Fourier-Series_FT_Laplace-Transform_Letures_Regular_F-for-Students_13.ppt

2
Example 56.2. Compute the DFT of the signal given by the
following four samples, x[n] = [2 1 -3 4].
Solution: We are going to use the DFT Eq. (56.c), although this is
not a particularly hard calculation. N, the length of the signal is 4,
hence the fundamental frequency is equal to 2/4 = /2. We write
the DFT as

3
Figure 6.7: The two ways of looking at a DFT The real and
imaginary parts of the DFT vs. bin number in (a) and (b), and in (c)
and (d) the magnitude and the phase for same bin numbers.

4
In Fig. 6.7 we see the computed DFT in both its common forms. In the top row,
we see the Real and Imaginary plots and in the second row, we have the
Magnitude and Phase plots. They are different but contain the same information.
Both forms have their particular use. Since this is a 4 point sequence, we get just
four values of the DFT, X[k]. It is hard to make sense of what is going on when
there are so few values. We should plot them against the DTFT to see if there is
a pattern. Where to get the DTFT? In this case it is pretty simple. We use Eq.
(6.1). The signal is simple enough that we go ahead and solve the DTFT in
closed form.
Note the DTFT equation we used to compute this DTFT. Now we can plot the
continuous DTFT from Eq. (??) along with the four computed points of the
DFT from Eq. (??) and we see where these points are coming from. First thing
to note, the DFT points are exactly the correct samples of the DTFT.

5
Figure 6.8: The phase and the magnitude of the signal when seen along with
the DTFT as its sample points.
We can see from this picture that the four discrete points the
DFT although accurate, missed the highs of the magnitude as
well as the lows of the phase. Since discrete points is all we get
from the DFT, we need more points to see the underlying
spectrum better. Butt here is a problem, if we are given the
sequence x[n] which consists of just 4 points and no more data
can be had, what can we do?

6
Figure 6.9: Zero-
padding increases N and
hence provides better
sampling of the DTFT.
There is something magical we can do at this point and it is called zero-
padding. This is about the most “fun” you can have in DSP for nothing. It’s a
very valuable tool. We add a bunch of zeros to the end of the signal, x[n] to
lengthen it artificially. Now we redo the DFT for this zero-padded and hence
lengthened signal (N + added zeros). Remember we said that the DFT counts
the zeros at the ends of a signal as real data. But what is truly amazing is what
we can extract from these zeros. In Fig. 6.9, we see the result of this zero-
padding. For 16 point signal, the signal x[n] is zero-padded with 12 zeros at
the end. Because we count these new points as part of the period, N is now 16
samples. Its DFT in Fig. 6.9 now has a frequency resolution of 2=16 instead
of 2=4 it had for the 4-point DFT. And of course we get 16 sampled data
points of the DTFT instead of 4. The new plot looks quite good!

7
Example 56.2. Compute the DFT of the signal given by the
following four samples, x[n] = [2 1 -3 4].
Solution: We are going to use the DFT Eq. (56.c), although this is
not a particularly hard calculation. N, the length of the signal is 4,
hence the fundamental frequency is equal to 2/4 = /2. We write
the DFT as

8
Figure 6.7: The two ways of looking at a DFT The real and
imaginary parts of the DFT vs. bin number in (a) and (b), and in (c)
and (d) the magnitude and the phase for same bin numbers.

9
In Fig. 6.7 we see the computed DFT in both its common forms. In the top row,
we see the Real and Imaginary plots and in the second row, we have the
Magnitude and Phase plots. They are different but contain the same information.
Both forms have their particular use. Since this is a 4 point sequence, we get just
four values of the DFT, X[k]. It is hard to make sense of what is going on when
there are so few values. We should plot them against the DTFT to see if there is
a pattern. Where to get the DTFT? In this case it is pretty simple. We use Eq.
(6.1). The signal is simple enough that we go ahead and solve the DTFT in
closed form.
Note the DTFT equation we used to compute this DTFT. Now we can plot the
continuous DTFT from Eq. (??) along with the four computed points of the
DFT from Eq. (??) and we see where these points are coming from. First thing
to note, the DFT points are exactly the correct samples of the DTFT.

