filter is an FIR filter with transfer function and impulse response:
H(z)= (1 + z−2
)3
= 1 + 3z−2
+ 3z−4
+ z−6
⇒ h = [1, 0, 3, 0, 3, 0, 1]
The causal impulse response of the third filter is obtained by the remove restore method, that is,
h(n)= w(n)−w(n − 4), where w(n)= (0.9)n
u(n)]
The first filter has zeros at z = ±j, each being a triple zero. The second filter has poles at
z = ±0.9j, and the third filter has zeros at the fourth roots of unity, i.e., z = ±1, ±j, and a pole
at z = 0.9. The pole/zero plots of the three filters are shown in Fig. P7.13 together with the
corresponding sketches of the magnitude responses |H(ω)| over the right half of the Nyquist
interval 0 ≤ ω ≤ π.
= zeros
= poles
j
-j
-0.9j
-j
0.9j
j
0.9
1-1
π
ω
|H(ω)| |H(ω)| |H(ω)|
ω ω
π π
Fig. P7.13 Pole/zero patterns of Problem 7.10.
Problem 7.11
Multiplying the transfer function factors, we obtain:
H(z)=
(1 −
√
2z−1
+ z−2
)(1 +
√
2z−1
+ z−2
)
(1 + 0.81z−2)(1 − 0.81z−2)
=
1 + z−4
1 − 0.6561z−4
Therefore, the zeros will be at the roots of 1 + z−4
= 0, that is,
z4
= −1 ⇒ z = ejπ(2k+1)/4
, k = 0, 1, 2, 3
that is, the 4th roots of unity rotated by 45o
. The poles are at z = ±0.9m, ±0.9j. Fig. P7.14 shows
the pole/zero locations and the corresponding magnitude response.
The direct and canonical realizations are shown in Fig. P7.15. Their sample processing algorithms
are:
118
= zeros
= poles
-0.9j
0.9j
0.9-0.9
|H(ω)|
ω
π
Fig. P7.14 Pole/zero plot of Problem 7.11.
for each input x do:
v0 = x
y = w0 = v0 + v4 + 0.6561w4
delay(4, v)
delay(4, w)
for each input x do:
w0 = x + 0.6561w4
y = w0 + w4
delay(4, w)
x y
z-1
z-1
z-1
w0
w1
w2
w3
w4
z-1
x y
z-1
z-1
v0
v1
v2
v3
v4
z-1
z-1
w0
w1
w2
w4
z-1
w3
z-1
z-1
z-1
0.6561 0.6561
Fig. P7.15 Direct and canonical realizations of Problem 7.11.
The cascade realization is shown in Fig. P7.16. Its sample processing algorithm is:
for each input x do:
v0 = x − 0.81v2
x1 = v0 −
√
2v1 + v2
delay(2, v)
w0 = x1 + 0.81w2
y = w0 +
√
2w1 + w2
delay(2, w)
Problem 7.12
The zeros are at the roots of 1 + 16z−4
= 0, that is,
119](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-60-320.jpg)
![x
0.0625
w0
w1
w2
y
z-1
z-1
v0
v1
x1
v2-0.25 4
-2 20.5 2
z-1
z-1
-0.25 4
2 2-0.5 2
Fig. P7.19 Cascade realization of Problem 7.12.
Problem 7.13
The following program implements this computer experiment:
/* iirfilt.c - IIR filtering experiments */
#include <stdio.h>
#include <stdlib.h>
void cas2can();
double dir(), can(), cas();
void main()
{
int i, j, K, M;
double x, ydir, ycan, ycas;
double *vdir, *wdir, *wcan, **W;
double **A, **B, *a, *b;
FILE *fpx;
fpx = fopen("x.dat", "r"); /* input data */
printf("no. sections K = ");
scanf("%d", &K);
M=2*K;
wdir = (double *) calloc(M+1, sizeof(double));
vdir = (double *) calloc(M+1, sizeof(double));
wcan = (double *) calloc(M+1, sizeof(double));
a = (double *) calloc(M+1, sizeof(double));
b = (double *) calloc(M+1, sizeof(double));
A = (double **) calloc(K, sizeof(double *));
B = (double **) calloc(K, sizeof(double *));
W = (double **) calloc(K, sizeof(double *));
for (i=0; i<K; i++) {
A[i] = (double *) calloc(3, sizeof(double));
B[i] = (double *) calloc(3, sizeof(double));
W[i] = (double *) calloc(3, sizeof(double));
}
for(i=0; i<K; i++) {
122
printf("enter A[%d] = ", i);
for(j=0; j<3; j++)
scanf("%lf", A[i]+j);
}
for(i=0; i<K; i++) {
printf("enter B[%d] = ", i);
for(j=0; j<3; j++)
scanf("%lf", B[i]+j);
}
cas2can(K, A, a);
cas2can(K, B, b);
printf("nn");
while(fscanf(fpx, "%lf", &x) != EOF) {
ydir = dir(M, a, M, b, vdir, wdir, x);
ycan = can(M, a, M, b, wcan, x);
ycas = cas(K, A, B, W, x);
printf("% .6f % .6f % .6f % .6fn", x, ydir, ycan, ycas);
}
printf("n a, b = n"); /* direct form coefficients */
for (i=0;i<=M;i++)
printf("% .6lf % .6lfn", a[i], b[i]);
}
The program must be linked with the routines cas2can, dir, can, cas, and also conv required
by cas2can and sos required by cas. The number of sections K and the corresponding rows of
the matrices A and B are entered interactively. For example, the user input for the given transfer
functions will be (starting with K = 4):
4
1 -0.373 0
1 -1.122 0.712
1 -0.891 0.360
1 -0.780 0.190
0.313 0.313 0
0.147 0.294 0.147
0.117 0.234 0.117
0.103 0.206 0.103
These inputs may also be saved in a file, say, input.dat, and piped into the program by
iirfilt < input.dat
The input signal xn to be filtered is assumed to be in the file x.dat. The program stops filtering
as soon as the end-of-file of x.dat is encountered. The first 10 of the input and output samples
are shown below:
123](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-62-320.jpg)
![x(n) ydir(n) ycan(n) ycas(n)
1.000000 0.000554 0.000554 0.000554
1.000000 0.006191 0.006191 0.006191
1.000000 0.032979 0.032979 0.032979
1.000000 0.112201 0.112201 0.112201
1.000000 0.275606 0.275606 0.275606
1.000000 0.524036 0.524036 0.524036
1.000000 0.807624 0.807624 0.807624
1.000000 1.043772 1.043772 1.043772
1.000000 1.164600 1.164600 1.164600
1.000000 1.157457 1.157457 1.157457
The outputs computed by dir, can, and cas are of course the same. Fig. P7.20 shows the full
output, plotted together with the input. The impulse response h(n), 0 ≤ n < 50 can be computed
by running this program on a file x.dat containing 1 followed by 49 zeros. It is shown also in
Fig. P7.20. The first 10 impulse response samples computed by the three realizations are shown
below:
x(n) hdir(n) hcan(n) hcas(n)
1.000000 0.000554 0.000554 0.000554
0.000000 0.005637 0.005637 0.005637
0.000000 0.026788 0.026788 0.026788
0.000000 0.079222 0.079222 0.079222
0.000000 0.163405 0.163405 0.163405
0.000000 0.248430 0.248430 0.248430
0.000000 0.283588 0.283588 0.283588
0.000000 0.236148 0.236148 0.236148
0.000000 0.120828 0.120828 0.120828
0.000000 −0.007142 −0.007142 −0.007142
The given filter is an example of a lowpass Butterworth filter and the normalization gains in each
section have been chosen to give unity gain at DC. Therefore, the steady-state DC response will
be
H(ω) ω=0 = H(z) z=1 = 1
This can be observed in the output which rises up to 1 before it dies out to zero. Calling cas2can
on the matrices A and B gives the direct form denominator and numerator coefficients:
a b
1.000000 0.000554
−3.166000 0.003881
4.873631 0.011644
−4.465987 0.019407
2.592519 0.019407
−0.941724 0.011644
0.196860 0.003881
−0.018165 0.000554
0.000000 0.000000
124
Fig. P7.20 Input, output, and impulse response of Problem 7.13.
Problem 7.14
An example of such program is given below. It is patterned after firfilt.c of Problem 4.10.
The program reads and allocates the filter coefficient arrays. The last while loop performs the
actual filtering:
/* canfilt.c - IIR filtering in canonical form */
#include <stdlib.h>
#include <stdio.h>
#define MAX 64 initial allocation size
double can();
void main(int argc, char **argv)
{
FILE *fpa, *fpb; coefficient files
double *a, *b, *w, x, y; coeffs., state, input, output
int L, M, K, i;
int max = MAX, dmax = MAX; initial allocation and increment
if (argc != 3) {
fprintf(stderr, "Usage: canfilt a.dat b.dat <x.dat >y.datn");
exit(0);
}
if ((fpa = fopen(argv[1], "r")) == NULL) {
fprintf(stderr, "Can’t open filter file: %sn", argv[1]);
exit(0);
}
if ((fpb = fopen(argv[2], "r")) == NULL) {
fprintf(stderr, "Can’t open filter file: %sn", argv[2]);
exit(0);
}
a = (double *) calloc(max + 1, sizeof(double));
for (M=0;; M++) {
if (M == max) {
125](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-63-320.jpg)
![max += dmax;
a = (double *) realloc((char *) a, (max + 1) * sizeof(double));
}
if (fscanf(fpa, "%lf", a + M) == EOF) break;
}
M--;
a = (double *) realloc((char *) a, (M + 1) * sizeof(double));
b = (double *) calloc(max + 1, sizeof(double));
for (L=0;; L++) {
if (L == max) {
max += dmax;
b = (double *) realloc((char *) b, (max + 1) * sizeof(double));
}
if (fscanf(fpb, "%lf", b + L) == EOF) break;
}
L--;
b = (double *) realloc((char *) b, (L + 1) * sizeof(double));
K = (L <= M) ? M : L;
w = (double *) calloc((K + 1), sizeof(double));
while(scanf("%lf", &x) != EOF) { process input samples
y = can(M, a, L, b, w, x);
printf("%lfn", y);
}
}
Problem 7.15
An example of such program is given below. The program reads and allocates the filter coefficient
matrices A and B. To facilitate the initial allocation of A, a temporary array a is used to read the
rows of A.dat in a concatenated form, and then it reshapes it into the K×3 matrix A. The last
while loop performs the actual filtering:
/* casfilt.c - IIR filtering in cascade form */
#include <stdlib.h>
#include <stdio.h>
#define MAX 64 initial allocation size
double cas();
void main(int argc, char **argv)
{
FILE *fpa, *fpb; coefficient files
double **A, **B, **W, x, y;
double *a; temporary coefficient array
int M, K, i, j, m;
int max = MAX, dmax = MAX; allocation and increment
if (argc != 3) {
fprintf(stderr, "Usage: casfilt A.dat B.dat <x.dat >y.datn");
exit(0);
}
126
if ((fpa = fopen(argv[1], "r")) == NULL) {
fprintf(stderr, "Can’t open filter file: %sn", argv[1]);
exit(0);
}
if ((fpb = fopen(argv[2], "r")) == NULL) {
fprintf(stderr, "Can’t open filter file: %sn", argv[2]);
exit(0);
}
a = (double *) calloc(max + 1, sizeof(double));
for (M=0;; M++) {
if (M == max) {
max += dmax;
a = (double *) realloc((char *) a, (max + 1) * sizeof(double));
}
if (fscanf(fpa, "%lf", a + M) == EOF) break;
}
a = (double *) realloc((char *) a, (M + 1) * sizeof(double));
if (M%3 != 0) {
fprintf(stderr, "all rows of A must be 3-dimensional");
exit(0);
}
else
K = M / 3;
A = (double **) calloc(K, sizeof(double *));
B = (double **) calloc(K, sizeof(double *));
W = (double **) calloc(K, sizeof(double *));
for (i=0; i<K; i++) {
A[i] = (double *) calloc(3, sizeof(double));
B[i] = (double *) calloc(3, sizeof(double));
W[i] = (double *) calloc(3, sizeof(double));
}
for(i=0; i<K; i++)
for(j=0; j<3; j++)
A[i][j] = a[3*i + j];
for(i=0; i<K; i++)
for(j=0; j<3; j++)
fscanf(fpb, "%lf", B[i] + j);
while(scanf("%lf", &x) != EOF) { process input samples
y = cas(K, A, B, W, x);
printf("%lfn", y);
}
}
The MATLAB version casfilt.m is given below. It calls cas.m N times, where N is the input
length:
% casfilt.m - IIR filtering in cascade form
%
% y = casfilt(B, A, x);
%
127](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-64-320.jpg)
![% B = Kx3 numerator matrix (K = number of sections)
% A = Kx3 denominator matrix
% x = row vector input
% y = row vector output
function y = casfilt(B, A, x)
[K, K1] = size(A);
[N1, N] = size(x);
W = zeros(K,3);
for n = 1:N,
[y(n), W] = cas(K, B, A, W, x(n));
end
Problem 7.16
The two filters are:
H1(z)= 0.70849 ·
1 + z−8
1 − 0.0625z−8
, H2(z)= 0.98021 ·
1 + z−8
1 + 0.94z−8
·
1 − 0.96z−8
1 − 0.98z−8
The scale factors, were chosen such that the magnitude responses are unity at ω = ω0 = 0.5π/8,
that is,
|H1(ω0)| = |H2(ω0)| = 1
The following program implements this computer experiment. The sample processing algorithms
of Examples 7.4.3 and 7.4.4 remain the same, except the input sample x gets multiplied by the
corresponding normalization gains:
/* combex.c - comb filtering examples */
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define A0 1.0
#define A1 0.5
#define A2 0.5
#define A3 0.5
#define G1 0.70849
#define G2 0.98021
void delay();
void main()
{
int n, N;
double s, x, y1, x1, y2;
double *w, *v, *u;
double pi = 4*atan(1.0), w0, w1, w2, w3;
FILE *fps, *fpx, *fpy1, *fpy2;
fps = fopen("s.dat", "w");
128
fpx = fopen("x.dat", "w");
fpy1 = fopen("y1.dat", "w");
fpy2 = fopen("y2.dat", "w");
w = (double *) calloc(9, sizeof(double)); /* filter 1 */
u = (double *) calloc(9, sizeof(double)); /* filter 2, section 1 */
v = (double *) calloc(9, sizeof(double)); /* filter 2, section 2 */
w0 = 0.50 * pi / 8;
w1 = 0.75 * pi / 8;
w2 = 1.00 * pi / 8;
w3 = 3.00 * pi / 8;
printf("enter input length N = ");
scanf("%d", &N);
for (n=0; n<N; n++) {
s = A0 * cos(w0*n) + A1 * cos(w1*n);
x = s + A2 * cos(w2*n) + A3 * cos(w3*n);
w[0] = G1 * x + 0.0625 * w[8]; /* filter 1 */
y1 = w[0] + w[8];
delay(8, w);
u[0] = G2 * x - 0.94 * u[8]; /* filter 2 */
x1 = u[0] + u[8]; /* section 1 output */
delay(8, u);
v[0] = x1 + 0.98 * v[8];
y2 = v[0] - 0.96 * v[8]; /* section 2 output */
delay(8, v);
fprintf(fps, "%.8lfn", s);
fprintf(fpx, "%.8lfn", x);
fprintf(fpy1, "%.8lfn", y1);
fprintf(fpy2, "%.8lfn", y2);
}
}
Fig. P7.21 shows the desired, noise-free, signal s(n) and the noisy signal x(n), which serves as
the input to the comb filters. The corresponding output signals y1(n) and y2(n) from H1(z)
and H2(z) are shown in Fig. P7.22.
The first filter has a short time constant, but does not remove the noise component well. The
second filter has a longer time constant, but it removes the noise very well.
Problem 7.17
Performing a partial fraction expansion on H1(z) and H2(z), we find (where the PFE coefficients
are given with 5-digit accuracy):
H1(z) = −11.33584 +
12.04433
1 − 0.0625z−8
H2(z) = 1.02150 −
0.06191
1 + 0.94z−8
+
0.02063
1 − 0.98z−8
Expanding in powers of z−8
and extracting coefficients, we find the impulse responses, for 0 ≤
n ≤ 16:
129](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-65-320.jpg)
![a = (double *) calloc(M+1, sizeof(double));
b = (double *) calloc(M+1, sizeof(double));
A = (double **) calloc(K, sizeof(double *));
B = (double **) calloc(K, sizeof(double *));
W = (double **) calloc(K, sizeof(double *));
P = (double **) calloc(K, sizeof(double *)); /* allocate */
for (i=0; i<K; i++) {
A[i] = (double *) calloc(3, sizeof(double));
B[i] = (double *) calloc(3, sizeof(double));
W[i] = (double *) calloc(3, sizeof(double));
P[i] = W[i]; /* initialize */
}
for(i=0; i<K; i++) {
printf("enter A[%d] = ", i);
for(j=0; j<3; j++)
scanf("%lf", A[i]+j);
}
for(i=0; i<K; i++) {
printf("enter B[%d] = ", i);
for(j=0; j<3; j++)
scanf("%lf", B[i]+j);
}
cas2can(K, A, a);
cas2can(K, B, b);
printf("nn");
while(fscanf(fpx, "%lf", &x) != EOF) {
ydir = dir(M, a, M, b, vdir, wdir, x);
ycan = ccan(M, a, b, wcan, &p, x);
ycas = ccas(K, A, B, W, P, x);
printf("% .6f % .6f % .6f % .6fn", x, ydir, ycan, ycas);
}
printf("a, b = n");
for (i=0;i<=M;i++)
printf("% .6lf % .6lfn", a[i], b[i]);
}
One minor difference in ccan is that it assumes the same order M for both vectors a and b;
Therefore its usage is:
ycan = ccan(M, a, b, wcan, &p, x);
whereas the usage of can was:
ycan = can(M, a, M, b, wcan, x);
Problem 7.19
The canonical realization is shown in Fig. P7.24.
The linear and circular buffer versions of the sample processing algorithm are as follows:
132
x
0.5 -2
6
y
z-3
s3w3
s0w0
Fig. P7.24 Canonical realization of Problem 7.19.
for each input x do:
w0 = x + 0.5w3
y = 6w0 − 2w3
delay(3, w)
for each input x do:
s3 = tap(3, w, p, 3)
s0 = x + 0.5s3
y = 6s0 − 2s3
∗p = s0
cdelay(3, w, &p)
The impulse response can be obtained by dividing out the numerator and writing
H(z)=
6 − 2z−3
1 − 0.5z−3
= 4 +
2
1 − 0.5z−3
Expanding in powers of z−3
, we have:
H(z)= 4 + 2 1 + (0.5)z−3
+ (0.5)2
z−6
+ (0.5)3
z−9
+ · · ·
which gives the impulse response:
h = 6, 0, 0, 2(0.5), 0, 0, 2(0.5)2
, 0, 0, 2(0.5)3
, 0, 0, . . . = [6, 0, 0, 1, 0, 0, 0.5, . . . ]
n x w0 w1 w2 w3 s0 s1 s2 s3 y = 6s0 − 2s3
0 1 1 0 0 0 1 0 0 0 6
1 0 1 0 0 0 0 1 0 0 0
2 0 1 0 0 0 0 0 1 0 0
3 0 1 0.5 0 0 0.5 0 0 1 1
4 0 0 0.5 0 0 0 0.5 0 0 0
5 0 0 0.5 0 0 0 0 0.5 0 0
6 0 0 0.5 0.52
0 0.52
0 0 0.5 0.5
The states si represent the delay-line outputs:
s(n)=
⎡
⎢
⎢
⎢
⎣
s0(n)
s1(n)
s2(n)
s3(n)
⎤
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎣
w(n)
w(n − 1)
w(n − 2)
w(n − 3)
⎤
⎥
⎥
⎥
⎦
that is, they are the successively delayed versions of w(n), which is the intermediate signal
obtained at the output of the input adder. It has z-transform:
W(z)=
1
D(z)
X(z)=
1
1 − 0.5z−3
X(z)
133](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-67-320.jpg)
![For a delta-function input, we have X(z)= 1, giving:
W(z)=
1
1 − 0.5z−3
= 1 + (0.5)z−3
+ (0.5)2
z−6
+ (0.5)3
z−9
+ · · ·
and in the time domain:
w(n)= [1, 0, 0, 0.5, 0, 0, 0.52
, 0, 0, . . . ]
For the input x = [1, 2, 3], we have:
n x w0 w1 w2 w3 s0 s1 s2 s3 y = 6s0 − 2s3
0 1 1 0 0 0 1 0 0 0 6
1 2 1 0 0 2 2 1 0 0 12
2 3 1 0 3 2 3 2 1 0 18
3 0 1 0.5 3 2 0.5 3 2 1 1
4 0 1 0.5 3 2 1 0.5 3 2 2
5 0 1 0.5 3 1.5 1.5 1 0.5 3 3
6 0 1 0.5 0.25 1.5 0.25 1.5 1 0.5 0.5
Problem 7.20
The full precision direct form denominator is:
a = [1, a1, a2, a3, a4]= [1, −3.502, 5.024, −3.464, 0.979] (P7.1)
The full precision cascade coefficients are:
a0 = [1, a01, a02]= [1, −1.8955, 0.9930]
a1 = [1, a11, a12]= [1, −1.6065, 0.9859]
(P7.2)
The poles are at:
p0 = R0ejω0
, R0 = 0.9965, ω0 = 0.1π
p1 = R1ejω1
, R1 = 0.9929, ω1 = 0.2π
(P7.3)
If we round the direct form coefficient vector a to 2-digit accuracy, we obtain:
ˆa = [1, ˆa1, ˆa2, ˆa3, ˆa4]= [1, −3.50, 5.02, −3.46, 0.98] (P7.4)
and the corresponding direct form transfer function:
ˆHdir(z)=
1
1 + ˆa1z−1 + ˆa2z−2 + ˆa3z−3 + ˆa4z−4
(P7.5)
This filter is unstable. Indeed, its four poles, the roots of the denominator polynomial, are:
ˆp0 = ˆR0ej ˆω0
, ˆR0 = 1.0045, ˆω0 = 0.1045π
ˆp1 = ˆR1ej ˆω1
, ˆR1 = 0.9860, ˆω1 = 0.1989π
(P7.6)
and their conjugates. Thus, the p0 pole has moved outside the unit circle, rendering the filter
unstable. The seemingly benign replacement of a by the rounded ˆa has made the direct form
realization unusable. The impulse responses of the two filters (7.7.1) and (P7.5) are shown in
Fig. P7.25. The latter is blowing up exponentially.
134
Fig. P7.25 Full and 2-digit precision filters.
The cascade realization, on the other hand, is more robust, in the sense that if we round its coeffi-
cients also to 2-digit accuracy, the filter will remain stable. Indeed, rounding the SOS coefficients
(P7.2) to 2-digits gives:
ˆa0 = [1, ˆa01, ˆa02]= [1, −1.90, 0.99]
ˆa1 = [1, ˆa11, ˆa12]= [1, −1.61, 0.99]
(P7.7)
The roots of these sections are:
ˆp0 = ˆR0ej ˆω0
, ˆR0 = 0.9950 ˆω0 = 0.0961π
ˆp1 = ˆR1ej ˆω1
, ˆR1 = 0.9950, ˆω1 = 0.2000π
(P7.8)
and their conjugates. They are both inside the unit circle, therefore, the filter remains stable. The
corresponding cascade transfer function will be:
ˆHcas(z)=
1
1 + ˆa01z−1 + ˆa02z−2
·
1
1 + ˆa11z−1 + ˆa12z−2
(P7.9)
The filters ˆHdir(z) and ˆHcas(z) are not equal. Rounding the cascade coefficients to 2-digit ac-
curacy is not equivalent to rounding the direct form coefficients, because rounding does not
preserve products—the rounded product of two numbers is not equal to the product of the two
rounded numbers. Therefore, rounding does not preserve convolution, that is,
ˆa = a0 ∗ a1 ≠ ˆa0 ∗ ˆa1
Indeed, carrying out the convolution of the two vectors in Eq. (P7.7) gives the following direct
form vector, which is not quite the same as that given by Eq. (P7.4):
ˆa0 ∗ ˆa1 = [1, −3.5100, 5.0390, −3.4749, 0.9801]
The impulse response and the magnitude response of the cascaded filter (P7.9) are shown in
Fig. P7.26, together with the exact magnitude response. The magnitude responses are plotted in
dB, that is, 20 log10 |H(ω)|, over 0 ≤ ω ≤ 0.5π, which is the right half of the Nyquist interval.
Thus, in the cascade case the rounding operation changed the magnitude response somewhat,
but has not affected the stability of the filter.
The sensitivity of the direct form and the robustness of the cascade form can be understood ana-
lytically in the following way. The zeros of a polynomial are indirect functions of the coefficients
135](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-68-320.jpg)
![Fig. P7.26 Impulse and magnitude responses of 2-digit-precision cascade form.
of the polynomial, therefore, small changes in the latter will induce changes in the former. For the
direct form, the induced changes in the zeros can be large, thus, causing them to move outside
the unit circle. In the cascade form, the induced changes remain small. For example, consider
the above fourth order denominator vector a. The relationship of the zeros to the coefficients is
established via the polynomial identity:
(1 − p0z−1
)(1 − p∗
0 z−1
)(1 − p1z−1
)(1 − p∗
1 z−1
)= 1 + a1z−1
+ a2z−2
+ a3z−3
+ a4z−4
A small change in the coefficients will induce a small change in the zeros p0 and p1, that is,
a → a + da ⇒ pi → pi + dpi, i = 0, 1
To determine dp0 and dp1 in terms of da, we take differentials of both sides of the above poly-
nomial identity to get:
− dp0z−1
(1 − p∗
0 z−1
)(1 − p1z−1
)(1 − p∗
1 z−1
)
− (1 − p0z−1
)dp∗
0 z−1
(1 − p1z−1
)(1 − p∗
1 z−1
)
− (1 − p0z−1
)(1 − p∗
0 z−1
)dp1z−1
(1 − p∗
1 z−1
)
− (1 − p0z−1
)(1 − p∗
0 z−1
)(1 − p1z−1
)dp∗
1 z−1
= da1z−1
+ da2z−2
+ da3z−3
+ da4z−4
which is still an identity in z. Setting z = p0 will remove all but the first term in the left-hand-side,
giving the relationship:
−dp0p−1
0 (1 − p∗
0 p−1
0 )(1 − p1p−1
0 )(1 − p∗
1 p−1
0 )= da1p−1
0 + da2p−2
0 + da3p−3
0 + da4p−4
0
Multiplying by p4
0 and solving for dp0, we obtain
dp0 = −
p3
0da1 + p2
0da2 + p0da3 + da4
(p0 − p∗
0 )(p0 − p1)(p0 − p∗
1 )
(P7.10)
Similarly, setting z = p1, we find for dp1:
dp1 = −
p3
1da1 + p2
1da2 + p1da3 + da4
(p1 − p∗
0 )(p1 − p0)(p1 − p∗
1 )
(P7.11)
136
Because, the denominator involves the differences between poles, that is, the distances from all
the other poles, it follows that if the poles are closely clustered, the induced changes in the poles
dpi will be large even if the coefficient changes dai are small. The above sensitivity formulas
generalize easily [2,3,246] to the case of an order-M filter a = [1, a1, a2, . . . , aM] with M zeros
p1, p2, . . . , pM:
dpi =
∂pi
∂a1
da1 +
∂pi
∂a2
da2 + · · · +
∂pi
∂aM
daM
where the partial derivatives are:
∂pi
∂am
= −
pM−m
i
M
j=1
j≠i
(pi − pj)
(P7.12)
In the cascade realization, p0 is associated with the first SOS and p1 with the second. The depen-
dence of p0 on a0 and p1 on a1 is established through the identities
(1 − p0z−1
)(1 − p∗
0 z−1
) = 1 + a01z−1
+ a02z−2
(1 − p1z−1
)(1 − p∗
1 z−1
) = 1 + a11z−1
+ a12z−2
Using the above procedure of taking differentials of both sides and setting z = p0 in the first and
z = p1 in the second, we obtain the induced changes in the poles due to small changes in the
coefficients of the sections:
dp0 = −
p0da01 + da02
(p0 − p∗
0 )
, dp1 = −
p1da11 + da12
(p1 − p∗
1 )
(P7.13)
The sensitivity of each pole depends only on the SOS it belongs to, not on the other sections. In
Eqs. (P7.10) and (P7.11), the sensitivity of each pole is coupled to the other poles, through the
denominator factors of the form (p0 − p1); if p0 is near p1, such factors cause large shifts in the
pole locations.
Even though the sensitivity formulas (P7.10) and (P7.11) involve differential changes, we may
actually apply them to the above example and estimate the pole shifts for the 2-digit case. Writing
the rounded ˆa of Eq. (P7.4) in the form ˆa = a + da, we obtain subtracting (P7.4) and (P7.1):
da = ˆa − a = [0, da1, da2, da3, da4]= [0, 0.002, −0.004, 0.004, 0.001]
Using the numerical values given in Eq. (P7.3), we have
p0 = 0.9965ej0.1π
= 0.9477 + j0.3079
p1 = 0.9929ej0.2π
= 0.8033 + j0.5836
With these values of the poles and dai, we can calculate the pole shifts using (P7.10) and (P7.11).
We find:
dp0 = 0.0023 + j0.0159, dp1 = −0.0033 − j0.0066
Therefore, the new pole locations will be:
ˆp0 = p0 + dp0 = 0.9501 + j0.3238 = 1.0037ej0.1046π
ˆp1 = p1 + dp1 = 0.7999 + j0.5770 = 0.9863ej0.1989π
which compare well with the actual values of Eq. (P7.6). A similar calculation can be made based
on Eq. (P7.13).
137](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-69-320.jpg)
![Problem 7.21
The numerical values of a are shown in the first column of Table P7.1. The second and third
columns of this table show the quantized versions of a, rounded to 5- and 4-digit accuracy.
a 5-digit ˆa 4-digit ˆa ˆacas
1.00000000 1.00000000 1.00000000 1.00000000
−5.56573395 −5.56573000 −5.56570000 −5.56570000
13.05062482 13.05062000 13.05060000 13.05046030
−16.49540451 −16.49540000 −16.49540000 −16.49508213
11.84993647 11.84994000 11.84990000 11.84961490
−4.58649943 −4.58650000 −4.58650000 −4.58633537
0.74713457 0.74713000 0.74710000 0.74710016
relative error 3.2580×10−7
1.6487×10−6
1.9240×10−5
Table P7.1 Exact, 5-digit, and 4-digit coefficients.
The fourth column of Table P7.1 shows the equivalent direct-form coefficient vector ˆacas arising
from the cascade of the quantized second order section coefficients, rounded to 4-digit accuracy.
The rounded ai are:
ˆa0 = [1, −1.8671, 0.9623]
ˆa1 = [1, −1.8468, 0.8992]
ˆa2 = [1, −1.8518, 0.8634]
(P7.14)
The table also shows the relative error in each case, defined as the ratio
relative error =
da
a
=
ˆa − a
a
where a denotes the length of the vector a. As expected, the error in the 5-digit case is less than
that in the 4-digit case. But, the error of the cascade case is larger. However, as we see below, the
4-digit cascade form gives an adequate approximation of the exact filter, whereas both the 5- and
4-digit direct forms fail. This behavior can be traced to the pole sensitivity formula Eq. (P7.12)
which can cause large shifts in the poles even though da is small.
Table P7.2 shows the poles of the exact and approximate filters, that is, the roots of the denomi-
nator polynomials a, 5- and 4-digit ˆa, and of the rounded SOS polynomials ˆa0, ˆa1, and ˆa2.
For the 5-digit direct form, the filter remains stable. But, for the 4-digit direct form, one root of ˆa
has moved outside the unit circle and the third conjugate pair of roots has been replaced by two
real-valued ones, one almost on the unit circle. Thus, the quantized coefficients ˆa correspond
to an unstable filter. By contrast, the roots of the 4-digit quantized cascaded form ˆacas remain
inside the unit circle, and in fact they are practically equal to the unquantized ones. Figure P7.27
shows the impulse responses of the exact filter based on a and the 4-digit-precision filter based
on ˆa, which is unstable.
The impulse responses of the 5-digit case and the 4-digit cascade case based on ˆacas are shown
in Fig. P7.28. The magnitude responses of the full precision, 5-digit direct, and 4-digit cascade
cases are shown in Fig. P7.29. Note that the 5-digit direct no longer meets the passband specs; it
has a passband ripple of almost 2 dB instead of the required 1 dB. The 4-digit cascade response,
on the other hand, meets the specifications.
138
full precision magnitude
0.933557 ± j0.301259 0.980962
0.923398 ± j0.215739 0.948265
0.925911 ± j0.078311 0.929217
5-digit direct magnitude
0.931610 ± j0.300414 0.978849
0.927787 ± j0.216576 0.952730
0.923469 ± j0.079185 0.926857
4-digit direct magnitude
0.967208 ± j0.284105 1.008071
0.888774 ± j0.266869 0.927976
0.999999 0.999999
0.853736 0.853736
4-digit cascade magnitude
0.933552 ± j0.301303 0.980971
0.923396 ± j0.215719 0.948258
0.925902 ± j0.078152 0.929194
Table P7.2 Filter poles.
Fig. P7.27 Impulse responses of full-precision and 4-digit-precision filters. Problem 7.21.
Problem 7.22
See the solution of Problem 7.20.
Problem 7.23
See the solution of Problem 7.20.
139](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-70-320.jpg)
![Fig. P7.28 Impulse responses of 5-digit direct and 4-digit cascade filters. Problem 7.21.
Fig. P7.29 Magnitude responses of exact, 5-digit direct, and 4-digit cascade. Problem 7.21.
Problem 7.24
The transfer function is:
H(z)=
b0 + b1z−1
+ b2z−2
+ b3z−3
1 + a1z−1 + a2z−2 + a3z−3
(P7.15)
Transposed realizations of FIR filters are obtained by the four transposition rules of exchanging
input with output, reversing all flows, replacing nodes by adders, and adders by nodes. These
rules can be applied to the canonical or direct realizations of IIR filters.
The I/O difference equations describing the time-domain operation of the realization and the
corresponding sample processing algorithm can be obtained by introducing as internal states the
contents of the three delay registers at time n: v1(n), v2(n), v3(n). For indexing convenience,
we also introduce the quantity v0(n) which is the overall output obtained at the top adder, that
is,
v0(n)= y(n)= b0x(n)+v1(n)
The output of each delay is the delayed version of the output of the adder below it, for example,
v1(n)= b1x(n − 1)−a1v0(n − 1)+v2(n − 1)= b1x(n − 1)−a1y(n − 1)+v2(n − 1)
140
or, advancing it to the next time instant:
v1(n + 1)= b1x(n)−a1v0(n)+v2(n)= b1x(n)−a1y(n)+v2(n)
Therefore, the overall I/O system, which includes the updating of the internal states, will be:
y(n)= v0(n)= b0x(n)+v1(n)
v1(n + 1)= b1x(n)−a1y(n)+v2(n)
v2(n + 1)= b2x(n)−a2y(n)+v3(n)
v3(n + 1)= b3x(n)−a3y(n)
(P7.16)
which leads to the sample processing algorithm:
for each input sample x do:
y = v0 = b0x + v1
v1 = b1x − a1y + v2
v2 = b2x − a2y + v3
v3 = b3x − a3y
Problem 7.25
In the case of a general transfer function of the type of Eq. (7.1.4), we have the sample processing
algorithm, where we assumed L = M for simplicity:
for each input sample x do:
y = v0 = b0x + v1
for i = 1, 2, . . . , M−1 do:
vi = bix − aiv0 + vi+1
vM = bMx − aMv0
(P7.17)
The following C function transp.c is an implementation.
/* transp.c - IIR filtering in transposed of canonical form */
double transp(M, a, b, v, x) usage: y = transp(M, a, b, v, x);
double *a, *b, *v, x; v = internal state vector
int M; a,b have same order M
{
int i;
v[0] = b[0] * x + v[1]; output y(n)= v0(n)
for (i=1; i<=M-1; i++) state updating
v[i] = b[i] * x - a[i] * v[0] + v[i+1];
v[M] = b[M] * x - a[M] * v[0];
return v[0];
}
Its usage is the same as that of can or dir. The (M+1)-dimensional array v = [v0, v1, . . . , vM]
must be declared and allocated in the main program, for example, by
141](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-71-320.jpg)
![double *v;
v = (double *) calloc(M+1, sizeof(double));
Transposed realizations are not intuitively obvious, but their overall implementation in terms of
Eq. (P7.17) has a certain simplicity and elegance. It corresponds to writing the transfer function
of the filter in a nested form, which is similar to H¨orner’s rule of evaluating a polynomial. For
example, the transfer function (P7.15) can be written as:
H(z)= −(a1z−1
+ a2z−2
+ a3z−3
)H(z)+b0 + b1z−1
+ b2z−2
+ b3z−3
which can be written in the nested form:
H = b0 + z−1
(b1 − a1H)+z−1
(b2 − a2H)+z−1
(b3 − a3H)
Multiplying both sides by the z-transform X(z) of the input x(n), and using Y(z)= H(z)X(z),
we obtain the following nested form for the computation of the output, which is basically the
transposed form:
Y = b0X + z−1
(b1X − a1Y)+z−1
(b2X − a2Y)+z−1
(b3X − a3Y)
The transposed form can also be used in the cascade realization, where each second order section
will be realized in its transposed form. The sample processing algorithm of a 2nd order section
is simply:
for each input sample x do:
y = v0 = b0x + v1
v1 = b1x − a1y + v2
v2 = b2x − a2y
which can replace Eq. (7.2.2).
Problem 7.26
The following program implements filtering in the transposed form, calling the routine transp.c
of Problem 7.25. It replaces canfilt.c of Problem 7.14. Because transp assumes that a and b
have the same length M, this program aborts if the lengths are unequal.
/* trspfilt.c - IIR filtering in transposed form */
#include <stdlib.h>
#include <stdio.h>
#define MAX 64 initial allocation size
double transp();
void main(int argc, char **argv)
{
FILE *fpa, *fpb; coefficient files
double *a, *b, *v, x, y; coeffs., state, input, output
int M, L, i;
int max = MAX, dmax = MAX; initial allocation and increment
if (argc != 3) {
142
fprintf(stderr, "Usage: trspfilt a.dat b.dat <x.dat >y.datn");
fprintf(stderr, "coefficient vectors a,b must have same length");
exit(0);
}
if ((fpa = fopen(argv[1], "r")) == NULL) {
fprintf(stderr, "Can’t open filter file: %sn", argv[1]);
exit(0);
}
if ((fpb = fopen(argv[2], "r")) == NULL) {
fprintf(stderr, "Can’t open filter file: %sn", argv[2]);
exit(0);
}
a = (double *) calloc(max + 1, sizeof(double));
for (M=0;; M++) {
if (M == max) {
max += dmax;
a = (double *) realloc((char *) a, (max + 1) * sizeof(double));
}
if (fscanf(fpa, "%lf", a + M) == EOF) break;
}
M--;
a = (double *) realloc((char *) a, (M + 1) * sizeof(double));
b = (double *) calloc(max + 1, sizeof(double));
for (L=0;; L++) {
if (L == max) {
max += dmax;
b = (double *) realloc((char *) b, (max + 1) * sizeof(double));
}
if (fscanf(fpb, "%lf", b + L) == EOF) break;
}
L--;
b = (double *) realloc((char *) b, (L + 1) * sizeof(double));
if (M != L) {
fprintf(stderr, "coefficient vectors a,b must have same length");
exit(0);
}
v = (double *) calloc(M + 1, sizeof(double));
while(scanf("%lf", &x) != EOF) { process input samples
y = transp(M, a, b, v, x);
printf("%lfn", y);
}
}
Problem 7.27
The transfer function is:
H(z)=
9 − 4z−2
+ 4z−4
1 + 0.84z−2 + 0.25z−4
=
3 − 4z−1
+ 2z−2
1 − 0.4z−1 + 0.5z−2
·
3 + 4z−1
+ 2z−2
1 + 0.4z−1 + 0.5z−2
The transpose of the canonical form is shown in Fig. P7.30. The I/O difference equations are:
143](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-72-320.jpg)
![v(n − 1) and y(n − 1) are available from the previous time step. Note that y(n) is computed
last and then put into the feedback delay to be used in the next time step.
Problem 7.30
The filter G(z) can be implemented by the input/output sample processing algorithm of its
canonical form:
for each x do:
y = 0.2x + 0.8w1
w1 = y
where w1 is its internal state, that is, w1(n)= y(n − 1) in the difference equation: y(n)=
0.2x(n)+0.8y(n − 1). Fig. P7.35 shows the three cases with explicitly assigned signal labels.
Identifying the proper input and output to the filter G(z) and using the above sample processing
algorithm, we obtain the implementation of the three cases:
for each x do:
u = 0.2v4 + 0.8w1
w1 = u
y = v0 = x + u
delay(4, v)
for each x do:
y = x + v4
v0 = 0.2y + 0.8w1
w1 = v0
delay(4, v)
for each x do:
y = u0 = x + v2
v0 = 0.2u2 + 0.8w1
w1 = v0
delay(2, v), delay(2, u)
z-4
G(z)
z-4
G(z)
z-2 z-2
G(z)
(a) (b) (c)
x
u
v0
v4
y x x
v0 v0
u0
u2
v4 v2
y y
Fig. P7.35 Feedback systems of Problem 7.30.
148
Chapter 8 Problems
Problem 8.1
The transfer function is obtained by the period-3 repetition of the sequence b(n)= [0, 1, 2], that
is, in the time- and z-domains:
h(n) = b(n)+b(n − 3)+b(n − 6)+ · · ·
H(z) = B(z)+z−3
B(z)+z−6
B(z)+ · · · =
B(z)
1 − z−3
=
z−1
+ 2z−2
1 − z−3
The direct and canonical realizations are shown in Fig. P8.1. Their sample processing algorithms
are:
for each input x do:
v0 = x
y = w0 = v1 + 2v2 + w3
delay(2, v)
delay(3, w)
for each input x do:
w0 = x + w3
y = w1 + 2w2
delay(3, w)
x y x y
z-1
z-1
z-1
z-1
z-1
z-1
w0 w0
w1 w1
w2 w2
w3 w3
z-1
z-1
v0
v1
v2
2 2
Fig. P8.1 Direct and canonical realizations of Problem 8.1.
For the direct form realization, if the input is a unit impulse so that x = 1 at the first iteration of
the algorithm and zero for all other iterations, then the first three iterations will fill the feedback
delay buffer w with the desired waveform samples. From then on, the feed-forward part of the
algorithm will be zero, giving rise to the following generation algorithm, expressed in its of the
linear and circular buffer versions:
repeat forever:
w0 = w3
delay(3, w)
repeat forever:
∗p = tap(3, w, p, 3)
cdelay(3, w, &p)
The following table shows the initialization and steady part of the linear buffer case. The columns
v0, v1, v2 are the input impulse and its delays: δ(n), δ(n − 1), δ(n − 2). At each iteration, w0 is
replaced by the value of w3 and only then is the row of w’s shifted to the right to give the values
of w1, w2, w3 of the next row. The w-columns are the successively delayed output signals y(n),
y(n − 1), y(n − 2), y(n − 3):
149](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-75-320.jpg)
![n v0 v1 v2 w0 w1 w2 w3 y = w0
0 1 0 0 0 0 0 0 0
1 0 1 0 1 0 0 0 1
2 0 0 1 2 1 0 0 2
3 0 0 0 0 2 1 0 0
4 0 0 0 1 0 2 1 1
5 0 0 0 2 1 0 2 2
6 0 0 0 0 2 1 0 0
7 0 0 0 1 0 2 1 1
8 0 0 0 2 1 0 2 2
The circular version is given in the following table. The buffer entry that contains the current
output sample is shown with an up-arrow symbol and it corresponds to the entry wq defined by
the circulating value of q:
n q w0 w1 w2 w3 y
0 0 ↑0 0 0 0 0
1 3 0 0 0 ↑1 1
2 2 0 0 ↑2 1 2
3 1 0 ↑0 2 1 0
n q w0 w1 w2 w3 y
4 0 ↑1 0 2 1 1
5 3 1 0 2 ↑2 2
6 2 1 0 ↑0 2 0
7 1 1 ↑1 0 2 1
n q w0 w1 w2 w3 y
8 0 ↑2 1 0 2 2
9 3 2 1 0 ↑0 0
10 2 2 1 ↑1 0 1
11 1 2 ↑2 1 0 2
n q w0 w1 w2 w3 y
12 0 ↑0 2 1 0 0
13 3 0 2 1 ↑1 1
14 2 0 2 ↑2 1 2
15 1 0 ↑0 2 1 0
The table entries circulate every D(D + 1)= 12 iterations. Thus, entry n = 3 and n = 15 will
be the same. We used 15 iterations because they contain one complete cycle of the table entries
after the first 3 initializing iterations.
Problem 8.2
Expanding the denominator in powers of z−5
, will cause the period-5 replication of the numerator
sequence b = [1, 2, 3, −4, −5]. The direct and canonical realizations are shown in Fig. P8.2. Their
sample processing algorithms are:
for each input x do:
v0 = x
w0 = v0 + 2v1 + 3v2 − 4v3 − 5v4 + w5
y = w0
delay(4, v)
delay(5, w)
for each input x do:
w0 = x + w5
y = w0 + 2w1 + 3w2 − 4w3 − 5w4
delay(5, w)
The steady parts of the linear and circular buffer generation algorithms are:
repeat forever:
w0 = w5
delay(5, w)
repeat forever:
∗p = tap(5, w, p, 5)
cdelay(5, w, &p)
150
x y x y
z-1
z-1
z-1
z-1
w5
w0
w1
w2
w3
w4
z-1
z-1
z-1
z-1
z-1
w5
w0
w1
w2
w3
w4
z-1
z-1
z-1
z-1
v0
v1
v2
v3
v4
z-1
-5 -5
-4 -4
2 2
3 3
Fig. P8.2 Direct and canonical realizations of Problem 8.2.
The following table shows the initialization and steady parts of the linear buffer case:
n v0 v1 v2 v3 v4 w0 w1 w2 w3 w4 w5 y
0 1 0 0 0 0 1 0 0 0 0 0 1
1 0 1 0 0 0 2 1 0 0 0 0 2
2 0 0 1 0 0 3 2 1 0 0 0 3
3 0 0 0 1 0 −4 3 2 1 0 0 −4
4 0 0 0 0 1 −5 −4 3 2 1 0 −5
5 0 0 0 0 0 1 −5 −4 3 2 1 1
6 0 0 0 0 0 2 1 −5 −4 3 2 2
7 0 0 0 0 0 3 2 1 −5 −4 3 3
8 0 0 0 0 0 −4 3 2 1 −5 −4 −4
9 0 0 0 0 0 −5 −4 3 2 1 −5 −5
10 0 0 0 0 0 1 −5 −4 3 2 1 1
11 0 0 0 0 0 2 1 −5 −4 3 2 2
12 0 0 0 0 0 3 2 1 −5 −4 3 3
13 0 0 0 0 0 −4 3 2 1 −5 −4 −4
14 0 0 0 0 0 −5 −4 3 2 1 −5 −5
The circular version is given in the following table. The buffer entry that contains the current
output sample is shown with an up-arrow symbol and it corresponds to the entry wq defined by
the circulating value of q:
n q w0 w1 w2 w3 w4 w5 y
0 0 1↑ 0 0 0 0 0 1
1 5 1 0 0 0 0 2↑ 2
2 4 1 0 0 0 3↑ 2 3
3 3 1 0 0 −4↑ 3 2 −4
4 2 1 0 −5↑ −4 3 2 −5
5 1 1 1↑ −5 −4 3 2 1
151](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-76-320.jpg)
![n q w0 w1 w2 w3 w4 w5 y
6 0 2↑ 1 −5 −4 3 2 2
7 5 2 1 −5 −4 3 3↑ 3
7 4 2 1 −5 −4 −4↑ 3 −4
9 3 2 1 −5 −5↑ −4 3 −5
10 2 2 1 1↑ −5 −4 3 1
11 1 2 2↑ 1 −5 −4 3 2
n q w0 w1 w2 w3 w4 w5 y
12 0 3↑ 2 1 −5 −4 3 3
13 5 3 2 1 −5 −4 −4↑ −4
14 4 3 2 1 −5 −5↑ −4 −5
15 3 3 2 1 1↑ −5 −4 1
16 2 3 2 2↑ 1 −5 −4 2
17 1 3 3↑ 2 1 −5 −4 3
n q w0 w1 w2 w3 w4 w5 y
18 0 −4↑ 3 2 1 −5 −4 −4
19 5 −4 3 2 1 −5 −5↑ −5
20 4 −4 3 2 1 1↑ −5 1
21 3 −4 3 2 2↑ 1 −5 2
22 2 −4 3 3↑ 2 1 −5 3
23 1 −4 −4↑ 3 2 1 −5 −4
n q w0 w1 w2 w3 w4 w5 y
24 0 −5↑ −4 3 2 1 −5 −5
25 5 −5 −4 3 2 1 1↑ 1
26 4 −5 −4 3 2 2↑ 1 2
27 3 −5 −4 3 3↑ 2 1 3
28 2 −5 −4 −4↑ 3 2 1 −4
29 1 −5 −5↑ −4 3 2 1 −5
n q w0 w1 w2 w3 w4 w5 y
30 0 1↑ −5 −4 3 2 1 1
31 5 1 −5 −4 3 2 2↑ 2
32 4 1 −5 −4 3 3↑ 2 3
33 3 1 −5 −4 −4↑ 3 2 −4
34 2 1 −5 −5↑ −4 3 2 −5
35 1 1 1↑ −5 −4 3 2 1
The table entries circulate every D(D + 1)= 30 iterations. Thus, entry n = 5 and n = 35 will
be the same. We used 35 iterations because they contain one complete cycle of the table entries
after the first 5 initializing iterations.
152
Problem 8.3
The shift is c = D/d = 8/5 = 1.6. The d = 5 possible values of the offset index q are obtained
by iterating Eq. (8.1.36):
q0 = 0
q1 = q0 − c = −
8
5
≡ 8 −
8
5
=
32
5
= 6
2
5
q2 = q1 − c =
32
5
−
8
5
=
24
5
= 4
4
5
q3 = q2 − c =
24
5
−
8
5
=
16
5
= 3
1
5
q4 = q3 − c =
16
5
−
8
5
=
8
5
= 1
3
5
Fig. P8.3 shows the relative locations of the q’s with respect to the circular buffer. The five
q-arrows are now at relative angles ω = 2π/5.
w0 q0
q4
w4
w5
w6
w7
w2
w3 w1
q1
q3
q2
b0
b7
b6
b5
b4
b3
b2
b1
Fig. P8.3 Successive positions of q when d = 5.
Down-truncation gives the following integer values for the q’s and corresponding buffer entries:
[q0, q1, q2, q3, q4]= [0, 6, 4, 3, 1]
w[0], w[6], w[4], w[3], w[1] = [b0, b2, b4, b5, b7]
Up-truncation gives:
[q0, q1, q2, q3, q4]= [0, 7, 5, 4, 2]
w[0], w[7], w[5], w[4], w[2] = [b0, b1, b3, b4, b6]
Rounding to the nearest integer gives:
[q0, q1, q2, q3, q4]= [0, 6, 5, 3, 2]
w[0], w[6], w[5], w[3], w[2] = [b0, b2, b3, b5, b6]
For the linear interpolation case, we have the outputs:
153](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-77-320.jpg)
![q0 = 0 ⇒ y0 = w[0]= b0
6 < q1 < 7 ⇒ y1 = w[6]+(q1 − 6)(w[7]−w[6])=
2
5
b1 +
3
5
b2
4 < q2 < 5 ⇒ y2 = w[4]+(q2 − 4)(w[5]−w[4])=
4
5
b3 +
1
5
b4
3 < q3 < 4 ⇒ y3 = w[3]+(q3 − 3)(w[4]−w[3])=
1
5
b4 +
4
5
b5
1 < q4 < 2 ⇒ y4 = w[1]+(q4 − 1)(w[2]−w[1])=
3
5
b6 +
2
5
b7
Thus, depending on the output method, the following period-5 subsequences will be produced:
[b0, b2, b4, b5, b7, b0, b2, b4, b5, b7, . . . ] (truncate down)
[b0, b1, b3, b4, b6, b0, b1, b3, b4, b6, . . . ] (truncate up)
[b0, b2, b3, b5, b6, b0, b2, b3, b5, b6, . . . ] (round)
[y0, y1, y2, y3, y4, y0, y1, y2, y3, y4, . . . ] (interpolation)
where the y’s were given above.
Problem 8.4
The shift is c = D/d = 8/6 = 4/3. The d = 6 possible values of the offset index q are obtained
by iterating Eq. (8.1.36):
q0 = 0
q1 = q0 − c = −
4
3
≡ 8 −
4
3
=
20
3
= 6
2
3
q2 = q1 − c =
20
3
−
4
3
=
16
3
= 5
1
3
q3 = q2 − c =
16
3
−
4
3
=
12
3
= 4
q4 = q3 − c =
12
3
−
4
3
=
8
3
= 2
2
3
q5 = q4 − c =
8
3
−
4
3
=
4
3
= 1
1
3
Fig. P8.4 shows the relative locations of the q’s with respect to the circular buffer. The six q-arrows
are now at relative angles ω = 2π/6.
Down-truncation gives the following integer values for the q’s and corresponding buffer entries:
[q0, q1, q2, q3, q4, q5]= [0, 6, 5, 4, 2, 1]
w[0], w[6], w[5], w[4], w[2], w[1] = [b0, b2, b3, b4, b6, b7]
Up-truncation gives:
[q0, q1, q2, q3, q4, q5]= [0, 7, 6, 4, 3, 2]
w[0], w[7], w[6], w[4], w[3], w[2] = [b0, b1, b2, b4, b5, b6]
154
w0 q0
q1
q2
q3
q4 q5
w4
w5
w6
w7
w2
w3 w1
b0
b7
b6
b5
b4
b3
b2
b1
Fig. P8.4 Successive positions of q when d = 6.
Rounding to the nearest integer gives:
[q0, q1, q2, q3, q4, q5]= [0, 7, 5, 4, 3, 1]
w[0], w[7], w[5], w[4], w[3], w[1] = [b0, b1, b3, b4, b5, b7]
For the linear interpolation case, we have the outputs:
q0 = 0 ⇒ y0 = w[0]= b0
6 < q1 < 7 ⇒ y1 = w[6]+(q1 − 6)(w[7]−w[6])=
2
3
b1 +
1
3
b2
5 < q2 < 6 ⇒ y2 = w[5]+(q2 − 5)(w[6]−w[5])=
1
3
b2 +
2
3
b3
q3 = 4 ⇒ y3 = w[4]= b4
2 < q4 < 3 ⇒ y4 = w[2]+(q4 − 2)(w[3]−w[2])=
2
3
b5 +
1
3
b6
1 < q5 < 2 ⇒ y5 = w[1]+(q5 − 1)(w[2]−w[1])=
1
3
b6 +
2
3
b7
Thus, depending on the output method, the following period-5 subsequences will be produced:
[b0, b2, b3, b4, b6, b7, b0, b2, b3, b4, b6, b7, . . . ] (truncate down)
[b0, b1, b2, b4, b5, b6, b0, b1, b2, b4, b5, b6, . . . ] (truncate up)
[b0, b1, b3, b4, b5, b7, b0, b1, b3, b4, b5, b7, . . . ] (round)
[y0, y1, y2, y3, y4, y5, y0, y1, y2, y3, y4, y5, . . . ] (interpolate)
where the y’s are given above.
Problem 8.5
This problem is similar to Example 8.1.2. Referring to Fig. 8.1.14, we have for the cases of trun-
cating down, up, and rounding:
[b0, b3, b6, b0, b3, b6, . . . ]= [1, 4, 7, 1, 4, 7, . . . ] (truncate down)
[b0, b2, b5, b0, b2, b5, . . . ]= [1, 3, 6, 1, 3, 6, . . . ] (truncate up)
[b0, b3, b5, b0, b3, b5, . . . ]= [1, 4, 6, 1, 4, 6, . . . ] (round)
155](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-78-320.jpg)
![If we interpolate linearly, we get the sequence:
y0 = b0 = 1
y1 =
1
3
b2 +
2
3
b3 =
1
3
3 +
2
3
4 =
11
3
= 3.667
y2 =
1
3
b5 +
2
3
b6 =
1
3
6 +
2
3
7 =
20
3
= 6.667
so that the period-3 sequence is:
[1, 3.667, 6.667, 1, 3.667, 6.667, . . . ]
Problem 8.6
This problem is similar to Problem 8.3. Inserting the numerical values for the wavetable contents:
b = [b0, b1, b2, b4, b5, b6, b7]= [1, 2, 3, 4, 5, 6, 7, 8]
we find for the period-5 sequences:
[b0, b2, b4, b5, b7, . . . ]= [1, 3, 4, 5, 7 . . . ] (truncate down)
[b0, b1, b3, b4, b6, . . . ]= [1, 2, 3, 5, 7, . . . ] (truncate up)
[b0, b2, b3, b5, b6, . . . ]= [1, 3, 4, 6, 7, . . . ] (round)
[y0, y1, y2, y3, y4, . . . ]= [1, 2.6, 4.2, 5.8, 7.4, . . . ] (interpolation)
where
y0 = b0 = 1
y1 =
2
5
b1 +
3
5
b2 =
2
5
2 +
3
5
3 = 2.6
y2 =
4
5
b3 +
1
5
b4 =
4
5
4 +
1
5
5 = 4.2
y3 =
1
5
b4 +
4
5
b5 =
1
5
5 +
1
5
6 = 5.8
y4 =
3
5
b6 +
2
5
b7 =
3
5
7 +
2
5
8 = 7.4
The case d = 6 is similar to Problem 8.4.
Problem 8.7
An example program that exercises the wavetable routines is listed below. It generates three
wavetables: sinusoidal, square wave, and trapezoidal.
/* wavgenex.c - circular wavetable example */
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
void gdelay2();
double wavgen();
double sine(), square(), trapez();
156
void main(int argc, char **argv)
{
int i, n, D;
double *w1, q1, *w2, q2, *w3, q3, c, A, F;
FILE *fp1, *fp2, *fp3;
if (argc != 4) {
fprintf(stderr, "usage: wavgenex A, c, Dn");
fprintf(stderr, "A = wavetable amplituden");
fprintf(stderr, "c = no. of periods in D samples, F=c/Dn");
fprintf(stderr, "D = wavetable sizen");
exit(0);
}
A = atof(argv[1]);
c = atof(argv[2]);
D = atoi(argv[3]);
F = c / D;
fp1 = fopen("y1.dat", "w");
fp2 = fopen("y2.dat", "w");
fp3 = fopen("y3.dat", "w");
w1 = (double *) calloc(D, sizeof(double));
w2 = (double *) calloc(D, sizeof(double));
w3 = (double *) calloc(D, sizeof(double));
q1 = 0;
q2 = 0;
q3 = 0;
for (i=0; i<D; i++) { /* load wavetables */
w1[q1] = sine(D, i); /* might need the cast w1[(int)q1] */
w2[q2] = square(D/2, i);
w3[q3] = trapez(D, 3*D/4, 0, i);
gdelay2(D-1, 1.0, &q1);
gdelay2(D-1, 1.0, &q2);
gdelay2(D-1, 1.0, &q3);
}
for (n=0; n<D; n++) {
fprintf(fp1, "%lfn", wavgen(D, w1, A, F, &q1));
fprintf(fp2, "%lfn", wavgen(D, w2, A, F, &q2));
fprintf(fp3, "%lfn", wavgen(D, w3, A, F, &q3));
}
}
Problem 8.8
Partial C code for this problem was given in the text, just before Figs. 8.1.20 and 8.1.21.
Problem 8.9
The zero patterns and magnitude responses |H(ω)| of the four filters are shown in Fig. 8.9. The
first filter has H(z)= 1 + z−8
. Therefore, its zeros are the solutions of:
157](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-79-320.jpg)
![Fig. P8.7 Recursive flanger.
double *w3, *p3;
double v[2] = {0., 0.};
double a[2] = {1, -0.5};
double b[2] = {0.3, 0.15};
double a0 = 0.75;
double x, y1, y2, y3, sD;
int D, Ntot, M=1, n;
FILE *fp1, *fp2, *fp3;
fp1 = fopen("y1.dat", "w");
fp2 = fopen("y2.dat", "w");
fp3 = fopen("y3.dat", "w");
if (argc != 3) {
fprintf(stderr, "usage: reverb D Ntot"); use D=20, Ntot=100
exit(0);
}
D = atoi(argv[1]);
Ntot = atoi(argv[2]);
w1 = (double *) calloc(D+1, sizeof(double));
p1 = w1;
w2 = (double *) calloc(D+1, sizeof(double));
p2 = w2;
w3 = (double *) calloc(D+1, sizeof(double));
p3 = w3;
for (n=0; n<Ntot; n++) {
triangular pulse
if (n<=5)
x=n;
else if (n<=10)
x = 10-n;
else
x=0;
160
y1 = plain(D, w1, &p1, a0, x);
y2 = allpass(D, w2, &p2, a0, x);
y3 = lowpass(D, w3, &p3, M, a, b, v, x);
fprintf(fp1, "%lfn", y1);
fprintf(fp2, "%lfn", y2);
fprintf(fp3, "%lfn", y3);
}
}
Problem 8.12
A typical program is as follows:
/* reverb.c - Schroeder’s plain/allpass reverb system */
#define D1 29
#define D2 37
#define D3 44
#define D4 50
#define D5 27
#define D6 31
#define a1 .75
#define a2 .75
#define a3 .75
#define a4 .75
#define a5 .75
#define a6 .75
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double plain(), allpass();
void main()
{
double *w1, *p1;
double *w2, *p2;
double *w3, *p3;
double *w4, *p4;
double *w5, *p5;
double *w6, *p6;
double y1, y2, y3, y4, y5, y, x;
int Ntot=500, n;
FILE *fpy;
fpy = fopen("y.dat", "w");
w1 = (double *) calloc(D1+1, sizeof(double));
p1 = w1;
w2 = (double *) calloc(D2+1, sizeof(double));
161](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-81-320.jpg)
![p2 = w2;
w3 = (double *) calloc(D3+1, sizeof(double));
p3 = w3;
w4 = (double *) calloc(D4+1, sizeof(double));
p4 = w4;
w5 = (double *) calloc(D5+1, sizeof(double));
p5 = w5;
w6 = (double *) calloc(D6+1, sizeof(double));
p6 = w6;
for (n=0; n<Ntot; n++) {
if(n==0) impulse
x = 1;
else
x = 0;
y1 = plain(D1, w1, &p1, a1, x);
y2 = plain(D2, w2, &p2, a2, x);
y3 = plain(D3, w3, &p3, a3, x);
y4 = plain(D4, w4, &p4, a4, x);
y = y1 + 0.9*y2 + 0.8*y3 + 0.7*y4;
y5 = allpass(D5, w5, &p5, a5, y);
y = allpass(D6, w6, &p6, a6, y5);
fprintf(fpy, "%lfn", y);
}
}
Problem 8.13
The difference equations are:
v(n) = y(n − D)−a1v(n − 1)
u(n) = b0v(n)+b1v(n − 1)
y(n) = x(n)+u(n)
The corresponding circular version of the sample processing algorithm is
for each input x do:
sD = tap(D, w, p, D)
v0 = sD − a1v1
u = b0v0 + b1v1
v1 = v0
y = x + u
∗p = y
cdelay(D, w, &p)
162
Problem 8.14
The response to a causal sinusoid will be of the form:
ejωn
u(n)−→ H(ω)ejωn
u(n)+
D+1
i=1
Bipn
i u(n)
where Bi = Aipi/(pi − ejω
), as shown in Problem 6.29. It follows that the response to a delayed
sinusoid will be:
ejω(n−L)
u(n − L)−→ H(ω)ejω(n−L)
u(n − L)+
D+1
i=1
Bipn−L
i u(n − L)
Multiplying both sides by the factor ejωL
and subtracting from the undelayed result, we find that
the response to the finite sinusoid:
x(n)= ejωn
u(n)−u(n − L)
is
y(n)= H(ω)ejωn
u(n)−u(n − L) +
D+1
i=1
Bipn
i u(n)−ejωL
p−L
i u(n − L)
Problem 8.15
A typical C program that utilizes the routines plain and lowpass is given below:
/* reverbex.c - plain and lowpass reverberation of sinusoid */
*
* run with:
* D=300, L=150, N=300, w=0.2pi, w=pi
*
*/
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double plain(), lowpass();
void main(int argc, char **argv)
{
double *w1, *p1;
double *w2, *p2;
double v[2] = {0., 0.};
double a[2] = {1, -0.5};
double b[2] = {0.3, 0.15};
double a0 = 0.75;
double w, x, y1, y2, sD;
double pi = 4*atan(1.0);
int D, L, N, M=1, n;
FILE *fp1, *fp2;
163](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-82-320.jpg)
![fp1 = fopen("ypl.dat", "w");
fp2 = fopen("ylp.dat", "w");
if (argc != 5) {
puts("usage: reverbex D L N wnn"
"D = delay periodn"
"L = time to turn off inputn"
"N = total duration of input n"
"w = in units of pi n");
exit(0);
}
D = atoi(argv[1]);
L = atoi(argv[2]);
N = atoi(argv[3]);
w = atof(argv[4]); in units of pi
w1 = (double *) calloc(D+1, sizeof(double));
p1 = w1;
w2 = (double *) calloc(D+1, sizeof(double));
p2 = w2;
for (n=0; n<N; n++) {
if (n < L) sinusoid on for L samples
x = cos(pi*w*n);
else
x = 0;
y1 = plain(D, w1, &p1, a0, x);
y2 = lowpass(D, w2, &p2, M, a, b, v, x);
fprintf(fp1, "%.16lfn", y1);
fprintf(fp2, "%.16lfn", y2);
}
}
The analytical expression for y(n) can be obtained as follows. Using the MATLAB function
lprevb.m given below, we determine the poles pi and residues Ai of the transfer function:
function [p, A] = lprvb(b, a, D)
% returns poles and residues of lowpass reverberator (delay D)
% [p, A] = lprvb(b, a, D)
%
% b=[b(1), b(2)]
% a=[1, a(2)]
c = [a, zeros(1,D-2), -b];
p=roots(c);
for i=1:D+1,
A(i) = 1 + a(2) / p(i);
for j=1:D+1,
if j ~= i,
A(i) = A(i) * p(i) / (p(i) - p(j));
end
164
end
end
Then, the following MATLAB function calculates the residues Bi and computes N samples of the
output signal y(n):
function [y, p, A] = yon(N, L, w, b, a, D)
% calculate y(n) for a finite-duration sinusoidal input.
[p, A] = lprvb(b, a, D);
z = exp(j * w);
H = 1 / (1 - z^(-D) * (b(1) + z^(-1)*b(2)) / (1 + a(2)*z^(-1)));
for i=1:D+1,
B(i) = A(i) * p(i) / (p(i) - z);
end
for n=0:N-1,
if (n<L),
y(n+1) = H * z^n;
for i=1:D+1,
y(n+1) = y(n+1) + B(i) * p(i)^n;
end
else
y(n+1) = 0;
for i=1:D+1,
y(n+1) = y(n+1) + B(i) * p(i)^n * (1 - z^L / p(i)^L);
end
end
end
Problem 8.16
A typical C program is as follows:
/* karplus.c - Karplus-Strong String Algorithm */
*
* run with:
* D=25, Ntot=500, iseed=1234567
* D=50, Ntot=500, iseed=1234567
*
/*
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double ran(), lowpass();
void main(int argc, char **argv)
{
double *w, *p;
double v[2] = {0., 0.};
double a[2] = {1, 0.0};
165](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-83-320.jpg)
![double b[2] = {0.5, 0.5};
double x, y;
int D, Ntot, M=1, n;
long iseed;
FILE *fpy;
fpy = fopen("y.dat", "w");
if (argc != 4) {
fprintf(stderr, "usage: karplus D Ntot iseed");
exit(0);
}
D = atoi(argv[1]);
Ntot = atoi(argv[2]);
iseed = atol(argv[2]);
w = (double *) calloc(D+1, sizeof(double));
for (n=0; n<=D; n++) initialize
w[n] = ran(&iseed) - 0.5;
p = w;
for (n=0; n<Ntot; n++) {
y = lowpass(D, w, &p, M, a, b, v, 0.0); x=0
fprintf(fpy, "%.16lfn", y);
}
}
The output signals corresponding to D = 25 and D = 50 are shown in Fig. P8.8. Note how
the initial random numbers get smoothed at each D-cycle resulting in a decaying quasi-periodic
signal tuned to the frequency f = fs/D.
Fig. P8.8 Karplus-Strong string algorithm outputs.
Problem 8.17
A block diagram realization is given in Fig. P8.9. The difference equations are the system:
166
w(n) = aw(n − D)+x(n)
y(n) = cx(n)+bw(n − D)
x y
b
c
a
z-D
w0 wD
Fig. P8.9 Prototypical delay effect.
The sample processing algorithm is in its linear and circular buffer forms:
for each input x do:
y = cx + bwD
w0 = awD + x
delay(D, w)
for each input x do:
sD = tap(D, w, p, D)
y = cx + bsD
∗p = asD + x
cdelay(D, w, &p)
Problem 8.18
The block diagrams and sample processing algorithms are essentially given in Section 8.2.4. The
C routines are:
/* plaindel.c - plain reverberating delay */
double tap();
void cdelay();
double plaindel(D, w, p, a, x) usage: y=plaindel(D,w,&p,a,x);
double *w, **p, a, x; p is passed by address
int D;
{
double y;
y = tap(D, w, *p, D); D-th tap delay output
**p = x + a * y; delay input
cdelay(D, w, p); update delay line
return y;
}
/* lpdel.c - lowpass reverberating delay */
double tap(), can();
void cdelay();
double lpdel(D, w, p, M, a, b, v, x)
double *w, **p, *a, *b, *v, x;
167](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-84-320.jpg)
![int D;
{
double y;
y = tap(D, w, *p, D); D-th tap delay output
**p = x + can(M, a, M, b, v, y); delay input
cdelay(D, w, p); update delay line
return y;
}
The routines differ from plain and lowpass in that the filter outputs are taken after the delay
z−D
, instead of before.
Problem 8.19
Typical C code for this problem is:
for (n=0; n<Ntot; n++) {
if (n<L) /* use if (n<1) for impulse response */
x = 1;
else
x = 0;
x1 = plaindel(D1, w1, &p1, a0, x); /* plain */
x2 = plaindel(D2, w2, &p2, a0, x1);
/* x1 = lpdel(D1, w1, &p1, 1, a, b, v1, x); */ /* lowpass */
/* x2 = lpdel(D2, w2, &p2, 1, a, b, v2, x1); */
y = b0 * x + b1 * x1 + b2 * x2; /* output */
fprintf(fpx, "%.16lfn", x);
fprintf(fpy, "%.16lfn", y);
}
where the initializations are:
double x, x1, x2, y, *w1, *w2, *p1, *p2;
double v1[2] = {0., 0.}; /* state of LP filter 1 */
double v2[2] = {0., 0.}; /* state of LP filter 2 */
double a[2] = {1, -0.5}; /* G(z) denominator */
double b[2] = {0.3, 0.15}; /* G(z) numerator */
double b0=1, b1=0.8, b2=0.6 /* feed-forward coefficients */
double a0 = 0.75; /* feedback for plain delay */
w1 = (double *) calloc(D1+1, sizeof(double)); p1 = w1;
w2 = (double *) calloc(D2+1, sizeof(double)); p2 = w2;
Typical output is as in Fig. P8.10.
Problem 8.20
The sample processing routine is as follows. It can easily be generalized to more than two multi-
tap delay segments:
168
Fig. P8.10 Multi-delay effects processor outputs.
/* mlttap.c - multi-tap delay line */
double tap();
void cdelay();
double mlttap(D1, D2, b, a, w, p, x)
int D1, D2;
double b[3], a[3], *w, **p, x;
{
double s[3], y;
s[1] = tap(D1+D2, w, *p, D1);
s[2] = tap(D1+D2, w, *p, D1+D2);
s[0] = x + a[1] * s[1] + a[2] * s[2];
y = b[0] * x + b[1] * s[1] + b[2] * s[2];
**p = s[0];
cdelay(D1+D2, w, p);
return y;
}
Typical output is shown in Fig. P8.11 and is obtained by the code segment:
for (n=0; n<Ntot; n++) {
if (n < L) /* use L=1 or L=200 */
x = 1;
else
x = 0;
y = mlttap(D1, D2, b, a, w, &p, x); /* p passed by reference */
fprintf(fpx, "%.8lfn", x);
fprintf(fpy, "%.8lfn", y);
}
Problem 8.21
From the pole equation, we have:
169](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-85-320.jpg)
![Fig. P8.11 Multi-Tap Delay Line.
zD1+D2
= a1zD2
+ a2 ⇒
|z|D1+D2
= |a1zD2
+ a2| ≤ |a1||z|D2
+ |a2|
(P8.1)
If a pole had |z| ≥ 1, then using the inequality |a1| + |a2| < 1, we would have:
|z|D1+D2
≥ |z|D2
> |a1||z|D2
+ |a2||z|D2
≥ |a1||z|D2
+ |a2|
which contradicts Eq. (P8.1). Thus, all poles must have |z| < 1.
Problem 8.22
Working in the z-domain and denoting by WL(z) and WR(z) the inputs to the delays z−L
and
z−R
, the outputs of these delays will be z−L
WL(z) and z−R
WR(z). Thus, the I/O equations of
the block diagram will be:
YL(z) = cLXL(z)+z−L
WL(z)
WL(z) = z−L
GL(z)WL(z)+bLXL(z)+dRz−R
WR(z)
YR(z) = cRXR(z)+z−R
WR(z)
WR(z) = z−R
GL(z)WR(z)+bRXR(z)+dLz−L
WL(z)
Solving the first and third for the Ws, we get:
WL(z)= zL
YL(z)−cLXL(z) , WR(z)= zR
YR(z)−cRXR(z)
Inserting them into the second and fourth, we get the system:
1 − z−L
GL(z) YL(z)−cLXL(z) − dRz−L
YR(z)−cRXR(z) = bLz−L
XL(z)
1 − z−R
GR(z) YR(z)−cRXR(z) − dLz−R
YL(z)−cLXL(z) = bRz−R
XR(z)
Solving for YL(z), YR(z) in terms of XL(z), XR(z), we get:
YL(z) = HLL(z)XL(z)+HLR(z)XR(z)
YR(z) = HRL(z)XL(z)+HRR(z)XR(z)
170
where
HLL(z) = cL +
1
D(z)
1 − z−R
GR(z) bLz−L
HLR(z) =
1
D(z)
dRbRz−L
z−R
HRL(z) =
1
D(z)
dLbLz−R
z−L
HRR(z) = cR +
1
D(z)
1 − z−L
GL(z) bRz−R
with
D(z)= 1 − z−R
GR(z) 1 − z−L
GL(z) − dLdRz−L
z−R
For the special case when the cross couplings are dL = 0 and dR = 0, we have:
D(z)= 1 − z−R
GR(z) 1 − z−L
GL(z)
and therefore, the transfer functions simplify to:
HLL(z) = cL +
bLz−L
1 − z−LGL(z)
HLR(z) = 0
HRL(z) =
dLbLz−L
z−R
1 − z−RGR(z) 1 − z−LGL(z)
HRR(z) = cR +
bRz−R
1 − z−RGR(z)
The corresponding I/O system is then:
YL(z) = HLL(z)XL(z)
YR(z) = HRL(z)XL(z)+HRR(z)XR(z)
which means that the left channel responds only to the left input (it does not receive any cross-
feedback from the right channel), but the right channel responds to both the right input and the
cross feedback from the left output. If both cross-feedbacks are zero, dL = dR = 0, then the left
and right channels decouple completely, responding independently to their respective inputs. For
the case GL(z)= aL, GR(z)= aR, the sample processing algorithm is straightforward. Denote
the left and right delay-line buffers by
wL = [wL0, wL1, . . . , wLL]= (L+1)-dimensional
wR = [wR0, wR1, . . . , wRR]= (R+1)-dimensional
Let pL and pR be the circular pointers circulating over them, and let the corresponding delay
outputs be sL and sR. Then, the computational sample processing algorithm will be:
171](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-86-320.jpg)
![for each input pair {xL, xR} do:
sL = tap(L, wL, pL, L)
sR = tap(R, wR, pR, R)
yL = cLxL + sL
yR = cRxR + sR
∗pL = bLxL + aLsL + dRsR
∗pR = bRxR + aRsR + dLsL
cdelay(L, wL, &pL)
cdelay(R, wR, &pR)
Problem 8.23
The sample processing algorithm is implemented by the C code segment:
int L = 30, R = 70;
double *wL, *wR, *pL, *pR;
wL = (double *) calloc(L+1, sizeof(double)); pL = wL;
wR = (double *) calloc(R+1, sizeof(double)); pR = wR;
for (n=0; n<Ntot; n++) {
if (n < P)
xL = 1;
else
xL = 0;
sL = tap(L, wL, pL, L);
sR = tap(R, wR, pR, R);
yL = cL * xL + sL;
yR = cR * xR + sR; /* use xR = 0 */
*pL = bL * xL + aL * sL + dR * sR;
*pR = bR * xR + aR * sR + dL * sL;
cdelay(L, wL, &pL);
cdelay(R, wR, &pR);
fprintf(fp1, "%.16lfn", yL);
fprintf(fp2, "%.16lfn", yR);
}
Figures P8.12 and P8.13 show typical outputs. The left output does not start until n = L, there-
fore, the first right output will begin at n = L + R. That output will be fed back into the left and
will come out at n = 2L + R. Similarly, the second left pulse at n = 2L will appear on the right
at n = 2L + R, and so on.
Problem 8.24
The sample processing algorithm is implemented by the C code segment:
if (L > 1) { /* L-point smoother */
h = (double *) calloc(L, sizeof(double));
v = (double *) calloc(L, sizeof(double));
for (n=0; n<L; n++)
h[n] = 1.0 / L;
172
Fig. P8.12 Stereo delays, with dL = dR = 0.3.
Fig. P8.13 Stereo delays, with dL = 0.3, dR = 0.
}
for (n=0; n<N; n++) { /* use N = 600 */
if (n < N/3) /* generate input */
x = A1 * cos(w0 * n); /* use w0 = 0.15 * pi */
else if (n < 2*N/3)
x = A2 * cos(w0 * n);
else
x = A3 * cos(w0 * n);
c = a * c1 + (1-a) * fabs(x); /* control signal */
c1 = c; /* update delay */
g = f(rho, c/c0); /* gain function */
if (L > 1)
G = fir(L-1, h, v, g); /* gain smoothing */
else
G = g; /* no smoothing */
y = G * x; /* compressor output */
173](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-87-320.jpg)
![fprintf(fpx, "%.16lfn", x); /* save input */
fprintf(fpy, "%.16lfn", y); /* save output */
fprintf(fpc, "%.16lfn", c); /* save control */
fprintf(fpg, "%.16lfn", G); /* save gain */
}
where the compressor function is defined by f(x)= xρ−1
, for x ≥ 1:
/* compressor function with 1:rho compressor ratio */
#include <math.h>
double f(rho, x)
double rho, x;
{
if (x >= 1)
return pow(x, rho-1);
else
return 1;
}
The compressing action takes place only above the threshold. For an expander, the gain function
is defined as follows, where now ρ > 1, but it takes effect only below the threshold:
/* expander gain function with 1:rho expander ratio */
#include <math.h>
double f(rho, x)
double rho, x;
{
if (x <= 1)
return pow(x, rho-1);
else
return 1;
}
Problem 8.25
Increasing the threshold to c0 will cause both A1 and A3 to be attenuated. However, A3 because
it is further from c0 than in the case of Fig. 8.2.38, it will be attenuated more. Fig. P8.14 shows
the results.
The noise gate with threshold c0 = 1.5, will suppress both A1 and A3, but A3 by much more, for
the same reasons explained above. Fig. P8.15 shows the results.
Problem 8.26
Assuming the causal sequence {x(0), x(1), . . . } is zero-mean white, then the terms in the convo-
lutional sum will be mutually uncorrelated and zero-mean:
y(n)= h0x(n)+h1x(n − 1)+ · · · + h(n)x(0)
Thus, the output will have zero mean and its variance will be the sum of the variances of the
successive terms:
σ2
y = E[y(n)2
]= h2
0σ2
x + h2
1σ2
x + · · · h2
nσ2
x = σ2
x
n
n=0
h2
m
174
Fig. P8.14 Expander output and gain with increased threshold.
Fig. P8.15 Noise gate output and gain with increased threshold.
Problem 8.27
Assuming a partial fraction expansion of the form:
H(z)=
M
i=1
Ai
1 − piz−1
we have for the impulse response:
hn =
M
i=1
Aipn
i u(n)
It follows that we have the bound:
|hn| =
M
i=1
Aipn
i ≤
M
i=1
|Ai||pi|n
≤
M
i=1
|Ai||pmax|n
= C|pmax|n
where C =
M
i=1 |Ai| and we used the inequality |pi| ≤ |pmax|. The NRR inequality follows now
by squaring and summing the above inequality for |hn| and applying the geometric series on the
right-hand side.
175](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-88-320.jpg)
![Problem 8.28
For ideal filter shapes, the NRR is computed most conveniently using its frequency response
definition:
NRR =
σ2
yv
σ2
v
=
π
−π
|H(ω)|2 dω
2π
For a bandpass filter, H(ω) is non-zero only over the interval ωa ≤ |ω| ≤ ωb, where it is unity.
Thus,
NRR =
σ2
yv
σ2
v
=
−ωa
−ωb
1 ·
dω
2π
+
ωb
ωa
1 ·
dω
2π
=
ωb − ωa
π
Problem 8.29
The following program is an implementation:
/* smooth1.c - 1st-order exponential smoother */
#include <stdio.h>
#include <math.h>
void main(int argc, char **argv)
{
double a, x, y, w1 = 0;
if (argc != 2) {
fprintf(stderr, "usage: smooth1 a <x.dat >y.datn");
exit(0);
}
a = atof(argv[1]);
fprintf(stderr, "%lfn", a);
while(scanf("%lf", &x) != EOF) {
y = a * w1 + (1-a) * x;
w1 = y;
printf("%lfn", y);
}
}
Problem 8.30
The mean of x(n) is mx = s = 5. The mean of y(n) is the same because it is preserved through
the filter. Indeed, taking means of both sides of
y(n)= ay(n − 1)+(1 − a)x(n)
gives
my = amy + (1 − a)mx ⇒ my = mx
The mean-square values may be determine recursively as follows: If we define
s2
n =
1
n
n−1
k=0
(xk − mx)2
176
then we have the recursion:
s2
n+1 = s2
n +
1
n + 1
(xn − mx)2
−s2
n
initialized with s2
0 = 0. The following program will calculate the input and output signals and
the experimental mean-square values. Any discrepancies are due to the filter transients and they
disappear with increasing L:
/* smoothex.c - filtering with 1st order smoother */
#include <stdio.h>
#include <math.h>
double gran();
void main(int argc, char **argv)
{
int n, L;
long iseed;
double s=5, x, y, w1, a, m=0, sigma=1;
double sx2=0, sy2=0;
FILE *fpx, *fpy;
if (argc != 4) {
puts("nUsage: smoothex a iseed L");
exit(0);
}
a = atof(argv[1]);
iseed = atol(argv[2]);
L = atoi(argv[3]);
fpx = fopen("ttx", "w");
fpy = fopen("tty", "w");
for (w1=0, n=0; n<L; n++) {
x = s + gran(m, sigma, &iseed);
y = (1 - a) * x + a * w1;
w1 = y;
fprintf(fpx, "%lfn", x);
fprintf(fpy, "%lfn", y);
sx2 += ((x-s)*(x-s) - sx2) / (n+1);
sy2 += ((y-s)*(y-s) - sy2) / (n+1);
}
printf("theoretical 1/NRR = %lfn", (1+a)/(1-a));
printf("experimental 1/NRR = %lfn", sx2/sy2);
}
Problem 8.31
The requirement that the high-frequency signal s(−1)n
go through unchanged is that the filter
have unity gain at ω = π or z = −1. This requires the value for the constant b:
b
1 − az−1
z=−1
=
b
1 + a
= 1 ⇒ b = 1 + a
177](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-89-320.jpg)
![Fig. P8.25 Filtered s(n) for Q = 400 and Q = 800. Problem 8.36.
Fig. P8.26 Magnitude responses for Q = 80 and Q = 800. Problem 8.36.
% ecgex1.m - simulated ECG + 60Hz noise example
%
% assume 2 beats/sec => 0.5 sec/beat => 500 samples/beat => 1000 samples/sec
% f0=60; Q=60; df=f0/Q;
%
% compute notch filter coefficients
% generate length-500 ecg by calling x0=ecg(500); normalize it to 1.
% replicate three beats x=[x0,x0,x0] and smooth them with x=sgfilt(0, 5, x);
% add 80 percent 60-Hz noise
% and filter it yv=filter(B,A,xv)
% finally, compute filter’s magnitude response.
f0 = 60/1000; Q = 60; df = f0/Q;
w0 = 2 * pi * f0; dw = 2 * pi * df;
b = 1 / (1+tan(dw/2));
B = b * [1, -2*cos(w0), 1];
A = [1, B(2), 2*b-1];
x0 = ecg(500);
x = [x0, x0, x0];
x = sgfilt(0, 5, x); m = max(x); x = x/m;
184
n = 0:1499;
xv = x + 0.5 * cos(w0 * n);
yv = filter(B, A, xv);
w = (0:499)*pi/500; z = exp(-i*w);
h = abs((B(1) + B(2)*z + B(3)*z.^2) ./ (1 + A(2)*z + A(3)*z.^2));
Problem 8.38
The following MATLAB m-file designs the 60 Hz comb filter, generates the noisy ECG and filters
it, and computes the filter’s magnitude response:
% ecgex2.m - simulated ECG + 60Hz square-wave noise example
%
% assume 1 beats/sec => 1 sec/beat => 600 samples/beat => 600 samples/sec
%
% compute notch filter coefficients
% generate length-600 ecg by calling x0=ecg(600); normalize it to 1.
% replicate three beats x=[x0,x0,x0] and smooth them with x=sgfilt(0, 9, x);
% add shifted square wave
% and filter it yv=filter(B,A,xv)
% finally, compute filter’s magnitude response.
fs = 600; f0 = 60; df = 0.75;
w0 = 2 * pi * f0 / fs;
dw = 2 * pi * df / fs;
beta = tan(dw * 10 / 4); % D=10 here
a = (1 - beta) / (1 + beta);
b = (1 + a) / 2;
B = b * [1 0 0 0 0 0 0 0 0 0 -1]; % numerator
A = [1 0 0 0 0 0 0 0 0 0 -a]; % denominator
x0 = ecg(600);
x = [x0, x0];
x = sgfilt(0, 9, x); m = max(x); x = x/m;
n = 0:1199;
v0 = [1 1 1 1 1 -1 -1 -1 -1 -1]; v = v0; % one noise period
for k=1:119, % 120 noise periods
v = [v, v0];
end
v = 2 + v; % noise baseline shift
xv = x + 0.5 * v; % noisy ECG
yv = filter(B, A, xv);
w = (0:500)*pi/500; z = exp(-i*w);
h = abs(b * (1 - z.^10) ./ (1 - a * z.^10));
185](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-93-320.jpg)
![Problem 8.39
Using partial fractions, we expand
H(z)= b
1 + z−D
1 − az−D
= A +
B
1 − az−D
= (A + B)+Baz−D
+ Bz2
z−2D
+ · · ·
where
A = −
b
a
, B = b 1 +
1
a
, A + B = b
Thus, the impulse response is
h = [b, 0, 0, . . . , 0, Ba, 0, 0, . . . , 0, Ba2
, 0, 0, . . . , 0, Ba3
, . . . ]
with the nonzero entries separated by D − 1 zeros. The sum of the squares is
NRR =
∞
n=0
h2
n = b2
+ B2
∞
m=1
a2m
= b2
+ B2 a2
1 − a2
=
b2
(1 − a2
)+b2
a2
(1 + a−2
)
1 − a2
=
2b2
(1 + a)
1 − a2
=
2(1 − a)2
(1 + a)
4(1 − a)(1 + a)
=
1 − a
2
In the filter design parameter β = tan(ΔωD/4), the tangent has argument:
ΔωD
4
=
Δω
4
·
2π
ω1
=
πΔω
2ω1
=
π
2Q
where we used ω1 = 2π/D, or D = 2π/ω1.
Problem 8.40
The following program designs the comb filter, constructs 40 periods of length-50 of the noisy
triangular waveform, and filters them using the circular buffer implementation of the canonical
realization of the filter:
/* combex.c - IIR comb filter with noisy periodic input */
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double gran(), tap();
void cdelay(), triang();
void main(int argc, char **argv)
{
int n, i, N, D;
long iseed;
double a, b, pi=4*atan(1.0), dw = 0.0008 * pi;
double *s, *w, *p;
double x, y, sD;
FILE *fpy, *fps, *fpx;
if (argc != 4) {
186
puts("nUsage: combex D N iseed");
exit(0);
}
D = atoi(argv[1]); period
N = atoi(argv[2]); number of periods
iseed = atol(argv[3]); initial seed
fpy = fopen("y.dat", "w"); filter output
fps = fopen("s.dat", "w"); noise-free input
fpx = fopen("x.dat", "w"); noisy input
a = (1 - tan(D*dw/4)) / (1 + tan(D*dw/4));
b = (1 - a) / 2;
s = (double *) calloc(D, sizeof(double));
triang(D, s); one period of triangular wave
w = (double *) calloc(D+1, sizeof(double));
p = w;
for (i=0; i<N; i++) generate N periods
for (n=0; n<D; n++) {
x = s[n] + 0.5 * gran(0., 1., &iseed);
sD = tap(D, w, p, D);
*p = b * x + a * sD;
y = (*p) + sD;
cdelay(D, w, &p);
fprintf(fps, "%lfn", s[n]);
fprintf(fpx, "%lfn", x);
fprintf(fpy, "%lfn", y);
}
}
/* -------------------------------------------------------- */
/* triang.c - triangular wave */
void triang(D, s)
int D;
double *s;
{
int i;
double D4 = D/4.0;
for (i = 0; i<D4; i++)
s[i] = i / D4;
for (i=D4; i<3*D4; i++)
s[i] = 1 + (D4 - i) / D4;
for (i=3*D4; i<D; i++)
s[i] = -1 + (i - 3*D4) / D4;
}
187](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-94-320.jpg)
![Problem 8.41
The complementarity properties may be verified for each pair. For example, for filter 2, we have
half-order M = 8/2 = 4. Thus, we must verify:
HLP(z)+HBP(z)=
1
16
(1 + z−2
)2
(−1 + 6z−2
− z−4
)+
1
16
(1 − z−2
)4
= z−4
which follows by expanding out the terms. Fig. P8.27 shows the frequency responses in the four
cases. The frequency scale is in MHz and extends only up to about 4.5 MHz. For reference, the
Nyquist frequency here is fs/2 = 2fsc = 7.159 MHz, so that the subcarrier frequency shown in
the figure is only one-quarter of the Nyquist interval.
Fig. P8.27 Lowpass/bandpass and highpass/bandstop responses. Problem 8.41.
Problem 8.42
See Table P8.1. Note that the circular pointer p points to the buffer entry w[q], whereas the D-th
tap output is the content of w[qD]. At each time instant, the content of w[q] is updated by
w[q]= w[qD]+x, where the scaling by N is done afterwards. Table P8.1 shows the results. The
advantage of the circular implementation is that only one entry of w, namely, w[q], is changed
at each time instant. This is efficient in hardware implementations and for large values of D.
Problem 8.43
The following is a complete C program for performing signal averaging. The inputs to the program
are the period D and the number N of periods to be averaged. The input data are taken from
stdin and the averaged period is output to the stdout:
/* sigav.c - signal averaging */
#include <stdio.h>
#include <stdlib.h>
void cdelay();
double tap();
188
void main(int argc, char **argv)
{
double x, *w, *p;
int D, N, i, k;
if (argc != 3) {
fprintf(stderr, "usage: sigav D N < x.dat > y.datn");
fprintf(stderr, "x.dat must contain N periods of length Dn");
exit(0);
}
D = atoi(argv[1]);
N = atoi(argv[2]);
w = (double *) calloc(D+1, sizeof(double));
p = w;
for (k=0; k<N; k++)
for (i=0; i<D; i++)
if (scanf("%lf", &x) != EOF) { /* keep reading input samples */
*p = tap(D, w, p, D) + x / N;
cdelay(D, w, &p);
}
else {
fprintf(stderr, "Input must have %d periods of length %dn", N, D);
exit(0);
}
for (i=0; i<D; i++)
printf("%.12lfn", tap(D, w, p, D-i));
}
n x qD q w0 w1 w2 w3 y = wq/4
0 s0 3 0 s0 0 0 0 s0/4
1 s1 2 3 s0 0 0 s1 s1/4
2 s2 1 2 s0 0 s2 s1 s2/4
3 s0 0 1 s0 2s0 s2 s1 2s0/4
4 s1 3 0 2s1 2s0 s2 s1 2s1/4
5 s2 2 3 2s1 2s0 s2 2s2 2s2/4
6 s0 1 2 2s1 2s0 3s0 2s2 3s0/4
7 s1 0 1 2s1 3s1 3s0 2s2 3s1/4
8 s2 3 0 3s2 3s1 3s0 2s2 3s2/4
9 s0 2 3 3s2 3s1 3s0 4s0 s0
10 s1 1 2 3s2 3s1 4s1 4s0 s1
11 s2 0 1 3s2 4s2 4s1 4s0 s2
12 − − − − 4s2 4s1 4s0 −
Table P8.1 Signal averager with circular delay for D = 3, N = 4.
