discrete-time-systems and discetre time fourier

UNIT - 8: Implementation of Discrete-time
Systems[?, ?, ?, ?]
Dr. Manjunatha. P
manjup.jnnce@gmail.com
Professor
Dept. of ECE
J.N.N. College of Engineering, Shimoga
October 18, 2016

Unit 8 Syllabus Introduction
Implementation of discrete-time systems::
[?, ?, ?, ?, ?, ?]
Slides are prepared to use in class room purpose, may be used as a
reference material
All the slides are prepared based on the reference material
Most of the figures/content used in this material are redrawn, some
of the figures/pictures are downloaded from the Internet.
This material is not for commercial purpose.
This material is prepared based on Digital Signal Processing for
ECE/TCE course as per Visvesvaraya Technological University (VTU)
syllabus (Karnataka State, India).
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[?, ?, ?, ?]
October 18, 2016 2 / 151

Unit 8 Syllabus Introduction
Unit 8: Implementation of Discrete-time Systems:
PART - B
Realization of FIR system
Direct form structure of FIR system
Linear phase FIR structure
Cascade form structure for FIR system
Frequency sampling structure for FIR system
Lattice structure for FIR system
Realization of IIR system
Direct form structure for IIR system
1 Direct form-I structure of IIR system
2 Direct form-II structure of IIR system
Cascade form structure for IIR system
Parallel form structure for IIR system
Lattice structure of IIR system
October 18, 2016 3 / 151

Implementation of Discrete-time Systems Introduction
Implementation of Discrete-time Systems
System: A system is a physical device which consists of interrelated and interdependent elements
which process the input signal and transform into output signal. Example: Filters, Amplifiers.
Linearity:
If input x1(n) produces response y1(n)
and if x2(n) produces response y2(n) and
ax1(n) + bx2(n) = ay1(n) + by2(n) then the
system is called Linear system.
System
1( )
x n 1( )
y n
System
2 ( )
x n 2 ( )
y n
System
1 2
( ) ( )
ax n bx n
+ 1 2
( ) ( )
ay n by n
+
Figure 1: Linearity
Time invariance: A system is said to be time invariant if its behavior and characteristics does
not change with time.
If input x(n) produces response y(n) then if x(n − n0) produces response y(n − n0), then the
system is called as time invariant
Causality: A system is causal if the output depends only on present and past, but not future
inputs. All memoryless systems are causal.
y[n] = x[n] + x[n − 1] + x[n − 2] . . .
Some of the examples for such analog systems are Oscillator, regulated power supply.,,
October 18, 2016 4 / 151

Realization of FIR system Realization of FIR system
Digital filters are discrete Linear Time Invariant (LTI) systems and described by difference
equations and are implemented in hardware or software.
The discrete time sytems can be of finite impulse response (FIR) or infinite impulse
response IIR type.
A FIR filter is a filter whose impulse response is of finite duration, because it settles to
zero in finite time, because there is no feedback in the FIR system.
The basic components of discrete-time system are, delay element, multiplier, and adder. The
details of these components and their symbols with its input output relationship is as shown in
Figure 2.
1
Z−
( )
x n ( ) ( 1)
y n x n
= −
Delay Element
a
( )
x n ( ) ( )
y n ax n
=
Multiplier
+
2 ( )
x n
1( )
x n
1 2
( ) ( ) ( )
y n x n x n
= +
Adder
Figure 2: Basic Elements
October 18, 2016 5 / 151

Realization of FIR system
The difference equation is a formula for computing an output sample at time n based on past
and present input samples and past output samples in the time domain. The general, causal,
LTI difference equation is as follows:
y(n) = b0x(n) + b1x(n − 1) + . . . bk x(n − k) − a1y(n − 1) − a2y(n − 2) . . . − ak y(n − k)
=
M
X
k=0
bk x(n − k) −
N
X
k=1
ak y(n − k)
where x is the input signal, y is the output signal ak and bk are called the coefficients.
The second term in this equation is usually termed as feedback for the system. This is the
equation used to represent Infinite Impulse Response (IIR) system.
If the feedback term is absent then this equation is used to represent Finite Impulse
Response (FIR) system.
y(n) =
M
X
k=0
bk x(n − k)
October 18, 2016 6 / 151