10
Figure 6.8: The phase and the magnitude of the signal when seen along with
the DTFT as its sample points.
We can see from this picture that the four discrete points the
DFT although accurate, missed the highs of the magnitude as
well as the lows of the phase. Since discrete points is all we get
from the DFT, we need more points to see the underlying
spectrum better. Butt here is a problem, if we are given the
sequence x[n] which consists of just 4 points and no more data
can be had, what can we do?

11
Figure 6.9: Zero-
padding increases N and
hence provides better
sampling of the DTFT.
There is something magical we can do at this point and it is called zero-
padding. This is about the most “fun” you can have in DSP for nothing. It’s a
very valuable tool. We add a bunch of zeros to the end of the signal, x[n] to
lengthen it artificially. Now we redo the DFT for this zero-padded and hence
lengthened signal (N + added zeros). Remember we said that the DFT counts
the zeros at the ends of a signal as real data. But what is truly amazing is what
we can extract from these zeros. In Fig. 6.9, we see the result of this zero-
padding. For 16 point signal, the signal x[n] is zero-padded with 12 zeros at
the end. Because we count these new points as part of the period, N is now 16
samples. Its DFT in Fig. 6.9 now has a frequency resolution of 2/16 instead
of 2/4 it had for the 4-point DFT. And of course we get 16 sampled data
points of the DTFT instead of 4. The new plot looks quite good!

12
N= 4 Point DFT Matlab Coding
xn=[1, 1, 1, 0]
Xk=fft (xn);
mag=abs (Xk)
ang=angle (Xk)
subplot (3, 1, 1)
stem (xn,'filled')
xlabel ('n'), ylabel ('x(n)')
subplot (3, 1, 2)
stem (abs (Xk),'filled')
xlabel ('k'), ylabel ('abs (X(k)')
subplot (3, 1, 3)
stem (angle (Xk),'filled')
xlabel ('k'), ylabel ('angle (X(k)')
N= 8 (4 Zero-Padding) Point DFT Matlab Coding
xn=[1, 1, 1, 0, 0, 0, 0, 0]
Xk=fft (xn);
mag=abs (Xk)
ang=angle (Xk)
subplot (3, 1, 1)
stem (xn,'filled')
subplot (3, 1, 2)
subplot (3, 1, 3)
N= 32 (28 Zero-Padding) Point DFT Matlab Coding
xn=[1, 1, 1, 0]
Xk=fft (xn,32);
mag=abs (Xk)
ang=angle (Xk)
subplot (3, 1, 1)
stem (xn,'filled')
subplot (3, 1, 2)
subplot (3, 1, 3)

13
DFT CALCULATION USING MATLAB
xn=input ('Enter the sequence x(n)=')
[N]=length (xn);
xn=xn.';
n=0:N-1;
for k=0:N-1
xk(k+1)=exp(-j*2*pi*k*n/N)*xn;
end
disp ('x(k)=');
disp (xk)

14
Computing DFT the Easier Way with Matrix Math
In previous example we calculated the DFT and the DTFT in closed
form. The DTFT of arbitrary signals using this method is a difficult
task, not one we want to do on paper. The DFT on the other hand, has a
reasonably easy to understand architecture amenable to digital
calculations. It can be implemented as a linear device as shown in Fig.
6.10 with N inputs and N outputs. It is completely lossless and
bidirectional. N numbers go in and N numbers come out, using only
addition and multiplication.
We can view the DFT as a linear input/output processor
consisting of N equations. We have collected L data symbols. If N is
less than L, then we select N samples out of the L and perform the DFT
calculations on these N points. If N is larger than the L collected
samples, then we have to do something to increase the number of
samples from L to N. Some algorithms of the DFT will automatically
zero-pad a time domain signal to make the length equal to N.

15
Figure 6.10: Architecture of a DFT Processor.
We can build the DFT calculating machine in two parts, one for the
forward or the DFT and the other for the inverse, or the iDFT. We feed N
time-domain samples into the DFT machine in Fig. 6.10. The device needs
all N points simultaneously before anything can happen. This is because
each sample makes a contribution to the each of the outputs, the harmonic
components X[k]. N samples go in and N samples of frequency response,
X[k] come out. Each X[k] gets a little contribution from each x[n].