189](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-95-320.jpg)
![Problem 8.44
The following program generates N periods of length D of the noisy signal, and averages these
periods using the circular buffer form of the averager. After processing the ND input samples,
it writes out the contents of the delay line in reverse order, so that the files y.dat and s.dat will
contain the averaged periods and the noise-free period:
/* sigavex.c - signal averaging example */
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double gran(), tap();
void cdelay();
void main(int argc, char **argv)
{
int n, i, N, D;
long iseed;
double *s, x, *w, *p;
double sD;
FILE *fpx, *fpy, *fps;
if (argc != 4) {
puts("nUsage: sigavex D N iseed");
exit(0);
}
D = atoi(argv[1]); period
N = atoi(argv[2]); number of periods
iseed = atol(argv[3]); noise seed
fpx = fopen("x.dat", "w");
fpy = fopen("y.dat", "w");
fps = fopen("s.dat", "w");
s = (double *) calloc(D, sizeof(double));
for (i=0; i<D; i++)
if (i < D/2)
s[i] = 1;
else
s[i] = -1;
w = (double *) calloc(D+1, sizeof(double));
p = w;
for (i=0; i<N; i++) accumulate N periods
for (n=0; n<D; n++) { each of length D
x = s[n] + gran(0., 1., &iseed); noisy input
sD = tap(D, w, p, D); D-th tap of averager
*p = sD + x; accumulate x
cdelay(D, w, &p); update delay line
fprintf(fpx, "%.12lfn", x); save input sample
}
for (n=0; n<D; n++) { write averaged period
190
fprintf(fps, "%lfn", s[n]);
fprintf(fpy, "%.12lfn", tap(D, w, p, D-n)/N);
}
}
Problem 8.45
The following MATLAB function generates the noisy ECG and smoothes it once and twice by
Savitzky-Golay filter of length N and smoothing order d:
% sgecgex.m - SG filtering of noisy ECG
L=500;
x0 = ecg(L)’; % generate ECG
x = sgfilt(0, 15, x0);
mx = max(x);
x = x/mx; % unity max
rand(’normal’); rand(’seed’, 10033);
v = 0.3 * rand(L, 1); % generate noise
xv = x + v; % generate noisy ECG
N=11; d=2;
y1 = sgfilt(d, N, xv); % first pass through SG
y2 = sgfilt(d, N, y2); % second pass
The filtering twice through the SG filter improves the smoothing operation but introduces longer
transients in the filters. The following figures should be compared with the single-pass cases of
Figs. 8.3.33 and 8.3.34.
Fig. P8.28 Double SG filtering; N = 11, 41 and d = 0.
191](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-96-320.jpg)
![Problem 9.4
The expected DFT index will be at
k = N
f
fs
= 16 ·
18
8
= 36
Because k is greater than N − 1 = 15, it may be reduced mod-N down to:
k = 36 − 2 × 16 = 4
It corresponds to the DFT frequency fk = kfs/N = 4 · 8/16 = 2 kHz. This is frequency within
the Nyquist interval that is aliased with f = 18 kHz; indeed, we have
falias = f mod fs = 18 mod 8 = 18 − 2 × 8 = 2 kHz
The −2 kHz or −18 kHz component will be represented by the DFT index N − k = 16 − 4 = 12.
Problem 9.5
To be able to resolve a peak, the width of the main lobe of the data window must be equal or
smaller than the width of the peak. This gives the condition
fs
L
≤ Δf ⇒ L ≥
fs
Δf
or, in our case, L ≥ 8000/20 = 400 samples. The duration of these 400 samples will be
TR = LT ≥
L
fs
=
1
Δf
=
1
20
= 50 msec
To avoid wrap-around errors in the inverse FFT, the FFT length must be at least L, i.e., N ≥ L =
400. The next power of two is N = 512. We must pad 112 zeros to the end of the length-400
signal before the FFT is taken.
Problem 9.6
The following MATLAB function generates the time data and computes the required DTFTs with
the help of the function dtft.m, included below, which replaces the C functions dtft.c and
dtftr.c:
% dtftexp.m - rectangular/hamming windowed sinsuoids
L = 200; N = 200;
w0 = 0.1*pi;
w = (0:N-1) * 0.2 * pi / N; % frequency range
n = 0:(L-1);
wh = 0.54 - 0.46 * cos(2*pi*n/(L-1)); % Hamming window
xrec = cos(w0 * n);
xham = wh .* xrec; % windowed data
Xrec = abs(dtft(xrec, w)); % DTFT of rectangular data
Xham = abs(dtft(xham, w)); % DTFT of Hamming data
save xrec.dat xrec /ascii;
save xham.dat xham /ascii;
save frec.dat Xrec /ascii;
save fham.dat Xham /ascii;
194
where
% dtft.m - DTFT of a signal at a frequency vector w
%
% X = dtft(x, w);
%
% x = row vector of time samples
% w = row vector of frequencies in rads/sample
% X = row vector of DTFT values
%
% based on and replaces both dtft.c and dtftr.c
function X = dtft(x, w)
[L1, L] = size(x);
z = exp(-j*w);
X = 0;
for n = L-1:-1:0,
X = x(n+1) + z .* X;
end
Problem 9.7
The following MATLAB function generates the time data and computes the required DTFTs:
% sinexp.m - 3 sinusoids of length L=10
L=10; N=256;
f1 = 0.2; f2 = 0.25; f3 = 0.3;
n = 0:(L-1);
x = cos(2*pi*f1*n) + cos(2*pi*f2*n) + cos(2*pi*f3*n);
w = 0.54 - 0.46 * cos(2*pi*n/(L-1));
xham = w.*x;
X = abs(fft(x,N));
Xham = abs(fft(xham,N));
For the other cases, use L = 10, 40, 100.
Problem 9.8
The following C program generates the length-L time signal and computes its N-point DFT, with
L = 10, 20, 100 and N = 256. Instead of the DFT routine dft.c, one could use fft.c. The
32-point DFT is extracted by keeping one out of every 256/32 = 8 points of the 256-point DFT.
/* dtftex.c - physical vs. computational resolution example */
#include <stdio.h>
#include <cmplx.h>
#include <stdlib.h>
void dtftr();
195](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-98-320.jpg)
![main()
{
int L, N, n, k;
double *x, pi = 4 * atan(1.0);
double f1 = 0.20, f2 = 0.25, f3 = 0.30;
complex *X;
FILE *fpdtft;
fpdtft= fopen("dtft.dat", "w"); DTFT file
printf("enter L, N: ");
scanf("%d %d", &L, &N);
x = (double *) calloc(L, sizeof(double));
X = (complex *) calloc(N, sizeof(complex));
for (n=0; n<L; n++)
x[n] = cos(2*pi*f1*n) + cos(2*pi*f2*n) + cos(2*pi*f3*n);
dft(L, x, N, X); N-point DFT
for (k=0; k<N; k++)
fprintf(fpdtft, "%lfn", cabs(X[k]));
}
Problem 9.9
See Example 9.2.1.
Problem 9.10
Using the relationship k = Nf/fs, we calculate the group-A and group-B DFT indices:
group A kA = 22.3, 24.6, 27.3, 30.1
group B kB = 38.7, 42.8, 47.3
and their rounded versions:
group A: kA = 22, 25, 27, 30
group B: kB = 39, 43, 47
Because each keypress generates one tone from group A and one from group B, the DTFT will
have peaks at the corresponding pair of DFT indices (and their negatives, that is, N − kA and
N − kB.)
Problem 9.11
The time signal and its DTFT are shown in Figs. P9.1 and P9.2, both for the rectangular and
Hamming windows.
196
Fig. P9.1 Rectangularly and Hamming windowed DTMF sinusoids.
Fig. P9.2 Spectra of DTMF sinusoids.
Problem 9.12
Acting on x by the 4×8 DFT matrix, we obtain:
X =
⎡
⎢
⎢
⎢
⎣
X(0)
X(1)
X(2)
X(3)
⎤
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎣
1 1 1 1 1 1 1 1
1 −j −1 j 1 −j −1 j
1 −1 1 −1 1 −1 1 −1
1 j −1 −j 1 j −1 −j
⎤
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1
2
2
1
2
1
1
2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎣
12
0
0
0
⎤
⎥
⎥
⎥
⎦
The mod-4 wrapped version of x is ˜x = [3, 3, 3, 3]. Its 4-point DFT is
X =
⎡
⎢
⎢
⎢
⎣
X(0)
X(1)
X(2)
X(3)
⎤
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎣
1 1 1 1
1 −j −1 j
1 −1 1 −1
1 j −1 −j
⎤
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎣
3
3
3
3
⎤
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎣
12
0
0
0
⎤
⎥
⎥
⎥
⎦
For the IDFT we use the formula IDFT(X)= [DFT(X∗
)]∗
/N:
197](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-99-320.jpg)
![x =
⎡
⎢
⎢
⎢
⎣
x(0)
x(1)
x(2)
x(3)
⎤
⎥
⎥
⎥
⎦
=
1
4
⎡
⎢
⎢
⎢
⎣
1 1 1 1
1 −j −1 j
1 −1 1 −1
1 j −1 −j
⎤
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎣
12
0
0
0
⎤
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎣
3
3
3
3
⎤
⎥
⎥
⎥
⎦
Problem 9.13
The 8-point FFT is:
x =
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
5
−1
−3
−1
5
−1
−3
−1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
8−FFT
−→ X =
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
0
0
16
0
8
0
16
0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
It follows that the FFT spectrum will show peaks at the frequencies:
X(ω2)= X(ω6)∗
= X(−ω2)∗
= 16, X(ω4)= 8
Problem 9.14
The IFFT is obtained by conjugating X, computing its FFT, and dividing by 8. The required FFT of
X∗
is:
X∗
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
0
4
4j
4
0
4
−4j
4
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
8−FFT
−→
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
16
8
0
−8
−16
8
0
−8
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⇒ x =
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
2
1
0
−1
−2
1
0
−1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
Using the inverse DFT formula, we express x(n) as a sum of sinusoids:
x(n) =
1
8
7
k=0
X(k)ejωkn
=
1
8
4ejω1n
− 4jejω2n
+ 4ejω3n
+ 4ejω5n
+ 4jejω6n
+ 4ejω7n
=
1
2
ejω1n
+
1
2j
ejω2n
+
1
2
ejω3n
+
1
2
e−jω3n
−
1
2j
e−jω2n
+
1
2
e−jω1n
= cos(ω1n)+ sin(ω2n)+ cos(ω3n)= cos(πn/4)+ sin(πn/2)+ cos(3πn/4)
which agrees with the above IFFT values at n = 0, 1, . . . , 7. We used the values of the DFT fre-
quencies:
ω1 =
2π
8
=
π
4
, ω2 =
2π2
8
=
π
2
, ω3 =
2π3
8
=
3π
4
The DFT frequencies ω7, ω6, ω5 are the negatives (modulo-2π) of ω1, ω2, ω3, respectively.
198
Problem 9.15
Figure P9.3 shows the procedure. The (4N)-dimensional time data vector is shuffled once into
two (2N)-dimensional subvectors by putting every other one of its entries on top and the remain-
ing entries on the bottom. Then, each of the (2N)-dimensional vectors is shuffled again into two
N-dimensional subvectors.
shuffling
N-FFT
N-FFT
N-FFT
N-FFT
DFT merging
4N 4N
2N 2N
2N 2N
N N
N N
N N
N N
W2N
k
W2N
k
W4N
k
Fig. P9.3 Performing large FFTs.
The four N-dimensional vectors are then shipped over to the FFT processor, one at a time, and
their FFTs are returned into four N-dimensional DFT vectors. These are then merged first into
(2N)-point DFTs and then into the desired (4N)-point DFT. The first merging stage require mul-
tiplication by the twiddle factors Wk
2N, k = 0, 1, . . . , N − 1. The second stage uses the factors
Wk
4N, k = 0, 1, . . . , 2N − 1. The butterfly operations can be done is small parts, for example, in
groups of N operations at a time so that they can be implemented by the DSP hardware.
The total cost is the cost of four N-point FFTs, the cost of two N-fold multiplications for the first
merging stage, and a (4N)-fold multiplication for the second stage:
cost = 4
1
2
N log2 N + 2N + 2N = 2N log2 N + 4N
If the (4N)-point FFT could be done in one pass, we would have cost:
cost =
1
2
(4N)log2(4N)= 2N log2(4N)
It is easily verified that the costs of the two methods are the same. The indirect method is of
course much slower because of all the I/O time for shipping data back and forth from secondary
storage to the FFT processor.
Problem 9.16
The ordinary convolution is
y = h ∗ x = [1, 3, 4, 6, 7, 6, 4, 3, 1]
199](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-100-320.jpg)
![Reducing this mod-4 gives: ˜y = [9, 9, 8, 9]. The alternative method is to reduce first x and h
mod-4:
˜h = [2, 2, 1, 2], ˜x = [2, 1, 1, 1],
then, compute their ordinary convolution:
˜h ∗ ˜x = [4, 6, 6, 9, 5, 3, 2],
and, finally, reduce that modulo 4 to get ˜y = [9, 9, 8, 9].
Problem 9.17
Carrying out the 8-point FFT gives:
x =
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
4
2
4
−6
4
2
4
−6
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
8−FFT
−→ X =
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
8
0
−16j
0
24
0
16j
0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
Taking every other entry of the result gives the 4-point FFT, namely,
X =
⎡
⎢
⎢
⎢
⎣
8
−16j
24
16j
⎤
⎥
⎥
⎥
⎦
And, taking every other entry of that gives the 2-point FFT:
X =
8
24
Can we get these results directly? Yes. To do the 4-point FFT, first reduce the signal modulo 4 to
get
˜x =
⎡
⎢
⎢
⎢
⎣
8
4
8
−12
⎤
⎥
⎥
⎥
⎦
4−FFT
−→ X =
⎡
⎢
⎢
⎢
⎣
8
−16j
24
16j
⎤
⎥
⎥
⎥
⎦
Putting the above 8-point FFT into the 8-point inverse FFT formula gives
x(n)=
1
8
7
k=0
X(k)ejωkn
=
1
8
8ejω0n
− 16jejω2n
+ 24ejω4n
+ 16jejω6n
Noting that for an 8-point FFT, ω0 = 0, ω2 = π/2, ω4 = π, and ω6 = 3π/2 ≡ −π/2, and using
Euler’s formula we obtain:
x(n)= 1 + 4 sin(
πn
2
)+3 cos(πn)
which agrees with the given x, for n = 0, 1, · · · , 7.
200
Problem 9.18
Consider the length-7 signal
[1 − a, 2 − b, 3, 4, 5, a, b]
It should be evident that, for any values of the parameters a and b, its mod-5 reduction is
[1, 2, 3, 4, 5].
Problem 9.19
Using the finite geometric series and the definition WN = e−2πj/N
, gives
1
N
N−1
n=0
Wkn
N =
1
N
1 − WNk
N
1 − Wk
N
=
1
N
1 − e−2πjk
1 − e−2πjk/N
If k = 0, the numerator vanishes but not the denominator. Thus, in this case the answer is zero.
If k = 0, we have a ratio of 0/0, but in this case, we can just go back to the series definition and
set k = 0. All of the N terms in the sum become unity, giving N/N = 1.
In conclusion, the answer is zero if k = 0, and unity if k = 0. Note, that these arguments break
down if k is allowed to go beyond the range 0 ≤ k ≤ N − 1, because whenever it is a multiple of
N, we would get unity as an answer (and zero otherwise).
Problem 9.20
For a general power p, we have:
W
p
pN = e−2πj/(pN)
p
= e−2πj/N
= WN
The DTFT of a length-L signal and its N-point and (pN)-point DFTs are:
X(ω)=
L−1
n=0
x(n)e−jωn
XN(k)=
L−1
n=0
x(n)Wkn
N
XpN(k)=
L−1
n=0
x(n)Wkn
pN
Replacing k by pk in the third and using part (a), we have:
XpN(pk)=
L−1
n=0
x(n)W
pkn
pN =
L−1
n=0
x(n)Wkn
N = XN(k)
Problem 9.21
This follows from the results of Problem 9.20 by setting N = 8 and p = 2, which implies:
X8(k)= X16(2k), k = 0, 1, . . . , 15
A geometrical way to understand this is to recognize that the 8th roots of unity are a subset of
the 16th roots of unity and are obtained by taking every other one of the latter.
201](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-101-320.jpg)
![Problem 9.22
Using a trig identity to write the product of sines as a sum of sines, we get the equivalent expres-
sion:
x(t)= cos(24πt)+2 sin(12πt)cos(8πt)= cos(24πt)+ sin(20πt)+ sin(4πt)
The frequencies of the three terms are 12, 10, and 2 kHz. Their aliased versions that lie inside
the Nyquist interval [−4, 4] kHz are, respectively, 4, 2, and 2 kHz. Thus, the aliased signal that
would be reconstructed by an ideal reconstructor will be:
x(t)= cos(8πt)+ sin(4πt)+ sin(4πt)= cos(8πt)+2 sin(4πt)
Setting t = nT = n/8, we get the time samples:
x(n)= cos(πn)+2 sin(πn/2)
Evaluating it at n = 0, 1, . . . , 7 gives the values, and their FFT:
x = [1, 1, 1, −3, 1, 1, 1, −3]
8−FFT
−→ X = [0, 0, −8j, 0, 8, 0, 8j, 0]
The same DFT values can be determined without any computation by comparing x(n) with the
inverse DFT. Using Euler’s formula, setting ω4 = π, ω2 = π/2, ω6 = −(π/2)mod(2π), and
multiplying and dividing by 8, we get:
x(n)= ejπn
− jejπn/2
+ je−jπn/2
=
1
8
−8jejω2n
+ 8ejω4n
+ 8jejω6n
Comparing with the 8-point IDFT, gives the above DFT vector:
x(n)=
1
8
7
k=0
X(k)ejωkn
Problem 9.23
Using Euler’s formula, we write:
x(n) = 1 + 2 sin(
πn
4
)−2 sin(
πn
2
)+2 sin(
3πn
4
)+3(−1)n
= 1 − jejπn/4
+ je−jπn/4
+ jejπn/2
− je−jπn/2
− jej3πn/4
+ je−j3πn/4
+ 3ejπn
=
1
8
8ejω0n
− 8jejω1n
+ 8jejω2n
− 8jejω3n
+ 24ejω4n
+ 8jejω5n
− 8jejω6n
+ 8jejω7n
where we reordered the terms according to increasing DFT index. Comparing with the IDFT
formula, we get:
X = [8, −8j, 8j, −8j, 24, 8j, −8j, 8j]
202
Problem 9.24
Using Euler’s formula, we write
x(n) = 0.5eπjn/2
+ 0.5e−πjn/2
+ eπjn/8
+ e−πjn/8
=
1
16
8eπjn/2
+ 8e3πjn/2
+ 16eπjn/8
+ 16e15πjn/8
where, we shifted the negative frequency exponents by adding 2πn to them, e.g., replaced −πn/8
by 2πn − πn/8 = 15πn/8. Now, comparing with the 16-point IDFT formula
x(n)=
1
16
15
k=0
X(k)ejωkn
and noting that π/2 = 8π/16 = ω4, 3π/2 = 24π/16 = ω12, π/8 = 2π/16 = ω1, and
15π/8 = 30π/16 = ω15, we may identify the (non-zero) values of X(k) as follows:
X(1)= 16, X(4)= 8, X(12)= 8, X(15)= 16, or,
X = [0, 16, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 16]
Problem 9.25
Arguing in a similar fashion, we find the (non-zero) values of the 32-point DFT:
X(2)= 32, X(8)= 16, X(24)= 16, X(30)= 32
Problem 9.26
Write x(n) in its complex sinusoidal form:
x(n)= 0.5ejω0n
− jejω4n
+ jωjω12n
+ 1.5ejω8n
(P9.1)
where ω0 = 0, ω4 = 2π4/16 = π/2, ω12 = −ω4mod(2π), ω8 = 2π8/16 = π are the signal
frequencies expressed as 16-point DFT frequencies. The DTFT of each complex sinusoid is given
by the shifted spectrum of the rectangular window W(ω) of length-16:
X(ω)= 0.5W(ω)−jW(ω − ω4)+jW(ω − ω12)+1.5W(ω − ω8)
where
W(ω)=
1 − e−16jω
1 − e−jω
=
sin(8ω)
sin(ω/2)
e−15jω/2
The magnitude spectrum |X(ω)| is plotted versus 0 ≤ ω ≤ 2π in Fig. P9.4. The spectral peaks
at the signal frequencies are evident. Dividing the Nyquist interval into 16 equal bins, gives the
16-point DFT.
It so happens that the zeros of the sidelobes of the W(ω − ωi) coincide with 16-point DFT fre-
quencies. Taking every other 16-DFT point gives the 8-point DFT. The peak values are correctly
identified by the 16- and 8-point DFTs because the peak frequencies coincide with DFT frequen-
cies.
203](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-102-320.jpg)
![ω
π0 2π
|X(ω)| = DTFT = 16-point DFT
= 8-point DFT24
16
8
Fig. P9.4 DTFT, 16-point and 8-point DFTs.
The 16-point DFT can be determined from the expression of x(n) by comparing it with 16-point
IDFT formula. Multiplying and dividing Eq. (P9.1) by 16, we may extract the DFT coefficients:
X = [8, 0, 0, 0, −16j, 0, 0, 0, 24, 0, 0, 0, 16j, 0, 0, 0]
Every other entry is the 8-point DFT:
X = [8, 0, −16j, 0, 24, 0, 16j, 0]
Problem 9.27
We will show that the product AA∗
is proportional to the identity matrix. The ij-th matrix element
of the product is:
(AA∗
)ij=
N−1
n=0
AinA∗
nj =
N−1
n=0
Win
N W
−jn
N =
N−1
n=0
W
(i−j)n
N
where we used the fact that W∗
N = W−1
N . Using the results of Problem 9.19, we note that the above
sum will be zero if i = j. And, it will be equal to N if i = j. Thus, we can write compactly:
(AA∗
)ij=
N−1
n=0
W
(i−j)n
N = Nδ(i − j)
which is equivalent to the matrix relationship
AA∗
= NI
where I is the N × N identity matrix.
Problem 9.28
Vectorially, the desired equation reads
x†
x =
1
N
X†
X
where † denotes conjugate transpose. Using X = Ax and the property
AA†
= A†
A = NIN, (IN = N × N identity matrix)
of the N × N DFT matrix, we obtain
X†
X = x†
A†
Ax = x†
NINx = Nx†
x
204
Problem 9.29
First, compute the ordinary linear convolution:
y = h ∗ x = [2, −1, −2, 2, −2, −1, 2]
The mod-4 reduction of this is y = [0, −2, 0, 2]. The mod-5 reduction is y = [1, 1, −2, 2, −2],
the mod-6 reduction is y = [4, −1, −2, 2, −2, −1]. Finally, the mod-7 and mod-8 reductions of
y leave it unchanged. One minor detail is that in the mod-8 case, the length should be N = 8, thus,
strictly speaking y is y with one zero padded at the end, that is, y = [2, −1, −2, 2, −2, −1, 2, 0].
The minimal N is N = Ly = Lx + Lh − 1 = 4 + 4 − 1 = 7.
Problem 9.30
First, compute the linear convolution of the given signals:
y = h ∗ x = [2, 1, 1, 1, 2, 1, 1, 1]∗[2, 1, 2, −3, 2, 1, 2, −3]
= [4, 4, 7, −1, 8, 7, 11, −2, 4, 2, 1, −1, 0, −1, −3]
and wrap it mod-8 to get
˜y = [8, 6, 8, −2, 8, 6, 8, −2]
Then, compute the FFTs of the two signals:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
2
1
1
1
2
1
1
1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
8−FFT
−→
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
10
0
2
0
2
0
2
0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
,
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
2
1
2
−3
2
1
2
−3
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
8−FFT
−→
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
4
0
−8j
0
12
0
8j
0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
Multiply them point-wise:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
10
0
2
0
2
0
2
0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
×
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
4
0
−8j
0
12
0
8j
0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
40
0
−16j
0
24
0
16j
0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
Divide by N = 8, conjugate every entry, and take an 8-point FFT:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
5
0
2j
0
3
0
−2j
0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
8−FFT
−→
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
8
6
8
−2
8
6
8
−2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
205](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-103-320.jpg)
![Problem 9.31
a. The 8-FFT algorithm gives
x = [6, 1, 0, 1, 6, 1, 0, 1] ⇒ X = [16, 0, 12, 0, 8, 0, 12, 0]
b. Using x(n)=
1
N
N−1
k=0
X(k)ejωkn
,
x(n)=
1
8
16 + 12ejπn/2
+ 8ejπn
+ 12e−jπn/2
= 2 + 3 cos
πn
2
+ (−1)n
c. With arbitrary values of a and b:
x = [6 − a, 1, 0, 1, 6, 1, 0, 1, a], x = [6 − a, 1 − b, 0, 1, 6, 1, 0, 1, a, b]
d. Reduce x mod-4 to get ˜x = [12, 2, 0, 2]. Its 4-FFT is easily found to be X = [16, 12, 8, 12].
It can also be obtained by picking every other entry of the 8-point DFT.
Problem 9.32
Start with
A(k)=
N−1
n=0
Wkn
N a(n)
and evaluate it at N − k
A(N − k)=
N−1
n=0
W
(N−k)n
N a(n)
Noting that WN
N = 1, it can be written as
A(N − k)=
N−1
n=0
W−kn
N a(n)
On the other hand, because a(n) is real
A(k)∗
=
N−1
n=0
Wkn
N
∗
a(n)
But,
Wkn
N
∗
= W−kn
N
Thus,
A(N − k)= A(k)∗
206
Problem 9.33
Using linearity, we have
X(k)= A(k)+jB(k) ⇒ X(N − k)∗
= A(N − k)∗
−jB(N − k)∗
Using the results of Problem 9.32, A(N − k)∗
= A(k) and B(N − k)∗
= B(k), we obtain X(N −
k)∗
= A(k)−jB(k). Thus, we have the system of equations:
X(k) = A(k)+jB(k)
X(N − k)∗
= A(k)−jB(k)
which can be solved for A(k) and B(k).
Problem 9.34
The expressions for G(k) and H(k) follow from the results of Problem 9.33, except that the di-
mension of the FFTs is N/2. The final expressions that construct X(k) are the standard Butterfly
operations for merging G(k) and H(k).
The additional multiplications to build X(k) from Y(k) are 3N/2, (that is, the divisions by 2
and 2j, and multiplications by Wk
N). Thus the total number of multiplications using this method
compared to a direct N-point FFT will be
1
2
N
2
log2(
N
2
)+
3N
2
1
2
N log2 N
=
1
2
+
2.5
log2 N
If one takes into account the multiplications required for computing the Wk
powers, then the
total becomes
N
2
log2(
N
2
)+
4N
2
N log2 N
=
1
2
+
1.5
log2 N
Typically, 65% of the standard cost for N = 1024.
Problem 9.35
/* fftreal.c - FFT of a real-valued signal */
#include <math.h>
#include <cmplx.h>
#include <stdlib.h>
void fft();
void fftreal(N, x, X)
int N;
double *x; time data
complex *X; FFT array
{
int n, k, Nhalf=N/2;
double pi=4*atan(1.0);
207](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-104-320.jpg)
![complex one, half, halfj, W;
complex *Y, temp, Wk, Gk, Hk;
one = cmplx(1.0, 0.0);
half = cmplx(0.5, 0.0);
halfj = cmplx(0.0, -0.5);
W = cexp(cmplx(0.0, -2*pi/N));
Y = (complex *) calloc(Nhalf, sizeof(complex));
for (n=0; n<Nhalf; n++)
Y[n] = cmplx(x[2*n], x[2*n+1]);
fft(Nhalf, Y);
for (Wk=one, k=0; k<Nhalf; k++) { Wk = Wk
if (k==0)
temp = conjg(Y[0]); note: Y(0) = Y(N/2)
else
temp = conjg(Y[Nhalf-k]);
Gk = cmul(half, cadd(Y[k], temp)); G(k)
Hk = cmul(halfj, csub(Y[k], temp)); H(k)
temp = cmul(Wk, Hk); Wk H(k)
Wk = cmul(W, Wk); next Wk = Wk+1
X[k] = cadd(Gk, temp); G(k) + Wk H(k)
X[k+Nhalf] = csub(Gk, temp); G(k) − Wk H(k)
}
}
Problem 9.36
An example will demonstrate the result. For N = 4, we have:
x = [x0, x1, x2, x3] ⇒ X(z)= x0 + x1z−1
+ x2z−2
+ x3z−3
xR = [x3, x2, x1, x0] ⇒ XR(z)= x3 + x2z−1
+ x1z−2
+ x0z−3
It follows that
XR(z)= z−3
[x3z3
+ x2z2
+ x1z + x0]= z−3
X(z−1
)
In the time domain, the operation of reflection leads to the z-transform pairs:
x(n)
Z
−→ X(z) ⇒ x(−n)
Z
−→ X(z−1
)
The operation of delay on the reflected signal then gives:
xR(n)= x(N − 1 − n)= x(−(n − N + 1))
Z
−→ z−(N−1)
X(z−1
)
The N-point DFT of the reversed signal is obtained by setting z = zk = e2πjk/N
= W−k
N . Using
the property WN
N = 1, we find:
XR(k)= z−N+1
k X(z−1
k )= W
(N−1)k
N X(−k)= W−k
N X(N − k)
where in the last step, we used X(z−1
k )= X(−ωk)= X(−k)= X(N − k) by the periodicity of the
N-point DFT.
208
Problem 9.37
In the definition of y, the reversed signal xR is delayed by N samples, that is, we have
x = [x, xR]= [x, 0, 0, . . . , 0
N zeros
]+[0, 0, . . . , 0
N zeros
, xR]
Thus, using linear superposition, the delay property of z-transforms, and the results of Problem
9.36, we have in the z-domain:
Y(z)= X(z)+z−N
XR(z)= X(z)+z−2N+1
X(z−1
)
where the factor z−N
represents the delay of xR by N samples. Replacing z by zk = ejωk , and
using the result z2N
k = 1, we get
Yk = Y(ωk)= X(ωk)+ejωk X(−ωk)= ejωk/2
e−jωk/2
X(ωk)+ejωk/2
X(−ωk)
= 2ejωk/2
Re e−jωk/2
X(ωk)
where we used the hermitian property X(−ω)= X(ω)∗
. The quantity under the real part is
now:
e−jωk/2
X(ωk)= e−jωk/2
N−1
n=0
xne−jωkn
=
N−1
n=0
xne−jωk(n+1/2)
Thus, its real part is the cosine transform:
Re e−jωk/2
X(ωk) =
N−1
n=0
xn cos ωk(n + 1/2) = Ck
It follows that
Yk = 2ejωk/2
Ck = 2ejπk/2N
Ck
Because Yk is the (2N)-point DFT of a real-valued signal, it will satisfy the hermitian property
Y2N−k = Y∗
k . In terms of Ck, we have:
Y2N−k = 2ejπ(2N−k)/2N
C2N−k = 2ejπ
e−jπk/2N
C2N−k = −2e−jπk/2N
C2N−k
Y∗
k = 2e−jπk/2N
Ck
Equating the right-hand sides and canceling the common factors gives the symmetry property
C2N−k = −Ck , k = 0, 1, . . . , 2N − 1
At k = N, it gives C2N−N = −CN or CN = −CN which implies CN = 0. Next, considering the
inverse DFT for Yk, we have:
yn =
1
2N
2N−1
k=0
Ykejωkn
, n = 0, 1, . . . , 2N − 1
But the first N samples of yn are precisely xn. Replacing Yk in terms of Ck gives for n =
0, 1, . . . , N − 1:
xn =
1
2N
2N−1
k=0
2ejωk/2
Ckejωkn
=
1
N
2N−1
k=0
Ckejωk(n+1/2)
209](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-105-320.jpg)
![Next, we split the summation into three terms:
xn =
1
N
C0 +
N−1
k=1
Ckejωk(n+1/2)
+
2N−1
k=N+1
Ckejωk(n+1/2)
where the second summation starts at k = N + 1 because CN = 0. If we change variables from k
to m = 2N − k, we can write
2N−1
k=N+1
Ckejωk(n+1/2)
=
N−1
m=1
C2N−mejω2N−m(n+1/2)
Using the symmetry property C2N−m = −Cm, and the trig identity
ejω2N−m(n+1/2)
= ejπ(2N−m)(n+1/2)/N
= e2jπ(n+1/2)
e−jπm(n+1/2)/N
= −e−jωm(n+1/2)
we have (with the two minus signs canceling):
N−1
m=1
C2N−mejω2N−m(n+1/2)
=
N−1
m=1
Cme−jωm(n+1/2)
Thus, the inverse DFT becomes (changing the summation from m back to k):
xn =
1
N
C0 +
N−1
k=1
Ckejωk(n+1/2)
+
N−1
k=1
Cke−jωk(n+1/2)
The two summations now combine with Euler’s formula to give:
xn =
1
N
C0 + 2
N−1
k=1
Ck cos ωk(n + 1/2)
Problem 9.38
By definition, we have
X2N(k)=
L−1
n=0
x(n)Wnk
2N ⇒ X2N(2k)=
L−1
n=0
x(n)W2nk
2N
But, W2nk
2N = exp −2πj2nk/2N = exp −2πjnk/N = Wnk
N . Thus,
X2N(2k)=
L−1
n=0
x(n)W2nk
2N =
L−1
n=0
x(n)Wnk
N = XN(k)
Problem 9.39
Observe that the five length-5 sequences are the mod-5 wrapped versions of the following “linearly
shifted” sequences:
x0 x1 x2 x3 x4 0 0 0 0 0
0 x0 x1 x2 x3 x4 0 0 0 0
0 0 x0 x1 x2 x3 x4 0 0 0
0 0 0 x0 x1 x2 x3 x4 0 0
0 0 0 0 x0 x1 x2 x3 x4 0
210
Their z-transforms differ by a simple delay factor, that is,
Xi(z)= z−i
X0(z), i = 1, 2, 3, 4
Setting z = zk = ejωk = W−k
5 , we obtain
Xi(zk)= z−i
k X0(zk)= Wik
5 X0(zk), i = 1, 2, 3, 4
Problem 9.40
Consider, for example, the first and the third. Their z-transforms are
X1(z) = x0 + x1z−1
+ x2z−2
+ x3z−3
+ x4z−4
+ x5z−5
+ x6z−6
+ x7z−7
X3(z) = x0 + x1z−1
+ (x2 + x6)z−2
+ (x3 + x7)z−3
+ x4z−4
+ x5z−5
Subtracting, we have
X3(z)−X1(z)= x6z−2
(1 − z−4
)+x7z−3
(1 − z−4
)= (x6z−2
+ x7z−3
)(1 − z−4
)
Evaluating at the 4th roots of unity, z = zk = e2πjk/4
, we obtain
X3(zk)−X1(zk)= 0, k = 0, 1, 2, 3
Problem 9.41
Using the convolution table, perform the ordinary linear convolution of the given signal and filter
to get
y = h ∗ x = [1, 0, −1, 0, 2, 0, −2, 0, −2, 0, 2, 1, 0, −1, 0, −1, 0, 1, 0, −1, 0, 1]
The overlap-save method divides the input signal into segments of length N = 8 that are over-
lapping by M = 3 points, as shown below:
x = 1, 1, 1, 1, 3, (3, 3, 3 , 1, 1, 1, 2, 2), 2, 2, (1, 1, 1 , 1, 0, 0, 0, 0)
Convolving these segments with the filter gives:
y0 = [1, −1, −1, 1]∗[1, 1, 1, 1, 3, 3, 3, 3] = [1, 0, −1, 0, 2, 0, −2, 0, −3, 0, 3]
y1 = [1, −1, −1, 1]∗[3, 3, 3, 1, 1, 1, 2, 2] = [3, 0, −3, −2, 0, 2, 1, 0, −3, 0, 2]
y2 = [1, −1, −1, 1]∗[1, 2, 2, 2, 2, 1, 1, 1] = [1, 1, −1, −1, 0, −1, 0, 1, −1, 0, 1]
y3 = [1, −1, −1, 1]∗[1, 1, 1, 1, 0, 0, 0, 0] = [1, 0, −1, 0, −1, 0, 1, 0, 0, 0, 0]
Reducing these mod-8 and ignoring their first M = 3 points (indicated by x), we obtain:
˜y0 = [x, x, x, 0, 2, 0, −2, 0]
˜y1 = [x, x, x, −2, 0, 2, 1, 0]
˜y2 = [x, x, x, −1, 0, −1, 0, 1]
˜y3 = [x, x, x, 0, −1, 0, 1, 0]
Abutting the rest of the points, we get
y = [x, x, x, 0, 2, 0, −2, 0][−2, 0, 2, 1, 0][−1, 0, −1, 0, 1][0, −1, 0, 1, 0]
With the exception of the first M points, the answer is correct.
211](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-106-320.jpg)
![Problem 9.42
At fs = 8 Hz, there are 8 samples per period. These samples can be evaluated by setting t =
nT = n/8 in the expression for x(t). We find:
x(n)=
1
4
{0, 0.5, 1, 0.5, 0, −0.5, −1, −0.5}
According to Eq. (9.7.4), the aliased coefficients can be obtained by computing the 8-point DFT
of one period and dividing by 8. This DFT can be done quickly by an 8-point FFT by hand, which
gives:
x =
1
4
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
0
0.5
1
0.5
0
−0.5
−1
−0.5
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
DFT
−→ X =
1
4
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
0
−j(2 +
√
2)
0
j(2 −
√
2)
0
−j(2 −
√
2)
0
j(2 +
√
2)
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
, b =
1
4
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
0
(2 +
√
2)/8j
0
−(2 −
√
2)/8j
0
(2 −
√
2)/8j
0
−(2 +
√
2)/8j
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
The coefficients b1 and b7 correspond to frequencies ±f1 = ±1 Hz, whereas b3 and b5 correspond
to ±f2 = ±3 Hz. Thus, we find from Eq. (9.7.5) and after using Euler’s formula:
xal(t)= 2jb1 sin(2πf1t)+2jb3 sin(2πf2t)=
2 +
√
2
16
sin(2πt)−
2 −
√
2
16
sin(6πt)
It can be verified easily that xal and x(t) agree at the sampling instants t = nT = n/8.
Problem 9.43
In the overlap-add method, the input blocks have length N − M and are non-overlapping. Thus,
if there are K blocks in the total length, we will have:
L = (N − M)K (P9.2)
In Fig. 9.9.1, the input blocks overlap by M points, but again if there are K segments, the total
length will be (N − M)K plus an extra M points at the end. Thus, we have in the overlap-save
method:
L = (N − M)K + M (N − M)K
which is essentially the same as in Eq. (P9.2) for large L. Thus, in both cases, the total number of
segments can be estimated to be:
K =
L
N − M
Per segment, we are doing two FFTs and N multiplications, namely,
N log2 N + N = N(1 + log2 N) per segment
Thus, for K segments, we have cost:
KN(1 + log2 N)=
LN(1 + log2 N)
N − M
212
This is to be compared with the cost of L(M + 1) of linear convolution of the entire input array
with the filter (here, we treat these MACs as complex-valued.) Thus, the relative cost of the fast
versus slow convolution will be:
fast
slow
=
LN(1 + log2 N)
N − M
L(M + 1)
=
N(1 + log2 N)
(N − M)(M + 1)
Problem 9.44
An example of such a program is given below:
/* ovsave.c - FIR filtering by overlap-save method and FFT */
#include <stdio.h>
#include <cmplx.h>
#include <stdlib.h> calloc(), realloc()
#define MAX 64 initial allocation size
void circonv(), complexify(), fft(), ifft();
void main(int argc, char **argv)
{
FILE *fph; filter file
double *h, *x, *xhead; x = length-N block
complex *H, *X, *Y; N-point DFTs
int M, N, n, i;
int max = MAX, dmax = MAX;
if (argc != 3) {
fprintf(stderr, "usage: fastconv hfile N < xfile > yfilen");
exit(0);
}
if ((fph = fopen(argv[1], "r")) == NULL) {
fprintf(stderr, "can’t open filter file: %sn", argv[1]);
exit(0);
}
N = atoi(argv[2]);
h = (double *) calloc(max + 1, sizeof(double)); preliminary allocation
for (M=0;; M++) { read h
if (M == max) { reallocate h, if necessary
max += dmax;
h = (double *) realloc((char *) h, (max + 1) * sizeof(double));
}
if (fscanf(fph, "%lf", h + M) == EOF) break;
}
M--; M = filter order
if (N <= 2*M) {
fprintf(stderr, "must have N > 2M = %d", 2*M);
exit(0);
}
213](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-107-320.jpg)
![h = (double *) realloc((char *) h, (M+1) * sizeof(double));
x = (double *) calloc(N, sizeof(double));
xhead = (double *) calloc(M, sizeof(double)); head/tail of input block
H = (complex *) calloc(N, sizeof(complex)); allocate DFT arrays
X = (complex *) calloc(N, sizeof(complex));
Y = (complex *) calloc(N, sizeof(complex));
for (n=0; n<N; n++) complexify/extend to length N
if (n <= M)
H[n] = cmplx(h[n], 0.0);
else
H[n] = cmplx(0.0, 0.0);
fft(N, H); H = FFT(h), done once
for (n=0; n<M; n++)
xhead[n] = 0; pad M zeros in front of first block
for (;;) { keep reading input in blocks x
for (n=0; n<M; n++) previous tail is head of present block x
x[n] = xhead[n];
for (n=M; n<N; n++) read rest of x from input stream
if (scanf("%lf", x + n) == EOF) goto lastblocks;
for (n=0; n<M; n++) tail of present block x
xhead[n] = x[n + N - M]; will be head of next block
complexify(N, x, X); X = cmplx(x, 0)
circonv(N, H, X, Y); done by fft
for (n=M; n<N; n++) discard first M points, save the rest
printf("%lfn", real(Y[n]));
}
lastblocks:
n = index where EOF occurred
for (i=n; i<N; i++) pad rest of x with zeros
x[i] = 0; if n <= N-M there is only
one last block to consider
complexify(N, x, X); otherwise, there are two last blocks
circonv(N, H, X, Y);
for (i=M; i<N; i++)
printf("%lfn", real(Y[i]));
if (n <= N-M) exit(0); exit, if only one last block
otherwise, process very last block
for (i=0; i<M; i++) if n > N-M, the tail of previous
xhead[i] = x[i + N - M]; block will be non-zero
for (i=0; i<N; i++)
if (i < M)
x[i] = xhead[i]; head of very last block
214
else
x[i] = 0; rest of block is zero
complexify(N, x, X);
circonv(N, H, X, Y);
for (i=M; i<N; i++)
printf("%lfn", real(Y[i]));
}
——————————————————————–
void complexify(N, x, X)
int N;
double *x;
complex *X;
{
int n;
for (n=0; n<N; n++)
X[n] = cmplx(x[n], 0.0);
}
——————————————————————–
void circonv(N, H, X, Y) circular convolution via FFT
int N;
complex *H, *X, *Y; H = FFT(h), must be done previously
{
int k;
fft(N, X);
for (k=0; k<N; k++)
Y[k] = cmul(H[k], X[k]);
ifft(N, Y);
}
215](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-108-320.jpg)
![Problem 10.4
Using the differentiation argument, we have in this case:
d(k)=
∂
∂k
−ωa
−ωb
ejωk dω
2π
+
ωb
ωa
ejωk dω
2π
The integrations give the bandpass filter case. Thus,
d(k)=
∂
∂k
sin(ωbk)− sin(ωak)
πk
which becomes the difference of two lowpass differentiators:
d(k)=
ωb cos(ωbk)
πk
−
sin(ωbk)
πk2
−
ωa cos(ωak)
πk
+
sin(ωak)
πk2
Problem 10.5
For the first-order differentiator, we have impulse response, transfer function, and frequency
response:
h = [1, −1] ⇒ H(z)= 1 − z−1
⇒ H(ω)= 1 − e−jω
Thus, the NRR will be
NRR = h2
0 + h2
1 = 1 + 1 = 2
Thus, the noise power doubles at the output. To see how closely it resembles the ideal differen-
tiator, we expand the frequency response at low frequencies:
H(ω)= 1 − e−jω
1 − (1 − jω)= jω for small ω
For the ideal lowpass differentiator case, the NRR can be computed as:
NRR =
ωc
−ωc
|D(ω)|2 dω
2π
=
ωc
−ωc
|jω|2 dω
2π
=
ω3
c
3π
Thus, it increases rapidly with ωc. Compared to the full-band maximum NRR obtained with
ωc = π, we have NRRmax = π3
/3π = π2
/3 and therefore,
NRR
NRRmax
=
ωc
π
3
Problem 10.6
Let E(ω)= D(ω)−D(ω) be the error in the designed filter. Then, we have
E(ω)= D(ω)−D(ω)=
∞
k=−∞
d(k)e−jωk
−
M
k=−M
d(k)e−jωk
=
|k|>M
d(k)e−jωk
because the coefficients d(k) cancel for |k| ≤ M. Thus, the time-domain coefficients of the error
are:
e(k)=
0 if |k| ≤ M
d(k) if |k| > M
218
Parseval’s identity applied to e(k) gives then
π
−π
|E(ω)|2 dω
2π
=
∞
k=−∞
e2
(k)=
|k|>M
d2
(k)
or,
EM =
π
−π
|D(ω)−D(ω)|2 dω
2π
=
|k|>M
d2
(k)
The right-hand side sum can be written with the help of Parseval’s identity applied to D(ω) itself:
|k|>M
d2
(k)=
∞
k=−∞
d2
(k)−
|k|≤M
d2
(k)=
π
−π
|D(ω)|2 dω
2π
−
M
k=−M
d(k)2
In the limit as M → ∞, the second term in the right-hand side tends to the Parseval limit, thus
canceling the first term and resulting in EM → 0.
Problem 10.7
From Problem 10.3, we have d(k)= (−1)k
/k, k = 0, for the differentiator filter. Using Parseval’s
identity we have
∞
k=−∞
d2
(k)=
π
−π
|D(ω)|2 dω
2π
=
π
−π
ω2 dω
2π
=
π2
3
But the left-hand side is twice the required sum:
∞
k=−∞
d2
(k)=
k=0
1
k2
= 2
∞
k=1
1
k2
Thus,
2
∞
k=1
1
k2
=
π2
3
⇒
∞
k=1
1
k2
=
π2
6
For the full-band Hilbert transformer, we have
d(k)=
1 − cos(πk)
πk
=
1 − (−1)k
πk
, k = 0
Thus, d(k) is nonzero only for odd k, and is in this case d(k)= 2/πk. Parseval’s identity applied
to this filter gives:
∞
k=−∞
d2
(k)=
π
−π
|D(ω)|2 dω
2π
=
π
−π
|jsign(ω)|2 dω
2π
=
π
−π
1
dω
2π
= 1
The left-hand side on the other hand becomes:
∞
k=−∞
d2
(k)= 2
∞
k=1
k=odd
4
π2k2
Thus, we obtain
219](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-110-320.jpg)
![2
∞
k=1
k=odd
4
π2k2
= 1 ⇒
∞
k=1
k=odd
1
k2
=
π2
8
If the sum for all integers k is π2
/6 and for odd ones π2
/8, it follows by subtraction that the
sum over even integers will be:
∞
k=2
k=even
1
k2
=
π2
6
−
π2
8
=
π2
24
Problem 10.8
The sinusoidal response of D(ω) in its steady-state form is:
ejωn D
−→ D(ω)ejωn
= −jsign(ω)ejωn
Replacing ω by −ω, and using the odd symmetry property of D(ω), we have:
e−jωn D
−→ D(−ω)e−jωn
= jsign(ω)e−jωn
But cos(ωn)= [ejωn
+ e−jωn
]/2. Thus, forming the same linear combination of outputs, we
have:
cos(ωn)
D
−→
1
2
−jsign(ω)ejωn
+ jsign(ω)e−jωn
= sign(ω)sin(ωn)
Similarly, because sin(ωn)= [ejωn
− e−jωn
]/2j, we have for the outputs:
sin(ωn)
D
−→
1
2j
−jsign(ω)ejωn
− jsign(ω)e−jωn
= −sign(ω)cos(ωn)
Problem 10.9
From the definition of D(z), we have:
D(z)=
M
k=−M
d(k)z−k
⇒ D(z−1
)=
M
k=−M
d(k)zk
=
M
k=−M
d(−k)z−k
where in the last step we changed summation from k to −k. In the symmetric/antisymmetric
cases, we have d(k)= ±d(k). Therefore,
D(z−1
)= ±
M
k=−M
d(k)z−k
= ±D(z)
From the reality of d(k), it follows that the zeros will come in conjugate pairs if they are complex.
Thus, if z0 is a complex zero, then z∗
0 is also a zero. From the symmetry or antisymmetry
properties, it follows that if z0 is a zero, then
D(z−1
0 )= ±D(z0)= 0
Thus, 1/z0 and 1/z∗
0 are both zeros. The four zeros {z0, z∗
0 ,
1
z0
,
1
z∗
0
} are shown in Fig. P10.1.
If z0 lies on the unit circle then because 1/z∗
0 = z0, the only other distinct zero is the complex
conjugate.
220
z0
1/z0
*
1/z0
z0
*
Fig. P10.1 Pattern of zeros of linear phase filters.
In the antisymmetric case, we have in particular for z = ±1,
D(±1)=
M
k=−M
d(k)(±1)k
= 0
This follows because d(k) is odd and summed over symmetric limits and because the factor
(±1)k
is even in k. The same conclusion follows from the frequency response. Using the anti-
symmetry property and splitting the sum in two parts, we get:
D(ω)=
M
k=−M
d(k)e−jωk
=
M
k=1
d(k) e−jωk
− ejωk
= −2j
M
k=1
d(k)sin(ωk)
Setting ω = 0, π, corresponding to z = ±1, we have sin(ωk)= 0, and therefore, D(ω) vanishes.
Finally, we note that if d(k) is windowed by a window w(k), that is, d (k)= w(k)w(k), then
because the window w(k) is symmetric with respect to the origin, the symmetry/antisymmetry
properties of d(k) will be preserved and will imply similar results on the locations of the zeros.
Problem 10.10
For the symmetric case, the following must necessarily be zeros:
z0 = 0.5 + 0.5j, z∗
0 = 0.5 − 0.5j,
1
z0
= 1 − j,
1
z∗
0
= 1 + j
Thus, the transfer function will be
H(z)= 1−(0.5+0.5j)z−1
1−(0.5−0.5j)z−1
1−(1−j)z−1
1−(1+j)z−1
= 1−3z−1
+4.5z−2
−3z−3
+z−4
In the time domain, we get the length-5 symmetric (about its middle) impulse response:
h = [1, −3, 4.5, −3, 1]
In the antisymmetric case, we have in addition two zeros at z = ±1 contributing a factor (1−z−2
)
to the transfer function. Thus,
H(z)= (1 − z−2
)(1 − 3z−1
+ 4.5z−2
− 3z−3
+ z−4
)= 1 − 3z−1
+ 3.5z−2
− 3.5z−4
+ 3z−5
− z−6
and in the time domain:
h = [1, −3, 3.5, 0, −3.5, 3, −1]
221](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-111-320.jpg)
![The frequency response in the symmetric case is obtained by factoring out a factor of z−2
and
using Euler’s formula:
H(z)= z−2
z2
+ z−2
− 3(z + z−1
)+4.5 ⇒
H(ω)= 2e−2jω
cos(2ω)−3 cos(ω)+4.5
For the antisymmetric case, we have:
H(z)= z−3
(z3
− z−3
)−3(z2
− z−2
)+3.5(z − z−1
) ⇒
H(ω)= −2je−3jω
sin(3ω)−3 sin(2ω)+3.5 sin(ω)
Problem 10.11
Proceeding as in the previous problem, we find the transfer function of the symmetric case:
H(z)= (1−0.5jz−1
)(1+0.5jz−1
)(1−2jz−1
)(1+2jz−1
)= 1+4.25z−2
+z−4
⇒ h = [1, 0, 4.25, 0, 1]
and for the antisymmetric case:
H(z)= (1−z−2
)(1+4.25z−2
+z−4
)= 1+3.25z−2
−3.25z−4
−z−6
⇒ h = [1, 0, 3.25, 0, −3.25, 0, −1]
Problem 10.12
We use the first-order Taylor series approximation:
ln(1 ± x) ±x
Changing to base-10, we have:
log10(1 ± x)=
ln(1 ± x)
ln(10)
= ±0.4343 x
It follows that
Apass = 20 log10
1 + δpass
1 − δpass
= 20 log10(1 + δpass)−20 log10(1 − δpass)
20 · 0.4343 δpass + 20 · 0.4343 δpass = 17.372 δpass
Problem 10.13
First, we calculate δpass and δstop from Eq. (10.2.5):
δpass =
100.1/20
− 1
100.1/20 + 1
= 0.0058, δstop = 10−80/20
= 0.0001
Therefore, δ = min(δpass, δstop)= δstop = 0.0001, which in dB is A = −20 log10 δ = Astop = 80.
The D and α parameters are computed by:
α = 0.1102(A − 8.7)= 0.1102(80 − 8.7)= 7.857, D =
A − 7.95
14.36
= 5.017
The transition width is Δf = fstop − fpass = 2 − 1.5 = 0.5 kHz; therefore, the filter length will be:
N = 1 +
Dfs
Δf
= 1 +
5.017 · 10
0.5
= 101.34 ⇒ N = 103
222
Problem 10.14
The transition width is inversely related to the filter length N:
Δf =
Dfs
N − 1
As in the previous problem, we find D = 5.017. Thus,
Δf =
Dfs
N − 1
=
5.017 · 10
251 − 1
= 0.2007 kHz
Problem 10.15
The window’s width factor is D = 5.017 as in the previous problem. Thus,
Δf =
Dfs
N − 1
=
5.017 · 44.1
129 − 1
= 1.729 kHz
Conversely, if Δf = 2 kHz, then we may solve for the D-factor and attenuation A:
D =
A − 7.95
14.36
= (N − 1)
Δf
fs
= 5.805 ⇒ A = 14.36D + 7.95 = 91.31 dB
The corresponding ripple will be
δ = 10−A/20
= 10−91.31/20
= 0.000027
which gives the passband attenuation:
Apass = 17.372 δ = 0.000472 dB
In the 4-times oversampling case, the transition width is required to be:
Δf = 24.55 − 19.55 = 5 kHz
This gives the width factor and attenuation:
D =
A − 7.95
14.36
= (N − 1)
Δf
4fs
= 3.6281 ⇒ A = 14.36D + 7.95 = 60.05 dB
Problem 10.16
Assuming that an N-tap filter requires approximately N instructions for processing each input
sample, the maximum filter length will correspond to the maximum number of instructions that
fit within the sampling interval T, that is,
Nmax =
T
Tinstr
=
finstr
fs
Inserting this into the Kaiser design formula, we find the minimum width (where we used N
instead of N − 1 in the denominator):
Δf =
Dfs
N
=
Dfs
2
finstr
223](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-112-320.jpg)
![Problem 10.17
Using N = finstr/fs for the maximum filter length, we obtain the maximum width parameter D
and attenuation A:
D =
A − 7.95
14.36
= N
Δf
fs
=
finstrΔf
fs
2 = FinstrΔF ⇒ A = 14.36FinstrΔF + 7.95
Problem 10.18
The following MATLAB program segment computes the rectangularly and Hamming windowed
impulse responses with the help of the MATLAB function dlh.m, and then it computes the cor-
responding frequency responses over NF equally spaced frequencies over the right half of the
Nyquist interval [0, π] with the help of the function dtft.m:
wc = 0.3 * pi; % cutoff frequency
N = 81; % filter length
w = 0.54 - 0.46 * cos(2 * pi * (0:N-1) / (N-1)); % Hamming window
hrec = dlh(1, wc, N); % ideal lowpass filter
hham = w .* hrec; % windowed filter
NF = 200; % 200 frequencies in rads/sample
omega = (0:NF-1) * pi / NF; % over the interval [0, π]
Hrec = dtft(hrec, omega); % frequency response
Hham = dtft(hham, omega);
Problem 10.19
As an example, the following MATLAB program segment computes the bandpass filter of Example
10.2.2 (alternative design):
fs = 20; in kHz
fpa = 4; fpb = 6; fsa = 3; fsb = 8;
Apass = 0.1; Astop = 80; in dB
h = kbp(fs, fpa, fpb, fsa, fsb, Apass, Astop, -1); % s = −1 for alternative design
NF = 200;
omega = (0:NF-1) * pi / NF;
H = dtft(h, omega);
Similarly, the highpass filter can be designed by:
fs = 20; in kHz
fpass = 5; fstop = 4;
Apass = 0.1; Astop = 80; in dB
h = klh(-1, fs, fpass, fstop, Apass, Astop); % s = −1 for highpass
NF = 200;
omega = (0:NF-1) * pi / NF;
H = dtft(h, omega);
224
Problem 10.20
Use the design MATLAB routine kdiff, with fs = 2 so that ω = 2πf/2 = πf, that is, the f’s are
the ω’s in units of π. The filter lengths depend only on the values of Δf and A, they are:
Δf A N
0.10 30 33
0.10 60 75
0.05 30 63
0.05 60 147
For example, in the case ωc = 0.4π, Δω = 0.05π, A = 60 dB, we have the MATLAB code:
h = kdiff(2, 0.4, 0.05, 60);
NF = 200;
omega = (0:NF-1) * pi / NF;
H = dtft(h, omega);
The magnitude responses are shown in Figs. P10.2 and P10.3.
Fig. P10.2 Lowpass differentiator with ωc = 0.4π.
225](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-113-320.jpg)
![Fig. P10.3 Lowpass differentiator with ωc = 0.8π.
Problem 10.21
For example, in the case ωc = 0.4π, Δω = 0.05π, A = 60 dB, we have the MATLAB segment that
computes the Kaiser differentiator and the Savitzky-Golay differentiators of polynomial order
d = 3:
h = kdiff(2, 0.4, 0.05, 60); % Kaiser differentiator
[B, S] = sg(3, N); % Savitzky-Golay smoothing filters
F = S’ * S; % polynomial basis matrix
G = S * F^(-1); % Savitzky-Golay differentiator filters
g = G(:,2)’; % first-order derivative filter
NF = 500;
omega = (0:NF-1) * pi / NF;
H = dtft(h, omega); % frequency responses
G = dtft(g, omega);
The magnitude responses are shown in Fig. P10.4.
Problem 10.22
Use the design MATLAB routine khilb, with fs = 2 so that ω = 2πf/2 = πf, that is, the f’s are
the ω’s in units of π. The filter lengths depend only on the values of Δf and A, they are the
226
Fig. P10.4 Comparison of Kaiser and SG differentiators.
same as in Problem 10.20. For example, the Hilbert transformer with ωc = 0.8π, Δω = 0.05π,
and A = 60 dB, can be designed by the MATLAB segment:
h = khilb(2, 0.8, 0.05, 60);
NF = 200;
omega = (0:NF-1) * pi / NF;
H = dtft(h, omega); % frequency responses
The magnitude responses are shown in Figs. P10.5 and P10.6.
Problem 10.23
The following MATLAB segment computes the Kaiser and Hamming spectra in dB. The Kaiser
window parameters are determined by the function kparm2.m:
F1 = 0.20; F2 = 0.25; F3 = 0.30; DF = (F2-F1)/3; % units of fs
A1 = 1; A2 = 10^(-50/20); A3 = 1; % amplitudes
R = 70; % sidelobe level in dB
[alpha, L] = kparm2(DF, R); % Kaiser parameters
n = (0:L-1);
x = A1 * cos(2*pi*F1*n) + A2 * cos(2*pi*F2*n) + A3 * cos(2*pi*F3*n);
xk = kwind(alpha, L) .* x; % Kaiser windowed signal
xh = (0.54 - 0.46 * cos(2*pi*n/(L-1))) .* x; % Hamming windowed signal
NF = 256;
omega = (0:NF-1) * pi / NF;
Xk = abs(dtft(xk, omega)); % DTFT spectrum
Xkmax = max(Xk);
Xk = 20 * log10(Xk / Xkmax); % normalized and in dB
Xh = abs(dtft(xh, omega));
227](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-114-320.jpg)
![Therefore, the boost and cut filters will be:
|Hboost(Ω)|2
= G2
0
(Ω2
− Ω2
0)2
+N(g, gB)γ2
Ω2
(Ω2 − Ω2
0)2+D(g, gB)γ2Ω2
, |Hcut(Ω)|2
= G2
0
(Ω2
− Ω2
0)2
+D(g, gB)γ2
Ω2
(Ω2 − Ω2
0)2+N(g, gB)γ2Ω2
Hence, their product will be identically equal to G4
0. The geometric-mean and weighted geometric-
mean choices for g2
B are special cases of the substitution {g, gB} → {g−1
, g−1
B }, and therefore,
satisfy the desired property. But, any other choice of gB will do, as long as one defines the boost
bandwidth to be gB and the cut bandwidth, g−1
B , that is, equal and opposite bandwidth gains in
dB. Note, also that the transfer functions themselves are inverses to each other, not just their
magnitudes. Indeed, we can write them in the form:
Hboost(s)= G0
s2
+ Ω2
0 + γs N(g, gB)
s2 + Ω2
0 + γs D(g, gB)
, Hcut(s)= G0
s2
+ Ω2
0 + γs N(g−1, g−1
B )
s2 + Ω2
0 + γs D(g−1, g−1
B )
= G0
s2
+ Ω2
0 + γs D(g, gB)
s2 + Ω2
0 + γs N(g, gB)
which implies for the analog and the bilinearly transformed digital transfer functions:
Hcut(s)=
G2
0
Hboost(s)
⇒ Hcut(z)=
G2
0
Hboost(z)
Problem 11.5
The following MATLAB code segment designs the three filters (two peaking and one complemen-
tary notch), computes their impulse responses, and their magnitude responses:
w0 = 0.35*pi; Dw = 0.1*pi; in rads/sample
[b1, a1, beta1] = parmeq(0, 1, 1/sqrt(2), w0, Dw); peaking 3-dB width
[b2, a2, beta2] = parmeq(0, 1, sqrt(0.1), w0, Dw); peaking 10-dB width
[b3, a3, beta3] = parmeq(1, 0, 1/sqrt(2), w0, Dw); notching 3-dB width
x = [1, zeros(1, 99)]; unit impulse
h1 = filter(b1, a1, x); impulse responses
h2 = filter(b2, a2, x);
h3 = filter(b3, a3, x);
w = (0:199)*pi/200; 200 frequencies over [0, π]
H1 = abs(dtft(b1, w) ./ dtft(a1, w)).^2; magnitude responses
H2 = abs(dtft(b2, w) ./ dtft(a2, w)).^2;
H3 = abs(dtft(b3, w) ./ dtft(a3, w)).^2;
The 5-percent time constants can be calculated as in Problem 11.1, with = 0.05.
Problem 11.6
For example, the following MATLAB code designs the first four filters of Example 11.4.1 and
computes their impulse response, frequency, and phase responses (in degrees):
[b1, a1, beta1] = parmeq(1, 10^(9/20), 10^(6/20), 0.35*pi, 0.1*pi);
[b2, a2, beta2] = parmeq(1, 10^(9/20), 10^(3/20), 0.35*pi, 0.1*pi);
[b3, a3, beta3] = parmeq(1, 10^(-9/20), 10^(-6/20), 0.60*pi, 0.2*pi);
[b4, a4, beta4] = parmeq(1, 10^(-9/20), 10^(-3/20), 0.60*pi, 0.2*pi);
x = [1, zeros(1, 49)]; length-50 unit impulse
232
h1 = filter(b1, a1, x); impulse responses
h2 = filter(b2, a2, x);
h3 = filter(b3, a3, x);
h4 = filter(b4, a4, x);
w = (0:199)*pi/200;
H1 = dtft(b1, w) ./ dtft(a1, w); argH1 = angle(H1) * 180 / pi;
H2 = dtft(b2, w) ./ dtft(a2, w); argH2 = angle(H2) * 180 / pi;
H3 = dtft(b3, w) ./ dtft(a3, w); argH3 = angle(H3) * 180 / pi;
H4 = dtft(b4, w) ./ dtft(a4, w); argH4 = angle(H4) * 180 / pi;
Figure P11.1 shows the phase responses. Note the rapid variations in the vicinities of the center
frequencies.
Fig. P11.1 Phase responses of Problem 11.6.
Problem 11.7
The following MATLAB function lheq.m implements the design of both the lowpass and highpass
shelving filters (the parameter s = ±1 selects the two cases):
% lheq.m - lowpass/highpass first-order shelving EQ filter design
%
% [b, a, beta] = lheq(s, G0, G, Gc, wc)
%
% s = 1, −1 = lowpass, highpass
% b = [b0, b1] = numerator coefficients
% a = [1, a1] = denominator coefficients
% G0, G, Gc = reference, boost/cut, and cutoff gains
% wc = cutoff frequency in [rads/sample]
% beta = design parameter β
function [b, a, beta] = lheq(s, G0, G, Gc, wc)
beta = (tan(wc/2))^s * sqrt(abs(Gc^2 - G0^2)) / sqrt(abs(G^2 - Gc^2));
b = [(G0 + G*beta), -s*(G0 - G*beta)] / (1+beta);
a = [1, -s*(1-beta)/(1+beta)];
The filters of Examples 11.2.1 and 11.2.2 may be designed by the following MATLAB code, which
also shows how to compute the magnitude responses of the two complementary cases:
233](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-117-320.jpg)
![[b1, a1, beta1] = lheq(1, 0, 1, 1/sqrt(2), 0.2*pi); lowpass cases
[b2, a2, beta2] = lheq(1, 0, 1, sqrt(0.9), 0.2*pi);
[b3, a3, beta3] = lheq(1, 0, 1, 1/sqrt(2), 0.7*pi);
[b4, a4, beta4] = lheq(1, 0, 1, sqrt(0.9), 0.7*pi);
[b5, a5, beta5] = lheq(-1, 0, 1, 1/sqrt(2), 0.2*pi); highpass cases
[b6, a6, beta6] = lheq(-1, 0, 1, sqrt(0.9), 0.2*pi);
omega = (0:199)*pi/200; 200 frequencies in [0, π]
H1 = abs(dtft(b1, omega)./dtft(a1, omega)).^2; magnitude response squared
H5 = abs(dtft(b5, omega)./dtft(a5, omega)).^2;
Problem 11.8
The following MATLAB code generates the first pair of comb and notch filters of Example 11.5.1,
that is, with unshifted peaks, and computes the corresponding magnitude and phase responses;
it also computes the impulse response of the first filter in two ways: (a) using MATLAB’s function
filter and (b) the customized function combfilt.m that uses a circular buffer:
D = 10;
wc = 0.025 * pi;
[a1, b1, c1, beta1] = combeq(1, 10^(9/20), 10^(3/20), D, wc, 1);
[a2, b2, c2, beta2] = combeq(1, 10^(-12/20), 10^(-3/20), D, wc, 1);
omega = (0:399)*2*pi/ 400; 400 frequencies in [0, 2π]
num1 = [b1, zeros(1, D-1), -c1]; numerator coefficients
den1 = [1, zeros(1, D-1), -a1]; denominator coefficients
H1 = dtft(num1, omega) ./ dtft(den1, omega); frequency response
magH1 = 20 * log10(abs(H1)); magnitude response
angH1 = angle(H1) * 180 / pi; phase response in degrees
num2 = [b2, zeros(1, D-1), -c2];
den2 = [1, zeros(1, D-1), -a2];
H2 = dtft(num2, omega) ./ dtft(den2, omega);
magH2 = 20 * log10(abs(H2));
angH2 = angle(H2) * 180 / pi;
x = [1, zeros(1, 199)]; length-200 unit impulse
h1 = filter(num1, den1, x); impulse response
w = zeros(1, D+1); (D+1)-dimensional state vector
q = 0; circular pointer index
for n=0:199,
[y1(n+1), w, q] = combfilt(a1, b1, c1, D, w, q, x(n+1)); note: h1=y1
end
The phase responses of the comb and notch filters and the impulse response of the comb are
shown in Fig. P11.2.
The canonical realization and the corresponding circular buffer version of the sample processing
algorithm of the transfer function H(z)=
b − cz−D
1 − az−D
are given below:
234
Fig. P11.2 Phase responses and impulse response of Problem 11.8.
for each input sample x do:
sD = tap(D, w, p, D)
s0 = x + asD
y = bs0 − csD
∗p = s0
cdelay(D, w, &p)
x yb
a -c
z-D
sD
s0
The following C function combfilt.c is an implementation:
/* combfilt.c - comb filtering with circular buffer */
void cdelay();
double tap();
double combfilt(a, b, c, D, w, p, x)
double a, b, c, *w, **p, x; p passed by address as in cfir.c
int D;
{
double s0, sD, y;
sD = tap(D, w, *p, D); Dth tap output
s0 = x + a * sD; input to D-fold delay
y = b * s0 - c * sD; output sample
**p = s0; update circular delay line
cdelay(D, w, p);
return y;
}
The MATLAB version is the function combfilt.m given below:
% combfilt.m - comb filtering with circular buffer
%
% [y, w, q] = combfilt(a, b, c, D, w, q, x);
%
% The filter parameters [a, b, c] can be obtained from combeq.m
235](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-118-320.jpg)
![%
% It uses cdelay2.m, wrap2.m, tap2.m
% based on combfilt.c
function [y, w, q] = combfilt(a, b, c, D, w, q, x)
sD = tap2(D, w, q, D);
s0 = x + a * sD;
y = b * s0 - c * sD;
w(q+1) = s0;
q = cdelay2(D, q);
It uses the following routine tap2.m that helps extract the components of the circular internal
state vector:
% tap2.m - read ith tap of circular delay-line buffer
%
% si = tap2(D, w, q, i)
%
% w = (D + 1)-dimensional
% q = circular index into w
% delay amount i = 0, 1, . . . , D
% based on tap2.c
function si = tap2(D, w, q, i)
si = w(rem(q+i, D+1)+1); % si is delayed output s(n − i)
Problem 11.9
In the FIR case of designing a lowpass filter, the limits 1 ± δpass and ±δstop represent the ripples
of the (real-valued) frequency response D(ω). These translate to the limits for the magnitude
response 1 ± δpass in the passband and [0, δstop] in the stopband.
In the IIR case, the definitions are motivated by the closed-form expressions of the magnitude
response squared in the Butterworth and Chebyshev cases.
Problem 11.10
Because Ωstop > Ωpass, and N ≥ Nexact, we have the inequality
Ωstop
Ωpass
2N
>
Ωstop
Ωpass
2Nexact
=
ε2
stop
ε2
pass
⇒ 1 + ε2
pass
Ωstop
Ωpass
2N
> 1 + ε2
stop
Thus, using Eq. (11.6.14) we have:
A(Ωstop)= 10 log10
⎡
⎣1 + ε2
pass
Ωstop
Ωpass
2N
⎤
⎦ > 10 log10(1 + ε2
stop)= Astop
Similarly, if we fix Ω0 by the stopband specification, then we have:
Ωpass
Ωstop
2N
<
Ωpass
Ωstop
2Nexact
=
ε2
pass
ε2
stop
⇒ 1 + ε2
stop
Ωpass
Ωstop
2N
< 1 + ε2
pass
Then, using Eq. (11.6.15), we have:
236
A(Ωpass)= 10 log10
⎡
⎣1 + ε2
stop
Ωpass
Ωstop
2N
⎤
⎦ < 10 log10(1 + ε2
pass)= Apass
Thus, in either case, the specifications are more than satisfied.
Problem 11.11
The call to the MATLAB function lhbutt.m:
[A, B, P] = lhbutt(1, 40, 10, 15, 3, 35);
produces the following cascade matrices
B =
⎡
⎢
⎣
0.5001 0.5001 0
0.3821 2 × 0.3821 0.3821
0.2765 2 × 0.2765 0.2765
⎤
⎥
⎦ , A =
⎡
⎢
⎣
1 0.0002 0
1 0.0007 0.5279
1 0.0005 0.1056
⎤
⎥
⎦
Thus, the designed transfer function will be:
H(z)=
0.5001(1 + z−1
)
1 + 0.0002z−1
·
0.3821(1 + z−1
)2
1 + 0.0007z−1 + 0.5279z−2
·
0.2765(1 + z−1
)2
1 + 0.0005z−1 + 0.1056z−2
The parameter vector P is:
P = [Ωpass, Ωstop, εpass, εstop, Nexact, N, Astop, Ω0, f0]
= [1, 2.4142, 0.9976, 56.2252, 4.57, 5, 38.26, 1.0005, 10.0030]
where Astop is the actual stopband attenuation realized by the designed filter, and f0 is the 3-
dB cutoff frequency. The apparent discrepancy between the computed 3-dB frequency and the
given 3-dB passband specification is accounted for from the fact that the 3-dB frequency Ω0 or
f0 is defined to be the frequency where the magnitude squared response drops to one-half its
DC value; this is almost a 3-dB drop, but not exactly. By hand, the design is as follows. First, we
prewarp the physical frequencies:
Ωpass = tan
ωpass
2
= tan
πfpass
fs
= tan
π10
40
= tan(0.25π)= 1
Ωstop = tan
ωstop
2
= tan
πfstop
fs
= tan
π15
40
= tan(0.375π)= 2.4142
Then, we compute the -parameters:
εpass = 10Apass/10
− 1 = 103/10 − 1 = 0.9976, εstop = 10Astop/10
− 1 = 1035/10 − 1 = 56.2252
and the exact and up-rounded value of N:
Nexact =
ln(εstop/εpass)
ln(Ωstop/Ωpass)
=
ln(56.2252/0.9976)
ln(2.4142/1.0000)
= 4.57 , N = 5
Then, we calculate the 3-dB Butterworth parameter Ω0 and invert it to get the 3-dB frequency f0:
Ω0 =
Ωpass
ε1/N
pass
=
1
(0.9976)1/5
= 1.0005 , f0 =
fs
π
arctan(Ω0)= 1.0030 kHz
237](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-119-320.jpg)
![The coefficients of the 2nd-order sections are computed by Eq. (11.6.24). Note that because of
the small discrepancy mentioned above, Ω0 is not exactly equal to unity, and therefore, the a1
coefficients of all the sections will not be exactly zero. The Butterworth angles are computed by
θi =
π
2N
(N − 1 + 2i)=
π
10
(4 + 2i), i = 1, 2
which gives
θ1 = 0.6π, θ2 = 0.8π
Problem 11.12
The call to lhbutt.m:
[A, B, P] = lhbutt(-1, 10, 3, 2, 0.5, 35);
produces the following cascade matrices
B =
0.3414 −2 × 0.3414 0.3414
0.2454 −2 × 0.2454 0.2454
, A =
1 0.0817 0.4471
1 0.0587 0.0404
Thus, the designed transfer function will be:
H(z)=
0.3414(1 − z−1
)2
1 + 0.0817z−1 + 0.4471z−2
·
0.2454(1 − z−1
)2
1 + 0.0587z−1 + 0.0404z−2
The parameter vector P is:
P = [Ωpass, Ωstop, εpass, εstop, Nexact, N, Astop, Ω0, f0]
= [0.7265, 1.3764, 0.3493, 3.0000, 3.37, 4, 13.27, 0.9451, 2.5899]
where Astop is the actual stopband attenuation realized by the designed filter, and f0 is the 3-dB
cutoff frequency in kHz. The required closed-form expression for part (c) is:
|H(ω)|2
=
1
1 +
cot(ω/2)
Ω0
2N
Problem 11.13
The real part of s can be expressed as follows:
Re s =
1
2
(s + s∗
)=
z2
− 2cz + 1
2(z2 − 1)
+
z∗2
− 2cz∗
+ 1
2(z∗2 − 1)
=
(z2
− 2cz + 1)(z∗2
− 1)+(z∗2
− 2cz∗
+ 1)(z2
− 1)
2(z2 − 1)(z2 − 1)∗
=
|z|4
− 1 − c(z + z∗
)(|z|2
− 1)
|z2 − 1|2
=
(|z|2
− 1)(|z|2
− c(z + z∗
)+1)
|z2 − 1|2
where in the last step we factored |z|4
− 1 = (|z|2
− 1)(|z|2
+ 1). Because |c| ≤ 1, the second
factor of the numerator is non-negative. Indeed, noting that (z + z∗
) is real-valued, we have:
|c(z + z∗
)| ≤ |z + z∗
| ≤ |z| + |z∗
| = 2|z| ⇒ −2|z| ≤ c(z + z∗
)≤ 2|z|
Adding |z|2
+ 1 to both sides we get the inequality:
|z|2
−2|z|+1 ≤ |z|2
−c(z+z∗
)+1 ≤ |z|2
+2|z|+1 ⇒ 0 ≤ (|z|−1)2
≤ |z|2
−c(z+z∗
)+1 ≤ (|z|+1)2
Therefore, the sign of Re s is the same as the sign of the factor (|z|2
− 1), and the desired result
follows.
238
Problem 11.14
Using a trig identity, we obtain:
| sin(ω1 + ω2)| = | sin ω1 cos ω2 + sin ω2 cos ω1| ≤ | sin ω1|| cos ω2| + | sin ω2|| cos ω1|
≤ | sin ω1| + | sin ω2| = sin ω1 + sin ω2
where we used the inequalities | cos ω1| ≤ 1 and | cos ω2| ≤ 1 and the fact that sin ω1 and sin ω2
are non-negative for values of their arguments in the range [0, π]. Thus, the above inequality
implies |c| ≤ 1.
Problem 11.15
Here, the Butterworth filter order is given to be N = 3. Thus, there remains only the parameter
Ω0 to be fixed. The normalized passband frequencies are
ωpa =
2πfpa
fs
=
2π2
20
= 0.2π, ωpb =
2πfpb
fs
=
2π8
20
= 0.8π
Because they add up to π, the bilinear transformation parameter c will be zero, as follows from
the equation
c =
sin(ωpa + ωpb)
sin ωpa + sin ωpb
= 0
Thus, the bilinear transformation will be here:
s =
1 + z−2
1 − z−2
The analog passband frequency will be:
Ωpass =
c − cos ωpb
sin ωpb
= − cot ωpb = 1.3764
The εpass parameter is calculated to be:
εpass = 10Apass/10
− 1 = 100.5/10 − 1 = 0.3493
which implies for Ω0:
Ω0 =
Ωpass
ε1/N
pass
=
1.3764
0.34931/3
= 1.9543
The coefficients of the second- and fourth-order digital filter sections are calculated from Eqs. (11.6.43)
and (11.6.41), where the there is only one Butterworth angle, namely,
θ1 =
π
2N
(N − 1 + 2)=
2π
3
⇒ cos θ1 = −0.5
We find for the second-order section:
G0 = 0.6615, a01 = 0, a02 = −0.3230
and the fourth-order section:
G1 = 0.5639, a11 = 0, a12 = −0.8325, a13 = 0, a14 = 0.4230
resulting in the transfer function:
H(z)=
0.6615(1 − z−2
)
1 − 0.3230z−2
·
0.5639(1 − z−2
)2
1 − 0.8325z−2 + 0.4230z−4
The 3-dB frequencies are calculated from Eq. (11.6.44) with c = 0. We find:
f0a = 1.5054 kHz, f0b = 8.4946 kHz
239](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-120-320.jpg)
![Problem 11.16
We have:
1
1 − 2
s
Ω0
cos θi +
s2
Ω2
0
=
Ω2
0
Ω2
0 − 2sΩ0 cos θi + s2
=
Ω2
0
Ω2
0 − 2
1 − z−1
1 + z−1
Ω0 cos θi +
1 − z−1
1 + z−1
2
=
Ω2
0(1 + z−1
)2
Ω2
0(1 + z−1)2−2(1 − z−1)(1 + z−1)Ω0 cos θi + (1 − z−1)2
=
Ω2
0(1 + z−1
)2
Ω2
0(1 + 2z−1 + z−2)−2(1 − z−2)Ω0 cos θi + (1 − 2z−1 + z−2)
=
Ω2
0(1 + z−1
)2
(1 − 2Ω0 cos θi + Ω2
0)+2(Ω2
0 − 1)z−1 + (1 + 2Ω0 cos θi + Ω2
0)z−2
=
Gi(1 + z−1
)2
1 + ai1z−1 + ai2z−2
where the final result was obtained by dividing numerator and denominator by the factor (1 −
2Ω0 cos θi + Ω2
0) in order to make the constant coefficient of the denominator unity.
The highpass case is obtained by replacing z−1
by −z−1
, which effectively changes the sign of the
ai1 coefficient. The bandpass case can be done in a similar, but more tedious, fashion.
Problem 11.17
The right 3-dB frequency is obtained by solving the equation
c − cos ω0b
sin ω0b
= Ω0
Using the trig identities
cos ω0b =
1 − tan2
(ω0b/2)
1 + tan2(ω0b/2)
, sin ω0b =
2 tan(ω0b/2)
1 + tan2(ω0b/2)
The above equation reduces to the quadratic equation in the variable t = tan(ω0b/2):
c −
1 − t2
1 + t2
2t
1 + t2
= Ω0 ⇒ (1 + c)t2
− 2Ω0t − (1 − c)= 0
which has positive solution:
t =
Ω0 + Ω2
0 + (1 − c)(1 + c)
1 + c
implying for the 3-dB frequency
tan
ω0b
2
= tan
πf0b
fs
=
Ω2
0 + 1 − c2 + Ω0
1 + c
⇒ f0b =
fs
π
arctan
⎛
⎜
⎝
Ω2
0 + 1 − c2 + Ω0
1 + c
⎞
⎟
⎠
The left 3-dB frequency is obtained by replacing Ω0 by −Ω0 in the above solution.
240
Problem 11.18
Matching the stopband specifications exactly means that the stopband frequencies ωsa, ωsb map
onto ±Ωstop and that at Ωstop the attenuation is exactly Astop. The first requirement implies
−Ωstop =
c − cos ωsa
sin ωsa
Ωstop =
c − cos ωsb
sin ωsb
which can be solve for c and Ωstop:
c =
sin(ωsa + ωsb)
sin ωsa + sin ωsb
, Ωstop =
c − cos ωsb
sin ωsb
Given the value of c, we compute next the two candidates for Ωpass:
Ωpa =
c − cos ωpa
sin ωpa
, Ωpb =
c − cos ωpb
sin ωpb
Ideally, we would like have Ωpb = Ωpass and Ωpa = −Ωpass. But this is impossible because c has
already been fixed. Because the Butterworth magnitude response is a monotonically decreasing
function of Ω, it is enough to choose the widest of the two passbands defined above. Thus, we
define:
Ωpass = max |Ωpa|, |Ωpb|
Once we have computed Ωpass and Ωstop, we design our Butterworth analog filter; that is, compute
N and Ω0 from
Nexact =
ln(εstop/εpass)
ln(Ωstop/Ωpass)
, N = Nexact , Ω0 =
Ωstop
ε1/N
stop
where the equation for Ω0 matches the stopband specification exactly. The digital filter coef-
ficients are then computed in terms of N and Ω0 as usual. The following MATLAB function
bpbutt2.m implements the above procedure and is an alternative to bpsbutt.m:
% bpbutt2.m - bandpass Butterworth digital filter design
%
% [A, B, P] = bpbutt2(fs, fpa, fpb, fsa, fsb, Apass, Astop)
%
% alternative to bpsbutt.m
% it matches stopbands and uses worst passband.
%
% design parameters:
% P = [Wpass, Wstop, Wpa, Wpb, c, fc, epass, estop, Nex, N, Apass, W0, f0a, f0b];
% A,B are Kx5 matrices of cascade of 4th/2nd order sections
function [A, B, P] = bpbutt2(fs, fpa, fpb, fsa, fsb, Apass, Astop)
c = sin(2*pi*(fsa + fsb) / fs) / (sin(2*pi*fsa / fs) + sin(2*pi*fsb / fs));
fc = 0.5 * (fs/pi) * acos(c);
Wstop = abs((c - cos(2*pi*fsb / fs)) / sin(2*pi*fsb / fs));
Wpa = (c - cos(2*pi*fpa / fs)) / sin(2*pi*fpa / fs);
Wpb = (c - cos(2*pi*fpb / fs)) / sin(2*pi*fpb / fs);
241](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-121-320.jpg)
![Wpass = max(abs(Wpa), abs(Wpb));
epass = sqrt(10.0^(Apass/10) - 1);
estop = sqrt(10.0^(Astop/10) - 1);
Nex = log(estop/epass) / log(Wstop/Wpass);
N = ceil(Nex);
r = rem(N,2);
K = (N - r) / 2;
W0 = Wstop * (estop^(-1/N));
Apass = 10 * log10(1 + (Wpass/W0)^(2*N));
f0a = (fs/pi) * atan((sqrt(W0^2 - c^2 + 1) - W0)/(c+1));
f0b = (fs/pi) * atan((sqrt(W0^2 - c^2 + 1) + W0)/(c+1));
P = [Wpass, Wstop, Wpa, Wpb, c, fc, epass, estop, Nex, N, Apass, W0, f0a, f0b];
if r==1,
G = W0 / (1 + W0);
a1 = -2 * c / (1 + W0);
a2 = (1 - W0) / (1 + W0);
B(1,:) = G * [1, 0, -1, 0, 0];
A(1,:) = [1, a1, a2, 0, 0];
else
B(1,:) = [1, 0, 0, 0, 0];
A(1,:) = [1, 0, 0, 0, 0];
end
for i=1:K,
th = pi * (N - 1 + 2 * i) / (2 * N);
D = 1 - 2 * W0 * cos(th) + W0^2;
G = W0^2 / D;
a1 = 4 * c * (W0 * cos(th) - 1) / D;
a2 = 2 * (2*c^2 + 1 - W0^2) / D;
a3 = - 4 * c * (W0 * cos(th) + 1) / D;
a4 = (1 + 2 * W0 * cos(th) + W0^2) / D;
B(i+1,:) = G * [1, 0, -2, 0, 1];
A(i+1,:) = [1, a1, a2, a3, a4];
end
Similarly, the function bpcheb2a.m is an alternative to the bandpass Chebyshev type-2 design
function bpcheb2.m:
% bpcheb2a.m - alternative bandpass Chebyshev type 2 digital filter design
%
% [A, B, P] = bpcheb2a(fs, fpa, fpb, fsa, fsb, Apass, Astop)
%
% alternative to bpcheb2.m
% matches stopbands instead of passbands
%
% design parameters:
% P = [Wpass, Wstop, Wpa, Wpb, c, epass, estop, Nex, N, f3a, f3b, a];
% A, B are Kx5 matrices of cascade 4th/2nd order sections
function [A, B, P] = bpcheb2a(fs, fpa, fpb, fsa, fsb, Apass, Astop)
242
c = sin(2*pi*(fsa + fsb) / fs) / (sin(2*pi*fsa / fs) + sin(2*pi*fsb / fs));
Wstop = abs((c - cos(2*pi*fsb / fs)) / sin(2*pi*fsb / fs));
Wpa = (c - cos(2*pi*fpa / fs)) / sin(2*pi*fpa / fs);
Wpb = (c - cos(2*pi*fpb / fs)) / sin(2*pi*fpb / fs);
Wpass = max(abs(Wpa), abs(Wpb));
epass = sqrt(10.0^(Apass/10) - 1);
estop = sqrt(10.0^(Astop/10) - 1);
Nex = acosh(estop/epass) / acosh(Wstop/Wpass);
N = ceil(Nex);
r = rem(N,2);
K = (N - r) / 2;
a = asinh(estop) / N;
W3 = Wstop / cosh(acosh(estop)/N);
f3a = (fs/pi) * atan((sqrt(W3^2 - c^2 + 1) - W3)/(c+1));
f3b = (fs/pi) * atan((sqrt(W3^2 - c^2 + 1) + W3)/(c+1));
P = [Wpass, Wstop, Wpa, Wpb, c, epass, estop, Nex, N, f3a, f3b, a];
W0 = sinh(a) / Wstop;
if r==1,
G = 1 / (1 + W0);
a1 = -2 * c * W0 / (1 + W0);
a2 = -(1 - W0) / (1 + W0);
B(1,:) = G * [1, 0, -1, 0, 0];
A(1,:) = [1, a1, a2, 0, 0];
else
B(1,:) = [1, 0, 0, 0, 0];
A(1,:) = [1, 0, 0, 0, 0];
end
for i=1:K,
th = pi * (N - 1 + 2 * i) / (2 * N);
Wi = sin(th) / Wstop; % reciprocal of text
D = 1 - 2 * W0 * cos(th) + W0^2 + Wi^2;
G = (1 + Wi^2) / D;
b1 = - 4 * c * Wi^2 / (1 + Wi^2);
b2 = 2 * (Wi^2 * (2*c*c+1) - 1) / (1 + Wi^2);
a1 = 4 * c * (W0 * cos(th) - W0^2 - Wi^2) / D;
a2 = 2 * ((2*c*c + 1)*(W0^2 + Wi^2) - 1) / D;
a3 = - 4 * c * (W0 * cos(th) + W0^2 + Wi^2) / D;
a4 = (1 + 2 * W0 * cos(th) + W0^2 + Wi^2) / D;
B(i+1,:) = G * [1, b1, b2, b1, 1];
A(i+1,:) = [1, a1, a2, a3, a4];
end
243](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-122-320.jpg)
![Problem 11.19
We work with the type-2 design since it has more complicated numerator. The transformation of
the analog second-order sections is as follows:
1 + Ωi
−2
s2
1 − 2Ω−1
0 cos θi s + (Ω−2
0 + Ωi
−2
)s2
=
(1 + z−1
)2
+Ωi
−2
(1 − z−1
)2
(1 + z−1)2−2Ω−1
0 cos θi(1 − z−1)(1 + z−1)+(Ω−2
0 + Ωi
−2
)(1 − z−1)2
=
(1 + Ωi
−2
)+2(1 − Ωi
−2
)z−1
+ (1 + Ωi
−2
)z−2
(1 − 2Ω−2
0 cos θi + Ω−2
0 + Ωi
−2
)+2(1 − Ω−2
0 − Ωi
−2
)z−1 + (1 + 2Ω−2
0 cos θi + Ω−2
0 + Ωi
−2
)z−2
=
Gi(1 + bi1z−1
+ z−2
)
1 + ai1z−1 + ai2z−2
where the final step follows by dividing the numerator by the factor (1 + Ωi
−2
) and the denomi-
nator by (1 − 2Ω−2
0 cos θi + Ω−2
0 + Ωi
−2
).