LTI systems are represented by the following difference equation.
y(n) = −
N
X
k=1
ak y(n − k) +
M
X
k=0
bk x(n − k)
By taking z-transform on both sides
Y (z) = −
N
X
k=1
ak z−k
Y (z) +
M
X
k=0
bk z−k
X(z)
Y (z)
"
1 +
N
X
k=1
ak z−k
#
=
M
X
k=0
bk z−k
X(z)
The system function H(z) is defined as
H(z) =
Y (z)
X(z)
H(z) =
M
P
k=0
bk z−k
1 +
N
P
k=1
ak z−k
October 18, 2016 7 / 151

An FIR system does not have feedback. Hence y(n − k) term is absent in the system. FIR
output is expressed as
y(n) =
M
X
k=0
bk x(n − k)
If there are M coefficients then
y(n) =
M−1
X
k=0
bk x(n − k)
Y (z) =
M
X
k=0
bk z−k
X(z)
System function H(z) is defined as
H(z) =
Y (z)
X(z)
=
M−1
X
k=0
bk z−k
By taking inverse z transform
h(n) =

bn for 0 ≤ n ≤ M − 1
0 otherwise
October 18, 2016 8 / 151

Realization of FIR system Direct form Structure of FIR System
Direct form Structure of FIR System
Structures in which the multiplier coefficients are precisely the coefficients of the transfer
function are called direct form structure.
Since h(n) = bn then y(n) is
y(n) =
M−1
X
k=0
h(k)x(n − k)
Expanding the summation
y(n) = h(0)x(n) + h(1)x(n − 1) + h(2)x(n − 2) + . . . h(M − 1)x(n − M + 1)
Z-1
h(0)
+
Z-1
h(1)
+
Z-1
h(M-2)
+
Z-1
h(2)
+
h(M-1)
( )
x n
+
( ) (0)
x n h
(1) ( 1)
h x n − (2) ( 2)
h x n −
( 1) ( 1)
h M x n M
− − +
( ) (0)
(1) ( 1)
x n h
h x n
+ −
( ) (0)
(1) ( 1)
(2) ( 2)
x n h
h x n
h x n
+ −
+ −
1
0
( ) ( ) ( )
M
k
y n h k x n k
−
=
= −
∑
Figure 3: Direct form realization of FIR system
October 18, 2016 9 / 151

Realize a direct form FIR filter for the following impulse response.
h(n) = δ(n) +
1
2
δ(n − 1) −
1
4
δ(n − 2) + δ(n − 4) +
1
2
δ(n − 3)
Solution: h(n) = δ(n) + 1
2
δ(n − 1) − 1
4
δ(n − 2) + 1
2
δ(n − 3) + δ(n − 4)
H(z) = 1 +
1
2
z−1
−
1
4
z−2
+
1
2
z−3
+ z−4
Y (z) = X(z)H(z) =

1 +
1
2
z−1
−
1
4
z−2
+
1
2
z−3
+ z−4

X(z)
= X(z) +
1
2
z−1
X(z) −
1
4
z−2
X(z) +
1
2
z−3
X(z) + z−4
X(z)
y(n) = x(n) +
1
2
x(n − 1) −
1
4
x(n − 2) +
1
2
x(n − 3) + x(n − 4)
Z-1
+
Z-1 Z-1
+
Z-1
+
y(n)
( )
x n
+
(0) 1
h = (4) 1
h =
1
(1)
2
h =
1
(2)
4
h
−
=
1
(3)
2
h =
( 1)
x n − ( 4)
x n −
( 3)
x n −
( 2)
x n −
October 18, 2016 10 / 151

DEC-2010 EE
Realize the system function H(z) = 1 + 3
2
z−1 + 4
5
z−2 + 5
9
z−3 + 1
9
z−4 using direct form II
Solution:
H(z) = 1 +
3
2
z−1
+
4
5
z−2
+
5
9
z−3
+
1
9
z−4
Y (z) = X(z)H(z) =

1 +
3
2
z−1
+
4
5
z−2
+
5
9
z−3
+
1
9
z−4

X(z)
= X(z) +
3
2
z−1
X(z) +
4
5
z−2
X(z) +
5
9
z−3
X(z) +
1
9
z−4
X(z)
y(n) = x(n) +
3
2
x(n − 1) +
4
5
x(n − 2) +
5
9
x(n − 3) +
1
9
x(n − 4)
Z-1
+
Z-1
+
Z-1
( )
x n
(0) 1
h = 3
(1)
2
h =
4
(2)
5
h =
Z-1
+
+
5
(3)
9
h =
1
(4)
9
h =
y(n)
October 18, 2016 11 / 151