16
Here is how each output point will be calculated. There are N of these
equations that the machine has to compute. Here we write out a few of
these equations, for k = 0, 1 and N.
…(56.n)
Computers were designed to do matrix math, so these equations can be easily
solved. The only problem comes as N increases. The number of calculations gets
very large. The matrix computation done this way requires a number of calculations
that are proportional to N2. If we take advantage of the symmetry as well the phase
relationship of the sine and cosine, the number of computations can come down to a
number more like N log2 N. This and other algorithmic innovations were first
proposed by J.W. Cooley and John Tukey in 1965. There are now many variations
on their original invention and they are all referred to as the Fast Fourier transform
or FFT. Their innovation was extremely important particularly for hardware
implementation. We use this algorithm today for calculating the DFT, so much so
that we often find ourselves saying “take a FFT” when we really mean the DFT.

17
We notice that all calculations require the manipulation of the
complex exponential. The complex exponential is hard to type, so
we simplify it by writing it like this.
We introduced this idea in Chapter 3, when discussing the discrete
Fourier series calculations. It is easy to see that WN is a function of
N only, the rest of the terms are all constants. Now we also define
the same constant in terms of other two variables, n and k. These
are the time and harmonic indexes.
N
N
j
W
e 


2
…(56.o)
nk
N
j
nk
N
j
nk
N e
e
W

 2
2











 …(56.p)

18
This form has the variables n and k in the exponent and so the
constant WN is written with all three indexes as WN
nk. Here we are
merely raising WN to power nk, or (e-j2/N)nk. Hence WN
nk is a matrix
of size n  k with each index ranging from 0 to N - 1. Now rewrite
the DFT using the term, WN
nk. This is somewhat easier to write than
the exponential, as it has less moving parts.
Now we write Eq. (6.9) in matrix form as




1
0
]
[
]
[
N
n
nk
N
W
n
x
k
X …(56.q)
…(56.r)
x
.
X
nk
N
W


19
X is the output Fourier transform vector of size N and x is the input
vector of the same size.
The WN
nk term in Eq. (6.15) is called the DFT matrix and is written
for all n and k indices as
…(56.s)

20
Note that the matrix size is n  k, but since both n and k are equal to N,
the DFT matrix size is equal to N  N, hence this matrix is always
square. The columns represent the index k of the harmonics and rows the
index n, time. The first row has index n = 0, so all the first row terms are
equal to 1, because WN
(n=0)k =e-j(2k/N).0 = 1
The first column has index k = 0, hence all WN
nk terms in the first
column have a zero exponent and so all are also equal to 1. The third
row, second term is equal to WN
2, for k = 1 and n = 2. The rest can be
understood the same way. The matrix is symmetrical about the right
diagonal such that
kn
N
nk
N W
W  …(56.t)

21
All of the terms in Eq.(56.s) are constant for given n and k integers. If
you stare at the equation long enough, I am sure just as Cooley and
Tukey, you will be able to see the advantage of matrix calculations. This
matrix can be pre-computed and stored for various value of N. That
saves computation time. We don’t need to compute the whole matrix as
well, because it is symmetrical. There are actually a lot of tricks that can
be used to make computations efficient. We can also show that the
inverse of this matrix is easy to compute because it is just the inverse of
the individual terms. This also makes reverse computation easier and
quicker.
   *
1 1 nk
N
nk
N W
N
W 

…(56.u)

22
Let’s write out the DFT matrix for N = 4. The fundamental frequency
for N = 4 sample signal is equal to 0 = 2/4 = /2. Hence we write
W4
nk=e-j(2k/4)n.
This is equal in Euler form to
j
j
j
W
W
j
j
W
W
j
j
j
W
W
j
j
W
W
j
nk
j
nk
e
e
W nk
nk
j
nk
j
nk










































































