Problem 11.20
The s-domain poles are
si = Ωpass sinh a cos θi + jΩpass cosh a sin θi = Ω0 cos θi + jΩi cosh a
It follows that
|si|2
= Ω2
0 cos2
θi + Ωi
2
cosh2
a = Ω2
0(1 − sin2
θi)+Ωi
2
(1 + sinh2
a)
= Ω2
0 + Ωi
2
+ (Ω2
0 sin2
θi − Ωi
2
sinh2
a)= Ω2
0 + Ωi
2
where the last two terms canceled because
Ω0 sin θi = Ωpass sinh a sin θi = Ωi sinh a
Problem 11.21
The following C functions sos4.c and cas4.c implement the sample-by-sample processing al-
gorithm of the cascade of fourth-order sections:
/* sos4.c - IIR filtering by single (canonical) fourth-order section */
double sos4(a, b, w, x) a, b, w are 5-dimensional
double *a, *b, *w, x; a[0]= 1 always
{
double y;
w[0] = x - a[1]*w[1] - a[2]*w[2] - a[3]*w[3] - a[4]*w[4];
y = b[0]*w[0] + b[1]*w[1] + b[2]*w[2] + b[3]*w[3] + b[4]*w[4];
w[4] = w[3]; could replace these updates by
w[3] = w[2]; the single call to
w[2] = w[1]; delay(4, w);
w[1] = w[0];
return y;
}
and
244
/* cas4.c - IIR filtering in cascade of fourth-order sections */
double sos4(); single 4th-order section
double cas4(K, A, B, W, x)
int K;
double **A, **B, **W, x; A, B, W are K×5 matrices
{
int i;
double y;
y = x; initial input to first SOS4
for (i=0; i<K; i++)
y = sos4(A[i], B[i], W[i], y); output of ith section
return y; final output from last SOS4
}
The K × 5 matrices A and B are generated by the filter design MATLAB functions, such as bps-
butt.m, bpcheb2.m, or bscheb2.m, and can be passed as inputs to the above C functions. This
can be done by saving the A and B outputs of the MATLAB functions into ASCII files and then
invoking the following filtering routine casfilt4.c which is analogous to casfilt.c of Problem
7.15 and which reads A and B dynamically from the command line:
/* casfilt4.c - IIR filtering in cascade of fourth-order sections */
#include <stdlib.h>
#include <stdio.h>
#define MAX 64 initial allocation size
double cas4();
void main(int argc, char **argv)
{
FILE *fpa, *fpb; coefficient files
double **A, **B, **W, x, y;
double *a; temporary coefficient array
int M, K, i, j, m;
int max = MAX, dmax = MAX; allocation and increment
if (argc != 3) {
fprintf(stderr, "Usage: casfilt4 A.dat B.dat <x.dat >y.datn");
exit(0);
}
if ((fpa = fopen(argv[1], "r")) == NULL) {
fprintf(stderr, "Can’t open filter file: %sn", argv[1]);
exit(0);
}
if ((fpb = fopen(argv[2], "r")) == NULL) {
fprintf(stderr, "Can’t open filter file: %sn", argv[2]);
exit(0);
}
a = (double *) calloc(max + 1, sizeof(double));
245](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-123-320.jpg)
![for (M=0;; M++) {
if (M == max) {
max += dmax;
a = (double *) realloc((char *) a, (max + 1) * sizeof(double));
}
if (fscanf(fpa, "%lf", a + M) == EOF) break;
}
a = (double *) realloc((char *) a, (M + 1) * sizeof(double));
if (M%5 != 0) {
fprintf(stderr, "all rows of A must be 5-dimensional");
exit(0);
}
else
K = M / 5; number of SOS4
A = (double **) calloc(K, sizeof(double *)); allocate K rows
B = (double **) calloc(K, sizeof(double *));
W = (double **) calloc(K, sizeof(double *)); states of each SOS4
for (i=0; i<K; i++) {
A[i] = (double *) calloc(5, sizeof(double)); each row is 5-dim
B[i] = (double *) calloc(5, sizeof(double));
W[i] = (double *) calloc(5, sizeof(double));
}
for(i=0; i<K; i++) form A matrix
for(j=0; j<5; j++)
A[i][j] = a[5*i + j];
for(i=0; i<K; i++) read B matrix
for(j=0; j<5; j++)
fscanf(fpb, "%lf", B[i] + j);
while(scanf("%lf", &x) != EOF) { process input samples
y = cas4(K, A, B, W, x);
printf("%lfn", y);
}
}
The MATLAB version of sos4 is as follows:
% sos4.m - fourth order section
%
% [y, w] = sos4(b, a, w, x)
%
% b = [b0, b1, b2, b3, b4] = 5-dim numerator (row vector)
% a = [1, a1, a2, a3, a4] = 5-dim denominator (row vector)
% w = [w0, w1, w2, w3, w4] = 5-dim filter state (row vector)
% x = scalar input
% y = scalar output
function [y, w] = sos4(b, a, w, x)
w(1) = x - a(2:5) * w(2:5)’;
y = b * w’;
w = delay(4, w);
The MATLAB version of cas4:
246
% cas4.m - cascade of fourth order sections
%
% [y, W] = cas4(K, B, A, W, x)
%
% B = Kx5 numerator matrix
% A = Kx5 denominator matrix
% W = Kx5 state matrix
% x = scalar input
% y = scalar output
function [y, W] = cas4(K, B, A, W, x)
y = x;
for i = 1:K,
[y, W(i,:)] = sos4(B(i,:), A(i,:), W(i,:), y);
end
The MATLAB version of casfilt4 for filtering of a long input vector is as follows. It is analogous
to casfilt.m of Problem 7.15:
% casfilt4.m - cascade of fourth-order sections
%
% y = casfilt4(B, A, x)
%
% B = Kx5 numerator matrix
% A = Kx5 denominator matrix
% x = row vector input
% y = row vector output
function y = casfilt4(B, A, x)
[K, K1] = size(A);
[N1, N] = size(x);
W = zeros(K,5);
for n = 1:N,
[y(n), W] = cas4(K, B, A, W, x(n));
end
Problem 11.22
The following MATLAB code designs the pairs of lowpass, highpass, bandpass, and bandstop
Butterworth filters, and shows how to compute the phase response and impulse response of one
of the filters (the first bandpass filter):
Apass = -10*log10(0.98); stringent case
Astop = -10*log10(0.02);
[A1, B1, P1] = lhbutt(1, 20, 4, 5, 0.5, 10); lowpass
[A2, B2, P2] = lhbutt(1, 20, 4, 5, Apass, Astop); lowpass
[A3, B3, P3] = lhbutt(-1, 20, 5, 4, 0.5, 10); highpass
[A4, B4, P4] = lhbutt(-1, 20, 5, 4, Apass, Astop); highpass
[A5, B5, P5] = bpsbutt(1, 20, 2, 4, 1.5, 4.5, 0.5, 10); bandpass
[A6, B6, P6] = bpsbutt(1, 20, 2, 4, 1.5, 4.5, Apass, Astop); bandpass
247](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-124-320.jpg)
![[A7, B7, P7] = bpsbutt(-1, 20, 1.5, 4.5, 2, 4, 0.5, 10); bandstop
[A8, B8, P8] = bpsbutt(-1, 20, 1.5, 4.5, 2, 4, Apass, Astop); bandstop
om = (0:199) * pi / 200; 200 frequencies in [0, π]
a5 = cas2can(A5); bandpass case
b5 = cas2can(B5);
H5 = dtft(b5, om)./dtft(a5, om); frequency response
argH5 = angle(H5) * 180 / pi; phase response (in degrees)
x = [1, zeros(1, 99)]; length-100 unit impulse
h5 = casfilt4(B5, A5, x); impulse response
y5 = filter(b5, a5, x); conventional method. Note, y5=h5
The computed phase and impulse responses are shown in Fig. P11.3. Notice how the phase
response is almost linear within its passband.
Fig. P11.3 Phase and impulse responses of (the less stringent) bandpass filter of Problem 11.22.
Problem 11.23
The following MATLAB code designs the pairs of lowpass, highpass, bandpass, and bandstop
Chebyshev filters, and shows how to compute the phase response and impulse response of one
of the filters (the first bandpass filter):
Apass = -10*log10(0.98); stringent case
Astop = -10*log10(0.02);
[A1, B1, P1] = lhcheb1(1, 20, 4, 5, 0.5, 10); LP type 1
[A2, B2, P2] = lhcheb1(1, 20, 4, 5, Apass, Astop); LP type 1
[A3, B3, P3] = lhcheb1(-1, 20, 5, 4, 0.5, 10); HP type 1
[A4, B4, P4] = lhcheb1(-1, 20, 5, 4, Apass, Astop); HP type 1
[A5, B5, P5] = lhcheb2(1, 20, 4, 5, 0.5, 10); LP type 2
[A6, B6, P6] = lhcheb2(1, 20, 4, 5, Apass, Astop); LP type 2
[A7, B7, P7] = bpcheb2(20, 2, 4, 1.5, 4.5, 0.5, 10); BP type 2
[A8, B8, P8] = bpcheb2(20, 2, 4, 1.5, 4.5, Apass, Astop); BP type 2
248
[A9, B9, P9] = bscheb2(20, 1.5, 4.5, 2, 4, 0.5, 10); BS type 2
[A10, B10, P10] = bscheb2(20, 1.5, 4.5, 2, 4, Apass, Astop); BS type 2
om = (0:199) * pi / 200; 200 frequencies in [0, π]
a7 = cas2can(A7); bandpass case
b7 = cas2can(B7);
H7 = dtft(b7, om) ./ dtft(a7, om); frequency response
argH7 = angle(H7) * 180 / pi; phase response (in degrees)
x = [1, zeros(1, 99)]; length-100 unit impulse
h7 = casfilt4(B7, A7, x); impulse response
y7 = filter(b7, a7, x); conventional method. Note, y7=h7
The computed phase and impulse responses are shown in Fig. P11.4. Note that the Chebyshev
phase response is not as linear within its passband.
Fig. P11.4 Phase and impulse responses of (the less stringent) bandpass filter of Problem 11.23.
249](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-125-320.jpg)
![Chapter 12 Problems
Problem 12.1
The low-rate transfer functions, Di(z) and Hi(z)= z−2
Di(z), corresponding to Eq. (12.4.2) will
be:
D0(z) = 1
D1(z) = −0.13z2
+ 0.30z + 0.90 − 0.18z−1
D2(z) = −0.21z2
+ 0.64z + 0.64 − 0.21z−1
D3(z) = −0.18z2
+ 0.90z + 0.30 − 0.13z−1
H0(z) = z−2
H1(z) = −0.13 + 0.30z−1
+ 0.90z−2
− 0.18z−3
H2(z) = −0.21 + 0.64z−1
+ 0.64z−2
− 0.21z−3
H3(z) = −0.18 + 0.90z−1
+ 0.30z−2
− 0.13z−3
Setting z = ζ4
and using Eq. (12.2.15), we obtain the transfer function D(ζ), which is recognized
as the double-sided ζ-transform of the sequence d given in Eq. (12.4.1):
D(ζ)=D0(ζ4
)+ζ−1
D1(ζ4
)+ζ−2
D2(ζ4
)+ζ−3
D3(ζ4
)
=1 + ζ−1
(−0.13ζ8
+ 0.30ζ4
+ 0.90 − 0.18ζ−4
)
+ ζ−2
(−0.21ζ8
+ 0.64ζ4
+ 0.64 − 0.21ζ−4
)
+ ζ−3
(−0.18ζ8
+ 0.90ζ4
+ 0.30 − 0.13ζ−4
)
=1 + 0.90(ζ + ζ−1
)+0.64(ζ2
+ ζ−2
)+0.30(ζ3
+ ζ−3
)
− 0.18(ζ5
+ ζ−5
)−0.21(ζ6
+ ζ−6
)−0.13(ζ7
+ ζ−7
)
Problem 12.2
In this case, L = 2 and we choose M = 2, so that N = 2LM + 1 = 9. The ideal impulse response
is:
d(k )=
sin(πk /2)
πk /2
, −4 ≤ k ≤ 4
or, numerically,
h = d = [0, −0.21, 0, 0.64, 1, 0.64, 0, −0.21, 0 ]
It is depicted in Fig. P12.1.
The two polyphase subfilters are defined by Eq. (12.2.10), that is, for i = 0, 1,
di(k)= d(2k + i), −2 ≤ k ≤ 1
They can be extracted from h by taking every second entry, starting with the ith entry:
h0 = d0 = [0, 0, 1, 0]
h1 = d1 = [−0.21, 0.64, 0.64, −0.21]
250
0
4-4
2-2 1-1
3-3
d(k′)
k′
1
0.640.64
-0.21-0.21
L=2, M=2, N=9
Fig. P12.1 Length-9 symmetric impulse response of 2-fold FIR interpolator.
The interpolated samples between x(n)= xup(2n) and x(n+1)= xup(2n+2) are calculated from
Eqs. (12.2.18). The two subfilters act on the time-advanced (by M = 2) low-rate input samples
{x(n + 2), x(n + 1), x(n), x(n − 1)}, or, {xup(2n + 4), xup(2n + 2), xup(2n), xup(2n − 2)}.
Equations (12.2.18) can be cast in a compact matrix form:
yup(2n)
yup(2n + 1)
=
0 0 1 0
−0.21 0.64 0.64 −0.21
⎡
⎢
⎢
⎢
⎣
xup(2n + 4)
xup(2n + 2)
xup(2n)
xup(2n − 2)
⎤
⎥
⎥
⎥
⎦
These filtering equations can also be obtained by superimposing the symmetric impulse response
of Fig. P12.1 on each of the contributing low-rate samples
{xup(2n + 4), xup(2n + 2), xup(2n), xup(2n − 2)}
and adding up their contributions at each of the intermediate sampling instants 2n + i, i = 0, 1,
as shown in Fig. P12.2.
xup(2n-2)
2n-2
xup(2n+2)
yup(2n+1)
2n+2
xup(2n+4)
2n+4
xup(2n)
2n
n′
Fig. P12.2 Superposition of impulse responses.
The Hamming windowed version of the filter is obtained by multiplying the full length-9 filter
response h by a length-9 Hamming window. The resulting impulse response will be:
h = [0, −0.05, 0, 0.55, 1, 0.55, 0, −0.05, 0 ]
The corresponding polyphase subfilters are obtained by extracting every second entry:
h0 = [0, 0, 1, 0]
h1 = [−0.05, 0.55, 0.55, −0.05]
251](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-126-320.jpg)
![The interpolation equations will be in this case:
yup(2n)
yup(2n + 1)
=
0 0 1 0
−0.05 0.55 0.55 −0.05
⎡
⎢
⎢
⎢
⎣
xup(2n + 4)
xup(2n + 2)
xup(2n)
xup(2n − 2)
⎤
⎥
⎥
⎥
⎦
Problem 12.3
For the case of a 3-fold interpolator, we have L = 3 and choose again M = 2, corresponding to
filter length N = 2LM + 1 = 13. The ideal impulse response is:
d(k )=
sin(πk /3)
πk /3
, −6 ≤ k ≤ 6
or, numerically,
h = d = [0, −0.17, −0.21, 0, 0.41, 0.83, 1, 0.83, 0.41, 0, −0.21, −0.17, 0]
where h is the causal version, with time origin at the left of the vector, and d is the symmetric
one, with time origin at the middle of the vector. This impulse response is shown in Fig. P12.3.
The three polyphase subfilters are defined by Eq. (12.2.10), that is, for i = 0, 1, 2,
di(k)= d(3k + i), −2 ≤ k ≤ 1
They can be extracted from h by taking every third entry, starting with the ith entry:
h0 = d0 = [0, 0, 1, 0]
h1 = d1 = [−0.17, 0.41, 0.83, −0.21]
h2 = d2 = [−0.21, 0.83, 0.41, −0.17]
The interpolated samples between x(n)= xup(3n) and x(n + 1)= xup(3n + 3) are calculated
from Eqs. (12.2.18).
0
5-5 4-4
2-2 1-1
6-6
3-3
d(k′)
k′
1
0.830.83
0.410.41
-0.17-0.17
-0.21-0.21
L=3, M=2, N=13
Fig. P12.3 Length-13 symmetric impulse response of 3-fold FIR interpolator.
All three subfilters act on the time-advanced (by M = 2) low-rate input samples {x(n+2), x(n+1),
x(n), x(n − 1)}, or, {xup(3n + 6), xup(3n + 3), xup(3n), xup(3n − 3)}. Equations (12.2.18) can
be cast in a compact matrix form:
⎡
⎢
⎣
yup(3n)
yup(3n + 1)
yup(3n + 2)
⎤
⎥
⎦ =
⎡
⎢
⎣
0 0 1 0
−0.17 0.41 0.83 −0.21
−0.21 0.83 0.41 −0.17
⎤
⎥
⎦
⎡
⎢
⎢
⎢
⎣
xup(3n + 6)
xup(3n + 3)
xup(3n)
xup(3n − 3)
⎤
⎥
⎥
⎥
⎦
252
These filtering equations can also be obtained by superimposing the symmetric impulse response
of Fig. P12.3 on each of the contributing low-rate samples
{xup(3n + 6), xup(3n + 3), xup(3n), xup(3n − 3)}
and adding up their contributions at each of the intermediate sampling instants 3n+i, i = 0, 1, 2,
as shown in Fig. P12.4.
xup(3n-3)
3n-3
xup(3n+3)
yup(3n+1)
3n+3
xup(3n+6)
3n+6
xup(3n)
3n
n′
Fig. P12.4 Superposition of impulse responses.
The Hamming windowed version of the filter is obtained by multiplying the full length-13 filter
response h by a length-13 Hamming window. The resulting impulse response will be:
h = [0, −0.02, −0.06, 0, 0.32, 0.78, 1, 0.78, 0.32, 0, −0.06, 0.02, 0]
The corresponding polyphase subfilters are obtained by extracting every third entry:
h0 = [0, 0, 1, 0]
h1 = [−0.02, 0.32, 0.78, −0.06]
h2 = [−0.06, 0.78, 0.32, −0.02]
The interpolation equations will be in this case:
⎡
⎢
⎣
yup(3n)
yup(3n + 1)
yup(3n + 2)
⎤
⎥
⎦ =
⎡
⎢
⎣
0 0 1 0
−0.02 0.32 0.78 −0.06
−0.06 0.78 0.32 −0.02
⎤
⎥
⎦
⎡
⎢
⎢
⎢
⎣
xup(3n + 6)
xup(3n + 3)
xup(3n)
xup(3n − 3)
⎤
⎥
⎥
⎥
⎦
Problem 12.4
For interpolation factor L = 4 and filter length N = 25, we find:
N = 2LM + 1 ⇒ M = (N − 1)/2L = (25 − 1)/8 = 3
Thus, we use 3 low-rate samples above and three below every interpolated value to be computed.
The length-25 ideal interpolation impulse response is calculated from:
d(k )=
sin(πk /4)
πk /4
, for − 12 ≤ k ≤ 12
This generates the numerical values:
253](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-127-320.jpg)
![d = [ 0.000, 0.082, 0.127, 0.100, 0.000, −0.129, −0.212, −0.180,
0.000, 0.300, 0.637, 0.900, 1.000, 0.900, 0.637, 0.300,
0.000, −0.180, −0.212, −0.129, 0.000, 0.100, 0.127, 0.082, 0.000 ]
This impulse response is shown in Fig. P12.5 (where the values have been rounded to 2-digit
accuracy for convenience.)
0 4-4 2-2 1-1 3-3
d(k′)
k′
1
0.900.90
0.640.64
0.30
-0.18 -0.21
-0.13 12-12
0.13
0.10 0.08
0.13
0.10 0.08
-0.13
-0.21 -0.18
0.30
L=4, M=3, N=25
Fig. P12.5 Length-25 symmetric impulse response of 4-fold FIR interpolator.
Given 2M = 6 low-rate samples {A, B, C, D, E, F} as shown in Fig. 12.1.11, the three values
{X, Y, Z} interpolated between C and D are calculated by convolving d with the upsampled
low-rate samples. Using the flip-and-slide form of convolution as in Fig. 12.4.3, we position the
impulse response d at the three successive positions, read off the values of d(k ) where it inter-
sects with the low-rate samples, and add up the results. This procedure is shown in Fig. P12.6.
The corresponding linear combinations of low-rate samples are precisely those of Eq. (12.1.3).
They may be rewritten in the polyphase matrix form (with the low-rate samples listed from the
latest down to the earliest):
⎡
⎢
⎣
X
Y
Z
⎤
⎥
⎦ =
⎡
⎢
⎣
0.082 −0.129 0.300 0.900 −0.180 0.100
0.127 −0.212 0.637 0.637 −0.212 0.127
0.100 −0.180 0.900 0.300 −0.129 0.082
⎤
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
F
E
D
C
B
A
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
Problem 12.5
The ideal 3-fold interpolation case with M = 2, so that N = 2LM + 1 = 2 · 3 · 2 + 1 = 13, was
discussed in Problem 12.3. For the linear interpolator, using Eq. (12.3.7), we have:
⎡
⎢
⎣
yup(3n)
yup(3n + 1)
yup(3n + 2)
⎤
⎥
⎦ =
⎡
⎢
⎣
0 0 1 0
0 1/3 2/3 0
0 2/3 1/3 0
⎤
⎥
⎦
⎡
⎢
⎢
⎢
⎣
x(n + 2)
x(n + 1)
x(n)
x(n − 1)
⎤
⎥
⎥
⎥
⎦
=
1/3 2/3
2/3 1/3
x(n + 1)
x(n)
Problem 12.6
The following complete C program interp.c implements the polyphase sample-by-sample pro-
cessing algorithm of Eq. (12.2.20) and the initialization procedure of (12.2.19):
254
n′
A
B
C
D
X
E
F
0.90
0.30
0.10 0.08
-0.13-0.18
n′
A
B
C
D
Y
E
F
0.64 0.64
0.13 0.13
-0.21 -0.21
n′
A
B
C
D
Z
E
F
0.30
0.90
0.08 0.10
-0.18-0.13
Fig. P12.6 Interpolated values are calculated by the flip-and-slide form of convolution.
/* interp.c - L-fold FIR interpolator (rectangular and Hamming window design)
*
* Usage: interp hamm L M < xlow.dat > yhigh.dat
*
* hamm = 0/1 for rectangular/Hamming window
* L = interpolation factor
* M = half-length of polyphase subfilters (length = 2M, order = P = 2M-
1)
* xlow.dat = file containing low-rate input samples
* yhigh.dat = file containing high-rate interpolated output samples
*/
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
void delay();
double dot();
double d(); ideal interpolation filter
void main(int argc, char ** argv)
{
int i, n, L, M, P, hamm;
double x, *y, *w, **h, pi = 4 * atan(1.0);
if (argc != 4) {
255](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-128-320.jpg)
![fprintf(stderr,"nUsage: interp hamm L M < xlow.dat > yhigh.datnn");
fprintf(stderr,"where: hamm = 0/1 for rectangular/hamm windown");
fprintf(stderr," L = interpolation factorn");
fprintf(stderr," M = half-length of polyphase subfiltersn");
fprintf(stderr," xlow.dat = file for low-rate input samplesn");
fprintf(stderr," yhigh.dat = file for high-rate interpolated output samplesn");
exit(0);
}
hamm = atoi(argv[1]);
L = atoi(argv[2]);
M = atoi(argv[3]);
P = 2*M - 1; polyphase filter order
h = (double **) calloc(L, sizeof(double *)); allocate L subfilters
for (i=0; i<L; i++)
h[i] = (double *) calloc(2*M, sizeof(double)); each of length 2M
y = (double *) calloc(L, sizeof(double)); allocate L polyphase outfprintf(stderr,
w = (double *) calloc(2*M, sizeof(double)); allocate low-rate internal state
for (i=0; i<L; i++) construct polyphase subfilters
for (n=0; n<2*M; n++)
if (hamm == 0) rectangular window
h[i][n] = d(L, L*n+i-L*M);
else Hamming window
h[i][n] = d(L, L*n+i-L*M) * (0.54 - 0.46*cos((L*n+i)*pi/(L*M)));
initialize by Eq. (12.2.19)
for (i=M; i>=1; i--) initialize state w by reading
scanf("%lf", w + i); first M low-rate input samples
process by Eq. (12.2.20)
while(scanf("%lf", &x) != EOF) { keep reading low-rate samples till EOF
w[0] = x;
for (i=0; i<L; i++) {
y[i] = dot(P, h[i], w); ith interpolated output sample
printf("%.12lfn", y[i]); and write it into stdout
}
delay(P, w); update low-rate delay line
}
for (n=0; n<M; n++) { input-off transients
w[0] = 0.0; zero input
for (i=0; i<L; i++) {
y[i] = dot(P, h[i], w); ith interpolated output sample
printf("%.12lfn", y[i]); and write it into stdout
}
delay(P, w); update low-rate delay line
}
}
/* -------------------------------------------------------- */
/* d(L, k) - ideal L-fold interpolation coefficients */
256
double d(L, k)
int L, k;
{
double t, pi = 4 * atan(1.0);
if (k == 0) d(0) = 1
return (1.0);
else {
if (k%L == 0) d(k) vanishes at non-zero multiples of L
return (0.0);
else {
t = pi * k / L;
return (sin(t)/t); typical sin x/x impulse response
}
}
}
The program has usage:
interp hamm L M < xlow.dat > yhigh.dat
where the parameter hamm specifies a Hamming or rectangular window, L is the interpolation
factor, and M is the half-length of the polyphase subfilters, so that the total filter length is
N = 2LM + 1.
The program designs the windowed length-N interpolation filter and then reads the low-rate in-
put samples sequentially from a file xlow.dat or stdin, computes the high-rate interpolated
samples, and writes them to the file yhigh.dat or to the stdout. Each low-rate input generates
L high-rate outputs. Inside the program, the rectangular and Hamming window design of the
impulse response may be replaced easily by a Kaiser design.
The program interp.c uses a linear-buffer version of the low-rate delay line used by all the
polyphase subfilters. The following program cinterp.c is the circular-buffer version of in-
terp.c:
/* cinterp.c - circular-buffer version of interp.c */
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
void cdelay(), wrap();
double cdot(); circular version of dot.c
double d(); ideal interpolation filter
void main(int argc, char ** argv)
{
int i, n, L, M, P, hamm;
double x, *y, *w, *p, **h, pi = 4 * atan(1.0);
if (argc != 4) {
fprintf(stderr,"nUsage: cinterp hamm L M < xlow.dat > yhigh.datnn");
fprintf(stderr,"where: hamm = 0/1 for rectangular/hamm windown");
fprintf(stderr," L = interpolation factorn");
fprintf(stderr," M = half-length of polyphase subfiltersn");
fprintf(stderr," xlow.dat = file for low-rate input samplesn");
257](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-129-320.jpg)
![fprintf(stderr," yhigh.dat = file for high-rate interpolated output samplesn");
exit(0);
}
hamm = atoi(argv[1]);
L = atoi(argv[2]);
M = atoi(argv[3]);
P = 2*M - 1; polyphase filter order
h = (double **) calloc(L, sizeof(double *)); allocate L subfilters
for (i=0; i<L; i++)
h[i] = (double *) calloc(2*M, sizeof(double)); each of length 2M
y = (double *) calloc(L, sizeof(double)); allocate L polyphase outfprintf(stderr,
w = (double *) calloc(2*M, sizeof(double)); allocate low-rate internal state
p = w; initialize circular pointer
for (i=0; i<L; i++) construct polyphase subfilters
for (n=0; n<2*M; n++)
if (hamm == 0) rectangular window
h[i][n] = d(L, L*n+i-L*M);
else Hamming window
h[i][n] = d(L, L*n+i-L*M) * (0.54 - 0.46*cos((L*n+i)*pi/(L*M)));
initialize by Eq. (P12.1)
for (i=M; i>=1; i--) { initialize state w by reading
scanf("%lf", p); first M low-rate input samples
cdelay(P, w, &p);
}
process by Eq. (P12.2)
while(scanf("%lf", &x) != EOF) { keep reading low-rate samples till EOF
*p = x; beginning of state vector
for (i=0; i<L; i++) {
y[i] = cdot(P, h[i], w, p); circular version of dot()
printf("%.12lfn", y[i]); and write it into stdout
}
cdelay(P, w, &p); update low-rate delay line
}
for (n=0; n<M; n++) { input-off transients
*p = 0.0; zero input
for (i=0; i<L; i++) {
y[i] = cdot(P, h[i], w, p); circular version of dot()
printf("%.12lfn", y[i]); and write it into stdout
}
cdelay(P, w, &p); update low-rate delay line
}
}
/* -------------------------------------------------------- */
/* d(L, k) - ideal L-fold interpolation coefficients */
double d(L, k)
int L, k;
258
{
double t, pi = 4 * atan(1.0);
if (k == 0) d(0) = 1
return (1.0);
else {
if (k%L == 0) d(k) vanishes at non-zero multiples of L
return (0.0);
else {
t = pi * k / L;
return (sin(t)/t); typical sin x/x impulse response
}
}
}
/* -------------------------------------------------------- */
/* cdot(P, h, w, p) - circular version of dot.c */
double tap();
double cdot(P, h, w, p) computes dot product of h with
double *h, *w, *p; current internal state vector
int P; pointed to by pointer p
{
int i;
double y;
for (y=0, i=0; i<=P; i++)
y += h[i] * tap(P, w, p, i);
return y;
}
It uses a circular pointer p that circulates over the low-rate buffer w. The circular-buffer version
of Eqs. (12.2.19) and (12.2.20), implemented by cinterp.c, can be stated as follows:
for m = M down to m = 1 do:
read low-rate input sample x
∗p = x
cdelay(P, w, &p)
(P12.1)
and
for each low-rate input sample x do:
∗p = x
for m = 0, 1, . . . , P determine internal states:
sm = tap(P, w, p, m)
for i = 0, 1, . . . , L − 1 compute output of ith filter:
yi = dot(P, hi, s)
cdelay(P, w, &p)
(P12.2)
The length-17 rectangular and Hamming windowed filters for this problem were derived in Section
12.4.1. The given inputs were processed by interp.c. The upsampled versions of these inputs,
xup(n ), are shown in Fig. P12.7 together with the corresponding interpolated output signals
yup(n ), computed by the length-17 Hamming windowed version of the interpolation filter.
259](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-130-320.jpg)
![N1 − 1 =
DΔF1
F0 − 1
=
5.0174 · 4
1.000
= 20.07 ⇒ N1 = 25, M1 = (N1 − 1)/(2L1)= 6
The polyphase computational rate of the two sections combined will be
Rmulti = N0fs + N1(2fs)= 159fs = 159 · 44.1 = 7011.9 MAC/msec = 7.0119 MIPS
Thus, the relative computational cost of the two-stage versus the single stage design will be:
Rmulti
Rsingle
=
159fs
217fs
= 0.73
Problem 12.9
The frequency responses can be computed by the MATLAB code:
f = (0:399) * 160 /400; 400 frequencies in [0,160] kHz
f1 = 28; normalization frequency for Butterworth
f2 = 16; normalization frequency for Bessel
H1 = 1 ./ (1 + 2*j*(f/f1) - 2*(f/f1).^2 - j*(f/f1).^3); Butterworth
H2 = 15 ./ (15 + 15*j*(f/f2) - 6*(f/f2).^2 - j*(f/f2).^3); Bessel
magH1 = 20 * log10(abs(H1));
magH2 = 20 * log10(abs(H2));
argH1 = angle(H1) * 180 / pi;
argH2 = angle(H2) * 180 / pi;
The phase and magnitude responses are shown in Figs. P12.15 and P12.16. The Bessel phase
response is essentially linear over the 0–20 kHz passband. But its magnitude response is not as
flat as the Butterworth’s.
Fig. P12.15 Phase responses of Bessel and Butterworth filters.
264
Fig. P12.16 Magnitude responses of Bessel and Butterworth filters.
Problem 12.10
The filter H0 has a cutoff frequency fc = fs/2. The linear, hold, and DAC magnitude responses
(normalized to 0 dB at DC) are in the general and the particular case of L0 = 4, L1 = L2 = 2:
|H1(f)| =
sin(πf/L0fs)
L1 sin(πf/L0L1fs)
2
=
sin(πf/4fs)
2 sin(πf/8fs)
2
|H2(f)| =
sin(πf/L0L1fs)
L2 sin(πf/L0L1L2fs)
=
sin(πf/8fs)
2 sin(πf/16fs)
|Hdac(f)| =
sin(πf/L0L1L2fs)
πf/L0L1L2fs
=
sin(πf/16fs)
πf/16fs
The filter H0 operates at rate 4fs and cancels all the images at multiples of fs which are not
multiples of 4fs. The linear interpolator H1 operates at rate 8fs and vanishes at multiples of 4fs
which are not multiples of 8fs. The filter H2 operates at rate 16fs and vanishes at multiples of 8fs
which are not multiples of 16fs. Finally, Hdac vanishes at all non-zero multiples of 16fs. Thus, the
combined effect of H0, H1, H2, and Hdac is to (partially) remove all spectral images at multiples
of fs.
We note that the combined effect of the two hold transfer functions is an effective DAC operating
at 8fs:
|H2(f)Hdac(f)| =
sin(πf/L0L1fs)
L2 sin(πf/L0L1L2fs)
·
sin(πf/L0L1L2fs)
πf/L0L1L2fs
=
sin(πf/L0L1fs)
πf/L0L1fs
=
sin(πf/8fs)
πf/8fs
The reason for using H2 is that the higher the L, the more bits can be saved via the noise-shaping
requantizer.
Problem 12.11
The proof of the result may be seen graphically in Fig. P12.17 for the case L = 4. The ideal inter-
polator has cutoff fs/2 and is periodic in f with period Lfs = 4fs. Thus, the passband replicates
at multiples of 4fs.
265](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-133-320.jpg)
![Problem 12.18
Here, we have the conversion ratio fs = 32 =
32
48
· 48 =
2
3
fs, so that L = 2, M = 3. Because
fs < fs, the filter’s cutoff frequency will be fc = fs /2 = 16 kHz. Assuming a ±1 kHz transition
width about fc, that is, Δf = 2 kHz, and a A = 60 dB stopband attenuation, we find the Kaiser
parameter D and filter length N:
D =
A − 7.95
14.36
=
60 − 7.95
14.36
= 3.6247, N−1 =
DLfs
Δf
=
3.6247 · 2 · 48
2
= 173.98 ⇒ N = 177
where we rounded N to the next odd integer of the form N = 2LK + 1. Thus, K = 44, and there
will be L = 2 polyphase subfilters of length 2K = 88. Fig. P12.19 shows the ideal specifications
of such a filter. The filter operates at the fast rate fs = 3fs = 2fs = 96 kHz and acts as an
antialiasing filter for the downsampler, that is, it removes the high frequencies from the input
in the range [fs 2, fs/2]= [16, 24] kHz, so that the downshifted replicas—caused by the 3-fold
downsampling operation—will not overlap.
0
0
fs′ 2fs′ 3fs′
fs′ 2fs′ 3fs′=fs′′
2fs=fs′′
0
fs 2fs
fs
0
filter
polyphase
selector
T′/3=T′′
T/2=T′′
6T′′=3T=2T′
T
x0
x1
x2
y0
y1
t
t
t
t
f
f
f
f
T′
min(fs, fs′)1
2
0 1i=
Fig. P12.19 Ideal SRC filter with conversion ratio 2/3.
Because T = T /2 and T = T /3, it follows that the basic block interval will be Tblock = 6T =
3T = 2T , containing 3 input-rate samples {x0, x1, x2} and 2 output-rate samples {y0, y1}. The
interpolated output y1 that lies halfway between x1 and x2 is obtained by the polyphase filter h1.
Indeed, the polyphase selector indices can be precomputed by
im = 3m % 2 = [0, 1], for m = 0, 1
The sample processing algorithm of the 2/3 sample rate converter will be then:
for each input block {x0, x1, x2} do:
w0 = x0
y0 = dot(P, h0, w)= x0
delay(P, w)
w0 = x1
y1 = dot(P, h1, w)
delay(P, w)
w0 = x2
delay(P, w)
270
where P = 2K − 1 is the order of the polyphase subfilters h0, h1, and w is the length-(P + 1)
input-rate delay-line buffer. Note that there is no interpolated output after x2 (the next two out-
puts come from the next group of three inputs), and therefore, x2 is simply read into the buffer
w and the delay line is updated.
For reference, we also show in Fig. P12.20 the reverse converter by a ratio 3/2, converting from
32 kHz up to 48 kHz. Here, the SRC filter acts as an anti-image filter for the upsampler.
polyphase
selector
T′/2=T′′
T/3=T′′
6T′′=2T=3T′
y0
y1
y2
T
x0
x1
t
t
t
t
f
f
f
f
T′
min(fs, fs′)1
2
0 2 1i= fs 2fs 3fs=fs′′0
fs 2fs 3fs
0
filter
0 fs′ 2fs′
0 2fs′=fs′′fs′
Fig. P12.20 Ideal SRC filter with conversion ratio 3/2.
In this case, we have fs = 3fs/2 = 3 · 32/2 = 48. The fast rate is fs = 2fs = 3fs = 96 kHz. The
input- and output-rate sampling intervals are T = 3T and T = 2T , and the basic time block
Tblock = 6T = 2T = 3T . Thus, every group of two input samples {x0, x1} generates a group of
three output samples {y0, y1, y2}. As can be seen in the figure, the polyphase selector sequence
is im = 2m % 3 = [0, 2, 1], for m = 0, 1, 2. Therefore, the three interpolated outputs {y0, y1, y2}
will be computed by the subfilters {h0, h2, h1}. The corresponding sample processing algorithm
will be:
for each input block {x0, x1} do:
w0 = x0
y0 = dot(P, h0, w)= x0
y1 = dot(P, h2, w)
delay(P, w)
w0 = x1
y2 = dot(P, h1, w)
delay(P, w)
Problem 12.19
For the case L/M = 7/4, we have fs = 7fs/4 and fs = 4fs = 7fs, T = T /4 = T/7, Tblock =
28T = 7T = 4T. Fig. P12.21 depicts the operation of the converter both in the time and
frequency domains.
In the time domain, each input block of 4 input-rate samples {x0, x1, x2, x3} generates 28 fast-rate
interpolated samples, out of which 7 output-rate samples {y0, y1, y2, y3, y4, y4, y6} are selected
and computed according to their polyphase indices:
271](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-136-320.jpg)
![0
0
0
0
filter
polyphase
selector
T′/4=T′′
T/7=T′′
28T′′=4T=7T′
y0
y1
y3 y5y2
y4
y6
x0
x1
x2 x3
t
t
t
t
f
f
f
f
T′
T
min(fs, fs′)1
2
0 4 1 25 6 3i= fs 2fs 3fs 4fs 5fs 6fs 7fs=fs′′
fs 2fs 3fs 4fs 5fs 6fs 7fs
4fs′=fs′′2fs′ 3fs′fs′
4fs′2fs′ 3fs′fs′
Fig. P12.21 Ideal SRC filter with conversion ratio 7/4.
im = 4m % 7 = [0, 4, 1, 5, 2, 6, 3], for m = 0, 1, . . . , 6
nm = (4m − im)/7 = [0, 0, 1, 1, 2, 2, 3]
whenever the index nm is repeated, the input-rate polyphase delay line is not updated. The
sample processing conversion algorithm is as follows:
for each input block {x0, x1, x2, x3} do:
w0 = x0
y0 = dot(P, h0, w)= x0
y1 = dot(P, h4, w)
delay(P, w)
w0 = x1
y2 = dot(P, h1, w)
y3 = dot(P, h5, w)
delay(P, w)
w0 = x2
y4 = dot(P, h2, w)
y5 = dot(P, h6, w)
delay(P, w)
w0 = x3
y6 = dot(P, h3, w)
delay(P, w)
For example, the two output samples y4, y5 lie between the input samples x2, x3 and correspond
to the i = 2 and i = 6 interpolated samples. The delay line is updated after both outputs are
computed, and then the next input sample x3 is read into the delay line, and so on.
In the frequency domain, the SRC filter acts as an anti-image postfilter for the upsampler, remov-
ing the L−1 = 6 replicas between multiples of the fast rate fs . The downshifting of the replicas
caused by the downsampling operation will position replicas at the 3 multiples of the output rate
fs , 2fs , and 3fs .
272
In the L/M = 4/7 case, we have fs = 4fs/7 and fs = 7fs = 4fs, T = T /7 = T/4, Tblock =
28T = 4T = 7T. Fig. P12.22 depicts the operation of the converter both in the time and
frequency domains.
0
0
fs′ 2fs′ 3fs′ 4fs′ 5fs′ 6fs′ 7fs′
fs′ 2fs′ 3fs′ 4fs′ 5fs′ 6fs′ 7fs′=fs′′
4fs=fs′′
0
4fs2fs 3fsfs
2fs 3fsfs
0
filter
polyphase
selector
T′/7=T′′
T/4=T′′
28T′′=7T=4T′
T
x0
x1
x3 x5x2
x4
x6
y0
y1
y2 y3
t
t
t
t
f
f
f
f
T′
min(fs, fs′)1
2
0 3 2 1i=
Fig. P12.22 Ideal SRC filter with conversion ratio 4/7.
In the time domain, each input block of 7 input-rate samples {x0, x1, x2, x3, x4, x5, x6} generates
28 fast-rate interpolated samples, out of which 4 output-rate samples {y0, y1, y2, y3} are selected
and computed according to their polyphase indices:
im = 7m % 4 = [0, 3, 2, 1], for m = 0, 1, . . . , 3
nm = (7m − im)/4 = [0, 1, 3, 5]
so that only the input samples x0, x1, x3, x5 will produce output samples. However, for the
remaining input samples the delay line must be updated. Thus, the sample processing conversion
algorithm will be as follows:
for each input block {x0, x1, x2, x3, x4, x5, x6} do:
w0 = x0
y0 = dot(P, h0, w)= x0
delay(P, w)
w0 = x1
y1 = dot(P, h3, w)
delay(P, w)
w0 = x2
delay(P, w)
w0 = x3
y2 = dot(P, h2, w)
delay(P, w) w0 = x4
delay(P, w)
w0 = x5
y3 = dot(P, h1, w)
delay(P, w)
w0 = x6
delay(P, w)
273](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-137-320.jpg)
![In the frequency domain, the SRC filter acts as an antialiasing prefilter for the downsampler, re-
moving the 3 replicas between multiples of the fast rate fs . The downshifting of the replicas
caused by the downsampling operation will position replicas at the 6 multiples of the output rate
fs , 2fs , . . . , 6fs , without overlapping.
As a design example, consider the 4/7 case and assume a Kaiser design with a stopband attenu-
ation of 80 dB and normalized transition width ΔF = Δf/fs = 0.1. The width parameter D and
filter length N will be:
D =
A − 7.95
14.36
=
80 − 7.95
14.36
= 5.0174 , N−1 =
DL
ΔF
=
5.0174 · 4
0.1
= 200.70 ⇒ N = 209
where N the smallest odd integer of the form N = 2LK + 1; here, K = 26.
Problem 12.20
A typical window depends on the time variable k through the ratio k/(N−1). The effective length
of the stretched impulse response is Nρ − 1 = 2LKρ = 2LK/ρ. A symmetric window of such
length will depend on the variable k /(2LKρ)= ρk /(2LK). Thus, the window of length 2LKρ
can be expressed in terms of the same window of length 2LK, via the stretching relationship
wρ(k )= w(ρk ). It follows that the corresponding windowed impulse responses will be
wρ(k )dρ(k )= ρw(ρk )d(ρk )
Problem 12.21
The polyphase Kaiser filter can be designed by the code fragment:
double *hd, **h, *w, *x, *y, wind, pi = 4 * atan(1.0);
D = (A - 7.95) / 14.36;
if (A > 50)
alpha = 0.1102 * (A - 8.7);
else if (A > 21)
alpha = 0.5842 * pow(A-21, 0.4) + 0.07886 * (A-21);
else
alpha = 0;
N = 1 + ceil(D*L / DF); Kaiser length
K = ceil((N - 1) / (2.0*L)); round up
N = 2*L*K + 1; final length
P = 2*K - 1; polyphase filter order
LK = L*K; filter delay
hd = (double *) calloc(N, sizeof(double)); allocate direct form
wc = pi / max(L, M); cutoff frequency
I0alpha = I0(alpha); window normalization
for (n=0; n<N; n++) compute filter
if (n==LK) middle value
hd[n] = L /(double) max(L, M);
else {
wind = I0(alpha * sqrt(n*(2.0*LK-n)) / LK) / I0alpha;
274
hd[n] = wind * L * (sin(wc*(n-LK))) / (pi*(n-LK));
}
h = (double **) calloc(L, sizeof(double *)); allocate polyphase
for (i = 0; i < L; i++)
h[i] = (double *) calloc(P+1, sizeof(double)); i-th polyphase
for (i=0; i<L; i++) define polyphase filters
for (n=0; n<=P; n++)
h[i][n] = hd[L*n+i];
Assuming the input data are being read from stdin, we allocate and initialize the polyphase
delay-line buffer w as follows:
w = (double *) calloc(P+1, sizeof(double)); allocate w
for (i=K; i>=1; i--) read K input samples
if (scanf("%lf", w+i) == EOF) {
printf("must have at least K = %d input samplesn", K);
exit(0);
}
Then, allocate the input/output blocks for the algorithm Eq. (12.6.13) and keep processing input
samples in groups of M. The computed outputs go into stdout. Upon encountering the end-
of-file of the inputs, we jump out of the for-ever loop and call Eq. (12.6.13) K/M more times
with zero input samples. Because each call generates M zero inputs, the K/M calls will generate
approximately M(K/M)= K zero inputs:
x = (double *) calloc(M, sizeof(double));
y = (double *) calloc(L, sizeof(double));
for (;;) { for-ever do:
for (n=0; n<M; n++) read M inputs
if (scanf("%lf", x+n) == EOF)
goto transients;
src(L, M, P, h, w, x, y); compute L outputs
for (m=0; m<L; m++) write L outputs
printf("%.12lfn", y[m]);
}
transients:
for (i=0; i <= K/M; i++) { last K inputs
for (n=0; n<M; n++)
x[n] = 0;
src(L, M, P, h, w, x, y);
for (m=0; m<L; m++)
printf("%.12lfn", y[m]);
}
The sample processing algorithm (12.6.13) is implemented by the routine:
/* src.c - sample rate converter processing algorithm */
#include <math.h>
double dot();
275](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-138-320.jpg)
![void delay();
void src(L, M, P, h, w, x, y) h is Lx(P+1) polyphase filter matrix
double **h, *w, *x, *y; w is (P+1)-dimensional delay buffer
{ x,y are M- and L-dimensional vectors
int n, m, i;
double R = L / (double) M; conversion ratio
for (n = 0; n < M; n++) {
w[0] = x[n];
for (m = ceil(R*n); m <= floor(R*(n+1)); m++) {
i = (M * m) % L; polyphase selector
y[m] = dot(P, h[i], w);
}
delay(P, w);
}
}
With the values A = 30 dB, ΔF = 0.1, the 5/3 filter has D = 1.536, α = 2.117, N = 81, K = 8,
P = 2K − 1 = 15. The reverse 3/5 filter has the same D, α, K, and P, but its length is N = 49,
because N = 2LK + 1, where now L = 3. The filter responses are shown in Fig. P12.23 both
in absolute and dB scales. The dc gains are 5 and 3, respectively. The dB responses have been
normalized to 0 dB at DC.
Fig. P12.23 Frequency responses of SRC filters.
Both filters have a cutoff frequency of fc = min(fs/2, fs /2)= 3/2 = 1.5 kHz, or, ωc = π/5. The
passband ripple is δ = 10−A/20
= 0.032. Their transition widths are different: the first filter has
Δf = 0.1×3 = 0.3 kHz, and the second Δf = 0.1×5 = 0.5 kHz. Were they to be redesigned to
have the same Δf, say 0.3 kHz, then for the 3/5 filter we would have ΔF = 0.06, resulting in the
lengths N = 79, K = 13, P = 25. In that case, the filters would be virtually identical (except for
the gain and the delay by 1 sample).
The impulse responses are obtained from Eq. (12.6.5), with LK = 5×8 = 40 and LK = 3×8 = 24
in the two cases, that is,
276
h5/3(n )= 5
I0(α n (80 − n )/80)
I0(α)
sin π(n − 40)/5
π(n − 40)
, n = 0, 1, . . . , 80
h3/5(n )= 3
I0(α n (48 − n )/48)
I0(α)
sin π(n − 24)/5
π(n − 24)
, n = 0, 1, . . . , 48
The frequency responses were computed by direct evaluation of:
H(f)=
N−1
n =0
h(n )e−2πjfn /fs
, with fs = 15 kHz
Figure P12.24 shows the output y(n ) of the 5/3 filter and compares it with the signal x (n ) that
would have been obtained had the analog signal been sampled at 3 kHz. On the right, it shows
the effect of the reverse 3/5 filter on y(n ), and compares it with the original x(n).
Fig. P12.24 Comparison of y(n ) and x (n ), and of x(n) and reconverted y(n ).
Problem 12.22
Replacing the quantizer Q by its equivalent noise model, we can write the ζ-domain input/output
equations:
V (ζ) = X (ζ)−ζ−1
Y (ζ)
W0(ζ) = V (ζ)+ζ−1
W0(ζ)
Y (ζ) = E (ζ)+W0(ζ)
Eliminating V (ζ) and W0(ζ), we obtain:
Y (ζ)= X (ζ)+(1 − ζ−1
)E (ζ)
Therefore, Hx(ζ)= 1, and HNS(ζ)= 1 − ζ−1
. The main advantage of this form is that the input
signal X (ζ) appears unchanged at the output (as opposed to being delayed as in the conventional
case).
277](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-139-320.jpg)
![y = Q(w1); ΔΣ quantizer output
v = x - y; delta part
w0 = w1 + v; sigma part
w1 = w0; update quantizer’s delay
yup[n] = fir(N-1, h, w, y); (upsampled) decimator output
}
where h and w are the N-dimensional impulse response and internal state arrays for the decima-
tor. In the Kaiser and the rectangular window designs, the filter length is N = 2LM + 1. In the
first-order comb case, it is N = 10, and the impulse response and transfer function are:
h =
1
10
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1] , H(ζ)=
1
10
1 − ζ−10
1 − ζ−1
For the second-order comb case, we obtain the impulse response by convolving the first-order
comb’s impulse response with itself. This gives a length N = 19 response and transfer function:
h =
1
100
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1] , H(ζ)=
1
10
1 − ζ−10
1 − ζ−1
2
The magnitude response of the second-order comb case is the square of that of the first-order
case:
|H(f)| =
sin(πf/fs)
L sin(πf/10fs)
2
It is plotted in dB in Fig. P12.26 together with the first-order comb case and the rectangular
window design. The upsampled output yup(n ) of the second-order comb decimator is plotted
in the right graph of Fig. P12.26.
Fig. P12.26 Length-19 second-order comb magnitude response and filter output.
Problem 12.27
The filter designs are the same as in the previous problems. The implementation of the second-
order delta-sigma quantizer is given by the sample processing algorithm discussed in Problem
12.24. The following for-loop generates the high-rate input, quantizes it by the 2nd-order quan-
tizer, and filters the quantized output with the decimation filter:
280
w1 = v1 = u = 0; initialize delays
for (n=0; n<Ntot; n++) {
x = 0.5 * sin(2 * pi * f0 * n); high-rate input sample
w0 = w1 + x - u; output of H1
v0 = v1 + w0 - u; output of H2
y = Q(v0); ΔΣ quantizer output
w1 = w0; update quantizer’s delays
v1 = v0; update quantizer’s delays
u = y; update quantizer’s delays
yup[n] = fir(N-1, h, w, y); (upsampled) decimator output
}
Figure P12.27 shows the signal x (n ) and the 2-level quantized output y (n ) of the second-order
quantizer.
Fig. P12.27 High-rate input and output signals of second-order ΔΣ quantizer.
The upsampled outputs yup(n ) of the averaging, second-order comb, rectangular, and Kaiser
window decimators are shown in Fig. P12.28.
Problem 12.28
Working in the ζ-domain, we obtain the I/O equation of the first stage. The input to H(ζ) is
X (ζ)−ζ−1
Y1(ζ), Thus, adding the noise component, we have:
Y1(ζ)= E1(ζ)+H(ζ) X (ζ)−ζ−1
Y1(ζ)
which can be rearranged as:
Y1(ζ)=
H(ζ)
1 + ζ−1H(ζ)
X (ζ)+
1
1 + ζ−1H(ζ)
E1(ζ)
Inserting H(ζ)= 1/(1 − ζ−1
), we obtain the transfer relationship of one stage:
Y1(ζ)= X (ζ)+(1 − ζ−1
)E1(ζ)= X (ζ)+D(ζ)E1(ζ) (P12.3)
where D(ζ)= 1−ζ−1
. The input to the second stage is obtained by forming the difference of the
signals around the first quantizer, that is, (Y − E1)−Y = −E1. Applying the basic I/O equation
(P12.3) to this input and corresponding noise E2, we obtain
281](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-141-320.jpg)
![Fig. P12.28 First- and second-order comb, and rectangular- and Kaiser-window decimator out-
puts.
Y2(ζ)= −E1(ζ)+D(ζ)E2(ζ) (P12.4)
The overall output is obtained by the linear combination:
Y (ζ) = Y1(ζ)+D(ζ)Y2(ζ)= X (ζ)+D(ζ)E1(ζ)+D(ζ) −E1(ζ)+D(ζ)E2(ζ)
= X (ζ)+D(ζ)E1(ζ)−D(ζ)E1(ζ)+D2
(ζ)E2(ζ)= X (ζ)+D2
(ζ)E2(ζ)
Thus, the quantization error from the first stage is canceled, whereas the error of the second
stage gets filtered by the second-order highpass filter D2
(ζ)= (1 − ζ−1
)2
.
Problem 12.29
Such an arrangement is shown in Fig. P12.29. The outputs of the three stages are:
Y1(ζ) = X (ζ)+D(ζ)E1(ζ)
Y2(ζ) = −E1(ζ)+D(ζ)E2(ζ)
Y3(ζ) = −E2(ζ)+D(ζ)E3(ζ)
The overall output is obtained by the linear combination:
282
Y (ζ) = Y1(ζ)+D(ζ) Y2(ζ)+D(ζ)Y3(ζ)
= X (ζ)+D(ζ)E1(ζ)+D(ζ) −E1(ζ)+D(ζ)E2(ζ)+D(ζ) −E2(ζ)+D(ζ)E3(ζ)
= X (ζ)+D3
(ζ)E3(ζ)
x +
−
ζ-1
H(ζ)
y1
y2
e1
−e1
−e2
+
+
+
−
−
−
ζ-1
H(ζ)
e2
y
1st stage
2nd stage
y3+
−
ζ-1
H(ζ)
e3
3d stage
D(ζ)
D(ζ)
Fig. P12.29 MASH architecture of third-order delta-sigma quantizer.
Problem 12.30
Such an architecture is depicted in Fig. P12.30. The I/O relationship of the first stage is
Y1(ζ)= X (ζ)+D(ζ)E1(ζ)
The input to the second-order quantizer is −E1(ζ). Therefore, its I/O relationship is (from
Problem 12.8.3):
Y2(ζ)= −E1(ζ)+D2
(ζ)E2(ζ)
The overall output is obtained by the linear combination:
Y (ζ) = Y1(ζ)+D(ζ)Y2(ζ)= X (ζ)+D(ζ)E1(ζ)+D(ζ) −E1(ζ)+D2
(ζ)E2(ζ)
= X (ζ)+D3
(ζ)E2(ζ)
Problem 12.31
The delay of (N − 1)/2 = LM introduced by the decimation filter must be taken into account in
both multiplexing and control applications. In feedback control applications, such a delay may
render the closed-loop system unstable. See [294] for further discussion.
283](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-142-320.jpg)
![x +
−
ζ-1
H(ζ)
y1
y2
e1
−e1
+
+ − −
−
ζ-1
H(ζ) H(ζ)
e2
D(ζ)
y
1st stage
2nd stage
Fig. P12.30 MASH architecture of third-order delta-sigma quantizer.
284
Appendix B Problems
Problem B.1
The input sequence is
x(n)=
i
uiδ(n − iD − d)
Its autocorrelation function is
Rxx(k)= E x(n + k)x(n) =
i j
D−1
d=0
E[uiuj]δ(n + k − iD − d)δ(n − jD − d)p(d)
where the expectation values were taken with respect to both ui and d. Because the ui are mutually
independent and zero mean and d is uniform, we have E[uiuj]= σ2
uδ(i − j) and p(d)= 1/D. It
follows that:
Rxx(k)=
σ2
u
D i
D−1
d=0
δ(n + k − iD − d)δ(n − iD − d)
Changing variables to m = n−iD−d and noting that m runs over all the integers, we can replace
the double summation over i and d by the summation over m:
Rxx(k)=
σ2
u
D m
δ(k + m)δ(m)=
σ2
u
D
δ(k)
Thus, x(n) is stationary and has a delta-function autocorrelation. Sending this sequence into the
interpolation filter will generate a stationary output sequence y(n) with a noise reduction ratio:
NRR =
σ2
y
σ2
x
=
n
h(n)2
For the particular case of a hold interpolator, h(n) consists of D ones:
NRR =
σ2
y
σ2
x
=
D−1
n=0
1 = D ⇒ σ2
y = Dσ2
x = D
σ2
u
D
= σ2
u
For the linear interpolator, we have h(n)= 1 − |n|/D, for |n| ≤ D. This gives:
NRR =
σ2
y
σ2
x
=
D
n=0
1 −
|n|
D
2
= 1 + 2
D
n=1
1 −
n
D
2
= 1 + 2
1
D2
D−1
m=1
m2
Using the identity
D−1
m=1 m2
=
1
6
D(D − 1)(2D − 1), we obtain:
NRR =
σ2
y
σ2
x
= 1 + 2
D−1
m=1
m2
= 1 +
D(D − 1)(2D − 1)
3D2
= 1 +
(D − 1)(2D − 1)
3D
=
2D2
+ 1
3D
Thus,
σ2
y =
2D2
+ 1
3D
σ2
x =
2D2
+ 1
3D2
σ2
u
285](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-143-320.jpg)
![Problem B.2
The power spectral density of the output sequence will be related to that of the input by:
Syy(z)= H(z)H(z−1
)Sxx(z)= H(z)H(z−1
)σ2
x = H(z)H(z−1
)
σ2
u
D
For the hold interpolator, we have:
H(z)=
1 − z−D
1 − z−1
, H(z−1
)=
1 − zD
1 − z
=
1 − z−D
1 − z−1
zD−1
Thus, the output psd becomes:
Syy(z)= H(z)H(z−1
)
σ2
u
D
=
1
D
1 − z−D
1 − z−1
2
zD−1
σ2
u
Taking inverse z-transforms with the help of Problem 5.11, we find:
Ryy(k)= 1 −
|k|
D
σ2
u
Problem B.3
The generation of the held signal and the computation of its autocorrelation is illustrated by the
following C program segment:
double *u, *y, *R;
long iseed;
y = (double *) calloc(N, sizeof(double)); N-dimensional
R = (double *) calloc(M+1, sizeof(double)); (M+1)-dimensional
u[0] = ran(&iseed)-0.5; initialize u
q = D * ran(&iseed); initialize q
for (n=0; n<N; n++)
y[n] = ranh(D, u, &q, &iseed); zero mean
corr(N, y, y, M, R); sample autocorrelation
A typical sample autocorrelation is shown in Fig. P14.1
Problem B.4
As was mentioned in the end of Section B.3, the averaged periodogram spectrum can be calculated
by the following loop:
for (k=0; k<K; k++) { generate K = 200 blocks
u[0] = ran(&iseed); initialize kth block
q = D * ran(&iseed);
for (n=0; n<N; n++) generate kth block
X[n] = cmplx(ranh(D, u, &q, &iseed), 0.0);
fft(N, X); compute its FFT
for(n=0; n<N; n++) accumulate its periodogram
S[i] += cabs(X[i]) * cabs(X[i]);
}
A typical averaged spectrum is shown in Fig. P14.2, for the case D = 5.
286
Fig. P14.1 Sample autocorrelation of length-100 hold noise.
Fig. P14.2 Averaged periodogram spectrum of held noise.
Problem B.5
The random number generator C routine is:
/* rani.c - low-frequency random number generator using interpolation */
double ran(), dot();
void delay(), cdelay2();
double rani(L, P, h, w, q, iseed) usage: y=rani(L, P, h, w, & q, & iseed);
int L, P, *q; q and iseed passed by reference
double **h, *w; h = polyphase filter matrix
long *iseed; required long by ran()
{
int i;
double y;
if (*q==0) every L calls get new random number
w[0] = ran(iseed) - 0.5; zero mean
i = (L - *q) % L; current polyphase index
y = dot(P, h[i], w); compute interpolated value
287](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-144-320.jpg)
![cdelay2(L-1, q); decrement q for next call
if (*q==0) every L calls update delay line
delay(P, w);
return y;
}
For the hold and linear interpolators, it generates the same output as ranh.c and ranl.c.
The following program raniex.c is an example C program that invokes rani. Most of it is
concerned with designing and allocating the polyphase filters. The very last for-loop generates
the desired numbers. The for-loop before the last one, initializes the delay line w in accordance
with Eq. (12.2.19):
/* raniex.c - interpolated random number generator example */
run with:
k = 0, 1, 2 = hold, linear, Kaiser interpolators
L = 10, Ntot = 150, iseed = 1234567
A = 30 dB, ΔF = 0.3
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
double ran(), rani(), I0(); I0.c was given in Ch.10
void main(int argc, char ** argv)
{
double pi = 4 * atan(1.0);
double **h, *w, *hd;
double I0al, wind, A, D, wc, DF, al;
int L, P, K, N, LK;
int q, i, k, n, Ntot;
long iseed;
if (argc != 5) {
puts("nUsage: raniex k L Ntot iseed > y.datn"
" k = 0,1,2 for hold,linear,general interpolatorn"
" L = upsampling ration"
" Ntot = total random numbersn"
" iseed = initial seedn");
exit(0);
}
k = atoi(argv[1]);
L = atoi(argv[2]);
Ntot = atoi(argv[3]);
iseed = atol(argv[4]);
if (k >= 2) { Kaiser interpolator
fprintf(stderr, "enter A, DF (in units of fs): " );
scanf("%lf %lf", &A, &DF);
D = (A - 7.95) / 14.36;
288
if (A > 50)
al = 0.1102 * (A - 8.7);
else if (A > 21)
al = 0.5842 * pow(A-21, 0.4) + 0.07886 * (A-21);
else
al = 0;
N = 1 + ceil(D * L / DF);
K = ceil((N - 1) / (2.0*L));
N = 2 * L * K + 1; filter length
P = 2 * K - 1; filter order
LK = L * K; filter delay
hd = (double *) calloc(N, sizeof(double)); direct form filter
wc = pi / L; ideal cutoff frequency
I0al = I0(al);
for (n=0; n<N; n++) windowed filter
if (n==LK)
hd[n] = 1;
else {
wind = I0(al * sqrt(n*(2.0*LK-n)) / LK) / I0al;
hd[n] = wind * L * (sin(wc*(n-LK)))/(pi*(n-LK));
}
}
if (k == 0) { hold interpolator
N = L;
K = 0;
P = 0;
for (n=0; n<N; n++)
hd[n] = 1;
}
if (k == 1) { linear interpolator
P = 1;
K = 1;
N = 2*L+1;
for (n=0; n<N; n++)
hd[n] = 1 - abs(n - L) / (double) L;
}
h = (double **) calloc(L, sizeof(double *)); polyphase subfilters
for (i = 0; i < L; i++) allocate L subfilters
h[i] = (double *) calloc(P+1, sizeof(double));
for (i=0; i<L; i++)
for (n=0; n<=P; n++)
h[i][n] = hd[L*n+i]; ith subfilter
w = (double *) calloc(P+1, sizeof(double)); low-rate delay-line buffer
for (i=K; i>=1; i--)
w[i] = ran(&iseed) - 0.5; initialize w
289](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-145-320.jpg)
![q = 0;
for (n=0; n<Ntot; n++) generate interpolated noise
printf("%.12lfn", rani(L, P, h, w, &q, &iseed));
}
With the choices A = 30 dB, ΔF = 0.3, the Kaiser parameters become D = 1.536, α = 2.117.
The generated sequences are shown in Fig. P14.3.
Fig. P14.3 Kaiser, hold, and linearly interpolated random sequences.
Problem B.6
Noting that the autocorrelation of a hold-interpolated sequence of period D is R(k)= (1 −
|k|/D)σ2
u, −D ≤ k ≤ D, we have:
Ryy(k)=
1
B2
B−1
b=0
Rybyb
(k)=
1
B2
B−1
b=0
1 −
|k|
2b
u(2b
− |k|) σ2
u (P14.1)
where the unit-step u(2b
−|k|) restricts the duration of the b-th term to be |k| ≤ 2b
, or |k| ≤ 2b
−1
for non-zero values of Ryy(k). The resulting autocorrelation function is piece-wise linear, where
the linear segments have geometrically increasing slopes. This is shown in Fig. P14.4 for the case
B = 4. At lag k = 0, Eq. (P14.1) gives the expected result:
σ2
y = Ryy(0)=
1
B2
· Bσ2
u =
σ2
u
B
The increasing linear slopes in Ryy(k) cause it to have long-range correlations. Indeed, because
of the restriction |k| ≤ 2b
− 1, the maximum lag will be:
kmax = 2B−1
− 1 (P14.2)
Given a lag k in the range |k| ≤ kmax, the expression of Eq. (P14.1) can be summed in closed form
as follows. The restriction |k| ≤ 2b
−1 can be turned around into a restriction for the summation
index b ≥ log2(|k| + 1). Denoting the ceiling of this quantity by
b(k)= log2(|k| + 1)
290
k
kmax=2B-1
-1
R(k)/R(0)
1 3 5 7 94 6 8 102
1.00
0.75
0.50
0.25
0
B=4
Fig. P14.4 Theoretical autocorrelation of 1/f-noise model.
we find
Ryy(k)=
1
B2
B−1
b=b(k)
1 − |k|2−b
σ2
u
and using the finite geometric series formula (e.g., Eq. (6.3.13)), we get the expression of Eq. (B.25).
Problem B.7
Figure P14.5 shows the theoretical autocorrelation (P14.1) or (B.25), together with the sample
autocorrelation of a (zero-mean) block of 1/f-noise samples of length N = 2000 for B = 8.
The maximum correlation length is in this case kmax = 2B−1
− 1 = 27
− 1 = 127. The sample
autocorrelation was computed by the program segment:
double *y, *R;
y = (double *) calloc(N, sizeof(double)); N-dimensional
R = (double *) calloc(M+1, sizeof(double)); (M+1)-dimensional
for (n=0; n<N; n++) use N=2000
y[n] = ran1f(B, u, q, &iseed); zero-mean y’s
corr(N, y, y, M, R); use M=150
Fig. P14.5 Theoretical and sample autocorrelations for B = 8.
291](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-146-320.jpg)
![The reason for using N = 2000 was because we wanted to evaluate the sample autocorrelation
at M = 150 lags. Thus, the number of lags is 150/2000 ≡ 7.5% of the block length, which falls
within the recommended range for getting statistically reliable lags using the routine corr, as
was mentioned in Section A.1.
Problem B.8
We look at the cases ωmin = 0.001π and ωmax = 0.1π, 0.3π, 0.6π. The designed model param-
eters are:
For ωmax = 0.1π, we have B = 3 and
c = 10, a = [0.6858, 0.9686, 0.9969]
For ωmax = 0.3π, we have B = 4 and
c = 6.6943, a = [0.0575, 0.8592, 0.9790, 0.9969]
For ωmax = 0.6π, we have B = 6 and
c = 3.5944, a = [−0.8850, 0.4756, 0.8541, 0.9594, 0.9887, 0.9969]
The spectra are shown in Figs. P14.6—P14.8.
Fig. P14.6 Model spectrum and periodogram estimate. ωmax = 0.1π.
Problem B.9
Solving the equation (0.99574)c
= 0.94791, we find c = 12.5309. Then, the filter of Eq. (B.30)
becomes:
H(z)= G
(1 − 0.98444z−1
)(1 − 0.82159z−1
)(1 − 0.08522z−1
)
(1 − 0.99574z−1)(1 − 0.94791z−1)(1 − 0.51153z−1)
The NRR = hT
h of the filter (B.29) without the factor G can be computed analytically or much
faster using MATLAB to compute a fairly long portion of the impulse response h (e.g., of length
2000) and then setting G = 1/
√
NRR = 1/ hT
h. The magnitude response squared of Eqs. (B.29)
and (B.30) are shown in Fig. P14.9. The cascade realization is shown in Fig. P14.10. The corre-
sponding sample processing algorithm will be:
292
Fig. P14.7 Model spectrum and periodogram estimate. ωmax = 0.3π.
Fig. P14.8 Model spectrum and periodogram estimate. ωmax = 0.6π.