June 2012 EC
A FIR filter is given by y(n) = x[n] + 2
5
x[n − 1] + 3
4
x[n − 2] + 1
3
x[n − 3] draw the direct form.
Solution:
Z-1
+
Z-1
+
Z-1
y(n)
( )
x n
+
(0) 1
h = 2
(1)
5
h =
3
(2)
4
h =
1
(3)
3
h =
October 18, 2016 12 / 151

Determine a direct form realization for the following linear phase filters
h(n) = [1, 2, 3, 4, 3, 2, 1]
Solution:
H(z) = 1 + 2z−1
+ 3z−2
+ 4z−3
+ 3z−4
+ 2z−5
+ 1z−6
]
Y (z) = X(z)H(z) =

1 + 2z−1
+ 3z−2
+ 4z−3
+ 3z−4
+ 2z−5
+ 1z−6

X(z)
= X(z) + 2z−1
X(z) + 3z−2
X(z) + 4z−3
X(z) + 3z−4
X(z) + 2z−5
X(z) + 1z−6
X(z)
y(n) = x(n) + 2x(n − 1) + 3x(n − 2) + 4x(n − 3) + 3x(n − 4) + 2x(n − 5) + 1x(n − 6)
Z-1
+
Z-1
+
Z-1
( )
x n
(0) 1
h = (1) 2
h = (2) 3
h =
Z-1
+
+
(3) 4
h =
Z-1
+
(5) 2
h =
(4) 3
h =
Z-1
+
y(n)
(6) 1
h =
October 18, 2016 13 / 151

Determine a direct form realization for the following linear phase filters
h(n) = [1, 2, 3, 3, 2, 1]
Solution:
H(z) = 1 + 2z−1
+ 3z−2
+ 3z−3
+ 2z−4
+ 1z−5
]
Y (z) = X(z)H(z) =

1 + 2z−1
+ 3z−2
+ 3z−3
+ 2z−4
+ 1z−5

X(z)
= X(z) + 2z−1
X(z) + 3z−2
X(z) + 3z−3
X(z) + 2z−4
X(z) + 1z−5
X(z)
y(n) = x(n) + 2x(n − 1) + 3x(n − 2) + 3x(n − 3) + 2x(n − 4) + 1x(n − 5)+
Z-1
+
Z-1
+
Z-1
( )
x n
(0) 1
h = (1) 2
h = (2) 3
h =
Z-1
+
+
(3) 3
h =
Z-1
+
(5) 1
h =
(4) 2
h =
y(n)
October 18, 2016 14 / 151

For the following FIR filter system function sketch a direct form
H(z) = 1 + 2.88z−1
+ 3.4048z−2
+ 1.74z−3
+ 0.4z−4
Solution:
Z-1
+
Z-1
+
Z-1
( )
x n
(0) 1
h = (1) 2.88
h = (2) 3.4048
h =
Z-1
+
+
(3) 1.74
h = (4) 0.4
h =
y(n)
October 18, 2016 15 / 151

Realize direct form FIR filter with impulse response h(n) is given
h(n) = 4δ(n) + 5δ(n − 1) + 6δ(n − 2) + 7δ(n − 3). With input x(n) = [1, 2, 3] calculate output
y(n)
Solution:
h(n) = 4δ(n) + 5δ(n − 1) + 6δ(n − 2) + 7δ(n − 3)
H(z) = 4 + 5z−1
+ 6z−2
+ 7z−3
Y (z) = X(z)H(z) =

4 + 5z−1
+ 6z−2
+ 7z−3

X(z)
= 4X(z) + 5z−1
X(z) + 6z−2
X(z) + 7z−3
X(z)
y(n) = 4x(n) + 5x(n − 1) + 6x(n − 2) + 7x(n − 3)
0 1 2 3
2
3
n
1
( )
x n
Figure 4: Input x(n) to the FIR filter
Z-1
+
Z-1
+
Z-1
( )
x n
(0) 4
h = (1) 5
h = (2) 6
h =
+
(3) 7
h =
y(n)
( 1)
x n − ( 3)
x n −
( 2)
x n −
October 18, 2016 16 / 151