)
1
.(
0
1
.
3
2
sin
1
.
3
2
cos
1
0
.
1
1
.
2
2
sin
1
.
2
2
cos
1
.
0
1
.
1
2
sin
1
.
1
2
cos
1
0
.
1
1
.
0
2
sin
1
.
0
2
cos
)
(
2
sin
2
cos
3
4
1
.
3
4
2
4
1
.
2
4
1
4
1
.
1
4
0
4
1
.
0
4
2
4
2
4












,
1
k
3,
n
For
,
1
k
2,
n
For
,
1
k
1,
n
For
,
1
k
0,
n
For
Method-1:

23
From this we can develop the WN
nk matrix as















































j
-
1
-
j
1
1
-
1
1
-
1
j
1
j
1
1
1
1
1
3
2
1
0
9
6
3
6
4
2
3
2
1
3
.
3
2
.
3
1
.
3
0
.
3
3
.
2
2
.
2
1
.
2
0
.
2
3
.
1
2
.
1
1
.
1
0
.
1
3
.
0
2
.
0
1
.
0
0
.
0
1
1
1
1
1
1
1
3
2
1
0
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
nk
N
W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
k
W
n
Method-2:
…(56.v)

24
…(56.w)
The inverse comes easily from Eq.(56.u)
[In case of Inverse/conjugate, only the sign of j
should be changed.]












j
1
j
1
1
1
1
1
j
1
j
1
1
1
1
1
Matrix
t
Coefficien














j
-
1
-
j
1
1
-
1
1
-
1
j
1
j
1
1
1
1
1
DFT
For













j
1
-
j
-
1
1
-
1
1
-
1
j
-
1
j
1
1
1
1
1
IDFT
For
Non-negative sign Non-negative sign
DFT and IDFT Coefficients
 















j
1
-
j
-
1
1
-
1
1
-
1
j
-
1
j
1
1
1
1
1
N
W
nk
N
1
1

25
We can now use this DFT matrix to calculate the DFT of any arbitrary
signal of 4 input points using the equation (56.q). Larger DFT matrices
would be needed to compute larger size DFT. What is nice is that these
matrices (of any given size) are always the same and can be pre-
computed and kept in memory to speedup calculation.
Notice from Eq.(56.v) and Eq.(56.w) that only the imaginary components
changed sign. All the real terms kept their sign. Some values of the
matrix are complex conjugate pairs so we don’t even need to calculate
them saving time. This is of course related to that fact that for the
forward transform we need e-j2nk/N and for the reverse we need ej2nk/N.
The difference between these is that the imaginary part changes sign.
(cos + j sin) vs. the (cos - j sin). Hence the Eq.(56.u) for the inverse
we are seeing the imaginary part changing sign but not the real part.

26
Example 56.3. Compute the DFT and IDFT for the signal, x[n] = [2 1 -3
4] using the matrix method.
Solution: We already computed the 4 point DFT. Now we write the
matrix equation using the DFT matrix for N = 4.









































































































j3
5
6
-
j3
-
5
4
2
2
2
2
4
3
-
1
2
j
-
1
-
j
1
1
-
1
1
-
1
j
1
j
1
1
1
1
1
x(3)
x(2)
x(1)
x(0)
j
-
1
-
j
1
1
-
1
1
-
1
j
1
j
1
1
1
1
1
X(3)
X(2)
X(1)
X(0)
4
3
4
3
1
4
3
4
3
1
j
j
j
j
We get the original values back. The DFT
matrix of size (4x4) can be used to determine
the DFT and the iDFT of any discrete signal
of length 4 samples. The generalization of
this idea can be used to compute the DFT and
the iDFT of any length signal using the
matrix method. This is exactly what happens
inside Matlab when asked to execute the
FFT/iFFT function.

27
Now let’s compute the inverse DFT.
1
-















































































 
4
3
-
1
2
j3
5
6
-
j3
-
5
4
j
1
-
j
-
1
1
-
1
1
-
1
j
-
1
j
1
1
1
1
1
X(3)
X(2)
X(1)
X(0)
j
-
1
-
j
1
1
-
1
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Fourier-Series_FT_Laplace-Transform_Letures_Regular_F-for-Students_13.ppt

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