for each white noise input sample x do:
u0 = 0.57534 x + 0.99574 u1
x1 = u0 − 0.98444 u1
u1 = u0
v0 = x1 + 0.94791 v1
x2 = v0 − 0.83392 v1
v1 = v0
w0 = x2 + 0.53568 w1
y = w0 − 0.07568 w1
w1 = w0
293](https://image.slidesharecdn.com/51628783-solution-signal-processing-160422182740/85/Introduction-to-Signal-Processing-Orfanidis-Solution-Manual-147-320.jpg)
Introduction to Signal Processing Orfanidis [Solution Manual]
- 1. INTRODUCTION TO Signal Processing Solutions Manual Sophocles J. Orfanidis Department of Electrical & Computer Engineering Rutgers University, Piscataway, NJ 08855 orfanidi@ece.rutgers.edu Copyright © 2010 by Sophocles J. Orfanidis Web page: www.ece.rutgers.edu/~orfanidi/i2sp Chapter 1 Problems Problem 1.1 The Nyquist interval is [−fs/2, fs/2]= [−4, 4] Hz. The 6 Hz frequency of the wheel lies outside it, therefore, it will be aliased with f − fs = 6 − 8 = −2 Hz. Thus, the wheel will appear to be turning at 2 Hz in the opposite direction. If fs = 12, the Nyquist interval is [−6, 6]. Points on the wheel will appear to be moving up and down, with either positive or negative sense of rotation. For the other two sampling frequencies, the Nyquist interval is [−8, 8] or [−12, 12] Hz, and therefore, the original 6 Hz frequency lies in it and no aliasing will be perceived. Problem 1.2 The three terms of the signal correspond to frequencies f1 = 1, f2 = 4, and f3 = 6 Hz. Of these, f2 and f3 lie outside the Nyquist interval [−2.5, 2.5]. Therefore, they will be aliased with f2 − fs = 4 − 5 = −1 and f3 − fs = 6 − 5 = 1 and the aliased signal will be: xa(t)= 10 sin(2πt)+10 sin(2π(−1)t)+5 sin(2πt)= 5 sin(2πt) To show that they have the same sample values, we set t = nT, with T = 1/fs = 1/5 sec. Then, x(nT)= 10 sin(2πn/5)+10 sin(8πn/5)+5 sin(12πn/5) But, sin(8πn/5)= sin(2πn − 2πn/5)= − sin(2πn/5) and sin(12πn/5)= sin(2πn + 2πn/5)= sin(2πn/5) Thus, x(nT) = 10 sin(2πn/5)−10 sin(2πn/5)+5 sin(2πn/5) = 5 sin(2πn/5)= xa(nT). If fs = 10 Hz, then the Nyquist interval is [−5, 5] Hz and only f3 lies outside it. It will be aliased with f3 − fs = 6 − 10 = −4 resulting in the aliased signal: xa(t)= 10 sin(2πt)+10 sin(8πt)+5 sin(2π(−4)t)= 10 sin(2πt)+5 sin(8πt) Problem 1.3 Using the trig identity 2 sin α sin β = cos(α − β)− cos(α + β), we find: x(t) = cos(5πt)+4 sin(2πt)sin(3πt)= cos(5πt)+2[cos(πt)− cos(5πt)] = 2 cos(πt)− cos(5πt) The frequencies in the signal are f1 = 0.5 and f2 = 2.5 kHz. The Nyquist interval is [−1.5, 1.5] kHz, and f2 lies outside it. Thus, it will be aliased with f2a = 2.5 − 3 = −0.5 giving rise to the signal: 1
- 2. xa(t)= 2 cos(2πf1t)− cos(2πf2at)= 2 cos(πt)− cos(−πt)= cos(πt) A class of signals aliased with x(t) and xa(t) is obtained by replacing f1 and f2 by their shifted versions: f1 + mfs, f2 + nfs resulting in: xmn(t)= 2 cos(πt + 6πmt)− cos(πt − 6πnt) Problem 1.4 Using a trig identity, we write x(t)= cos(8πt)+ cos(10πt)+ cos(2πt). The frequencies con- tained in x(t) are thus, 1 Hz, 4 Hz, and 5 Hz. If the sampling rate is 5 Hz, then the Nyquist interval is [−2.5, 2.5] Hz, and therefore, the 4 Hz and 5 Hz components lie outside it and will be aliased with the frequencies 4 − 5 = −1 Hz and 5 − 5 = 0 Hz, etc. Problem 1.5 Using trig identities we find: x(t)= sin(6πt) 1 + 2 cos(4πt) = sin(6πt)+ sin(10πt)+ sin(2πt) with frequency content: f1 = 3, f2 = 5, f3 = 1 kHz. The Nyquist interval is [−2, 2] kHz, and the aliased frequencies are: f1a = f1 − fs = 3 − 4 = −1, f2a = f2 − fs = 5 − 4 = 1, f3a = f3 = 1 Thus, xa(t)= sin(−2πt)+ sin(2πt)+ sin(2πt)= sin(2πt) Problem 1.6 Use the trigonometric identity 2 cos a cos b = cos(a + b)+ cos(a − b) three times to get x(t) = 2[cos(10πt)+ cos(6πt)]cos(12πt) = cos(22πt)+ cos(2πt)+ cos(18πt)+ cos(6πt) The frequencies present in this signal are f1 = 11, f2 = 1, f3 = 9, and f4 = 3 Hz. With a sampling rate of 10 Hz, only f1 and f3 lie outside the Nyquist interval [−5, 5] Hz, and they will be aliased with f1 − fs = 11 − 10 = 1 Hz and f3 − fs = 9 − 10 = −1 Hz. The aliased signal will be: xa(t) = cos(2π(1)t)+ cos(2πt)+ cos(2π(−1)t)+ cos(6πt) = 3 cos(2πt)+ cos(6πt) To prove the equality of the samples, replace t = nT = n/10, because T = 1/fs = 1/10. Then, x(nT)= cos(22πn/10)+ cos(2πn/10)+ cos(18πn/10)+ cos(6πn/10) But, cos(22πn/10)= cos(2πn+2πn/10)= cos(2πn/10) and similarly, cos(18πn/10)= cos(2πn− 2πn/10)= cos(2πn/10). Therefore, the sample values become 2 x(nT) = cos(2πn/10)+ cos(2πn/10)+ cos(2πn/10)+ cos(6πn/10) = 3 cos(2πn/10)+ cos(6πn/10)= xa(nT) If fs = 12 Hz, then f1 and f3 will lie outside of [−6, 6] and will be aliased with 11 − 12 = −1 and 9 − 12 = −3. The aliased signal will be: xa(t) = cos(2π(−1)t)+ cos(2πt)+ cos(2π(−3)t)+ cos(6πt) = 2 cos(2πt)+2 cos(6πt) Problem 1.7 We use the same technique as in the square-wave Example 1.4.6. At a sampling rate of 8 Hz, the signal frequencies of {1, 3, 5, 7, 9, 11, 13, 15, . . . } will be aliased with: {1, 3, −3, −1, 1, 3, −3, −1, . . . } Therefore only sin(2πt) and sin(6πt) terms will appear in the aliased signal. Thus, we write it in the form: xa(t)= B sin(2πt)+C sin(6πt) To determine B and C, we demand that xa(t) and x(t) agree at the sampling instants t = nT = n/8, because T = 1/fs = 1/8 sec. Therefore, we demand B sin(2πn/8)+C sin(6πn/8)= x(n/8) Setting n = 1, 2, we get two equations B sin(2π/8)+C sin(6π/8)= x(1/8)= 0.5 B sin(4π/8)+C sin(12π/8)= x(2/8)= 1 ⇒ B 1 √ 2 + C 1 √ 2 = 0.5 B − C = 1 The values for x(1/8) and x(2/8) were obtained by inspecting the triangular waveform. Solving for B and C, we find: B = √ 2 + 2 4 , C = √ 2 − 2 4 Problem 1.8 For fs = 5 kHz, we have: xa(t)= sin(2πf1t) For fs = 10 kHz, we have: xa(t)= 2 sin(2πf1t)+ sin(2πf2t) 3
- 3. Fig. P1.1 Parts (a,b) of Problem 1.8. Problem 1.9 The audible and inaudible parts are: x(t)= sin(10πt)+ sin(20πt) audible + sin(60πt)+ sin(90πt) inaudible The frequencies of the four terms and their aliased versions are: fA = 5 fB = 10 fC = 30 fD = 45 ⇒ fAa = 5 fBa = 10 fCa = 30 − 40 = −10 fDa = 45 − 40 = 5 a. When there is no prefilter the aliased signal at the output of the reconstructor will be: ya(t)= sin(10πt)+ sin(20πt)+ sin(−20πt)+ sin(10πt)= 2 sin(10πt) b. When there is a perfect prefilter, the fC, fD components are removed prior to sampling thus, the sampled and reconstructed signal will remain unchanged, that is, the audible part of x(t): ya(t)= sin(10πt)+ sin(20πt) c. The attenuations introduced by the practical prefilter shown in Fig. P1.2 are obtained by determining the number of octaves to the frequencies fC, fD and multiplying by the filter’s attenuation of 48 dB/octave: log2 fC fs/2 = log2 30 20 = 0.585 ⇒ AC = 48 × 0.585 = 28.08 dB log2 fD fs/2 = log2 45 20 = 1.170 ⇒ AD = 48 × 1.170 = 56.16 dB and in absolute units: 4 |HC| = 10−AC/20 = 10−28.08/20 = 0.0394 |HD| = 10−AD/20 = 10−56.16/20 = 0.0016 Thus, the resulting reconstructed signal will be: ya(t) = sin(10πt)+ sin(20πt)+|HC| sin(−20πt)+|HD| sin(10πt) = 1.0016 sin(10πt)+0.9606 sin(20πt) which agrees closely with the audible part of x(t). 10 3020 45 48 dB/octave Analog Prefilter kHz f H(f) 0 A B C D Fig. P1.2 Prefilter specifications of Problem 1.9. Problem 1.10 The Fourier series expansion of the periodic function s(t) is: s(t)= ∞ m=−∞ cme2πjmt/T , cm = 1 T T/2 −T/2 s(t)e−2πjmt/T dt But within the basic period [−T/2, T/2], s(t) equals δ(t). Thus, cm = 1 T T/2 −T/2 δ(t)e−2πjmt/T dt = 1 T Using the delta function integral representation: ∞ −∞ e−2πjft dt = δ(f) we find: S(f) = ∞ −∞ s(t)e−2πjft dt = 1 T ∞ m=−∞ ∞ −∞ s(t)e−2πj(f−mfs)t dt = 1 T ∞ m=−∞ δ(f − mfs) Problem 1.11 This is simply the Fourier series expansion of the periodic function X(f) in the variable f, and x(nT) are the Fourier series expansion coefficients calculated by integrating over one period, that is, over [−fs/2, fs/2]. 5
- 4. Problem 1.12 Write x(t)= 0.5e2πjf0t + 0.5e−2πjf0t , with Fourier transform: X(f)= 0.5δ(f − f0)+0.5δ(f + f0) Its replication gives the spectrum of the sampled signal: X(f) = 1 T ∞ m=−∞ X(f − mfs)= 1 2T ∞ m=−∞ δ(f − mfs − f0)+δ(f − mfs + f0) = 1 2T ∞ m=−∞ δ(f − mfs − f0)+δ(f + mfs + f0) where in the second term, we changed the summation index from m to −m. Problem 1.13 We have for the complex sinusoid, the Laplace, Fourier, and magnitude spectra: x(t)= e−at e2πjf0t = e−(a−2πjf0)t , t ≥ 0 X(s)= 1 s + a − 2πjf0 X(f)= 1 a + 2πj(f − f0) ⇒ |X(f)|2 = 1 a2 + 4π2(f − f0)2 The length-L sampled signal x(nT)= e−(a−2πjf0)nT , n = 0, 1, . . . , L − 1 has spectrum: XL(f) = L−1 n=0 x(nT)e−2πjfnT = L−1 n=0 e−(a+2πj(f−f0))nT = 1 − e−(a+2πj(f−f0))LT 1 − e−(a+2πj(f−f0))T with magnitude spectrum: |XL(f)|2 = 1 − 2e−aLT cos 2π(f − f0)LT + e−2aLT 1 − 2e−aT cos 2π(f − f0)T + e−2aT In the limit L → ∞, we have e−aLT → 0 and obtain the sampled spectrum of the infinitely long signal: X(f)= ∞ n=0 x(nT)e−2πjfnT = 1 1 − e−(a+2πj(f−f0))T with magnitude spectrum: |X(f)|2 = 1 1 − 2e−aT cos 2π(f − f0)T + e−2aT The continuous time limit T → 0 of X(f) gives rise to the analog spectrum: 6 lim T→0 TX(f)= lim T→0 T 1 − e−(a+2πj(f−f0))T = 1 a + 2πj(f − f0) = X(f) Fig. P1.3 shows the required spectra. From the left graph, we see that as fs increases, the sampled spectrum agrees more closely with the analog spectrum within the Nyquist interval. From the right graph, we see that the spectrum of the truncated sampled signal has ripples that follow the general shape of X(f). As L → ∞, the sampled spectrum XL(f) will tend to X(f), which is only an approximation of the true spectrum even within the Nyquist interval. Fig. P1.3 Spectra of analog, sampled, and windowed signals of Problem 1.13. In the case of real-valued signals, we will also get the spectral peaks at the negative frequency −f0. If f0 is small, the peaks at ±f0 will influence each other somewhat, but apart from that the same conclusions can be drawn about the sampled signals. Problem 1.14 This is the same as Example 1.5.4. In units of dB per decade, the attenuations due to the signal and filter combine to give: A(f)= α log10 f fmax + β log10 f fmax = γ log10 f fmax where γ = α + β. The requirement A(fs − fmax)≥ A gives: γ log10 fs − fmax fmax ≥ A ⇒ fs ≥ fmax + fmax10A/γ Problem 1.15 The first replica of Xin(f), shifted to fs, will have spectrum: |Xin(f − fs)| = 1 1 + 0.1(f − fs) 8 Its value at the passband frequency fpass = fmax = 20 kHz will be: 7
- 5. |Xin(fmax − fs)| = 1 1 + 0.1(fmax − fs) 8 This must be suppressed by A = 60 dB relative to |Xin(fmax)|, thus, we require: −10 log10 Xin(fmax − fs) Xin(fmax) 2 = A ⇒ 10 log10 1 + 0.1(fmax − fs) 8 1 + (0.1fmax)8 = 60 which can be solved for fs to give fs = 132.52 kHz. Using the prefilter, we require that its passband attenuation be Apass = 1 dB at fpass = fmax = 20 kHz, that is, −10 log10 |H(fmax)|2 = 10 log10 1 + (fmax/f0)6 = Apass which can be solved for f0 to give f0 = 25.05 kHz. The prefiltered signal that gets sampled is now |X(f)| = |H(f)Xin(f)| = 1 1 + (f/f0)6 1 1 + (0.1f)8 Its first replica |X(f − fs)| is required to be suppressed by A = 60 dB at fmax −10 log10 X(fmax − fs) X(fmax) 2 = A, ffs fs-fmaxfmax A |X(fmax)| 0 |X(fs-fmax)| |X(fmax-fs)|= which gives the condition 10 log10 1 + (fmax − fs)/f0 6 1 + (fmax/f0)6 · 1 + 0.1(fmax − fs) 8 1 + (0.1fmax)8 = A (P1.1) It can be solved approximately by ignoring the +1 in the numerators, that is, 10 log10 (fmax − fs)/f0 6 1 + (fmax/f0)6 · 0.1(fmax − fs) 8 1 + (0.1fmax)8 = 60 dB (P1.2) which gives fs = 80.1 kHz. The exact solution of Eq. (P1.1) is fs = 80.0633 kHz. The first factor in Eq. (P1.2) represents the stopband attenuation due to the filter, and the second the attenuation due to the signal. With fs = 80.1 kHz, we have 10 log10 1 + (fmax − fs)/f0 6 1 + (fmax/f0)6 = Astop = 21.8 dB 10 log10 1 + 0.1(fmax − fs) 8 1 + (0.1fmax)8 = A − Astop = 38.2 dB Thus, the use of the prefilter enables us to reduce the sampling rate from 132 kHz to 80 kHz. Note also, that for large f, the spectrum |X(f)| follows the power law |X(f)|2 = 1 1 + (f/f0)6 · 1 1 + (0.1f)8 ≈ const. f6 · const. f8 = const. f14 giving rise to an effective attenuation 8 −10 log10 |X(f)|2 = γ log10 f + const. where γ = 140 dB/decade, of which 60 dB/decade is due to the filter and 80 dB/decade due to the signal. The approximate method of Example 1.5.4 or Problem 1.14 would have given here fs = fmax + 10A/γ fmax = 20 + 20 · 1060/140 = 73.65 kHz, which is fairly close to the more accurate value of 80 kHz. Problem 1.16 The passband condition at f = fmax = 20 kHz is: 10 log10 1 + (fmax/f0)2N = Apass (P1.3) which can be solved for f0 in terms of N as follows: f2N 0 = f2N max r − 1 , where r = 10Apass/10 The antialiasing condition, replacing Eq. (P1.1), is now 10 log10 1 + (fmax − fs)/f0 2N 1 + (fmax/f0)2N · 1 + 0.1(fmax − fs) 8 1 + (0.1fmax)8 = A or, 1 + (fmax − fs)/f0 2N 1 + (fmax/f0)2N · 1 + 0.1(fmax − fs) 8 1 + (0.1fmax)8 = 10A/10 Replacing f2N 0 by its expression above, we obtain 1 + (fmax − fs)/fmax 2N (r − 1) r · 1 + 0.1(fmax − fs) 8 1 + (0.1fmax)8 = 10A/10 and solving for N, we have N = ln 10A/10 rR(fs)−1 r − 1 2 ln fs − fmax fmax (P1.4) where R(fs) is defined by R(fs)= 1 + (0.1fmax)8 1 + 0.1(fmax − fs) 8 Eq. (P1.4) gives the minimum prefilter order for a desired suppression level A and rate fs. With fs = 70 kHz, A = 60 dB, Apass = 1 dB, we find r = 10Apass/10 = 100.1 = 1.259, 10A/10 R(fs)= 657.918 and N = ln(3194.281) 2 ln(50/20) = 4.403 ⇒ N = 5 We calculate also f0 = fmax/(r − 1)1/2N = 22.9 kHz. 9
- 6. Fig. P1.4 Butterworth order versus sampling rate. Problem 1.17 Figure P1.4 shows a plot of N given by Eq. (P1.4) as a function of fs for different choices of A. Notice the two data points on the A = 60 curve at fs = 70 and fs = 80 kHz that were computed in this and the last example. It is evident from this graph that the complexity of the prefilter increases with A, at any fixed fs; and that it decreases with fs, for fixed A. Problem 1.18 Using the results of Problem 1.14, we solve for the combined attenuation γ of signal and prefilter: A = γ log10 fs − fmax fmax ⇒ where A = 60 dB, fs = 30 kHz and fmax = 10 kHz. We find: γ = A log10 f fmax = 60 log10 30 − 10 10 = 200 Because the signal has α = 80 dB/decade, we find the filter’s attenuation: β = γ − α = 200 − 80 = 120 dB/decade which translates into a filter order: β = 20N ⇒ N = β/20 = 120/20 = 6 If we increase the sampling rate to 50 kHz, the combined attenuation will be γ = 100, which gives β = 100 − 80 = 20 dB/decade and order N = β/20 = 1. Problem 1.19 This problem is the generalization of Prob. 1.15. If we do not use a prefilter, the requirement that the aliased components (of the 1st replica) be suppressed by A dB relative to the signal components can be expressed in the form: 10 |Xin(fs − fmax)|2 |Xin(fmax)|2 = 10−A/10 Inserting the given expression for |Xin(f)|, we get the desired equation for fs in terms of A and fmax: 1 + (fmax/fa)2Na 1 + (fs − fmax)/fa 2Na = 10−A/10 If we use a Butterworth prefilter of order N: |H(f)|2 = 1 1 + (f/f0)2N then, the 3-dB frequency f0 is determined by requiring that at the edge of the passband the attenuation be B dB: |H(fmax)|2 = 1 1 + (fmax/f0)2N = 10−B/10 The filtered input that gets sampled will now have spectrum X(f)= H(f)Xin(f), which gives: |X(f)|2 = |H(f)|2 |Xin(f)|2 = 1 1 + (f/f0)2N · 1 1 + (f/fa)2Na The antialiasing condition |X(fs − fmax)|2 = 10−A/10 |X(f)|2 gives now: 1 1 + (fs − fmax)/f0 2N · 1 1 + (fs − fmax)/fa 2Na = 10−A/10 · 1 1 + (fmax/f0)2N · 1 1 + (fmax/fa)2Na Using the asymptotic forms of the spectra, valid for large f’s, that is, |Xin(f)|2 1 (f/fa)2Na , |H(f)|2 1 (f/f0)2N the antialiasing condition would read: (fs − fmax)2N f2N 0 · (fs − fmax)2Na f 2Na a = 10−A/10 f2N max f2N 0 · f 2Na max f 2Na a which simplifies into: fs − fmax fmax 2(N+Na) = 10A/10 or, fs = fmax + fmax10A/20(N+Na) It agrees with the results of Problem 1.14, with γ = α + β = 20Na + 20N. For any fixed desired value of A, the limit N → ∞ gives fs = fmax + fmax10A/20(N+Na) → fmax + fmax = 2fmax In this case, the Butterworth filter resembles more and more a brick-wall filter with cutoff fmax. 11
- 7. Problem 1.20 The linear relationship between N and A follows from the approximation of Problem 1.19: fs − fmax fmax 2(N+Na) = 10A/10 or, 2(N + Na)log10 fs − fmax fmax = A 10 Problem 1.21 The presence of P(f) is called the aperture effect. The replicas of X(f) embedded in X(f) will still be present, but they will be weighted by P(f). The need for using antialiasing prefilters to avoid overlapping of replicas remains the same. For a pulse of width τ, we have: |P(f)| = τ sin(πfτ) πfτ Therefore, the narrower the width τ the flatter the spectrum P(f). If τ T, then P(f) will be essentially flat within the Nyquist interval, thus, leaving the central replica almost unchanged. Fig. P1.5 shows P(f) relative to the replicas of X(f) for the case τ = T/5. |P(f)| fs 5fs-5fs -fs 0 f Fig. P1.5 Butterworth order versus sampling rate. Problem 1.22 The stopband frequency of the postfilter is: fstop = fs − fc = 6 − 1 = 5 kHz. At this frequency, the attenuation of the staircase DAC is ADAC = −20 log10 sin(πfstop/fs) πfstop/fs = 14.4 dB Thus, the stopband attenuation of the postfilter at f = fstop must be: Astop = A − ADAC = 40 − 14.4 = 25.6 dB Problem 1.23 Using convolution in the time domain, we get: xa(t) = ∞ −∞ h(t − t )ˆx(t ) dt = n x(nT) ∞ −∞ h(t − t )δ(t − nT) dt = n x(nT)h(t − nT) 12 Working in the frequency domain we have: Xa(f)= H(f)X(f)= n x(nT) e−2πjfnT H(f) Using the delay theorem of Fourier transforms, we recognize the term in the bracket as the FT of h(t − nT). Thus, going to the time domain, we get: xa(t)= n x(nT)h(t − nT) Problem 1.24 Start with a complex sinusoid x(t)= e2πjf0t . Its sampled spectrum will be: X(f)= 1 T ∞ m=−∞ X(f − mfs)= 1 T ∞ m=−∞ δ(f − f0 − mfs) The spectrum of the reconstructed analog signal will be then: Xrec(f)= H(f)X(f)= 1 T ∞ m=−∞ H(f)δ(f − f0 − mfs)= 1 T ∞ m=−∞ H(f0 + mfs)δ(f − f0 − mfs) Taking inverse Fourier transforms, we find: xrec(t)= 1 T ∞ m=−∞ H(f0 + mfs)e2πj(f0+mfs)t Thus, fm = f0 + mfs. Writing 1 T H(fm)= Amejθm , so that Am = 1 T |H(fm)|, θm = arg H(fm) we have: xrec(t)= ∞ m=−∞ Ame2πj(f0+mfs)t+jθm Taking imaginary parts, we find for the real sinusoidal case: xrec(t)= ∞ m=−∞ Am sin 2π(f0 + mfs)t + θm Assuming |f0| < fs/2, for the staircase reconstructor we have: 1 T H(fm)= sin(πfmT) πfmT e−jπfmT and because πfmT = πf0T + mπ, we have: 1 T H(fm)= sin(πf0T) πfmT e−jπf0T Thus, Am = sin(πf0T) πfmT , θm = −πf0T For the ideal reconstructor, we have θm = 0 and Am = 0 except at m = 0 where A0 = 1, that is, Am = δm. Thus, xrec(t)= ∞ m=−∞ δm sin 2π(f0 + mfs)t = sin(2πf0t) 13
- 8. Problem 1.25 Take the linear combination of the results of the single sinusoid case of Eq. (1.6.12). 14 Chapter 2 Problems Problem 2.1 The quantized values are: analog value quantized value DAC output 2.9 2 0 0 1 3.1 4 0 1 0 3.7 4 0 1 0 4 4 0 1 0 -2.9 -2 1 1 1 -3.1 -4 1 1 0 -3.7 -4 1 1 0 -4 -4 1 1 0 For the offset binary case, complement the first bit in this table. Problem 2.2 Filling the table, we have n x(n) xQ(n) offset binary 2’s complement 0 0.000 0.000 1 0 0 0 0 0 0 0 1 1.243 1.000 1 0 0 1 0 0 0 1 2 2.409 2.000 1 0 1 0 0 0 1 0 3 3.423 3.000 1 0 1 1 0 0 1 1 4 4.222 4.000 1 1 0 0 0 1 0 0 5 4.755 5.000 1 1 0 1 0 1 0 1 6 4.990 5.000 1 1 0 1 0 1 0 1 7 4.911 5.000 1 1 0 1 0 1 0 1 8 4.524 5.000 1 1 0 1 0 1 0 1 9 3.853 4.000 1 1 0 0 0 1 0 0 10 2.939 3.000 1 0 1 1 0 0 1 1 11 1.841 2.000 1 0 1 0 0 0 1 0 12 0.627 1.000 1 0 0 1 0 0 0 1 13 -0.627 -1.000 0 1 1 1 1 1 1 1 14 -1.841 -2.000 0 1 1 0 1 1 1 0 15 -2.939 -3.000 0 1 0 1 1 1 0 1 16 -3.853 -4.000 0 1 0 0 1 1 0 0 17 -4.524 -5.000 0 0 1 1 1 0 1 1 18 -4.911 -5.000 0 0 1 1 1 0 1 1 19 -4.990 -5.000 0 0 1 1 1 0 1 1 Problem 2.3 With R = 10 volts and erms ≤ 10−3 volts, we have using Q = √ 12erms: 2B = R Q = R √ 12erms ≥ 10 √ 12 · 10−3 which gives 15
- 9. B ≥ log2 10 √ 12 · 10−3 = 11.495 Therefore, B = 12. The actual rms error will be erms = R 2B √ 12 = 10 212 √ 12 = 0.705 mvolt. The dynamic range will be 6B = 6 · 12 = 72 dB. Problem 2.4 At 44.1 kHz sampling rate, one minute of music has 60 sec containing 60 × 44100 samples. And, for two stereo channels, 2 × 60 × 44100 samples. If each sample is quantized with 16 bits, the total number of bits will be 2 × 60 × 44100 × 16 bits or, dividing by 8 we get the total number of bytes: 2 × 60 × 44100 × 16 8 bytes Dividing by 1024, we get kbytes and by another 1024, Mbytes: 2 × 60 × 44100 × 16 8 × 1024 × 1024 = 10.09 Mbytes Problem 2.5 Following similar steps as in the previous problem, we have for part (a): (3 × 60 sec)(16 channels)(48000 samples/sec)(20 bits/sample) (8 bits/byte)(10242 bytes/Mbyte) = 330 Mbytes that is, practically the entire hard disk of a typical home PC. For part (b), if in each channel we must perform NMAC = 35 operations on each sample, then these operations must be finished during the sampling interval T = 1/fs between samples: NMACTMAC ≤ T ⇒ TMAC ≤ T NMAC = 1 NMACfs which gives for the time per MAC: TMAC = 1 35 · 48 kHz = 0.595 · 10−3 msec = 595 nsec This is plenty of time for today’s DSP chips, which can perform a MAC operation within 30–80 nsec. If a single DSP chip handles all 16 channels, then the total number of MACs that must be performed within the sampling interval T are 16NMAC. This gives: TMAC = 1 16 · 35 · 48 kHz = 595 16 = 37 nsec which is just within the capability of the fastest DSP chips. In practice, a typical professional digital recording system would require the processing and mixing of at least 32 channels. The above results suggest that one use at least two DSPs to multiplex the required operations. 16 Problem 2.6 The mean is me = 0 −Q ep(e) de = 1 Q 0 −Q e de = − Q 2 and the variance: σ2 e = E[(e − me)2 ]= 0 −Q (e − me)2 p(e) de = 1 Q 0 −Q e + Q 2 2 de = Q2 12 Problem 2.7 Solving Eq. (2.2.10) for L in terms of ΔB, we find: L = 22ΔB π2p 2p + 1 1 2p+1 (P2.1) With ΔB = 15 and p = 1, 2, 3, we find: L = 1523, 116, 40 The oversampled audio rates will be, with fs = 44.1 kHz: Lfs = 67.164, 5.114, 1.764 MHz Problem 2.8 Using Eq. (P2.1) with ΔB = 7 bits and p = 1, 2, 3, we find: L = 38, 13, 8 The oversampled speech rates will be, with fs = 8 kHz: Lfs = 304, 104, 48 kHz Problem 2.9 Replacing b1 = 1 − b1, we have: xQ = R(b12−1 + b22−2 + · · · + bB2−B − 0.5) = R (1 − b1)2−1 + b22−2 + · · · + bB2−B − 0.5 = R −b12−1 + b22−2 + · · · + bB2−B where the first 2−1 canceled the last −0.5. Problem 2.10 As an example, consider the polynomial of degree M = 3: B(z)= b0 +b1z+b2z2 +b3z3 . Iterating the algorithm for i = 3, 2, 1, 0, we find: p = 0 p = ap + b3 = b3 p = ap + b2 = ab3 + b2 p = ap + b1 = a(ab3 + b2)+b1 = a2 b3 + ab2 + b1 p = ap + b0 = a(a2 b3 + ab2 + b1)+b0 = a3 b3 + a2 b2 + ab1 + b0 = B(a) 17
- 10. Problem 2.11 /* pol.c - polynomial evaluator */ double pol(M, b, a) int M; double *b, a; int i; double p = 0; for (i=M; i>=0; i--) p = a * p + b[i]; return p; Problem 2.12 Consider the division of B(z) by (z − a): B(z)= (z − a)Q(z)+R where the quotient polynomial Q(z) must have degree M − 1, and the remainder R must be a constant. Indeed, setting z = a, gives R = b(a). So the objective of the algorithm is to compute R iteratively. Inserting the polynomial expressions we have: bMzM + bM−1zM−1 + · · · + b1z + b0 = (z − a)(qM−1zM−1 + qM−2zM−2 + · · · + q1z + q0)+R Equating like powers of z, we find: bM = qM−1 bM−1 = qM−2 − aqM−1 bM−2 = qM−3 − aqM−2 · · · b1 = q0 − aq1 b0 = R − aq0 ⇒ qM−1 = bM qM−2 = aqM−1 + bM−1 qM−3 = aqM−2 + bM−2 · · · q0 = aq1 + b1 R = aq0 + b0 Problem 2.13 For a polynomial of degree M = 3, B(z)= b1z + b2z2 + b3z3 , we iterate the algorithm for i = 3, 2, 1: p = 0 p = a(p + b3)= ab3 p = a(p + b2)= a(ab3 + b2)= a2 b3 + ab2 p = a(p + b1)= a(a2 b3 + ab2 + b1)= a3 b3 + a2 b2 + ab1 = B(a) 18 natural binary offset binary 2’s C b1b2b3b4 m xQ = Qm m xQ = Qm b1b2b3b4 — 16 8.0 8 4.0 — 1 1 1 1 15 7.5 7 3.5 0 1 1 1 1 1 1 0 14 7.0 6 3.0 0 1 1 0 1 1 0 1 13 6.5 5 2.5 0 1 0 1 1 1 0 0 12 6.0 4 2.0 0 1 0 0 1 0 1 1 11 5.5 3 1.5 0 0 1 1 1 0 1 0 10 5.0 2 1.0 0 0 1 0 1 0 0 1 9 4.5 1 0.5 0 0 0 1 1 0 0 0 8 4.0 0 0.0 0 0 0 0 0 1 1 1 7 3.5 −1 −0.5 1 1 1 1 0 1 1 0 6 3.0 −2 −1.0 1 1 1 0 0 1 0 1 5 2.5 −3 −1.5 1 1 0 1 0 1 0 0 4 2.0 −4 −2.0 1 1 0 0 0 0 1 1 3 1.5 −5 −2.5 1 0 1 1 0 0 1 0 2 1.0 −6 −3.0 1 0 1 0 0 0 0 1 1 0.5 −7 −3.5 1 0 0 1 0 0 0 0 0 0.0 −8 −4.0 1 0 0 0 Table P2.1 Converter codes for B = 4 bits, R = 8 volts. Problem 2.14 The quantization width is Q = R/2B = 8/24 = 0.5 volts. Table P2.1 shows the quantization levels and their binary codes. To convert x = 1.2 by rounding, we shift it by half the quantization spacing y = x + Q/2 = 1.2 + 0.25 = 1.45. The following tables show the successive approximation tests for the natural and offset binary cases. The 2’s complement case is obtained by complementing the MSB of the offset case. test b1b2b3b4 xQ C = u(y − xQ) b1 1 0 0 0 4.0 0 b2 0 1 0 0 2.0 0 b3 0 0 1 0 1.0 1 b4 0 0 1 1 1.5 0 0 0 1 0 1.0 test b1b2b3b4 xQ C = u(y − xQ) b1 1 0 0 0 0.0 1 b2 1 1 0 0 2.0 0 b3 1 0 1 0 1.0 1 b4 1 0 1 1 1.5 0 1 0 1 0 1.0 The natural binary and 2’C cases agree because x lies in the positive range of both quantizers. For x = 5.2, we have y = x + Q/2 = 5.45. The tests are shown below: test b1b2b3b4 xQ C = u(y − xQ) b1 1 0 0 0 4.0 1 b2 1 1 0 0 6.0 0 b3 1 0 1 0 5.0 1 b4 1 0 1 1 5.5 0 1 0 1 0 5.0 test b1b2b3b4 xQ C = u(y − xQ) b1 1 0 0 0 0.0 1 b2 1 1 0 0 2.0 1 b3 1 1 1 0 3.0 1 b4 1 1 1 1 3.5 1 1 1 1 1 3.5 For the 2’C case, the quantizer saturates to its maximum positive value. Finally, for x = −3.2, we have y = −2.95, and the following testing tables: 19
- 11. test b1b2b3b4 xQ C = u(y − xQ) b1 1 0 0 0 4.0 0 b2 0 1 0 0 2.0 0 b3 0 0 1 0 1.0 0 b4 0 0 0 1 0.5 0 0 0 0 0 0.0 test b1b2b3b4 xQ C = u(y − xQ) b1 1 0 0 0 0.0 0 b2 0 1 0 0 −2.0 0 b3 0 0 1 0 −3.0 1 b4 0 0 1 1 −2.5 0 0 0 1 0 −3.0 In this case, the natural binary scale saturates to its minimum positive value. Problem 2.15 Similar to the testing tables of Examples 2.4.1 and 2.4.2. Problem 2.16 These versions are straight forward. To implement truncation instead of rounding, replace the statement within adc: y = x + Q/2; by y = x; Problem 2.17 This problem is basically the same as Example 2.4.5. The same program segment of that example applies here for generating the codes and quantized values. Only the full-scale range is different, R = 32 volts. In part (c), the amplitude of the sinusoid A = 20 exceeds the quantizer range R/2 = 16, and therefore both the positive and negative peaks of the sinusoid will saturate to the positive and negative maxima of the quantizer. Problem 2.18 The following C program carries out all the required operations. It uses the routine corr of Appendix A.1 to compute all the sample autocorrelations and cross-correlations. /* exmpl246.c - quantization noise model * * default values: * N=10 bits * R=1024 * L=1000, iseed=10, M=50 */ #include <stdio.h> #include <math.h> #include <stdlib.h> void adc(), corr(); double dac(); 20 double ran(); void main() { int n, i, L, N, M, *b; long iseed; double *x, xq, *e, R, Q, *Ree, *Rex, *Rxx; double normee, normex; double v, pi = 4 * atan(1.0); FILE *fpe, *fpx; FILE *fpRee, *fpRex, *fpRxx; printf("enter N, R = "); scanf("%d %lf", &N, &R); printf("enter L, iseed, M = "); scanf("%d %ld %d", &L, &iseed, &M); fpe = fopen("e", "w"); fpx = fopen("x", "w"); fpRee = fopen("Ree", "w"); fpRex = fopen("Rex", "w"); fpRxx = fopen("Rxx", "w"); b = (int *) calloc(N, sizeof(int)); x = (double *) calloc(L, sizeof(double)); e = (double *) calloc(L, sizeof(double)); Ree = (double *) calloc(M+1, sizeof(double)); Rex = (double *) calloc(M+1, sizeof(double)); Rxx = (double *) calloc(M+1, sizeof(double)); Q = R / (1 << N); for (n=0; n<L; n++) { v = 0.25 * R * (2 * ran(&iseed) - 1); x[n] = 0.25 * R * cos(2 * pi * n / sqrt(131.)) + v; adc(x[n], b, N, R); xq = dac(b, N, R); e[n] = x[n] - xq; fprintf(fpx, "% lfn", x[n]); fprintf(fpe, "% lfn", e[n]); } corr(L, e, e, M, Ree); corr(L, e, x, M, Rex); corr(L, x, x, M, Rxx); normee = Ree[0]; normex = sqrt(Ree[0] * Rxx[0]); for (i=0; i<=M; i++) { fprintf(fpRee, "%lfn", Ree[i]/normee); fprintf(fpRex, "%lfn", Rex[i]/normex); fprintf(fpRxx, "%lfn", Rxx[i]/Rxx[0]); } 21
- 12. printf("erms = Q/sqrt(12) = %lfn", Q/sqrt(12.0)); printf("Ree(0) = %lfn", sqrt(normee)); } Problem 2.19 The mean of v is zero because E[u]= 0.5: E[v]= 0.5R(E[u]−0.5)= 0 The variance of v is related to the variance of u as follows: σ2 v = E[v2 ]= (0.5R)2 E[(u − 0.5)2 ]= (0.5R)2 σ2 u = (0.5R)2 1 12 = R2 48 Problem 2.20 Inserting x(n)= A cos(2πf0n + φ)+v(n) into the definition of the autocorrelation function, we find: Rxx(k) = E[x(n + k)x(n)] = A2 E[cos 2πf0(n + k)+φ cos(2πf0n + φ)] + AE[cos 2πf0(n + k)+φ v(n)]+AE[v(n + k)cos(2πf0n + φ)] + E[v(n + k)v(n)] assuming φ and v(n) are independent random variables, the cross terms will vanish. Indeed, by the independence assumption the expectation value of the product becomes the product of the expectation values: E[v(n + k)cos(2πf0n + φ)]= E[v(n + k)]·E[cos(2πf0n + φ)]= 0 where the last zero follows either from the zero-mean property of v(n) and/or from the zero mean property of E[cos(2πf0n + φ)]= 0. The latter follows from uniformity of φ whose prob- ability density will be: p(φ)= 1/(2π). We have: E[cos(a + φ)]= 2π 0 cos(a + φ)p(φ) dφ = 1 2π 2π 0 cos(a + φ) dφ = 0 Similarly, we have the property: E[cos(a + φ)cos(b + φ)] = 2π 0 cos(a + φ)cos(b + φ)p(φ) dφ = 1 2π 2π 0 cos(a + φ)cos(b + φ) dφ = 1 2 cos(a − b) Thus, we find E[cos 2πf0(n + k)+φ cos(2πf0n + φ)]= 1 2 cos(2πf0(n + k)−2πf0n)= 1 2 cos(2πf0k) Finally, because v(n) is assumed to be zero-mean white noise, we will have: 22 E[v(n + k)v(n)]= σ2 vδ(k) It follows that the autocorrelation of the noisy sinusoid will be: Rxx(k)= 1 2 A2 cos(2πf0k)+σ2 vδ(k) At lag k = 0, we have: Rxx(0)= 1 2 A2 + σ2 v Therefore, the normalized autocorrelation function becomes: ρxx(k)= Rxx(k) Rxx(0) = A2 /2 A2/2 + σ2 v cos(2πf0k)+ σ2 v A2/2 + σ2 v δ(k) Defining the parameter: a = A2 /2 A2/2 + σ2 v ⇒ 1 − a = σ2 v A2/2 + σ2 v we finally have: ρxx(k)= a cos(2πf0k)+(1 − a)δ(k) Defining the signal to noise ratio as the relative strength of the sinusoid versus the noise in the autocorrelation function, that is, SNR = A2 /2 σ2 v we may express the parameter a as: a = SNR SNR + 1 ⇒ 1 − a = 1 SNR + 1 Problem 2.21 The quantization width of the first ADC will be Q1 = R1/2B1 . The residual signal formed at the output of the first DAC, x2 = x−x1, is the quantization error, and therefore, it varies of the limits − Q1 2 ≤ x2 ≤ Q1 2 These limits must serve as the full scale range of the second ADC. Thus, we must choose: R2 = Q1 = R1 2B1 , Q2 = R2 2B2 = R1 2B1+B2 The following program segment illustrates the sequence of operations: 23
- 13. R2 = R1 / (1 << B1); b1 = (int *) calloc(B1, sizeof(int)); b2 = (int *) calloc(B2, sizeof(int)); for (n=0; n<L; n++) { x = A * cos(2 * pi * f0 * n); adc(x, b1, B1, R1); x1 = dac(b1, B1, R1); x2 = x - x1; adc(x2, b2, B2, R2); } When there is a third B3-bit stage, the total number of bits will be B = B1 + B2 + B3 The B2-bit output of the second ADC must be passed into a second DAC to get the corresponding analog value and subtracted from the input x − x1 to get the third error output, which is then quantized to B3 bits with a full scale range and quantization width: R3 = Q2 = R2 2B2 , Q3 = R3 2B3 = R1 2B1+B2+B3 The following loop shows these operations: R2 = R1 / (1 << B1); R3 = R2 / (1 << B2); b1 = (int *) calloc(B1, sizeof(int)); b2 = (int *) calloc(B2, sizeof(int)); b3 = (int *) calloc(B3, sizeof(int)); for (n=0; n<L; n++) { x = A * cos(2 * pi * f0 * n); adc(x, b1, B1, R1); x1 = dac(b1, B1, R1); adc(x-x1, b2, B2, R2); x2 = dac(b2, B2, R2); x3 = x - x1 - x2; adc(x3, b3, B3, R3); } Problem 2.22 The quantized triangularly-dithered signals were generated by the for-loop given in Example 2.5.1. 24 Chapter 3 Problems Problem 3.1 (a) If the input is doubled the output is not doubled. Thus, the system is not linear. Indeed, the output due to x1(n)= 2x(n) will be y1(n)= 3x1(n)+5 = 6x(n)+5 = 2(3x(n)+5). The system is, however, time-invariant since it has constant-in-time coefficients. (b) The quadratic term x2 (n − 1) breaks linearity because it quadruples whenever the input doubles. The term x(2n) will break time-invariance. Indeed, let xD(n)= x(n − D) be a delayed input. The corresponding output yD(n) can be computed from the given I/O equation: yD(n)= x2 D(n − 1)+xD(2n)= x2 (n − 1 − D)+x(2n − D) but, this is not the same as the delayed (shifted) y(n) obtained by replacing n by n − D: y(n − D)= x2 (n − D − 1)+x(2(n − D))= x2 (n − D − 1)+x(2n − 2D) Thus, the system is neither linear nor time-invariant. (c) System is time-invariant, but not linear since the output does not double whenever the input doubles. (d) The system is linear, but the term nx(n − 3) breaks time-invariance. (e) The time-dependent term n breaks both time invariance and linearity. Problem 3.2 Picking out the coefficients of the x-terms, we find: (a) h = [3, −2, 0, 4] (b) h = [4, 1, 0, −3] (c) h = [1, 0, 0, −1] Problem 3.3 (a,b) These problems are of the general form: y(n)= ay(n − 1)+x(n) with a = −0.9 for (a), and a = 0.9 for (b). Setting x(n)= δ(n) we find the difference equation for the impulse response: h(n)= ah(n − 1)+δ(n) Iterating forward in time with zero initial condition h(−1)= 0, we obtain the causal solu- tion: h(0)= ah(−1)+δ(0)= a · 0 + 1 = 1 h(1)= ah(0)+δ(1)= a · 1 + 0 = a h(2)= ah(1)+δ(2)= a · a + 0 = a2 h(3)= ah(2)+δ(3)= a · a2 + 0 = a3 · · · h(n)= ah(n − 1)+δ(n)= a · an−1 = an , for n ≥ 1 25
- 14. Thus, h(n) will be the geometric series: h = [1, a, a2 , a3 , . . . ] (c,d) These problems are similar to y(n)= ay(n − 2)+x(n) with a = 0.64 = (0.8)2 for (c), and a = −0.81 = −(0.9)2 for (d). The impulse response will satisfy the difference equation: h(n)= ah(n − 2)+δ(n) A few iterations (with zero initial conditions) give: h(0)= 1 h(1)= 0 h(2)= ah(0)= a h(3)= ah(1)= 0 h(4)= ah(2)= a2 h(5)= ah(3)= 0, etc. Thus, all the odd-indexed h(n) are zero: h = [1, 0, a, 0, a2 , 0, a3 , 0, a4 , 0, . . . ] (e) The impulse response satisfies: h(n)= 0.5h(n − 1)+4δ(n)+δ(n − 1) A few iterations give: h(0)= 4 h(1)= 0.5h(0)+1 = 3 h(2)= 0.5h(1)= 3(0.5) h(3)= 0.5h(2)= 3(0.5)2 , etc. Thus, h(n)= ⎧ ⎨ ⎩ 4 if n = 0 3(0.5)n−1 if n ≥ 1 Problem 3.4 We have seen that if h(n)= an u(n), then the I/O difference equation is y(n)= ay(n − 1)+x(n) In cases (a) and (b), we apply this result with a = 0.9 and a = −0.9. In cases (c) and (d), we apply this result twice and verify that h(n)= an u(n)+(−a)n u(n) leads to the difference equation y(n)= a2 y(n − 2)+2x(n) 26 Thus, for (c) we have a = 0.9, a2 = 0.81, and y(n)= 0.81y(n − 2)+2x(n) And, for (d) we have a = 0.9j, a2 = −0.81, and y(n)= −0.81y(n − 2)+2x(n) A systematic and simpler way of solving this type of problem will be presented after we cover z-transforms. Problem 3.5 The impulse response sequence is explicitly: h = [4, 3, 3(0.5), 3(0.5)2 , 3(0.5)3 , . . . ] Replacing these values into the convolutional equation, we get: yn = h0xn + h1xn−1 + h2xn−2 + h3xn−3 + h4xn−4 + · · · = 4xn + 3 xn−1 + 0.5xn−2 + 0.52 xn−3 + 0.53 xn−4 + · · · It follows that: 0.5yn−1 = 2xn−1 + 3 0.5xn−2 + 0.52 xn−3 + 0.53 xn−4 · · · Subtracting, we have: yn − 0.5yn−1 = 4xn + 3xn−1 − 2xn−1 or, yn = 0.5yn−1 = 4xn + xn−1 Problem 3.6 We may write the given impulse response in the form: h(n)= 5δ(n)+6(0.8)n−1 u(n − 1)= [5, 6, 6(0.8), 6(0.8)2 , . . . ] Proceeding as in the previous problem we have: yn = 5xn + 6 xn−1 + 0.8xn−2 + 0.82 xn−3 + · · · 0.8yn−1 = 4xn−1 + 6 0.8xn−2 + 0.82 xn−3 + · · · which gives yn − 0.8yn−1 = 5xn + 2xn−1 27
- 15. Problem 3.7 It is evident from the definition that for n ≥ 2, h(n) satisfies the recursions h(n)= ah(n − 1). Instead of manipulating the I/O convolutional equation into a difference equation as we did above, let us determine directly the difference equation satisfied by h(n) and from that determine the difference equation for y(n). The given expression for h(n) can be rewritten in the more compact form, which is valid for all n: h(n)= c0δ(n)+c1an−1 u(n − 1) where the shifted unit-step u(n − 1) vanishes for n = 0 and equals one for all n ≥ 1. Because h(n) satisfies h(n)= ah(n − 1) for n ≥ 2, we are led to consider the delayed function h(n − 1) and multiply it by a: h(n − 1)= c0δ(n − 1)+c1an−2 u(n − 2) and ah(n − 1)= ac0δ(n − 1)+c1an−1 u(n − 2) Subtracting, we have h(n)−ah(n − 1)= c0δ(n)−ac0δ(n − 1)+c1an−1 u(n − 1)−u(n − 2) But the difference u(n−1)−u(n−2)= δ(n−1) as follows from the standard result u(n)−u(n− 1)= δ(n), which was essentially derived in Example 3.4.4. Therefore, we have h(n)−ah(n − 1)= c0δ(n)−ac0δ(n − 1)+c1an−1 δ(n − 1) But note that an−1 δ(n − 1)= δ(n − 1) because it is nonzero only at n = 1 for which we have a1−1 δ(0)= a0 δ(0)= δ(0). Then, h(n)−ah(n − 1)= c0δ(n)−ac0δ(n − 1)+c1δ(n − 1)= c0δ(n)+(c1 − ac0)δ(n − 1) Or, setting b0 = c0 and b1 = c1 − ac0 and solving for h(n) we find h(n)= ah(n − 1)+b0δ(n)+b1δ(n − 1) Example 3.4.7 had b0 = c0 = 2 and b1 = c1 − ac0 = 4 − 2 · 0.5 = 3. Next, we map this difference equation for h(n) into a difference equation for y(n). To do this we start with the convolutional equation and replace h(m) by its difference equation expression, that is, y(n) = m h(m)x(n − m) = m ah(m − 1)+b0δ(m)+b1δ(m − 1) x(n − m) = a m h(m − 1)x(n − m)+ m b0δ(m)+b1δ(m − 1) x(n − m) In the second sum, the presence of δ(m) extracts only the m = 0 term, that is, b0x(n), whereas the presence of δ(m − 1) extracts only the m = 1 term, that is, b1x(n − 1). Moreover, with a change of variables of summation from m to k = m − 1 or m = k + 1, the a-term is recognized to be ay(n − 1); indeed, 28 a m h(m − 1)x(n − m)= a k h(k)x(n − k − 1)= ay(n − 1) the last equation following from the convolution equation by replacing n by n − 1: y(n)= k h(k)x(n − k) ⇒ y(n − 1)= k h(k)x(n − 1 − k) We finally find the I/O difference equation for y(n): y(n)= ay(n − 1)+b0x(n)+b1x(n − 1) which is exactly the same as that satisfied by h(n). Problem 3.8 First, note that if fn = an u(n) then for n ≥ 1 (1 − aD)fn = fn − afn−1 = an − aan−1 = 0 where D is the delay operator defined by (Df)n= fn−1. It follows that the first term pn 1 u(n) of hn will be annihilated by (1 − p1D), the second term of hn will be annihilated by (1 − p2D), etc. Therefore, all the terms in hn will be annihilated by the product (1 − p1D)(1 − p2D)· · · (1 − pMD)= 1 + a1D + a2D2 + · · · + aMDM Problem 3.9 The Cipn i terms are annihilated by the same product as in the previous problem. However, the δ(n) term will contain the delayed terms: aMδ(n − M)+aM−1δ(n − M + 1)+ · · · But, if n ≥ M + 1, then these terms vanish too. Thus, the difference equation satisfied by hn will be: hn + a1hn−1 + · · · + aMhn−M = 0, n ≥ M + 1 Problem 3.10 Define the polynomial with roots p1, p2: 1 + a1z−1 + a2z−2 = (1 − p1z−1 )(1 − p2z−1 ) so that a1 = −(p1 + p2), a2 = p1p2 Then, the pn i terms in hn will be annihilated by the delay polynomial: hn + a1hn−1 + a2hn−2 = (1 + a1D + a2D2 )hn = (1 − p1D)(1 − p2D)hh 29
- 16. for n ≥ 2. The coefficients b0, b1 may be found by explicitly calculating the right-hand side of this difference equation using the expression for hn, for n = 0, 1: b0 = h0 + a1h−1 + a2h−2 = h0 = C1 + C2 b1 = h1 + a1h0 + a2h−1 = h1 + a1h0 = C1p1 + C2p2 + a1(C1 + C2) Using the expression for a1, we find: b0 = C1 + C2 b1 = C1p1 + C2p2 − (p1 + p2)(C1 + C2)= −(C1p2 + C2p1) To summarize, hn will satisfy the difference equation: hn + a1hn−1 + a2hn−2 = b0δ(n)+b1δ(n − 1) (P3.1) Inserting this into the convolutional equation, we obtain the I/O difference equation: yn + a1yn−1 + a2yn−2 = b0xn + b1xn−1 These results can be obtained quickly using z-transforms: H(z)= C1 1 − p1z−1 + C2 1 − p2z−1 = (C1 + C2)−(C1p2 + C2p1)z−1 (1 − p1z−1)(1 − p2z−1) which may be rewritten as H(z)= b0 + b1z−1 1 + a1z−1 + a2z−2 and leads to the desired difference equations for hn and yn. Problem 3.11 Now the difference equation Eq. (P3.1) will be satisfied for n ≥ 3. To determine the coefficients b0, b1, b2, we evaluate the right hand side for n = 0, 1, 2: b0 = h0 b1 = h1 + a1h0 b2 = h2 + a1h1 + a2h0 which give: b0 = C0 + C1 + C2 b1 = C1p1 + C2p2 + a1(C0 + C1 + C2)= −(p1 + p2)C0 − (C1p2 + C2p1) b2 = C1p2 1 + C2p2 2 + a1(C1p1 + C2p2)+a2(C0 + C1 + C2)= C0p1p2 In b2, the coefficient of C1 is p2 1 + a1p1 + a2, which is zero because p1 is a root of the polynomial z2 + a1z + a2. Similarly, the coefficient of C2 is zero. Again, the results can be justified quickly using z-transforms: H(z) = C0 + C1 1 − p1z−1 + C2 1 − p2z−1 = (C0 + C1 + C2)− C1p2 + C2p1 + C0(p1 + p2) z−1 + C0p1p2z−2 (1 − p1z−1)(1 − p2z−1) 30 Problem 3.12 Define the polynomial with roots p1, p2, p3: 1 + a1z−1 + a2z−2 + a3z−3 = (1 − p1z−1 )(1 − p2z−1 )(1 − p3z−1 ) so that a1 = −(p1 + p2 + p3), a2 = p1p2 + p2p3 + p3p1, a3 = −p1p2p3 Then, the pn i terms in hn will be annihilated by the delay polynomial: (1 + a1D + a2D2 + a3D3 )hn = (1 − p1D)(1 − p2D)(1 − p3D)hh = 0 for n ≥ 3, or, hn + a1hn−1 + a2hn−2 + a3hn−3 = 0, n ≥ 3 The coefficients b0, b1, b2 may be found by explicitly calculating the right-hand side of this dif- ference equation using the expression for hn, for n = 0, 1, 2: b0 = h0 b1 = h1 + a1h0 b2 = h2 + a1h1 + a2h0 or, b0 = C1 + C2 + C3 b1 = C1p1 + C2p2 + C3p3 + a1(C1 + C2 + C3) b2 = C1p2 1 + C2p2 2 + C3p2 3 + a1(C1p1 + C2p2 + C3p3)+a2(C1 + C2 + C3) which simplify into: b0 = C1 + C2 + C3 b1 = − C1(p2 + p3)+C2(p3 + p1)+C3(p1 + p2) b2 = C1p2p3 + C2p3p1 + C3p1p2 Problem 3.13 For (a), we have with n ≥ 2: (1 − 0.5D)(1 − 0.8D)= 1 − 1.3D + 0.4D2 ⇒ hn − 1.3hn−1 + 0.4hn−2 = 0 For (b), we have with n ≥ 2: (1 − 0.5jD)(1 + 0.5jD)= 1 + 0.25D2 ⇒ hn + 0.25hn−2 = 0 For (c), we have with n ≥ 3: (1 − 0.4D)(1 − 0.5D)(1 + 0.5D)= 1 − 0.4D − 0.25D2 + 0.1D3 ⇒ hn − 0.4hn−1 − 0.25hn−2 + 0.1hn−3 = 0 31
- 17. Problem 3.14 From the convolutional I/O equation we get: |y(n)| = | m h(m)x(n − m)| ≤ m |h(m)||x(n − m)| ≤ m |h(m)|B = AB Problem 3.15 Starting with y(n)= m h(m)x(n − m) we obtain at n = 0: y(0)= m h(m)x(−m) But, x(m)= sign h(−m) . Therefore, x(−m)= sign h(m) . It follows: y(0)= m h(m)x(−m)= m h(m)sign h(m) Recognizing that in general xsign(x)= |x|, we find y(0)= m |h(m)| Because x(n) was bounded, y(n) and, hence y(0), must be bounded. If at some m, h(m)= 0, then this term would not contribute to the convolutional sum. Problem 3.16 Working with the convolutional I/O equation, we have: yD(n) = m hD(m)x(n − m)= m h(m − D)x(n − m) = k h(k)x(n − D − k)= y(n − D) where we changed summation variables from m to k = m − D, and evaluated y(n) at n − D, that is, y(n)= k h(k)x(n − k) ⇒ y(n − D)= k h(k)x(n − D − k) Problem 3.17 We have: m hD(m)xA(n − m)= m h(m − D)x(n − m + D)= k h(k)x(n − k) where we changed variables of summation from m to k = m − D. 32 Problem 3.18 Because ˜h(n) agrees with h(n) for n ≥ −D, we have: y(n) = ∞ m=−∞ h(m)x(n − m) ˜y(n) = ∞ m=−D ˜h(m)x(n − m)= ∞ m=−D h(m)x(n − m) Subtracting, the terms m ≥ −D cancel and we get: y(n)−˜y(n)= −D−1 m=−∞ h(m)x(n − m) Assuming x(n) is bounded by A, we have: |y(n)−˜y(n)| = | −D−1 m=−∞ h(m)x(n − m)| ≤ −D−1 m=−∞ |h(m)||x(n − m)| ≤ −D−1 m=−∞ |h(m)|A This summation is finite because it is a subsum of the stability condition: −D−1 m=−∞ |h(m)| ≤ ∞ m=−∞ |h(m)| < ∞ In the limit of large D, the number of terms shrinks to zero giving: lim D→∞ −D−1 m=−∞ |h(m)| → 0 33
- 18. Chapter 4 Problems Problem 4.1 The convolution table is: hx 1 2 1 1 2 1 1 1 1 1 2 1 1 2 1 1 1 1 1 2 1 1 2 1 1 1 2 2 4 2 2 4 2 2 2 1 1 2 1 1 2 1 1 1 Folding the table, we get y = [1, 3, 5, 7, 7, 6, 7, 6, 4, 3, 1] The first and last three samples are the input and output transients, and the middle 5 samples are the steady-state outputs. The LTI table is: n 0 1 2 3 4 5 6 7 8 9 10 xh 1 1 2 1 partial output 1 1 1 2 1 x(0)h(n − 0) 2 2 2 4 2 x(1)h(n − 1) 1 1 1 2 1 x(2)h(n − 2) 1 1 1 2 1 x(3)h(n − 3) 2 2 2 4 2 x(4)h(n − 4) 1 1 1 2 1 x(5)h(n − 5) 1 1 1 2 1 x(6)h(n − 6) 1 1 1 2 1 x(7)h(n − 7) y(n) 1 3 5 7 7 6 7 6 4 3 1 m x(m)h(n − m) For the overlap-add method, the input is divided into the following three contiguous blocks: x = [ 1, 2, 1 x0 , 1, 2, 1 x1 , 1, 1, 0 x2 ] where we padded an extra zero at the end to get a length-3 block. Convolving each block separately with h gives: y0 = h ∗ x0 = [1, 3, 5, 6, 4, 1] y1 = h ∗ x1 = [1, 3, 5, 6, 4, 1] y2 = h ∗ x2 = [1, 2, 3, 3, 1, 0] These convolutions can be done by separately folding the three convolution subtables: block 0 block 1 block 2 hx 1 2 1 1 2 1 1 1 0 1 1 2 1 1 2 1 1 1 0 1 1 2 1 1 2 1 1 1 0 2 2 4 2 2 4 2 2 2 0 1 1 2 1 1 2 1 1 1 0 34 The three subblocks begin at the absolute times n = 0, 3, 6, respectively. It follows from time- invariance that the corresponding output blocks will also begin at the same absolute times. Thus, aligning the output blocks according to their absolute timings and adding them up gives the final result: n 0 1 2 3 4 5 6 7 8 9 10 y0 1 3 5 6 4 1 y1 1 3 5 6 4 1 y2 1 2 3 3 1 y 1 3 5 7 7 6 7 6 4 3 1 In practice this method is implemented efficiently by computing the individual block convolutions using the FFT instead of time-domain convolution. For an FIR filter of order M and an FFT of length N (which is a power of two), the length of each x-block is chosen to be N1 = N − M. The computational gain of this “fast” convolution method versus the conventional time-domain “slow” method is approximately fast slow = log2 N M If the input is divided into length-5 subblocks, the last subblock will have length 3: x = [ 1, 2, 1, 1, 2 x0 , 1, 1, 1 x1 ] The convolutions of the subblocks with the filter are: y0 = h ∗ x0 = [1, 3, 5, 7, 7, 5, 5, 2] y1 = h ∗ x1 = [1, 2, 4, 4, 3, 1] Aligning them at multiples of n = 5, we have: n 0 1 2 3 4 5 6 7 8 9 10 y0 1 3 5 7 7 5 5 2 y1 1 2 4 4 3 1 y 1 3 5 7 7 6 7 6 4 3 1 Problem 4.2 The convolution table is: hx 2 2 0 1 -1 0 1 2 2 4 4 0 2 -2 0 2 4 -2 -4 -4 0 -2 2 0 -2 -4 -1 -2 -2 0 -1 1 0 -1 -2 1 2 2 0 1 -1 0 1 2 Folding the table, we get y = [4, 0, −6, 2, −2, 1, 4, 1, −5, −1, 2] The first and last three samples are the input and output transients, and the middle 5 samples are the steady-state outputs. The LTI table is: 35
- 19. n 0 1 2 3 4 5 6 7 8 9 10 xh 2 -2 -1 1 partial output 2 4 -4 -2 2 x(0)h(n − 0) 2 4 -4 -2 2 x(1)h(n − 1) 0 0 0 0 0 x(2)h(n − 2) 1 2 -2 -1 1 x(3)h(n − 3) -1 -2 2 1 -1 x(4)h(n − 4) 0 0 0 0 0 x(5)h(n − 5) 1 2 -2 -1 1 x(6)h(n − 6) 2 4 -4 -2 2 x(7)h(n − 7) y(n) 4 0 -6 2 -2 1 4 1 -5 -1 2 m x(m)h(n − m) For the overlap-add method, the input is divided into the following three contiguous blocks: x = [ 2, 2, 0 x0 , 1, −1, 0 x1 , 1, 2, 0 x2 ] where we padded an extra zero at the end to get a length-3 block. Convolving each block separately with h gives: y0 = h ∗ x0 = [4, 0, −6, 0, 2] y1 = h ∗ x1 = [2, −4, 1, 2, −1, 0] y2 = h ∗ x2 = [2, −2, −5, −1, 2] These convolutions can be done by separately folding the three convolution subtables: block 0 block 1 block 2 hx 2 2 0 1 -1 0 1 2 2 4 4 0 2 -2 0 2 4 -2 -4 -4 0 -2 2 0 -2 -4 -1 -2 -2 0 -1 1 0 -1 -2 1 2 2 0 1 -1 0 1 2 Aligning the output subblocks at multiples of n = 3, we get: n 0 1 2 3 4 5 6 7 8 9 10 y0 4 0 -6 0 2 1 y1 2 -4 1 2 -1 0 y2 2 2 -5 -1 2 y 4 0 -6 2 -2 1 4 1 -5 -1 2 If the input is divided into length-5 subblocks, the last subblock will have length 3: x = [ 2, 2, 0, 1, −1 x0 , 0, 1, 2 x1 ] The convolutions of the subblocks with the filter are: y0 = h ∗ x0 = [4, 0, −6, 2, −2, 1, 2, −1] y1 = h ∗ x1 = [0, 2, 2, −5, −1, 2] Aligning them at multiples of n = 5, we have: 36 n 0 1 2 3 4 5 6 7 8 9 10 y0 4 0 -6 2 -2 1 2 -1 y1 0 2 2 -5 -1 2 y 4 0 -6 2 -2 1 4 1 -5 -1 2 Problem 4.3 Figure P4.1 shows the filter, input, and computed output signals. 0 3 6 10 13 16 20 23 26 0 1 1 2 3 4 3 6 10 13 16 20 23 26 n n h(n) x(n) y(n) on steady off Fig. P4.1 h(n), x(n), and y(n) signals of Problem 4.3. For the direct form, the indices of hm and xn−m are restricted to lie within their respective ranges: 3 ≤ m ≤ 6 10 ≤ n − m ≤ 20 (P4.1) Solving the second with respect to n, we have: 10 + m ≤ n ≤ 20 + m Using the first of Eq. (P4.1), we extend the range to: 10 + 3 ≤ 10 + m ≤ n ≤ 20 + m ≤ 20 + 6 ⇒ 13 ≤ n ≤ 26 This is the range of index n of the output yn. To find the limits of summation over m, we change the sign of the second of Eq. (P4.1) and solve for m: −20 ≤ m − n ≤ −10 ⇒ n − 20 ≤ m ≤ n − 10 Thus, the inequalities in Eq. (P4.1) are equivalent to 3 ≤ m ≤ 6 n − 20 ≤ m ≤ n − 10 ⇒ max(3, n − 20)≤ m ≤ min(6, n − 10) The final expression for the direct form will be: 37
- 20. yn = min(6,n−10) m=max(3,n−20) hmxn−m , for 13 ≤ n ≤ 26 The transition from the input-on transients to the steady state takes place at time 6 = n − 10 or n = 16 at which the upper limit switches from n − 10 to 6. Similarly, the transition from steady state to the input-off transients occurs at 3 = n − 20 or n = 23 when the lower limit switches from 3 to n − 10. In summary, the three ranges are: input-on: 13 ≤ n ≤ 15 yn = n−10 m=3 hmxn−m steady-state: 16 ≤ n ≤ 23 yn = 6 m=3 hmxn−m input-off: 24 ≤ n ≤ 26 yn = 6 m=n−20 hmxn−m In particular, if the filter and input signals are unity over the ranges (P4.1), the output can be expressed compactly as: yn = min(6,n−10) m=max(3,n−20) 1 · 1 = min(6, n − 10)− max(3, n − 20)+1 (P4.2) for 13 ≤ n ≤ 26. Or, more explicitly: input-on: 13 ≤ n ≤ 15 yn = n − 10 − 3 + 1 = n − 12 steady-state: 16 ≤ n ≤ 23 yn = 6 − 3 + 1 = 4 input-off: 24 ≤ n ≤ 26 yn = 6 − (n − 20)+1 = 27 − n With numerical values: y = {1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 3, 2, 1} The first and last M = 3 samples are the input on/off transients. The middle samples are the steady samples. Note that because the input is constant, the steady-state DC output can be predicted in advance from the DC gain: yDC = m hm = 1 + 1 + 1 + 1 = 4 Note also that because the filter is delayed by 3 samples, the output will not begin to come out until 3 time instants after the input begins, that is at n = 13. Also, the input is cutoff at n = 20 but this is not felt by the filter until 3 instants later, that is, at n = 23. For the LTI form, Eq. (P4.1) is replaced by 10 ≤ m ≤ 20 3 ≤ n − m ≤ 6 This leads to the same range for the output index n, but to the summation limits: 38 yn = min(20,n−3) m=max(10,n−6) xmhn−m , for 13 ≤ n ≤ 26 For part (b), we have: yn = min(20,n−3) m=max(10,n−6) 1 · 1 = min(10, n − 6)− max(20, n − 3)+1 (P4.3) Note that Eq. (P4.3) is equivalent to Eq. (P4.2). Problem 4.4 Because both the input and filter are causal and have infinite duration, we use the formula: y(n)= n m=0 h(m)x(n − m)= n m=0 am u(m)u(n − m) or, y(n)= n m=0 am = 1 − an+1 1 − a The steady-state response is the large-n limit of this formula; that is, y(n)−→ 1 1 − a for n −→ ∞ If x(n)= (−1)n u(n), we have y(n)= n m=0 am (−1)n−m = (−1)n n m=0 (−a)m = (−1)n 1 − (−a)n+1 1 + a Problem 4.5 Consider the IIR filter h(n)= an u(n), where 0 < a < 1. The square pulse x(n)= u(n)−u(n−L) of duration L is applied as input. Using the time-domain convolution formula determine a closed- form expression for the output signal y(n) for the two time ranges: 0 ≤ n ≤ L − 1 and n ≥ L. Because the filter is infinite (M = ∞), Eq. (4.1.16) becomes: y(n)= n m=max(0,n−L+1) h(m)x(n − m), 0 ≤ n < ∞ Inserting the given signals for h(n) and x(n), we find: y(n)= n m=max(0,n−L+1) am · 1 Using the geometric series identity M2 m=M1 am = aM1 − aM2+1 1 − a we find: 39
- 21. y(n)= amax(0,n−L+1) − an+1 1 − a In particular, for 0 ≤ n ≤ L − 1, we have max(0, n − L + 1)= 0, and y(n)= 1 − an+1 1 − a (input-on transients and steady state) whereas for n ≥ L, we have max(0, n − L + 1)= n − L + 1, and y(n)= an−L+1 − an+1 1 − a = an−L+1 1 − aL 1 − a = an−L+1 y(L − 1), (input-off) Thus, after the input is turned off, the output decays exponentially to zero. Problem 4.6 From the previous problem, we have at n = 0, y(0)−ay(−1)= 1 − a0+1 1 − a − a · 0 = 1 = x(0) Then, for 1 ≤ n ≤ L − 1, we have x(n)= 1 and y(n)−ay(n − 1)= 1 − an+1 1 − a − a 1 − an 1 − a = 1 − an+1 − a + an+1 1 − a = 1 = x(n) Finally, for n ≥ L, we have x(n)= 0 and y(n)−ay(n − 1)= an−L+1 yL−1 − aan−1−L+1 yL−1 = 0 = x(n) Problem 4.7 For example, the LTI form is obtained by interchanging the roles played by x and h and their lengths L and M + 1. The following routine is an implementation of Eq. (4.1.19): /* convlti.c - convolution in LTI form */ #include <stdlib.h> defines max( ) and min( ) void convlti(M, h, L, x, y) double *h, *x, *y; h=filter, x=input block, y=output block int M, L; M=filter order, L=input length { int n, m; for (n = 0; n < L+M; n++) for (y[n] = 0, m = max(0, n-M); m <= min(n, L-1); m++) y[n] += x[m] * h[n-m]; } 40 Problem 4.8 The following C program computes the output signals of the two filters (a) and (b) of Example 4.1.8: /* convex.c - convolution example 4.1.8 */ #include <stdio.h> #include <stdlib.h> #include <math.h> #define Ma 14 #define Mb 5 void conv(); void main() { int K = 50, L = 200, n; double *x, *ya, *yb; double ha[Ma+1], hb[Mb+1] = {0.2, -1, 2, -2, 1, -0.2}; x = (double *) calloc(L, sizeof(double)); ya = (double *) calloc(L+Ma, sizeof(double)); Ex. 4.1.8 (a) yb = (double *) calloc(L+Mb, sizeof(double)); Ex. 4.1.8.(b) for (n=0; n<=Ma; n++) define filter (a) ha[n] = 0.25 * pow(0.75, n); for (n=0; n<L; n++) define input if (n%K < K/2) x[n] = 1; else x[n] = 0; conv(Ma, ha, L, x, ya); compute outputs conv(Mb, hb, L, x, yb); for (n=0; n<L+Ma; n++) printf("%lfn", ya[n]); printf("nn"); for (n=0; n<L+Mb; n++) printf("%lfn", yb[n]); } Problem 4.9 A complete such program is included below. /* blkfilt.c - FIR filtering by block convolution */ #include <stdio.h> #include <stdlib.h> #define MAX 64 memory allocation size 41
- 22. void blockcon(); void main(int argc, char **argv) { FILE *fph; filter file double *h, *x, *y, *ytemp; int M, N, L, i; M = filter order, L = blocksize int max = MAX, dmax = MAX; initial allocation & increment if (argc != 3) { fprintf(stderr, "usage: blkfilt hfile L <xfile >yfilen"); exit(0); } if ((fph = fopen(argv[1], "r")) == NULL) { fprintf(stderr, "can’t open filter file: %sn", argv[1]); exit(0); } L = atoi(argv[2]); block length h = (double *) calloc(max + 1, sizeof(double)); for (M=0;; M++) { read h if (M == max) { if necessary, reallocate h max += dmax; h = (double *) realloc((char *) h, (max + 1) * sizeof(double)); } if (fscanf(fph, "%lf", h + M) == EOF) break; } M--; M is filter order if (L < M) { fprintf(stderr, "blocksize L must be at least %dn", M); exit(0); } h = (double *) realloc((char *) h, (M + 1) * sizeof(double)); final allocation x = (double *) calloc(L, sizeof(double)); allocate input block y = (double *) calloc(L + M, sizeof(double)); allocate output block ytemp = (double *) calloc(M, sizeof(double)); initialized to zero start filtering: for (;;) { read input block for (N=0; N<L; N++) N=L except for last if (scanf("%lf", x+N) == EOF) goto last; blockcon(M, h, L, x, y, ytemp); process block for (i=0; i<L; i++) write output block printf("%lfn", y[i]); } last: process last block blockcon(M, h, N, x, y, ytemp); last block has N<=L 42 for (i=0; i<N+M; i++) write entire y block printf("%lfn", y[i]); } The dimension of the filter array h is determined on the fly by reading the file of filter coeffi- cients. Because the length of this file is not known in advance, the program keeps reallocating the length of h in increments of MAX coefficients until the final length is determined and the final reallocation is made. The filtering portion of the program is identical to the program segment discussed at the end of Section 4.1.10. The program can receive its input from stdin with the user entering the input samples one at a time. However, only after a group of L samples has been entered followed by a <CR> (carriage return), will the program output the first group of L outputs. Then, the next group of L input samples is entered and the program outputs the corresponding group of L outputs, and so on. At the end, one must enter a <CTRL-Z> (for MSDOS) in order for the program to process the input-off transients. Problem 4.10 A complete such program is included below. /* firfilt.c - FIR filtering by sample-by-sample processing */ #include <stdlib.h> #include <stdio.h> #define MAX 64 initial allocation size double fir(); void main(int argc, char **argv) { FILE *fph; filter file double *h, *w, x, y; filter, states, input, output samples int M, i; int max = MAX, dmax = MAX; allocation for h and increment if (argc != 2) { fprintf(stderr, "usage: firfilt hfile <xfile >yfilen"); exit(0); } if ((fph = fopen(argv[1], "r")) == NULL) { fprintf(stderr, "can’t open filter file: %sn", argv[1]); exit(0); } h = (double *) calloc(max + 1, sizeof(double)); preliminary allocation for (M=0;; M++) { read h if (M == max) { reallocate h, if necessary max += dmax; h = (double *) realloc((char *) h, (max + 1) * sizeof(double)); } if (fscanf(fph, "%lf", h + M) == EOF) break; 43
- 23. } M--; M is filter order h = (double *) realloc((char *) h, (M + 1) * sizeof(double)); final allocation w = (double *) calloc(M + 1, sizeof(double)); start filtering: while(scanf("%lf", &x) != EOF) { keep reading input samples y = fir(M, h, w, x); compute output sample printf("%lfn", y); write output sample } for (i=0; i<M; i++) { input-off transients y = fir(M, h, w, 0.0); printf("%lfn", y); } } As in the case of blkfilt.c, the dimension of the filter array h is determined on the fly by reading the file of filter coefficients. The program keeps reallocating the length of h in increments of MAX coefficients until the final length is determined and the final reallocation is made. The filtering portion of the program is identical to the program segment discussed in Section 4.2.3. The program can receive its input from stdin with the user entering the input one sample at a time. After each input is entered followed by a <CR>, the program computes and outputs the corresponding output sample. (One can also enter a few input samples at a time, followed by a <CR>, and the program will produce the corresponding outputs.) At the end, one must enter a <CTRL-Z> (for MSDOS) in order for the program to process the input-off transients. Problem 4.11 A complete such program is included below. /* cfirfilt.c - FIR filtering using circular delay-line buffer */ #include <stdlib.h> #include <stdio.h> #define MAX 64 initial allocation size double cfir(); void main(int argc, char **argv) { FILE *fph; filter file double *h, *w, *p; filter, states, circular pointer double x, y; input and output samples int M, i; int max = MAX, dmax = MAX; allocation for h and increment if (argc != 2) { fprintf(stderr, "usage: cfirfilt hfile <xfile >yfilen"); exit(0); } 44 if ((fph = fopen(argv[1], "r")) == NULL) { fprintf(stderr, "can’t open filter file: %sn", argv[1]); exit(0); } h = (double *) calloc(max + 1, sizeof(double)); preliminary allocation for (M=0;; M++) { read h if (M == max) { reallocate h, if necessary max += dmax; h = (double *) realloc((char *) h, (max + 1) * sizeof(double)); } if (fscanf(fph, "%lf", h + M) == EOF) break; } M--; M is filter order h = (double *) realloc((char *) h, (M + 1) * sizeof(double)); w = (double *) calloc(M + 1, sizeof(double)); p = w; initialize p start filtering: while(scanf("%lf", &x) != EOF) { keep reading input samples y = cfir(M, h, w, &p, x); compute output sample printf("%lfn", y); write output sample } for (i=0; i<M; i++) { input-off transients y = cfir(M, h, w, &p, 0); printf("%lfn", y); } } It is identical to firfilt.c, except it uses the circular version cfir instead of fir to do the fil- tering. In addition, the circular pointer p is defined and initialized by the program. This program must be compiled and linked with cfir.c and wrap.c. The version using the routine cfir1 is identical. The following version is based on the routine cfir2 and uses the circular pointer index q instead of the pointer p. It must be linked with the routines cfir2 and wrap2. /* cfirfil2.c - FIR filtering using circular delay-line buffer */ #include <stdlib.h> #include <stdio.h> #define MAX 64 initial allocation size double cfir2(); void main(int argc, char **argv) { FILE *fph; filter file double *h, *w; filter, states, circular pointer double x, y; input and output samples int M, i, q; circular pointer index int max = MAX, dmax = MAX; allocation for h and increment 45
- 24. if (argc != 2) { fprintf(stderr, "usage: cfirfil2 hfile <xfile >yfilen"); exit(0); } if ((fph = fopen(argv[1], "r")) == NULL) { fprintf(stderr, "can’t open filter file: %sn", argv[1]); exit(0); } h = (double *) calloc(max + 1, sizeof(double)); preliminary allocation for (M=0;; M++) { read h if (M == max) { reallocate h, if necessary max += dmax; h = (double *) realloc((char *) h, (max + 1) * sizeof(double)); } if (fscanf(fph, "%lf", h + M) == EOF) break; } M--; M is filter order h = (double *) realloc((char *) h, (M + 1) * sizeof(double)); w = (double *) calloc(M + 1, sizeof(double)); q = 0; initialize q start filtering: while(scanf("%lf", &x) != EOF) { keep reading input samples y = cfir2(M, h, w, &q, x); compute output sample printf("%lfn", y); write output sample } for (i=0; i<M; i++) { input-off transients y = cfir2(M, h, w, &q, 0); printf("%lfn", y); } } Problem 4.12 The following C program is an implementation: /* cdelfilt.c - Implementation of delay using circular buffer */ #include <stdlib.h> #include <stdio.h> double cdelay(); double tap(); void main(int argc, char **argv) { double *w, *p; linear buffer & circular pointer double x, y; input and output samples int D, i, j; 46 if (argc != 3) { fprintf(stderr, "usage: cdelfilt i D <xfile >yfilen"); exit(0); } i = atoi(argv[1]); D = atoi(argv[2]); if (i>D) { fprintf(stderr, "i must be in the range {0, 1, ..., D}n"); exit(0); } w = (double *) calloc(D + 1, sizeof(double)); p = w; initialize p while(scanf("%lf", &x) != EOF) { keep reading input samples *p = x; store in circular buffer y = tap(D, w, p, i); y(n) = x(n-i) = i-th delay cdelay(D, w, &p); update delay line printf("%lfn", y); write output sample } for (j=0; j<i; j++) { input-off transients *p = 0; zero input y = tap(D, w, p, i); cdelay(D, w, &p); printf("%lfn", y); } } Problem 4.13 Consider a 3d order filter with internal states wi(n)= x(n − i), i = 0, 1, 2, 3. They are delayed replicas of each other: w3(n) = w2(n − 1) w2(n) = w1(n − 1) w1(n) = w0(n − 1) w0(n) = x(n) Thus, the current states are obtained from the previous ones by shifting the previous contents of the delay line and reading x(n) into w0(n). Once this shift is done, the filter’s output is computed by the dot product: y(n)= h0w0(n)+h1w1(n)+h2w2(n)+h3w3(n) The following routine firalt.c implements this alternative procedure: /* firalt.c - Alternative version of FIR filter in direct form */ double firalt(M, h, w, x) Usage: y=firalt(M, h, w, x); double *h, *w, x; h=filter, w=state, x=input int M; M=filter order { 47
- 25. int i; double y; y=output sample for (i=M; i>=1; i--) update states from previous call w[i] = w[i-1]; done in reverse order w[0] = x; read current input sample x for (y=0, i=0; i<=M; i++) y += h[i] * w[i]; process current output sample return y; } Note that the previous state vector is updated first into its current value, the current input is read into w0, and only then is the output y computed. Upon exit, the states are left at their current values and not updated. They will be updated by the next call. The vector w must be initialized to zero prior to the first call, just like the fir.c case. Problem 4.14 Introduce the internal states: w0(n) = x(n) w1(n) = x(n − 1)= w0(n − 1) w2(n) = x(n − 2)= w1(n − 1) w3(n) = x(n − 3)= w2(n − 1) Then, the I/O equation together with the state-updating equations will read: w0(n)= x(n) y(n)= w0(n)+w1(n)+2w2(n)+w3(n) and w3(n + 1) = w2(n) w2(n + 1) = w1(n) w1(n + 1) = w0(n) This leads to the following sample processing algorithm and block diagram: x y w0 z-1 z-1 z-1 w1 w2 w3 2 for each input sample x do: w0 = x y = w0 + w1 + 2w2 + w3 w3 = w2 w2 = w1 w1 = w0 The sample processing algorithm generates the following output samples: 48 n x w0 w1 w2 w3 y = w0 + w1 + 2w2 + w3 0 1 1 0 0 0 1 1 2 2 1 0 0 3 2 1 1 2 1 0 5 3 1 1 1 2 1 7 4 2 2 1 1 2 7 5 1 1 2 1 1 6 6 1 1 1 2 1 7 7 1 1 1 1 2 6 8 0 0 1 1 1 4 9 0 0 0 1 1 3 10 0 0 0 0 1 1 The first three outputs correspond to the input-on transients (the internal delay registers are still filling up). The period 3 ≤ n ≤ 7 corresponds to steady state (all delays are filled). The last three outputs, in general the last M outputs for an M-th order FIR filter, are the input-off (x = 0) transients (the delays gradually empty out their contents). A C routine and main program implementing this algorithm are given below: #include <stdio.h> #include <malloc.h> declares calloc() double x[8] = {1,2,1,1,2,1,1,1}; double filter2(); void main() { int n; double y, *w; w = (double *) calloc(4, sizeof(double)); allocates & initializes w to zero for (n=0; n<8; n++) { input-on transients & steady-state y = filter2(x[n], w); printf("%lfn", y); } for (n=8; n<11; n++) { input-off transients y = filter2(0.0, w); printf("%lfn", y); } } double filter2(x, w) usage: y = filter2(x, w); double x, *w; { double y; w[0] = x; read input sample y = w[0] + w[1] + 2 * w[2] + w[3]; compute output sample w[3] = w[2]; update internal states w[2] = w[1]; w[1] = w[0]; 49
- 26. return y; } Problem 4.15 We define the state-variables wi(n), i = 0, 1, 2, 3 as follows: w0(n) = x(n) w1(n) = x(n − 1)= w0(n − 1) w2(n) = x(n − 2)= w1(n − 1) w3(n) = x(n − 3)= w2(n − 1) Then, the I/O equation y(n)= x(n)−x(n − 3) together with the state-updating equations will read: w0(n)= x(n) y(n)= w0(n)−w3(n) and w3(n + 1) = w2(n) w2(n + 1) = w1(n) w1(n + 1) = w0(n) Note that the output y(n) at time n and the next internal states at time n + 1 are computable from the knowledge of the present input x(n) and the present states at time n. This leads to the following sample processing algorithm and block diagram realization: x y w0 z-1 z-1 z-1 w1 w2 w3 -1 for each input sample x do: w0 = x y = w0 − w3 w3 = w2 w2 = w1 w1 = w0 The sample processing algorithm can be applied to the given input samples to generate the output samples: n x w0 w1 w2 w3 y = w0 − w3 0 1 1 0 0 0 1 1 1 1 1 0 0 1 2 2 2 1 1 0 2 3 2 2 2 1 1 1 4 4 4 2 2 1 3 Problem 4.16 We define the internal state: w1(n)= y(n − 1) 50 Then, the difference equation y(n)= 0.8y(n − 1)+x(n) and the state updating equation will read: y(n)= 0.8w1(n)+x(n) and w1(n + 1)= y(n) This leads to the following sample processing algorithm and block diagram realization: x 0.8 y y z-1 w1 for each input sample x do: y = 0.8w1 + x w1 = y The table of computations is: n x w1 y = 0.8w1 + x 0 1 0.0000 1.0000 1 1 1.0000 1.8000 2 2 1.8000 3.4400 3 2 3.4400 4.7520 4 4 4.7520 7.8016 Problem 4.17 We define the internal states: w1(n) = y(n − 1) w2(n) = y(n − 2)= w1(n − 1) ⇒ w1(n + 1) = y(n) w2(n + 1) = y(n − 1)= w1(n) The difference equation y(n)= 0.25y(n − 2)+x(n) and state updating equations become: y(n)= 0.25w2(n)+x(n) and w2(n + 1) = w1(n) w1(n + 1) = y(n) This leads to the following sample processing algorithm and block diagram realization: x 0.25 y y z-1 z-1 w2 w1 for each input sample x do: y = 0.25w2 + x w2 = w1 w1 = y General versions of such routines will be considered in Chapter 7. The sample processing algo- rithm can be cranked on the given input samples to generate the output: n x y = 0.25w2 + x w1 w2 0 1 1 0 0 1 1 1 1 0 2 2 2.25 1 1 3 2 2.25 2.25 1 4 4 4.5625 2.25 2.25 51
- 27. Notice how the computed y becomes the next w1 and w1 shifts into w2. Problem 4.18 The impulse response is read off from the coefficients of the I/O equation: h = [1, 0, −1, 2] The LTI form is implemented by the following table: n 0 1 2 3 4 5 6 7 8 9 10 xh 1 0 -1 2 partial output 1 1 0 -1 2 x(0)h(n − 0) 1 1 0 -1 2 x(1)h(n − 1) 2 2 0 -2 4 x(2)h(n − 2) 2 2 0 -2 4 x(3)h(n − 3) 2 2 0 -2 4 x(4)h(n − 4) 2 2 0 -2 4 x(5)h(n − 5) 1 1 0 -1 2 x(6)h(n − 6) 1 1 0 -1 2 x(7)h(n − 7) y(n) 1 1 1 3 2 4 3 3 3 1 2 m x(m)h(n − m) For the overlap-add method, the input is divided into the following length-4 contiguous blocks: x = [ 1, 1, 2, 2 x0 , 2, 2, 1, 1 x1 , ] Convolving each block separately with h gives: y0 = h ∗ x0 = [1, 1, 1, 3, 0, 2, 4] y1 = h ∗ x1 = [2, 2, −1, 3, 3, 1, 2] These convolutions can be done by separately folding the two convolution subtables: block 0 block 1 hx 1 1 2 2 2 2 1 1 1 1 2 2 2 2 2 1 1 0 0 0 0 0 0 0 0 0 -1 -1 -1 -2 -2 -2 -2 -1 -1 2 2 2 4 4 4 4 2 2 The two subblocks begin at the absolute times n = 0, 4, respectively. It follows from time- invariance that the corresponding output blocks will also begin at the same absolute times. Thus, aligning the output blocks according to their absolute timings and adding them up gives the final result: n 0 1 2 3 4 5 6 7 8 9 10 y0 1 1 1 3 0 2 4 y1 2 2 -1 3 3 1 2 y 1 1 1 3 2 4 3 3 3 1 2 52 The block diagram and sample processing algorithm are: x y w0 z-1 z-1 z-1 w1 w2 w3 2 -1 for each input sample x do: w0 = x y = w0 − w2 + 2w3 w3 = w2 w2 = w1 w1 = w0 The corresponding circular-buffer version will be: for each input sample x do: s0 = ∗p = x s2 = tap(3, w, p, 2) s3 = tap(3, w, p, 3) y = s0 − s2 + 2s3 cdelay(3, w, &p) where tap-1 is not needed, but tap-2 and tap-3 are. Tap-0 s0 is the content of the current location pointed to by p, which receives the input sample x. After y is computed, the call to cdelay circularly decrements the pointer p. Problem 4.19 Figure P4.2 shows the contents of the registers w = [w0, w1, w1] at successive time instants. Each time the circular pointer p = pin points to the w-register that receives the current input, whereas the pointer pout points to the w-register containing the current output, that is, y(n)= x(n − 2). Therefore, pout is shifted clockwise by 2 with respect to pin: pout = pin + 2 (modulo 3) As pin gets decremented circularly, so does pout. If pin points to w[qin] and pout to w[qout], then pin = w + qin pout = w + qout where qout = (qin + 2)% 3 Thus, the input and output circular indices take on the sequence of values: qin = 0, 2, 1, 0, 2, 1, 0, 2, 1, 0, 2 qout = 2, 1, 0, 2, 1, 0, 2, 1, 0, 2, 1 The following table lists the contents of w for both the linear and circular buffer implementa- tions. In the linear case, w0, w1 and w2 are simply the input signal delayed by i = 0, 1, 2 units. In the circular case, the down-arrow indicates the w-register receiving the current input, that is, w[qin], whereas the up-arrow indicates the corresponding output sample, that is, w[qout]. In the circular case, only one w-register changes value at each time instant. 53
- 28. pin w0 w1 w2 pout 0 0 n=0 pin pout pin pout pin pout pin pout 0 n=1 pin pout pin pout pin pout n=2 pin pout pin pout pin pout n=3 n=4 n=5 n=6 n=7 n=8 n=9 n=10 8 8 8 7 7 7 6 6 6 5 5 5 4 4 4 3 3 3 2 2 2 1 1 1 0 0 0 0 0 0 Fig. P4.2 Circular delay-line buffer contents. linear circular n x w0 w1 w2 w0 w1 w2 qin qout y 0 8 8 0 0 8↓ 0 0↑ 0 2 0 1 7 7 8 0 8 0↑ 7↓ 2 1 0 2 6 6 7 8 8↑ 6↓ 7 1 0 8 3 5 5 6 7 5↓ 6 7↑ 0 2 7 4 4 4 5 6 5 6↑ 4↓ 2 1 6 5 3 3 4 5 5↑ 3↓ 4 1 0 5 6 2 2 3 4 2↓ 3 4↑ 0 2 4 7 1 1 2 3 2 3↑ 1↓ 2 1 3 8 0 0 1 2 2↑ 0↓ 1 1 0 2 9 0 0 0 1 0↓ 0 1↑ 0 2 1 10 0 0 0 0 0 0↑ 0↓ 2 1 0 Problem 4.20 The transposed form is obtained by the following rules: 1. Exchange input with output. 2. Reverse all signal flows. 4. Replace all adders by nodes. 3. Replace all nodes by adders. For example, in Fig. 4.2.7, the output adder will be replaced by a node with the signals flowing in reversed directions, and the nodes at the inputs of the delays will be replaced by adders, and x(n) will be exchanged with y(n). The resulting block diagram is shown in Fig. P4.3. We can draw it reversed, as in Fig. 4.3.1. 54 x(n)y(n) v1(n) v2(n) v3(n) z-1 z-1 z-1 h0 h1 h2 h3 Fig. P4.3 Transposed direct form realization. Let v1(n), v2(n), and v3(n) be the contents of the three delays at time n. They can be thought of as the internal states of the realization. Because of the intervening adders between the delays, the vi(n) are no longer the delayed versions of each other. Rather, each is a delayed version of its input. These inputs will become the contents of the delays at the next time instant. Therefore, the I/O equations describing the filter are in this case: y(n)= v1(n)+h0x(n) and v1(n + 1) = v2(n)+h1x(n) v2(n + 1) = v3(n)+h2x(n) v3(n + 1) = h3x(n) This system is equivalent to the original I/O equation (4.2.6). Indeed, down-shifting v1(n + 1), we have v1(n)= v2(n − 1)+h1x(n − 1) Therefore, y(n)= h0x(n)+v1(n)= h0x(n)+ h1x(n − 1)+v2(n − 1) Similarly, down-shifting v2(n+1) twice, that is, v2(n−1)= v3(n−2)+h2x(n−2), and v3(n+1) three times, that is, v3(n − 2)= h3x(n − 3), we get y(n) = h0x(n)+h1x(n − 1)+v2(n − 1) = h0x(n)+h1x(n − 1)+ h2x(n − 2)+v3(n − 2) = h0x(n)+h1x(n − 1)+h2x(n − 2)+h3x(n − 3) which is equivalent to Eq. (4.2.6). The I/O system leads then to the following sample by sample processing algorithm: 55
- 29. for each input sample x do: y = v1 + h0x v1 = v2 + h1x v2 = v3 + h2x v3 = h3x For Example 4.2.1, Table P4.1 shows the contents of the vi registers at each time instant. The vi are initialized to zero. The v1 and v2 entries in each row are obtained by forming v2 + h1x and v3 + h2x of the previous row. The v3 column is the delayed (down-shifted) version of the h3x column. The final result is the same as in Example 4.2.1. n x h0x h1x h2x h3x v1 v2 v3 y = v1 + h0x 0 1 1 2 −1 1 0 0 0 1 1 1 1 2 −1 1 2 −1 1 3 2 2 2 4 −2 2 1 0 1 3 3 1 1 2 −1 1 4 −1 2 5 4 2 2 4 −2 2 1 1 1 3 5 2 2 4 −2 2 5 −1 2 7 6 1 1 2 −1 1 3 0 2 4 7 1 1 2 −1 1 2 1 1 3 8 0 0 0 0 0 3 0 1 3 9 0 0 0 0 0 0 1 0 0 10 0 0 0 0 0 1 0 0 1 Table P4.1 Computations in Problem 4.20. Problem 4.21 The generalization of the transposed algorithm to order M is straightforward: for each input sample x do: for i = 0, 1, . . . , M − 1 do: vi = vi+1 + hix vM = hMx y = v0 where for indexing convenience we introduced v0(n)= y(n)= v1(n)+h0x(n). The following C routine firtr.c is an implementation: /* firtr.c - FIR filter in transposed direct form */ double firtr(M, h, v, x) Usage: y=firtr(M, h, v, x); double *h, *v, x; h=filter, v=state, x=input int M; M=filter order { int i; 56 for (i=0; i<=M-1; i++) v[i] = v[i+1] + h[i] * x; output & update states v[M] = h[M] * x; last state return v[0]; y = v0 = v1 + h0x } The following stand-alone C program firfiltr.c is the corresponding version of firfilt.c. The only difference is that it uses the routine firtr.c instead of fir.c: /* firfiltr.c - FIR filtering in transposed form */ #include <stdlib.h> #include <stdio.h> #define MAX 64 initial allocation size double firtr(); void main(int argc, char **argv) { FILE *fph; filter file double *h, *v, x, y; filter array, input, output samples int M, i; int max = MAX, dmax = MAX; allocation for h and increment if (argc != 2) { fprintf(stderr, "usage: firfilt hfile <xfile >yfilen"); exit(0); } if ((fph = fopen(argv[1], "r")) == NULL) { fprintf(stderr, "can’t open filter file: %sn", argv[1]); exit(0); } h = (double *) calloc(max + 1, sizeof(double)); preliminary allocation for (M=0;; M++) { read h if (M == max) { reallocate h, if necessary max += dmax; h = (double *) realloc((char *) h, (max + 1) * sizeof(double)); } if (fscanf(fph, "%lf", h + M) == EOF) break; } M--; M is filter order h = (double *) realloc((char *) h, (M + 1) * sizeof(double)); final allocation v = (double *) calloc((M + 1), sizeof(double)); while(scanf("%lf", &x) != EOF) { keep reading input samples y = firtr(M, h, v, x); compute output sample printf("%lfn", y); write output sample } for (i=0; i<M; i++) { input-off transients 57
- 30. y = firtr(M, h, v, 0.0); compute output sample printf("%lfn", y); } } Transposed forms are convenient in some applications, such as in the implementation of FIR decimators and interpolators in multirate applications. MATLAB uses this form to implement its built-in filtering routines. 58 Chapter 5 Problems Problem 5.1 The linearity property follows from: ∞ n=−∞ [a1x1(n)+a2x2(n)]z−n = a1 ∞ n=−∞ x1(n)z−n + a2 ∞ n=−∞ x2(n)z−n The delay property follows from: XD(z) = ∞ n=−∞ xD(n)z−n = ∞ n=−∞ x(n − D)z−n = = ∞ k=−∞ x(k)z−(k+D) = z−D ∞ k=−∞ x(k)z−k = z−D X(z) where we changed summation index from n to k = n − D, so that n = k + D. The convolution property follows from the linearity and delay properties. Writing y(n) in terms of the delayed replicas of x(n), we have: y(n)= ∞ m=−∞ h(m)x(n − m)= ∞ m=−∞ h(m)xm(n) Taking z-transforms of both sides and using the linearity and delay properties, we get: Y(z)= ∞ m=−∞ h(m)Xm(z)= ∞ m=−∞ h(m)z−m X(z)= ∞ m=−∞ h(m)z−m X(z) or, Y(z)= H(z)X(z) Problem 5.2 a. From the delay property of z-transforms, X(z)= z−5 Δ(z), where Δ(z)= 1 is the z- transform of δ(n). Thus, X(z)= z−5 . The ROC is the entire z-plane with the exception of the point z = 0. b. Similarly, X(z)= z5 Δ(z)= z5 , but now the ROC must only exclude the point at infinity z = ∞. c. The unit-step has z-transform: U(z)= 1/(1 − z−1 ). Thus, X(z)= z−5 U(z)= z−5 1 − z−1 with ROC |z| > 1. d. We may work with the definition: X(z) = ∞ n=−∞ u(−n + 5)z−n = 5 n=−∞ 1 · z−n = z−5 + z−4 + z−3 + z−2 + z−1 + 1 + z + z2 + z3 + · · · = z−5 1 + z + z2 + z3 + · · · = z−5 1 − z = z−5 U(z−1 ) 59
- 31. The convergence of the series requires the ROC |z| < 1. Alternatively, we recognize that x(n) is the delayed version of u(−n), that is, x(n)= u(−n + 5)= u(−(n − 5)). Using the general property that g(n) Z −→ G(z) ⇒ g(−n) Z −→ G(z−1 ) we find X(z)= z−5 U(z−1 ). Problem 5.3 a. Using the general result: an u(n)−→ 1/(1 − az−1 ), ROC |z| > |a|, we obtain: X(z)= 1/(1 − (−0.5)z−1 )= 1/(1 + 0.5z−1 ), with ROC |z| > | − 0.5| = 0.5. b. Method 1: Let w(n)= u(n)−u(n − 10) with z-transform W(z). Using the linearity and delay properties, we have W(z)= U(z)−z−10 U(z)= (1 − z−10 )U(z)= 1 − z−10 1 − z−1 Using the modulation property of z-transforms, we have an w(n)−→ W(z/a)= 1 − (z/a)−10 1 − (z/a)−1 = 1 − a10 z−10 1 − az−1 With a = −0.5, we have X(z)= (1 − (−0.5)10 z−10 )/(1 + 0.5z−1 ). ROC is the whole z-plane minus the point z = 0. Method 2: We recognize that x(n) is the finite sequence x(n)= [1, a, a2 , a3 , a4 , a5 , a6 , a7 , a8 , a9 , 0, 0, 0, · · · ] and use the definition of z-transforms: X(z)= 1 + az−1 + a2 z−2 + · · · + a9 z−9 = 1 − a10 z−10 1 − az−1 where in the last step we used the finite geometric series: 1 + x + x2 + · · · + xm = 1 − xm+1 1 − x , with x = az−1 and m = 9 c. X(z)= 1 1 − 0.5z−1 + 1 1 + 0.5z−1 = 2 (1 − 0.5z−1)(1 + 0.5z−1) = 2 1 − 0.25z−2 , ROC |z| > 0.5. Problem 5.4 Using Euler, 2 cos θ = ejθ +e−jθ , with θ = πn/2 and also the fact that ejπ/2 = j and e−jπ/2 = −j, we rewrite 2(0.8)n cos(πn/2) = (0.8)n [ejπn/2 + e−jπn/2 ] = (0.8)n [jn + (−j)n ]= (0.8j)n +(−0.8j)n Thus, the signals in questions (a) and (b) are the same. Their z-transform is 60 X(z)= 1 1 − 0.8jz−1 + 1 1 + 0.8jz−1 = 2 (1 − 0.8jz−1)(1 + 0.8jz−1) = 2 1 + 0.64z−2 ROC is |z| > |0.8j| = 0.8. An alternative method is to list a few of the signal values and then use the infinite geometric series to sum them up, that is, with a = 0.8: x(n)= 2[1, 0, −a2 , 0, a4 , 0, −a6 , 0, a8 , 0, −a10 , 0, · · · ] we obtain X(z)= 2[1 − a2 z−2 + a4 z−4 − a6 z−6 + a8 z−8 − a10 z−10 + · · · ]= 2 1 + a2z−2 where we applied the infinite geometric series 1 + x + x2 + x3 + · · · = 1 1 − x with x = −a2 z−2 . Convergence of the geometric series requires that |x| < 1 or | − a2 z−2 | < 1 or |z| > |a| or in our case |z| > 0.8. Problem 5.5 The three signals in parts (a,b,c) all have the same z-transform, namely, X(z)= 1 1 − 0.25z−1 + 1 1 − 4z−1 = 2 − 4.25z−1 1 − 4.25z−1 + z−2 They differ only in their region of convergence: In case (a), both terms are causal, therefore |z| > 0.25 and |z| > 4. Thus, ROC is |z| > 4. In case (b), the first term requires |z| > 0.25 while the second |z| < 4. Combining, we find ROC 0.25 < |z| < 4. In case (c), both terms are anticausal, thus, |z| < 0.25 and |z| < 4. Thus, ROC is |z| < 0.25. The signal in (d) is unstable from both sides of the time axis. The first term in the z-transform expansion X(z)= − −1 n=−∞ (0.25)n z−n + ∞ n=0 4n z−n would require ROC |z| < 0.25 and the second |z| > 4. The intersection of the two ROC sets is empty, that is, there is no set of z’s for which the z-transform would converge. Problem 5.6 Method 1: List the values of x(n) and sum the geometric series: x(n) = [1, 0, −1, 0, 1, 0, −1, 0, 1, 0, −1, 0, . . . ] X(z) = 1 − z−2 + z−4 − z−6 + z−8 − · · · = 1 + x + x2 + x3 + x4 + · · · x=−z−2 = 1 1 − x x=−z−2 = 1 1 + z−2 where convergence of the geometric series requires: 61
- 32. |x| < 1 ⇒ | − z−2 | < 1 ⇒ |z| > 1 Method 2: Use Euler’s formula to split x(n) as the sum of two terms of the form an u(n): x(n)= cos(πn/2)u(n)= 1 2 eπjn/2 + e−πjn/2 u(n)= 1 2 [jn + (−j)n ]u(n) where we used eπj/2 = j. Taking z-transforms, we find: X(z)= 1/2 1 − jz−1 + 1/2 1 + jz−1 = 1 (1 − jz−1)(1 + jz−1) = 1 1 + z−2 where we applied the result an u(n) Z −→ 1 1 − az−1 for a = j and a = −j. Method 3: Recognize that x(n) is periodic with period 4. Defining the length-4 sequence g = [1, 0, −1, 0], we see that x(n) will be the periodic replication of g(n): x = [g, g, g, . . . ] or, x(n)= g(n)+g(n − 4)+g(n − 8)+g(n − 12)+ · · · Taking z-transforms of both sides gives X(z)= G(z)+z−4 G(z)+z8 G(z)+ · · · = 1 + z−4 + z−8 + z−12 + · · · G(z) Summing up the series and using G(z)= 1 − z−2 , we get: X(z)= G(z) 1 − z−4 = 1 − z−2 1 − z−4 = 1 − z−2 (1 − z−2)(1 + z−2) = 1 1 + z−2 Problem 5.7 Whenever X(z) is a polynomial, the best method to invert is to just pick out the coefficients of the polynomial. For (a), we have: X(z) = (1 − 4z−2 )(1 + 3z−1 )= 1 + 3z−1 − 4z−2 − 12z−3 ⇒ x(n) = [1, 3, −4, −12] For (b), the signal is non-zero only at n = −3, 0, 2: x(n)= 5δ(n)+3δ(n + 3)+2δ(n − 2) 62 Problem 5.8 a. Carry out a partial fraction expansion: X(z)= 3(1 + 0.3z−1 ) 1 − 0.81z−2 = 3(1 + 0.3z−1 ) (1 − 0.9z−1)(1 + 0.9z−1) = A 1 − 0.9z−1 + B 1 + 0.9z−1 where A = 3(1 + 0.3z−1 ) 1 + 0.9z−1 z=0.9 = 2, B = 3(1 + 0.3z−1 ) 1 − 0.9z−1 z=−0.9 = 1 The first ROC is |z| > |0.9| = | − 0.9|, thus both terms will be inverted causally: x(n)= A(0.9)n u(n)+B(−0.9)n u(n) As expected, the answer is stable because the ROC contains the unit circle. The second ROC is |z| < |0.9| = | − 0.9|, thus both terms will be inverted anticausally: x(n)= −A(0.9)n u(−n − 1)−B(−0.9)n u(−n − 1) The answer is unstable, because the ROC does not contain the unit circle. b. Ordinary partial fraction expansion is not valid in this case because the degree of the numerator is the same as the degree of the denominator. However, we may still have an expansion of the form: X(z) = 6 − 3z−1 − 2z−2 1 − 0.25z−2 = 6 − 3z−1 − 2z−2 (1 − 0.5z−1)(1 + 0.5z−1) = A + B 1 − 0.5z−1 + C 1 + 0.5z−1 where B and C are determined in the usual manner and A is determined by evaluating X(z) at z = 0: A = 6 − 3z−1 − 2z−2 1 − 0.25z−2 z=0 = 6z2 − 3z − 2 z2 − 0.25 z=0 = −2 −0.25 = 8 B = 6 − 3z−1 − 2z−2 1 + 0.5z−1 z=0.5 = −4, C = 6 − 3z−1 − 2z−2 1 − 0.5z−1 z=−0.5 = 2 For the first ROC, |z| > 0.5, the last two terms will be inverted causally: x(n)= Aδ(n)+B(0.5)n u(n)+C(−0.5)n u(n) For the second ROC, |z| < 0.5, the two terms will be inverted anticausally: x(n)= Aδ(n)−B(0.5)n u(−n − 1)−C(−0.5)n u(−n − 1) As expected, only the first inverse is stable because its ROC contains the unit circle. 63
- 33. c. The degree of the numerator is strictly greater than the degree of the denominator. The simplest approach in such cases is to use the “remove/restore” method, that is, ignore the numerator completely, do a partial fraction on the denominator, get its inverse z- transform, and finally restore the effect of the numerator. To show these steps, write X(z) = 6 + z−5 1 − 0.64z−2 = (6 + z−5 )· 1 1 − 0.64z−2 ≡ (6 + z−5 )W(z)= 6W(z)+z−5 W(z) In the time-domain, we have then x(n)= 6w(n)+w(n − 5) Thus, the problem is reduced to the problem of finding w(n). Doing a partial fraction expansion on W(z), we find W(z)= 1 1 − 0.64z−2 = 1 (1 − 0.8z−1)(1 + 0.8z−1) = A 1 − 0.8z−1 + B 1 + 0.8z−1 where A = B = 0.5. For the ROC |z| > 0.8, the two terms are inverted causally: w(n)= A(0.8)n u(n)+B(−0.8)n u(n) Inserting in x(n)= 6w(n)+w(n − 5), we find x(n) = 6A(0.8)n u(n)+6B(−0.8)n u(n)+A(0.8)n−5 u(n − 5) + B(−0.8)n−5 u(n − 5) Note that the last two terms are active only for n ≥ 5. For the ROC |z| < 0.8, we have the anticausal/unstable answer: w(n)= −A(0.8)n u(−n − 1)−B(−0.8)n u(−n − 1) which gives for x(n): x(n) = −6A(0.8)n u(−n − 1)−6B(−0.8)n u(−n − 1) − A(0.8)n−5 u(−n + 4)−B(−0.8)n−5 u(−n + 4) The u(−n+4) was obtained as u(−(n−5)−1). Note also that the last two terms are now active for −n + 4 ≥ 0 or n ≤ 4, that is, x(n) has a slightly causal part extending to the right up to n = 4. This happened because the strictly anticausal signal w(n) was delayed (shifted to the right) by 5 time units by the term w(n − 5). d. The minor new feature of this problem is that the poles are complex-valued. When the poles are complex, they come in conjugate pairs. In this case, the corresponding residues are complex conjugates, too. Thus, only half of the residues need be computed: X(z) = 10 + z−2 1 + 0.25z−2 = 10 + z−2 (1 − 0.5jz−1)(1 + 0.5jz−1) = A + B 1 − 0.5jz−1 + B∗ 1 + 0.5jz−1 64 Again, the A-term is needed because the degrees of numerator and denominator polyno- mials are equal. We find A = 10 + z−2 10.25z−2 z=0 = 10z2 + 1 z2 + 0.25 z=0 = 1 0.25 = 4 B = 10 + z−2 1 + 0.5jz−1 z=0.5j = 3 We only needed to calculate B, and use B∗ for the conjugate pole. For the causal case, we have x(n)= Aδ(n)+B(0.5j)n u(n)+B∗ (−0.5j)n u(n) Now, because the last two terms are complex conjugates of each other, we may use the general identity 2Re(z)= z + z∗ to write B(0.5j)n u(n)+B∗ (−0.5j)n u(n)= 2Re[B(0.5j)n u(n)]= 6(0.5)n Re[jn ]u(n) But, Re[jn ]= Re[ejπn/2 ]= cos(πn/2) Thus, B(0.5j)n u(n)+B∗ (−0.5j)n u(n)= 6(0.5)n cos(πn/2)u(n) and the final result is x(n)= 4δ(n)+6(0.5)n cos(πn/2)u(n) For the anticausal case, we obtain the anticausal version of the second term, namely, x(n)= 4δ(n)−6(0.5)n cos(πn/2)u(−n − 1) e. The partial fraction expansion is: X(z)= 6 − 2z−1 − z−2 (1 − z−1)(1 − 0.25z−2) = 4 1 − z−1 + 1 1 − 0.5z−1 + 1 1 + 0.5z−1 Because ROC is |z| > 1 > 0.5, all terms are inverted causally to give: x(n)= 4u(n)+(0.5)n u(n)+(−0.5)n u(n) The answer is marginally stable because the pole z = 1 is on the unit circle. f. The PF expansion gives: X(z) = −4 + 1 1 + 4z−2 = −4 + 1 (1 − 2jz−1)(1 + 2jz−1) = −4 + 1/2 1 − 2jz−1 + 1/2 1 + 2jz−1 The two regions of convergence are: |z| > 2 and |z| < 2 65
- 34. They correspond to the causal and anticausal inverses: x1(n) = −4δ(n)+0.5(2j)n u(n)+0.5(−2j)n u(n) x2(n) = −4δ(n)−0.5(2j)n u(−n − 1)+0.5(−2j)n u(−n − 1) Using Euler’s formula, we may write: 0.5 (2j)n +(−2j)n ] = 2n 1 2 [jn + (−j)n ]= 2n 1 2 [ejπn/2 + e−jπn/2 ] = 2n cos(πn/2) Thus, we can rewrite: x1(n) = −4δ(n)+2n cos(πn/2)u(n) x2(n) = −4δ(n)−2n cos(πn/2)u(−n − 1) Only x2(n) is stable because its ROC contains the unit circle, or, because in the second term n is effectively negative and causes it to decay exponentially for large negative n, that is, writing n = −|n|, we have: 2n cos(πn/2)u(−n − 1)= 2−|n| cos(πn/2)u(−n − 1)→ 0 as n → −∞ g. The PF expansion is in this case: X(z)= 4 − 0.6z−1 + 0.2z−2 (1 − 0.5z−1)(1 + 0.4z−1) = A + B 1 − 0.5z−1 + C 1 + 0.4z−1 where A = 4 − 0.6z−1 + 0.2z−2 (1 − 0.5z−1)(1 + 0.4z−1) z=0 = 0.2 −0.5 · 0.4 = −1 B = 4 − 0.6z−1 + 0.2z−2 1 + 0.4z−1 z=0.5 = 2 C = 4 − 0.6z−1 + 0.2z−2 1 − 0.5z−1 z=−0.4 = 3 The three ROC’s are: |z| > 0.5, 0.5 > |z| > 0.4, 0.4 > |z| The corresponding inverses are: x1(n) = Aδ(n)+B(0.5)n u(n)+C(−0.4)n u(n) x2(n) = Aδ(n)−B(0.5)n u(−n − 1)+C(−0.4)n u(n) x3(n) = Aδ(n)−B(0.5)n u(−n − 1)−C(−0.4)n u(−n − 1) Only x1(n) is stable. Its ROC contains the unit circle. The B-term in x2(n) and both the B and C terms of x3(n) diverge exponentially for large negative n. 66 Problem 5.9 Consider the z-transform pair: xa(n)= an u(n) Xa(z)= 1 1 − az−1 Applying the derivative operator ∂/∂a to the pair, derive the z-transform of the sequence x(n)= nan u(n). Noting that ∂ ∂a xa(n)= nan−1 u(n) we get ∂ ∂a Xa(z)= ∂ ∂a 1 1 − az−1 = z−1 (1 − az−1)2 Thus, nan−1 u(n) Z −→ z−1 (1 − az−1)2 Multiplying by a, we get: nan u(n) Z −→ az−1 (1 − az−1)2 Problem 5.10 We note that each application of D brings down a factor of n: D(an ) = a ∂ ∂a (an )= anan−1 = nan D2 (an ) = D(D(an ))= D(nan )= nD(an )= n(nan )= n2 an D3 (an ) = D(D2 (an ))= D(n2 an )= n2 D(an )= n2 (nan )= n3 an · · · Dk (an ) = nk an Thus, in the z-domain: nk an u(n) Z −→ a ∂ ∂a k 1 1 − az−1 In particular, for k = 1, we have: a ∂ ∂a 1 1 − az−1 = az−1 (1 − az−1)2 Therefore, nan u(n) Z −→ az−1 (1 − az−1)2 67
- 35. which agrees with the previous problem. For k = 2, we apply D again to get: a ∂ ∂a 2 1 1 − az−1 = a ∂ ∂a az−1 (1 − az−1)2 = az−1 1 (1 − az)2 + 2az−1 (1 − az−1)3 = az−1 (1 + az−1 ) (1 − az−1)3 Therefore, n2 an u(n) Z −→ az−1 (1 + az−1 ) (1 − az−1)3 Adding the results of the k = 1 and k = 2 cases, we also get: n(n + 1)an u(n) Z −→ 2az−1 (1 − az−1)3 For the cases k = 3, 4, we find after some tedious algebra: n3 an u(n) Z −→ az−1 (1 + 4az−1 + a2 z−2 ) (1 − az−1)4 n4 an u(n) Z −→ az−1 (1 + 11az−1 + 11a2 z−2 + a3 z−3 ) (1 − az−1)5 Problem 5.11 Defining the square pulses g1(n) = u(n)−u(n − L) g2(n) = g1(n + L − 1)= u(n + L − 1)−u(n − 1) and using the z-transform u(n) Z −→ U(z)= 1 1 − z−1 we find the z-transforms: G1(z) = U(z)−z−L U(z)= 1 − z−L 1 − z−1 G2(z) = U(z)zL−1 − z−1 U(z)= 1 − z−L 1 − z−1 zL−1 Therefore, the right-hand-side of the desired z-transform is the product: X(z)= 1 L 1 − z−L 1 − z−1 2 zL−1 = 1 L G1(z)G2(z) Thus, in the time domain we have the convolution: x(n)= 1 L g1(n)∗g2(n)= 1 L m g1(m)g2(n − m) 68 Because g1(n) is non-zero only over 0 ≤ n ≤ L − 1 and g2(n) over −(L − 1)≤ n ≤ 0, we obtain the restrictions on the convolution summation indices: 0 ≤ m ≤ L − 1 −(L − 1) ≤ n − m ≤ 0 By adding them, we obtain the overall range of the index n of x(n), that is, −(L − 1)≤ n ≤ L − 1 For any n in this range, we obtain the range of the summation in m: 0 ≤ m ≤ L − 1 0 ≤ m − n ≤ L − 1 ⇒ n ≤ m ≤ n + L − 1 which combine into the single inequality: max(0, n)≤ m ≤ min(L − 1, n + L − 1) The convolution sum becomes then: x(n)= 1 L min(L−1,n+L−1) m=max(0,n) g1(m)g2(n − m), −(L − 1)≤ n ≤ L − 1 Because over this range, both g1(n) and g2(n) are unity, we obtain: x(n)= 1 L min(L−1,n+L−1) m=max(0,n) 1 · 1 = 1 L max(L − 1, n + L − 1)− min(0, n)+1 By checking separately the ranges −(L − 1)≤ n ≤ 0 and 0 ≤ n ≤ L − 1, it is easily verified that the above expression is equivalent to: x(n)= 1 L max(L − 1, n + L − 1)− min(0, n)+1 = 1 − |n| L , −(L − 1)≤ n ≤ L − 1 At all other n, including the end-points n = ±L, x(n) is zero. Problem 5.12 Start with the result an u(n) Z −→ 1 1 − az−1 and apply it for a = Rejω0 and a∗ = Re−jω0 : Rn ejω0n u(n) Z −→ 1 1 − Rejω0 z−1 Rn e−jω0n u(n) Z −→ 1 1 − Re−jω0 z−1 The cosine and sine signals are obtained from the above as the following linear combinations: Rn cos(ω0n)u(n) = Rn 1 2 ejω0n u(n)+e−jω0n u(n) Rn sin(ω0n)u(n) = Rn 1 2j ejω0n u(n)−e−jω0n u(n) 69
- 36. Therefore, the corresponding linear combinations of z-transforms will be: Rn cos(ω0n)u(n) Z −→ 1 2 1 1 − Rejω0 z−1 + 1 1 − Re−jω0 z−1 Rn sin(ω0n)u(n) Z −→ 1 2j 1 1 − Rejω0 z−1 − 1 1 − Re−jω0 z−1 which simplify into: Rn cos(ω0n)u(n) Z −→ 1 − R cos ω0 z−1 1 − 2R cos ω0 z−1 + R2z−2 Rn sin(ω0n)u(n) Z −→ R sin ω0 z−1 1 − 2R cos ω0 z−1 + R2z−2 Problem 5.13 Define the z-transform of one period: A(z)= a0 + a1z−1 + a2z−2 + a3z−3 Then, the z-transform of the periodic signal is recognized as the sum of the delayed replicas of A(z), that is, X(z) = a0 + a1z−1 + a2z−2 + a3z−3 + a0z−4 + a1z−5 + a2z−6 + a3z−7 + a0z−8 + a1z−9 + a2z−10 + a3z−11 + · · · = (a0 + a1z−1 + a2z−2 + a3z−3 )+z−4 (a0 + a1z−1 + a2z−2 + a3z−3 ) + z−8 (a0 + a1z−1 + a2z−2 + a3z−3 )+ · · · = A(z)+z−4 A(z)+z−8 A(z)+ · · · = (1 + z−4 + z−8 + · · · )A(z)= A(z) 1 − z−4 = a0 + a1z−1 + a2z−2 + a3z−3 1 − z−4 where the convergence of the geometric series requires the ROC |z| > 1. Problem 5.14 Factoring the denominator 1 − z−4 , we obtain the PF expansion: X(z) = a0 + a1z−1 + a2z−2 + a3z−3 1 − z−4 = a0 + a1z−1 + a2z−2 + a3z−3 (1 − z−1)(1 + z)(1 − jz−1)(1 + jz−1) = A 1 − z−1 + B 1 + z−1 + C 1 − jz−1 + C∗ 1 + jz−1 where the coefficients C and C∗ are conjugates because they correspond to the conjugate poles ±j (this follows from the fact that the denominator has real coefficients.) The PFE coefficients are determined as follows: 70 A = a0 + a1z−1 + a2z−2 + a3z−3 (1 + z)(1 − jz−1)(1 + jz−1) z=1 = a0 + a1 + a2 + a3 4 B = a0 + a1z−1 + a2z−2 + a3z−3 (1 − z−1)(1 − jz−1)(1 + jz−1) z=−1 = a0 − a1 + a2 − a3 4 C = a0 + a1z−1 + a2z−2 + a3z−3 (1 − z−1)(1 + z)(1 + jz−1) z=j = (a0 − a2)−j(a1 − a3) 4 ≡ CR − jCI Taking (causal) inverse z-transforms of the PF expansion, we obtain for n ≥ 0: x(n) = A + B(−1)n +Cjn + C∗ (−j)n = A + B(−1)n +Cejπn/2 + C∗ e−jπn/2 = A + B(−1)n +2Re Cejπn/2 = A + B(−1)n +2Re (CR − jCI)ejπn/2 = A + B(−1)n +2CR cos(πn/2)+2CI sin(πn/2) So that finally: x(n) = a0 + a1 + a2 + a3 4 + a0 − a1 + a2 − a3 4 (−1)n + a0 − a2 2 cos(πn/2)+ a1 − a3 2 sin(πn/2) Setting n = 0, 1, 2, 3, we can verify the first four values {a0, a1, a2, a3}. Problem 5.15 Summing the geometric series, we have: X(z)= 1 − z−2 + z−4 − z−6 + z−8 − · · · = 1 1 + z−2 Alternatively, we have: X(z)= 1−z−2 +z−4 −z−6 +z−8 −· · · = 1−z−2 (1−z−2 +z−4 −z−6 +z−8 −· · · )= 1−z−2 X(z) which gives (1 + z−2 )X(z)= 1 ⇒ X(z)= 1 1 + z−2 The inverse z-transform is causal and can be obtained from the PFE: X(z)= 1 1 + z−2 = 0.5 1 − jz−1 + 0.5 1 + jz−1 which gives: x(n)= 0.5jn + 0.5(−j)n u(n)= cos(πn/2)u(n) 71
- 37. Problem 5.16 In all cases, we expand in the appropriate geometric series and pick out the coefficients of the expansion. a. The z-transform and its inverse are: X(z) = 1 1 + z−4 = 1 − z−4 + z−8 − z−12 + z−16 − · · · x(n) = [1, 0, 0, 0, −1, 0, 0, 0, 1, 0, 0, 0, −1, 0, 0, 0, 1, . . . ] b. The z-transform and its inverse are: X(z) = 1 1 − z−4 = 1 + z−4 + z−8 + z−12 + z−16 + · · · x(n) = [1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, . . . ] c. The z-transform and its inverse are: X(z) = 1 1 + z−8 = 1 − z−8 + z−16 − z−24 + · · · x(n) = [1, 0, 0, 0, 0, 0, 0, 0, −1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, −1, . . . ] d. The z-transform and its inverse are: X(z) = 1 1 − z−8 = 1 + z−8 + z−16 + z−24 + · · · x(n) = [1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, . . . ] Problem 5.17 For case (a), the roots of the denominator are the solutions of 1 + z−4 = 0 or, z4 = −1. They are obtained as follows: z4 = −1 = ejπ = ejπ+2jπk ⇒ z = e(jπ+2jπk)/4 , k = 0, 1, 2, 3 that is, the four numbers z = ejπ/4 , e3jπ/4 , e5jπ/4 , e7jπ/4 which are the conjugate pairs: z = e±jπ/4 , e±3jπ/4 Thus, the z-transform factors as follows and has the PF expansion: X(z) = 1 1 + z−3 = 1 (1 − ejπ/4z−1)(1 − e−jπ/4z−1)(1 − e3jπ/4z−1)(1 − e−3jπ/4z−1) = A 1 − ejπ/4z−1 + A∗ 1 − e−jπ/4z−1 + B 1 − e3jπ/4z−1 + B∗ 1 − e−3jπ/4z−1 where the coefficients are: 72 A = 1 (1 − e−jπ/4z−1)(1 − e3jπ/4z−1)(1 − e−3jπ/4z−1) z=eπj/4 = 1 4 B = 1 (1 − ejπ/4z−1)(1 − e−jπ/4z−1)(1 − e−3jπ/4z−1) z=e3πj/4 = 1 4 Thus, we find the causal inverse for n ≥ 0: x(n) = 1 4 ejπn/4 + 1 4 e−jπn/4 + 1 4 e3jπn/4 + 1 4 e−jπn/4 = 1 2 cos(πn/4)+ 1 2 cos(3πn/4) Cases (b,c,d) are done in a similar fashion. See the solution of Problem 5.19 for the denominator zeros. Problem 5.18 Let Q(z) and R(z) be the quotient and remainder polynomials of the division of N(z) by D(z). Then, Q(z) will have degree L − M, R(z) degree M − 1. Thus, N(z)= Q(z)D(z)+R(z) It follows that: H(z)= N(z) D(z) = Q(z)D(z)+R(z) D(z) = Q(z)+ R(z) D(z) Because the degree of R(z) is strictly less than the degree of D(z), the term R(z)/D(z) will admit a PF expansion of the form: R(z) D(z) = M i=1 Ai 1 − piz−1 where pi are the M zeros of D(z) or the poles of H(z). Any truly complex poles will come in conjugate pairs, that is, of the type: Ai 1 − piz−1 + A∗ i 1 − p∗ i z−1 which can be assembled into the 2nd order section: Ai 1 − piz−1 + A∗ i 1 − p∗ i z−1 = (Ai + A∗ i )−(Aip∗ i + A∗ i pi)z−1 1 − (pi + p∗ i )z−1 + pip∗ i z−2 where the numerator and denominator coefficients are all real. 73
- 38. Problem 5.19 The roots of the polynomial 1 − z−D are the solutions of the equation zD = 1, that is, the D-th roots of unity. To find them, replace 1 by e2πjk = 1, where k is an integer, and take D-th roots of both sides: zD = 1 = e2πjk ⇒ zk = ej2πk/D , k = 0, 1, . . . D − 1 They lie on the unit circle at the angles ωk = 2πk/D, k = 0, 1, . . . , D−1 obtained by dividing the circle into D equal parts. The product of the 1st order zero factors must make up the polynomial 1 − z−D : D−1 k=0 (1 − zkz−1 )= 1 − z−D This identity follows from a theorem of algebra that states that if two polynomials have the same set of roots and unity constant coefficient, then they must be identically equal to each other. For the other case, we have the solutions of 1+z−D = 0 or zD = −1. Now, we replace −1 = eπj = ejπ+j2πk with any integer k, and find the roots zD = −1 = ejπ(2k+1) ⇒ zk = ejπ(2k+1)/D , k = 0, 1, . . . , D − 1 They also lie on the unit circle at D equal divisions, but they are shifted as a whole with respect to the standard D-th roots of unity by a rotation angle of π/D. Fig. P5.1 shows the two cases for D = 8. z8 = 1 z8 = -1 Fig. P5.1 Roots of 1 − z−8 and 1 + z−8 . Problem 5.20 The roots of 1 − az−D are the solutions of the equation zD = a = aej2πk ⇒ zk = a1/D ej2πk/D , k = 0, 1, . . . , D − 1 They all lie on a circle of radius a1/D at the D-th root of unity angles. Similarly, the roots of 1 + az−D , lie on the same circle, but at the shifted D-th root of unity angles. Problem 5.21 Let y(n)= an x(n). Its z-transform is: Y(z)= n y(n)z−n = n an x(n)z−n = n x(n)(z/a)−n 74 Comparing with X(z)= n x(n)z−n we see that Y(z)= X(z/a) Since the spectrum X(ω) is obtained by replacing z = ejω , the scaled z will be z/a = ejω /ejω0 = ej(ω−ω0) Thus, the relationship Y(ejω )= X(ej(ω−ω0) ) may be written as Y(ω)= X(ω − ω0) Alternatively, we may work directly with the definition of the DTFT: Y(ω)= n y(n)e−jωn = n ejω0n x(n)e−jωn = n x(n)e−j(ω−ω0)n = X(ω − ω0) Problem 5.22 We use the identity 1 2π π −π ejω(n−m) dω = δ(n − m) for n, m integers. Indeed, for n = m the left hand side is unity. And, for n = m, we have: 1 2π π −π ejω(n−m) dω = cos π(n − m) − cos −π(n − m) 2πj(n − m) = 0 Inserting the definition of the DTFT into the inversion formula, we get 1 2π π −π X(ω)ejωn dω = 1 2π π −π m x(m)e−jωm ejωn dω = m x(m) 1 2π π −π ejω(n−m) dω = m x(m)δ(n − m) = x(n) where we interchanged the series with the integral (this step is justified by assuming that the series is convergent, that is, the unit circle lies in the ROC or that the signal x(n) is strictly stable). Problem 5.23 We may assume in general that x(n) is complex-valued. Then, we have: X(ω) = n x(n)e−jωn X∗ (ω) = n x∗ (n)ejωn 75
- 39. Using the inverse DTFT formula, it follows that 1 2π π −π X(ω)X∗ (ω) dω = 1 2π π −π X(ω) n x∗ (n)ejωn dω = n x∗ (n) 1 2π π −π X(ω)ejωn dω = n x∗ (n)x(n) Problem 5.24 For a real-valued signal x(n), we have: X(ω) = n x(n)e−jωn X∗ (ω) = n x(n)ejωn X∗ (−ω) = n x(n)e−jωn where in the third equation, we replaced ω by −ω. Comparing the first and third equations, we obtain: X∗ (−ω)= X(ω) Writing X(ω) in its polar form, e.g., X(ω)= |X(ω)|ej arg X(ω) , the hermitian property reads: |X(−ω)|ej arg X(−ω) ∗ = |X(ω)|ej arg X(ω) or, |X(−ω)|e−j arg X(−ω) = |X(ω)|ej arg X(ω) Equating magnitudes and phases, we get the conditions: |X(−ω)| = |X(ω)| , arg X(−ω)= − arg X(ω) that is, the magnitude spectrum is even in ω and the phase spectrum is odd. 76 Chapter 6 Problems Problem 6.1 a. From Y(z)= −0.8z−1 Y(z)+X(z) it follows: H(z)= 1 1 + 0.8z−1 ⇒ H(ω)= 1 1 + 0.8e−jω The causal inverse z-transform of H(z) is h(n)= (−0.8)n u(n). There is only one pole at z = −0.8, that is, near the “high frequency” part of the unit circle. Thus, the filter will tend to enhance high frequencies, i.e., it will behave as a high pass filter: = pole z-plane -0.8 1 π ω 0 5 0.55 |H(ω)| Block diagram realization and the sample-by-sample processing algorithm: x -0.8 y y z-1 w1 for each input sample x do: y = −0.8w1 + x w1 = y b. Change −0.8 to 0.8 in the above problem. Now the pole at z = 0.8 is in the “low frequency” part of the unit circle and the filter is acting as a low pass filter. c. The I/O equation Y(z)= 0.8z−1 Y(z)+X(z)+z−1 X(z) gives H(z)= 1 + z−1 1 − 0.8z−1 = A + B 1 − 0.8z−1 ⇒ h(n)= Aδ(n)+B(0.8)n u(n) where A = 1 + z−1 1 − 0.8z−1 z=0 = −1.25, B = 1 + z−1 z=0.8 = 2.25 The filter enhances low frequencies for two reasons: first, it has a pole in the low frequency range, z = 0.8, and it has a zero in the high frequency range, z = −1. Its frequency response will be H(ω)= 1 + e−jω 1 − 0.8e−jω 77
- 40. z-plane 0.8-1 = poles = zeros π ω 0 10 exact zero |H(ω)| The canonical form realization and its sample processing algorithm are: x 0.8 y w0 z-1 w1 for each input sample x do: w0 = 0.8w1 + x y = w0 + w1 w1 = w0 The direct form realization and its sample processing algorithm are: x 0.8 y z-1 z-1 v1 w1 for each input sample x do: y = 0.8w1 + x + v1 v1 = x w1 = y d. The I/O equation Y(z)= 0.8z−1 Y(z)+X(z)−0.5z−1 X(z) gives H(z)= 1 − 0.5z−1 1 − 0.8z−1 = A + B 1 − 0.8z−1 ⇒ h(n)= Aδ(n)+B(0.8)n u(n) where A = 1 − 0.5z−1 1 − 0.8z−1 z=0 = 0.625, B = 1 − 0.5z−1 z=0.8 = 0.375 The filter has a zero at z = 0.5 and a pole at z = 0.8 — both in the low frequency range. Thus, their effect will be to cancel each other, and whichever is closest to the unit circle ultimately wins. Here, the pole is nearer. Thus, the filter will tend to act as a lowpass filter. Indeed, its response at ω = 0 or z = 1 is (1 − 0.5)/(1 − 0.8)= 2.5, whereas its response at ω = π or z = −1 is (1 + 0.5)/(1 + 0.8)= 0.833. z-plane 0.8 0.5 = poles = zeros π ω 0 2.5 0.833 |H(ω)| The canonical form realization and its sample processing algorithm are: 78 x 0.8 -0.5 y w0 z-1 w1 for each input sample x do: w0 = 0.8w1 + x y = w0 − 0.5w1 w1 = w0 The direct form realization and its sample processing algorithm are: x 0.8-0.5 y z-1 z-1 v1 w1 for each input sample x do: y = 0.8w1 + x − 0.5v1 v1 = x w1 = y e. Y(z)= 0.8z−1 Y(z)+X(z)+0.25z−2 X(z) ⇒ H(z)= 1 + 0.25z−2 1 − 0.8z−1 . Using the “remove/restore numerator” method of problem (2h), we find h(n)= (0.8)n u(n)+0.25(0.8)n−2 u(n − 2) The filter has two conjugate zeros at midrange, z = ±0.5j = 0.5e±jπ/2 , and a low frequency pole at z = 0.8. Thus, the filter will enhance low frequencies and suppress midrange frequencies: z-plane 0.8 0.5j -0.5j = poles = zeros π ω 0 |H(ω)| The canonical form realization and its sample processing algorithm are: x 0.8 0.25 y w0 z-1 z-1 w1 w2 for each input sample x do: w0 = 0.8w1 + x y = w0 + 0.25w2 w2 = w1 w1 = w0 The direct form realization and its sample processing algorithm are: 79
- 41. x 0.8 0.25 y z-1 z-1 z-1 v1 v2 w1 for each input sample x do: y = x + 0.25v2 + 0.8w1 v2 = v1 v1 = x w1 = y f. Y(z)= 0.9z−1 Y(z)−0.2z−2 Y(z)+X(z)+z−1 X(z)−6z−2 X(z), which implies H(z)= 1 + z−1 − 6z−2 1 − 0.9z−1 + 0.2z−2 Factor numerator and denominator and expand in partial fractions: H(z)= (1 + 3z−1 )(1 − 2z−1 ) (1 − 0.4z−1)(1 − 0.5z−1) = A + B 1 − 0.4z−1 + C 1 − 0.5z−1 where A = (1 + 3z−1 )(1 − 2z−1 ) (1 − 0.4z−1)(1 − 0.5z−1) z=0 = −30 B = (1 + 3z−1 )(1 − 2z−1 ) 1 − 0.5z−1 z=0.4 = 136 C = (1 + 3z−1 )(1 − 2z−1 ) 1 − 0.4z−1 z=0.5 = −105 Thus, h(n)= Aδ(n)+B(0.4)n u(n)+C(0.5)n u(n). The two zeros at z = −3 and z = 2 are too far from the unit circle to have any significant effect on the magnitude response. The two poles at z = 0.4, 0.5 are both in the low frequency range. Thus, the filter will be a low pass filter. The value of the magnitude response at ω = 0 or z = 1 is |1+1−6|/|1−0.5||1− 0.4| = 13.33, whereas its value at ω = π or z = −1 is |1 −1 −6|/|1 +0.5||1 +0.4| = 2.86. z-plane 0.5 0.4 -3 2 = poles = zeros π ω 0 13.33 2.86 |H(ω)| The canonical form realization and its sample processing algorithm are: 80 x 0.9 -6-0.2 y w0 z-1 z-1 w1 w2 for each input sample x do: w0 = x + 0.9w1 − 0.2w2 y = w0 + w1 − 6w2 w2 = w1 w1 = w0 The direct form realization and its sample processing algorithm are: x 0.9 -6 -0.2 y z-1 z-1 z-1 z-1 v1 v2 w1 w2 for each input sample x do: y = x + v1 − 6v2 + 0.9w1 − 0.2w2 v2 = v1 v1 = x w2 = w1 w1 = y Problem 6.2 a. Setting x(n)= u(n) in the I/O difference equation, we find y(n)= u(n)+6u(n − 1)+11u(n − 2)+6u(n − 3) The 4th term is active only for n ≥ 3, the 3d term is active only n ≥ 2, and the 2nd term only for n ≥ 2. Thus, evaluating at a few n’s we find: y(n)= [1, 7, 18, 24, 24, 24, 24, . . . ] The first 3 outputs are the initial transients, the remaining constant values are the steady part. Note that when you send in a unit step, the output always settles to a constant value (for a stable filter). That constant value can be easily precalculated as H(1). In the present case, H(1)= 1 + 6 + 11 + 6 = 24. For the alternating step, we have: Y(z) = 1 + 6z−1 + 11z−2 + 6z−3 1 + z−1 = 1 + 5z−1 + 6z−2 y(n) = [1, 5, 6, 0, 0, 0, · · · ] There is a pole zero cancellation: the filter has a zero at z = −1, cutting off the high- frequency input. b. Noting that the input z-transform is X(z)= 1/(1 − z−1 ), we find Y(z)= H(z)X(z)= 1 − z−4 1 − z−1 = 1 + z−1 + z−2 + z−3 where we used the finite geometric series. Inverting, we get 81
- 42. y(n)= [1, 1, 1, 1, 0, 0, 0, 0, . . . ] The first four samples are the transients, the remaining zeros are the steady state. In the steady-state, the filter cuts off the unit step from going through because the filter has a zero at z = 1. The pole of the input canceled the zero of the filter. For the alternating step: Y(z)= 1 − z−4 1 + z−1 = 1 − z−1 + z−2 − z−3 ⇒ y(n)= [1, −1, 1, −1, 0, 0, 0, · · · ] Again, the filter has a high frequency zero at z = −1, cutting off the high frequency input. Problem 6.3 For filter (a), we have: Y(z) = 1 (1 − z−1)(1 − 0.25z−2) = 1 (1 − z−1)(1 − 0.5z−1)(1 + 0.5z−1) = A 1 − z−1 + B 1 − 0.5z−1 + C 1 + 0.5z−1 y(n) = Au(n)+B(0.5)n u(n)+C(−0.5)n u(n) where A = H(1)= 4/3, B = −0.5, C = 1/6. The A-term is the steady part. For the alternating step, we have: Y(z) = 1 (1 + z−1)(1 − 0.25z−2) = 1 (1 + z−1)(1 − 0.5z−1)(1 + 0.5z−1) = A 1 + z−1 + B 1 − 0.5z−1 + C 1 + 0.5z−1 y(n) = A(−1)n u(n)+B(0.5)n u(n)+C(−0.5)n u(n) where A = H(−1)= 4/3, B = 1/6, C = −0.5. The A-term is the steady part. For (b), we have: Y(z) = 1 (1 − z−1)(1 + 0.25z−2) = 1 (1 − z−1)(1 − 0.5jz−1)(1 + 0.5jz−1) = A 1 − z−1 + B 1 − 0.5jz−1 + B∗ 1 + 0.5jz−1 y(n) = Au(n)+B(0.5j)n u(n)+B∗ (−0.5j)n u(n) where A = H(1)= 4/5 and B = 1 (1 − z−1)(1 + 0.5jz−1) z=0.5j = 0.1 − 0.2j And, for the alternating step: Y(z) = 1 (1 + z−1)(1 + 0.25z−2) = 1 (1 + z−1)(1 − 0.5jz−1)(1 + 0.5jz−1) = A 1 + z−1 + B 1 − 0.5jz−1 + B∗ 1 + 0.5jz−1 y(n) = A(−1)n u(n)+B(0.5j)n u(n)+B∗ (−0.5j)n u(n) 82 where A = H(−1)= 4/5 and B = 1 (1 + z−1)(1 + 0.5jz−1) z=0.5j = 0.1 + 0.2j Problem 6.4 For a unit-step input and the filter (a) of Problem 6.1, we have: Y(z) = H(z)X(z)= 1 (1 + 0.8z−1)(1 − z−1) = A 1 + 0.8z−1 + B 1 − z−1 y(n) = A(−0.8)n u(n)+Bu(n) where A = 1/2.25, B = H(1)= 1/1.8. The B-term represents the steady part. For filter (b): Y(z) = H(z)X(z)= 1 (1 − 0.8z−1)(1 − z−1) = A 1 − 0.8z−1 + B 1 − z−1 y(n) = A(0.8)n u(n)+Bu(n) where A = −4, B = H(1)= 5. The B-term represents the steady part. For filter (c): Y(z) = 1 + z−1 (1 − z−1)(1 − 0.8z−1) = A 1 − z−1 + B 1 − 0.8z−1 y(n) = Au(n)+B(0.8)n u(n) with A = H(1)= 10, B = −9. For filter (d): Y(z) = 1 − 0.5z−1 (1 − z−1)(1 − 0.8z−1) = A 1 − z−1 + B 1 − 0.8z−1 y(n) = Au(n)+B(0.8)n u(n) with A = H(1)= 2.5, B = −1.5. For filter (e): Y(z) = 1 + 0.25z−2 (1 − z−1)(1 − 0.8z−1) = A + B 1 − z−1 + C 1 − 0.8z−1 y(n) = Aδ(n)+Bu(n)+C(0.8)n u(n) where A = Y(0)= 0.3125, B = H(1)= 6.25, C = −5.5625. The B-term is the steady part. For filter (f): Y(z) = 1 + z−1 − 6z−2 (1 − z−1)(1 − 0.4z−1)(1 − 0.5z−1) = A 1 − z−1 + B 1 − 0.4z−1 + C 1 − 0.5z−1 y(n) = Au(n)+B(0.4)n u(n)+C(0.5)n u(n) For the alternating step and filter (a) of Problem 6.1, we have: Y(z) = 1 (1 + 0.8z−1)(1 + z−1) = A 1 + 0.8z−1 + B 1 + z−1 y(n) = A(−0.8)n u(n)+B(−1)n u(n) 83
- 43. where A = −4, B = H(−1)= 5. The B-term represents the steady part. For filter (b): Y(z) = 1 (1 − 0.8z−1)(1 + z−1) = A 1 − 0.8z−1 + B 1 + z−1 y(n) = A(0.8)n u(n)+B(−1)n u(n) where A = 1/2.25, B = H(−1)= 1/1.8. The B-term represents the steady part. For filter (c): Y(z) = 1 + z−1 (1 + z−1)(1 − 0.8z−1) = 1 1 − 0.8z−1 y(n) = (0.8)n u(n) The high frequency zero at z = −1 of the filter canceled the high frequency input signal. The output is only transient — it decays to zero exponentially. For filter (d): Y(z) = 1 − 0.5z−1 (1 + z−1)(1 − 0.8z−1) = A 1 + z−1 + B 1 − 0.8z−1 y(n) = A(−1)n u(n)+B(0.8)n u(n) For filter (e): Y(z) = 1 + 0.25z−2 (1 + z−1)(1 − 0.8z−1) = A + B 1 + z−1 + C 1 − 0.8z−1 y(n) = Aδ(n)+B(−1)n u(n)+C(0.8)n u(n) The B-term is the steady part. For filter (f): Y(z) = 1 + z−1 − 6z−2 (1 + z−1)(1 − 0.4z−1)(1 − 0.5z−1) = A 1 + z−1 + B 1 − 0.4z−1 + C 1 − 0.5z−1 y(n) = A(−1)n u(n)+B(0.4)n u(n)+C(0.5)n u(n) For question (c), the input z-transform is X(z)= 1/(1 − 0.5z−1 ). Thus, applied to Problem 6.1(d) gives: Y(z) = 1 − 0.5z−1 (1 − 0.5z−1)(1 − 0.8z−1) = 1 1 − 0.8z−1 y(n) = (0.8)n u(n) The filter zero canceled the signal pole. For question (d), we have x(n)= (0.5)n cos(πn/2)u(n) Z −→ X(z)= 1/(1 + 0.25z−2 ). Thus, Y(z) = 1 + 0.25z−2 (1 + 0.25z−2)(1 − 0.8z−1) = 1 1 − 0.8z−1 y(n) = (0.8)n u(n) Again, the signal poles were canceled by the filter zeros. For question (e), the input has X(z)= 1/(1 − 2z−1 ). Therefore, 84 Y(z) = (1 + 3z−1 )(1 − 2z−1 ) (1 − 2z−1)(1 − 0.4z−1)(1 − 0.5z−1) = 1 + 3z−1 (1 − 0.4z−1)(1 − 0.5z−1) = A 1 − 0.4z−1 + B 1 − 0.5z−1 y(n) = A(0.4)n u(n)+B(0.5)n u(n) The unstable input was canceled by a filter zero to give a stable output. Problem 6.5 Filter (a): H(z)= z−5 with I/O equation Y(z)= H(z)X(z)= z−5 X(z) and in the time domain y(n)= x(n − 5). H(ω)= e−5jω and |H(ω)| = 1, i.e., flat in ω. Block diagram realization and sample-by-sample processing algorithm: x y z-1 z-1 z-1 z-1 z-1 w1 w2 w5 w3 w4 for each input sample x do: y = w5 w5 = w4 w4 = w3 w3 = w2 w2 = w1 w1 = x Filter (b): H(z)= z−5 U(z)= z−5 /(1 − z−1 ) and the I/O equation gives Y(z)= H(z)X(z)= z−5 X(z) 1 − z−1 ⇒ (1 − z−1 )Y(z)= z−5 X(z) ⇒ Y(z)= z−1 Y(z)+z−5 X(z) ⇒ y(n)= y(n − 1)+x(n − 5) Its frequency response is H(ω)= e−5jω 1 − e−jω ⇒ |H(ω)| = 1 2| sin(ω/2)| The pole at z = 1 or ω = 0 dominates the frequency response. The filter acts as a lowpass filter (it is actually an accumulator with delay). 85
- 44. z-plane 1 = poles π ω 0 0.5 |H(ω)| The canonical form realization and its sample processing algorithm are: x y w0 z-1 z-1 z-1 z-1 z-1 w1 w2 w5 w3 w4 for each input sample x do: w0 = x + w1 y = w5 w5 = w4 w4 = w3 w3 = w2 w2 = w1 w1 = w0 The direct form realization and its sample processing algorithm are: x y z-1 z-1 z-1 z-1 z-1 z-1 v1 w1 v2 v5 v3 v4 for each input sample x do: y = w1 + v5 w1 = y v5 = v4 v4 = v3 v3 = v2 v2 = v1 v1 = x Filter (c): H(z)= 1/(1 − 0.8z−1 ) and the I/O equation becomes: Y(z)= H(z)X(z)= X(z) 1 − 0.8z−1 ⇒ (1 − 0.8z−1 )Y(z)= X(z) ⇒ Y(z)= 0.8z−1 Y(z)+X(z) 86 and in the time domain y(n)= 0.8y(n − 1)+x(n) The rest is as in problem (3d). Filter (d): H(z)= 1/(1 + 0.8z−1 ) and the I/O equation becomes: Y(z)= H(z)X(z)= X(z) 1 + 0.8z−1 ⇒ (1 + 0.8z−1 )Y(z)= X(z) ⇒ Y(z)= −0.8z−1 Y(z)+X(z) ⇒ y(n)= −0.8y(n − 1)+x(n) The rest is as in problem (3c). Filter (e): With a = −0.8, the impulse response is finite: h(n)= [1, a, a2 , a3 , a4 , a5 , a6 , a7 , 0, 0, 0, · · · ] Its z-transform is H(z)= 1 + az−1 + a2 z−2 + a3 z−3 + a4 z−4 + a5 z−5 + a6 z−6 + a7 z−7 = 1 − a8 z−8 1 − az−1 The point z = a is not really a pole of H(z) because it is canceled by a numerator zero — H(z) has seven zeros. To find them, solve the equation z8 = a8 and exclude the solution z = a, that is, z = ae2πjk/8 = −0.8e2πjk/8 = 0.8ejπ e2πjk/8 , k = 1, 2, 3, 4, 5, 6, 7 They are shown below z-plane = zeros z=a=-0.8 1 The I/O equation in the time domain is y(n)= x(n)+ax(n − 1)+a2 x(n − 2)+ · · · + a7 x(n − 7) Sample-by-sample processing algorithm and block diagram realization: for each input sample x do: y = x + aw1 + a2 w2 + a3 w3 + a4 w4 + a5 w5 + a6 w6 + a7 w7 w7 = w6 w6 = w5 w5 = w4 w4 = w3 w3 = w2 w2 = w1 w1 = x 87
- 45. x y a0 a1 w1 a2 w2 a3 w3 a4 w4 a5 w5 a6 w6 a7 w7 z-1 z-1 z-1 z-1 z-1 z-1 z-1 Filter (f): H(z)= 1 1 − 0.8z−1 + 1 1 + 0.8z−1 = 2 1 − 0.64z−2 Y(z) = H(z)X(z)= 2X(z) 1 − 0.64z−2 ⇒ Y(z)= 0.64z−2 Y(z)+2X(z) y(n) = 0.64y(n − 2)+2x(n) There is a low frequency pole at z = 0.8 and a high frequency pole at z = −0.8. Thus, the filter will emphasize both the low and high frequencies: z-plane 0.8-0.8 = poles ππ/2 ω 0 |H(ω)| The direct form realization and its sample processing algorithm are: x 0.64 y y z-1 z-1 w2 w1 2 for each input sample x do: y = 2x + 0.64w2 w2 = w1 w1 = y Filters (g,h): h(n) = 2(0.8)n cos(πn/2)u(n)= (0.8)n 2Re[jn ]u(n) = (0.8j)n u(n)+(−0.8j)n u(n) H(z) = 1 1 − 0.8jz−1 + 1 1 + 0.8jz−1 = 2 1 + 0.64z−2 Y(z) = H(z)X(z)= 2X(z) 1 + 0.64z−2 ⇒ Y(z)= −0.64z−2 Y(z)+2X(z) y(n) = −0.64y(n − 2)+2x(n) 88 There is a conjugate pair of poles at midfrequency, z = ±0.8j = 0.8e±π/2 . Thus, the filter will emphasize mid frequencies: z-plane 0.8j -0.8j = poles ππ/2 ω 0 |H(ω)| The direct form realization and its sample processing algorithm are: x -0.64 y y z-1 z-1 w2 w1 2 for each input sample x do: y = 2x − 0.64w2 w2 = w1 w1 = y Problem 6.6 Taking z-transforms, we find: H(z)= 1 + 0.5z−8 + 0.52 z−16 + 0.53 z−24 + · · · = 1 1 − 0.5z−8 where convergence requires |z|8 > 0.5, or, |z| > (0.5)1/8 . The block diagram and sample pro- cessing algorithm are: x 0.5 y z-8 w8 w0 for each input sample x do: y = w0 = x + 0.5w8 delay(8, w) where the call to delay updates the 9-dimensional state vector of the eightfold delay w = [w0, w1, . . . , w8]. Problem 6.7 Because z = ejω , we can deduce the transfer function from H(ω): H(z)= −0.5 + z−8 1 − 0.5z−8 It can be written in a form that can be expanded in powers of z−8 : H(z)= −0.5 + 0.75z−8 1 − 0.5z−8 = −0.5 + 0.75z−8 1 + 0.5z−8 + 0.52 z−16 + 0.53 z−24 + · · · 89
- 46. Thus, the impulse response will be h = [−0.5, 0, 0, 0, 0, 0, 0, 0, 0.75, 0, 0, 0, 0, 0, 0, 0, 0.75(0.5), 0, 0, 0, 0, 0, 0, 0, 0.75(0.5)2 , . . . ] It is depicted below: -0.5 0 0.75 0.75/2 0.75/4 n h(n) ... Problem 6.8 a. The impulse response h(n) satisfies the difference equation: h(n)= h(n − 1)+h(n − 2)+δ(n − 1) (P6.1) Indeed, for n ≥ 2 the δ(n − 1) term is absent and each h(n) is the sum of the previous two. The sequence is properly initialized at n = 0, 1: h(0) = h(−1)+h(−2)+δ(−1)= 0 + 0 + 0 = 0 h(1) = h(0)+h(−1)+δ(0)= 0 + 0 + 1 = 1 Next, we take z-transforms of both sides of Eq. (P6.1) to get H(z): H(z)= z−1 H(z)+z−2 H(z)+z−1 because the z-transform of δ(n − 1) is z−1 . Solving for H(z): H(z)= z−1 1 − z−1 − z−2 b. The poles of H(z) are the roots of 1 − z−1 − z−2 = 0 or z2 − z − 1 = 0 ⇒ z = 1 ± √ 5 2 Denoting the first one by φ, the other one will be: φ = 1 + √ 5 2 , − 1 φ = 1 − √ 5 2 Indeed, we have − 1 φ = − 2 1 + √ 5 = − 2(1 − √ 5) (1 + √ 5)(1 − √ 5) = − 2(1 − √ 5) 1 − 5 = 1 − √ 5 2 Performing a partial fraction expansion on H(z), we get: H(z)= z−1 1 − z−1 − z−2 = z−1 (1 − φz−1)(1 + φ−1z−1) = A 1 − φz−1 + B 1 + φ−1z−1 90 where A = z−1 1 + φ−1z−1 z=φ = φ−1 1 + φ−2 = 1 √ 5 B = z−1 1 − φz−1 z=−φ−1 = − φ 1 + φ2 = − 1 √ 5 Taking the causal inverse z-transform, we obtain a closed-form solution for the nth Fi- bonacci number: h(n)= Aφn + B(−φ)−n , n ≥ 0 It is unstable because φ is greater than 1 (note, φ 1.618). c. For large n, the exponentially increasing term φn dominates the expression. Therefore, the ratio of two successive Fibonacci numbers will tend to: h(n + 1) h(n) → Aφn+1 Aφn = φ This can also be seen from the difference equation of h(n). For n ≥ 1, we have h(n + 1)= h(n)+h(n − 1) Dividing by h(n − 1), we have: h(n + 1) h(n − 1) = h(n) h(n − 1) + 1 or, h(n + 1) h(n) · h(n) h(n − 1) = h(n) h(n − 1) + 1 If we denote the ratio r(n)= h(n + 1)/h(n), we may write this as: r(n)r(n − 1)= r(n − 1)+1 The limiting value r(n)→ r will satisfy the limiting equation: r2 = r + 1 which is the same as that satisfied by φ. Thus, r = φ because they are the positive solutions. d. The sequence y(n) satisfies the recursion y(n)= y(n − 1)+y(n − 2) for n ≥ 3. Indeed, using the property φ2 = φ + 1 and successively multiplying it by powers of φ, we get y(3) = φ2 = φ + 1 = y(2)+y(1) y(4) = φ3 = φ2 + φ = y(3)+y(2) y(5) = φ4 = φ3 + φ2 = y(4)+y(3) · · · 91
- 47. The z-transform of y(n) will be the once-delayed z-transform of the geometric series, that is, Y(z)= z−1 1 − φz−1 Thinking of Y(z) as the output of H(z) for a particular input X(z), we have Y(z)= H(z)X(z), which may be solved for X(z): X(z)= Y(z) H(z) = z−1 1 − φz−1 z−1 (1 − φz−1)(1 + φ−1z−1) = 1 + φ−1 z−1 which gives the sequence: x(n)= δ(n)+φ−1 δ(n − 1) or x = [1, φ−1 , 0, 0, 0, 0 . . . ] Problem 6.9 As in the previous problem, the impulse response satisfies the difference equation: h(n)= 2h(n − 1)+h(n − 2)+δ(n − 1) with zero initial conditions. That is, the two start-up values are: h(0) = 2h(−1)+h(−2)+δ(−1)= 0 + 0 + 0 = 0 h(1) = 2h(0)+h(−1)+δ(0)= 0 + 0 + 1 = 1 Taking z-transforms, we find H(z): H(z)= 2z−1 H(z)+z−2 H(z)+z−1 ⇒ H(z)= z−1 1 − 2z−1 − z−2 The roots of the denominator polynomial are the solutions of z2 = 2z + 1 ⇒ z = 1 ± √ 2 = {θ, −θ−1 } The PF expansion is then H(z)= z−1 1 − 2z−1 − z−2 = z−1 (1 − θz−1)(1 + θ−1z−1) = A 1 − θz−1 + B 1 + θ−1z−1 where A = −B = θ θ2 + 1 = 1 2 √ 2 Thus, the nth Pell number will be: h(n)= Aθn + B(−θ)−n , n ≥ 0 Because the exponentially diverging term θn dominates the sum, the ratio of two successive numbers will converge to 92 h(n + 1) h(n) → Aθn+1 Aθn = θ The y(n) sequence has z-transform Y(z)= z−1 1 − θz−1 = z−1 (1 − θz−1)(1 + θ−1z−1) · (1 + θ−1 z−1 )= H(z)(1 + θ−1 z−1 ) and therefore, it may be thought of as the output when the input is: X(z)= 1 + θ−1 z−1 or, x(n)= δ(n)+θ−1 δ(n − 1) ⇒ x = [1, θ−1 , 0, 0, 0, 0 . . . ] Problem 6.10 The input and output z-transforms are: X1(z)= 1 1 − 0.5z−1 , Y1(z)= 1 1 − 0.5z−1 + 1 1 − 0.4z−1 = 2(1 − 0.45z−1 ) (1 − 0.5z−1)(1 − 0.4z−1) It follows that transfer function is: H(z)= Y1(z) X1(z) = 2(1 − 0.45z−1 ) 1 − 0.4z−1 Therefore, the second output, having z-transform Y2(z)= 1 1 − 0.4z−1 will be produced by X2(z)= Y2(z) H(z) = 0.5 1 − 0.45z−1 ⇒ x2(n)= 0.5(0.45)n u(n) Problem 6.11 The input and output z-transforms are: X1(z) = 1 1 − az−1 Y1(z) = 1 1 − az−1 + 1 1 − bz−1 = 2 − (a + b)z−1 ) (1 − az−1)(1 − bz−1) = 2(1 − cz−1 ) (1 − az−1)(1 − bz−1) where c = (a + b)/2. Therefore, the transfer function is H(z)= Y1(z) X1(z) = 2(1 − cz−1 ) 1 − bz−1 If the input x2(n)= cn u(n) is applied it will have z-transform X2(z)= 1/(1 − cz−1 ) and will cause the output: Y2(z)= H(z)X2(z)= 2(1 − cz−1 ) 1 − bz−1 · 1 1 − cz−1 = 2 1 − bz−1 and in the time domain: y2(n)= 2bn u(n) 93
- 48. Problem 6.12 The input and output z-transforms are: X(z) = 1 1 − 0.7z−1 Y(z) = 1 1 − 0.7z−1 + 1 1 − 0.5z−1 = 2(1 − 0.6z−1 ) (1 − 0.7z−1)(1 − 0.5z−1) Therefore the transfer function will be: H(z)= Y(z) X(z) = 2(1 − 0.6z−1 ) 1 − 0.5z−1 Expanding in partial fractions, we have: H(z)= 2(1 − 0.6z−1 ) 1 − 0.5z−1 = 2.4 − 0.4 1 − 0.5z−1 which has the causal inverse: h(n)= 2.4δ(n)−0.4(0.5)n u(n) Problem 6.13 The normalized peak and width frequencies are: ω0 = 2πf0 fs = 2π250 5000 = 0.1π [rads/sample] Δω = 2πΔf fs = 2π20 5000 = 0.008π [rads/sample] Therefore, the Q-factor of the filter will be: Q = ω0 Δω = 0.1π 0.008π = 12.5 We take the filter poles to be at locations: p = Rejω0 , p∗ = Re−jω0 The pole radius R may be determined from the approximate relationship Δω = 2(1 − R) which gives: R = 1 − 1 2 Δω = 1 − 1 2 0.008π = 0.9874 The denominator polynomial will be: 1 + a1z−1 + a2z−2 = (1 − pz−1 )(1 − p∗ z−1 )= 1 − (p + p∗ )z−1 + p∗ pz−2 which gives: a1 = −(p + p∗ )= −2R cos ω0 = −2 · 0.9874 · cos(0.1π)= −1.8781 a2 = p∗ p = R2 = 0.9750 94 Therefore, the designed filter will be: H(z)= 1 1 − 1.8781z−1 + 0.9750z−2 The time constant of the resonator is given in terms of the maximum pole radius (there is only one radius here): neff = ln ln R = ln(0.01) ln(0.9874) = 363.18 The approximate expression is obtained by replacing R in terms of Δω: neff = ln ln R = ln ln(1 − Δω/2) ln −Δω/2 = − 2 ln Δω where we used the small-x approximation ln(1 − x) −x. The approximate numerical value of the time constant will be: neff = − 2 ln Δω = − 2 ln(0.01) 0.008π = 366.47 which compares well with the exact value. Problem 6.14 The -level time constant of a filter is (in sampling instants): neff = ln ln ρ where ρ = maxi |pi| is the maximum pole radius. In seconds, the time constant is: τ = neffT = ln ln ρ T where T = 1/fs is the interval between samples. It follows that the 1 and 2 level time constants will be τ1 = ln 1 ln ρ T , τ2 = ln 2 ln ρ T And therefore, τ1 = ln 1 ln 2 τ2 In particular, for the 60 versus 40 dB time constants, the scale factor is: ln 1 ln 2 = ln(0.001) ln(0.01) = 3 2 = 1.5 Problem 6.15 Using the approximate expression of Problem 6.13 and = 0.001, we have: τ = neffT = − 2 ln Δω T = − 2 ln 2πΔfT T = − ln πΔf which gives: τ = − ln πΔf − ln(0.001) πΔf = 2.2 Δf 95
- 49. Problem 6.16 The periodic impulse response may be thought of as the sum of the delayed replicas of one period, that is, h(n)= g(n)+g(n − 8)+g(n − 16)+ · · · where g(n)= [1, 2, 3, 4, 0, 0, 0, 0] is the basic period of length 8. Taking z-transforms we obtain: H(z)= G(z)+z−8 G(z)+z−16 G(z)+ · · · = (1 + z−8 + z−16 + · · · )G(z) where G(z)= 1 + 2z−1 + 3z−2 + 4z−3 . Using the geometric series on the first factor we obtain: H(z)= G(z) 1 − z−8 = 1 + 2z−1 + 3z−2 + 4z−3 1 − z−8 The direct and canonical realizations are shown in Fig. P6.1. Note that in the canonical realization the total number of delays is 8, but the last 5 are lumped together into a fivefold delay z−5 . The sample processing algorithm for the canonical case is: x y 2 3 4 z-1 z-1 z-1 x y 2 3 4 w0 w1 w2 w3 w4 z-1 z-1 z-1 z-1 w5 w6 w7 w8 z-1 z-1 z-1 z-1 z-1 z-1 z-1 z-1 z-1 z-1 z-1 z-1 Fig. P6.1 Filter realizations of Problem 6.16. for each input x do: w0 = x + w8 y = w0 + 2w1 + 3w2 + 4w3 delay(8, w) where the call to delay(8, w) implements the updating of the delay line, that is, w8 = w7, w7 = w6, . . . , w1 = w0. Problem 6.17 The solution is similar to that of Problem 6.16. Here, the repetition period is 4 and we have: h(n)= g(n)+g(n − 4)+g(n − 8)+g(n − 12)+ · · · 96 where g(n)= [0, 1, 2, 3]. In the z-domain, we have: H(z)= G(z)+z−4 G(z)+z−8 G(z)+ · · · = (1 + z−4 + z−8 + · · · )G(z)= G(z) 1 − z−4 where G(z)= z−1 + 2z−2 + 3z−3 . Thus, H(z)= z−1 + 2z−2 + 3z−3 1 − z−4 The direct and canonical realizations are shown in Fig. P6.2. x y 2 3 z-1 z-1 z-1 x y 2 3 w0 w1 w2 w3 w4 z-1 z-1 z-1 z-1 z-1 z-1 z-1 z-1 Fig. P6.2 Direct and canonical realizations of Problem 6.17. The sample processing algorithm for the canonical case is: for each input x do: w0 = x + w4 y = w1 + 2w2 + 3w3 delay(4, w) where the call to delay(4, w) implements the updating of the delay line, that is, w4 = w3, w3 = w2, . . . , w1 = w0. The cascade realization is obtained by factoring the numerator and denominator into (up to) 2nd order polynomials: H(z)= z−1 + 2z−2 + 3z−3 1 − z−4 = z−1 1 − z−2 1 + 2z−1 + 3z−2 1 + z−2 ≡ H0(z)H1(z) The block diagram is shown in Fig. P6.3. The corresponding sample processing algorithm is: for each input x do: w0 = x + w2 x1 = w1 w2 = w1 w1 = w0 v0 = x1 − v2 y = v0 + 2v1 + 3v3 v2 = v1 v1 = v0 Using partial fractions, we may write: 97
- 50. x w0 w1 w2 z-1 z-1 x1 y 3 -1 2 v0 v1 v2 z-1 z-1 H0(z) H1(z) Fig. P6.3 Cascade realization of Problem 6.17. H(z) = z−1 + 2z−2 + 3z−3 1 − z−4 = z−1 + 2z−2 + 3z−3 (1 − z−1)(1 + z−1)(1 − jz−1)(1 + jz−1) = A 1 − z−1 + B 1 + z−1 + C − jD 1 − jz−1 + C + jD 1 + jz−1 where the PFE coefficients C ± jD are conjugates. The numerical values are: A = z−1 + 2z−2 + 3z−3 (1 + z−1)(1 − jz−1)(1 + jz−1) z=1 = 1.5 B = z−1 + 2z−2 + 3z−3 (1 − z−1)(1 − jz−1)(1 + jz−1) z=−1 = −0.5 C − jD = z−1 + 2z−2 + 3z−3 (1 − z−1)(1 + z−1)(1 + jz−1) z=j = −0.5 + j0.5 Taking (causal) inverse z-transforms, we find for n ≥ 0: h(n) = A + B(−1)n +(C − jD)jn + (C + jD)(−j)n = A + B(−1)n +2Re (C − jD)ejπn/2 = A + B(−1)n +2C cos(πn/2)+2D sin(πn/2) = 1.5 − 0.5(−1)n − cos(πn/2)− sin(πn/2) Evaluating at the first couple of periods, we find: h(n)= [0, 1, 2, 3, 0, 1, 2, 3, . . . ] Problem 6.18 Expand the denominator in powers of z−7 and pick out the coefficients of the powers of z−1 . We have: H(z)= 1 + z−1 + z−2 + z−3 1 − z−7 = (1 + z−1 + z−2 + z−3 )(1 + z−7 + z−14 + · · · ) The powers of z−7 cause the period-7 replication of the basic period G(z)= 1 + z−1 + z−2 + z−3 : h(n)= [1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, . . . ] 98 x y w0 w1 w2 w3 w4 z-1 z-1 z-1 z-1 w5 w6 w7 z-1 z-1 z-1 Fig. P6.4 Canonical realizations of Problem 6.18. The canonical realization is shown in Fig. P6.4. The sample processing algorithm is: for each input x do: w0 = x + w7 y = w0 + w1 + w2 + w3 delay(7, w) where the call to delay implements the delay-line updates w7 = w6, w6 = w5, . . . , w1 = w0. The output due to the input x = [3, 2, 1] may be obtained by using the LTI form of convolution, that is, summing up the delayed replicas of the impulse response, scaled by the input samples: x0hn = [3, 3, 3, 3, 0, 0, 0, 3, 3, 3, 3, 0, 0, 0, 3, 3, 3, 3, 0, 0, 0, . . . ] x0hn−1 = [0, 2, 2, 2, 2, 0, 0, 0, 2, 2, 2, 2, 0, 0, 0, 2, 2, 2, 2, 0, 0, . . . ] x0hn−2 = [0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, . . . ] yn = [3, 5, 6, 6, 3, 1, 0, 3, 5, 6, 6, 3, 1, 0, 3, 5, 6, 6, 3, 1, 0, . . . ] Thus, the output also has period 7. The same conclusion can be reached working with z-transforms. We have for the output z-transform: Y(z)= H(z)X(z) = (1 + z−1 + z−2 + z−3 )(3 + 2z−1 + z−2 ) 1 − z−7 = 3 + 5z−1 + 6z−2 + 6z−3 + 3z−4 + z−5 1 − z−7 The expansion of the denominator in powers of z−7 will cause the period-7 replication of the numerator. Problem 6.19 The expansion of the denominator, will cause the period-2 replication of the numerator. Because the numerator has length 4, its period-2 replicas will overlap with each other and must be added together: 99
- 51. H(z)= 1 + z−1 + z−2 + z−3 1 − z−2 = (1 + z−1 + z−2 + z−3 )(1 + z−2 + z−4 + · · · ) or, in the time domain: h =[1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, . . . ] +[0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, . . . ] +[0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, . . . ]+ · · · =[1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, . . . ] The canonical realization and sample processing algorithm are: x y w0 w1 w2 w3 z-1 z-1 z-1 for each input x do: w0 = x + w2 y = w0 + w1 + w2 + w3 w3 = w2 w2 = w1 w1 = w0 Problem 6.20 Writing the sample processing algorithm in terms of the time index n, we have w0(n) = x(n)+w1(n) y(n) = w0(n)+w2(n) w2(n + 1) = w1(n) w1(n + 1) = w0(n) or, in the z-domain W0(z) = X(z)+W1(z) Y(z) = W0(z)+W2(z) zW2(z) = W1(z) zW1(z) = W0(z) Eliminating W0(z), W1(z), W2(z) in favor of Y(z) and X(z), we obtain: W1(z)= z−1 W0(z), W2(z)= z−1 W1(z)= z−2 W0(z) W0(z)= X(z)+z−1 W0(z) ⇒ W0(z)= X(z) 1 − z−1 Y(z)= W0(z)+z−2 W0(z)= (1 + z−2 )W0(z)= (1 + z−2 ) X(z) 1 − z−1 100 which gives: H(z)= Y(z) X(z) = 1 + z−2 1 − z−1 Multiplying numerator and denominator by 1 + z−1 , we get: H(z)= (1 + z−2 )(1 + z−1 ) (1 − z−1)(1 + z−1) = 1 + z−1 + z−2 + z−3 1 − z−2 which is the same as that of Problem 6.19. Problem 6.21 The number of harmonics fitting within the Nyquist interval are: fs f1 = 240 60 = 4 Therefore, the harmonics will lie at the fourth roots of unity around the unit circle: ωk = kω1 = 2πk 4 , k = 0, 1, 2, 3 The numerator polynomial 1 − z−4 will have ejωk as roots. Putting the poles just behind the zeros, we get the desired multinotch filter: H(z)= 1 − z−4 1 − az−4 where 0 < a < 1 must be chosen to be very near 1. Problem 6.22 The zeros are at the 16th roots of unity, that is, zk = ejωk = e2πjk/16 , k = 0, 1, . . . , 15 The poles are just behind the zeros at the same angles: pk = a1/16 ejωk = a1/16 e2πjk/16 , k = 0, 1, . . . , 15 The zeros and poles are shown in Fig. P6.5. If a is very near 1, the magnitude response will vanish at the harmonics ωk = kω1 = 2πk/16, and be essentially flat between zeros. Fig. P6.6 shows a sketch of |H(ω)|. The impulse response is obtained by expanding H(z) into powers of z−16 . We have with A = 1/a and B = 1 − A = 1 − 1/a: H(z)= 1 − z−16 1 − az−16 = A + B 1 − az−16 = A + B(1 + az−16 + a2 z−32 + · · · ) Thus, the causal impulse response will be h(n): h(n)= ⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩ A + B = 1, if n = 0 Ban/16 , if n is a non-zero multiple of 16 0, otherwise 101
- 52. z-plane = zeros = zeros 1 Fig. P6.5 Pole/zero pattern of Problem 6.22. ω1 ω1 = 2π/16 ω 2ω1 3ω1 4ω1 15ω1 16ω1=2π0 ... |H(ω)| Fig. P6.6 Magnitude response of Problem 6.22. The canonical realization and sample processing algorithm are: yx a -1 w0 w16 z-16 for each input x do: w0 = x + w16 y = w0 − aw16 delay(16, w) where the call to delay updates the 17-dimensional delay line. Problem 6.23 H(z)= −0.3 + 0.6 1 − 0.5z−1 . It follows that h(n)= −0.3δ(n)+0.6(0.5)n u(n). Problem 6.24 The gain factor 1 − a normalizes the magnitude response to unity at DC. The direct form I/O difference equation is obtained from: H(z)= Y(z) X(z) = 1 − a 1 − az−1 ⇒ (1 − az−1 )Y(z)= (1 − a)X(z) or, Y(z)= az−1 Y(z)+(1 − a)X(z)= X(z)+a(z−1 Y(z)−X(z)) and in the time domain: 102 y(n)= x(n)+a(y(n − 1)−x(n)) A realization based on this expression, as well as the canonical realization, are shown in Fig. 6.24. The corresponding sample processing algorithm will be: for each input x do: y = x + a(w1 − x) w1 = y yx a w1 z-1 yx a 1−a w1 w0 z-1 − + Fig. P6.7 Realizations of Problem 6.24. Problem 6.25 The filter poles are at z = 2, −0.5. Thus, H(z)= 3 − 3z−1 − z−2 1 − 1.5z−1 − z−2 = 3 − 3z−1 − z−2 (1 − 2z−1)(1 + 0.5z−1) Expanding in partial fractions, we have: H(z)= 3 − 3z−1 − z−2 (1 − 2z−1)(1 + 0.5z−1) = A + B 1 − 2z−1 + C 1 + 0.5z−1 where A = 3 − 3z−1 − z−2 1 − 1.5z−1 − z−2 z=0 = 1 B = 3 − 3z−1 − z−2 1 + 0.5z−1 z=2 = 1 C = 3 − 3z−1 − z−2 1 − 2z−1 z=−0.5 = 1 The three possible ROCs and corresponding inverses are: |z| > 2 , h(n)= Aδ(n)+B2n u(n)+C(−0.5)n u(n) 2 > |z| > 0.5, h(n)= Aδ(n)−B2n u(−n − 1)+C(−0.5)n u(n) |z| < 0.5, h(n)= Aδ(n)−B2n u(−n − 1)−C(−0.5)n u(−n − 1) Only the second one is stable. Its ROC contains the unit circle. 103
- 53. Problem 6.26 The transfer function is obtained from the I/O difference equation by taking z-transforms of both sides and solving for the ratio Y(z)/X(z): Y(z)= 2.5z−1 Y(z)−z−2 Y(z)+3X(z)+3z−2 X(z) ⇒ H(z)= Y(z) X(z) = 3(1 + z−2 ) 1 − 2.5z−1 + z−2 Factoring the denominator into its poles and expanding in partial fractions, we get: H(z)= 3(1 + z−2 ) 1 − 2.5z−1 + z−2 = 3(1 + z−2 ) (1 − 2z−1)(1 − 0.5z−1) = A + B 1 − 2z−1 C 1 − 0.5z−1 where A = 3(1 + z−2 ) 1 − 2.5z−1 + z−2 z=0 = 3 B = 3(1 + z−2 ) 1 − 0.5z−1 z=2 = 5 C = 3(1 + z−2 ) 1 − 2z−1 z=0.5 = −5 Thus, the ROCs and corresponding inverses are: |z| > 2 , h(n)= Aδ(n)+B2n u(n)+C0.5n u(n) 2 > |z| > 0.5, h(n)= Aδ(n)−B2n u(−n − 1)+C0.5n u(n) |z| < 0.5, h(n)= Aδ(n)−B2n u(−n − 1)−C0.5n u(−n − 1) The z-transform of the signal g(n) is: g(n)= cos(πn/2)u(n) Z −→ G(z)= 1 1 + z−2 Therefore, the z-transform of the input x(n) will be: x(n)= g(n)−2g(n − 1) Z −→ X(z)= G(z)−2z−1 G(z)= (1 − 2z−1 )G(z)= 1 − 2z−1 1 + z−2 Thus, the output z-transform will be: Y(z)= H(z)X(z)= 3(1 + z−2 ) (1 − 2z−1)(1 − 0.5z−1) · 1 − 2z−1 1 + z−2 = 3 1 − 0.5z−1 Inverting causally, we find: y(n)= 3(0.5)n u(n) Problem 6.27 Using the inverse DTFT formula of Eq. (6.3.5), we have: y(n)= 1 2π π −π H(ω)X(ω)ejωn dω Similarly, for the input signal, assuming its spectrum is restricted to be only over the range [−ωc, ωc]: 104 x(n)= 1 2π ωc −ωc X(ω)ejωn dω Using d(ω)= D, the definition of H(ω) over one Nyquist interval will be: H(ω)= ⎧ ⎨ ⎩ Ge−jDω , for 0 ≤ |ω| ≤ ωc 0, for ωc < |ω| ≤ π where G is its passband gain. Thus, the integration range for y(n) also collapses to [−ωc, ωc], giving: y(n)= 1 2π ωc −ωc Ge−jDω X(ω)ejωn dω = G 1 2π ωc −ωc X(ω)ejω(n−D) dω Comparing this with the expression of x(n), we recognize that y(n)= Gx(n − D) Problem 6.28 The PF expansion of H(z) is of the form: H(z)= N(z) M i=1 (1 − piz−1 ) = M i=1 Ai 1 − piz−1 where the i-th coefficient is obtained by deleting the factor (1 − piz−1 ) from the left-hand side and evaluating the remainder at z = pi, that is, Ai = ⎡ ⎢ ⎢ ⎢ ⎣ N(z) j=i (1 − pjz−1 ) ⎤ ⎥ ⎥ ⎥ ⎦ z=pi = N(pi) j=i (1 − pjp−1 i ) Taking causal inverse z-transforms, we find for h(n): h(n)= M i=1 Aipn i u(n) Problem 6.29 The z-transforms of the causal and anticausal sinusoidal inputs are the same up to a negative sign: ejω0n u(n) Z −→ 1 1 − ejω0 z−1 ejω0n u(−n − 1) Z −→ − 1 1 − ejω0 z−1 with ROCs |z| > 1 and |z| < 1, respectively. The output z-transform will be Y(z)= H(z)X(z). Thus, in the two cases: 105
- 54. Y(z)= ± N(z) (1 − ejω0 z−1 ) M i=1 (1 − piz−1 ) The partial fraction expansion leads to Y(z)= ± C 1 − ejω0 z−1 ± M i=1 Bi 1 − piz−1 where C = H(ω0), as we saw in Eq. (6.3.10), and Bi are obtained by deleting the factor (1−piz−1 ) and evaluating the rest at z = pi: Bi = ⎡ ⎢ ⎢ ⎢ ⎣ N(z) (1 − ejω0 z−1 ) j=i (1 − pjz−1 ) ⎤ ⎥ ⎥ ⎥ ⎦ z=pi = N(pi) (1 − ejω0 p−1 i ) M j=i (1 − pjp−1 i ) Comparing with the result of the previous problem, we have: Bi = Ai 1 − ejω0 p−1 i To get the inverse z-transform y(n), we must assume a particular ROC. For the causal case, we take the ROC to be |z| > 1, so that all the PFE terms will be inverted causally. For the anticausal case, we take the filter part to be causal and the input anticausal, that is, the ROC will be the annular region between the maximum pole and the unit circle: max i |pi| < |z| < 1 In this case the C term will be inverted anticausally and the Bi terms causally. These choices for the ROCs, give: y(n) = H(ω0)ejω0n u(n)+ M i=1 Bipn i u(n) y(n) = H(ω0)ejω0n u(−n − 1)− M i=1 Bipn i u(n) Adding up the right-sided and left-sided sinusoidal inputs gives rise to the double-sided input: ejω0n u(n)+ejω0n u(−n − 1)= ejω0n u(n)+u(−n − 1) = ejω0n , −∞ < n < ∞ where, we used the property u(n)+u(−n − 1)= 1 for all n. It follows from the linearity of the filter that the corresponding output will be the sum of the two outputs obtained above. This gives: y(n) = H(ω0)ejω0n u(n)+ M i=1 Bipn i u(n)+H(ω0)ejω0n u(−n − 1)− M i=1 Bipn i u(n) = H(ω0)ejω0n u(n)+H(ω0)ejω0n u(−n − 1) = H(ω0)ejω0n which is recognized as the standard steady-state sinusoidal response of the filter. 106 Problem 6.30 The PF expansion of H(z) gives: H(z)= 3 − 5z−1 + z−2 (1 − 0.5z−1)(1 − 2z−1) = 1 + 1 1 − 0.5z−1 + 1 1 − 2z−1 The stable ROC containing the unit circle will be 0.5 < |z| < 2. Therefore, the 0.5-pole term will be inverted causally and the 2-pole anticausally: h(n)= δ(n)+0.5n u(n)−2n u(−n − 1) The truncated impulse response agrees with h(n) for n ≥ −D and is zero otherwise, that is: h(n)= ⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩ δ(n)+0.5n , n ≥ 0 −2n , −D ≤ n ≤ −1 0, n ≤ −D − 1 The corresponding z-transform is by definition: H(z)= ∞ n=0 [δ(n)+0.5n ]z−n − −1 n=−D 2n z−n The last term can be summed by changing summation variables from n to m = −n: −1 n=−D 2n z−n = D m=1 2−m zm = 2−1 z(1 − 2−D zD ) 1 − 2−1z = − 1 − 2−D zD 1 − 2z−1 where we used the finite geometric series: D m=1 xm = x + x2 + · · · xD = x(1 + x + x2 + · · · xD−1 )= x(1 − xD ) 1 − x Thus, H(z)= 1 + 1 1 − 0.5z−1 + 1 − 2−D zD 1 − 2z−1 Comparing with the exact H(z) we have: H(z)= H(z)− 2−D zD 1 − 2z−1 or, the error transfer function: H(z)−H(z)= 2−D zD 1 − 2z−1 The given input has z-transform: x(n)= δ(n)−2δ(n − 1) Z −→ X(z)= 1 − 2z−1 The error output will be E(z)= Y(z)−Y(z)= H(z)X(z)−H(z)X(z)= H(z)−H(z) X(z) or, E(z)= 2−D zD 1 − 2z−1 · (1 − 2z−1 )= 2−D zD and in the time domain: e(n)= 2−D δ(n + D) By choosing D large enough, the factor 2−D can be made as small as desired. 107
- 55. Chapter 7 Problems Problem 7.1 Expanding the denominator in powers of z−5 will replicate the numerator periodically with period 5. But because the numerator itself has length 5, the replicas will not overlap, and we will get the repetition of the numerator coefficients: h = [0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, . . . ] The direct and canonical realizations are shown in Fig. P7.1. x y x y z-1 z-1 z-1 z-1 z-1 z-1 2 3 4 z-1 z-1 z-1 w5 w0 w1 w2 w3 w4 w5 w0 w1 w2 w3 w4 v0 v1 v2 v3 v4 z-1 2 3 4 z-1 z-1 z-1 z-1 Fig. P7.1 Direct and canonical realizations of Problem 7.1. The sample processing algorithms are expressed with the help of the routine delay, which updates the filter’s internal state vectors. for each input x do: v0 = x w0 = w5 + v1 + 2v2 + 3v3 + 4v4 y = w0 delay(4, v) delay(5, w) for each input x do: w0 = x + w5 y = w1 + 2w2 + 3w3 + 4w4 delay(5, w) where v = [v0, v1, v2, v4], w = [w0, w1, w2, w3, w4.w5] are the 5- and 6-dimensional state vec- tors needed for the numerator and denominator polynomials. The calls to the delay routines update the delay line, e.g., the call delay(5, w) is equivalent to the shifts: w5 = w4, w4 = w3, w3 = w2, w2 = w1, w1 = w0. Factoring the denominator as 1−z−5 = (1−z−1 )(1+z−1 +z−2 +z−3 +z−4 ), we get the cascaded transfer functions: H(z)= z−1 + 2z−2 + 3z−3 + 4z−4 1 − z−5 = z−1 1 − z−1 · 1 + 2z−1 + 3z−2 + 4z−3 1 + z−1 + z−2 + z−3 + z−4 Figure P7.2 shows the cascade of these two filters, with each filter realized in its canonical form. The corresponding sample processing algorithm is as follows: 108 x1 y 2-1 -1 -1 -1 3 4 z-1 z-1 w0 w1 w2 z-1 w3 z-1 w4 x z-1 v0 v1 Fig. P7.2 Cascade realization of Problem 7.1. for each input x do: v0 = x + v1 x1 = v1 delay(1, v) w0 = x1 − w1 − w2 − w3 − w4 y = w0 + 2w1 + 3w2 + 4w3 delay(4, w) where v = [v0, v1], w = [w0, w1, w2, w3, w4] are the internal state vectors for the two filter sections. Problem 7.2 In the z-domain, the input to H1(z) is X(z)−Y(z). Therefore, its output will be H1(z) Y(z) −X(z) . The input and output of H2(z) will be then : H1(z) Y(z)−X(z) − Y(z)−→ H2(z) H1(z) Y(z)−X(z) − Y(z) Adding the noise component E(z) to the output of H2(z) will generate Y(z), resulting in: Y(z)= H2(z) H1(z) Y(z)−X(z) − Y(z) + E(z) Moving the Y-dependent terms to the left, we get: 1 + H2(z)+H1(z)H2(z) Y(z)= H1(z)H2(z)X(z)+E(z) and solving for Y(z): Y(z)= H1(z)H2(z) 1 + H2(z)+H1(z)H2(z) X(z)+ 1 1 + H2(z)+H1(z)H2(z) E(z) Thus, the desired transfer functions are identified to be: Hx(z) = H1(z)H2(z) 1 + H2(z)+H1(z)H2(z) He(z) = 1 1 + H2(z)+H1(z)H2(z) 109
- 56. The conditions that Hx(z) be a plain delay and He(z) a double differentiator, gives the two equations: H1(z)H2(z) 1 + H2(z)+H1(z)H2(z) = z−1 1 1 + H2(z)+H1(z)H2(z) = (1 − z−1 )2 which can be solved for H1(z) and H2(z) giving: H1(z) = 1 1 − z−1 H2(z) = z−1 1 − z−1 They are both integrators. The presence of z−1 in the numerator of H2(z) is necessary to make the overall closed loop computable. See also Problems 12.24 and 12.25. Problem 7.3 Multiplying the numerator factors, we get the transfer function: H(z)= z−1 (1 + 2z−2 )(1 + 3z−2 ) 1 − z−6 = z−1 + 5z−3 + 6z−5 1 − z−6 The direct and canonical realizations are shown in Fig. P7.3. x y x y z-1 z-1 z-1 z-1 z-1 w5 w6 w0 w1 w2 w3 w4 z-1 z-1 z-1 z-1 z-1 z-1 w5 w6 w0 w1 w2 w3 w4 z-1 z-1 z-1 z-1 z-1 v5 v0 v1 v2 v3 v4 z-1 5 6 5 6 Fig. P7.3 Direct and canonical realizations of Problem 7.3. The sample processing algorithms are expressed with the help of the routine delay, which updates the filter’s internal state vectors. for each input x do: v0 = x w0 = v1 + 5v3 + 6v5 + w6 y = w0 delay(5, v) delay(6, w) for each input x do: w0 = x + w6 y = w1 + 5w3 + 6w5 delay(6, w) 110 where v = [v0, v1, v2, v4, v5], w = [w0, w1, w2, w3.w4, w5, w6] are the 6- and 7-dimensional state vectors needed for the numerator and denominator polynomials. The calls to the delay routines update the delay line, e.g., the call delay(6, w) is equivalent to the shifts: w6 = w5, w5 = w4, w4 = w3, w3 = w2, w2 = w1, w1 = w0. Factoring the denominator as 1 − z−6 = (1 − z−2 )(1 + z−2 + z−4 )= (1 − z−2 )(1 + z−1 + z−2 )(1 − z−1 + z−2 ) we get the cascade transfer function: H(z)= z−1 1 − z−2 · 1 + 2z−2 1 + z−1 + z−2 · 1 + 3z−2 1 − z−1 + z−2 Figure P7.4 shows the cascade of these three filters, with each filter realized in its canonical form. z-1 z-1 u0 u1 u2 x x1 x2 y z-1 z-1 v0 v1 v2 2-1 -1 z-1 z-1 w0 w1 w2 3-1 Fig. P7.4 Cascade realization of Problem 7.3. The corresponding sample processing algorithm is as follows: for each input x do: u0 = x + u2 x1 = u1 delay(2, u) v0 = x1 − v1 − v2 x2 = v0 + 2v2 delay(2, v) w0 = x2 + w1 − w2 y = w0 + 3w2 delay(2, w) where u = [u0, u1, u2], v = [v0, v1, v2], w = [w0, w1, w2], are the internal state vectors for the three filter sections. To determine the impulse response, we note that the numerator corresponds to the length-6 signal b = [0, 1, 0, 5, 0, 6]. Expanding the denominator in powers of z−6 will replicate the numerator periodically with period 6. But because the numerator itself has length 6, the replicas will not overlap, and we will get the repetition of the numerator coefficients: h = [0, 1, 0, 3, 0, 5, 0, 1, 0, 3, 0, 5, 0, 1, 0, 3, 0, 5, . . . ] Problem 7.4 The z-domain versions of the time domain difference equations are: v(n) = x(n)+v(n − 1) y(n) = v(n)+v(n − 2)+v(n − 4) ⇒ V(z) = X(z)+z−1 V(z) Y(z) = V(z)+z−2 V(z)+z−4 V(z) 111
- 57. Eliminating V(z) in favor of Y(z), X(z), gives: V(z)= X(z) 1 − z−1 , Y(z)= (1 + z−2 + z−4 )V(z)= 1 + z−2 + z−4 1 − z−1 X(z) Therefore, the transfer function is: H(z)= 1 + z−2 + z−4 1 − z−1 = 1 + z−1 + z−2 1 − z−1 · (1 − z−1 + z−2 ) where we factored the numerator as 1 + z−2 + z−4 = (1 + z−1 + z−2 )(1 − z−1 + z−2 ) The direct, canonical, and cascade realizations are shown in Fig. P7.5. x y z-1 z-1 z-1 w0 w1 w2 w3 w4 z-1 x y z-1 w0 w1 z-1 z-1 z-1 v0 v1 v2 v3 v4 z-1 x x1 z-1 z-1 v0 v1 v2 y z-1 z-1 w0 w1 w2 -1 Fig. P7.5 Direct, canonical, and cascade realizations of Problem 7.4. The direct and canonical sample processing algorithms are: for each input x do: v0 = x w0 = v0 + v2 + v4 + w1 y = w0 delay(4, v) delay(1, w) for each input x do: w0 = x + w1 y = w0 + w2 + w4 delay(4, w) And for the cascade realization: 112 for each input x do: v0 = x + v1 x1 = v0 + v1 + v2 delay(2, v) w0 = x1 y = w0 − w1 + w2 delay(2, w) Problem 7.5 Factoring the denominator, we may write the transfer function as: H(z)= 2 − 3z−1 1 − 0.5z−2 · 1 + z−2 1 + 0.5z−2 = 2 − 3z−1 + 2z−2 − 3z−3 1 − 0.25z−4 The direct and canonical realizations are shown in Fig. P7.6. Their sample processing algorithms are: for each input x do: v0 = x w0 = 2v0 − 3v1 + 2v2 − 3v3 + 0.25w4 y = w0 delay(3, v) delay(4, w) for each input x do: w0 = x + 0.25w4 y = 2w0 − 3w1 + 2w2 − 3w3 delay(4, w) x y z-1 z-1 z-1 w0 w1 w2 w3 w4 z-1 x y z-1 z-1 z-1 v0 v1 v2 v3 z-1 z-1 z-1 w0 w1 w2 w3 w4 z-1 2 2 -3 -3 2 2 -3 -3 0.25 0.25 Fig. P7.6 Direct and canonical realizations of Problem 7.5. The cascade realization is shown in Fig. P7.7. Its sample processing algorithm is: for each input x do: v0 = x + 0.5v2 x1 = 2v0 − 3v1 delay(2, v) w0 = x1 − 0.5w2 y = w0 + w2 delay(2, w) 113
- 58. x z-1 z-1 v0 v1 x1 v2 2 -3 0.5 y z-1 z-1 w0 w1 w2-0.5 Fig. P7.7 Cascade realization of Problem 7.5. Problem 7.6 The direct and canonical realizations are shown in Fig. P7.8. Their sample processing algorithms are: for each input x do: v0 = x w0 = v0 − 2.25v2 + 0.0625w4 y = w0 delay(2, v) delay(4, w) for each input x do: w0 = x + 0.0625w4 y = w0 − 2.25w2 delay(4, w) x y z-1 z-1 z-1 w0 w1 w2 w3 w4 z-1 x y z-1 z-1 v0 v1 v2 z-1 z-1 z-1 w0 w1 w2 w3 w4 z-1 -2.25 -2.25 0.0625 0.0625 Fig. P7.8 Direct and canonical realizations of Problem 7.6. The cascade realization is shown in Fig. P7.9. Its sample processing algorithm is: for each input x do: v0 = x − 0.25v2 x1 = v0 − 2.25v2 delay(2, v) y = w0 = x1 + 0.25w2 delay(2, w) Expanding H(z) in partial fractions, we have: H(z)= 2.5 1 − 0.5jz−1 + 2.5 1 + 0.5jz−1 + 2 1 − 0.5z−1 + 2 1 + 0.5z−1 114 x z-1 z-1 v0 v1 x1 v2 -2.25-0.25 y z-1 z-1 w0 w1 w20.25 Fig. P7.9 Cascade realization of Problem 7.6. The four poles z = ±0.5, ±0.5j, define only two ROCs, namely, the causal case |z| > 0.5 and the anticausal one |z| < 0.5. For example, in the causal case, the impulse response will be: h(n)= 2.5(0.5j)n u(n)+2.5(−0.5j)n u(n)+2(0.5)n u(n)+2(−0.5)n u(n) Writing jn + (−j)n = ejπn/2 + e−jπn/2 = 2 cos(πn/2), we have: h(n)= 5(0.5)n cos(2πn/2)+2(0.5)n u(n)+2(−0.5)n u(n) Problem 7.7 The direct and canonical realizations are shown in Fig. P7.10. Their sample processing algorithms are: for each input x do: v0 = x y = w0 = v0 − 2v2 + v4 + 0.4096w4 delay(4, v) delay(4, w) for each input x do: w0 = x + 0.4096w4 y = w0 − 2w2 + w4 delay(4, w) x y z-1 z-1 z-1 w0 w1 w2 w3 w4 z-1 x y z-1 z-1 v0 v1 v2 v3 v4 z-1 z-1 w0 w1 w2 w4 z-1 w3 z-1 z-1 z-1 -2 -2 0.4096 0.4096 Fig. P7.10 Direct and canonical realizations of Problem 7.7. The cascade realization is shown in Fig. P7.11. It corresponds the factorization: H(z)= 1 − 2z−2 + z−4 1 − 0.4096z−4 = 1 − z−2 1 − 0.64z−2 · 1 − z−2 1 + 0.64z−2 115
- 59. Note that the numerator could also have been factored as: 1 − 2z−2 + z−4 = (1 − z−2 )2 = (1 − z−1 )2 (1 + z−1 )2 = (1 − 2z−1 + z−2 )(1 + 2z−1 + z−2 ) But the chosen one has the simplest coefficients. Its sample processing algorithm is: for each input x do: v0 = x + 0.64v2 x1 = v0 − v2 delay(2, v) w0 = x1 − 0.64w2 y = w0 − w2 delay(2, w) x z-1 z-1 v0 v1 x1 v2 -10.64 z-1 z-1 w0 w1 w2 -1-0.64 y Fig. P7.11 Cascade realization of Problem 7.7. Problem 7.8 The given difference equations describe the canonical realization form of the filter: H(z)= −0.75 + z−10 1 − 0.75z−10 Its block diagram realization and sample processing algorithm are given below: x z-10 w0 w10 0.75 -0.75 y for each input sample x do: w0 = x + 0.75w10 y = −0.75w0 + w10 delay(10, w) The C version of the algorithm is as follows: double reverb(x, w) double x, w[11]; { int i; double y; w[0] = x + 0.75 * w[10]; y = - 0.75 * w[0] + w[10]; 116 for (i=10; i>=1; i--) w[i] = w[i-1]; return y; } From the transfer function, we obtain the frequency response: H(ω)= −0.75 + e−10jω 1 − 0.75e−10jω = 1 − 0.75e10jω 1 − 0.75e−10jω e−10jω = (1 − 0.75e−10jω )∗ (1 − 0.75e−10jω) e−10jω Noting that the ratio of two conjugate complex numbers has unit magnitude, it follows that H(ω) will have unit magnitude. Problem 7.9 The given sample processing algorithm corresponds to the canonical realization of the filter: H(z)= 1 + z−3 1 − 0.64z−4 Factoring the numerator in the form: 1 + z−4 = (1 + z−1 )(1 − z−1 + z−2 ) we obtain the cascade factors: H(z)= 1 + z−3 1 − 0.64z−4 = 1 + z−1 1 − 0.8z−2 · 1 − z−1 + z−2 1 + 0.8z−2 The cascade realization is shown in Fig. P7.12. Its sample processing algorithm is: for each input x do: v0 = x + 0.8v2 x1 = v0 + v1 delay(2, v) w0 = x1 − 0.8w2 y = w0 − w1 + w2 delay(2, w) x z-1 z-1 v0 v1 x1 v2 0.8 z-1 z-1 w0 w1 w2-0.8 -1 y Fig. P7.12 Cascade realization of Problem 7.9. 117
- 60. Problem 7.10 The first filter is an FIR filter with transfer function and impulse response: H(z)= (1 + z−2 )3 = 1 + 3z−2 + 3z−4 + z−6 ⇒ h = [1, 0, 3, 0, 3, 0, 1] The causal impulse response of the third filter is obtained by the remove restore method, that is, h(n)= w(n)−w(n − 4), where w(n)= (0.9)n u(n)] The first filter has zeros at z = ±j, each being a triple zero. The second filter has poles at z = ±0.9j, and the third filter has zeros at the fourth roots of unity, i.e., z = ±1, ±j, and a pole at z = 0.9. The pole/zero plots of the three filters are shown in Fig. P7.13 together with the corresponding sketches of the magnitude responses |H(ω)| over the right half of the Nyquist interval 0 ≤ ω ≤ π. = zeros = poles j -j -0.9j -j 0.9j j 0.9 1-1 π ω |H(ω)| |H(ω)| |H(ω)| ω ω π π Fig. P7.13 Pole/zero patterns of Problem 7.10. Problem 7.11 Multiplying the transfer function factors, we obtain: H(z)= (1 − √ 2z−1 + z−2 )(1 + √ 2z−1 + z−2 ) (1 + 0.81z−2)(1 − 0.81z−2) = 1 + z−4 1 − 0.6561z−4 Therefore, the zeros will be at the roots of 1 + z−4 = 0, that is, z4 = −1 ⇒ z = ejπ(2k+1)/4 , k = 0, 1, 2, 3 that is, the 4th roots of unity rotated by 45o . The poles are at z = ±0.9m, ±0.9j. Fig. P7.14 shows the pole/zero locations and the corresponding magnitude response. The direct and canonical realizations are shown in Fig. P7.15. Their sample processing algorithms are: 118 = zeros = poles -0.9j 0.9j 0.9-0.9 |H(ω)| ω π Fig. P7.14 Pole/zero plot of Problem 7.11. for each input x do: v0 = x y = w0 = v0 + v4 + 0.6561w4 delay(4, v) delay(4, w) for each input x do: w0 = x + 0.6561w4 y = w0 + w4 delay(4, w) x y z-1 z-1 z-1 w0 w1 w2 w3 w4 z-1 x y z-1 z-1 v0 v1 v2 v3 v4 z-1 z-1 w0 w1 w2 w4 z-1 w3 z-1 z-1 z-1 0.6561 0.6561 Fig. P7.15 Direct and canonical realizations of Problem 7.11. The cascade realization is shown in Fig. P7.16. Its sample processing algorithm is: for each input x do: v0 = x − 0.81v2 x1 = v0 − √ 2v1 + v2 delay(2, v) w0 = x1 + 0.81w2 y = w0 + √ 2w1 + w2 delay(2, w) Problem 7.12 The zeros are at the roots of 1 + 16z−4 = 0, that is, 119
- 61. x z-1 z-1 v0 v1 x1 v2-0.81 z-1 z-1 w0 w1 w20.81 y - 2 2 Fig. P7.16 Cascade realization of Problem 7.11. z4 = −16 ⇒ z = 2ejπ(2k+1)/4 , k = 0, 1, 2, 3 that is, the 4th roots of unity rotated by 45o and scaled in magnitude by a factor of 2. Similarly, the poles are at: z4 = − 1 16 ⇒ z = 0.5ejπ(2k+1)/4 , k = 0, 1, 2, 3 Fig. P7.17 shows the pole/zero locations. = zeros = poles |H(ω)| ω π unit circle 1 Fig. P7.17 Pole/zero plot of Problem 7.12. The magnitude response is constant in ω. Indeed, setting c = 1/16, we have: H(ω)= c + e−4jω 1 + ce−4jω = e−4jω (1 + ce4jω ) 1 + ce−4jω = e−4jω (1 + ce−4jω )∗ (1 + ce−4jω) The ratio of the two conjugate factors will have unity magnitude, giving |H(ω)| = |e−4jω | · (1 + ce−4jω )∗ (1 + ce−4jω) = 1 · 1 = 1 To get the cascade form, we may factor the numerator and denominator using the identity: 1 + a4 z−4 = (1 − √ 2az−1 + a2 z−2 )(1 + √ 2az−1 + a2 z−2 ) Thus the numerator becomes: 1 16 + z−4 = 1 16 (1 + 24 z−4 )= 1 16 (1 − 2 √ 2z−1 + 4z−2 )(1 + 2 √ 2z−1 + 4z−2 ) 120 Noting that 1/16 = 0.0625, the transfer function is then: H(z)= 0.0625 + z−4 1 + 0.0625z−4 = 0.0625 · 1 − 2 √ 2z−1 + 4z−2 1 − 0.5 √ 2z−1 + 0.25z−2 · 1 + 2 √ 2z−1 + 4z−2 1 + 0.5 √ 2z−1 + 0.25z−2 The direct and canonical realizations are shown in Fig. P7.18. Their sample processing algorithms are: for each input x do: v0 = x y = w0 = 0.0625v0 + v4 − 0.0625w4 delay(4, v) delay(4, w) for each input x do: w0 = x − 0.0625w4 y = 0.0625w0 + w4 delay(4, w) x y z-1 z-1 z-1 w0 w1 w2 w3 w4 z-1 x y z-1 z-1 v0 v1 v2 v3 v4 z-1 z-1 w0 w1 w2 w4 z-1 w3 z-1 z-1 z-1 -0.0625 -0.0625 0.06250.0625 Fig. P7.18 Direct and canonical realizations of Problem 7.12. The cascade realization is shown in Fig. P7.19. Its sample processing algorithm is: for each input x do: v0 = 0.0625x + 0.5 √ 2v1 − 0.25v2 x1 = v0 − 2 √ 2v1 + 4v2 delay(2, v) w0 = x1 − 0.5 √ 2w1 − 0.25w2 y = w0 + 2 √ 2w1 + 4w2 delay(2, w) Finally, the impulse response may be obtained by writing: H(z)= c + z−4 1 + cz−4 = c + (1 − c2 )z−4 1 + cz−4 = c + (1 − c2 ) z−4 − cz−8 + c2 z−12 − c3 z−16 + · · · Extracting the coefficients of z−1 , we find: h(n)= ⎧ ⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎩ c if n = 0 −(1 − c2 )c(n−4)/4 if n is an even multiple of 4 (1 − c2 )c(n−4)/4 if n is an odd multiple of 4 0 otherwise 121
- 62. x 0.0625 w0 w1 w2 y z-1 z-1 v0 v1 x1 v2-0.25 4 -2 20.5 2 z-1 z-1 -0.25 4 2 2-0.5 2 Fig. P7.19 Cascade realization of Problem 7.12. Problem 7.13 The following program implements this computer experiment: /* iirfilt.c - IIR filtering experiments */ #include <stdio.h> #include <stdlib.h> void cas2can(); double dir(), can(), cas(); void main() { int i, j, K, M; double x, ydir, ycan, ycas; double *vdir, *wdir, *wcan, **W; double **A, **B, *a, *b; FILE *fpx; fpx = fopen("x.dat", "r"); /* input data */ printf("no. sections K = "); scanf("%d", &K); M=2*K; wdir = (double *) calloc(M+1, sizeof(double)); vdir = (double *) calloc(M+1, sizeof(double)); wcan = (double *) calloc(M+1, sizeof(double)); a = (double *) calloc(M+1, sizeof(double)); b = (double *) calloc(M+1, sizeof(double)); A = (double **) calloc(K, sizeof(double *)); B = (double **) calloc(K, sizeof(double *)); W = (double **) calloc(K, sizeof(double *)); for (i=0; i<K; i++) { A[i] = (double *) calloc(3, sizeof(double)); B[i] = (double *) calloc(3, sizeof(double)); W[i] = (double *) calloc(3, sizeof(double)); } for(i=0; i<K; i++) { 122 printf("enter A[%d] = ", i); for(j=0; j<3; j++) scanf("%lf", A[i]+j); } for(i=0; i<K; i++) { printf("enter B[%d] = ", i); for(j=0; j<3; j++) scanf("%lf", B[i]+j); } cas2can(K, A, a); cas2can(K, B, b); printf("nn"); while(fscanf(fpx, "%lf", &x) != EOF) { ydir = dir(M, a, M, b, vdir, wdir, x); ycan = can(M, a, M, b, wcan, x); ycas = cas(K, A, B, W, x); printf("% .6f % .6f % .6f % .6fn", x, ydir, ycan, ycas); } printf("n a, b = n"); /* direct form coefficients */ for (i=0;i<=M;i++) printf("% .6lf % .6lfn", a[i], b[i]); } The program must be linked with the routines cas2can, dir, can, cas, and also conv required by cas2can and sos required by cas. The number of sections K and the corresponding rows of the matrices A and B are entered interactively. For example, the user input for the given transfer functions will be (starting with K = 4): 4 1 -0.373 0 1 -1.122 0.712 1 -0.891 0.360 1 -0.780 0.190 0.313 0.313 0 0.147 0.294 0.147 0.117 0.234 0.117 0.103 0.206 0.103 These inputs may also be saved in a file, say, input.dat, and piped into the program by iirfilt < input.dat The input signal xn to be filtered is assumed to be in the file x.dat. The program stops filtering as soon as the end-of-file of x.dat is encountered. The first 10 of the input and output samples are shown below: 123
- 63. x(n) ydir(n) ycan(n) ycas(n) 1.000000 0.000554 0.000554 0.000554 1.000000 0.006191 0.006191 0.006191 1.000000 0.032979 0.032979 0.032979 1.000000 0.112201 0.112201 0.112201 1.000000 0.275606 0.275606 0.275606 1.000000 0.524036 0.524036 0.524036 1.000000 0.807624 0.807624 0.807624 1.000000 1.043772 1.043772 1.043772 1.000000 1.164600 1.164600 1.164600 1.000000 1.157457 1.157457 1.157457 The outputs computed by dir, can, and cas are of course the same. Fig. P7.20 shows the full output, plotted together with the input. The impulse response h(n), 0 ≤ n < 50 can be computed by running this program on a file x.dat containing 1 followed by 49 zeros. It is shown also in Fig. P7.20. The first 10 impulse response samples computed by the three realizations are shown below: x(n) hdir(n) hcan(n) hcas(n) 1.000000 0.000554 0.000554 0.000554 0.000000 0.005637 0.005637 0.005637 0.000000 0.026788 0.026788 0.026788 0.000000 0.079222 0.079222 0.079222 0.000000 0.163405 0.163405 0.163405 0.000000 0.248430 0.248430 0.248430 0.000000 0.283588 0.283588 0.283588 0.000000 0.236148 0.236148 0.236148 0.000000 0.120828 0.120828 0.120828 0.000000 −0.007142 −0.007142 −0.007142 The given filter is an example of a lowpass Butterworth filter and the normalization gains in each section have been chosen to give unity gain at DC. Therefore, the steady-state DC response will be H(ω) ω=0 = H(z) z=1 = 1 This can be observed in the output which rises up to 1 before it dies out to zero. Calling cas2can on the matrices A and B gives the direct form denominator and numerator coefficients: a b 1.000000 0.000554 −3.166000 0.003881 4.873631 0.011644 −4.465987 0.019407 2.592519 0.019407 −0.941724 0.011644 0.196860 0.003881 −0.018165 0.000554 0.000000 0.000000 124 Fig. P7.20 Input, output, and impulse response of Problem 7.13. Problem 7.14 An example of such program is given below. It is patterned after firfilt.c of Problem 4.10. The program reads and allocates the filter coefficient arrays. The last while loop performs the actual filtering: /* canfilt.c - IIR filtering in canonical form */ #include <stdlib.h> #include <stdio.h> #define MAX 64 initial allocation size double can(); void main(int argc, char **argv) { FILE *fpa, *fpb; coefficient files double *a, *b, *w, x, y; coeffs., state, input, output int L, M, K, i; int max = MAX, dmax = MAX; initial allocation and increment if (argc != 3) { fprintf(stderr, "Usage: canfilt a.dat b.dat <x.dat >y.datn"); exit(0); } if ((fpa = fopen(argv[1], "r")) == NULL) { fprintf(stderr, "Can’t open filter file: %sn", argv[1]); exit(0); } if ((fpb = fopen(argv[2], "r")) == NULL) { fprintf(stderr, "Can’t open filter file: %sn", argv[2]); exit(0); } a = (double *) calloc(max + 1, sizeof(double)); for (M=0;; M++) { if (M == max) { 125
- 64. max += dmax; a = (double *) realloc((char *) a, (max + 1) * sizeof(double)); } if (fscanf(fpa, "%lf", a + M) == EOF) break; } M--; a = (double *) realloc((char *) a, (M + 1) * sizeof(double)); b = (double *) calloc(max + 1, sizeof(double)); for (L=0;; L++) { if (L == max) { max += dmax; b = (double *) realloc((char *) b, (max + 1) * sizeof(double)); } if (fscanf(fpb, "%lf", b + L) == EOF) break; } L--; b = (double *) realloc((char *) b, (L + 1) * sizeof(double)); K = (L <= M) ? M : L; w = (double *) calloc((K + 1), sizeof(double)); while(scanf("%lf", &x) != EOF) { process input samples y = can(M, a, L, b, w, x); printf("%lfn", y); } } Problem 7.15 An example of such program is given below. The program reads and allocates the filter coefficient matrices A and B. To facilitate the initial allocation of A, a temporary array a is used to read the rows of A.dat in a concatenated form, and then it reshapes it into the K×3 matrix A. The last while loop performs the actual filtering: /* casfilt.c - IIR filtering in cascade form */ #include <stdlib.h> #include <stdio.h> #define MAX 64 initial allocation size double cas(); void main(int argc, char **argv) { FILE *fpa, *fpb; coefficient files double **A, **B, **W, x, y; double *a; temporary coefficient array int M, K, i, j, m; int max = MAX, dmax = MAX; allocation and increment if (argc != 3) { fprintf(stderr, "Usage: casfilt A.dat B.dat <x.dat >y.datn"); exit(0); } 126 if ((fpa = fopen(argv[1], "r")) == NULL) { fprintf(stderr, "Can’t open filter file: %sn", argv[1]); exit(0); } if ((fpb = fopen(argv[2], "r")) == NULL) { fprintf(stderr, "Can’t open filter file: %sn", argv[2]); exit(0); } a = (double *) calloc(max + 1, sizeof(double)); for (M=0;; M++) { if (M == max) { max += dmax; a = (double *) realloc((char *) a, (max + 1) * sizeof(double)); } if (fscanf(fpa, "%lf", a + M) == EOF) break; } a = (double *) realloc((char *) a, (M + 1) * sizeof(double)); if (M%3 != 0) { fprintf(stderr, "all rows of A must be 3-dimensional"); exit(0); } else K = M / 3; A = (double **) calloc(K, sizeof(double *)); B = (double **) calloc(K, sizeof(double *)); W = (double **) calloc(K, sizeof(double *)); for (i=0; i<K; i++) { A[i] = (double *) calloc(3, sizeof(double)); B[i] = (double *) calloc(3, sizeof(double)); W[i] = (double *) calloc(3, sizeof(double)); } for(i=0; i<K; i++) for(j=0; j<3; j++) A[i][j] = a[3*i + j]; for(i=0; i<K; i++) for(j=0; j<3; j++) fscanf(fpb, "%lf", B[i] + j); while(scanf("%lf", &x) != EOF) { process input samples y = cas(K, A, B, W, x); printf("%lfn", y); } } The MATLAB version casfilt.m is given below. It calls cas.m N times, where N is the input length: % casfilt.m - IIR filtering in cascade form % % y = casfilt(B, A, x); % 127
- 65. % B = Kx3 numerator matrix (K = number of sections) % A = Kx3 denominator matrix % x = row vector input % y = row vector output function y = casfilt(B, A, x) [K, K1] = size(A); [N1, N] = size(x); W = zeros(K,3); for n = 1:N, [y(n), W] = cas(K, B, A, W, x(n)); end Problem 7.16 The two filters are: H1(z)= 0.70849 · 1 + z−8 1 − 0.0625z−8 , H2(z)= 0.98021 · 1 + z−8 1 + 0.94z−8 · 1 − 0.96z−8 1 − 0.98z−8 The scale factors, were chosen such that the magnitude responses are unity at ω = ω0 = 0.5π/8, that is, |H1(ω0)| = |H2(ω0)| = 1 The following program implements this computer experiment. The sample processing algorithms of Examples 7.4.3 and 7.4.4 remain the same, except the input sample x gets multiplied by the corresponding normalization gains: /* combex.c - comb filtering examples */ #include <stdio.h> #include <stdlib.h> #include <math.h> #define A0 1.0 #define A1 0.5 #define A2 0.5 #define A3 0.5 #define G1 0.70849 #define G2 0.98021 void delay(); void main() { int n, N; double s, x, y1, x1, y2; double *w, *v, *u; double pi = 4*atan(1.0), w0, w1, w2, w3; FILE *fps, *fpx, *fpy1, *fpy2; fps = fopen("s.dat", "w"); 128 fpx = fopen("x.dat", "w"); fpy1 = fopen("y1.dat", "w"); fpy2 = fopen("y2.dat", "w"); w = (double *) calloc(9, sizeof(double)); /* filter 1 */ u = (double *) calloc(9, sizeof(double)); /* filter 2, section 1 */ v = (double *) calloc(9, sizeof(double)); /* filter 2, section 2 */ w0 = 0.50 * pi / 8; w1 = 0.75 * pi / 8; w2 = 1.00 * pi / 8; w3 = 3.00 * pi / 8; printf("enter input length N = "); scanf("%d", &N); for (n=0; n<N; n++) { s = A0 * cos(w0*n) + A1 * cos(w1*n); x = s + A2 * cos(w2*n) + A3 * cos(w3*n); w[0] = G1 * x + 0.0625 * w[8]; /* filter 1 */ y1 = w[0] + w[8]; delay(8, w); u[0] = G2 * x - 0.94 * u[8]; /* filter 2 */ x1 = u[0] + u[8]; /* section 1 output */ delay(8, u); v[0] = x1 + 0.98 * v[8]; y2 = v[0] - 0.96 * v[8]; /* section 2 output */ delay(8, v); fprintf(fps, "%.8lfn", s); fprintf(fpx, "%.8lfn", x); fprintf(fpy1, "%.8lfn", y1); fprintf(fpy2, "%.8lfn", y2); } } Fig. P7.21 shows the desired, noise-free, signal s(n) and the noisy signal x(n), which serves as the input to the comb filters. The corresponding output signals y1(n) and y2(n) from H1(z) and H2(z) are shown in Fig. P7.22. The first filter has a short time constant, but does not remove the noise component well. The second filter has a longer time constant, but it removes the noise very well. Problem 7.17 Performing a partial fraction expansion on H1(z) and H2(z), we find (where the PFE coefficients are given with 5-digit accuracy): H1(z) = −11.33584 + 12.04433 1 − 0.0625z−8 H2(z) = 1.02150 − 0.06191 1 + 0.94z−8 + 0.02063 1 − 0.98z−8 Expanding in powers of z−8 and extracting coefficients, we find the impulse responses, for 0 ≤ n ≤ 16: 129
- 66. Fig. P7.21 Desired and noisy inputs of Problem 7.16. Fig. P7.22 Outputs of comb filters 1 and 2 of Problem 7.16. n h1(n) h2(n) 0 0.70849 0.98021 1 0 0 2 0 0 3 0 0 4 0 0 5 0 0 6 0 0 7 0 0 8 0.75277 0.07841 9 0 0 10 0 0 11 0 0 12 0 0 13 0 0 14 0 0 15 0 0 16 0.04705 −0.03489 130 The impulse responses of the two filters can be computed by the same program as above, but with changing the input x(n) into an impulse δ(n). The computed impulse responses are shown in Fig. P7.23 Fig. P7.23 Impulse responses of comb filters 1 and 2 of Problem 7.17. Problem 7.18 The following program replaces iirfilt.c of Problem 7.13. For comparison, dir.c is left in the program as before. /* iirfilt.c - IIR filtering experiments - circular buffer versions */ #include <stdio.h> #include <stdlib.h> void cas2can(); double dir(), ccan(), ccas(); void main() { int i, j, K, M; double x, ydir, ycan, ycas; double *vdir, *wdir, *wcan, **W; double **A, **B, *a, *b; double *p, **P; /* circular pointers */ FILE *fpx; fpx = fopen("x.dat", "r"); printf("no. sections K = "); scanf("%d", &K); M=2*K; wdir = (double *) calloc(M+1, sizeof(double)); vdir = (double *) calloc(M+1, sizeof(double)); wcan = (double *) calloc(M+1, sizeof(double)); p = wcan; /* initialize */ 131
- 67. a = (double *) calloc(M+1, sizeof(double)); b = (double *) calloc(M+1, sizeof(double)); A = (double **) calloc(K, sizeof(double *)); B = (double **) calloc(K, sizeof(double *)); W = (double **) calloc(K, sizeof(double *)); P = (double **) calloc(K, sizeof(double *)); /* allocate */ for (i=0; i<K; i++) { A[i] = (double *) calloc(3, sizeof(double)); B[i] = (double *) calloc(3, sizeof(double)); W[i] = (double *) calloc(3, sizeof(double)); P[i] = W[i]; /* initialize */ } for(i=0; i<K; i++) { printf("enter A[%d] = ", i); for(j=0; j<3; j++) scanf("%lf", A[i]+j); } for(i=0; i<K; i++) { printf("enter B[%d] = ", i); for(j=0; j<3; j++) scanf("%lf", B[i]+j); } cas2can(K, A, a); cas2can(K, B, b); printf("nn"); while(fscanf(fpx, "%lf", &x) != EOF) { ydir = dir(M, a, M, b, vdir, wdir, x); ycan = ccan(M, a, b, wcan, &p, x); ycas = ccas(K, A, B, W, P, x); printf("% .6f % .6f % .6f % .6fn", x, ydir, ycan, ycas); } printf("a, b = n"); for (i=0;i<=M;i++) printf("% .6lf % .6lfn", a[i], b[i]); } One minor difference in ccan is that it assumes the same order M for both vectors a and b; Therefore its usage is: ycan = ccan(M, a, b, wcan, &p, x); whereas the usage of can was: ycan = can(M, a, M, b, wcan, x); Problem 7.19 The canonical realization is shown in Fig. P7.24. The linear and circular buffer versions of the sample processing algorithm are as follows: 132 x 0.5 -2 6 y z-3 s3w3 s0w0 Fig. P7.24 Canonical realization of Problem 7.19. for each input x do: w0 = x + 0.5w3 y = 6w0 − 2w3 delay(3, w) for each input x do: s3 = tap(3, w, p, 3) s0 = x + 0.5s3 y = 6s0 − 2s3 ∗p = s0 cdelay(3, w, &p) The impulse response can be obtained by dividing out the numerator and writing H(z)= 6 − 2z−3 1 − 0.5z−3 = 4 + 2 1 − 0.5z−3 Expanding in powers of z−3 , we have: H(z)= 4 + 2 1 + (0.5)z−3 + (0.5)2 z−6 + (0.5)3 z−9 + · · · which gives the impulse response: h = 6, 0, 0, 2(0.5), 0, 0, 2(0.5)2 , 0, 0, 2(0.5)3 , 0, 0, . . . = [6, 0, 0, 1, 0, 0, 0.5, . . . ] n x w0 w1 w2 w3 s0 s1 s2 s3 y = 6s0 − 2s3 0 1 1 0 0 0 1 0 0 0 6 1 0 1 0 0 0 0 1 0 0 0 2 0 1 0 0 0 0 0 1 0 0 3 0 1 0.5 0 0 0.5 0 0 1 1 4 0 0 0.5 0 0 0 0.5 0 0 0 5 0 0 0.5 0 0 0 0 0.5 0 0 6 0 0 0.5 0.52 0 0.52 0 0 0.5 0.5 The states si represent the delay-line outputs: s(n)= ⎡ ⎢ ⎢ ⎢ ⎣ s0(n) s1(n) s2(n) s3(n) ⎤ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ w(n) w(n − 1) w(n − 2) w(n − 3) ⎤ ⎥ ⎥ ⎥ ⎦ that is, they are the successively delayed versions of w(n), which is the intermediate signal obtained at the output of the input adder. It has z-transform: W(z)= 1 D(z) X(z)= 1 1 − 0.5z−3 X(z) 133
- 68. For a delta-function input, we have X(z)= 1, giving: W(z)= 1 1 − 0.5z−3 = 1 + (0.5)z−3 + (0.5)2 z−6 + (0.5)3 z−9 + · · · and in the time domain: w(n)= [1, 0, 0, 0.5, 0, 0, 0.52 , 0, 0, . . . ] For the input x = [1, 2, 3], we have: n x w0 w1 w2 w3 s0 s1 s2 s3 y = 6s0 − 2s3 0 1 1 0 0 0 1 0 0 0 6 1 2 1 0 0 2 2 1 0 0 12 2 3 1 0 3 2 3 2 1 0 18 3 0 1 0.5 3 2 0.5 3 2 1 1 4 0 1 0.5 3 2 1 0.5 3 2 2 5 0 1 0.5 3 1.5 1.5 1 0.5 3 3 6 0 1 0.5 0.25 1.5 0.25 1.5 1 0.5 0.5 Problem 7.20 The full precision direct form denominator is: a = [1, a1, a2, a3, a4]= [1, −3.502, 5.024, −3.464, 0.979] (P7.1) The full precision cascade coefficients are: a0 = [1, a01, a02]= [1, −1.8955, 0.9930] a1 = [1, a11, a12]= [1, −1.6065, 0.9859] (P7.2) The poles are at: p0 = R0ejω0 , R0 = 0.9965, ω0 = 0.1π p1 = R1ejω1 , R1 = 0.9929, ω1 = 0.2π (P7.3) If we round the direct form coefficient vector a to 2-digit accuracy, we obtain: ˆa = [1, ˆa1, ˆa2, ˆa3, ˆa4]= [1, −3.50, 5.02, −3.46, 0.98] (P7.4) and the corresponding direct form transfer function: ˆHdir(z)= 1 1 + ˆa1z−1 + ˆa2z−2 + ˆa3z−3 + ˆa4z−4 (P7.5) This filter is unstable. Indeed, its four poles, the roots of the denominator polynomial, are: ˆp0 = ˆR0ej ˆω0 , ˆR0 = 1.0045, ˆω0 = 0.1045π ˆp1 = ˆR1ej ˆω1 , ˆR1 = 0.9860, ˆω1 = 0.1989π (P7.6) and their conjugates. Thus, the p0 pole has moved outside the unit circle, rendering the filter unstable. The seemingly benign replacement of a by the rounded ˆa has made the direct form realization unusable. The impulse responses of the two filters (7.7.1) and (P7.5) are shown in Fig. P7.25. The latter is blowing up exponentially. 134 Fig. P7.25 Full and 2-digit precision filters. The cascade realization, on the other hand, is more robust, in the sense that if we round its coeffi- cients also to 2-digit accuracy, the filter will remain stable. Indeed, rounding the SOS coefficients (P7.2) to 2-digits gives: ˆa0 = [1, ˆa01, ˆa02]= [1, −1.90, 0.99] ˆa1 = [1, ˆa11, ˆa12]= [1, −1.61, 0.99] (P7.7) The roots of these sections are: ˆp0 = ˆR0ej ˆω0 , ˆR0 = 0.9950 ˆω0 = 0.0961π ˆp1 = ˆR1ej ˆω1 , ˆR1 = 0.9950, ˆω1 = 0.2000π (P7.8) and their conjugates. They are both inside the unit circle, therefore, the filter remains stable. The corresponding cascade transfer function will be: ˆHcas(z)= 1 1 + ˆa01z−1 + ˆa02z−2 · 1 1 + ˆa11z−1 + ˆa12z−2 (P7.9) The filters ˆHdir(z) and ˆHcas(z) are not equal. Rounding the cascade coefficients to 2-digit ac- curacy is not equivalent to rounding the direct form coefficients, because rounding does not preserve products—the rounded product of two numbers is not equal to the product of the two rounded numbers. Therefore, rounding does not preserve convolution, that is, ˆa = a0 ∗ a1 ≠ ˆa0 ∗ ˆa1 Indeed, carrying out the convolution of the two vectors in Eq. (P7.7) gives the following direct form vector, which is not quite the same as that given by Eq. (P7.4): ˆa0 ∗ ˆa1 = [1, −3.5100, 5.0390, −3.4749, 0.9801] The impulse response and the magnitude response of the cascaded filter (P7.9) are shown in Fig. P7.26, together with the exact magnitude response. The magnitude responses are plotted in dB, that is, 20 log10 |H(ω)|, over 0 ≤ ω ≤ 0.5π, which is the right half of the Nyquist interval. Thus, in the cascade case the rounding operation changed the magnitude response somewhat, but has not affected the stability of the filter. The sensitivity of the direct form and the robustness of the cascade form can be understood ana- lytically in the following way. The zeros of a polynomial are indirect functions of the coefficients 135
- 69. Fig. P7.26 Impulse and magnitude responses of 2-digit-precision cascade form. of the polynomial, therefore, small changes in the latter will induce changes in the former. For the direct form, the induced changes in the zeros can be large, thus, causing them to move outside the unit circle. In the cascade form, the induced changes remain small. For example, consider the above fourth order denominator vector a. The relationship of the zeros to the coefficients is established via the polynomial identity: (1 − p0z−1 )(1 − p∗ 0 z−1 )(1 − p1z−1 )(1 − p∗ 1 z−1 )= 1 + a1z−1 + a2z−2 + a3z−3 + a4z−4 A small change in the coefficients will induce a small change in the zeros p0 and p1, that is, a → a + da ⇒ pi → pi + dpi, i = 0, 1 To determine dp0 and dp1 in terms of da, we take differentials of both sides of the above poly- nomial identity to get: − dp0z−1 (1 − p∗ 0 z−1 )(1 − p1z−1 )(1 − p∗ 1 z−1 ) − (1 − p0z−1 )dp∗ 0 z−1 (1 − p1z−1 )(1 − p∗ 1 z−1 ) − (1 − p0z−1 )(1 − p∗ 0 z−1 )dp1z−1 (1 − p∗ 1 z−1 ) − (1 − p0z−1 )(1 − p∗ 0 z−1 )(1 − p1z−1 )dp∗ 1 z−1 = da1z−1 + da2z−2 + da3z−3 + da4z−4 which is still an identity in z. Setting z = p0 will remove all but the first term in the left-hand-side, giving the relationship: −dp0p−1 0 (1 − p∗ 0 p−1 0 )(1 − p1p−1 0 )(1 − p∗ 1 p−1 0 )= da1p−1 0 + da2p−2 0 + da3p−3 0 + da4p−4 0 Multiplying by p4 0 and solving for dp0, we obtain dp0 = − p3 0da1 + p2 0da2 + p0da3 + da4 (p0 − p∗ 0 )(p0 − p1)(p0 − p∗ 1 ) (P7.10) Similarly, setting z = p1, we find for dp1: dp1 = − p3 1da1 + p2 1da2 + p1da3 + da4 (p1 − p∗ 0 )(p1 − p0)(p1 − p∗ 1 ) (P7.11) 136 Because, the denominator involves the differences between poles, that is, the distances from all the other poles, it follows that if the poles are closely clustered, the induced changes in the poles dpi will be large even if the coefficient changes dai are small. The above sensitivity formulas generalize easily [2,3,246] to the case of an order-M filter a = [1, a1, a2, . . . , aM] with M zeros p1, p2, . . . , pM: dpi = ∂pi ∂a1 da1 + ∂pi ∂a2 da2 + · · · + ∂pi ∂aM daM where the partial derivatives are: ∂pi ∂am = − pM−m i M j=1 j≠i (pi − pj) (P7.12) In the cascade realization, p0 is associated with the first SOS and p1 with the second. The depen- dence of p0 on a0 and p1 on a1 is established through the identities (1 − p0z−1 )(1 − p∗ 0 z−1 ) = 1 + a01z−1 + a02z−2 (1 − p1z−1 )(1 − p∗ 1 z−1 ) = 1 + a11z−1 + a12z−2 Using the above procedure of taking differentials of both sides and setting z = p0 in the first and z = p1 in the second, we obtain the induced changes in the poles due to small changes in the coefficients of the sections: dp0 = − p0da01 + da02 (p0 − p∗ 0 ) , dp1 = − p1da11 + da12 (p1 − p∗ 1 ) (P7.13) The sensitivity of each pole depends only on the SOS it belongs to, not on the other sections. In Eqs. (P7.10) and (P7.11), the sensitivity of each pole is coupled to the other poles, through the denominator factors of the form (p0 − p1); if p0 is near p1, such factors cause large shifts in the pole locations. Even though the sensitivity formulas (P7.10) and (P7.11) involve differential changes, we may actually apply them to the above example and estimate the pole shifts for the 2-digit case. Writing the rounded ˆa of Eq. (P7.4) in the form ˆa = a + da, we obtain subtracting (P7.4) and (P7.1): da = ˆa − a = [0, da1, da2, da3, da4]= [0, 0.002, −0.004, 0.004, 0.001] Using the numerical values given in Eq. (P7.3), we have p0 = 0.9965ej0.1π = 0.9477 + j0.3079 p1 = 0.9929ej0.2π = 0.8033 + j0.5836 With these values of the poles and dai, we can calculate the pole shifts using (P7.10) and (P7.11). We find: dp0 = 0.0023 + j0.0159, dp1 = −0.0033 − j0.0066 Therefore, the new pole locations will be: ˆp0 = p0 + dp0 = 0.9501 + j0.3238 = 1.0037ej0.1046π ˆp1 = p1 + dp1 = 0.7999 + j0.5770 = 0.9863ej0.1989π which compare well with the actual values of Eq. (P7.6). A similar calculation can be made based on Eq. (P7.13). 137
- 70. Problem 7.21 The numerical values of a are shown in the first column of Table P7.1. The second and third columns of this table show the quantized versions of a, rounded to 5- and 4-digit accuracy. a 5-digit ˆa 4-digit ˆa ˆacas 1.00000000 1.00000000 1.00000000 1.00000000 −5.56573395 −5.56573000 −5.56570000 −5.56570000 13.05062482 13.05062000 13.05060000 13.05046030 −16.49540451 −16.49540000 −16.49540000 −16.49508213 11.84993647 11.84994000 11.84990000 11.84961490 −4.58649943 −4.58650000 −4.58650000 −4.58633537 0.74713457 0.74713000 0.74710000 0.74710016 relative error 3.2580×10−7 1.6487×10−6 1.9240×10−5 Table P7.1 Exact, 5-digit, and 4-digit coefficients. The fourth column of Table P7.1 shows the equivalent direct-form coefficient vector ˆacas arising from the cascade of the quantized second order section coefficients, rounded to 4-digit accuracy. The rounded ai are: ˆa0 = [1, −1.8671, 0.9623] ˆa1 = [1, −1.8468, 0.8992] ˆa2 = [1, −1.8518, 0.8634] (P7.14) The table also shows the relative error in each case, defined as the ratio relative error = da a = ˆa − a a where a denotes the length of the vector a. As expected, the error in the 5-digit case is less than that in the 4-digit case. But, the error of the cascade case is larger. However, as we see below, the 4-digit cascade form gives an adequate approximation of the exact filter, whereas both the 5- and 4-digit direct forms fail. This behavior can be traced to the pole sensitivity formula Eq. (P7.12) which can cause large shifts in the poles even though da is small. Table P7.2 shows the poles of the exact and approximate filters, that is, the roots of the denomi- nator polynomials a, 5- and 4-digit ˆa, and of the rounded SOS polynomials ˆa0, ˆa1, and ˆa2. For the 5-digit direct form, the filter remains stable. But, for the 4-digit direct form, one root of ˆa has moved outside the unit circle and the third conjugate pair of roots has been replaced by two real-valued ones, one almost on the unit circle. Thus, the quantized coefficients ˆa correspond to an unstable filter. By contrast, the roots of the 4-digit quantized cascaded form ˆacas remain inside the unit circle, and in fact they are practically equal to the unquantized ones. Figure P7.27 shows the impulse responses of the exact filter based on a and the 4-digit-precision filter based on ˆa, which is unstable. The impulse responses of the 5-digit case and the 4-digit cascade case based on ˆacas are shown in Fig. P7.28. The magnitude responses of the full precision, 5-digit direct, and 4-digit cascade cases are shown in Fig. P7.29. Note that the 5-digit direct no longer meets the passband specs; it has a passband ripple of almost 2 dB instead of the required 1 dB. The 4-digit cascade response, on the other hand, meets the specifications. 138 full precision magnitude 0.933557 ± j0.301259 0.980962 0.923398 ± j0.215739 0.948265 0.925911 ± j0.078311 0.929217 5-digit direct magnitude 0.931610 ± j0.300414 0.978849 0.927787 ± j0.216576 0.952730 0.923469 ± j0.079185 0.926857 4-digit direct magnitude 0.967208 ± j0.284105 1.008071 0.888774 ± j0.266869 0.927976 0.999999 0.999999 0.853736 0.853736 4-digit cascade magnitude 0.933552 ± j0.301303 0.980971 0.923396 ± j0.215719 0.948258 0.925902 ± j0.078152 0.929194 Table P7.2 Filter poles. Fig. P7.27 Impulse responses of full-precision and 4-digit-precision filters. Problem 7.21. Problem 7.22 See the solution of Problem 7.20. Problem 7.23 See the solution of Problem 7.20. 139
- 71. Fig. P7.28 Impulse responses of 5-digit direct and 4-digit cascade filters. Problem 7.21. Fig. P7.29 Magnitude responses of exact, 5-digit direct, and 4-digit cascade. Problem 7.21. Problem 7.24 The transfer function is: H(z)= b0 + b1z−1 + b2z−2 + b3z−3 1 + a1z−1 + a2z−2 + a3z−3 (P7.15) Transposed realizations of FIR filters are obtained by the four transposition rules of exchanging input with output, reversing all flows, replacing nodes by adders, and adders by nodes. These rules can be applied to the canonical or direct realizations of IIR filters. The I/O difference equations describing the time-domain operation of the realization and the corresponding sample processing algorithm can be obtained by introducing as internal states the contents of the three delay registers at time n: v1(n), v2(n), v3(n). For indexing convenience, we also introduce the quantity v0(n) which is the overall output obtained at the top adder, that is, v0(n)= y(n)= b0x(n)+v1(n) The output of each delay is the delayed version of the output of the adder below it, for example, v1(n)= b1x(n − 1)−a1v0(n − 1)+v2(n − 1)= b1x(n − 1)−a1y(n − 1)+v2(n − 1) 140 or, advancing it to the next time instant: v1(n + 1)= b1x(n)−a1v0(n)+v2(n)= b1x(n)−a1y(n)+v2(n) Therefore, the overall I/O system, which includes the updating of the internal states, will be: y(n)= v0(n)= b0x(n)+v1(n) v1(n + 1)= b1x(n)−a1y(n)+v2(n) v2(n + 1)= b2x(n)−a2y(n)+v3(n) v3(n + 1)= b3x(n)−a3y(n) (P7.16) which leads to the sample processing algorithm: for each input sample x do: y = v0 = b0x + v1 v1 = b1x − a1y + v2 v2 = b2x − a2y + v3 v3 = b3x − a3y Problem 7.25 In the case of a general transfer function of the type of Eq. (7.1.4), we have the sample processing algorithm, where we assumed L = M for simplicity: for each input sample x do: y = v0 = b0x + v1 for i = 1, 2, . . . , M−1 do: vi = bix − aiv0 + vi+1 vM = bMx − aMv0 (P7.17) The following C function transp.c is an implementation. /* transp.c - IIR filtering in transposed of canonical form */ double transp(M, a, b, v, x) usage: y = transp(M, a, b, v, x); double *a, *b, *v, x; v = internal state vector int M; a,b have same order M { int i; v[0] = b[0] * x + v[1]; output y(n)= v0(n) for (i=1; i<=M-1; i++) state updating v[i] = b[i] * x - a[i] * v[0] + v[i+1]; v[M] = b[M] * x - a[M] * v[0]; return v[0]; } Its usage is the same as that of can or dir. The (M+1)-dimensional array v = [v0, v1, . . . , vM] must be declared and allocated in the main program, for example, by 141
- 72. double *v; v = (double *) calloc(M+1, sizeof(double)); Transposed realizations are not intuitively obvious, but their overall implementation in terms of Eq. (P7.17) has a certain simplicity and elegance. It corresponds to writing the transfer function of the filter in a nested form, which is similar to H¨orner’s rule of evaluating a polynomial. For example, the transfer function (P7.15) can be written as: H(z)= −(a1z−1 + a2z−2 + a3z−3 )H(z)+b0 + b1z−1 + b2z−2 + b3z−3 which can be written in the nested form: H = b0 + z−1 (b1 − a1H)+z−1 (b2 − a2H)+z−1 (b3 − a3H) Multiplying both sides by the z-transform X(z) of the input x(n), and using Y(z)= H(z)X(z), we obtain the following nested form for the computation of the output, which is basically the transposed form: Y = b0X + z−1 (b1X − a1Y)+z−1 (b2X − a2Y)+z−1 (b3X − a3Y) The transposed form can also be used in the cascade realization, where each second order section will be realized in its transposed form. The sample processing algorithm of a 2nd order section is simply: for each input sample x do: y = v0 = b0x + v1 v1 = b1x − a1y + v2 v2 = b2x − a2y which can replace Eq. (7.2.2). Problem 7.26 The following program implements filtering in the transposed form, calling the routine transp.c of Problem 7.25. It replaces canfilt.c of Problem 7.14. Because transp assumes that a and b have the same length M, this program aborts if the lengths are unequal. /* trspfilt.c - IIR filtering in transposed form */ #include <stdlib.h> #include <stdio.h> #define MAX 64 initial allocation size double transp(); void main(int argc, char **argv) { FILE *fpa, *fpb; coefficient files double *a, *b, *v, x, y; coeffs., state, input, output int M, L, i; int max = MAX, dmax = MAX; initial allocation and increment if (argc != 3) { 142 fprintf(stderr, "Usage: trspfilt a.dat b.dat <x.dat >y.datn"); fprintf(stderr, "coefficient vectors a,b must have same length"); exit(0); } if ((fpa = fopen(argv[1], "r")) == NULL) { fprintf(stderr, "Can’t open filter file: %sn", argv[1]); exit(0); } if ((fpb = fopen(argv[2], "r")) == NULL) { fprintf(stderr, "Can’t open filter file: %sn", argv[2]); exit(0); } a = (double *) calloc(max + 1, sizeof(double)); for (M=0;; M++) { if (M == max) { max += dmax; a = (double *) realloc((char *) a, (max + 1) * sizeof(double)); } if (fscanf(fpa, "%lf", a + M) == EOF) break; } M--; a = (double *) realloc((char *) a, (M + 1) * sizeof(double)); b = (double *) calloc(max + 1, sizeof(double)); for (L=0;; L++) { if (L == max) { max += dmax; b = (double *) realloc((char *) b, (max + 1) * sizeof(double)); } if (fscanf(fpb, "%lf", b + L) == EOF) break; } L--; b = (double *) realloc((char *) b, (L + 1) * sizeof(double)); if (M != L) { fprintf(stderr, "coefficient vectors a,b must have same length"); exit(0); } v = (double *) calloc(M + 1, sizeof(double)); while(scanf("%lf", &x) != EOF) { process input samples y = transp(M, a, b, v, x); printf("%lfn", y); } } Problem 7.27 The transfer function is: H(z)= 9 − 4z−2 + 4z−4 1 + 0.84z−2 + 0.25z−4 = 3 − 4z−1 + 2z−2 1 − 0.4z−1 + 0.5z−2 · 3 + 4z−1 + 2z−2 1 + 0.4z−1 + 0.5z−2 The transpose of the canonical form is shown in Fig. P7.30. The I/O difference equations are: 143
- 73. y(n)= 9x(n)+v1(n) v1(n + 1)= v2(n) v2(n + 1)= −4x(n)−0.84y(n)+v3(n) v3(n + 1)= v4(n) v4(n + 1)= 4x(n)−0.25y(n) and the corresponding sample processing algorithm: for each input sample x do: y = 9x + v1 v1 = v2 v2 = −4x − 0.84y + v3 v3 = v4 v4 = 4x − 0.25y The cascade form with both 2nd order sections realized in their transposed form is shown in Fig. P7.31. The I/O difference equations are in this case: x1(n)= 3x(n)+v01(n) v01(n + 1)= −4x(n)+0.4x1(n)+v02(n) v02(n + 1)= 2x(n)−0.5x1(n) y(n)= 3x1(n)+v11(n) v11(n + 1)= 4x1(n)−0.4y(n)+v12(n) v12(n + 1)= 2x1(n)−0.5y(n) The sample processing algorithm will be: for each input sample x do: x1 = 3x + v01 v01 = −4x + 0.4x1 + v02 v02 = 2x − 0.5x1 y = 3x1 + v11 v11 = 4x1 − 0.4y + v12 v12 = 2x1 − 0.5y Problem 7.28 The overall, closed-loop, transfer function of the feedback system can be worked out easily using z-transforms. The general approach to analyzing this, and any other, block diagram is to assign names to all signal lines that do not already have names and then set up the various adder and filtering equations. Consider case A, with signal names assigned as in Fig. P7.32 Then, the adder and filtering equa- tions are in the time and z domains: 144 v1(n) v0(n) z-1 x(n) 9 -0.84 -0.25 -4 4 y(n) v2(n) z-1 v3(n) z-1 v4(n) z-1 Fig. P7.30 Transposed of canonical form. x(n) x1(n) v01(n) v00(n) v02(n) z-1 z-1 3 0.4 -0.5 -4 2 y(n) v11(n) v10(n) v12(n) z-1 z-1 3 -0.4 -0.5 4 2 Fig. P7.31 Cascade form with transposed sections. x(n) e(n) f(n) g(n) y(n) z-1 H1(z) H2(z) Fig. P7.32 Case A of Problem 7.28 e(n) = x(n)+f(n) g(n) = e(n − 1) y(n) = h1(n)∗g(n) f(n) = h2(n)∗y(n) ⇒ E(z) = X(z)+F(z) G(z) = z−1 E(z) Y(z) = H1(z)G(z) F(z) = H2(z)Y(z) 145
- 74. Eliminating F(z) and E(z) in favor of X(z) and Y(z) gives: Y(z)= z−1 H1(z)E(z)= z−1 H1(z) X(z)+F(z) = z−1 H1(z) X(z)+H2(z)Y(z) and 1 − z−1 H1(z)H2(z) Y(z)= z−1 H1(z)X(z) Solving for the closed-loop transfer function H(z)= Y(z)/X(z), we find: H(z)= z−1 H1(z) 1 − z−1H1(z)H2(z) Case B has the same transfer function H(z). Cases C and D have: H(z)= H1(z) 1 − z−1H1(z)H2(z) Problem 7.29 The overall transfer function of this system can be computed using the results of Problem 7.28 and the individual transfer functions: H1(z)= 1 + z−1 1 − 0.5z−1 , H2(z)= z−1 0.4(1 − z−1 ) 1 − 0.4z−1 H(z)= H1(z) 1 − H1(z)H2(z) = 1 + z−1 1 − 0.5z−1 1 − 1 + z−1 1 − 0.5z−1 · 0.4(1 − z−1 )z−1 1 − 0.4z−1 which gives H(z)= 1 + 0.6z−1 − 0.4z−2 1 − 1.3z−1 + 0.2z−2 + 0.4z−3 In case A, we assign internal state variables as shown in Fig. P7.33. The difference equations and sample processing algorithm, written in the right computational order, are: e(n) = x(n)+f(n − 1) w(n) = 0.5w(n − 1)+e(n) y(n) = w(n)+w(n − 1) v(n) = 0.4v(n − 1)+y(n) f(n) = 0.4v(n)−0.4v(n − 1) for each input sample x do: e = x + f1 w0 = 0.5w1 + e y = w0 + w1 v0 = 0.4v1 + y f0 = 0.4(v0 − v1) f1 = f0 (update delays) w1 = w0 v1 = v0 Notice how the extra delay factor z−1 in H2(z) delays the output f(n) and allows the start of the computational cycle with e(n)= x(n)+f(n − 1). The output f(n) is computed last. If this delay were not there, we would have e(n)= x(n)+f(n), which is not possible because f(n) is not yet 146 x(n) e(n) w(n) w0 f0 v0 w1 v1 f1 v(n) w(n-1) v(n-1) f(n)f(n-1) y(n) 0.5 z-1 0.4 z-1 z-1 -0.4 0.4 Fig. P7.33 Case A of Problem 7.29. computed. In case B, shown in Fig. P7.34, the overall transfer function remains the same. The change causes a re-ordering of the computational cycle. The difference equations and corresponding sample processing algorithm are now: v(n) = 0.4v(n − 1)+y(n − 1) f(n) = 0.4v(n)−0.4v(n − 1) e(n) = x(n)+f(n) w(n) = 0.5w(n − 1)+e(n) y(n) = w(n)+w(n − 1) for each input sample x do: v0 = 0.4v1 + y1 f = 0.4(v0 − v1) e = x + f w0 = 0.5w1 + e y = w0 + w1 y1 = y (update delays) w1 = w0 v1 = v0 x(n) e(n) f(n) w(n) w0 v0 w1 v1 y1 v(n) w(n-1) v(n-1) y(n-1) y(n) 0.5 z-1 0.4 z-1 -0.4 0.4 z-1 Fig. P7.34 Case B of Problem 7.29. Now, the start of the computational cycle is v(n)= 0.4v(n−1)+y(n−1), where both quantities 147
- 75. v(n − 1) and y(n − 1) are available from the previous time step. Note that y(n) is computed last and then put into the feedback delay to be used in the next time step. Problem 7.30 The filter G(z) can be implemented by the input/output sample processing algorithm of its canonical form: for each x do: y = 0.2x + 0.8w1 w1 = y where w1 is its internal state, that is, w1(n)= y(n − 1) in the difference equation: y(n)= 0.2x(n)+0.8y(n − 1). Fig. P7.35 shows the three cases with explicitly assigned signal labels. Identifying the proper input and output to the filter G(z) and using the above sample processing algorithm, we obtain the implementation of the three cases: for each x do: u = 0.2v4 + 0.8w1 w1 = u y = v0 = x + u delay(4, v) for each x do: y = x + v4 v0 = 0.2y + 0.8w1 w1 = v0 delay(4, v) for each x do: y = u0 = x + v2 v0 = 0.2u2 + 0.8w1 w1 = v0 delay(2, v), delay(2, u) z-4 G(z) z-4 G(z) z-2 z-2 G(z) (a) (b) (c) x u v0 v4 y x x v0 v0 u0 u2 v4 v2 y y Fig. P7.35 Feedback systems of Problem 7.30. 148 Chapter 8 Problems Problem 8.1 The transfer function is obtained by the period-3 repetition of the sequence b(n)= [0, 1, 2], that is, in the time- and z-domains: h(n) = b(n)+b(n − 3)+b(n − 6)+ · · · H(z) = B(z)+z−3 B(z)+z−6 B(z)+ · · · = B(z) 1 − z−3 = z−1 + 2z−2 1 − z−3 The direct and canonical realizations are shown in Fig. P8.1. Their sample processing algorithms are: for each input x do: v0 = x y = w0 = v1 + 2v2 + w3 delay(2, v) delay(3, w) for each input x do: w0 = x + w3 y = w1 + 2w2 delay(3, w) x y x y z-1 z-1 z-1 z-1 z-1 z-1 w0 w0 w1 w1 w2 w2 w3 w3 z-1 z-1 v0 v1 v2 2 2 Fig. P8.1 Direct and canonical realizations of Problem 8.1. For the direct form realization, if the input is a unit impulse so that x = 1 at the first iteration of the algorithm and zero for all other iterations, then the first three iterations will fill the feedback delay buffer w with the desired waveform samples. From then on, the feed-forward part of the algorithm will be zero, giving rise to the following generation algorithm, expressed in its of the linear and circular buffer versions: repeat forever: w0 = w3 delay(3, w) repeat forever: ∗p = tap(3, w, p, 3) cdelay(3, w, &p) The following table shows the initialization and steady part of the linear buffer case. The columns v0, v1, v2 are the input impulse and its delays: δ(n), δ(n − 1), δ(n − 2). At each iteration, w0 is replaced by the value of w3 and only then is the row of w’s shifted to the right to give the values of w1, w2, w3 of the next row. The w-columns are the successively delayed output signals y(n), y(n − 1), y(n − 2), y(n − 3): 149
- 76. n v0 v1 v2 w0 w1 w2 w3 y = w0 0 1 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 1 2 0 0 1 2 1 0 0 2 3 0 0 0 0 2 1 0 0 4 0 0 0 1 0 2 1 1 5 0 0 0 2 1 0 2 2 6 0 0 0 0 2 1 0 0 7 0 0 0 1 0 2 1 1 8 0 0 0 2 1 0 2 2 The circular version is given in the following table. The buffer entry that contains the current output sample is shown with an up-arrow symbol and it corresponds to the entry wq defined by the circulating value of q: n q w0 w1 w2 w3 y 0 0 ↑0 0 0 0 0 1 3 0 0 0 ↑1 1 2 2 0 0 ↑2 1 2 3 1 0 ↑0 2 1 0 n q w0 w1 w2 w3 y 4 0 ↑1 0 2 1 1 5 3 1 0 2 ↑2 2 6 2 1 0 ↑0 2 0 7 1 1 ↑1 0 2 1 n q w0 w1 w2 w3 y 8 0 ↑2 1 0 2 2 9 3 2 1 0 ↑0 0 10 2 2 1 ↑1 0 1 11 1 2 ↑2 1 0 2 n q w0 w1 w2 w3 y 12 0 ↑0 2 1 0 0 13 3 0 2 1 ↑1 1 14 2 0 2 ↑2 1 2 15 1 0 ↑0 2 1 0 The table entries circulate every D(D + 1)= 12 iterations. Thus, entry n = 3 and n = 15 will be the same. We used 15 iterations because they contain one complete cycle of the table entries after the first 3 initializing iterations. Problem 8.2 Expanding the denominator in powers of z−5 , will cause the period-5 replication of the numerator sequence b = [1, 2, 3, −4, −5]. The direct and canonical realizations are shown in Fig. P8.2. Their sample processing algorithms are: for each input x do: v0 = x w0 = v0 + 2v1 + 3v2 − 4v3 − 5v4 + w5 y = w0 delay(4, v) delay(5, w) for each input x do: w0 = x + w5 y = w0 + 2w1 + 3w2 − 4w3 − 5w4 delay(5, w) The steady parts of the linear and circular buffer generation algorithms are: repeat forever: w0 = w5 delay(5, w) repeat forever: ∗p = tap(5, w, p, 5) cdelay(5, w, &p) 150 x y x y z-1 z-1 z-1 z-1 w5 w0 w1 w2 w3 w4 z-1 z-1 z-1 z-1 z-1 w5 w0 w1 w2 w3 w4 z-1 z-1 z-1 z-1 v0 v1 v2 v3 v4 z-1 -5 -5 -4 -4 2 2 3 3 Fig. P8.2 Direct and canonical realizations of Problem 8.2. The following table shows the initialization and steady parts of the linear buffer case: n v0 v1 v2 v3 v4 w0 w1 w2 w3 w4 w5 y 0 1 0 0 0 0 1 0 0 0 0 0 1 1 0 1 0 0 0 2 1 0 0 0 0 2 2 0 0 1 0 0 3 2 1 0 0 0 3 3 0 0 0 1 0 −4 3 2 1 0 0 −4 4 0 0 0 0 1 −5 −4 3 2 1 0 −5 5 0 0 0 0 0 1 −5 −4 3 2 1 1 6 0 0 0 0 0 2 1 −5 −4 3 2 2 7 0 0 0 0 0 3 2 1 −5 −4 3 3 8 0 0 0 0 0 −4 3 2 1 −5 −4 −4 9 0 0 0 0 0 −5 −4 3 2 1 −5 −5 10 0 0 0 0 0 1 −5 −4 3 2 1 1 11 0 0 0 0 0 2 1 −5 −4 3 2 2 12 0 0 0 0 0 3 2 1 −5 −4 3 3 13 0 0 0 0 0 −4 3 2 1 −5 −4 −4 14 0 0 0 0 0 −5 −4 3 2 1 −5 −5 The circular version is given in the following table. The buffer entry that contains the current output sample is shown with an up-arrow symbol and it corresponds to the entry wq defined by the circulating value of q: n q w0 w1 w2 w3 w4 w5 y 0 0 1↑ 0 0 0 0 0 1 1 5 1 0 0 0 0 2↑ 2 2 4 1 0 0 0 3↑ 2 3 3 3 1 0 0 −4↑ 3 2 −4 4 2 1 0 −5↑ −4 3 2 −5 5 1 1 1↑ −5 −4 3 2 1 151
- 77. n q w0 w1 w2 w3 w4 w5 y 6 0 2↑ 1 −5 −4 3 2 2 7 5 2 1 −5 −4 3 3↑ 3 7 4 2 1 −5 −4 −4↑ 3 −4 9 3 2 1 −5 −5↑ −4 3 −5 10 2 2 1 1↑ −5 −4 3 1 11 1 2 2↑ 1 −5 −4 3 2 n q w0 w1 w2 w3 w4 w5 y 12 0 3↑ 2 1 −5 −4 3 3 13 5 3 2 1 −5 −4 −4↑ −4 14 4 3 2 1 −5 −5↑ −4 −5 15 3 3 2 1 1↑ −5 −4 1 16 2 3 2 2↑ 1 −5 −4 2 17 1 3 3↑ 2 1 −5 −4 3 n q w0 w1 w2 w3 w4 w5 y 18 0 −4↑ 3 2 1 −5 −4 −4 19 5 −4 3 2 1 −5 −5↑ −5 20 4 −4 3 2 1 1↑ −5 1 21 3 −4 3 2 2↑ 1 −5 2 22 2 −4 3 3↑ 2 1 −5 3 23 1 −4 −4↑ 3 2 1 −5 −4 n q w0 w1 w2 w3 w4 w5 y 24 0 −5↑ −4 3 2 1 −5 −5 25 5 −5 −4 3 2 1 1↑ 1 26 4 −5 −4 3 2 2↑ 1 2 27 3 −5 −4 3 3↑ 2 1 3 28 2 −5 −4 −4↑ 3 2 1 −4 29 1 −5 −5↑ −4 3 2 1 −5 n q w0 w1 w2 w3 w4 w5 y 30 0 1↑ −5 −4 3 2 1 1 31 5 1 −5 −4 3 2 2↑ 2 32 4 1 −5 −4 3 3↑ 2 3 33 3 1 −5 −4 −4↑ 3 2 −4 34 2 1 −5 −5↑ −4 3 2 −5 35 1 1 1↑ −5 −4 3 2 1 The table entries circulate every D(D + 1)= 30 iterations. Thus, entry n = 5 and n = 35 will be the same. We used 35 iterations because they contain one complete cycle of the table entries after the first 5 initializing iterations. 152 Problem 8.3 The shift is c = D/d = 8/5 = 1.6. The d = 5 possible values of the offset index q are obtained by iterating Eq. (8.1.36): q0 = 0 q1 = q0 − c = − 8 5 ≡ 8 − 8 5 = 32 5 = 6 2 5 q2 = q1 − c = 32 5 − 8 5 = 24 5 = 4 4 5 q3 = q2 − c = 24 5 − 8 5 = 16 5 = 3 1 5 q4 = q3 − c = 16 5 − 8 5 = 8 5 = 1 3 5 Fig. P8.3 shows the relative locations of the q’s with respect to the circular buffer. The five q-arrows are now at relative angles ω = 2π/5. w0 q0 q4 w4 w5 w6 w7 w2 w3 w1 q1 q3 q2 b0 b7 b6 b5 b4 b3 b2 b1 Fig. P8.3 Successive positions of q when d = 5. Down-truncation gives the following integer values for the q’s and corresponding buffer entries: [q0, q1, q2, q3, q4]= [0, 6, 4, 3, 1] w[0], w[6], w[4], w[3], w[1] = [b0, b2, b4, b5, b7] Up-truncation gives: [q0, q1, q2, q3, q4]= [0, 7, 5, 4, 2] w[0], w[7], w[5], w[4], w[2] = [b0, b1, b3, b4, b6] Rounding to the nearest integer gives: [q0, q1, q2, q3, q4]= [0, 6, 5, 3, 2] w[0], w[6], w[5], w[3], w[2] = [b0, b2, b3, b5, b6] For the linear interpolation case, we have the outputs: 153
- 78. q0 = 0 ⇒ y0 = w[0]= b0 6 < q1 < 7 ⇒ y1 = w[6]+(q1 − 6)(w[7]−w[6])= 2 5 b1 + 3 5 b2 4 < q2 < 5 ⇒ y2 = w[4]+(q2 − 4)(w[5]−w[4])= 4 5 b3 + 1 5 b4 3 < q3 < 4 ⇒ y3 = w[3]+(q3 − 3)(w[4]−w[3])= 1 5 b4 + 4 5 b5 1 < q4 < 2 ⇒ y4 = w[1]+(q4 − 1)(w[2]−w[1])= 3 5 b6 + 2 5 b7 Thus, depending on the output method, the following period-5 subsequences will be produced: [b0, b2, b4, b5, b7, b0, b2, b4, b5, b7, . . . ] (truncate down) [b0, b1, b3, b4, b6, b0, b1, b3, b4, b6, . . . ] (truncate up) [b0, b2, b3, b5, b6, b0, b2, b3, b5, b6, . . . ] (round) [y0, y1, y2, y3, y4, y0, y1, y2, y3, y4, . . . ] (interpolation) where the y’s were given above. Problem 8.4 The shift is c = D/d = 8/6 = 4/3. The d = 6 possible values of the offset index q are obtained by iterating Eq. (8.1.36): q0 = 0 q1 = q0 − c = − 4 3 ≡ 8 − 4 3 = 20 3 = 6 2 3 q2 = q1 − c = 20 3 − 4 3 = 16 3 = 5 1 3 q3 = q2 − c = 16 3 − 4 3 = 12 3 = 4 q4 = q3 − c = 12 3 − 4 3 = 8 3 = 2 2 3 q5 = q4 − c = 8 3 − 4 3 = 4 3 = 1 1 3 Fig. P8.4 shows the relative locations of the q’s with respect to the circular buffer. The six q-arrows are now at relative angles ω = 2π/6. Down-truncation gives the following integer values for the q’s and corresponding buffer entries: [q0, q1, q2, q3, q4, q5]= [0, 6, 5, 4, 2, 1] w[0], w[6], w[5], w[4], w[2], w[1] = [b0, b2, b3, b4, b6, b7] Up-truncation gives: [q0, q1, q2, q3, q4, q5]= [0, 7, 6, 4, 3, 2] w[0], w[7], w[6], w[4], w[3], w[2] = [b0, b1, b2, b4, b5, b6] 154 w0 q0 q1 q2 q3 q4 q5 w4 w5 w6 w7 w2 w3 w1 b0 b7 b6 b5 b4 b3 b2 b1 Fig. P8.4 Successive positions of q when d = 6. Rounding to the nearest integer gives: [q0, q1, q2, q3, q4, q5]= [0, 7, 5, 4, 3, 1] w[0], w[7], w[5], w[4], w[3], w[1] = [b0, b1, b3, b4, b5, b7] For the linear interpolation case, we have the outputs: q0 = 0 ⇒ y0 = w[0]= b0 6 < q1 < 7 ⇒ y1 = w[6]+(q1 − 6)(w[7]−w[6])= 2 3 b1 + 1 3 b2 5 < q2 < 6 ⇒ y2 = w[5]+(q2 − 5)(w[6]−w[5])= 1 3 b2 + 2 3 b3 q3 = 4 ⇒ y3 = w[4]= b4 2 < q4 < 3 ⇒ y4 = w[2]+(q4 − 2)(w[3]−w[2])= 2 3 b5 + 1 3 b6 1 < q5 < 2 ⇒ y5 = w[1]+(q5 − 1)(w[2]−w[1])= 1 3 b6 + 2 3 b7 Thus, depending on the output method, the following period-5 subsequences will be produced: [b0, b2, b3, b4, b6, b7, b0, b2, b3, b4, b6, b7, . . . ] (truncate down) [b0, b1, b2, b4, b5, b6, b0, b1, b2, b4, b5, b6, . . . ] (truncate up) [b0, b1, b3, b4, b5, b7, b0, b1, b3, b4, b5, b7, . . . ] (round) [y0, y1, y2, y3, y4, y5, y0, y1, y2, y3, y4, y5, . . . ] (interpolate) where the y’s are given above. Problem 8.5 This problem is similar to Example 8.1.2. Referring to Fig. 8.1.14, we have for the cases of trun- cating down, up, and rounding: [b0, b3, b6, b0, b3, b6, . . . ]= [1, 4, 7, 1, 4, 7, . . . ] (truncate down) [b0, b2, b5, b0, b2, b5, . . . ]= [1, 3, 6, 1, 3, 6, . . . ] (truncate up) [b0, b3, b5, b0, b3, b5, . . . ]= [1, 4, 6, 1, 4, 6, . . . ] (round) 155
- 79. If we interpolate linearly, we get the sequence: y0 = b0 = 1 y1 = 1 3 b2 + 2 3 b3 = 1 3 3 + 2 3 4 = 11 3 = 3.667 y2 = 1 3 b5 + 2 3 b6 = 1 3 6 + 2 3 7 = 20 3 = 6.667 so that the period-3 sequence is: [1, 3.667, 6.667, 1, 3.667, 6.667, . . . ] Problem 8.6 This problem is similar to Problem 8.3. Inserting the numerical values for the wavetable contents: b = [b0, b1, b2, b4, b5, b6, b7]= [1, 2, 3, 4, 5, 6, 7, 8] we find for the period-5 sequences: [b0, b2, b4, b5, b7, . . . ]= [1, 3, 4, 5, 7 . . . ] (truncate down) [b0, b1, b3, b4, b6, . . . ]= [1, 2, 3, 5, 7, . . . ] (truncate up) [b0, b2, b3, b5, b6, . . . ]= [1, 3, 4, 6, 7, . . . ] (round) [y0, y1, y2, y3, y4, . . . ]= [1, 2.6, 4.2, 5.8, 7.4, . . . ] (interpolation) where y0 = b0 = 1 y1 = 2 5 b1 + 3 5 b2 = 2 5 2 + 3 5 3 = 2.6 y2 = 4 5 b3 + 1 5 b4 = 4 5 4 + 1 5 5 = 4.2 y3 = 1 5 b4 + 4 5 b5 = 1 5 5 + 1 5 6 = 5.8 y4 = 3 5 b6 + 2 5 b7 = 3 5 7 + 2 5 8 = 7.4 The case d = 6 is similar to Problem 8.4. Problem 8.7 An example program that exercises the wavetable routines is listed below. It generates three wavetables: sinusoidal, square wave, and trapezoidal. /* wavgenex.c - circular wavetable example */ #include <stdio.h> #include <stdlib.h> #include <math.h> void gdelay2(); double wavgen(); double sine(), square(), trapez(); 156 void main(int argc, char **argv) { int i, n, D; double *w1, q1, *w2, q2, *w3, q3, c, A, F; FILE *fp1, *fp2, *fp3; if (argc != 4) { fprintf(stderr, "usage: wavgenex A, c, Dn"); fprintf(stderr, "A = wavetable amplituden"); fprintf(stderr, "c = no. of periods in D samples, F=c/Dn"); fprintf(stderr, "D = wavetable sizen"); exit(0); } A = atof(argv[1]); c = atof(argv[2]); D = atoi(argv[3]); F = c / D; fp1 = fopen("y1.dat", "w"); fp2 = fopen("y2.dat", "w"); fp3 = fopen("y3.dat", "w"); w1 = (double *) calloc(D, sizeof(double)); w2 = (double *) calloc(D, sizeof(double)); w3 = (double *) calloc(D, sizeof(double)); q1 = 0; q2 = 0; q3 = 0; for (i=0; i<D; i++) { /* load wavetables */ w1[q1] = sine(D, i); /* might need the cast w1[(int)q1] */ w2[q2] = square(D/2, i); w3[q3] = trapez(D, 3*D/4, 0, i); gdelay2(D-1, 1.0, &q1); gdelay2(D-1, 1.0, &q2); gdelay2(D-1, 1.0, &q3); } for (n=0; n<D; n++) { fprintf(fp1, "%lfn", wavgen(D, w1, A, F, &q1)); fprintf(fp2, "%lfn", wavgen(D, w2, A, F, &q2)); fprintf(fp3, "%lfn", wavgen(D, w3, A, F, &q3)); } } Problem 8.8 Partial C code for this problem was given in the text, just before Figs. 8.1.20 and 8.1.21. Problem 8.9 The zero patterns and magnitude responses |H(ω)| of the four filters are shown in Fig. 8.9. The first filter has H(z)= 1 + z−8 . Therefore, its zeros are the solutions of: 157
- 80. z8 = −1 = ejπ e2πjk ⇒ zk = ejπ/8 ej2πk/8 , k = 0, 1, . . . , 7 That is, the 8th roots of unity rotated by π/8 ≡= 22.5o . The second filter has H(z)= 1 − z−8 and thus, its zeros are the 8th roots of unity. 2 2 ω 2π 2π |H(ω)| ω 3 2π ω 3 2π ω Fig. P8.5 Comb filters of Problem 8.9. The third and fourth filters can be written as follows, using the finite geometric series: H(z) = 1 − z−8 + z−16 = 1 + z−24 1 + z−8 H(z) = 1 + z−8 + z−16 = 1 − z−24 1 − z−8 The third has roots the 24th roots of unity shifted by π/24 = 7.5o , but among them the roots of the denominator 1 + z−8 must be excluded. This results in 16 zeros. Similarly, the last filter’s zeros are the 24th roots of unity with the 8th roots of unity excluded. Problem 8.10 Typical C code for generating Fig. 8.2.9 and 8.2.11 is given in the text. For the double chorus case, we have the code segment: for (n=0; n<Ntot; n++) { d1 = D * (0.5 + ranl(Dran, u1, &q1, &iseed1)); /* u1 is 2-dim */ d2 = D * (0.5 + ranl(Dran, u2, &q2, &iseed2)); /* mean = D/2 */ a1 = 1 + ranl(Dran, u3, &q3, &iseed3); /* mean = 1 */ a2 = 1 + ranl(Dran, u4, &q4, &iseed4); x = cos(2 * pi * F * n); /* input x(n) */ xd1 = tapi(D, w, p, d1); /* delay x(n-d1) */ xd2 = tapi(D, w, p, d2); y = (x + a1 * xd1 + a2 * xd2) / 3; /* output */ *p = x; cdelay(D, w, &p); /* update delay */ fprintf(fpx, "%lfn", x); /* declaration of fpx */ 158 fprintf(fpy, "%lfn", y); /* is not shown */ fprintf(fpd, "%lfn", 2*d1/D); fprintf(fpa, "%lfn", a1); } Typical output is shown in Fig. P8.6. For the recursive flanger, we have the code fragment: Fig. P8.6 Double chorusing of sinusoidal signal. for (n=0; n<Ntot; n++) { d = 0.50 * D * (1 - cos(2 * pi * Fd * n)); x = cos(2 * pi * F * n); yd = tapi(D, w, p, d); y = a * yd + x; /* yd is y(n-d) */ *p = y; cdelay(D, w, &p); fprintf(fpx, "%lfn", x); fprintf(fpy, "%lfn", y); fprintf(fpd, "%lfn", d/D); } Typical output for D = 100 and D = 10 is shown in Fig. P8.7. Problem 8.11 A typical program is as follows: /* reverb.c - plain, allpass, and lowpass reverb */ #include <stdio.h> #include <stdlib.h> #include <math.h> double plain(), allpass(), lowpass(); void main(int argc, char **argv) { double *w1, *p1; double *w2, *p2; 159
- 81. Fig. P8.7 Recursive flanger. double *w3, *p3; double v[2] = {0., 0.}; double a[2] = {1, -0.5}; double b[2] = {0.3, 0.15}; double a0 = 0.75; double x, y1, y2, y3, sD; int D, Ntot, M=1, n; FILE *fp1, *fp2, *fp3; fp1 = fopen("y1.dat", "w"); fp2 = fopen("y2.dat", "w"); fp3 = fopen("y3.dat", "w"); if (argc != 3) { fprintf(stderr, "usage: reverb D Ntot"); use D=20, Ntot=100 exit(0); } D = atoi(argv[1]); Ntot = atoi(argv[2]); w1 = (double *) calloc(D+1, sizeof(double)); p1 = w1; w2 = (double *) calloc(D+1, sizeof(double)); p2 = w2; w3 = (double *) calloc(D+1, sizeof(double)); p3 = w3; for (n=0; n<Ntot; n++) { triangular pulse if (n<=5) x=n; else if (n<=10) x = 10-n; else x=0; 160 y1 = plain(D, w1, &p1, a0, x); y2 = allpass(D, w2, &p2, a0, x); y3 = lowpass(D, w3, &p3, M, a, b, v, x); fprintf(fp1, "%lfn", y1); fprintf(fp2, "%lfn", y2); fprintf(fp3, "%lfn", y3); } } Problem 8.12 A typical program is as follows: /* reverb.c - Schroeder’s plain/allpass reverb system */ #define D1 29 #define D2 37 #define D3 44 #define D4 50 #define D5 27 #define D6 31 #define a1 .75 #define a2 .75 #define a3 .75 #define a4 .75 #define a5 .75 #define a6 .75 #include <stdio.h> #include <stdlib.h> #include <math.h> double plain(), allpass(); void main() { double *w1, *p1; double *w2, *p2; double *w3, *p3; double *w4, *p4; double *w5, *p5; double *w6, *p6; double y1, y2, y3, y4, y5, y, x; int Ntot=500, n; FILE *fpy; fpy = fopen("y.dat", "w"); w1 = (double *) calloc(D1+1, sizeof(double)); p1 = w1; w2 = (double *) calloc(D2+1, sizeof(double)); 161
- 82. p2 = w2; w3 = (double *) calloc(D3+1, sizeof(double)); p3 = w3; w4 = (double *) calloc(D4+1, sizeof(double)); p4 = w4; w5 = (double *) calloc(D5+1, sizeof(double)); p5 = w5; w6 = (double *) calloc(D6+1, sizeof(double)); p6 = w6; for (n=0; n<Ntot; n++) { if(n==0) impulse x = 1; else x = 0; y1 = plain(D1, w1, &p1, a1, x); y2 = plain(D2, w2, &p2, a2, x); y3 = plain(D3, w3, &p3, a3, x); y4 = plain(D4, w4, &p4, a4, x); y = y1 + 0.9*y2 + 0.8*y3 + 0.7*y4; y5 = allpass(D5, w5, &p5, a5, y); y = allpass(D6, w6, &p6, a6, y5); fprintf(fpy, "%lfn", y); } } Problem 8.13 The difference equations are: v(n) = y(n − D)−a1v(n − 1) u(n) = b0v(n)+b1v(n − 1) y(n) = x(n)+u(n) The corresponding circular version of the sample processing algorithm is for each input x do: sD = tap(D, w, p, D) v0 = sD − a1v1 u = b0v0 + b1v1 v1 = v0 y = x + u ∗p = y cdelay(D, w, &p) 162 Problem 8.14 The response to a causal sinusoid will be of the form: ejωn u(n)−→ H(ω)ejωn u(n)+ D+1 i=1 Bipn i u(n) where Bi = Aipi/(pi − ejω ), as shown in Problem 6.29. It follows that the response to a delayed sinusoid will be: ejω(n−L) u(n − L)−→ H(ω)ejω(n−L) u(n − L)+ D+1 i=1 Bipn−L i u(n − L) Multiplying both sides by the factor ejωL and subtracting from the undelayed result, we find that the response to the finite sinusoid: x(n)= ejωn u(n)−u(n − L) is y(n)= H(ω)ejωn u(n)−u(n − L) + D+1 i=1 Bipn i u(n)−ejωL p−L i u(n − L) Problem 8.15 A typical C program that utilizes the routines plain and lowpass is given below: /* reverbex.c - plain and lowpass reverberation of sinusoid */ * * run with: * D=300, L=150, N=300, w=0.2pi, w=pi * */ #include <stdio.h> #include <stdlib.h> #include <math.h> double plain(), lowpass(); void main(int argc, char **argv) { double *w1, *p1; double *w2, *p2; double v[2] = {0., 0.}; double a[2] = {1, -0.5}; double b[2] = {0.3, 0.15}; double a0 = 0.75; double w, x, y1, y2, sD; double pi = 4*atan(1.0); int D, L, N, M=1, n; FILE *fp1, *fp2; 163
- 83. fp1 = fopen("ypl.dat", "w"); fp2 = fopen("ylp.dat", "w"); if (argc != 5) { puts("usage: reverbex D L N wnn" "D = delay periodn" "L = time to turn off inputn" "N = total duration of input n" "w = in units of pi n"); exit(0); } D = atoi(argv[1]); L = atoi(argv[2]); N = atoi(argv[3]); w = atof(argv[4]); in units of pi w1 = (double *) calloc(D+1, sizeof(double)); p1 = w1; w2 = (double *) calloc(D+1, sizeof(double)); p2 = w2; for (n=0; n<N; n++) { if (n < L) sinusoid on for L samples x = cos(pi*w*n); else x = 0; y1 = plain(D, w1, &p1, a0, x); y2 = lowpass(D, w2, &p2, M, a, b, v, x); fprintf(fp1, "%.16lfn", y1); fprintf(fp2, "%.16lfn", y2); } } The analytical expression for y(n) can be obtained as follows. Using the MATLAB function lprevb.m given below, we determine the poles pi and residues Ai of the transfer function: function [p, A] = lprvb(b, a, D) % returns poles and residues of lowpass reverberator (delay D) % [p, A] = lprvb(b, a, D) % % b=[b(1), b(2)] % a=[1, a(2)] c = [a, zeros(1,D-2), -b]; p=roots(c); for i=1:D+1, A(i) = 1 + a(2) / p(i); for j=1:D+1, if j ~= i, A(i) = A(i) * p(i) / (p(i) - p(j)); end 164 end end Then, the following MATLAB function calculates the residues Bi and computes N samples of the output signal y(n): function [y, p, A] = yon(N, L, w, b, a, D) % calculate y(n) for a finite-duration sinusoidal input. [p, A] = lprvb(b, a, D); z = exp(j * w); H = 1 / (1 - z^(-D) * (b(1) + z^(-1)*b(2)) / (1 + a(2)*z^(-1))); for i=1:D+1, B(i) = A(i) * p(i) / (p(i) - z); end for n=0:N-1, if (n<L), y(n+1) = H * z^n; for i=1:D+1, y(n+1) = y(n+1) + B(i) * p(i)^n; end else y(n+1) = 0; for i=1:D+1, y(n+1) = y(n+1) + B(i) * p(i)^n * (1 - z^L / p(i)^L); end end end Problem 8.16 A typical C program is as follows: /* karplus.c - Karplus-Strong String Algorithm */ * * run with: * D=25, Ntot=500, iseed=1234567 * D=50, Ntot=500, iseed=1234567 * /* #include <stdio.h> #include <stdlib.h> #include <math.h> double ran(), lowpass(); void main(int argc, char **argv) { double *w, *p; double v[2] = {0., 0.}; double a[2] = {1, 0.0}; 165
- 84. double b[2] = {0.5, 0.5}; double x, y; int D, Ntot, M=1, n; long iseed; FILE *fpy; fpy = fopen("y.dat", "w"); if (argc != 4) { fprintf(stderr, "usage: karplus D Ntot iseed"); exit(0); } D = atoi(argv[1]); Ntot = atoi(argv[2]); iseed = atol(argv[2]); w = (double *) calloc(D+1, sizeof(double)); for (n=0; n<=D; n++) initialize w[n] = ran(&iseed) - 0.5; p = w; for (n=0; n<Ntot; n++) { y = lowpass(D, w, &p, M, a, b, v, 0.0); x=0 fprintf(fpy, "%.16lfn", y); } } The output signals corresponding to D = 25 and D = 50 are shown in Fig. P8.8. Note how the initial random numbers get smoothed at each D-cycle resulting in a decaying quasi-periodic signal tuned to the frequency f = fs/D. Fig. P8.8 Karplus-Strong string algorithm outputs. Problem 8.17 A block diagram realization is given in Fig. P8.9. The difference equations are the system: 166 w(n) = aw(n − D)+x(n) y(n) = cx(n)+bw(n − D) x y b c a z-D w0 wD Fig. P8.9 Prototypical delay effect. The sample processing algorithm is in its linear and circular buffer forms: for each input x do: y = cx + bwD w0 = awD + x delay(D, w) for each input x do: sD = tap(D, w, p, D) y = cx + bsD ∗p = asD + x cdelay(D, w, &p) Problem 8.18 The block diagrams and sample processing algorithms are essentially given in Section 8.2.4. The C routines are: /* plaindel.c - plain reverberating delay */ double tap(); void cdelay(); double plaindel(D, w, p, a, x) usage: y=plaindel(D,w,&p,a,x); double *w, **p, a, x; p is passed by address int D; { double y; y = tap(D, w, *p, D); D-th tap delay output **p = x + a * y; delay input cdelay(D, w, p); update delay line return y; } /* lpdel.c - lowpass reverberating delay */ double tap(), can(); void cdelay(); double lpdel(D, w, p, M, a, b, v, x) double *w, **p, *a, *b, *v, x; 167
- 85. int D; { double y; y = tap(D, w, *p, D); D-th tap delay output **p = x + can(M, a, M, b, v, y); delay input cdelay(D, w, p); update delay line return y; } The routines differ from plain and lowpass in that the filter outputs are taken after the delay z−D , instead of before. Problem 8.19 Typical C code for this problem is: for (n=0; n<Ntot; n++) { if (n<L) /* use if (n<1) for impulse response */ x = 1; else x = 0; x1 = plaindel(D1, w1, &p1, a0, x); /* plain */ x2 = plaindel(D2, w2, &p2, a0, x1); /* x1 = lpdel(D1, w1, &p1, 1, a, b, v1, x); */ /* lowpass */ /* x2 = lpdel(D2, w2, &p2, 1, a, b, v2, x1); */ y = b0 * x + b1 * x1 + b2 * x2; /* output */ fprintf(fpx, "%.16lfn", x); fprintf(fpy, "%.16lfn", y); } where the initializations are: double x, x1, x2, y, *w1, *w2, *p1, *p2; double v1[2] = {0., 0.}; /* state of LP filter 1 */ double v2[2] = {0., 0.}; /* state of LP filter 2 */ double a[2] = {1, -0.5}; /* G(z) denominator */ double b[2] = {0.3, 0.15}; /* G(z) numerator */ double b0=1, b1=0.8, b2=0.6 /* feed-forward coefficients */ double a0 = 0.75; /* feedback for plain delay */ w1 = (double *) calloc(D1+1, sizeof(double)); p1 = w1; w2 = (double *) calloc(D2+1, sizeof(double)); p2 = w2; Typical output is as in Fig. P8.10. Problem 8.20 The sample processing routine is as follows. It can easily be generalized to more than two multi- tap delay segments: 168 Fig. P8.10 Multi-delay effects processor outputs. /* mlttap.c - multi-tap delay line */ double tap(); void cdelay(); double mlttap(D1, D2, b, a, w, p, x) int D1, D2; double b[3], a[3], *w, **p, x; { double s[3], y; s[1] = tap(D1+D2, w, *p, D1); s[2] = tap(D1+D2, w, *p, D1+D2); s[0] = x + a[1] * s[1] + a[2] * s[2]; y = b[0] * x + b[1] * s[1] + b[2] * s[2]; **p = s[0]; cdelay(D1+D2, w, p); return y; } Typical output is shown in Fig. P8.11 and is obtained by the code segment: for (n=0; n<Ntot; n++) { if (n < L) /* use L=1 or L=200 */ x = 1; else x = 0; y = mlttap(D1, D2, b, a, w, &p, x); /* p passed by reference */ fprintf(fpx, "%.8lfn", x); fprintf(fpy, "%.8lfn", y); } Problem 8.21 From the pole equation, we have: 169
- 86. Fig. P8.11 Multi-Tap Delay Line. zD1+D2 = a1zD2 + a2 ⇒ |z|D1+D2 = |a1zD2 + a2| ≤ |a1||z|D2 + |a2| (P8.1) If a pole had |z| ≥ 1, then using the inequality |a1| + |a2| < 1, we would have: |z|D1+D2 ≥ |z|D2 > |a1||z|D2 + |a2||z|D2 ≥ |a1||z|D2 + |a2| which contradicts Eq. (P8.1). Thus, all poles must have |z| < 1. Problem 8.22 Working in the z-domain and denoting by WL(z) and WR(z) the inputs to the delays z−L and z−R , the outputs of these delays will be z−L WL(z) and z−R WR(z). Thus, the I/O equations of the block diagram will be: YL(z) = cLXL(z)+z−L WL(z) WL(z) = z−L GL(z)WL(z)+bLXL(z)+dRz−R WR(z) YR(z) = cRXR(z)+z−R WR(z) WR(z) = z−R GL(z)WR(z)+bRXR(z)+dLz−L WL(z) Solving the first and third for the Ws, we get: WL(z)= zL YL(z)−cLXL(z) , WR(z)= zR YR(z)−cRXR(z) Inserting them into the second and fourth, we get the system: 1 − z−L GL(z) YL(z)−cLXL(z) − dRz−L YR(z)−cRXR(z) = bLz−L XL(z) 1 − z−R GR(z) YR(z)−cRXR(z) − dLz−R YL(z)−cLXL(z) = bRz−R XR(z) Solving for YL(z), YR(z) in terms of XL(z), XR(z), we get: YL(z) = HLL(z)XL(z)+HLR(z)XR(z) YR(z) = HRL(z)XL(z)+HRR(z)XR(z) 170 where HLL(z) = cL + 1 D(z) 1 − z−R GR(z) bLz−L HLR(z) = 1 D(z) dRbRz−L z−R HRL(z) = 1 D(z) dLbLz−R z−L HRR(z) = cR + 1 D(z) 1 − z−L GL(z) bRz−R with D(z)= 1 − z−R GR(z) 1 − z−L GL(z) − dLdRz−L z−R For the special case when the cross couplings are dL = 0 and dR = 0, we have: D(z)= 1 − z−R GR(z) 1 − z−L GL(z) and therefore, the transfer functions simplify to: HLL(z) = cL + bLz−L 1 − z−LGL(z) HLR(z) = 0 HRL(z) = dLbLz−L z−R 1 − z−RGR(z) 1 − z−LGL(z) HRR(z) = cR + bRz−R 1 − z−RGR(z) The corresponding I/O system is then: YL(z) = HLL(z)XL(z) YR(z) = HRL(z)XL(z)+HRR(z)XR(z) which means that the left channel responds only to the left input (it does not receive any cross- feedback from the right channel), but the right channel responds to both the right input and the cross feedback from the left output. If both cross-feedbacks are zero, dL = dR = 0, then the left and right channels decouple completely, responding independently to their respective inputs. For the case GL(z)= aL, GR(z)= aR, the sample processing algorithm is straightforward. Denote the left and right delay-line buffers by wL = [wL0, wL1, . . . , wLL]= (L+1)-dimensional wR = [wR0, wR1, . . . , wRR]= (R+1)-dimensional Let pL and pR be the circular pointers circulating over them, and let the corresponding delay outputs be sL and sR. Then, the computational sample processing algorithm will be: 171
- 87. for each input pair {xL, xR} do: sL = tap(L, wL, pL, L) sR = tap(R, wR, pR, R) yL = cLxL + sL yR = cRxR + sR ∗pL = bLxL + aLsL + dRsR ∗pR = bRxR + aRsR + dLsL cdelay(L, wL, &pL) cdelay(R, wR, &pR) Problem 8.23 The sample processing algorithm is implemented by the C code segment: int L = 30, R = 70; double *wL, *wR, *pL, *pR; wL = (double *) calloc(L+1, sizeof(double)); pL = wL; wR = (double *) calloc(R+1, sizeof(double)); pR = wR; for (n=0; n<Ntot; n++) { if (n < P) xL = 1; else xL = 0; sL = tap(L, wL, pL, L); sR = tap(R, wR, pR, R); yL = cL * xL + sL; yR = cR * xR + sR; /* use xR = 0 */ *pL = bL * xL + aL * sL + dR * sR; *pR = bR * xR + aR * sR + dL * sL; cdelay(L, wL, &pL); cdelay(R, wR, &pR); fprintf(fp1, "%.16lfn", yL); fprintf(fp2, "%.16lfn", yR); } Figures P8.12 and P8.13 show typical outputs. The left output does not start until n = L, there- fore, the first right output will begin at n = L + R. That output will be fed back into the left and will come out at n = 2L + R. Similarly, the second left pulse at n = 2L will appear on the right at n = 2L + R, and so on. Problem 8.24 The sample processing algorithm is implemented by the C code segment: if (L > 1) { /* L-point smoother */ h = (double *) calloc(L, sizeof(double)); v = (double *) calloc(L, sizeof(double)); for (n=0; n<L; n++) h[n] = 1.0 / L; 172 Fig. P8.12 Stereo delays, with dL = dR = 0.3. Fig. P8.13 Stereo delays, with dL = 0.3, dR = 0. } for (n=0; n<N; n++) { /* use N = 600 */ if (n < N/3) /* generate input */ x = A1 * cos(w0 * n); /* use w0 = 0.15 * pi */ else if (n < 2*N/3) x = A2 * cos(w0 * n); else x = A3 * cos(w0 * n); c = a * c1 + (1-a) * fabs(x); /* control signal */ c1 = c; /* update delay */ g = f(rho, c/c0); /* gain function */ if (L > 1) G = fir(L-1, h, v, g); /* gain smoothing */ else G = g; /* no smoothing */ y = G * x; /* compressor output */ 173
- 88. fprintf(fpx, "%.16lfn", x); /* save input */ fprintf(fpy, "%.16lfn", y); /* save output */ fprintf(fpc, "%.16lfn", c); /* save control */ fprintf(fpg, "%.16lfn", G); /* save gain */ } where the compressor function is defined by f(x)= xρ−1 , for x ≥ 1: /* compressor function with 1:rho compressor ratio */ #include <math.h> double f(rho, x) double rho, x; { if (x >= 1) return pow(x, rho-1); else return 1; } The compressing action takes place only above the threshold. For an expander, the gain function is defined as follows, where now ρ > 1, but it takes effect only below the threshold: /* expander gain function with 1:rho expander ratio */ #include <math.h> double f(rho, x) double rho, x; { if (x <= 1) return pow(x, rho-1); else return 1; } Problem 8.25 Increasing the threshold to c0 will cause both A1 and A3 to be attenuated. However, A3 because it is further from c0 than in the case of Fig. 8.2.38, it will be attenuated more. Fig. P8.14 shows the results. The noise gate with threshold c0 = 1.5, will suppress both A1 and A3, but A3 by much more, for the same reasons explained above. Fig. P8.15 shows the results. Problem 8.26 Assuming the causal sequence {x(0), x(1), . . . } is zero-mean white, then the terms in the convo- lutional sum will be mutually uncorrelated and zero-mean: y(n)= h0x(n)+h1x(n − 1)+ · · · + h(n)x(0) Thus, the output will have zero mean and its variance will be the sum of the variances of the successive terms: σ2 y = E[y(n)2 ]= h2 0σ2 x + h2 1σ2 x + · · · h2 nσ2 x = σ2 x n n=0 h2 m 174 Fig. P8.14 Expander output and gain with increased threshold. Fig. P8.15 Noise gate output and gain with increased threshold. Problem 8.27 Assuming a partial fraction expansion of the form: H(z)= M i=1 Ai 1 − piz−1 we have for the impulse response: hn = M i=1 Aipn i u(n) It follows that we have the bound: |hn| = M i=1 Aipn i ≤ M i=1 |Ai||pi|n ≤ M i=1 |Ai||pmax|n = C|pmax|n where C = M i=1 |Ai| and we used the inequality |pi| ≤ |pmax|. The NRR inequality follows now by squaring and summing the above inequality for |hn| and applying the geometric series on the right-hand side. 175
- 89. Problem 8.28 For ideal filter shapes, the NRR is computed most conveniently using its frequency response definition: NRR = σ2 yv σ2 v = π −π |H(ω)|2 dω 2π For a bandpass filter, H(ω) is non-zero only over the interval ωa ≤ |ω| ≤ ωb, where it is unity. Thus, NRR = σ2 yv σ2 v = −ωa −ωb 1 · dω 2π + ωb ωa 1 · dω 2π = ωb − ωa π Problem 8.29 The following program is an implementation: /* smooth1.c - 1st-order exponential smoother */ #include <stdio.h> #include <math.h> void main(int argc, char **argv) { double a, x, y, w1 = 0; if (argc != 2) { fprintf(stderr, "usage: smooth1 a <x.dat >y.datn"); exit(0); } a = atof(argv[1]); fprintf(stderr, "%lfn", a); while(scanf("%lf", &x) != EOF) { y = a * w1 + (1-a) * x; w1 = y; printf("%lfn", y); } } Problem 8.30 The mean of x(n) is mx = s = 5. The mean of y(n) is the same because it is preserved through the filter. Indeed, taking means of both sides of y(n)= ay(n − 1)+(1 − a)x(n) gives my = amy + (1 − a)mx ⇒ my = mx The mean-square values may be determine recursively as follows: If we define s2 n = 1 n n−1 k=0 (xk − mx)2 176 then we have the recursion: s2 n+1 = s2 n + 1 n + 1 (xn − mx)2 −s2 n initialized with s2 0 = 0. The following program will calculate the input and output signals and the experimental mean-square values. Any discrepancies are due to the filter transients and they disappear with increasing L: /* smoothex.c - filtering with 1st order smoother */ #include <stdio.h> #include <math.h> double gran(); void main(int argc, char **argv) { int n, L; long iseed; double s=5, x, y, w1, a, m=0, sigma=1; double sx2=0, sy2=0; FILE *fpx, *fpy; if (argc != 4) { puts("nUsage: smoothex a iseed L"); exit(0); } a = atof(argv[1]); iseed = atol(argv[2]); L = atoi(argv[3]); fpx = fopen("ttx", "w"); fpy = fopen("tty", "w"); for (w1=0, n=0; n<L; n++) { x = s + gran(m, sigma, &iseed); y = (1 - a) * x + a * w1; w1 = y; fprintf(fpx, "%lfn", x); fprintf(fpy, "%lfn", y); sx2 += ((x-s)*(x-s) - sx2) / (n+1); sy2 += ((y-s)*(y-s) - sy2) / (n+1); } printf("theoretical 1/NRR = %lfn", (1+a)/(1-a)); printf("experimental 1/NRR = %lfn", sx2/sy2); } Problem 8.31 The requirement that the high-frequency signal s(−1)n go through unchanged is that the filter have unity gain at ω = π or z = −1. This requires the value for the constant b: b 1 − az−1 z=−1 = b 1 + a = 1 ⇒ b = 1 + a 177
- 90. Thus the filter transfer function and impulse response will be: H(z)= 1 + a 1 − az−1 , h(n)= (1 + a)an u(n) The NRR is: NRR = ∞ n=0 h(n)2 = (1 + a)2 1 1 − a2 = 1 + a 1 − a Thus, the NRR is magnified if 0 < a < 1. This can be understood in the frequency domain by the Fig. P8.16. π ω 0 1 input noise spectrum output noise spectrum 1+a 1-a( ( 2 Fig. P8.16 Lowpass filter for extracting high-frequency signal. The magnitude response is normalized to unity at high frequencies, and therefore it must be larger at low frequencies because it is a lowpass filter. Indeed, its values at DC and AC are: |H(0)|2 = 1 + a 1 − a 2 , |H(π)|2 = 1 Problem 8.32 For an FIR filter of length N, its frequency response will be: H(ω)= N−1 n=0 hne−jωn The condition that the gain is unity at AC is: H(π)= N−1 n=0 hn(−1)n = 1 where we used the fact ejπn = (−1)n . Thus, the NRR must be minimized subject to this AC condition: NRR = N−1 n=0 h2 n = min, subject to N−1 n=0 hn(−1)n = 1 This can be solved by Lagrange multipliers or more simply by defining the quantities gn = hn(−1)n . Because g2 n = h2 n, the problem reduces to the following minimization problem: 178 NRR = N−1 n=0 g2 n = min, subject to N−1 n=0 gn = 1 whose solution was found in Example 8.3.4 to be gn = 1/N, n = 0, 1, . . . , N−1. Thus, the solution of the highpass problem will be: hn = (−1)n gn = 1 N (−1)n , n = 0, 1, . . . , N − 1 The Lagrange multiplier method is as follows. Introducing a Lagrange multiplier, say λ, that enforces the constraint, we modify the NRR to be minimized into the effective performance index: J = N−1 n=0 h2 n + 2λ 1 − N−1 n=0 hn(−1)n The minimization condition with respect to the unknowns hn are: ∂J ∂hn = 2hn − 2λ(−1)n = 0, n = 0, 1, . . . , N − 1 This gives hn = λ(−1)n , and the AC constraint then fixes λ to be: N−1 n=0 hn(−1)n = λ N−1 n=0 (−1)2n = λN = 1 ⇒ λ = 1 N Problem 8.33 Expand the transfer function into partial fractions: H(z)= G (1 − pz−1)(1 − p∗z−1) = A 1 − pz−1 + A∗ 1 − p∗z−1 where A = G 1 − p∗/p = Gp p − p∗ = − jGejω0 2 sin ω0 The corresponding impulse response and NRR will be then hn = Apn + a∗ p∗n , n ≥ 0 NRR = ∞ n=0 h2 n = ∞ n=0 A2 p2n + A∗2 p∗2n + 2AA∗ |p|2n which sums up to NRR = A2 1 − p2 + A∗2 1 − p∗2 + 2|A|2 1 − |p|2 Inserting p = Rejω0 , and A as given above, and replacing G2 by its normalized value G2 = (1−R)2 1−2R cos(2ω0)+R2 gives the desired expression for NRR after some tedious algebra. 179
- 91. Problem 8.34 For the first filter, we find the transfer function: H(z)= 0.0730(1 − z−2 ) 1 − 1.7633z−1 + 0.8541z−2 Its pole radius is R = √ a2 = √ 0.8541 = 0.9242, which gives the 5% time constant: neff = ln(0.05) ln(R) = 38 Its NRR can be computed by carrying out a partial fraction expansion of the form: H(z)= B + A 1 − pz−1 + A∗ 1 − p∗z−1 which gives the impulse response: hn = Bδn + Apn un + A∗ p∗n un from which we can calculate the NRR: NRR = ∞ n=0 h2 n = B2 + ∞ n=1 (Apn + A∗ p∗n )2 = B2 + ∞ n=1 A2 p2n + A∗2 p∗2n + 2|A|2 |p|2n = B2 + A2 p2 1 − p2 + A∗2 p∗2 1 − p∗2 + 2|A|2 |p|2 1 − |p|2 The parameters B and A are: B = − G |p|2 , A = − G(1 − p2 ) p2 − |p|2 The numerical value of NRR is NRR = 1/13.4. For the second filter, we start with neff from which we may solve for R = (0.05)1/neff = 0.9901 where R is related to the 3-dB width by: R2 = a2 = 2b − 1 = 1 − tan(Δω/2) 1 + tan(Δω/2) which gives Δω = 0.0064π. Then, the design equations give: H(z)= 0.0099(1 − z−2 ) 1 − 1.8833z−1 + 0.9802z−2 and NRR = 1/101.1. The required graphs are shown in Figs. P8.17 and P8.18. Problem 8.35 For Q = 6, the transfer function is designed using Eqs. (8.2.22) and (8.2.23): H(z)= 0.96953 1 − 1.85955z−1 + z−2 1 − 1.80289z−1 + 0.93906z−2 Its 1% time constant is neff = 146. For Q = 60, the filter is identical to that of Example 8.3.8. Figures P8.19–P8.21 show the input signal, filter responses, and output signals. 180 Fig. P8.17 Magnitude responses and noisy input of Problem 8.34. Fig. P8.18 Filter outputs of Problem 8.34. Fig. P8.19 Noisy input and desired signal of Problem 8.35. Problem 8.36 The designed filter has the form: 181
- 92. Fig. P8.20 Magnitude responses, for Q = 6 and Q = 60. Problem 8.35. Fig. P8.21 Filter outputs, for Q = 6 and Q = 60. Problem 8.35. H(z)= b 1 − z−D 1 − az−D The time constant is computed as: neff = D ln ln a where = 0.01 for the 1% time constant. The filter parameters are in the four cases: Q a b neff 80 0.961481 0.980741 1172.4 200 0.984414 0.992207 2931.6 400 0.992177 0.996088 5863.7 800 0.996081 0.998040 11727.8 The relevant graphs are shown in Figs. P8.22–P8.26. 182 Fig. P8.22 Desired signal and noisy input. Problem 8.36. Fig. P8.23 Filtered output y(n) when Q = 80. Problem 8.36. Fig. P8.24 Filtered s(n) for Q = 80 and Q = 200. Problem 8.36. Problem 8.37 The following MATLAB m-file designs the 60 Hz notch filter, generates the noisy ECG and filters it, and computes the filter’s magnitude response: 183
- 93. Fig. P8.25 Filtered s(n) for Q = 400 and Q = 800. Problem 8.36. Fig. P8.26 Magnitude responses for Q = 80 and Q = 800. Problem 8.36. % ecgex1.m - simulated ECG + 60Hz noise example % % assume 2 beats/sec => 0.5 sec/beat => 500 samples/beat => 1000 samples/sec % f0=60; Q=60; df=f0/Q; % % compute notch filter coefficients % generate length-500 ecg by calling x0=ecg(500); normalize it to 1. % replicate three beats x=[x0,x0,x0] and smooth them with x=sgfilt(0, 5, x); % add 80 percent 60-Hz noise % and filter it yv=filter(B,A,xv) % finally, compute filter’s magnitude response. f0 = 60/1000; Q = 60; df = f0/Q; w0 = 2 * pi * f0; dw = 2 * pi * df; b = 1 / (1+tan(dw/2)); B = b * [1, -2*cos(w0), 1]; A = [1, B(2), 2*b-1]; x0 = ecg(500); x = [x0, x0, x0]; x = sgfilt(0, 5, x); m = max(x); x = x/m; 184 n = 0:1499; xv = x + 0.5 * cos(w0 * n); yv = filter(B, A, xv); w = (0:499)*pi/500; z = exp(-i*w); h = abs((B(1) + B(2)*z + B(3)*z.^2) ./ (1 + A(2)*z + A(3)*z.^2)); Problem 8.38 The following MATLAB m-file designs the 60 Hz comb filter, generates the noisy ECG and filters it, and computes the filter’s magnitude response: % ecgex2.m - simulated ECG + 60Hz square-wave noise example % % assume 1 beats/sec => 1 sec/beat => 600 samples/beat => 600 samples/sec % % compute notch filter coefficients % generate length-600 ecg by calling x0=ecg(600); normalize it to 1. % replicate three beats x=[x0,x0,x0] and smooth them with x=sgfilt(0, 9, x); % add shifted square wave % and filter it yv=filter(B,A,xv) % finally, compute filter’s magnitude response. fs = 600; f0 = 60; df = 0.75; w0 = 2 * pi * f0 / fs; dw = 2 * pi * df / fs; beta = tan(dw * 10 / 4); % D=10 here a = (1 - beta) / (1 + beta); b = (1 + a) / 2; B = b * [1 0 0 0 0 0 0 0 0 0 -1]; % numerator A = [1 0 0 0 0 0 0 0 0 0 -a]; % denominator x0 = ecg(600); x = [x0, x0]; x = sgfilt(0, 9, x); m = max(x); x = x/m; n = 0:1199; v0 = [1 1 1 1 1 -1 -1 -1 -1 -1]; v = v0; % one noise period for k=1:119, % 120 noise periods v = [v, v0]; end v = 2 + v; % noise baseline shift xv = x + 0.5 * v; % noisy ECG yv = filter(B, A, xv); w = (0:500)*pi/500; z = exp(-i*w); h = abs(b * (1 - z.^10) ./ (1 - a * z.^10)); 185
- 94. Problem 8.39 Using partial fractions, we expand H(z)= b 1 + z−D 1 − az−D = A + B 1 − az−D = (A + B)+Baz−D + Bz2 z−2D + · · · where A = − b a , B = b 1 + 1 a , A + B = b Thus, the impulse response is h = [b, 0, 0, . . . , 0, Ba, 0, 0, . . . , 0, Ba2 , 0, 0, . . . , 0, Ba3 , . . . ] with the nonzero entries separated by D − 1 zeros. The sum of the squares is NRR = ∞ n=0 h2 n = b2 + B2 ∞ m=1 a2m = b2 + B2 a2 1 − a2 = b2 (1 − a2 )+b2 a2 (1 + a−2 ) 1 − a2 = 2b2 (1 + a) 1 − a2 = 2(1 − a)2 (1 + a) 4(1 − a)(1 + a) = 1 − a 2 In the filter design parameter β = tan(ΔωD/4), the tangent has argument: ΔωD 4 = Δω 4 · 2π ω1 = πΔω 2ω1 = π 2Q where we used ω1 = 2π/D, or D = 2π/ω1. Problem 8.40 The following program designs the comb filter, constructs 40 periods of length-50 of the noisy triangular waveform, and filters them using the circular buffer implementation of the canonical realization of the filter: /* combex.c - IIR comb filter with noisy periodic input */ #include <stdio.h> #include <stdlib.h> #include <math.h> double gran(), tap(); void cdelay(), triang(); void main(int argc, char **argv) { int n, i, N, D; long iseed; double a, b, pi=4*atan(1.0), dw = 0.0008 * pi; double *s, *w, *p; double x, y, sD; FILE *fpy, *fps, *fpx; if (argc != 4) { 186 puts("nUsage: combex D N iseed"); exit(0); } D = atoi(argv[1]); period N = atoi(argv[2]); number of periods iseed = atol(argv[3]); initial seed fpy = fopen("y.dat", "w"); filter output fps = fopen("s.dat", "w"); noise-free input fpx = fopen("x.dat", "w"); noisy input a = (1 - tan(D*dw/4)) / (1 + tan(D*dw/4)); b = (1 - a) / 2; s = (double *) calloc(D, sizeof(double)); triang(D, s); one period of triangular wave w = (double *) calloc(D+1, sizeof(double)); p = w; for (i=0; i<N; i++) generate N periods for (n=0; n<D; n++) { x = s[n] + 0.5 * gran(0., 1., &iseed); sD = tap(D, w, p, D); *p = b * x + a * sD; y = (*p) + sD; cdelay(D, w, &p); fprintf(fps, "%lfn", s[n]); fprintf(fpx, "%lfn", x); fprintf(fpy, "%lfn", y); } } /* -------------------------------------------------------- */ /* triang.c - triangular wave */ void triang(D, s) int D; double *s; { int i; double D4 = D/4.0; for (i = 0; i<D4; i++) s[i] = i / D4; for (i=D4; i<3*D4; i++) s[i] = 1 + (D4 - i) / D4; for (i=3*D4; i<D; i++) s[i] = -1 + (i - 3*D4) / D4; } 187
- 95. Problem 8.41 The complementarity properties may be verified for each pair. For example, for filter 2, we have half-order M = 8/2 = 4. Thus, we must verify: HLP(z)+HBP(z)= 1 16 (1 + z−2 )2 (−1 + 6z−2 − z−4 )+ 1 16 (1 − z−2 )4 = z−4 which follows by expanding out the terms. Fig. P8.27 shows the frequency responses in the four cases. The frequency scale is in MHz and extends only up to about 4.5 MHz. For reference, the Nyquist frequency here is fs/2 = 2fsc = 7.159 MHz, so that the subcarrier frequency shown in the figure is only one-quarter of the Nyquist interval. Fig. P8.27 Lowpass/bandpass and highpass/bandstop responses. Problem 8.41. Problem 8.42 See Table P8.1. Note that the circular pointer p points to the buffer entry w[q], whereas the D-th tap output is the content of w[qD]. At each time instant, the content of w[q] is updated by w[q]= w[qD]+x, where the scaling by N is done afterwards. Table P8.1 shows the results. The advantage of the circular implementation is that only one entry of w, namely, w[q], is changed at each time instant. This is efficient in hardware implementations and for large values of D. Problem 8.43 The following is a complete C program for performing signal averaging. The inputs to the program are the period D and the number N of periods to be averaged. The input data are taken from stdin and the averaged period is output to the stdout: /* sigav.c - signal averaging */ #include <stdio.h> #include <stdlib.h> void cdelay(); double tap(); 188 void main(int argc, char **argv) { double x, *w, *p; int D, N, i, k; if (argc != 3) { fprintf(stderr, "usage: sigav D N < x.dat > y.datn"); fprintf(stderr, "x.dat must contain N periods of length Dn"); exit(0); } D = atoi(argv[1]); N = atoi(argv[2]); w = (double *) calloc(D+1, sizeof(double)); p = w; for (k=0; k<N; k++) for (i=0; i<D; i++) if (scanf("%lf", &x) != EOF) { /* keep reading input samples */ *p = tap(D, w, p, D) + x / N; cdelay(D, w, &p); } else { fprintf(stderr, "Input must have %d periods of length %dn", N, D); exit(0); } for (i=0; i<D; i++) printf("%.12lfn", tap(D, w, p, D-i)); } n x qD q w0 w1 w2 w3 y = wq/4 0 s0 3 0 s0 0 0 0 s0/4 1 s1 2 3 s0 0 0 s1 s1/4 2 s2 1 2 s0 0 s2 s1 s2/4 3 s0 0 1 s0 2s0 s2 s1 2s0/4 4 s1 3 0 2s1 2s0 s2 s1 2s1/4 5 s2 2 3 2s1 2s0 s2 2s2 2s2/4 6 s0 1 2 2s1 2s0 3s0 2s2 3s0/4 7 s1 0 1 2s1 3s1 3s0 2s2 3s1/4 8 s2 3 0 3s2 3s1 3s0 2s2 3s2/4 9 s0 2 3 3s2 3s1 3s0 4s0 s0 10 s1 1 2 3s2 3s1 4s1 4s0 s1 11 s2 0 1 3s2 4s2 4s1 4s0 s2 12 − − − − 4s2 4s1 4s0 − Table P8.1 Signal averager with circular delay for D = 3, N = 4. 189
- 96. Problem 8.44 The following program generates N periods of length D of the noisy signal, and averages these periods using the circular buffer form of the averager. After processing the ND input samples, it writes out the contents of the delay line in reverse order, so that the files y.dat and s.dat will contain the averaged periods and the noise-free period: /* sigavex.c - signal averaging example */ #include <stdio.h> #include <stdlib.h> #include <math.h> double gran(), tap(); void cdelay(); void main(int argc, char **argv) { int n, i, N, D; long iseed; double *s, x, *w, *p; double sD; FILE *fpx, *fpy, *fps; if (argc != 4) { puts("nUsage: sigavex D N iseed"); exit(0); } D = atoi(argv[1]); period N = atoi(argv[2]); number of periods iseed = atol(argv[3]); noise seed fpx = fopen("x.dat", "w"); fpy = fopen("y.dat", "w"); fps = fopen("s.dat", "w"); s = (double *) calloc(D, sizeof(double)); for (i=0; i<D; i++) if (i < D/2) s[i] = 1; else s[i] = -1; w = (double *) calloc(D+1, sizeof(double)); p = w; for (i=0; i<N; i++) accumulate N periods for (n=0; n<D; n++) { each of length D x = s[n] + gran(0., 1., &iseed); noisy input sD = tap(D, w, p, D); D-th tap of averager *p = sD + x; accumulate x cdelay(D, w, &p); update delay line fprintf(fpx, "%.12lfn", x); save input sample } for (n=0; n<D; n++) { write averaged period 190 fprintf(fps, "%lfn", s[n]); fprintf(fpy, "%.12lfn", tap(D, w, p, D-n)/N); } } Problem 8.45 The following MATLAB function generates the noisy ECG and smoothes it once and twice by Savitzky-Golay filter of length N and smoothing order d: % sgecgex.m - SG filtering of noisy ECG L=500; x0 = ecg(L)’; % generate ECG x = sgfilt(0, 15, x0); mx = max(x); x = x/mx; % unity max rand(’normal’); rand(’seed’, 10033); v = 0.3 * rand(L, 1); % generate noise xv = x + v; % generate noisy ECG N=11; d=2; y1 = sgfilt(d, N, xv); % first pass through SG y2 = sgfilt(d, N, y2); % second pass The filtering twice through the SG filter improves the smoothing operation but introduces longer transients in the filters. The following figures should be compared with the single-pass cases of Figs. 8.3.33 and 8.3.34. Fig. P8.28 Double SG filtering; N = 11, 41 and d = 0. 191
- 97. Fig. P8.29 Double SG filtering; N = 11, 41 and d = 2. Fig. P8.30 Double SG filtering; N = 11, 41 and d = 4. 192 Chapter 9 Problems Problem 9.1 Use the formula TR = LT = L/fs to solve for L = fsTR, or in words no. of samples = (no. of samples per second) x (no. of seconds) Thus, L = 8 kHz×128 msec = 1024 samples. The frequency spacing is Δf = fs/N = 8000/256 = 31.25 Hz. If the DFT is done directly, we would require L · N = 1024 · 256 = 262144 mul- tiplications. If the signal is reduced mod-256, then the number of multiplications would be N · N = 256 · 256 = 65536. And if the FFT is used, N log2 N/2 = 256 · 8/2 = 1024. The cost of performing the mod-256 reduction is N(M − 1)= 256 · 3 = 768 additions, where M = L/N = 1024/256 = 4 is the number of segments. This cost may be added to the costs of FFT or reduced DFT. Problem 9.2 The sinusoidal signal is something like x(t)= cos(2πft), where f = 10 kHz. Its spectrum will have peaks at ±f. Is there a DFT index k such that the kth DFT frequency fk equals ±f? ±f = fk = fs N k ⇒ k = ±N f fs = ±64 10 80 = ±8 Thus, we expect to see a peak at k = 8 and another one at k = −8. Because the DFT X(k) is periodic in k with period N and because we always list the indices in the positive range k = 0, 1, · · · , N − 1, we shift the negative index by one period N to get the positive index: −k → N − k = 64 − 8 = 56 Thus, the second peak will be at k = 56. Problem 9.3 The number of samples in one period is given by: D = fs f = 40 5 = 8 samples Therefore, if k periods are collected , the total signal length will be N = kD = k fs f which gives for k = 16, N = 16 · 8 = 128 samples. If we perform an N-point DFT, then because of the choice of N, the k-th DFT frequency will coincide with f, that is, fk = k fs N = f Thus, we expect to see a peak at the kth DFT bin. We will also get a peak at the negative index −k, which translates mod-N to N − k. In summary, we expect to see peaks at the DFT indices: k = 16 and N − k = 128 − 16 = 112 193
- 98. Problem 9.4 The expected DFT index will be at k = N f fs = 16 · 18 8 = 36 Because k is greater than N − 1 = 15, it may be reduced mod-N down to: k = 36 − 2 × 16 = 4 It corresponds to the DFT frequency fk = kfs/N = 4 · 8/16 = 2 kHz. This is frequency within the Nyquist interval that is aliased with f = 18 kHz; indeed, we have falias = f mod fs = 18 mod 8 = 18 − 2 × 8 = 2 kHz The −2 kHz or −18 kHz component will be represented by the DFT index N − k = 16 − 4 = 12. Problem 9.5 To be able to resolve a peak, the width of the main lobe of the data window must be equal or smaller than the width of the peak. This gives the condition fs L ≤ Δf ⇒ L ≥ fs Δf or, in our case, L ≥ 8000/20 = 400 samples. The duration of these 400 samples will be TR = LT ≥ L fs = 1 Δf = 1 20 = 50 msec To avoid wrap-around errors in the inverse FFT, the FFT length must be at least L, i.e., N ≥ L = 400. The next power of two is N = 512. We must pad 112 zeros to the end of the length-400 signal before the FFT is taken. Problem 9.6 The following MATLAB function generates the time data and computes the required DTFTs with the help of the function dtft.m, included below, which replaces the C functions dtft.c and dtftr.c: % dtftexp.m - rectangular/hamming windowed sinsuoids L = 200; N = 200; w0 = 0.1*pi; w = (0:N-1) * 0.2 * pi / N; % frequency range n = 0:(L-1); wh = 0.54 - 0.46 * cos(2*pi*n/(L-1)); % Hamming window xrec = cos(w0 * n); xham = wh .* xrec; % windowed data Xrec = abs(dtft(xrec, w)); % DTFT of rectangular data Xham = abs(dtft(xham, w)); % DTFT of Hamming data save xrec.dat xrec /ascii; save xham.dat xham /ascii; save frec.dat Xrec /ascii; save fham.dat Xham /ascii; 194 where % dtft.m - DTFT of a signal at a frequency vector w % % X = dtft(x, w); % % x = row vector of time samples % w = row vector of frequencies in rads/sample % X = row vector of DTFT values % % based on and replaces both dtft.c and dtftr.c function X = dtft(x, w) [L1, L] = size(x); z = exp(-j*w); X = 0; for n = L-1:-1:0, X = x(n+1) + z .* X; end Problem 9.7 The following MATLAB function generates the time data and computes the required DTFTs: % sinexp.m - 3 sinusoids of length L=10 L=10; N=256; f1 = 0.2; f2 = 0.25; f3 = 0.3; n = 0:(L-1); x = cos(2*pi*f1*n) + cos(2*pi*f2*n) + cos(2*pi*f3*n); w = 0.54 - 0.46 * cos(2*pi*n/(L-1)); xham = w.*x; X = abs(fft(x,N)); Xham = abs(fft(xham,N)); For the other cases, use L = 10, 40, 100. Problem 9.8 The following C program generates the length-L time signal and computes its N-point DFT, with L = 10, 20, 100 and N = 256. Instead of the DFT routine dft.c, one could use fft.c. The 32-point DFT is extracted by keeping one out of every 256/32 = 8 points of the 256-point DFT. /* dtftex.c - physical vs. computational resolution example */ #include <stdio.h> #include <cmplx.h> #include <stdlib.h> void dtftr(); 195
- 99. main() { int L, N, n, k; double *x, pi = 4 * atan(1.0); double f1 = 0.20, f2 = 0.25, f3 = 0.30; complex *X; FILE *fpdtft; fpdtft= fopen("dtft.dat", "w"); DTFT file printf("enter L, N: "); scanf("%d %d", &L, &N); x = (double *) calloc(L, sizeof(double)); X = (complex *) calloc(N, sizeof(complex)); for (n=0; n<L; n++) x[n] = cos(2*pi*f1*n) + cos(2*pi*f2*n) + cos(2*pi*f3*n); dft(L, x, N, X); N-point DFT for (k=0; k<N; k++) fprintf(fpdtft, "%lfn", cabs(X[k])); } Problem 9.9 See Example 9.2.1. Problem 9.10 Using the relationship k = Nf/fs, we calculate the group-A and group-B DFT indices: group A kA = 22.3, 24.6, 27.3, 30.1 group B kB = 38.7, 42.8, 47.3 and their rounded versions: group A: kA = 22, 25, 27, 30 group B: kB = 39, 43, 47 Because each keypress generates one tone from group A and one from group B, the DTFT will have peaks at the corresponding pair of DFT indices (and their negatives, that is, N − kA and N − kB.) Problem 9.11 The time signal and its DTFT are shown in Figs. P9.1 and P9.2, both for the rectangular and Hamming windows. 196 Fig. P9.1 Rectangularly and Hamming windowed DTMF sinusoids. Fig. P9.2 Spectra of DTMF sinusoids. Problem 9.12 Acting on x by the 4×8 DFT matrix, we obtain: X = ⎡ ⎢ ⎢ ⎢ ⎣ X(0) X(1) X(2) X(3) ⎤ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ 1 1 1 1 1 1 1 1 1 −j −1 j 1 −j −1 j 1 −1 1 −1 1 −1 1 −1 1 j −1 −j 1 j −1 −j ⎤ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 2 2 1 2 1 1 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ 12 0 0 0 ⎤ ⎥ ⎥ ⎥ ⎦ The mod-4 wrapped version of x is ˜x = [3, 3, 3, 3]. Its 4-point DFT is X = ⎡ ⎢ ⎢ ⎢ ⎣ X(0) X(1) X(2) X(3) ⎤ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ 1 1 1 1 1 −j −1 j 1 −1 1 −1 1 j −1 −j ⎤ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎣ 3 3 3 3 ⎤ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ 12 0 0 0 ⎤ ⎥ ⎥ ⎥ ⎦ For the IDFT we use the formula IDFT(X)= [DFT(X∗ )]∗ /N: 197
- 100. x = ⎡ ⎢ ⎢ ⎢ ⎣ x(0) x(1) x(2) x(3) ⎤ ⎥ ⎥ ⎥ ⎦ = 1 4 ⎡ ⎢ ⎢ ⎢ ⎣ 1 1 1 1 1 −j −1 j 1 −1 1 −1 1 j −1 −j ⎤ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎣ 12 0 0 0 ⎤ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ 3 3 3 3 ⎤ ⎥ ⎥ ⎥ ⎦ Problem 9.13 The 8-point FFT is: x = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 5 −1 −3 −1 5 −1 −3 −1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 8−FFT −→ X = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 0 0 16 0 8 0 16 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ It follows that the FFT spectrum will show peaks at the frequencies: X(ω2)= X(ω6)∗ = X(−ω2)∗ = 16, X(ω4)= 8 Problem 9.14 The IFFT is obtained by conjugating X, computing its FFT, and dividing by 8. The required FFT of X∗ is: X∗ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 0 4 4j 4 0 4 −4j 4 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 8−FFT −→ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 16 8 0 −8 −16 8 0 −8 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⇒ x = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 2 1 0 −1 −2 1 0 −1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Using the inverse DFT formula, we express x(n) as a sum of sinusoids: x(n) = 1 8 7 k=0 X(k)ejωkn = 1 8 4ejω1n − 4jejω2n + 4ejω3n + 4ejω5n + 4jejω6n + 4ejω7n = 1 2 ejω1n + 1 2j ejω2n + 1 2 ejω3n + 1 2 e−jω3n − 1 2j e−jω2n + 1 2 e−jω1n = cos(ω1n)+ sin(ω2n)+ cos(ω3n)= cos(πn/4)+ sin(πn/2)+ cos(3πn/4) which agrees with the above IFFT values at n = 0, 1, . . . , 7. We used the values of the DFT fre- quencies: ω1 = 2π 8 = π 4 , ω2 = 2π2 8 = π 2 , ω3 = 2π3 8 = 3π 4 The DFT frequencies ω7, ω6, ω5 are the negatives (modulo-2π) of ω1, ω2, ω3, respectively. 198 Problem 9.15 Figure P9.3 shows the procedure. The (4N)-dimensional time data vector is shuffled once into two (2N)-dimensional subvectors by putting every other one of its entries on top and the remain- ing entries on the bottom. Then, each of the (2N)-dimensional vectors is shuffled again into two N-dimensional subvectors. shuffling N-FFT N-FFT N-FFT N-FFT DFT merging 4N 4N 2N 2N 2N 2N N N N N N N N N W2N k W2N k W4N k Fig. P9.3 Performing large FFTs. The four N-dimensional vectors are then shipped over to the FFT processor, one at a time, and their FFTs are returned into four N-dimensional DFT vectors. These are then merged first into (2N)-point DFTs and then into the desired (4N)-point DFT. The first merging stage require mul- tiplication by the twiddle factors Wk 2N, k = 0, 1, . . . , N − 1. The second stage uses the factors Wk 4N, k = 0, 1, . . . , 2N − 1. The butterfly operations can be done is small parts, for example, in groups of N operations at a time so that they can be implemented by the DSP hardware. The total cost is the cost of four N-point FFTs, the cost of two N-fold multiplications for the first merging stage, and a (4N)-fold multiplication for the second stage: cost = 4 1 2 N log2 N + 2N + 2N = 2N log2 N + 4N If the (4N)-point FFT could be done in one pass, we would have cost: cost = 1 2 (4N)log2(4N)= 2N log2(4N) It is easily verified that the costs of the two methods are the same. The indirect method is of course much slower because of all the I/O time for shipping data back and forth from secondary storage to the FFT processor. Problem 9.16 The ordinary convolution is y = h ∗ x = [1, 3, 4, 6, 7, 6, 4, 3, 1] 199
- 101. Reducing this mod-4 gives: ˜y = [9, 9, 8, 9]. The alternative method is to reduce first x and h mod-4: ˜h = [2, 2, 1, 2], ˜x = [2, 1, 1, 1], then, compute their ordinary convolution: ˜h ∗ ˜x = [4, 6, 6, 9, 5, 3, 2], and, finally, reduce that modulo 4 to get ˜y = [9, 9, 8, 9]. Problem 9.17 Carrying out the 8-point FFT gives: x = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 4 2 4 −6 4 2 4 −6 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 8−FFT −→ X = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 8 0 −16j 0 24 0 16j 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Taking every other entry of the result gives the 4-point FFT, namely, X = ⎡ ⎢ ⎢ ⎢ ⎣ 8 −16j 24 16j ⎤ ⎥ ⎥ ⎥ ⎦ And, taking every other entry of that gives the 2-point FFT: X = 8 24 Can we get these results directly? Yes. To do the 4-point FFT, first reduce the signal modulo 4 to get ˜x = ⎡ ⎢ ⎢ ⎢ ⎣ 8 4 8 −12 ⎤ ⎥ ⎥ ⎥ ⎦ 4−FFT −→ X = ⎡ ⎢ ⎢ ⎢ ⎣ 8 −16j 24 16j ⎤ ⎥ ⎥ ⎥ ⎦ Putting the above 8-point FFT into the 8-point inverse FFT formula gives x(n)= 1 8 7 k=0 X(k)ejωkn = 1 8 8ejω0n − 16jejω2n + 24ejω4n + 16jejω6n Noting that for an 8-point FFT, ω0 = 0, ω2 = π/2, ω4 = π, and ω6 = 3π/2 ≡ −π/2, and using Euler’s formula we obtain: x(n)= 1 + 4 sin( πn 2 )+3 cos(πn) which agrees with the given x, for n = 0, 1, · · · , 7. 200 Problem 9.18 Consider the length-7 signal [1 − a, 2 − b, 3, 4, 5, a, b] It should be evident that, for any values of the parameters a and b, its mod-5 reduction is [1, 2, 3, 4, 5]. Problem 9.19 Using the finite geometric series and the definition WN = e−2πj/N , gives 1 N N−1 n=0 Wkn N = 1 N 1 − WNk N 1 − Wk N = 1 N 1 − e−2πjk 1 − e−2πjk/N If k = 0, the numerator vanishes but not the denominator. Thus, in this case the answer is zero. If k = 0, we have a ratio of 0/0, but in this case, we can just go back to the series definition and set k = 0. All of the N terms in the sum become unity, giving N/N = 1. In conclusion, the answer is zero if k = 0, and unity if k = 0. Note, that these arguments break down if k is allowed to go beyond the range 0 ≤ k ≤ N − 1, because whenever it is a multiple of N, we would get unity as an answer (and zero otherwise). Problem 9.20 For a general power p, we have: W p pN = e−2πj/(pN) p = e−2πj/N = WN The DTFT of a length-L signal and its N-point and (pN)-point DFTs are: X(ω)= L−1 n=0 x(n)e−jωn XN(k)= L−1 n=0 x(n)Wkn N XpN(k)= L−1 n=0 x(n)Wkn pN Replacing k by pk in the third and using part (a), we have: XpN(pk)= L−1 n=0 x(n)W pkn pN = L−1 n=0 x(n)Wkn N = XN(k) Problem 9.21 This follows from the results of Problem 9.20 by setting N = 8 and p = 2, which implies: X8(k)= X16(2k), k = 0, 1, . . . , 15 A geometrical way to understand this is to recognize that the 8th roots of unity are a subset of the 16th roots of unity and are obtained by taking every other one of the latter. 201
- 102. Problem 9.22 Using a trig identity to write the product of sines as a sum of sines, we get the equivalent expres- sion: x(t)= cos(24πt)+2 sin(12πt)cos(8πt)= cos(24πt)+ sin(20πt)+ sin(4πt) The frequencies of the three terms are 12, 10, and 2 kHz. Their aliased versions that lie inside the Nyquist interval [−4, 4] kHz are, respectively, 4, 2, and 2 kHz. Thus, the aliased signal that would be reconstructed by an ideal reconstructor will be: x(t)= cos(8πt)+ sin(4πt)+ sin(4πt)= cos(8πt)+2 sin(4πt) Setting t = nT = n/8, we get the time samples: x(n)= cos(πn)+2 sin(πn/2) Evaluating it at n = 0, 1, . . . , 7 gives the values, and their FFT: x = [1, 1, 1, −3, 1, 1, 1, −3] 8−FFT −→ X = [0, 0, −8j, 0, 8, 0, 8j, 0] The same DFT values can be determined without any computation by comparing x(n) with the inverse DFT. Using Euler’s formula, setting ω4 = π, ω2 = π/2, ω6 = −(π/2)mod(2π), and multiplying and dividing by 8, we get: x(n)= ejπn − jejπn/2 + je−jπn/2 = 1 8 −8jejω2n + 8ejω4n + 8jejω6n Comparing with the 8-point IDFT, gives the above DFT vector: x(n)= 1 8 7 k=0 X(k)ejωkn Problem 9.23 Using Euler’s formula, we write: x(n) = 1 + 2 sin( πn 4 )−2 sin( πn 2 )+2 sin( 3πn 4 )+3(−1)n = 1 − jejπn/4 + je−jπn/4 + jejπn/2 − je−jπn/2 − jej3πn/4 + je−j3πn/4 + 3ejπn = 1 8 8ejω0n − 8jejω1n + 8jejω2n − 8jejω3n + 24ejω4n + 8jejω5n − 8jejω6n + 8jejω7n where we reordered the terms according to increasing DFT index. Comparing with the IDFT formula, we get: X = [8, −8j, 8j, −8j, 24, 8j, −8j, 8j] 202 Problem 9.24 Using Euler’s formula, we write x(n) = 0.5eπjn/2 + 0.5e−πjn/2 + eπjn/8 + e−πjn/8 = 1 16 8eπjn/2 + 8e3πjn/2 + 16eπjn/8 + 16e15πjn/8 where, we shifted the negative frequency exponents by adding 2πn to them, e.g., replaced −πn/8 by 2πn − πn/8 = 15πn/8. Now, comparing with the 16-point IDFT formula x(n)= 1 16 15 k=0 X(k)ejωkn and noting that π/2 = 8π/16 = ω4, 3π/2 = 24π/16 = ω12, π/8 = 2π/16 = ω1, and 15π/8 = 30π/16 = ω15, we may identify the (non-zero) values of X(k) as follows: X(1)= 16, X(4)= 8, X(12)= 8, X(15)= 16, or, X = [0, 16, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 16] Problem 9.25 Arguing in a similar fashion, we find the (non-zero) values of the 32-point DFT: X(2)= 32, X(8)= 16, X(24)= 16, X(30)= 32 Problem 9.26 Write x(n) in its complex sinusoidal form: x(n)= 0.5ejω0n − jejω4n + jωjω12n + 1.5ejω8n (P9.1) where ω0 = 0, ω4 = 2π4/16 = π/2, ω12 = −ω4mod(2π), ω8 = 2π8/16 = π are the signal frequencies expressed as 16-point DFT frequencies. The DTFT of each complex sinusoid is given by the shifted spectrum of the rectangular window W(ω) of length-16: X(ω)= 0.5W(ω)−jW(ω − ω4)+jW(ω − ω12)+1.5W(ω − ω8) where W(ω)= 1 − e−16jω 1 − e−jω = sin(8ω) sin(ω/2) e−15jω/2 The magnitude spectrum |X(ω)| is plotted versus 0 ≤ ω ≤ 2π in Fig. P9.4. The spectral peaks at the signal frequencies are evident. Dividing the Nyquist interval into 16 equal bins, gives the 16-point DFT. It so happens that the zeros of the sidelobes of the W(ω − ωi) coincide with 16-point DFT fre- quencies. Taking every other 16-DFT point gives the 8-point DFT. The peak values are correctly identified by the 16- and 8-point DFTs because the peak frequencies coincide with DFT frequen- cies. 203
- 103. ω π0 2π |X(ω)| = DTFT = 16-point DFT = 8-point DFT24 16 8 Fig. P9.4 DTFT, 16-point and 8-point DFTs. The 16-point DFT can be determined from the expression of x(n) by comparing it with 16-point IDFT formula. Multiplying and dividing Eq. (P9.1) by 16, we may extract the DFT coefficients: X = [8, 0, 0, 0, −16j, 0, 0, 0, 24, 0, 0, 0, 16j, 0, 0, 0] Every other entry is the 8-point DFT: X = [8, 0, −16j, 0, 24, 0, 16j, 0] Problem 9.27 We will show that the product AA∗ is proportional to the identity matrix. The ij-th matrix element of the product is: (AA∗ )ij= N−1 n=0 AinA∗ nj = N−1 n=0 Win N W −jn N = N−1 n=0 W (i−j)n N where we used the fact that W∗ N = W−1 N . Using the results of Problem 9.19, we note that the above sum will be zero if i = j. And, it will be equal to N if i = j. Thus, we can write compactly: (AA∗ )ij= N−1 n=0 W (i−j)n N = Nδ(i − j) which is equivalent to the matrix relationship AA∗ = NI where I is the N × N identity matrix. Problem 9.28 Vectorially, the desired equation reads x† x = 1 N X† X where † denotes conjugate transpose. Using X = Ax and the property AA† = A† A = NIN, (IN = N × N identity matrix) of the N × N DFT matrix, we obtain X† X = x† A† Ax = x† NINx = Nx† x 204 Problem 9.29 First, compute the ordinary linear convolution: y = h ∗ x = [2, −1, −2, 2, −2, −1, 2] The mod-4 reduction of this is y = [0, −2, 0, 2]. The mod-5 reduction is y = [1, 1, −2, 2, −2], the mod-6 reduction is y = [4, −1, −2, 2, −2, −1]. Finally, the mod-7 and mod-8 reductions of y leave it unchanged. One minor detail is that in the mod-8 case, the length should be N = 8, thus, strictly speaking y is y with one zero padded at the end, that is, y = [2, −1, −2, 2, −2, −1, 2, 0]. The minimal N is N = Ly = Lx + Lh − 1 = 4 + 4 − 1 = 7. Problem 9.30 First, compute the linear convolution of the given signals: y = h ∗ x = [2, 1, 1, 1, 2, 1, 1, 1]∗[2, 1, 2, −3, 2, 1, 2, −3] = [4, 4, 7, −1, 8, 7, 11, −2, 4, 2, 1, −1, 0, −1, −3] and wrap it mod-8 to get ˜y = [8, 6, 8, −2, 8, 6, 8, −2] Then, compute the FFTs of the two signals: ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 2 1 1 1 2 1 1 1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 8−FFT −→ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 10 0 2 0 2 0 2 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ , ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 2 1 2 −3 2 1 2 −3 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 8−FFT −→ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 4 0 −8j 0 12 0 8j 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Multiply them point-wise: ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 10 0 2 0 2 0 2 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ × ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 4 0 −8j 0 12 0 8j 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 40 0 −16j 0 24 0 16j 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Divide by N = 8, conjugate every entry, and take an 8-point FFT: ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 5 0 2j 0 3 0 −2j 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 8−FFT −→ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 8 6 8 −2 8 6 8 −2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 205
- 104. Problem 9.31 a. The 8-FFT algorithm gives x = [6, 1, 0, 1, 6, 1, 0, 1] ⇒ X = [16, 0, 12, 0, 8, 0, 12, 0] b. Using x(n)= 1 N N−1 k=0 X(k)ejωkn , x(n)= 1 8 16 + 12ejπn/2 + 8ejπn + 12e−jπn/2 = 2 + 3 cos πn 2 + (−1)n c. With arbitrary values of a and b: x = [6 − a, 1, 0, 1, 6, 1, 0, 1, a], x = [6 − a, 1 − b, 0, 1, 6, 1, 0, 1, a, b] d. Reduce x mod-4 to get ˜x = [12, 2, 0, 2]. Its 4-FFT is easily found to be X = [16, 12, 8, 12]. It can also be obtained by picking every other entry of the 8-point DFT. Problem 9.32 Start with A(k)= N−1 n=0 Wkn N a(n) and evaluate it at N − k A(N − k)= N−1 n=0 W (N−k)n N a(n) Noting that WN N = 1, it can be written as A(N − k)= N−1 n=0 W−kn N a(n) On the other hand, because a(n) is real A(k)∗ = N−1 n=0 Wkn N ∗ a(n) But, Wkn N ∗ = W−kn N Thus, A(N − k)= A(k)∗ 206 Problem 9.33 Using linearity, we have X(k)= A(k)+jB(k) ⇒ X(N − k)∗ = A(N − k)∗ −jB(N − k)∗ Using the results of Problem 9.32, A(N − k)∗ = A(k) and B(N − k)∗ = B(k), we obtain X(N − k)∗ = A(k)−jB(k). Thus, we have the system of equations: X(k) = A(k)+jB(k) X(N − k)∗ = A(k)−jB(k) which can be solved for A(k) and B(k). Problem 9.34 The expressions for G(k) and H(k) follow from the results of Problem 9.33, except that the di- mension of the FFTs is N/2. The final expressions that construct X(k) are the standard Butterfly operations for merging G(k) and H(k). The additional multiplications to build X(k) from Y(k) are 3N/2, (that is, the divisions by 2 and 2j, and multiplications by Wk N). Thus the total number of multiplications using this method compared to a direct N-point FFT will be 1 2 N 2 log2( N 2 )+ 3N 2 1 2 N log2 N = 1 2 + 2.5 log2 N If one takes into account the multiplications required for computing the Wk powers, then the total becomes N 2 log2( N 2 )+ 4N 2 N log2 N = 1 2 + 1.5 log2 N Typically, 65% of the standard cost for N = 1024. Problem 9.35 /* fftreal.c - FFT of a real-valued signal */ #include <math.h> #include <cmplx.h> #include <stdlib.h> void fft(); void fftreal(N, x, X) int N; double *x; time data complex *X; FFT array { int n, k, Nhalf=N/2; double pi=4*atan(1.0); 207
- 105. complex one, half, halfj, W; complex *Y, temp, Wk, Gk, Hk; one = cmplx(1.0, 0.0); half = cmplx(0.5, 0.0); halfj = cmplx(0.0, -0.5); W = cexp(cmplx(0.0, -2*pi/N)); Y = (complex *) calloc(Nhalf, sizeof(complex)); for (n=0; n<Nhalf; n++) Y[n] = cmplx(x[2*n], x[2*n+1]); fft(Nhalf, Y); for (Wk=one, k=0; k<Nhalf; k++) { Wk = Wk if (k==0) temp = conjg(Y[0]); note: Y(0) = Y(N/2) else temp = conjg(Y[Nhalf-k]); Gk = cmul(half, cadd(Y[k], temp)); G(k) Hk = cmul(halfj, csub(Y[k], temp)); H(k) temp = cmul(Wk, Hk); Wk H(k) Wk = cmul(W, Wk); next Wk = Wk+1 X[k] = cadd(Gk, temp); G(k) + Wk H(k) X[k+Nhalf] = csub(Gk, temp); G(k) − Wk H(k) } } Problem 9.36 An example will demonstrate the result. For N = 4, we have: x = [x0, x1, x2, x3] ⇒ X(z)= x0 + x1z−1 + x2z−2 + x3z−3 xR = [x3, x2, x1, x0] ⇒ XR(z)= x3 + x2z−1 + x1z−2 + x0z−3 It follows that XR(z)= z−3 [x3z3 + x2z2 + x1z + x0]= z−3 X(z−1 ) In the time domain, the operation of reflection leads to the z-transform pairs: x(n) Z −→ X(z) ⇒ x(−n) Z −→ X(z−1 ) The operation of delay on the reflected signal then gives: xR(n)= x(N − 1 − n)= x(−(n − N + 1)) Z −→ z−(N−1) X(z−1 ) The N-point DFT of the reversed signal is obtained by setting z = zk = e2πjk/N = W−k N . Using the property WN N = 1, we find: XR(k)= z−N+1 k X(z−1 k )= W (N−1)k N X(−k)= W−k N X(N − k) where in the last step, we used X(z−1 k )= X(−ωk)= X(−k)= X(N − k) by the periodicity of the N-point DFT. 208 Problem 9.37 In the definition of y, the reversed signal xR is delayed by N samples, that is, we have x = [x, xR]= [x, 0, 0, . . . , 0 N zeros ]+[0, 0, . . . , 0 N zeros , xR] Thus, using linear superposition, the delay property of z-transforms, and the results of Problem 9.36, we have in the z-domain: Y(z)= X(z)+z−N XR(z)= X(z)+z−2N+1 X(z−1 ) where the factor z−N represents the delay of xR by N samples. Replacing z by zk = ejωk , and using the result z2N k = 1, we get Yk = Y(ωk)= X(ωk)+ejωk X(−ωk)= ejωk/2 e−jωk/2 X(ωk)+ejωk/2 X(−ωk) = 2ejωk/2 Re e−jωk/2 X(ωk) where we used the hermitian property X(−ω)= X(ω)∗ . The quantity under the real part is now: e−jωk/2 X(ωk)= e−jωk/2 N−1 n=0 xne−jωkn = N−1 n=0 xne−jωk(n+1/2) Thus, its real part is the cosine transform: Re e−jωk/2 X(ωk) = N−1 n=0 xn cos ωk(n + 1/2) = Ck It follows that Yk = 2ejωk/2 Ck = 2ejπk/2N Ck Because Yk is the (2N)-point DFT of a real-valued signal, it will satisfy the hermitian property Y2N−k = Y∗ k . In terms of Ck, we have: Y2N−k = 2ejπ(2N−k)/2N C2N−k = 2ejπ e−jπk/2N C2N−k = −2e−jπk/2N C2N−k Y∗ k = 2e−jπk/2N Ck Equating the right-hand sides and canceling the common factors gives the symmetry property C2N−k = −Ck , k = 0, 1, . . . , 2N − 1 At k = N, it gives C2N−N = −CN or CN = −CN which implies CN = 0. Next, considering the inverse DFT for Yk, we have: yn = 1 2N 2N−1 k=0 Ykejωkn , n = 0, 1, . . . , 2N − 1 But the first N samples of yn are precisely xn. Replacing Yk in terms of Ck gives for n = 0, 1, . . . , N − 1: xn = 1 2N 2N−1 k=0 2ejωk/2 Ckejωkn = 1 N 2N−1 k=0 Ckejωk(n+1/2) 209
- 106. Next, we split the summation into three terms: xn = 1 N C0 + N−1 k=1 Ckejωk(n+1/2) + 2N−1 k=N+1 Ckejωk(n+1/2) where the second summation starts at k = N + 1 because CN = 0. If we change variables from k to m = 2N − k, we can write 2N−1 k=N+1 Ckejωk(n+1/2) = N−1 m=1 C2N−mejω2N−m(n+1/2) Using the symmetry property C2N−m = −Cm, and the trig identity ejω2N−m(n+1/2) = ejπ(2N−m)(n+1/2)/N = e2jπ(n+1/2) e−jπm(n+1/2)/N = −e−jωm(n+1/2) we have (with the two minus signs canceling): N−1 m=1 C2N−mejω2N−m(n+1/2) = N−1 m=1 Cme−jωm(n+1/2) Thus, the inverse DFT becomes (changing the summation from m back to k): xn = 1 N C0 + N−1 k=1 Ckejωk(n+1/2) + N−1 k=1 Cke−jωk(n+1/2) The two summations now combine with Euler’s formula to give: xn = 1 N C0 + 2 N−1 k=1 Ck cos ωk(n + 1/2) Problem 9.38 By definition, we have X2N(k)= L−1 n=0 x(n)Wnk 2N ⇒ X2N(2k)= L−1 n=0 x(n)W2nk 2N But, W2nk 2N = exp −2πj2nk/2N = exp −2πjnk/N = Wnk N . Thus, X2N(2k)= L−1 n=0 x(n)W2nk 2N = L−1 n=0 x(n)Wnk N = XN(k) Problem 9.39 Observe that the five length-5 sequences are the mod-5 wrapped versions of the following “linearly shifted” sequences: x0 x1 x2 x3 x4 0 0 0 0 0 0 x0 x1 x2 x3 x4 0 0 0 0 0 0 x0 x1 x2 x3 x4 0 0 0 0 0 0 x0 x1 x2 x3 x4 0 0 0 0 0 0 x0 x1 x2 x3 x4 0 210 Their z-transforms differ by a simple delay factor, that is, Xi(z)= z−i X0(z), i = 1, 2, 3, 4 Setting z = zk = ejωk = W−k 5 , we obtain Xi(zk)= z−i k X0(zk)= Wik 5 X0(zk), i = 1, 2, 3, 4 Problem 9.40 Consider, for example, the first and the third. Their z-transforms are X1(z) = x0 + x1z−1 + x2z−2 + x3z−3 + x4z−4 + x5z−5 + x6z−6 + x7z−7 X3(z) = x0 + x1z−1 + (x2 + x6)z−2 + (x3 + x7)z−3 + x4z−4 + x5z−5 Subtracting, we have X3(z)−X1(z)= x6z−2 (1 − z−4 )+x7z−3 (1 − z−4 )= (x6z−2 + x7z−3 )(1 − z−4 ) Evaluating at the 4th roots of unity, z = zk = e2πjk/4 , we obtain X3(zk)−X1(zk)= 0, k = 0, 1, 2, 3 Problem 9.41 Using the convolution table, perform the ordinary linear convolution of the given signal and filter to get y = h ∗ x = [1, 0, −1, 0, 2, 0, −2, 0, −2, 0, 2, 1, 0, −1, 0, −1, 0, 1, 0, −1, 0, 1] The overlap-save method divides the input signal into segments of length N = 8 that are over- lapping by M = 3 points, as shown below: x = 1, 1, 1, 1, 3, (3, 3, 3 , 1, 1, 1, 2, 2), 2, 2, (1, 1, 1 , 1, 0, 0, 0, 0) Convolving these segments with the filter gives: y0 = [1, −1, −1, 1]∗[1, 1, 1, 1, 3, 3, 3, 3] = [1, 0, −1, 0, 2, 0, −2, 0, −3, 0, 3] y1 = [1, −1, −1, 1]∗[3, 3, 3, 1, 1, 1, 2, 2] = [3, 0, −3, −2, 0, 2, 1, 0, −3, 0, 2] y2 = [1, −1, −1, 1]∗[1, 2, 2, 2, 2, 1, 1, 1] = [1, 1, −1, −1, 0, −1, 0, 1, −1, 0, 1] y3 = [1, −1, −1, 1]∗[1, 1, 1, 1, 0, 0, 0, 0] = [1, 0, −1, 0, −1, 0, 1, 0, 0, 0, 0] Reducing these mod-8 and ignoring their first M = 3 points (indicated by x), we obtain: ˜y0 = [x, x, x, 0, 2, 0, −2, 0] ˜y1 = [x, x, x, −2, 0, 2, 1, 0] ˜y2 = [x, x, x, −1, 0, −1, 0, 1] ˜y3 = [x, x, x, 0, −1, 0, 1, 0] Abutting the rest of the points, we get y = [x, x, x, 0, 2, 0, −2, 0][−2, 0, 2, 1, 0][−1, 0, −1, 0, 1][0, −1, 0, 1, 0] With the exception of the first M points, the answer is correct. 211
- 107. Problem 9.42 At fs = 8 Hz, there are 8 samples per period. These samples can be evaluated by setting t = nT = n/8 in the expression for x(t). We find: x(n)= 1 4 {0, 0.5, 1, 0.5, 0, −0.5, −1, −0.5} According to Eq. (9.7.4), the aliased coefficients can be obtained by computing the 8-point DFT of one period and dividing by 8. This DFT can be done quickly by an 8-point FFT by hand, which gives: x = 1 4 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 0 0.5 1 0.5 0 −0.5 −1 −0.5 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ DFT −→ X = 1 4 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 0 −j(2 + √ 2) 0 j(2 − √ 2) 0 −j(2 − √ 2) 0 j(2 + √ 2) ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ , b = 1 4 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 0 (2 + √ 2)/8j 0 −(2 − √ 2)/8j 0 (2 − √ 2)/8j 0 −(2 + √ 2)/8j ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ The coefficients b1 and b7 correspond to frequencies ±f1 = ±1 Hz, whereas b3 and b5 correspond to ±f2 = ±3 Hz. Thus, we find from Eq. (9.7.5) and after using Euler’s formula: xal(t)= 2jb1 sin(2πf1t)+2jb3 sin(2πf2t)= 2 + √ 2 16 sin(2πt)− 2 − √ 2 16 sin(6πt) It can be verified easily that xal and x(t) agree at the sampling instants t = nT = n/8. Problem 9.43 In the overlap-add method, the input blocks have length N − M and are non-overlapping. Thus, if there are K blocks in the total length, we will have: L = (N − M)K (P9.2) In Fig. 9.9.1, the input blocks overlap by M points, but again if there are K segments, the total length will be (N − M)K plus an extra M points at the end. Thus, we have in the overlap-save method: L = (N − M)K + M (N − M)K which is essentially the same as in Eq. (P9.2) for large L. Thus, in both cases, the total number of segments can be estimated to be: K = L N − M Per segment, we are doing two FFTs and N multiplications, namely, N log2 N + N = N(1 + log2 N) per segment Thus, for K segments, we have cost: KN(1 + log2 N)= LN(1 + log2 N) N − M 212 This is to be compared with the cost of L(M + 1) of linear convolution of the entire input array with the filter (here, we treat these MACs as complex-valued.) Thus, the relative cost of the fast versus slow convolution will be: fast slow = LN(1 + log2 N) N − M L(M + 1) = N(1 + log2 N) (N − M)(M + 1) Problem 9.44 An example of such a program is given below: /* ovsave.c - FIR filtering by overlap-save method and FFT */ #include <stdio.h> #include <cmplx.h> #include <stdlib.h> calloc(), realloc() #define MAX 64 initial allocation size void circonv(), complexify(), fft(), ifft(); void main(int argc, char **argv) { FILE *fph; filter file double *h, *x, *xhead; x = length-N block complex *H, *X, *Y; N-point DFTs int M, N, n, i; int max = MAX, dmax = MAX; if (argc != 3) { fprintf(stderr, "usage: fastconv hfile N < xfile > yfilen"); exit(0); } if ((fph = fopen(argv[1], "r")) == NULL) { fprintf(stderr, "can’t open filter file: %sn", argv[1]); exit(0); } N = atoi(argv[2]); h = (double *) calloc(max + 1, sizeof(double)); preliminary allocation for (M=0;; M++) { read h if (M == max) { reallocate h, if necessary max += dmax; h = (double *) realloc((char *) h, (max + 1) * sizeof(double)); } if (fscanf(fph, "%lf", h + M) == EOF) break; } M--; M = filter order if (N <= 2*M) { fprintf(stderr, "must have N > 2M = %d", 2*M); exit(0); } 213
- 108. h = (double *) realloc((char *) h, (M+1) * sizeof(double)); x = (double *) calloc(N, sizeof(double)); xhead = (double *) calloc(M, sizeof(double)); head/tail of input block H = (complex *) calloc(N, sizeof(complex)); allocate DFT arrays X = (complex *) calloc(N, sizeof(complex)); Y = (complex *) calloc(N, sizeof(complex)); for (n=0; n<N; n++) complexify/extend to length N if (n <= M) H[n] = cmplx(h[n], 0.0); else H[n] = cmplx(0.0, 0.0); fft(N, H); H = FFT(h), done once for (n=0; n<M; n++) xhead[n] = 0; pad M zeros in front of first block for (;;) { keep reading input in blocks x for (n=0; n<M; n++) previous tail is head of present block x x[n] = xhead[n]; for (n=M; n<N; n++) read rest of x from input stream if (scanf("%lf", x + n) == EOF) goto lastblocks; for (n=0; n<M; n++) tail of present block x xhead[n] = x[n + N - M]; will be head of next block complexify(N, x, X); X = cmplx(x, 0) circonv(N, H, X, Y); done by fft for (n=M; n<N; n++) discard first M points, save the rest printf("%lfn", real(Y[n])); } lastblocks: n = index where EOF occurred for (i=n; i<N; i++) pad rest of x with zeros x[i] = 0; if n <= N-M there is only one last block to consider complexify(N, x, X); otherwise, there are two last blocks circonv(N, H, X, Y); for (i=M; i<N; i++) printf("%lfn", real(Y[i])); if (n <= N-M) exit(0); exit, if only one last block otherwise, process very last block for (i=0; i<M; i++) if n > N-M, the tail of previous xhead[i] = x[i + N - M]; block will be non-zero for (i=0; i<N; i++) if (i < M) x[i] = xhead[i]; head of very last block 214 else x[i] = 0; rest of block is zero complexify(N, x, X); circonv(N, H, X, Y); for (i=M; i<N; i++) printf("%lfn", real(Y[i])); } ——————————————————————– void complexify(N, x, X) int N; double *x; complex *X; { int n; for (n=0; n<N; n++) X[n] = cmplx(x[n], 0.0); } ——————————————————————– void circonv(N, H, X, Y) circular convolution via FFT int N; complex *H, *X, *Y; H = FFT(h), must be done previously { int k; fft(N, X); for (k=0; k<N; k++) Y[k] = cmul(H[k], X[k]); ifft(N, Y); } 215
- 109. Chapter 10 Problems Problem 10.1 Start with D(ω)= ∞ k=−∞ d(k)e−jωk and assume that d(k) is real and even in k. Taking conjugates and replacing ω by −ω, gives D(−ω)∗ = ∞ k=−∞ d(k)ejωk ∗ = ∞ k=−∞ d(k)e−jωk = D(ω) This result uses only the reality of d(k) and is the usual hermitian property for transfer functions of real-valued impulse responses. Because d(k)= d(−k), we can change the summation from k to −k to get: D(ω)= ∞ k=−∞ d(k)e−jωk = ∞ k=−∞ d(−k)e−jω(−k) = ∞ k=−∞ d(k)e−j(−ω)k = D(−ω) Thus, the reality and evenness conditions imply together: D(ω)= D(−ω)∗ = D(−ω) which states that D(ω) is both real and even in ω. In the antisymmetric case, we have d(−k)= −d(k), and therefore D(ω)= ∞ k=−∞ d(k)e−jωk = ∞ k=−∞ d(−k)e−jω(−k) = − ∞ k=−∞ d(k)e−j(−ω)k = −D(−ω) Together with the reality condition, we get D(ω)= D(−ω)∗ = −D(−ω) which implies that D(ω) is pure imaginary and odd in ω. Separating the negative- and positive-k terms, we can rewrite D(ω) in the form: D(ω)= d(0)+ ∞ k=1 d(k)e−jωk + −1 k=−∞ d(k)e−jωk = d(0)+ ∞ k=1 d(k)e−jωk + ∞ k=1 d(−k)ejωk In the symmetric case, we set d(−k)= d(k) and combine the two sums with Euler’s formula to get: D(ω)= d(0)+2 ∞ k=1 d(k)cos(ωk) (symmetric case) For the antisymmetric case, d(0)= 0 because it must satisfy d(−0)= −d(0). Setting d(−k)= −d(k) in the sum, we obtain: D(ω)= −2j ∞ k=1 d(k)sin(ωk) (antisymmetric case) 216 Problem 10.2 For the bandpass example, we use the fact that D(ω)= 1 over the passband ωa ≤ |ω| ≤ ωb to get: d(k) = π −π D(ω)ejωk dω 2π = −ωa −ωb D(ω)ejωk dω 2π + ωb ωa D(ω)ejωk dω 2π = = −ωa −ωb ejωk dω 2π + ωb ωa ejωk dω 2π = ejωk 2πjk −ωa −ωb + ejωk 2πjk ωb ωa = e−jωak − e−jωbk 2πjk + ejωbk − ejωak 2πjk = sin(ωbk)− sin(ωak) πk Setting ωa = 0, gives the lowpass case, and ωb = π the highpass one. In the latter, we have: d(k)= sin(πk)− sin(ωbk) πk The first term is zero for all non-zero k, and its equal to unity for k = 0. Thus, we may replace it by δ(k): d(k)= δ(k)− sin(ωak) πk For the differentiator and Hilbert transformer, see the more general cases of Problem 10.3. Problem 10.3 For the lowpass differentiator, we have d(k)= π −π D(ω)ejωk dω 2π = ωc −ωc jωejωk dω 2π It can be evaluated either by parts, or by recognizing the integrand as the derivative with respect to k (treating k initially as a real variable and at the end restricting it to integer values.) d(k)= ∂ ∂k ωc −ωc ejωk dω 2π = ∂ ∂k sin(ωck) πk where the ω-integration gave the standard lowpass filter. Performing the differentiation, we get the required expression for d(k): d(k)= ωc cos(ωck) πk − sin(ωck) πk2 Setting ωc = π gives the full-band differentiator: d(k)= cos(πk) k − sin(πk) πk2 Because d(k) belongs to the antisymmetric class, it will have d(0)= 0. The second term vanishes for nonzero integer ks. Thus, we have for k = 0: d(k)= cos(πk) k = (−1)k k (differentiator) The Hilbert transformer case is similar. 217
- 110. Problem 10.4 Using the differentiation argument, we have in this case: d(k)= ∂ ∂k −ωa −ωb ejωk dω 2π + ωb ωa ejωk dω 2π The integrations give the bandpass filter case. Thus, d(k)= ∂ ∂k sin(ωbk)− sin(ωak) πk which becomes the difference of two lowpass differentiators: d(k)= ωb cos(ωbk) πk − sin(ωbk) πk2 − ωa cos(ωak) πk + sin(ωak) πk2 Problem 10.5 For the first-order differentiator, we have impulse response, transfer function, and frequency response: h = [1, −1] ⇒ H(z)= 1 − z−1 ⇒ H(ω)= 1 − e−jω Thus, the NRR will be NRR = h2 0 + h2 1 = 1 + 1 = 2 Thus, the noise power doubles at the output. To see how closely it resembles the ideal differen- tiator, we expand the frequency response at low frequencies: H(ω)= 1 − e−jω 1 − (1 − jω)= jω for small ω For the ideal lowpass differentiator case, the NRR can be computed as: NRR = ωc −ωc |D(ω)|2 dω 2π = ωc −ωc |jω|2 dω 2π = ω3 c 3π Thus, it increases rapidly with ωc. Compared to the full-band maximum NRR obtained with ωc = π, we have NRRmax = π3 /3π = π2 /3 and therefore, NRR NRRmax = ωc π 3 Problem 10.6 Let E(ω)= D(ω)−D(ω) be the error in the designed filter. Then, we have E(ω)= D(ω)−D(ω)= ∞ k=−∞ d(k)e−jωk − M k=−M d(k)e−jωk = |k|>M d(k)e−jωk because the coefficients d(k) cancel for |k| ≤ M. Thus, the time-domain coefficients of the error are: e(k)= 0 if |k| ≤ M d(k) if |k| > M 218 Parseval’s identity applied to e(k) gives then π −π |E(ω)|2 dω 2π = ∞ k=−∞ e2 (k)= |k|>M d2 (k) or, EM = π −π |D(ω)−D(ω)|2 dω 2π = |k|>M d2 (k) The right-hand side sum can be written with the help of Parseval’s identity applied to D(ω) itself: |k|>M d2 (k)= ∞ k=−∞ d2 (k)− |k|≤M d2 (k)= π −π |D(ω)|2 dω 2π − M k=−M d(k)2 In the limit as M → ∞, the second term in the right-hand side tends to the Parseval limit, thus canceling the first term and resulting in EM → 0. Problem 10.7 From Problem 10.3, we have d(k)= (−1)k /k, k = 0, for the differentiator filter. Using Parseval’s identity we have ∞ k=−∞ d2 (k)= π −π |D(ω)|2 dω 2π = π −π ω2 dω 2π = π2 3 But the left-hand side is twice the required sum: ∞ k=−∞ d2 (k)= k=0 1 k2 = 2 ∞ k=1 1 k2 Thus, 2 ∞ k=1 1 k2 = π2 3 ⇒ ∞ k=1 1 k2 = π2 6 For the full-band Hilbert transformer, we have d(k)= 1 − cos(πk) πk = 1 − (−1)k πk , k = 0 Thus, d(k) is nonzero only for odd k, and is in this case d(k)= 2/πk. Parseval’s identity applied to this filter gives: ∞ k=−∞ d2 (k)= π −π |D(ω)|2 dω 2π = π −π |jsign(ω)|2 dω 2π = π −π 1 dω 2π = 1 The left-hand side on the other hand becomes: ∞ k=−∞ d2 (k)= 2 ∞ k=1 k=odd 4 π2k2 Thus, we obtain 219
- 111. 2 ∞ k=1 k=odd 4 π2k2 = 1 ⇒ ∞ k=1 k=odd 1 k2 = π2 8 If the sum for all integers k is π2 /6 and for odd ones π2 /8, it follows by subtraction that the sum over even integers will be: ∞ k=2 k=even 1 k2 = π2 6 − π2 8 = π2 24 Problem 10.8 The sinusoidal response of D(ω) in its steady-state form is: ejωn D −→ D(ω)ejωn = −jsign(ω)ejωn Replacing ω by −ω, and using the odd symmetry property of D(ω), we have: e−jωn D −→ D(−ω)e−jωn = jsign(ω)e−jωn But cos(ωn)= [ejωn + e−jωn ]/2. Thus, forming the same linear combination of outputs, we have: cos(ωn) D −→ 1 2 −jsign(ω)ejωn + jsign(ω)e−jωn = sign(ω)sin(ωn) Similarly, because sin(ωn)= [ejωn − e−jωn ]/2j, we have for the outputs: sin(ωn) D −→ 1 2j −jsign(ω)ejωn − jsign(ω)e−jωn = −sign(ω)cos(ωn) Problem 10.9 From the definition of D(z), we have: D(z)= M k=−M d(k)z−k ⇒ D(z−1 )= M k=−M d(k)zk = M k=−M d(−k)z−k where in the last step we changed summation from k to −k. In the symmetric/antisymmetric cases, we have d(k)= ±d(k). Therefore, D(z−1 )= ± M k=−M d(k)z−k = ±D(z) From the reality of d(k), it follows that the zeros will come in conjugate pairs if they are complex. Thus, if z0 is a complex zero, then z∗ 0 is also a zero. From the symmetry or antisymmetry properties, it follows that if z0 is a zero, then D(z−1 0 )= ±D(z0)= 0 Thus, 1/z0 and 1/z∗ 0 are both zeros. The four zeros {z0, z∗ 0 , 1 z0 , 1 z∗ 0 } are shown in Fig. P10.1. If z0 lies on the unit circle then because 1/z∗ 0 = z0, the only other distinct zero is the complex conjugate. 220 z0 1/z0 * 1/z0 z0 * Fig. P10.1 Pattern of zeros of linear phase filters. In the antisymmetric case, we have in particular for z = ±1, D(±1)= M k=−M d(k)(±1)k = 0 This follows because d(k) is odd and summed over symmetric limits and because the factor (±1)k is even in k. The same conclusion follows from the frequency response. Using the anti- symmetry property and splitting the sum in two parts, we get: D(ω)= M k=−M d(k)e−jωk = M k=1 d(k) e−jωk − ejωk = −2j M k=1 d(k)sin(ωk) Setting ω = 0, π, corresponding to z = ±1, we have sin(ωk)= 0, and therefore, D(ω) vanishes. Finally, we note that if d(k) is windowed by a window w(k), that is, d (k)= w(k)w(k), then because the window w(k) is symmetric with respect to the origin, the symmetry/antisymmetry properties of d(k) will be preserved and will imply similar results on the locations of the zeros. Problem 10.10 For the symmetric case, the following must necessarily be zeros: z0 = 0.5 + 0.5j, z∗ 0 = 0.5 − 0.5j, 1 z0 = 1 − j, 1 z∗ 0 = 1 + j Thus, the transfer function will be H(z)= 1−(0.5+0.5j)z−1 1−(0.5−0.5j)z−1 1−(1−j)z−1 1−(1+j)z−1 = 1−3z−1 +4.5z−2 −3z−3 +z−4 In the time domain, we get the length-5 symmetric (about its middle) impulse response: h = [1, −3, 4.5, −3, 1] In the antisymmetric case, we have in addition two zeros at z = ±1 contributing a factor (1−z−2 ) to the transfer function. Thus, H(z)= (1 − z−2 )(1 − 3z−1 + 4.5z−2 − 3z−3 + z−4 )= 1 − 3z−1 + 3.5z−2 − 3.5z−4 + 3z−5 − z−6 and in the time domain: h = [1, −3, 3.5, 0, −3.5, 3, −1] 221
- 112. The frequency response in the symmetric case is obtained by factoring out a factor of z−2 and using Euler’s formula: H(z)= z−2 z2 + z−2 − 3(z + z−1 )+4.5 ⇒ H(ω)= 2e−2jω cos(2ω)−3 cos(ω)+4.5 For the antisymmetric case, we have: H(z)= z−3 (z3 − z−3 )−3(z2 − z−2 )+3.5(z − z−1 ) ⇒ H(ω)= −2je−3jω sin(3ω)−3 sin(2ω)+3.5 sin(ω) Problem 10.11 Proceeding as in the previous problem, we find the transfer function of the symmetric case: H(z)= (1−0.5jz−1 )(1+0.5jz−1 )(1−2jz−1 )(1+2jz−1 )= 1+4.25z−2 +z−4 ⇒ h = [1, 0, 4.25, 0, 1] and for the antisymmetric case: H(z)= (1−z−2 )(1+4.25z−2 +z−4 )= 1+3.25z−2 −3.25z−4 −z−6 ⇒ h = [1, 0, 3.25, 0, −3.25, 0, −1] Problem 10.12 We use the first-order Taylor series approximation: ln(1 ± x) ±x Changing to base-10, we have: log10(1 ± x)= ln(1 ± x) ln(10) = ±0.4343 x It follows that Apass = 20 log10 1 + δpass 1 − δpass = 20 log10(1 + δpass)−20 log10(1 − δpass) 20 · 0.4343 δpass + 20 · 0.4343 δpass = 17.372 δpass Problem 10.13 First, we calculate δpass and δstop from Eq. (10.2.5): δpass = 100.1/20 − 1 100.1/20 + 1 = 0.0058, δstop = 10−80/20 = 0.0001 Therefore, δ = min(δpass, δstop)= δstop = 0.0001, which in dB is A = −20 log10 δ = Astop = 80. The D and α parameters are computed by: α = 0.1102(A − 8.7)= 0.1102(80 − 8.7)= 7.857, D = A − 7.95 14.36 = 5.017 The transition width is Δf = fstop − fpass = 2 − 1.5 = 0.5 kHz; therefore, the filter length will be: N = 1 + Dfs Δf = 1 + 5.017 · 10 0.5 = 101.34 ⇒ N = 103 222 Problem 10.14 The transition width is inversely related to the filter length N: Δf = Dfs N − 1 As in the previous problem, we find D = 5.017. Thus, Δf = Dfs N − 1 = 5.017 · 10 251 − 1 = 0.2007 kHz Problem 10.15 The window’s width factor is D = 5.017 as in the previous problem. Thus, Δf = Dfs N − 1 = 5.017 · 44.1 129 − 1 = 1.729 kHz Conversely, if Δf = 2 kHz, then we may solve for the D-factor and attenuation A: D = A − 7.95 14.36 = (N − 1) Δf fs = 5.805 ⇒ A = 14.36D + 7.95 = 91.31 dB The corresponding ripple will be δ = 10−A/20 = 10−91.31/20 = 0.000027 which gives the passband attenuation: Apass = 17.372 δ = 0.000472 dB In the 4-times oversampling case, the transition width is required to be: Δf = 24.55 − 19.55 = 5 kHz This gives the width factor and attenuation: D = A − 7.95 14.36 = (N − 1) Δf 4fs = 3.6281 ⇒ A = 14.36D + 7.95 = 60.05 dB Problem 10.16 Assuming that an N-tap filter requires approximately N instructions for processing each input sample, the maximum filter length will correspond to the maximum number of instructions that fit within the sampling interval T, that is, Nmax = T Tinstr = finstr fs Inserting this into the Kaiser design formula, we find the minimum width (where we used N instead of N − 1 in the denominator): Δf = Dfs N = Dfs 2 finstr 223
- 113. Problem 10.17 Using N = finstr/fs for the maximum filter length, we obtain the maximum width parameter D and attenuation A: D = A − 7.95 14.36 = N Δf fs = finstrΔf fs 2 = FinstrΔF ⇒ A = 14.36FinstrΔF + 7.95 Problem 10.18 The following MATLAB program segment computes the rectangularly and Hamming windowed impulse responses with the help of the MATLAB function dlh.m, and then it computes the cor- responding frequency responses over NF equally spaced frequencies over the right half of the Nyquist interval [0, π] with the help of the function dtft.m: wc = 0.3 * pi; % cutoff frequency N = 81; % filter length w = 0.54 - 0.46 * cos(2 * pi * (0:N-1) / (N-1)); % Hamming window hrec = dlh(1, wc, N); % ideal lowpass filter hham = w .* hrec; % windowed filter NF = 200; % 200 frequencies in rads/sample omega = (0:NF-1) * pi / NF; % over the interval [0, π] Hrec = dtft(hrec, omega); % frequency response Hham = dtft(hham, omega); Problem 10.19 As an example, the following MATLAB program segment computes the bandpass filter of Example 10.2.2 (alternative design): fs = 20; in kHz fpa = 4; fpb = 6; fsa = 3; fsb = 8; Apass = 0.1; Astop = 80; in dB h = kbp(fs, fpa, fpb, fsa, fsb, Apass, Astop, -1); % s = −1 for alternative design NF = 200; omega = (0:NF-1) * pi / NF; H = dtft(h, omega); Similarly, the highpass filter can be designed by: fs = 20; in kHz fpass = 5; fstop = 4; Apass = 0.1; Astop = 80; in dB h = klh(-1, fs, fpass, fstop, Apass, Astop); % s = −1 for highpass NF = 200; omega = (0:NF-1) * pi / NF; H = dtft(h, omega); 224 Problem 10.20 Use the design MATLAB routine kdiff, with fs = 2 so that ω = 2πf/2 = πf, that is, the f’s are the ω’s in units of π. The filter lengths depend only on the values of Δf and A, they are: Δf A N 0.10 30 33 0.10 60 75 0.05 30 63 0.05 60 147 For example, in the case ωc = 0.4π, Δω = 0.05π, A = 60 dB, we have the MATLAB code: h = kdiff(2, 0.4, 0.05, 60); NF = 200; omega = (0:NF-1) * pi / NF; H = dtft(h, omega); The magnitude responses are shown in Figs. P10.2 and P10.3. Fig. P10.2 Lowpass differentiator with ωc = 0.4π. 225
- 114. Fig. P10.3 Lowpass differentiator with ωc = 0.8π. Problem 10.21 For example, in the case ωc = 0.4π, Δω = 0.05π, A = 60 dB, we have the MATLAB segment that computes the Kaiser differentiator and the Savitzky-Golay differentiators of polynomial order d = 3: h = kdiff(2, 0.4, 0.05, 60); % Kaiser differentiator [B, S] = sg(3, N); % Savitzky-Golay smoothing filters F = S’ * S; % polynomial basis matrix G = S * F^(-1); % Savitzky-Golay differentiator filters g = G(:,2)’; % first-order derivative filter NF = 500; omega = (0:NF-1) * pi / NF; H = dtft(h, omega); % frequency responses G = dtft(g, omega); The magnitude responses are shown in Fig. P10.4. Problem 10.22 Use the design MATLAB routine khilb, with fs = 2 so that ω = 2πf/2 = πf, that is, the f’s are the ω’s in units of π. The filter lengths depend only on the values of Δf and A, they are the 226 Fig. P10.4 Comparison of Kaiser and SG differentiators. same as in Problem 10.20. For example, the Hilbert transformer with ωc = 0.8π, Δω = 0.05π, and A = 60 dB, can be designed by the MATLAB segment: h = khilb(2, 0.8, 0.05, 60); NF = 200; omega = (0:NF-1) * pi / NF; H = dtft(h, omega); % frequency responses The magnitude responses are shown in Figs. P10.5 and P10.6. Problem 10.23 The following MATLAB segment computes the Kaiser and Hamming spectra in dB. The Kaiser window parameters are determined by the function kparm2.m: F1 = 0.20; F2 = 0.25; F3 = 0.30; DF = (F2-F1)/3; % units of fs A1 = 1; A2 = 10^(-50/20); A3 = 1; % amplitudes R = 70; % sidelobe level in dB [alpha, L] = kparm2(DF, R); % Kaiser parameters n = (0:L-1); x = A1 * cos(2*pi*F1*n) + A2 * cos(2*pi*F2*n) + A3 * cos(2*pi*F3*n); xk = kwind(alpha, L) .* x; % Kaiser windowed signal xh = (0.54 - 0.46 * cos(2*pi*n/(L-1))) .* x; % Hamming windowed signal NF = 256; omega = (0:NF-1) * pi / NF; Xk = abs(dtft(xk, omega)); % DTFT spectrum Xkmax = max(Xk); Xk = 20 * log10(Xk / Xkmax); % normalized and in dB Xh = abs(dtft(xh, omega)); 227
- 115. Fig. P10.5 Lowpass Hilbert transformer with ωc = 0.8π. Xhmax = max(Xh); Xh = 20 * log10(Xh / Xhmax); 228 Fig. P10.6 Full-band Hilbert transformer. 229
- 116. Chapter 11 Problems Problem 11.1 Here, we assume that the parameter β = tan(Δω/2) remains positive and less than one, 0 ≤ β ≤ 1. This is guaranteed if the bandwidth is 0 ≤ Δω ≤ π/2 in rads/sample, or, 0 ≤ Δf ≤ fs/4 in Hz. This range covers most practical applications. This condition implies that the poles are complex conjugate and have radius R2 = 1 − β 1 + β ⇒ R = 1 − β 1 + β (If β > 1, which happens when π/2 < Δω ≤ π, then there are two real poles.) It follows that the -level time constant will be neff = ln ln R Problem 11.2 For small values of Δω, we may use the first-order Taylor expansion tan(x) x to write approx- imately: β = tan Δω 2 Δω 2 Then, expanding R to first-order in β and thus to first-order in Δω, we have: R = 1 − β 1 + β (1 − 1 2 β)(1 − 1 2 β) 1 − β 1 − Δω 2 where we used the Taylor expansions (1 ∓ x)±1/2 1 − x/2 and (1 − x/2)2 1 − x. Solving for Δω, we obtain the approximate expression Δω = 2(1 − R). Problem 11.3 For the 3-dB case, the parameter b is the same for both filters. Adding the two transfer functions, we recognize that the two numerators add up to become the denominator: Hnotch(z)+Hpeak(z) = b 1 − 2 cos ω0 z−1 + z−2 1 − 2b cos ω0 z−1 + (2b − 1)z−2 + (1 − b) 1 − z−2 1 − 2b cos ω0 z−1 + (2b − 1)z−2 = b(1 − 2 cos ω0 z−1 + z−2 )+(1 − b)(1 − z−2 ) 1 − 2b cos ω0 z−1 + (2b − 1)z−2 = 1 − 2b cos ω0 z−1 + (2b − 1)z−2 1 − 2b cos ω0 z−1 + (2b − 1)z−2 = 1 For the magnitude responses squared we may use the equivalent analog expressions to get: |Hnotch(ω)|2 +|Hpeak(ω)|2 = |Hnotch(Ω)|2 +|Hpeak(Ω)|2 = (Ω2 − Ω2 0)2 (Ω2 − Ω2 0)2+α2Ω2 + α2 Ω2 (Ω2 − Ω2 0)2+α2Ω2 = 1 230 Problem 11.4 The analog transfer function and magnitude response squared are in the general boost or cut case: H(s)= G0(s2 + Ω2 0)+Gαs s2 + αs + Ω2 0 ⇒ |H(Ω)|2 = G2 0(Ω2 − Ω2 0)2 +G2 α2 Ω2 (Ω2 − Ω2 0)2+α2Ω2 where α = (1 + Ω2 0)β = (1 + Ω2 0)tan Δω 2 G2 B − G2 0 G2 − G2 B ≡ γ G2 B − G2 0 G2 − G2 B where the last step defines the parameter γ, which does not depend on the gains. Normalizing everything to the reference gain G0, we have |H(Ω)|2 = G2 0 (Ω2 − Ω2 0)2 +g2 α2 Ω2 (Ω2 − Ω2 0)2+α2Ω2 , α = γ g2 B − 1 g2 − g2 B , where g = G G0 , gB = GB G0 Then, we may express the magnitude squared response in the following form that explicitly shows the gains: |H(Ω)|2 = G2 0 (Ω2 − Ω2 0)2 +N(g, gB)γ2 Ω2 (Ω2 − Ω2 0)2+D(g, gB)γ2Ω2 where the numerator and denominator gain functions are defined by N(g, gB)= g2 g2 B − 1 g2 − g2 B , D(g, gB)= g2 B − 1 g2 − g2 B For the arithmetic, geometric, and harmonic mean choices for g2 B the functions N and D simplify as follows: g2 B = 1 + g2 2 , N(g, gB)= g2 , D(g, gB)= 1 , (arithmetic) g2 B = g , N(g, gB)= g , D(g, gB)= 1 g , (geometric) g2 B = 2g2 1 + g2 , N(g, gB)= 1 , D(g, gB)= 1 g2 , (harmonic) Now, if we consider an equal and opposite boost and cut by, say by A dB (with respect to the reference), then the boost and cut gains will be g = 10±A/20 . Thus, the substitution g → g−1 changes a boost into an equal and opposite cut, and vice versa. Even though the boost and cut gains are equal and opposite, the bandwidth levels g2 B may be defined arbitrarily in the two cases. However, if we take them also to be related by the substitution gB → g−1 B , then we can show that the boost and cut filters are inverses to each other. Indeed, under the substitutions {g, gB} → {g−1 , g−1 B }, the boost and cut magnitude responses are: |Hboost(Ω)|2 = G2 0 (Ω2 − Ω2 0)2 +N(g, gB)γ2 Ω2 (Ω2 − Ω2 0)2+D(g, gB)γ2Ω2 , |Hcut(Ω)|2 = G2 0 (Ω2 − Ω2 0)2 +N(g−1 , g−1 B )γ2 Ω2 (Ω2 − Ω2 0)2+D(g−1, g−1 B )γ2Ω2 Under this substitution, the functions N and D interchange roles, that is, one can show easily that N(g−1 , g−1 B )= D(g, gB) , D(g−1 , g−1 B )= N(g, gB) 231
- 117. Therefore, the boost and cut filters will be: |Hboost(Ω)|2 = G2 0 (Ω2 − Ω2 0)2 +N(g, gB)γ2 Ω2 (Ω2 − Ω2 0)2+D(g, gB)γ2Ω2 , |Hcut(Ω)|2 = G2 0 (Ω2 − Ω2 0)2 +D(g, gB)γ2 Ω2 (Ω2 − Ω2 0)2+N(g, gB)γ2Ω2 Hence, their product will be identically equal to G4 0. The geometric-mean and weighted geometric- mean choices for g2 B are special cases of the substitution {g, gB} → {g−1 , g−1 B }, and therefore, satisfy the desired property. But, any other choice of gB will do, as long as one defines the boost bandwidth to be gB and the cut bandwidth, g−1 B , that is, equal and opposite bandwidth gains in dB. Note, also that the transfer functions themselves are inverses to each other, not just their magnitudes. Indeed, we can write them in the form: Hboost(s)= G0 s2 + Ω2 0 + γs N(g, gB) s2 + Ω2 0 + γs D(g, gB) , Hcut(s)= G0 s2 + Ω2 0 + γs N(g−1, g−1 B ) s2 + Ω2 0 + γs D(g−1, g−1 B ) = G0 s2 + Ω2 0 + γs D(g, gB) s2 + Ω2 0 + γs N(g, gB) which implies for the analog and the bilinearly transformed digital transfer functions: Hcut(s)= G2 0 Hboost(s) ⇒ Hcut(z)= G2 0 Hboost(z) Problem 11.5 The following MATLAB code segment designs the three filters (two peaking and one complemen- tary notch), computes their impulse responses, and their magnitude responses: w0 = 0.35*pi; Dw = 0.1*pi; in rads/sample [b1, a1, beta1] = parmeq(0, 1, 1/sqrt(2), w0, Dw); peaking 3-dB width [b2, a2, beta2] = parmeq(0, 1, sqrt(0.1), w0, Dw); peaking 10-dB width [b3, a3, beta3] = parmeq(1, 0, 1/sqrt(2), w0, Dw); notching 3-dB width x = [1, zeros(1, 99)]; unit impulse h1 = filter(b1, a1, x); impulse responses h2 = filter(b2, a2, x); h3 = filter(b3, a3, x); w = (0:199)*pi/200; 200 frequencies over [0, π] H1 = abs(dtft(b1, w) ./ dtft(a1, w)).^2; magnitude responses H2 = abs(dtft(b2, w) ./ dtft(a2, w)).^2; H3 = abs(dtft(b3, w) ./ dtft(a3, w)).^2; The 5-percent time constants can be calculated as in Problem 11.1, with = 0.05. Problem 11.6 For example, the following MATLAB code designs the first four filters of Example 11.4.1 and computes their impulse response, frequency, and phase responses (in degrees): [b1, a1, beta1] = parmeq(1, 10^(9/20), 10^(6/20), 0.35*pi, 0.1*pi); [b2, a2, beta2] = parmeq(1, 10^(9/20), 10^(3/20), 0.35*pi, 0.1*pi); [b3, a3, beta3] = parmeq(1, 10^(-9/20), 10^(-6/20), 0.60*pi, 0.2*pi); [b4, a4, beta4] = parmeq(1, 10^(-9/20), 10^(-3/20), 0.60*pi, 0.2*pi); x = [1, zeros(1, 49)]; length-50 unit impulse 232 h1 = filter(b1, a1, x); impulse responses h2 = filter(b2, a2, x); h3 = filter(b3, a3, x); h4 = filter(b4, a4, x); w = (0:199)*pi/200; H1 = dtft(b1, w) ./ dtft(a1, w); argH1 = angle(H1) * 180 / pi; H2 = dtft(b2, w) ./ dtft(a2, w); argH2 = angle(H2) * 180 / pi; H3 = dtft(b3, w) ./ dtft(a3, w); argH3 = angle(H3) * 180 / pi; H4 = dtft(b4, w) ./ dtft(a4, w); argH4 = angle(H4) * 180 / pi; Figure P11.1 shows the phase responses. Note the rapid variations in the vicinities of the center frequencies. Fig. P11.1 Phase responses of Problem 11.6. Problem 11.7 The following MATLAB function lheq.m implements the design of both the lowpass and highpass shelving filters (the parameter s = ±1 selects the two cases): % lheq.m - lowpass/highpass first-order shelving EQ filter design % % [b, a, beta] = lheq(s, G0, G, Gc, wc) % % s = 1, −1 = lowpass, highpass % b = [b0, b1] = numerator coefficients % a = [1, a1] = denominator coefficients % G0, G, Gc = reference, boost/cut, and cutoff gains % wc = cutoff frequency in [rads/sample] % beta = design parameter β function [b, a, beta] = lheq(s, G0, G, Gc, wc) beta = (tan(wc/2))^s * sqrt(abs(Gc^2 - G0^2)) / sqrt(abs(G^2 - Gc^2)); b = [(G0 + G*beta), -s*(G0 - G*beta)] / (1+beta); a = [1, -s*(1-beta)/(1+beta)]; The filters of Examples 11.2.1 and 11.2.2 may be designed by the following MATLAB code, which also shows how to compute the magnitude responses of the two complementary cases: 233
- 118. [b1, a1, beta1] = lheq(1, 0, 1, 1/sqrt(2), 0.2*pi); lowpass cases [b2, a2, beta2] = lheq(1, 0, 1, sqrt(0.9), 0.2*pi); [b3, a3, beta3] = lheq(1, 0, 1, 1/sqrt(2), 0.7*pi); [b4, a4, beta4] = lheq(1, 0, 1, sqrt(0.9), 0.7*pi); [b5, a5, beta5] = lheq(-1, 0, 1, 1/sqrt(2), 0.2*pi); highpass cases [b6, a6, beta6] = lheq(-1, 0, 1, sqrt(0.9), 0.2*pi); omega = (0:199)*pi/200; 200 frequencies in [0, π] H1 = abs(dtft(b1, omega)./dtft(a1, omega)).^2; magnitude response squared H5 = abs(dtft(b5, omega)./dtft(a5, omega)).^2; Problem 11.8 The following MATLAB code generates the first pair of comb and notch filters of Example 11.5.1, that is, with unshifted peaks, and computes the corresponding magnitude and phase responses; it also computes the impulse response of the first filter in two ways: (a) using MATLAB’s function filter and (b) the customized function combfilt.m that uses a circular buffer: D = 10; wc = 0.025 * pi; [a1, b1, c1, beta1] = combeq(1, 10^(9/20), 10^(3/20), D, wc, 1); [a2, b2, c2, beta2] = combeq(1, 10^(-12/20), 10^(-3/20), D, wc, 1); omega = (0:399)*2*pi/ 400; 400 frequencies in [0, 2π] num1 = [b1, zeros(1, D-1), -c1]; numerator coefficients den1 = [1, zeros(1, D-1), -a1]; denominator coefficients H1 = dtft(num1, omega) ./ dtft(den1, omega); frequency response magH1 = 20 * log10(abs(H1)); magnitude response angH1 = angle(H1) * 180 / pi; phase response in degrees num2 = [b2, zeros(1, D-1), -c2]; den2 = [1, zeros(1, D-1), -a2]; H2 = dtft(num2, omega) ./ dtft(den2, omega); magH2 = 20 * log10(abs(H2)); angH2 = angle(H2) * 180 / pi; x = [1, zeros(1, 199)]; length-200 unit impulse h1 = filter(num1, den1, x); impulse response w = zeros(1, D+1); (D+1)-dimensional state vector q = 0; circular pointer index for n=0:199, [y1(n+1), w, q] = combfilt(a1, b1, c1, D, w, q, x(n+1)); note: h1=y1 end The phase responses of the comb and notch filters and the impulse response of the comb are shown in Fig. P11.2. The canonical realization and the corresponding circular buffer version of the sample processing algorithm of the transfer function H(z)= b − cz−D 1 − az−D are given below: 234 Fig. P11.2 Phase responses and impulse response of Problem 11.8. for each input sample x do: sD = tap(D, w, p, D) s0 = x + asD y = bs0 − csD ∗p = s0 cdelay(D, w, &p) x yb a -c z-D sD s0 The following C function combfilt.c is an implementation: /* combfilt.c - comb filtering with circular buffer */ void cdelay(); double tap(); double combfilt(a, b, c, D, w, p, x) double a, b, c, *w, **p, x; p passed by address as in cfir.c int D; { double s0, sD, y; sD = tap(D, w, *p, D); Dth tap output s0 = x + a * sD; input to D-fold delay y = b * s0 - c * sD; output sample **p = s0; update circular delay line cdelay(D, w, p); return y; } The MATLAB version is the function combfilt.m given below: % combfilt.m - comb filtering with circular buffer % % [y, w, q] = combfilt(a, b, c, D, w, q, x); % % The filter parameters [a, b, c] can be obtained from combeq.m 235
- 119. % % It uses cdelay2.m, wrap2.m, tap2.m % based on combfilt.c function [y, w, q] = combfilt(a, b, c, D, w, q, x) sD = tap2(D, w, q, D); s0 = x + a * sD; y = b * s0 - c * sD; w(q+1) = s0; q = cdelay2(D, q); It uses the following routine tap2.m that helps extract the components of the circular internal state vector: % tap2.m - read ith tap of circular delay-line buffer % % si = tap2(D, w, q, i) % % w = (D + 1)-dimensional % q = circular index into w % delay amount i = 0, 1, . . . , D % based on tap2.c function si = tap2(D, w, q, i) si = w(rem(q+i, D+1)+1); % si is delayed output s(n − i) Problem 11.9 In the FIR case of designing a lowpass filter, the limits 1 ± δpass and ±δstop represent the ripples of the (real-valued) frequency response D(ω). These translate to the limits for the magnitude response 1 ± δpass in the passband and [0, δstop] in the stopband. In the IIR case, the definitions are motivated by the closed-form expressions of the magnitude response squared in the Butterworth and Chebyshev cases. Problem 11.10 Because Ωstop > Ωpass, and N ≥ Nexact, we have the inequality Ωstop Ωpass 2N > Ωstop Ωpass 2Nexact = ε2 stop ε2 pass ⇒ 1 + ε2 pass Ωstop Ωpass 2N > 1 + ε2 stop Thus, using Eq. (11.6.14) we have: A(Ωstop)= 10 log10 ⎡ ⎣1 + ε2 pass Ωstop Ωpass 2N ⎤ ⎦ > 10 log10(1 + ε2 stop)= Astop Similarly, if we fix Ω0 by the stopband specification, then we have: Ωpass Ωstop 2N < Ωpass Ωstop 2Nexact = ε2 pass ε2 stop ⇒ 1 + ε2 stop Ωpass Ωstop 2N < 1 + ε2 pass Then, using Eq. (11.6.15), we have: 236 A(Ωpass)= 10 log10 ⎡ ⎣1 + ε2 stop Ωpass Ωstop 2N ⎤ ⎦ < 10 log10(1 + ε2 pass)= Apass Thus, in either case, the specifications are more than satisfied. Problem 11.11 The call to the MATLAB function lhbutt.m: [A, B, P] = lhbutt(1, 40, 10, 15, 3, 35); produces the following cascade matrices B = ⎡ ⎢ ⎣ 0.5001 0.5001 0 0.3821 2 × 0.3821 0.3821 0.2765 2 × 0.2765 0.2765 ⎤ ⎥ ⎦ , A = ⎡ ⎢ ⎣ 1 0.0002 0 1 0.0007 0.5279 1 0.0005 0.1056 ⎤ ⎥ ⎦ Thus, the designed transfer function will be: H(z)= 0.5001(1 + z−1 ) 1 + 0.0002z−1 · 0.3821(1 + z−1 )2 1 + 0.0007z−1 + 0.5279z−2 · 0.2765(1 + z−1 )2 1 + 0.0005z−1 + 0.1056z−2 The parameter vector P is: P = [Ωpass, Ωstop, εpass, εstop, Nexact, N, Astop, Ω0, f0] = [1, 2.4142, 0.9976, 56.2252, 4.57, 5, 38.26, 1.0005, 10.0030] where Astop is the actual stopband attenuation realized by the designed filter, and f0 is the 3- dB cutoff frequency. The apparent discrepancy between the computed 3-dB frequency and the given 3-dB passband specification is accounted for from the fact that the 3-dB frequency Ω0 or f0 is defined to be the frequency where the magnitude squared response drops to one-half its DC value; this is almost a 3-dB drop, but not exactly. By hand, the design is as follows. First, we prewarp the physical frequencies: Ωpass = tan ωpass 2 = tan πfpass fs = tan π10 40 = tan(0.25π)= 1 Ωstop = tan ωstop 2 = tan πfstop fs = tan π15 40 = tan(0.375π)= 2.4142 Then, we compute the -parameters: εpass = 10Apass/10 − 1 = 103/10 − 1 = 0.9976, εstop = 10Astop/10 − 1 = 1035/10 − 1 = 56.2252 and the exact and up-rounded value of N: Nexact = ln(εstop/εpass) ln(Ωstop/Ωpass) = ln(56.2252/0.9976) ln(2.4142/1.0000) = 4.57 , N = 5 Then, we calculate the 3-dB Butterworth parameter Ω0 and invert it to get the 3-dB frequency f0: Ω0 = Ωpass ε1/N pass = 1 (0.9976)1/5 = 1.0005 , f0 = fs π arctan(Ω0)= 1.0030 kHz 237
- 120. The coefficients of the 2nd-order sections are computed by Eq. (11.6.24). Note that because of the small discrepancy mentioned above, Ω0 is not exactly equal to unity, and therefore, the a1 coefficients of all the sections will not be exactly zero. The Butterworth angles are computed by θi = π 2N (N − 1 + 2i)= π 10 (4 + 2i), i = 1, 2 which gives θ1 = 0.6π, θ2 = 0.8π Problem 11.12 The call to lhbutt.m: [A, B, P] = lhbutt(-1, 10, 3, 2, 0.5, 35); produces the following cascade matrices B = 0.3414 −2 × 0.3414 0.3414 0.2454 −2 × 0.2454 0.2454 , A = 1 0.0817 0.4471 1 0.0587 0.0404 Thus, the designed transfer function will be: H(z)= 0.3414(1 − z−1 )2 1 + 0.0817z−1 + 0.4471z−2 · 0.2454(1 − z−1 )2 1 + 0.0587z−1 + 0.0404z−2 The parameter vector P is: P = [Ωpass, Ωstop, εpass, εstop, Nexact, N, Astop, Ω0, f0] = [0.7265, 1.3764, 0.3493, 3.0000, 3.37, 4, 13.27, 0.9451, 2.5899] where Astop is the actual stopband attenuation realized by the designed filter, and f0 is the 3-dB cutoff frequency in kHz. The required closed-form expression for part (c) is: |H(ω)|2 = 1 1 + cot(ω/2) Ω0 2N Problem 11.13 The real part of s can be expressed as follows: Re s = 1 2 (s + s∗ )= z2 − 2cz + 1 2(z2 − 1) + z∗2 − 2cz∗ + 1 2(z∗2 − 1) = (z2 − 2cz + 1)(z∗2 − 1)+(z∗2 − 2cz∗ + 1)(z2 − 1) 2(z2 − 1)(z2 − 1)∗ = |z|4 − 1 − c(z + z∗ )(|z|2 − 1) |z2 − 1|2 = (|z|2 − 1)(|z|2 − c(z + z∗ )+1) |z2 − 1|2 where in the last step we factored |z|4 − 1 = (|z|2 − 1)(|z|2 + 1). Because |c| ≤ 1, the second factor of the numerator is non-negative. Indeed, noting that (z + z∗ ) is real-valued, we have: |c(z + z∗ )| ≤ |z + z∗ | ≤ |z| + |z∗ | = 2|z| ⇒ −2|z| ≤ c(z + z∗ )≤ 2|z| Adding |z|2 + 1 to both sides we get the inequality: |z|2 −2|z|+1 ≤ |z|2 −c(z+z∗ )+1 ≤ |z|2 +2|z|+1 ⇒ 0 ≤ (|z|−1)2 ≤ |z|2 −c(z+z∗ )+1 ≤ (|z|+1)2 Therefore, the sign of Re s is the same as the sign of the factor (|z|2 − 1), and the desired result follows. 238 Problem 11.14 Using a trig identity, we obtain: | sin(ω1 + ω2)| = | sin ω1 cos ω2 + sin ω2 cos ω1| ≤ | sin ω1|| cos ω2| + | sin ω2|| cos ω1| ≤ | sin ω1| + | sin ω2| = sin ω1 + sin ω2 where we used the inequalities | cos ω1| ≤ 1 and | cos ω2| ≤ 1 and the fact that sin ω1 and sin ω2 are non-negative for values of their arguments in the range [0, π]. Thus, the above inequality implies |c| ≤ 1. Problem 11.15 Here, the Butterworth filter order is given to be N = 3. Thus, there remains only the parameter Ω0 to be fixed. The normalized passband frequencies are ωpa = 2πfpa fs = 2π2 20 = 0.2π, ωpb = 2πfpb fs = 2π8 20 = 0.8π Because they add up to π, the bilinear transformation parameter c will be zero, as follows from the equation c = sin(ωpa + ωpb) sin ωpa + sin ωpb = 0 Thus, the bilinear transformation will be here: s = 1 + z−2 1 − z−2 The analog passband frequency will be: Ωpass = c − cos ωpb sin ωpb = − cot ωpb = 1.3764 The εpass parameter is calculated to be: εpass = 10Apass/10 − 1 = 100.5/10 − 1 = 0.3493 which implies for Ω0: Ω0 = Ωpass ε1/N pass = 1.3764 0.34931/3 = 1.9543 The coefficients of the second- and fourth-order digital filter sections are calculated from Eqs. (11.6.43) and (11.6.41), where the there is only one Butterworth angle, namely, θ1 = π 2N (N − 1 + 2)= 2π 3 ⇒ cos θ1 = −0.5 We find for the second-order section: G0 = 0.6615, a01 = 0, a02 = −0.3230 and the fourth-order section: G1 = 0.5639, a11 = 0, a12 = −0.8325, a13 = 0, a14 = 0.4230 resulting in the transfer function: H(z)= 0.6615(1 − z−2 ) 1 − 0.3230z−2 · 0.5639(1 − z−2 )2 1 − 0.8325z−2 + 0.4230z−4 The 3-dB frequencies are calculated from Eq. (11.6.44) with c = 0. We find: f0a = 1.5054 kHz, f0b = 8.4946 kHz 239
- 121. Problem 11.16 We have: 1 1 − 2 s Ω0 cos θi + s2 Ω2 0 = Ω2 0 Ω2 0 − 2sΩ0 cos θi + s2 = Ω2 0 Ω2 0 − 2 1 − z−1 1 + z−1 Ω0 cos θi + 1 − z−1 1 + z−1 2 = Ω2 0(1 + z−1 )2 Ω2 0(1 + z−1)2−2(1 − z−1)(1 + z−1)Ω0 cos θi + (1 − z−1)2 = Ω2 0(1 + z−1 )2 Ω2 0(1 + 2z−1 + z−2)−2(1 − z−2)Ω0 cos θi + (1 − 2z−1 + z−2) = Ω2 0(1 + z−1 )2 (1 − 2Ω0 cos θi + Ω2 0)+2(Ω2 0 − 1)z−1 + (1 + 2Ω0 cos θi + Ω2 0)z−2 = Gi(1 + z−1 )2 1 + ai1z−1 + ai2z−2 where the final result was obtained by dividing numerator and denominator by the factor (1 − 2Ω0 cos θi + Ω2 0) in order to make the constant coefficient of the denominator unity. The highpass case is obtained by replacing z−1 by −z−1 , which effectively changes the sign of the ai1 coefficient. The bandpass case can be done in a similar, but more tedious, fashion. Problem 11.17 The right 3-dB frequency is obtained by solving the equation c − cos ω0b sin ω0b = Ω0 Using the trig identities cos ω0b = 1 − tan2 (ω0b/2) 1 + tan2(ω0b/2) , sin ω0b = 2 tan(ω0b/2) 1 + tan2(ω0b/2) The above equation reduces to the quadratic equation in the variable t = tan(ω0b/2): c − 1 − t2 1 + t2 2t 1 + t2 = Ω0 ⇒ (1 + c)t2 − 2Ω0t − (1 − c)= 0 which has positive solution: t = Ω0 + Ω2 0 + (1 − c)(1 + c) 1 + c implying for the 3-dB frequency tan ω0b 2 = tan πf0b fs = Ω2 0 + 1 − c2 + Ω0 1 + c ⇒ f0b = fs π arctan ⎛ ⎜ ⎝ Ω2 0 + 1 − c2 + Ω0 1 + c ⎞ ⎟ ⎠ The left 3-dB frequency is obtained by replacing Ω0 by −Ω0 in the above solution. 240 Problem 11.18 Matching the stopband specifications exactly means that the stopband frequencies ωsa, ωsb map onto ±Ωstop and that at Ωstop the attenuation is exactly Astop. The first requirement implies −Ωstop = c − cos ωsa sin ωsa Ωstop = c − cos ωsb sin ωsb which can be solve for c and Ωstop: c = sin(ωsa + ωsb) sin ωsa + sin ωsb , Ωstop = c − cos ωsb sin ωsb Given the value of c, we compute next the two candidates for Ωpass: Ωpa = c − cos ωpa sin ωpa , Ωpb = c − cos ωpb sin ωpb Ideally, we would like have Ωpb = Ωpass and Ωpa = −Ωpass. But this is impossible because c has already been fixed. Because the Butterworth magnitude response is a monotonically decreasing function of Ω, it is enough to choose the widest of the two passbands defined above. Thus, we define: Ωpass = max |Ωpa|, |Ωpb| Once we have computed Ωpass and Ωstop, we design our Butterworth analog filter; that is, compute N and Ω0 from Nexact = ln(εstop/εpass) ln(Ωstop/Ωpass) , N = Nexact , Ω0 = Ωstop ε1/N stop where the equation for Ω0 matches the stopband specification exactly. The digital filter coef- ficients are then computed in terms of N and Ω0 as usual. The following MATLAB function bpbutt2.m implements the above procedure and is an alternative to bpsbutt.m: % bpbutt2.m - bandpass Butterworth digital filter design % % [A, B, P] = bpbutt2(fs, fpa, fpb, fsa, fsb, Apass, Astop) % % alternative to bpsbutt.m % it matches stopbands and uses worst passband. % % design parameters: % P = [Wpass, Wstop, Wpa, Wpb, c, fc, epass, estop, Nex, N, Apass, W0, f0a, f0b]; % A,B are Kx5 matrices of cascade of 4th/2nd order sections function [A, B, P] = bpbutt2(fs, fpa, fpb, fsa, fsb, Apass, Astop) c = sin(2*pi*(fsa + fsb) / fs) / (sin(2*pi*fsa / fs) + sin(2*pi*fsb / fs)); fc = 0.5 * (fs/pi) * acos(c); Wstop = abs((c - cos(2*pi*fsb / fs)) / sin(2*pi*fsb / fs)); Wpa = (c - cos(2*pi*fpa / fs)) / sin(2*pi*fpa / fs); Wpb = (c - cos(2*pi*fpb / fs)) / sin(2*pi*fpb / fs); 241
- 122. Wpass = max(abs(Wpa), abs(Wpb)); epass = sqrt(10.0^(Apass/10) - 1); estop = sqrt(10.0^(Astop/10) - 1); Nex = log(estop/epass) / log(Wstop/Wpass); N = ceil(Nex); r = rem(N,2); K = (N - r) / 2; W0 = Wstop * (estop^(-1/N)); Apass = 10 * log10(1 + (Wpass/W0)^(2*N)); f0a = (fs/pi) * atan((sqrt(W0^2 - c^2 + 1) - W0)/(c+1)); f0b = (fs/pi) * atan((sqrt(W0^2 - c^2 + 1) + W0)/(c+1)); P = [Wpass, Wstop, Wpa, Wpb, c, fc, epass, estop, Nex, N, Apass, W0, f0a, f0b]; if r==1, G = W0 / (1 + W0); a1 = -2 * c / (1 + W0); a2 = (1 - W0) / (1 + W0); B(1,:) = G * [1, 0, -1, 0, 0]; A(1,:) = [1, a1, a2, 0, 0]; else B(1,:) = [1, 0, 0, 0, 0]; A(1,:) = [1, 0, 0, 0, 0]; end for i=1:K, th = pi * (N - 1 + 2 * i) / (2 * N); D = 1 - 2 * W0 * cos(th) + W0^2; G = W0^2 / D; a1 = 4 * c * (W0 * cos(th) - 1) / D; a2 = 2 * (2*c^2 + 1 - W0^2) / D; a3 = - 4 * c * (W0 * cos(th) + 1) / D; a4 = (1 + 2 * W0 * cos(th) + W0^2) / D; B(i+1,:) = G * [1, 0, -2, 0, 1]; A(i+1,:) = [1, a1, a2, a3, a4]; end Similarly, the function bpcheb2a.m is an alternative to the bandpass Chebyshev type-2 design function bpcheb2.m: % bpcheb2a.m - alternative bandpass Chebyshev type 2 digital filter design % % [A, B, P] = bpcheb2a(fs, fpa, fpb, fsa, fsb, Apass, Astop) % % alternative to bpcheb2.m % matches stopbands instead of passbands % % design parameters: % P = [Wpass, Wstop, Wpa, Wpb, c, epass, estop, Nex, N, f3a, f3b, a]; % A, B are Kx5 matrices of cascade 4th/2nd order sections function [A, B, P] = bpcheb2a(fs, fpa, fpb, fsa, fsb, Apass, Astop) 242 c = sin(2*pi*(fsa + fsb) / fs) / (sin(2*pi*fsa / fs) + sin(2*pi*fsb / fs)); Wstop = abs((c - cos(2*pi*fsb / fs)) / sin(2*pi*fsb / fs)); Wpa = (c - cos(2*pi*fpa / fs)) / sin(2*pi*fpa / fs); Wpb = (c - cos(2*pi*fpb / fs)) / sin(2*pi*fpb / fs); Wpass = max(abs(Wpa), abs(Wpb)); epass = sqrt(10.0^(Apass/10) - 1); estop = sqrt(10.0^(Astop/10) - 1); Nex = acosh(estop/epass) / acosh(Wstop/Wpass); N = ceil(Nex); r = rem(N,2); K = (N - r) / 2; a = asinh(estop) / N; W3 = Wstop / cosh(acosh(estop)/N); f3a = (fs/pi) * atan((sqrt(W3^2 - c^2 + 1) - W3)/(c+1)); f3b = (fs/pi) * atan((sqrt(W3^2 - c^2 + 1) + W3)/(c+1)); P = [Wpass, Wstop, Wpa, Wpb, c, epass, estop, Nex, N, f3a, f3b, a]; W0 = sinh(a) / Wstop; if r==1, G = 1 / (1 + W0); a1 = -2 * c * W0 / (1 + W0); a2 = -(1 - W0) / (1 + W0); B(1,:) = G * [1, 0, -1, 0, 0]; A(1,:) = [1, a1, a2, 0, 0]; else B(1,:) = [1, 0, 0, 0, 0]; A(1,:) = [1, 0, 0, 0, 0]; end for i=1:K, th = pi * (N - 1 + 2 * i) / (2 * N); Wi = sin(th) / Wstop; % reciprocal of text D = 1 - 2 * W0 * cos(th) + W0^2 + Wi^2; G = (1 + Wi^2) / D; b1 = - 4 * c * Wi^2 / (1 + Wi^2); b2 = 2 * (Wi^2 * (2*c*c+1) - 1) / (1 + Wi^2); a1 = 4 * c * (W0 * cos(th) - W0^2 - Wi^2) / D; a2 = 2 * ((2*c*c + 1)*(W0^2 + Wi^2) - 1) / D; a3 = - 4 * c * (W0 * cos(th) + W0^2 + Wi^2) / D; a4 = (1 + 2 * W0 * cos(th) + W0^2 + Wi^2) / D; B(i+1,:) = G * [1, b1, b2, b1, 1]; A(i+1,:) = [1, a1, a2, a3, a4]; end 243
- 123. Problem 11.19 We work with the type-2 design since it has more complicated numerator. The transformation of the analog second-order sections is as follows: 1 + Ωi −2 s2 1 − 2Ω−1 0 cos θi s + (Ω−2 0 + Ωi −2 )s2 = (1 + z−1 )2 +Ωi −2 (1 − z−1 )2 (1 + z−1)2−2Ω−1 0 cos θi(1 − z−1)(1 + z−1)+(Ω−2 0 + Ωi −2 )(1 − z−1)2 = (1 + Ωi −2 )+2(1 − Ωi −2 )z−1 + (1 + Ωi −2 )z−2 (1 − 2Ω−2 0 cos θi + Ω−2 0 + Ωi −2 )+2(1 − Ω−2 0 − Ωi −2 )z−1 + (1 + 2Ω−2 0 cos θi + Ω−2 0 + Ωi −2 )z−2 = Gi(1 + bi1z−1 + z−2 ) 1 + ai1z−1 + ai2z−2 where the final step follows by dividing the numerator by the factor (1 + Ωi −2 ) and the denomi- nator by (1 − 2Ω−2 0 cos θi + Ω−2 0 + Ωi −2 ). Problem 11.20 The s-domain poles are si = Ωpass sinh a cos θi + jΩpass cosh a sin θi = Ω0 cos θi + jΩi cosh a It follows that |si|2 = Ω2 0 cos2 θi + Ωi 2 cosh2 a = Ω2 0(1 − sin2 θi)+Ωi 2 (1 + sinh2 a) = Ω2 0 + Ωi 2 + (Ω2 0 sin2 θi − Ωi 2 sinh2 a)= Ω2 0 + Ωi 2 where the last two terms canceled because Ω0 sin θi = Ωpass sinh a sin θi = Ωi sinh a Problem 11.21 The following C functions sos4.c and cas4.c implement the sample-by-sample processing al- gorithm of the cascade of fourth-order sections: /* sos4.c - IIR filtering by single (canonical) fourth-order section */ double sos4(a, b, w, x) a, b, w are 5-dimensional double *a, *b, *w, x; a[0]= 1 always { double y; w[0] = x - a[1]*w[1] - a[2]*w[2] - a[3]*w[3] - a[4]*w[4]; y = b[0]*w[0] + b[1]*w[1] + b[2]*w[2] + b[3]*w[3] + b[4]*w[4]; w[4] = w[3]; could replace these updates by w[3] = w[2]; the single call to w[2] = w[1]; delay(4, w); w[1] = w[0]; return y; } and 244 /* cas4.c - IIR filtering in cascade of fourth-order sections */ double sos4(); single 4th-order section double cas4(K, A, B, W, x) int K; double **A, **B, **W, x; A, B, W are K×5 matrices { int i; double y; y = x; initial input to first SOS4 for (i=0; i<K; i++) y = sos4(A[i], B[i], W[i], y); output of ith section return y; final output from last SOS4 } The K × 5 matrices A and B are generated by the filter design MATLAB functions, such as bps- butt.m, bpcheb2.m, or bscheb2.m, and can be passed as inputs to the above C functions. This can be done by saving the A and B outputs of the MATLAB functions into ASCII files and then invoking the following filtering routine casfilt4.c which is analogous to casfilt.c of Problem 7.15 and which reads A and B dynamically from the command line: /* casfilt4.c - IIR filtering in cascade of fourth-order sections */ #include <stdlib.h> #include <stdio.h> #define MAX 64 initial allocation size double cas4(); void main(int argc, char **argv) { FILE *fpa, *fpb; coefficient files double **A, **B, **W, x, y; double *a; temporary coefficient array int M, K, i, j, m; int max = MAX, dmax = MAX; allocation and increment if (argc != 3) { fprintf(stderr, "Usage: casfilt4 A.dat B.dat <x.dat >y.datn"); exit(0); } if ((fpa = fopen(argv[1], "r")) == NULL) { fprintf(stderr, "Can’t open filter file: %sn", argv[1]); exit(0); } if ((fpb = fopen(argv[2], "r")) == NULL) { fprintf(stderr, "Can’t open filter file: %sn", argv[2]); exit(0); } a = (double *) calloc(max + 1, sizeof(double)); 245
- 124. for (M=0;; M++) { if (M == max) { max += dmax; a = (double *) realloc((char *) a, (max + 1) * sizeof(double)); } if (fscanf(fpa, "%lf", a + M) == EOF) break; } a = (double *) realloc((char *) a, (M + 1) * sizeof(double)); if (M%5 != 0) { fprintf(stderr, "all rows of A must be 5-dimensional"); exit(0); } else K = M / 5; number of SOS4 A = (double **) calloc(K, sizeof(double *)); allocate K rows B = (double **) calloc(K, sizeof(double *)); W = (double **) calloc(K, sizeof(double *)); states of each SOS4 for (i=0; i<K; i++) { A[i] = (double *) calloc(5, sizeof(double)); each row is 5-dim B[i] = (double *) calloc(5, sizeof(double)); W[i] = (double *) calloc(5, sizeof(double)); } for(i=0; i<K; i++) form A matrix for(j=0; j<5; j++) A[i][j] = a[5*i + j]; for(i=0; i<K; i++) read B matrix for(j=0; j<5; j++) fscanf(fpb, "%lf", B[i] + j); while(scanf("%lf", &x) != EOF) { process input samples y = cas4(K, A, B, W, x); printf("%lfn", y); } } The MATLAB version of sos4 is as follows: % sos4.m - fourth order section % % [y, w] = sos4(b, a, w, x) % % b = [b0, b1, b2, b3, b4] = 5-dim numerator (row vector) % a = [1, a1, a2, a3, a4] = 5-dim denominator (row vector) % w = [w0, w1, w2, w3, w4] = 5-dim filter state (row vector) % x = scalar input % y = scalar output function [y, w] = sos4(b, a, w, x) w(1) = x - a(2:5) * w(2:5)’; y = b * w’; w = delay(4, w); The MATLAB version of cas4: 246 % cas4.m - cascade of fourth order sections % % [y, W] = cas4(K, B, A, W, x) % % B = Kx5 numerator matrix % A = Kx5 denominator matrix % W = Kx5 state matrix % x = scalar input % y = scalar output function [y, W] = cas4(K, B, A, W, x) y = x; for i = 1:K, [y, W(i,:)] = sos4(B(i,:), A(i,:), W(i,:), y); end The MATLAB version of casfilt4 for filtering of a long input vector is as follows. It is analogous to casfilt.m of Problem 7.15: % casfilt4.m - cascade of fourth-order sections % % y = casfilt4(B, A, x) % % B = Kx5 numerator matrix % A = Kx5 denominator matrix % x = row vector input % y = row vector output function y = casfilt4(B, A, x) [K, K1] = size(A); [N1, N] = size(x); W = zeros(K,5); for n = 1:N, [y(n), W] = cas4(K, B, A, W, x(n)); end Problem 11.22 The following MATLAB code designs the pairs of lowpass, highpass, bandpass, and bandstop Butterworth filters, and shows how to compute the phase response and impulse response of one of the filters (the first bandpass filter): Apass = -10*log10(0.98); stringent case Astop = -10*log10(0.02); [A1, B1, P1] = lhbutt(1, 20, 4, 5, 0.5, 10); lowpass [A2, B2, P2] = lhbutt(1, 20, 4, 5, Apass, Astop); lowpass [A3, B3, P3] = lhbutt(-1, 20, 5, 4, 0.5, 10); highpass [A4, B4, P4] = lhbutt(-1, 20, 5, 4, Apass, Astop); highpass [A5, B5, P5] = bpsbutt(1, 20, 2, 4, 1.5, 4.5, 0.5, 10); bandpass [A6, B6, P6] = bpsbutt(1, 20, 2, 4, 1.5, 4.5, Apass, Astop); bandpass 247
- 125. [A7, B7, P7] = bpsbutt(-1, 20, 1.5, 4.5, 2, 4, 0.5, 10); bandstop [A8, B8, P8] = bpsbutt(-1, 20, 1.5, 4.5, 2, 4, Apass, Astop); bandstop om = (0:199) * pi / 200; 200 frequencies in [0, π] a5 = cas2can(A5); bandpass case b5 = cas2can(B5); H5 = dtft(b5, om)./dtft(a5, om); frequency response argH5 = angle(H5) * 180 / pi; phase response (in degrees) x = [1, zeros(1, 99)]; length-100 unit impulse h5 = casfilt4(B5, A5, x); impulse response y5 = filter(b5, a5, x); conventional method. Note, y5=h5 The computed phase and impulse responses are shown in Fig. P11.3. Notice how the phase response is almost linear within its passband. Fig. P11.3 Phase and impulse responses of (the less stringent) bandpass filter of Problem 11.22. Problem 11.23 The following MATLAB code designs the pairs of lowpass, highpass, bandpass, and bandstop Chebyshev filters, and shows how to compute the phase response and impulse response of one of the filters (the first bandpass filter): Apass = -10*log10(0.98); stringent case Astop = -10*log10(0.02); [A1, B1, P1] = lhcheb1(1, 20, 4, 5, 0.5, 10); LP type 1 [A2, B2, P2] = lhcheb1(1, 20, 4, 5, Apass, Astop); LP type 1 [A3, B3, P3] = lhcheb1(-1, 20, 5, 4, 0.5, 10); HP type 1 [A4, B4, P4] = lhcheb1(-1, 20, 5, 4, Apass, Astop); HP type 1 [A5, B5, P5] = lhcheb2(1, 20, 4, 5, 0.5, 10); LP type 2 [A6, B6, P6] = lhcheb2(1, 20, 4, 5, Apass, Astop); LP type 2 [A7, B7, P7] = bpcheb2(20, 2, 4, 1.5, 4.5, 0.5, 10); BP type 2 [A8, B8, P8] = bpcheb2(20, 2, 4, 1.5, 4.5, Apass, Astop); BP type 2 248 [A9, B9, P9] = bscheb2(20, 1.5, 4.5, 2, 4, 0.5, 10); BS type 2 [A10, B10, P10] = bscheb2(20, 1.5, 4.5, 2, 4, Apass, Astop); BS type 2 om = (0:199) * pi / 200; 200 frequencies in [0, π] a7 = cas2can(A7); bandpass case b7 = cas2can(B7); H7 = dtft(b7, om) ./ dtft(a7, om); frequency response argH7 = angle(H7) * 180 / pi; phase response (in degrees) x = [1, zeros(1, 99)]; length-100 unit impulse h7 = casfilt4(B7, A7, x); impulse response y7 = filter(b7, a7, x); conventional method. Note, y7=h7 The computed phase and impulse responses are shown in Fig. P11.4. Note that the Chebyshev phase response is not as linear within its passband. Fig. P11.4 Phase and impulse responses of (the less stringent) bandpass filter of Problem 11.23. 249
- 126. Chapter 12 Problems Problem 12.1 The low-rate transfer functions, Di(z) and Hi(z)= z−2 Di(z), corresponding to Eq. (12.4.2) will be: D0(z) = 1 D1(z) = −0.13z2 + 0.30z + 0.90 − 0.18z−1 D2(z) = −0.21z2 + 0.64z + 0.64 − 0.21z−1 D3(z) = −0.18z2 + 0.90z + 0.30 − 0.13z−1 H0(z) = z−2 H1(z) = −0.13 + 0.30z−1 + 0.90z−2 − 0.18z−3 H2(z) = −0.21 + 0.64z−1 + 0.64z−2 − 0.21z−3 H3(z) = −0.18 + 0.90z−1 + 0.30z−2 − 0.13z−3 Setting z = ζ4 and using Eq. (12.2.15), we obtain the transfer function D(ζ), which is recognized as the double-sided ζ-transform of the sequence d given in Eq. (12.4.1): D(ζ)=D0(ζ4 )+ζ−1 D1(ζ4 )+ζ−2 D2(ζ4 )+ζ−3 D3(ζ4 ) =1 + ζ−1 (−0.13ζ8 + 0.30ζ4 + 0.90 − 0.18ζ−4 ) + ζ−2 (−0.21ζ8 + 0.64ζ4 + 0.64 − 0.21ζ−4 ) + ζ−3 (−0.18ζ8 + 0.90ζ4 + 0.30 − 0.13ζ−4 ) =1 + 0.90(ζ + ζ−1 )+0.64(ζ2 + ζ−2 )+0.30(ζ3 + ζ−3 ) − 0.18(ζ5 + ζ−5 )−0.21(ζ6 + ζ−6 )−0.13(ζ7 + ζ−7 ) Problem 12.2 In this case, L = 2 and we choose M = 2, so that N = 2LM + 1 = 9. The ideal impulse response is: d(k )= sin(πk /2) πk /2 , −4 ≤ k ≤ 4 or, numerically, h = d = [0, −0.21, 0, 0.64, 1, 0.64, 0, −0.21, 0 ] It is depicted in Fig. P12.1. The two polyphase subfilters are defined by Eq. (12.2.10), that is, for i = 0, 1, di(k)= d(2k + i), −2 ≤ k ≤ 1 They can be extracted from h by taking every second entry, starting with the ith entry: h0 = d0 = [0, 0, 1, 0] h1 = d1 = [−0.21, 0.64, 0.64, −0.21] 250 0 4-4 2-2 1-1 3-3 d(k′) k′ 1 0.640.64 -0.21-0.21 L=2, M=2, N=9 Fig. P12.1 Length-9 symmetric impulse response of 2-fold FIR interpolator. The interpolated samples between x(n)= xup(2n) and x(n+1)= xup(2n+2) are calculated from Eqs. (12.2.18). The two subfilters act on the time-advanced (by M = 2) low-rate input samples {x(n + 2), x(n + 1), x(n), x(n − 1)}, or, {xup(2n + 4), xup(2n + 2), xup(2n), xup(2n − 2)}. Equations (12.2.18) can be cast in a compact matrix form: yup(2n) yup(2n + 1) = 0 0 1 0 −0.21 0.64 0.64 −0.21 ⎡ ⎢ ⎢ ⎢ ⎣ xup(2n + 4) xup(2n + 2) xup(2n) xup(2n − 2) ⎤ ⎥ ⎥ ⎥ ⎦ These filtering equations can also be obtained by superimposing the symmetric impulse response of Fig. P12.1 on each of the contributing low-rate samples {xup(2n + 4), xup(2n + 2), xup(2n), xup(2n − 2)} and adding up their contributions at each of the intermediate sampling instants 2n + i, i = 0, 1, as shown in Fig. P12.2. xup(2n-2) 2n-2 xup(2n+2) yup(2n+1) 2n+2 xup(2n+4) 2n+4 xup(2n) 2n n′ Fig. P12.2 Superposition of impulse responses. The Hamming windowed version of the filter is obtained by multiplying the full length-9 filter response h by a length-9 Hamming window. The resulting impulse response will be: h = [0, −0.05, 0, 0.55, 1, 0.55, 0, −0.05, 0 ] The corresponding polyphase subfilters are obtained by extracting every second entry: h0 = [0, 0, 1, 0] h1 = [−0.05, 0.55, 0.55, −0.05] 251
- 127. The interpolation equations will be in this case: yup(2n) yup(2n + 1) = 0 0 1 0 −0.05 0.55 0.55 −0.05 ⎡ ⎢ ⎢ ⎢ ⎣ xup(2n + 4) xup(2n + 2) xup(2n) xup(2n − 2) ⎤ ⎥ ⎥ ⎥ ⎦ Problem 12.3 For the case of a 3-fold interpolator, we have L = 3 and choose again M = 2, corresponding to filter length N = 2LM + 1 = 13. The ideal impulse response is: d(k )= sin(πk /3) πk /3 , −6 ≤ k ≤ 6 or, numerically, h = d = [0, −0.17, −0.21, 0, 0.41, 0.83, 1, 0.83, 0.41, 0, −0.21, −0.17, 0] where h is the causal version, with time origin at the left of the vector, and d is the symmetric one, with time origin at the middle of the vector. This impulse response is shown in Fig. P12.3. The three polyphase subfilters are defined by Eq. (12.2.10), that is, for i = 0, 1, 2, di(k)= d(3k + i), −2 ≤ k ≤ 1 They can be extracted from h by taking every third entry, starting with the ith entry: h0 = d0 = [0, 0, 1, 0] h1 = d1 = [−0.17, 0.41, 0.83, −0.21] h2 = d2 = [−0.21, 0.83, 0.41, −0.17] The interpolated samples between x(n)= xup(3n) and x(n + 1)= xup(3n + 3) are calculated from Eqs. (12.2.18). 0 5-5 4-4 2-2 1-1 6-6 3-3 d(k′) k′ 1 0.830.83 0.410.41 -0.17-0.17 -0.21-0.21 L=3, M=2, N=13 Fig. P12.3 Length-13 symmetric impulse response of 3-fold FIR interpolator. All three subfilters act on the time-advanced (by M = 2) low-rate input samples {x(n+2), x(n+1), x(n), x(n − 1)}, or, {xup(3n + 6), xup(3n + 3), xup(3n), xup(3n − 3)}. Equations (12.2.18) can be cast in a compact matrix form: ⎡ ⎢ ⎣ yup(3n) yup(3n + 1) yup(3n + 2) ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ 0 0 1 0 −0.17 0.41 0.83 −0.21 −0.21 0.83 0.41 −0.17 ⎤ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎣ xup(3n + 6) xup(3n + 3) xup(3n) xup(3n − 3) ⎤ ⎥ ⎥ ⎥ ⎦ 252 These filtering equations can also be obtained by superimposing the symmetric impulse response of Fig. P12.3 on each of the contributing low-rate samples {xup(3n + 6), xup(3n + 3), xup(3n), xup(3n − 3)} and adding up their contributions at each of the intermediate sampling instants 3n+i, i = 0, 1, 2, as shown in Fig. P12.4. xup(3n-3) 3n-3 xup(3n+3) yup(3n+1) 3n+3 xup(3n+6) 3n+6 xup(3n) 3n n′ Fig. P12.4 Superposition of impulse responses. The Hamming windowed version of the filter is obtained by multiplying the full length-13 filter response h by a length-13 Hamming window. The resulting impulse response will be: h = [0, −0.02, −0.06, 0, 0.32, 0.78, 1, 0.78, 0.32, 0, −0.06, 0.02, 0] The corresponding polyphase subfilters are obtained by extracting every third entry: h0 = [0, 0, 1, 0] h1 = [−0.02, 0.32, 0.78, −0.06] h2 = [−0.06, 0.78, 0.32, −0.02] The interpolation equations will be in this case: ⎡ ⎢ ⎣ yup(3n) yup(3n + 1) yup(3n + 2) ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ 0 0 1 0 −0.02 0.32 0.78 −0.06 −0.06 0.78 0.32 −0.02 ⎤ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎣ xup(3n + 6) xup(3n + 3) xup(3n) xup(3n − 3) ⎤ ⎥ ⎥ ⎥ ⎦ Problem 12.4 For interpolation factor L = 4 and filter length N = 25, we find: N = 2LM + 1 ⇒ M = (N − 1)/2L = (25 − 1)/8 = 3 Thus, we use 3 low-rate samples above and three below every interpolated value to be computed. The length-25 ideal interpolation impulse response is calculated from: d(k )= sin(πk /4) πk /4 , for − 12 ≤ k ≤ 12 This generates the numerical values: 253
- 128. d = [ 0.000, 0.082, 0.127, 0.100, 0.000, −0.129, −0.212, −0.180, 0.000, 0.300, 0.637, 0.900, 1.000, 0.900, 0.637, 0.300, 0.000, −0.180, −0.212, −0.129, 0.000, 0.100, 0.127, 0.082, 0.000 ] This impulse response is shown in Fig. P12.5 (where the values have been rounded to 2-digit accuracy for convenience.) 0 4-4 2-2 1-1 3-3 d(k′) k′ 1 0.900.90 0.640.64 0.30 -0.18 -0.21 -0.13 12-12 0.13 0.10 0.08 0.13 0.10 0.08 -0.13 -0.21 -0.18 0.30 L=4, M=3, N=25 Fig. P12.5 Length-25 symmetric impulse response of 4-fold FIR interpolator. Given 2M = 6 low-rate samples {A, B, C, D, E, F} as shown in Fig. 12.1.11, the three values {X, Y, Z} interpolated between C and D are calculated by convolving d with the upsampled low-rate samples. Using the flip-and-slide form of convolution as in Fig. 12.4.3, we position the impulse response d at the three successive positions, read off the values of d(k ) where it inter- sects with the low-rate samples, and add up the results. This procedure is shown in Fig. P12.6. The corresponding linear combinations of low-rate samples are precisely those of Eq. (12.1.3). They may be rewritten in the polyphase matrix form (with the low-rate samples listed from the latest down to the earliest): ⎡ ⎢ ⎣ X Y Z ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ 0.082 −0.129 0.300 0.900 −0.180 0.100 0.127 −0.212 0.637 0.637 −0.212 0.127 0.100 −0.180 0.900 0.300 −0.129 0.082 ⎤ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ F E D C B A ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Problem 12.5 The ideal 3-fold interpolation case with M = 2, so that N = 2LM + 1 = 2 · 3 · 2 + 1 = 13, was discussed in Problem 12.3. For the linear interpolator, using Eq. (12.3.7), we have: ⎡ ⎢ ⎣ yup(3n) yup(3n + 1) yup(3n + 2) ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ 0 0 1 0 0 1/3 2/3 0 0 2/3 1/3 0 ⎤ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎣ x(n + 2) x(n + 1) x(n) x(n − 1) ⎤ ⎥ ⎥ ⎥ ⎦ = 1/3 2/3 2/3 1/3 x(n + 1) x(n) Problem 12.6 The following complete C program interp.c implements the polyphase sample-by-sample pro- cessing algorithm of Eq. (12.2.20) and the initialization procedure of (12.2.19): 254 n′ A B C D X E F 0.90 0.30 0.10 0.08 -0.13-0.18 n′ A B C D Y E F 0.64 0.64 0.13 0.13 -0.21 -0.21 n′ A B C D Z E F 0.30 0.90 0.08 0.10 -0.18-0.13 Fig. P12.6 Interpolated values are calculated by the flip-and-slide form of convolution. /* interp.c - L-fold FIR interpolator (rectangular and Hamming window design) * * Usage: interp hamm L M < xlow.dat > yhigh.dat * * hamm = 0/1 for rectangular/Hamming window * L = interpolation factor * M = half-length of polyphase subfilters (length = 2M, order = P = 2M- 1) * xlow.dat = file containing low-rate input samples * yhigh.dat = file containing high-rate interpolated output samples */ #include <stdio.h> #include <math.h> #include <stdlib.h> void delay(); double dot(); double d(); ideal interpolation filter void main(int argc, char ** argv) { int i, n, L, M, P, hamm; double x, *y, *w, **h, pi = 4 * atan(1.0); if (argc != 4) { 255
- 129. fprintf(stderr,"nUsage: interp hamm L M < xlow.dat > yhigh.datnn"); fprintf(stderr,"where: hamm = 0/1 for rectangular/hamm windown"); fprintf(stderr," L = interpolation factorn"); fprintf(stderr," M = half-length of polyphase subfiltersn"); fprintf(stderr," xlow.dat = file for low-rate input samplesn"); fprintf(stderr," yhigh.dat = file for high-rate interpolated output samplesn"); exit(0); } hamm = atoi(argv[1]); L = atoi(argv[2]); M = atoi(argv[3]); P = 2*M - 1; polyphase filter order h = (double **) calloc(L, sizeof(double *)); allocate L subfilters for (i=0; i<L; i++) h[i] = (double *) calloc(2*M, sizeof(double)); each of length 2M y = (double *) calloc(L, sizeof(double)); allocate L polyphase outfprintf(stderr, w = (double *) calloc(2*M, sizeof(double)); allocate low-rate internal state for (i=0; i<L; i++) construct polyphase subfilters for (n=0; n<2*M; n++) if (hamm == 0) rectangular window h[i][n] = d(L, L*n+i-L*M); else Hamming window h[i][n] = d(L, L*n+i-L*M) * (0.54 - 0.46*cos((L*n+i)*pi/(L*M))); initialize by Eq. (12.2.19) for (i=M; i>=1; i--) initialize state w by reading scanf("%lf", w + i); first M low-rate input samples process by Eq. (12.2.20) while(scanf("%lf", &x) != EOF) { keep reading low-rate samples till EOF w[0] = x; for (i=0; i<L; i++) { y[i] = dot(P, h[i], w); ith interpolated output sample printf("%.12lfn", y[i]); and write it into stdout } delay(P, w); update low-rate delay line } for (n=0; n<M; n++) { input-off transients w[0] = 0.0; zero input for (i=0; i<L; i++) { y[i] = dot(P, h[i], w); ith interpolated output sample printf("%.12lfn", y[i]); and write it into stdout } delay(P, w); update low-rate delay line } } /* -------------------------------------------------------- */ /* d(L, k) - ideal L-fold interpolation coefficients */ 256 double d(L, k) int L, k; { double t, pi = 4 * atan(1.0); if (k == 0) d(0) = 1 return (1.0); else { if (k%L == 0) d(k) vanishes at non-zero multiples of L return (0.0); else { t = pi * k / L; return (sin(t)/t); typical sin x/x impulse response } } } The program has usage: interp hamm L M < xlow.dat > yhigh.dat where the parameter hamm specifies a Hamming or rectangular window, L is the interpolation factor, and M is the half-length of the polyphase subfilters, so that the total filter length is N = 2LM + 1. The program designs the windowed length-N interpolation filter and then reads the low-rate in- put samples sequentially from a file xlow.dat or stdin, computes the high-rate interpolated samples, and writes them to the file yhigh.dat or to the stdout. Each low-rate input generates L high-rate outputs. Inside the program, the rectangular and Hamming window design of the impulse response may be replaced easily by a Kaiser design. The program interp.c uses a linear-buffer version of the low-rate delay line used by all the polyphase subfilters. The following program cinterp.c is the circular-buffer version of in- terp.c: /* cinterp.c - circular-buffer version of interp.c */ #include <stdio.h> #include <math.h> #include <stdlib.h> void cdelay(), wrap(); double cdot(); circular version of dot.c double d(); ideal interpolation filter void main(int argc, char ** argv) { int i, n, L, M, P, hamm; double x, *y, *w, *p, **h, pi = 4 * atan(1.0); if (argc != 4) { fprintf(stderr,"nUsage: cinterp hamm L M < xlow.dat > yhigh.datnn"); fprintf(stderr,"where: hamm = 0/1 for rectangular/hamm windown"); fprintf(stderr," L = interpolation factorn"); fprintf(stderr," M = half-length of polyphase subfiltersn"); fprintf(stderr," xlow.dat = file for low-rate input samplesn"); 257
- 130. fprintf(stderr," yhigh.dat = file for high-rate interpolated output samplesn"); exit(0); } hamm = atoi(argv[1]); L = atoi(argv[2]); M = atoi(argv[3]); P = 2*M - 1; polyphase filter order h = (double **) calloc(L, sizeof(double *)); allocate L subfilters for (i=0; i<L; i++) h[i] = (double *) calloc(2*M, sizeof(double)); each of length 2M y = (double *) calloc(L, sizeof(double)); allocate L polyphase outfprintf(stderr, w = (double *) calloc(2*M, sizeof(double)); allocate low-rate internal state p = w; initialize circular pointer for (i=0; i<L; i++) construct polyphase subfilters for (n=0; n<2*M; n++) if (hamm == 0) rectangular window h[i][n] = d(L, L*n+i-L*M); else Hamming window h[i][n] = d(L, L*n+i-L*M) * (0.54 - 0.46*cos((L*n+i)*pi/(L*M))); initialize by Eq. (P12.1) for (i=M; i>=1; i--) { initialize state w by reading scanf("%lf", p); first M low-rate input samples cdelay(P, w, &p); } process by Eq. (P12.2) while(scanf("%lf", &x) != EOF) { keep reading low-rate samples till EOF *p = x; beginning of state vector for (i=0; i<L; i++) { y[i] = cdot(P, h[i], w, p); circular version of dot() printf("%.12lfn", y[i]); and write it into stdout } cdelay(P, w, &p); update low-rate delay line } for (n=0; n<M; n++) { input-off transients *p = 0.0; zero input for (i=0; i<L; i++) { y[i] = cdot(P, h[i], w, p); circular version of dot() printf("%.12lfn", y[i]); and write it into stdout } cdelay(P, w, &p); update low-rate delay line } } /* -------------------------------------------------------- */ /* d(L, k) - ideal L-fold interpolation coefficients */ double d(L, k) int L, k; 258 { double t, pi = 4 * atan(1.0); if (k == 0) d(0) = 1 return (1.0); else { if (k%L == 0) d(k) vanishes at non-zero multiples of L return (0.0); else { t = pi * k / L; return (sin(t)/t); typical sin x/x impulse response } } } /* -------------------------------------------------------- */ /* cdot(P, h, w, p) - circular version of dot.c */ double tap(); double cdot(P, h, w, p) computes dot product of h with double *h, *w, *p; current internal state vector int P; pointed to by pointer p { int i; double y; for (y=0, i=0; i<=P; i++) y += h[i] * tap(P, w, p, i); return y; } It uses a circular pointer p that circulates over the low-rate buffer w. The circular-buffer version of Eqs. (12.2.19) and (12.2.20), implemented by cinterp.c, can be stated as follows: for m = M down to m = 1 do: read low-rate input sample x ∗p = x cdelay(P, w, &p) (P12.1) and for each low-rate input sample x do: ∗p = x for m = 0, 1, . . . , P determine internal states: sm = tap(P, w, p, m) for i = 0, 1, . . . , L − 1 compute output of ith filter: yi = dot(P, hi, s) cdelay(P, w, &p) (P12.2) The length-17 rectangular and Hamming windowed filters for this problem were derived in Section 12.4.1. The given inputs were processed by interp.c. The upsampled versions of these inputs, xup(n ), are shown in Fig. P12.7 together with the corresponding interpolated output signals yup(n ), computed by the length-17 Hamming windowed version of the interpolation filter. 259
- 131. Fig. P12.7 Interpolation of triangular and sinusoidal signals. Problem 12.7 We start with fs = 40 kHz, Δf = 5 kHz, and A = 80 for all the stages. The effective single stage filter will have Kaiser length: N − 1 = DL ΔF = 321.08 which is rounded up to: N = 337 = 2LM + 1 = 16M + 1 ⇒ M = 21 The 2×4 multistage filters, shown in Fig. P12.8, have Kaiser lengths: N0 − 1 = L0D ΔF = 80.27, N1 − 1 = DF1 F0 − 1 = 8D 2 − 1 = 40.14 and are rounded up to: N0 = 85 = 2L0M0 + 1 = 4M0 + 1 ⇒ M0 = 21 N1 = 49 = 2L1M1 + 1 = 8M1 + 1 ⇒ M1 = 6 The relative computational rate will be Rmulti Rsingle = N0 + N1F0 N = 0.54 260 H0 H1 2fs 8fsfs 2 4 Fig. P12.8 2x4=8-fold oversampling filter. which compares well with the approximation: Rmulti Rsingle = 1 L1 + L0 L0 − 1 ΔF = 0.50 Thus, the multistage realization requires only 54 percent the cost of the single-stage one. The combined filter Htot(f)= H0(f)H1(f) and the superimposed plots of the filters H0(f), H1(f) are shown in Fig. P12.9. Similarly, for the 4×2 case, shown in Fig. P12.10, we find the filter lengths: N0 = 169, N1 = 17 Fig. P12.9 Combined filter H0(f)H1(f), and individual filters H0(f), H1(f). H0 H1 4fs 8fsfs 24 Fig. P12.10 4x2=8-fold oversampling filter. The relative computational rate is in this case: Rmulti Rsingle = N0 + N1F0 N = 0.70 The combined filter Htot(f)= H0(f)H1(f) and the superimposed plots of the filters H0(f), H1(f) are shown in Fig. P12.11. Initially, we designed the filters with stopband attenuations A0 = A1 = 80 dB. However, we found that in order to keep the overall stopband attenuation 261
- 132. below 80 dB, the attenuation of the second filter had to be increased to A1 = 84 dB. This did not change the filter lengths, but it changed slightly the Kaiser parameters D1 and α1 for that factor. Finally, in the 2×2×2 case, shown in Fig. P12.12, we find the filter lengths Fig. P12.11 Combined filter H0(f)H1(f), and individual filters H0(f), H1(f). N0 = 85, N1 = 25, N2 = 17 resulting in the relative computational rate Rmulti Rsingle = N0 + N1F0 + N2F1 N = 0.60 H0 H1 H2 2fs 4fsfs 8fs 2 2 2 Fig. P12.12 2x2x2=8-fold oversampling filter. The combined filter Htot(f)= H0(f)H1(f)H2(f) and individual filters H0(f), H1(f), H2(f) are shown in Fig. P12.13. For reference, we also show in Fig. P12.14 the magnitude response of the single stage 8-fold oversampling filter. Again, the nominal attenuation of the last factor H2 had to be increased to A2 = 84 dB, whereas the first two were A0 = A1 = 80 dB. Problem 12.8 In units of fs = 44.1 kHz, the transition width from the edge of the passband to the edge of the stopband will be: ΔF = Δf fs = 24.1 − 20 44.1 = 0.0930 The Kaiser window width parameter is D = (A − 7.95)/14.36 = (80 − 7.95)/14.36 = 5.0174. Thus, for a single-stage design with oversampling ratio L = 4, we find the Kaiser length: 262 Fig. P12.13 Combined filter H0(f)H1(f)H2(f), and individual filters H0(f), H1(f), H2(f). Fig. P12.14 Single-stage 8-fold oversampling filter. N − 1 = DL ΔF = 5.0174 · 4 0.0930 = 215.803 ⇒ N = 217, M = (N − 1)/(2L)= 27 The corresponding computational rate of the polyphase realization is Rsingle = Nfs = 217 · 44.1 = 9569.7 MAC/msec = 9.5697 MIPS (assuming that each MAC is done with one instruction.) Thus, a 20-MIPS DSP chip can easily handle it. For the two-stage case, the design specifications are depicted in Fig. 12.2.9 of Section 12.2.5. Stage H0 operating at 2fs will have the narrow transition width Δf = 4.1 kHz, or ΔF = 0.0930. This gives the Kaiser length: N0 − 1 = DL0 ΔF = 5.0174 · 2 0.0930 = 107.9 ⇒ N0 = 109, M0 = (N0 − 1)/(2L0)= 27 The second stage H1 will have a wider transition width Δf1 = fs, or ΔF1 = 1. Thus, the filter length will be: 263
- 133. N1 − 1 = DΔF1 F0 − 1 = 5.0174 · 4 1.000 = 20.07 ⇒ N1 = 25, M1 = (N1 − 1)/(2L1)= 6 The polyphase computational rate of the two sections combined will be Rmulti = N0fs + N1(2fs)= 159fs = 159 · 44.1 = 7011.9 MAC/msec = 7.0119 MIPS Thus, the relative computational cost of the two-stage versus the single stage design will be: Rmulti Rsingle = 159fs 217fs = 0.73 Problem 12.9 The frequency responses can be computed by the MATLAB code: f = (0:399) * 160 /400; 400 frequencies in [0,160] kHz f1 = 28; normalization frequency for Butterworth f2 = 16; normalization frequency for Bessel H1 = 1 ./ (1 + 2*j*(f/f1) - 2*(f/f1).^2 - j*(f/f1).^3); Butterworth H2 = 15 ./ (15 + 15*j*(f/f2) - 6*(f/f2).^2 - j*(f/f2).^3); Bessel magH1 = 20 * log10(abs(H1)); magH2 = 20 * log10(abs(H2)); argH1 = angle(H1) * 180 / pi; argH2 = angle(H2) * 180 / pi; The phase and magnitude responses are shown in Figs. P12.15 and P12.16. The Bessel phase response is essentially linear over the 0–20 kHz passband. But its magnitude response is not as flat as the Butterworth’s. Fig. P12.15 Phase responses of Bessel and Butterworth filters. 264 Fig. P12.16 Magnitude responses of Bessel and Butterworth filters. Problem 12.10 The filter H0 has a cutoff frequency fc = fs/2. The linear, hold, and DAC magnitude responses (normalized to 0 dB at DC) are in the general and the particular case of L0 = 4, L1 = L2 = 2: |H1(f)| = sin(πf/L0fs) L1 sin(πf/L0L1fs) 2 = sin(πf/4fs) 2 sin(πf/8fs) 2 |H2(f)| = sin(πf/L0L1fs) L2 sin(πf/L0L1L2fs) = sin(πf/8fs) 2 sin(πf/16fs) |Hdac(f)| = sin(πf/L0L1L2fs) πf/L0L1L2fs = sin(πf/16fs) πf/16fs The filter H0 operates at rate 4fs and cancels all the images at multiples of fs which are not multiples of 4fs. The linear interpolator H1 operates at rate 8fs and vanishes at multiples of 4fs which are not multiples of 8fs. The filter H2 operates at rate 16fs and vanishes at multiples of 8fs which are not multiples of 16fs. Finally, Hdac vanishes at all non-zero multiples of 16fs. Thus, the combined effect of H0, H1, H2, and Hdac is to (partially) remove all spectral images at multiples of fs. We note that the combined effect of the two hold transfer functions is an effective DAC operating at 8fs: |H2(f)Hdac(f)| = sin(πf/L0L1fs) L2 sin(πf/L0L1L2fs) · sin(πf/L0L1L2fs) πf/L0L1L2fs = sin(πf/L0L1fs) πf/L0L1fs = sin(πf/8fs) πf/8fs The reason for using H2 is that the higher the L, the more bits can be saved via the noise-shaping requantizer. Problem 12.11 The proof of the result may be seen graphically in Fig. P12.17 for the case L = 4. The ideal inter- polator has cutoff fs/2 and is periodic in f with period Lfs = 4fs. Thus, the passband replicates at multiples of 4fs. 265
- 134. 4fs f0 3fs2fs-2fs-3fs-4fs fs-fs D(f-fs) D(f) D(f-2fs) D(f-3fs) D(f) + D(f-fs) + D(f-2fs) + D(f-3fs) 4fs f0 3fs2fs-2fs-3fs-4fs fs-fs 4fs f0 3fs2fs-2fs-3fs-4fs fs-fs 4fs f0 3fs2fs-2fs-3fs-4fs fs-fs 4fs f0 3fs2fs-2fs-3fs-4fs fs-fs Fig. P12.17 Sum of successive shifts of ideal lowpass interpolator with L = 4 in Problem 12.11. The terms D(f − mfs) correspond to the right shifts of D(f) centered at mfs. It is evident from the figure that the these shifted terms fill the gaps between the original replicas, resulting in a constant spectrum. The 1/L factor is needed to make the value of passband equal to unity. Thus, in the L = 4 case we have: 1 4 D(f)+D(f − fs)+D(f − 2fs)+D(f − 3fs) = 1, for all f Problem 12.12 Writing k = Lk + m, m = 0, 1, . . . , L − 1, corresponds to the quotient and remainder of the division of k by L, and as such the integers {k , m} are uniquely determined by k. Therefore, we may change the summation over k into a double summation over k and m: X(f) = 1 T ∞ k=−∞ Xa(f − kfs)= 1 T L−1 m=0 ∞ k =−∞ Xa(f − (k L + m)fs) = 1 T L−1 m=0 ∞ k =−∞ Xa f − (k L + m)fs = 1 LT L−1 m=0 ∞ k =−∞ Xa(f − mfs − k fs ) where we replaced T = LT and Lfs = fs . If in the definition of X (f) we replace f by f − mfs, we have: X (f − mfs)= 1 T ∞ k =−∞ Xa(f − mfs − k fs ) Thus the k summation in the above double sum may be replaced by X (f −mfs) resulting in the desired result: 266 X(f)= 1 L L−1 m=0 X (f − mfs) If we use the variables ω and ω , then X(f) and X (f) are given by X(ω)= n x(n)e−jωn , X (ω )= n x (n )e−jω n The shifted frequency f − mfs is in units of ω : 2π(f − mfs) fs = 2πf fs − 2πmfs Lfs = ω − 2πm L Thus, using ω = ω/L, we may write X(ω)= 1 L L−1 m=0 X ω − 2πm L = 1 L L−1 m=0 X ω − 2πm)/L Problem 12.13 The discrete-time sampling function s (n ) consists of a periodic train of unit impulses at mul- tiples of L, that is, at n = nL, and separated by L − 1 zeros between. Thus, the product of s (n ) with x (n ) will insert such zeros. But at the sampling times n = nL we will have xup(n )= x (n )= x (nL)= x(n). Because of its periodicity in n with period L, the sampling function admits a DFS expansion, that is, and inverse DFT expansion of the form of Eq. (9.7.1): s (n )= 1 L L−1 m=0 S (ωm)ejωmn , ωm = 2π(mfs) fs = 2π(mfs /L) fs = 2πm L = mth DFT frequency where S (ωm) is the L-point DFT of s (n ) calculated from one period of s (n as follows: S (ωm)= L−1 n =0 s (n )e−jωmn But within the first period 0 ≤ n ≤ L − 1, the sampling function acts as a unit impulse, s (n )= δ(n ). Therefore, the n -summation collapses to the term n = 0, which is unity: S (ωm)= 1, m = 0, 1, . . . , L − 1 which gives for the inverse DFT: s (n )= 1 L L−1 m=0 ejωmn = 1 L L−1 m=0 e2πjmn /L To transform the relationship xup(n )= s (n )x (n ) to the frequency domain, we replace s (n ) by its IDFT expression and use the modulation theorem of DTFTs to get: xup(n )= 1 L L−1 m=0 ejωmn x (n ) ⇒ Xup(ω )= 1 L L−1 m=0 X (ω − ωm) But we have seen in Eq. (12.2.21) that the upsampled spectrum is the same as the original low-rate spectrum, that is, Xup(f)= X(f), or in units of ω and ω , Xup(ω )= X(ω). It follows that X(ω)= Xup(ω )= 1 L L−1 m=0 X (ω − ωm)= 1 L L−1 m=0 X ω − 2πm L 267
- 135. Problem 12.14 Starting with the filtering equation X (f)= D(f)X(f), we form the sum of shifted high-rate spectra: 1 L L−1 m=0 X (f − mfs)= 1 L L−1 m=0 D(f − mfs)X(f − mfs)= 1 L L−1 m=0 D(f − mfs)X(f)= X(f) where we used the periodicity of X(f) with period fs to write X(f − mfs)= X(f), and then we used the result of Problem 12.11 to replace the sum of shifted D(f)’s by unity. Problem 12.15 The attenuation of an Nth order Butterworth filter in dB is given by A(F)= 10 log10 1 + F F0 2N where we used the normalized frequencies F = f/fs and F0 = f0/fs. Given a required attenuation of Apass dB over the passband Fpass = fpass/fs = (fs/2)/fs = 0.5, we obtain an equation for the 3-dB frequency F0: Apass = 10 log10 1 + 0.5 F0 2N ⇒ F0 = 0.5 10Apass/10 − 1 1/2N The prefilter’s stopband must begin at Fstop = fstop/fs = (Lfs − fs/2)/fs = L − 0.5. Thus, the stopband attenuation will be Astop = 10 log10 1 + L − 0.5 F0 2N ⇒ L = 0.5 + F0 10Astop/10 − 1 1/2N For large Astop we may ignore the −1 to get L = 0.5 + F010Astop/20N With the values Apass = 0.1 and N = 3, we find F0 = 0.94. Fig. P12.18 shows a plot of L versus Astop. Fig. P12.18 Decimation ratio versus stopband attenuation in Problem 12.15. 268 Problem 12.16 Using the expression of the attenuation of Problem 12.15, we get at the passband and stopband normalized frequencies Fpass = 0.5, Fstop = L − 0.5: Apass = 10 log10 1 + 0.5 F0 2N Astop = 10 log10 1 + L − 0.5 F0 2N ⇒ L − 0.5 0.5 2N = 10Astop/10 − 1 10Apass/10 − 1 which may be solved for N, giving: N = ln 10Astop/10 − 1 10Apass/10 − 1 2 ln(2L − 1) Let N0 = ceil(N). In order for the filter order to remain fixed at N0, the computed quantity N must be in the range N0 − 1 < ln 10Astop/10 − 1 10Apass/10 − 1 2 ln(2L − 1) Solving this inequality for L, we find the limits on L: 0.5 + 0.5 10Astop/10 − 1 10Apass/10 − 1 1/2N0 ≤ L < 0.5 + 0.5 10Astop/10 − 1 10Apass/10 − 1 1/2(N0−1) ≤ N0 For the given numerical values, we find N0 = 3 and the range 9.86 ≤ L < 40.97, which gives the integer range 10 ≤ L ≤ 40. Problem 12.17 The normalized transition width is ΔF = Δf/fs = (24.41 − 20)/44.1 = 0.1. Using a Kaiser design, the transition width parameter D and filter length N will be: D = A − 7.95 14.36 = 95 − 7.95 14.36 = 6.0620, N−1 = DL ΔF = 6.062 · 160 0.1 = 9699.2 ⇒ N = 9920 where the final N is of the form N = 2LK + 1 = 320K + 1 with K = 31. Thus, the interpola- tion/decimation filter operating at rate fs = 160fs = 147fs = 7.056 MHz can be implemented with 160 polyphase subfilters, each of length 2K = 62 and operating at a rate fs = 44.1 kHz. The overall sample rate conversion cost in MAC/sec can be calculated from Eq. (12.6.9): R = 2Kfs = 61fs = 61 · 48 = 2928 kMAC/sec = 2.928 MIPS The computational speed is easily handled by today’s DSP chips. Memory, however, to store the N FIR coefficients is high—of the order of 10k words. 269
- 136. Problem 12.18 Here, we have the conversion ratio fs = 32 = 32 48 · 48 = 2 3 fs, so that L = 2, M = 3. Because fs < fs, the filter’s cutoff frequency will be fc = fs /2 = 16 kHz. Assuming a ±1 kHz transition width about fc, that is, Δf = 2 kHz, and a A = 60 dB stopband attenuation, we find the Kaiser parameter D and filter length N: D = A − 7.95 14.36 = 60 − 7.95 14.36 = 3.6247, N−1 = DLfs Δf = 3.6247 · 2 · 48 2 = 173.98 ⇒ N = 177 where we rounded N to the next odd integer of the form N = 2LK + 1. Thus, K = 44, and there will be L = 2 polyphase subfilters of length 2K = 88. Fig. P12.19 shows the ideal specifications of such a filter. The filter operates at the fast rate fs = 3fs = 2fs = 96 kHz and acts as an antialiasing filter for the downsampler, that is, it removes the high frequencies from the input in the range [fs 2, fs/2]= [16, 24] kHz, so that the downshifted replicas—caused by the 3-fold downsampling operation—will not overlap. 0 0 fs′ 2fs′ 3fs′ fs′ 2fs′ 3fs′=fs′′ 2fs=fs′′ 0 fs 2fs fs 0 filter polyphase selector T′/3=T′′ T/2=T′′ 6T′′=3T=2T′ T x0 x1 x2 y0 y1 t t t t f f f f T′ min(fs, fs′)1 2 0 1i= Fig. P12.19 Ideal SRC filter with conversion ratio 2/3. Because T = T /2 and T = T /3, it follows that the basic block interval will be Tblock = 6T = 3T = 2T , containing 3 input-rate samples {x0, x1, x2} and 2 output-rate samples {y0, y1}. The interpolated output y1 that lies halfway between x1 and x2 is obtained by the polyphase filter h1. Indeed, the polyphase selector indices can be precomputed by im = 3m % 2 = [0, 1], for m = 0, 1 The sample processing algorithm of the 2/3 sample rate converter will be then: for each input block {x0, x1, x2} do: w0 = x0 y0 = dot(P, h0, w)= x0 delay(P, w) w0 = x1 y1 = dot(P, h1, w) delay(P, w) w0 = x2 delay(P, w) 270 where P = 2K − 1 is the order of the polyphase subfilters h0, h1, and w is the length-(P + 1) input-rate delay-line buffer. Note that there is no interpolated output after x2 (the next two out- puts come from the next group of three inputs), and therefore, x2 is simply read into the buffer w and the delay line is updated. For reference, we also show in Fig. P12.20 the reverse converter by a ratio 3/2, converting from 32 kHz up to 48 kHz. Here, the SRC filter acts as an anti-image filter for the upsampler. polyphase selector T′/2=T′′ T/3=T′′ 6T′′=2T=3T′ y0 y1 y2 T x0 x1 t t t t f f f f T′ min(fs, fs′)1 2 0 2 1i= fs 2fs 3fs=fs′′0 fs 2fs 3fs 0 filter 0 fs′ 2fs′ 0 2fs′=fs′′fs′ Fig. P12.20 Ideal SRC filter with conversion ratio 3/2. In this case, we have fs = 3fs/2 = 3 · 32/2 = 48. The fast rate is fs = 2fs = 3fs = 96 kHz. The input- and output-rate sampling intervals are T = 3T and T = 2T , and the basic time block Tblock = 6T = 2T = 3T . Thus, every group of two input samples {x0, x1} generates a group of three output samples {y0, y1, y2}. As can be seen in the figure, the polyphase selector sequence is im = 2m % 3 = [0, 2, 1], for m = 0, 1, 2. Therefore, the three interpolated outputs {y0, y1, y2} will be computed by the subfilters {h0, h2, h1}. The corresponding sample processing algorithm will be: for each input block {x0, x1} do: w0 = x0 y0 = dot(P, h0, w)= x0 y1 = dot(P, h2, w) delay(P, w) w0 = x1 y2 = dot(P, h1, w) delay(P, w) Problem 12.19 For the case L/M = 7/4, we have fs = 7fs/4 and fs = 4fs = 7fs, T = T /4 = T/7, Tblock = 28T = 7T = 4T. Fig. P12.21 depicts the operation of the converter both in the time and frequency domains. In the time domain, each input block of 4 input-rate samples {x0, x1, x2, x3} generates 28 fast-rate interpolated samples, out of which 7 output-rate samples {y0, y1, y2, y3, y4, y4, y6} are selected and computed according to their polyphase indices: 271
- 137. 0 0 0 0 filter polyphase selector T′/4=T′′ T/7=T′′ 28T′′=4T=7T′ y0 y1 y3 y5y2 y4 y6 x0 x1 x2 x3 t t t t f f f f T′ T min(fs, fs′)1 2 0 4 1 25 6 3i= fs 2fs 3fs 4fs 5fs 6fs 7fs=fs′′ fs 2fs 3fs 4fs 5fs 6fs 7fs 4fs′=fs′′2fs′ 3fs′fs′ 4fs′2fs′ 3fs′fs′ Fig. P12.21 Ideal SRC filter with conversion ratio 7/4. im = 4m % 7 = [0, 4, 1, 5, 2, 6, 3], for m = 0, 1, . . . , 6 nm = (4m − im)/7 = [0, 0, 1, 1, 2, 2, 3] whenever the index nm is repeated, the input-rate polyphase delay line is not updated. The sample processing conversion algorithm is as follows: for each input block {x0, x1, x2, x3} do: w0 = x0 y0 = dot(P, h0, w)= x0 y1 = dot(P, h4, w) delay(P, w) w0 = x1 y2 = dot(P, h1, w) y3 = dot(P, h5, w) delay(P, w) w0 = x2 y4 = dot(P, h2, w) y5 = dot(P, h6, w) delay(P, w) w0 = x3 y6 = dot(P, h3, w) delay(P, w) For example, the two output samples y4, y5 lie between the input samples x2, x3 and correspond to the i = 2 and i = 6 interpolated samples. The delay line is updated after both outputs are computed, and then the next input sample x3 is read into the delay line, and so on. In the frequency domain, the SRC filter acts as an anti-image postfilter for the upsampler, remov- ing the L−1 = 6 replicas between multiples of the fast rate fs . The downshifting of the replicas caused by the downsampling operation will position replicas at the 3 multiples of the output rate fs , 2fs , and 3fs . 272 In the L/M = 4/7 case, we have fs = 4fs/7 and fs = 7fs = 4fs, T = T /7 = T/4, Tblock = 28T = 4T = 7T. Fig. P12.22 depicts the operation of the converter both in the time and frequency domains. 0 0 fs′ 2fs′ 3fs′ 4fs′ 5fs′ 6fs′ 7fs′ fs′ 2fs′ 3fs′ 4fs′ 5fs′ 6fs′ 7fs′=fs′′ 4fs=fs′′ 0 4fs2fs 3fsfs 2fs 3fsfs 0 filter polyphase selector T′/7=T′′ T/4=T′′ 28T′′=7T=4T′ T x0 x1 x3 x5x2 x4 x6 y0 y1 y2 y3 t t t t f f f f T′ min(fs, fs′)1 2 0 3 2 1i= Fig. P12.22 Ideal SRC filter with conversion ratio 4/7. In the time domain, each input block of 7 input-rate samples {x0, x1, x2, x3, x4, x5, x6} generates 28 fast-rate interpolated samples, out of which 4 output-rate samples {y0, y1, y2, y3} are selected and computed according to their polyphase indices: im = 7m % 4 = [0, 3, 2, 1], for m = 0, 1, . . . , 3 nm = (7m − im)/4 = [0, 1, 3, 5] so that only the input samples x0, x1, x3, x5 will produce output samples. However, for the remaining input samples the delay line must be updated. Thus, the sample processing conversion algorithm will be as follows: for each input block {x0, x1, x2, x3, x4, x5, x6} do: w0 = x0 y0 = dot(P, h0, w)= x0 delay(P, w) w0 = x1 y1 = dot(P, h3, w) delay(P, w) w0 = x2 delay(P, w) w0 = x3 y2 = dot(P, h2, w) delay(P, w) w0 = x4 delay(P, w) w0 = x5 y3 = dot(P, h1, w) delay(P, w) w0 = x6 delay(P, w) 273
- 138. In the frequency domain, the SRC filter acts as an antialiasing prefilter for the downsampler, re- moving the 3 replicas between multiples of the fast rate fs . The downshifting of the replicas caused by the downsampling operation will position replicas at the 6 multiples of the output rate fs , 2fs , . . . , 6fs , without overlapping. As a design example, consider the 4/7 case and assume a Kaiser design with a stopband attenu- ation of 80 dB and normalized transition width ΔF = Δf/fs = 0.1. The width parameter D and filter length N will be: D = A − 7.95 14.36 = 80 − 7.95 14.36 = 5.0174 , N−1 = DL ΔF = 5.0174 · 4 0.1 = 200.70 ⇒ N = 209 where N the smallest odd integer of the form N = 2LK + 1; here, K = 26. Problem 12.20 A typical window depends on the time variable k through the ratio k/(N−1). The effective length of the stretched impulse response is Nρ − 1 = 2LKρ = 2LK/ρ. A symmetric window of such length will depend on the variable k /(2LKρ)= ρk /(2LK). Thus, the window of length 2LKρ can be expressed in terms of the same window of length 2LK, via the stretching relationship wρ(k )= w(ρk ). It follows that the corresponding windowed impulse responses will be wρ(k )dρ(k )= ρw(ρk )d(ρk ) Problem 12.21 The polyphase Kaiser filter can be designed by the code fragment: double *hd, **h, *w, *x, *y, wind, pi = 4 * atan(1.0); D = (A - 7.95) / 14.36; if (A > 50) alpha = 0.1102 * (A - 8.7); else if (A > 21) alpha = 0.5842 * pow(A-21, 0.4) + 0.07886 * (A-21); else alpha = 0; N = 1 + ceil(D*L / DF); Kaiser length K = ceil((N - 1) / (2.0*L)); round up N = 2*L*K + 1; final length P = 2*K - 1; polyphase filter order LK = L*K; filter delay hd = (double *) calloc(N, sizeof(double)); allocate direct form wc = pi / max(L, M); cutoff frequency I0alpha = I0(alpha); window normalization for (n=0; n<N; n++) compute filter if (n==LK) middle value hd[n] = L /(double) max(L, M); else { wind = I0(alpha * sqrt(n*(2.0*LK-n)) / LK) / I0alpha; 274 hd[n] = wind * L * (sin(wc*(n-LK))) / (pi*(n-LK)); } h = (double **) calloc(L, sizeof(double *)); allocate polyphase for (i = 0; i < L; i++) h[i] = (double *) calloc(P+1, sizeof(double)); i-th polyphase for (i=0; i<L; i++) define polyphase filters for (n=0; n<=P; n++) h[i][n] = hd[L*n+i]; Assuming the input data are being read from stdin, we allocate and initialize the polyphase delay-line buffer w as follows: w = (double *) calloc(P+1, sizeof(double)); allocate w for (i=K; i>=1; i--) read K input samples if (scanf("%lf", w+i) == EOF) { printf("must have at least K = %d input samplesn", K); exit(0); } Then, allocate the input/output blocks for the algorithm Eq. (12.6.13) and keep processing input samples in groups of M. The computed outputs go into stdout. Upon encountering the end- of-file of the inputs, we jump out of the for-ever loop and call Eq. (12.6.13) K/M more times with zero input samples. Because each call generates M zero inputs, the K/M calls will generate approximately M(K/M)= K zero inputs: x = (double *) calloc(M, sizeof(double)); y = (double *) calloc(L, sizeof(double)); for (;;) { for-ever do: for (n=0; n<M; n++) read M inputs if (scanf("%lf", x+n) == EOF) goto transients; src(L, M, P, h, w, x, y); compute L outputs for (m=0; m<L; m++) write L outputs printf("%.12lfn", y[m]); } transients: for (i=0; i <= K/M; i++) { last K inputs for (n=0; n<M; n++) x[n] = 0; src(L, M, P, h, w, x, y); for (m=0; m<L; m++) printf("%.12lfn", y[m]); } The sample processing algorithm (12.6.13) is implemented by the routine: /* src.c - sample rate converter processing algorithm */ #include <math.h> double dot(); 275
- 139. void delay(); void src(L, M, P, h, w, x, y) h is Lx(P+1) polyphase filter matrix double **h, *w, *x, *y; w is (P+1)-dimensional delay buffer { x,y are M- and L-dimensional vectors int n, m, i; double R = L / (double) M; conversion ratio for (n = 0; n < M; n++) { w[0] = x[n]; for (m = ceil(R*n); m <= floor(R*(n+1)); m++) { i = (M * m) % L; polyphase selector y[m] = dot(P, h[i], w); } delay(P, w); } } With the values A = 30 dB, ΔF = 0.1, the 5/3 filter has D = 1.536, α = 2.117, N = 81, K = 8, P = 2K − 1 = 15. The reverse 3/5 filter has the same D, α, K, and P, but its length is N = 49, because N = 2LK + 1, where now L = 3. The filter responses are shown in Fig. P12.23 both in absolute and dB scales. The dc gains are 5 and 3, respectively. The dB responses have been normalized to 0 dB at DC. Fig. P12.23 Frequency responses of SRC filters. Both filters have a cutoff frequency of fc = min(fs/2, fs /2)= 3/2 = 1.5 kHz, or, ωc = π/5. The passband ripple is δ = 10−A/20 = 0.032. Their transition widths are different: the first filter has Δf = 0.1×3 = 0.3 kHz, and the second Δf = 0.1×5 = 0.5 kHz. Were they to be redesigned to have the same Δf, say 0.3 kHz, then for the 3/5 filter we would have ΔF = 0.06, resulting in the lengths N = 79, K = 13, P = 25. In that case, the filters would be virtually identical (except for the gain and the delay by 1 sample). The impulse responses are obtained from Eq. (12.6.5), with LK = 5×8 = 40 and LK = 3×8 = 24 in the two cases, that is, 276 h5/3(n )= 5 I0(α n (80 − n )/80) I0(α) sin π(n − 40)/5 π(n − 40) , n = 0, 1, . . . , 80 h3/5(n )= 3 I0(α n (48 − n )/48) I0(α) sin π(n − 24)/5 π(n − 24) , n = 0, 1, . . . , 48 The frequency responses were computed by direct evaluation of: H(f)= N−1 n =0 h(n )e−2πjfn /fs , with fs = 15 kHz Figure P12.24 shows the output y(n ) of the 5/3 filter and compares it with the signal x (n ) that would have been obtained had the analog signal been sampled at 3 kHz. On the right, it shows the effect of the reverse 3/5 filter on y(n ), and compares it with the original x(n). Fig. P12.24 Comparison of y(n ) and x (n ), and of x(n) and reconverted y(n ). Problem 12.22 Replacing the quantizer Q by its equivalent noise model, we can write the ζ-domain input/output equations: V (ζ) = X (ζ)−ζ−1 Y (ζ) W0(ζ) = V (ζ)+ζ−1 W0(ζ) Y (ζ) = E (ζ)+W0(ζ) Eliminating V (ζ) and W0(ζ), we obtain: Y (ζ)= X (ζ)+(1 − ζ−1 )E (ζ) Therefore, Hx(ζ)= 1, and HNS(ζ)= 1 − ζ−1 . The main advantage of this form is that the input signal X (ζ) appears unchanged at the output (as opposed to being delayed as in the conventional case). 277
- 140. Problem 12.23 Using the signal labels indicated on the block diagram, we have the sample processing algorithm: for each input x do: v = x − u w0 = v + w1 y = Q(w0) w1 = w0 u = y Iterating this algorithm 10 times, we get for x = 0.4 and x = −0.2: x y 0.40 1 0.40 −1 0.40 1 0.40 1 0.40 1 0.40 −1 0.40 1 0.40 1 0.40 −1 0.40 1 x y −0.20 −1 −0.20 1 −0.20 −1 −0.20 1 −0.20 −1 −0.20 −1 −0.20 1 −0.20 −1 −0.20 1 −0.20 −1 The averages of the y values are 0.4 and −0.2, respectively. Problem 12.24 Working in the ζ-domain, we note that the input and output of the filter H1(ζ) will be: X (ζ)−ζ−1 Y (ζ) and H1(ζ) X (ζ)−ζ−1 Y (ζ) Similarly, the input and output of H2(ζ) are: H1(ζ) X (ζ)−ζ−1 Y (ζ) −ζ−1 Y (ζ) and H2(ζ) H1(ζ) X (ζ)−ζ−1 Y (ζ) −ζ−1 Y (ζ) Adding E (ζ) to this output will generate Y (ζ). Thus, Y (ζ = E (ζ)+H2(ζ) H1(ζ) X (ζ)−ζ−1 Y (ζ) − ζ−1 Y (ζ) Moving the Y (ζ) to the left-hand side and solving for Y (ζ) gives the transfer relationship: Y (ζ)= H1(ζ)H2(ζ) 1 + ζ−1H2(ζ)+ζ−1H1(ζ)H2(ζ) X (ζ)+ 1 1 + ζ−1H2(ζ)+ζ−1H1(ζ)H2(ζ) E (ζ) Thus, we identify: Hx(ζ)= H1(ζ)H2(ζ) 1 + ζ−1H2(ζ)+ζ−1H1(ζ)H2(ζ) , HNS(ζ)= 1 1 + ζ−1H2(ζ)+ζ−1H1(ζ)H2(ζ) The requirement that Hx(ζ)= 1 and HNS(ζ)= (1 − ζ−1 )2 is satisfied by the choices: H1(ζ)= H2(ζ)= 1 1 − ζ−1 278 A block diagram realization is shown in Fig. P12.25. The corresponding sample processing algo- rithm can be stated as follows: for each input x do: w0 = w1 + (x − u) v0 = v1 + (w0 − u) y = Q(v0) w1 = w0 v1 = v0 u = y + + − − H2(ζ)H1(ζ) ζ-1 ζ-1 ζ-1 x′ w0 v0 w1 v1 u y′ quantizer Q Fig. P12.25 Discrete-time model of second-order delta-sigma quantizer. Problem 12.25 Without the feedback delay, we get a similar transfer relationship as in the previous problem, but without the ζ−1 delays in the denominators, that is, Y (ζ)= H1(ζ)H2(ζ) 1 + H2(ζ)+H1(ζ)H2(ζ) X (ζ)+ 1 1 + H2(ζ)+H1(ζ)H2(ζ) E (ζ) Thus, we identify: Hx(ζ)= H1(ζ)H2(ζ) 1 + H2(ζ)+1(ζ)H2(ζ) , HNS(ζ)= 1 1 + H2(ζ)+H1(ζ)H2(ζ) The requirement that Hx(ζ)= ζ−1 and HNS(ζ)= (1 − ζ−1 )2 is satisfied by the choices: H1(ζ)= 1 1 − ζ−1 , H2(ζ)= ζ−1 1 − ζ−1 (This problem is similar to Problem 7.2.) Problem 12.26 Once the decimation filters are designed, the 2-level quantization of the input by the delta-sigma quantizer, and the subsequent filtering by the decimation filter can be carried out by the following for-loop (which does not discard the interpolated outputs, as discussed in Example 12.7.2): w1 = 0; initialize delay for (n=0; n<Ntot; n++) { x = 0.5 * sin(2 * pi * f0 * n); high-rate input sample 279
- 141. y = Q(w1); ΔΣ quantizer output v = x - y; delta part w0 = w1 + v; sigma part w1 = w0; update quantizer’s delay yup[n] = fir(N-1, h, w, y); (upsampled) decimator output } where h and w are the N-dimensional impulse response and internal state arrays for the decima- tor. In the Kaiser and the rectangular window designs, the filter length is N = 2LM + 1. In the first-order comb case, it is N = 10, and the impulse response and transfer function are: h = 1 10 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1] , H(ζ)= 1 10 1 − ζ−10 1 − ζ−1 For the second-order comb case, we obtain the impulse response by convolving the first-order comb’s impulse response with itself. This gives a length N = 19 response and transfer function: h = 1 100 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1] , H(ζ)= 1 10 1 − ζ−10 1 − ζ−1 2 The magnitude response of the second-order comb case is the square of that of the first-order case: |H(f)| = sin(πf/fs) L sin(πf/10fs) 2 It is plotted in dB in Fig. P12.26 together with the first-order comb case and the rectangular window design. The upsampled output yup(n ) of the second-order comb decimator is plotted in the right graph of Fig. P12.26. Fig. P12.26 Length-19 second-order comb magnitude response and filter output. Problem 12.27 The filter designs are the same as in the previous problems. The implementation of the second- order delta-sigma quantizer is given by the sample processing algorithm discussed in Problem 12.24. The following for-loop generates the high-rate input, quantizes it by the 2nd-order quan- tizer, and filters the quantized output with the decimation filter: 280 w1 = v1 = u = 0; initialize delays for (n=0; n<Ntot; n++) { x = 0.5 * sin(2 * pi * f0 * n); high-rate input sample w0 = w1 + x - u; output of H1 v0 = v1 + w0 - u; output of H2 y = Q(v0); ΔΣ quantizer output w1 = w0; update quantizer’s delays v1 = v0; update quantizer’s delays u = y; update quantizer’s delays yup[n] = fir(N-1, h, w, y); (upsampled) decimator output } Figure P12.27 shows the signal x (n ) and the 2-level quantized output y (n ) of the second-order quantizer. Fig. P12.27 High-rate input and output signals of second-order ΔΣ quantizer. The upsampled outputs yup(n ) of the averaging, second-order comb, rectangular, and Kaiser window decimators are shown in Fig. P12.28. Problem 12.28 Working in the ζ-domain, we obtain the I/O equation of the first stage. The input to H(ζ) is X (ζ)−ζ−1 Y1(ζ), Thus, adding the noise component, we have: Y1(ζ)= E1(ζ)+H(ζ) X (ζ)−ζ−1 Y1(ζ) which can be rearranged as: Y1(ζ)= H(ζ) 1 + ζ−1H(ζ) X (ζ)+ 1 1 + ζ−1H(ζ) E1(ζ) Inserting H(ζ)= 1/(1 − ζ−1 ), we obtain the transfer relationship of one stage: Y1(ζ)= X (ζ)+(1 − ζ−1 )E1(ζ)= X (ζ)+D(ζ)E1(ζ) (P12.3) where D(ζ)= 1−ζ−1 . The input to the second stage is obtained by forming the difference of the signals around the first quantizer, that is, (Y − E1)−Y = −E1. Applying the basic I/O equation (P12.3) to this input and corresponding noise E2, we obtain 281
- 142. Fig. P12.28 First- and second-order comb, and rectangular- and Kaiser-window decimator out- puts. Y2(ζ)= −E1(ζ)+D(ζ)E2(ζ) (P12.4) The overall output is obtained by the linear combination: Y (ζ) = Y1(ζ)+D(ζ)Y2(ζ)= X (ζ)+D(ζ)E1(ζ)+D(ζ) −E1(ζ)+D(ζ)E2(ζ) = X (ζ)+D(ζ)E1(ζ)−D(ζ)E1(ζ)+D2 (ζ)E2(ζ)= X (ζ)+D2 (ζ)E2(ζ) Thus, the quantization error from the first stage is canceled, whereas the error of the second stage gets filtered by the second-order highpass filter D2 (ζ)= (1 − ζ−1 )2 . Problem 12.29 Such an arrangement is shown in Fig. P12.29. The outputs of the three stages are: Y1(ζ) = X (ζ)+D(ζ)E1(ζ) Y2(ζ) = −E1(ζ)+D(ζ)E2(ζ) Y3(ζ) = −E2(ζ)+D(ζ)E3(ζ) The overall output is obtained by the linear combination: 282 Y (ζ) = Y1(ζ)+D(ζ) Y2(ζ)+D(ζ)Y3(ζ) = X (ζ)+D(ζ)E1(ζ)+D(ζ) −E1(ζ)+D(ζ)E2(ζ)+D(ζ) −E2(ζ)+D(ζ)E3(ζ) = X (ζ)+D3 (ζ)E3(ζ) x + − ζ-1 H(ζ) y1 y2 e1 −e1 −e2 + + + − − − ζ-1 H(ζ) e2 y 1st stage 2nd stage y3+ − ζ-1 H(ζ) e3 3d stage D(ζ) D(ζ) Fig. P12.29 MASH architecture of third-order delta-sigma quantizer. Problem 12.30 Such an architecture is depicted in Fig. P12.30. The I/O relationship of the first stage is Y1(ζ)= X (ζ)+D(ζ)E1(ζ) The input to the second-order quantizer is −E1(ζ). Therefore, its I/O relationship is (from Problem 12.8.3): Y2(ζ)= −E1(ζ)+D2 (ζ)E2(ζ) The overall output is obtained by the linear combination: Y (ζ) = Y1(ζ)+D(ζ)Y2(ζ)= X (ζ)+D(ζ)E1(ζ)+D(ζ) −E1(ζ)+D2 (ζ)E2(ζ) = X (ζ)+D3 (ζ)E2(ζ) Problem 12.31 The delay of (N − 1)/2 = LM introduced by the decimation filter must be taken into account in both multiplexing and control applications. In feedback control applications, such a delay may render the closed-loop system unstable. See [294] for further discussion. 283
- 143. x + − ζ-1 H(ζ) y1 y2 e1 −e1 + + − − − ζ-1 H(ζ) H(ζ) e2 D(ζ) y 1st stage 2nd stage Fig. P12.30 MASH architecture of third-order delta-sigma quantizer. 284 Appendix B Problems Problem B.1 The input sequence is x(n)= i uiδ(n − iD − d) Its autocorrelation function is Rxx(k)= E x(n + k)x(n) = i j D−1 d=0 E[uiuj]δ(n + k − iD − d)δ(n − jD − d)p(d) where the expectation values were taken with respect to both ui and d. Because the ui are mutually independent and zero mean and d is uniform, we have E[uiuj]= σ2 uδ(i − j) and p(d)= 1/D. It follows that: Rxx(k)= σ2 u D i D−1 d=0 δ(n + k − iD − d)δ(n − iD − d) Changing variables to m = n−iD−d and noting that m runs over all the integers, we can replace the double summation over i and d by the summation over m: Rxx(k)= σ2 u D m δ(k + m)δ(m)= σ2 u D δ(k) Thus, x(n) is stationary and has a delta-function autocorrelation. Sending this sequence into the interpolation filter will generate a stationary output sequence y(n) with a noise reduction ratio: NRR = σ2 y σ2 x = n h(n)2 For the particular case of a hold interpolator, h(n) consists of D ones: NRR = σ2 y σ2 x = D−1 n=0 1 = D ⇒ σ2 y = Dσ2 x = D σ2 u D = σ2 u For the linear interpolator, we have h(n)= 1 − |n|/D, for |n| ≤ D. This gives: NRR = σ2 y σ2 x = D n=0 1 − |n| D 2 = 1 + 2 D n=1 1 − n D 2 = 1 + 2 1 D2 D−1 m=1 m2 Using the identity D−1 m=1 m2 = 1 6 D(D − 1)(2D − 1), we obtain: NRR = σ2 y σ2 x = 1 + 2 D−1 m=1 m2 = 1 + D(D − 1)(2D − 1) 3D2 = 1 + (D − 1)(2D − 1) 3D = 2D2 + 1 3D Thus, σ2 y = 2D2 + 1 3D σ2 x = 2D2 + 1 3D2 σ2 u 285
- 144. Problem B.2 The power spectral density of the output sequence will be related to that of the input by: Syy(z)= H(z)H(z−1 )Sxx(z)= H(z)H(z−1 )σ2 x = H(z)H(z−1 ) σ2 u D For the hold interpolator, we have: H(z)= 1 − z−D 1 − z−1 , H(z−1 )= 1 − zD 1 − z = 1 − z−D 1 − z−1 zD−1 Thus, the output psd becomes: Syy(z)= H(z)H(z−1 ) σ2 u D = 1 D 1 − z−D 1 − z−1 2 zD−1 σ2 u Taking inverse z-transforms with the help of Problem 5.11, we find: Ryy(k)= 1 − |k| D σ2 u Problem B.3 The generation of the held signal and the computation of its autocorrelation is illustrated by the following C program segment: double *u, *y, *R; long iseed; y = (double *) calloc(N, sizeof(double)); N-dimensional R = (double *) calloc(M+1, sizeof(double)); (M+1)-dimensional u[0] = ran(&iseed)-0.5; initialize u q = D * ran(&iseed); initialize q for (n=0; n<N; n++) y[n] = ranh(D, u, &q, &iseed); zero mean corr(N, y, y, M, R); sample autocorrelation A typical sample autocorrelation is shown in Fig. P14.1 Problem B.4 As was mentioned in the end of Section B.3, the averaged periodogram spectrum can be calculated by the following loop: for (k=0; k<K; k++) { generate K = 200 blocks u[0] = ran(&iseed); initialize kth block q = D * ran(&iseed); for (n=0; n<N; n++) generate kth block X[n] = cmplx(ranh(D, u, &q, &iseed), 0.0); fft(N, X); compute its FFT for(n=0; n<N; n++) accumulate its periodogram S[i] += cabs(X[i]) * cabs(X[i]); } A typical averaged spectrum is shown in Fig. P14.2, for the case D = 5. 286 Fig. P14.1 Sample autocorrelation of length-100 hold noise. Fig. P14.2 Averaged periodogram spectrum of held noise. Problem B.5 The random number generator C routine is: /* rani.c - low-frequency random number generator using interpolation */ double ran(), dot(); void delay(), cdelay2(); double rani(L, P, h, w, q, iseed) usage: y=rani(L, P, h, w, & q, & iseed); int L, P, *q; q and iseed passed by reference double **h, *w; h = polyphase filter matrix long *iseed; required long by ran() { int i; double y; if (*q==0) every L calls get new random number w[0] = ran(iseed) - 0.5; zero mean i = (L - *q) % L; current polyphase index y = dot(P, h[i], w); compute interpolated value 287
- 145. cdelay2(L-1, q); decrement q for next call if (*q==0) every L calls update delay line delay(P, w); return y; } For the hold and linear interpolators, it generates the same output as ranh.c and ranl.c. The following program raniex.c is an example C program that invokes rani. Most of it is concerned with designing and allocating the polyphase filters. The very last for-loop generates the desired numbers. The for-loop before the last one, initializes the delay line w in accordance with Eq. (12.2.19): /* raniex.c - interpolated random number generator example */ run with: k = 0, 1, 2 = hold, linear, Kaiser interpolators L = 10, Ntot = 150, iseed = 1234567 A = 30 dB, ΔF = 0.3 #include <stdio.h> #include <math.h> #include <stdlib.h> double ran(), rani(), I0(); I0.c was given in Ch.10 void main(int argc, char ** argv) { double pi = 4 * atan(1.0); double **h, *w, *hd; double I0al, wind, A, D, wc, DF, al; int L, P, K, N, LK; int q, i, k, n, Ntot; long iseed; if (argc != 5) { puts("nUsage: raniex k L Ntot iseed > y.datn" " k = 0,1,2 for hold,linear,general interpolatorn" " L = upsampling ration" " Ntot = total random numbersn" " iseed = initial seedn"); exit(0); } k = atoi(argv[1]); L = atoi(argv[2]); Ntot = atoi(argv[3]); iseed = atol(argv[4]); if (k >= 2) { Kaiser interpolator fprintf(stderr, "enter A, DF (in units of fs): " ); scanf("%lf %lf", &A, &DF); D = (A - 7.95) / 14.36; 288 if (A > 50) al = 0.1102 * (A - 8.7); else if (A > 21) al = 0.5842 * pow(A-21, 0.4) + 0.07886 * (A-21); else al = 0; N = 1 + ceil(D * L / DF); K = ceil((N - 1) / (2.0*L)); N = 2 * L * K + 1; filter length P = 2 * K - 1; filter order LK = L * K; filter delay hd = (double *) calloc(N, sizeof(double)); direct form filter wc = pi / L; ideal cutoff frequency I0al = I0(al); for (n=0; n<N; n++) windowed filter if (n==LK) hd[n] = 1; else { wind = I0(al * sqrt(n*(2.0*LK-n)) / LK) / I0al; hd[n] = wind * L * (sin(wc*(n-LK)))/(pi*(n-LK)); } } if (k == 0) { hold interpolator N = L; K = 0; P = 0; for (n=0; n<N; n++) hd[n] = 1; } if (k == 1) { linear interpolator P = 1; K = 1; N = 2*L+1; for (n=0; n<N; n++) hd[n] = 1 - abs(n - L) / (double) L; } h = (double **) calloc(L, sizeof(double *)); polyphase subfilters for (i = 0; i < L; i++) allocate L subfilters h[i] = (double *) calloc(P+1, sizeof(double)); for (i=0; i<L; i++) for (n=0; n<=P; n++) h[i][n] = hd[L*n+i]; ith subfilter w = (double *) calloc(P+1, sizeof(double)); low-rate delay-line buffer for (i=K; i>=1; i--) w[i] = ran(&iseed) - 0.5; initialize w 289
- 146. q = 0; for (n=0; n<Ntot; n++) generate interpolated noise printf("%.12lfn", rani(L, P, h, w, &q, &iseed)); } With the choices A = 30 dB, ΔF = 0.3, the Kaiser parameters become D = 1.536, α = 2.117. The generated sequences are shown in Fig. P14.3. Fig. P14.3 Kaiser, hold, and linearly interpolated random sequences. Problem B.6 Noting that the autocorrelation of a hold-interpolated sequence of period D is R(k)= (1 − |k|/D)σ2 u, −D ≤ k ≤ D, we have: Ryy(k)= 1 B2 B−1 b=0 Rybyb (k)= 1 B2 B−1 b=0 1 − |k| 2b u(2b − |k|) σ2 u (P14.1) where the unit-step u(2b −|k|) restricts the duration of the b-th term to be |k| ≤ 2b , or |k| ≤ 2b −1 for non-zero values of Ryy(k). The resulting autocorrelation function is piece-wise linear, where the linear segments have geometrically increasing slopes. This is shown in Fig. P14.4 for the case B = 4. At lag k = 0, Eq. (P14.1) gives the expected result: σ2 y = Ryy(0)= 1 B2 · Bσ2 u = σ2 u B The increasing linear slopes in Ryy(k) cause it to have long-range correlations. Indeed, because of the restriction |k| ≤ 2b − 1, the maximum lag will be: kmax = 2B−1 − 1 (P14.2) Given a lag k in the range |k| ≤ kmax, the expression of Eq. (P14.1) can be summed in closed form as follows. The restriction |k| ≤ 2b −1 can be turned around into a restriction for the summation index b ≥ log2(|k| + 1). Denoting the ceiling of this quantity by b(k)= log2(|k| + 1) 290 k kmax=2B-1 -1 R(k)/R(0) 1 3 5 7 94 6 8 102 1.00 0.75 0.50 0.25 0 B=4 Fig. P14.4 Theoretical autocorrelation of 1/f-noise model. we find Ryy(k)= 1 B2 B−1 b=b(k) 1 − |k|2−b σ2 u and using the finite geometric series formula (e.g., Eq. (6.3.13)), we get the expression of Eq. (B.25). Problem B.7 Figure P14.5 shows the theoretical autocorrelation (P14.1) or (B.25), together with the sample autocorrelation of a (zero-mean) block of 1/f-noise samples of length N = 2000 for B = 8. The maximum correlation length is in this case kmax = 2B−1 − 1 = 27 − 1 = 127. The sample autocorrelation was computed by the program segment: double *y, *R; y = (double *) calloc(N, sizeof(double)); N-dimensional R = (double *) calloc(M+1, sizeof(double)); (M+1)-dimensional for (n=0; n<N; n++) use N=2000 y[n] = ran1f(B, u, q, &iseed); zero-mean y’s corr(N, y, y, M, R); use M=150 Fig. P14.5 Theoretical and sample autocorrelations for B = 8. 291
- 147. The reason for using N = 2000 was because we wanted to evaluate the sample autocorrelation at M = 150 lags. Thus, the number of lags is 150/2000 ≡ 7.5% of the block length, which falls within the recommended range for getting statistically reliable lags using the routine corr, as was mentioned in Section A.1. Problem B.8 We look at the cases ωmin = 0.001π and ωmax = 0.1π, 0.3π, 0.6π. The designed model param- eters are: For ωmax = 0.1π, we have B = 3 and c = 10, a = [0.6858, 0.9686, 0.9969] For ωmax = 0.3π, we have B = 4 and c = 6.6943, a = [0.0575, 0.8592, 0.9790, 0.9969] For ωmax = 0.6π, we have B = 6 and c = 3.5944, a = [−0.8850, 0.4756, 0.8541, 0.9594, 0.9887, 0.9969] The spectra are shown in Figs. P14.6—P14.8. Fig. P14.6 Model spectrum and periodogram estimate. ωmax = 0.1π. Problem B.9 Solving the equation (0.99574)c = 0.94791, we find c = 12.5309. Then, the filter of Eq. (B.30) becomes: H(z)= G (1 − 0.98444z−1 )(1 − 0.82159z−1 )(1 − 0.08522z−1 ) (1 − 0.99574z−1)(1 − 0.94791z−1)(1 − 0.51153z−1) The NRR = hT h of the filter (B.29) without the factor G can be computed analytically or much faster using MATLAB to compute a fairly long portion of the impulse response h (e.g., of length 2000) and then setting G = 1/ √ NRR = 1/ hT h. The magnitude response squared of Eqs. (B.29) and (B.30) are shown in Fig. P14.9. The cascade realization is shown in Fig. P14.10. The corre- sponding sample processing algorithm will be: 292 Fig. P14.7 Model spectrum and periodogram estimate. ωmax = 0.3π. Fig. P14.8 Model spectrum and periodogram estimate. ωmax = 0.6π. for each white noise input sample x do: u0 = 0.57534 x + 0.99574 u1 x1 = u0 − 0.98444 u1 u1 = u0 v0 = x1 + 0.94791 v1 x2 = v0 − 0.83392 v1 v1 = v0 w0 = x2 + 0.53568 w1 y = w0 − 0.07568 w1 w1 = w0 293
- 148. Fig. P14.9 1/f noise filters for Problem B.9. z-1 x u0 v0 w0 u1 x1 x2 v1 w1 y z-1 z-10.57534 0.99574 -0.98444 0.94791 -0.83392 0.53568 -0.07568 Fig. P14.10 Cascade realization of 1/f noise filter. 294