y(n) = 4x(n) + 5x(n − 1) + 6x(n − 2) + 7x(n − 3)
y(0) = 4x(0) + 5x(0 − 1) + 6x(0 − 2) + 7x(0 − 3) = 4 × 1 = 4
y(1) = 4x(1) + 5x(0) = 4 × 2 + 5 × 1 = 13
y(2) = 4x(2) + 5x(1) + 6x(0) = 4 × 3 + 5 × 2 + 6 × 1 = 28
y(3) = 4x(3) + 5x(2) + 6x(1) + 7x(0) = 4 × 0 + 5 × 3 + 6 × 2 + 7 × 1 = 34
y(4) = 4x(4) + 5x(3) + 6x(2) + 7x(1) = 0 + 0 + 6 × 3 + 7 × 2 = 32
y(5) = 4x(5) + 5x(4) + 6x(3) + 7x(2) = 0 + 0 + 0 + 7 × 3 = 21
y(n) = [4, 13, 28, 34, 32, 21]
0 1 2 3 4 5
n
( )
y n
4
13
21
34 32
28
Figure 5: Output y(n) of the FIR
filter
Z-1
+
Z-1
+
Z-1
( )
x n
(0) 4
h = (1) 5
h = (2) 6
h =
+
(3) 7
h =
y(n)
( 1)
x n − ( 3)
x n −
( 2)
x n −
October 18, 2016 17 / 151

June 2015 Obtain the direct form realization of linear phase FIR system given by
H(z) = 1 +
2
3
z−1
+
15
8
z−2
Solution:
Z-1
+
Z-1
+
( )
x n
(0) 1
h = 2
(1)
3
h =
15
(2)
8
h =
y(n)
October 18, 2016 18 / 151

Realization of FIR system Linear Phase FIR structure
Linear Phase FIR structure
Linear phase is a property of a
filter, where the phase response
of the filter is a linear function
of frequency. The result is that
all frequency components of the
input signal are shifted in time
(usually delayed) by the same
constant amount, which is
referred to as the phase delay.
And consequently, there is no
phase distortion due to the time
delay of frequencies relative to
one another.
Linear-phase filters have a
symmetric impulse response.
The FIR filter has linear phase if
its unit sample response satisfies
the following condition:
h(n) = h(M−1−n) n = 0, 1, 2, . . . , M−1
0 1 2 3 4 5 6 7 8 n
h[n]
-2
-1
0
1
2
3
4
Center of Symmetry
0 1 2 3 4 5 6 7 8 n
h[n]
-2
-1
0
1
2
3
4
Center of Symmetry
0 1 2 3 4 5 6 7 8 n
h[n]
-2
-1
0
1
2
3
4
Center of Symmetry
0 1 2 3 4 5 6 7 8 n
h[n]
-2
-1
0
1
2
3
4
Center of Symmetry
Symmetry: h(n)=h(M-1-n) Odd M Symmetry: h(n)=h(M-1-n) Even M
Antisymmetry: h(n)=-h(M-1-n) Odd M Antisymmetry: h(n)=-h(M-1-n) Even M
October 18, 2016 19 / 151

The Z transform of the unit sample response is expressed as H(z) =
M−1
P
n=0
h(n)z−n
For even M h(n) = h(M − 1 − n)
H(z) =
M/2−1
X
n=0
h(n)
h
z−n
+ z−(M−1−n)
i
The system expression is
H(z) =
Y (z)
X(z)
Y (z)
X(z)
=
M/2−1
X
n=0
h(n)
h
z−n
+ z−(M−1−n)
i
Y (z) =
M/2−1
X
n=0
h(n)
h
z−n
+ z−(M−1−n)
i
X(z)
By expanding the summation
Y (z) = h(0)
h
1 + z−(M−1)
i
X(z) + h(1)
h
z−1
+ z−(M−2)
i
X(z) + . . .
+ h(M/2 − 1)
h
z−(M/2−1)
+ z−(M/2)
i
X(z)
October 18, 2016 20 / 151

y(n) = h(0)[x(n) + x(n − (M − 1))] + h(1)[x(n − 1) + x(n − (M − 2))] + . . .
+ h(M/2 − 1)x [n − (M/2 − 1)] + x [n − (M/2)]
By considering M=8 then y(n) is
y(n) = h(0)[x(n) + x(n − 7)] + h(1)[x(n − 1) + x(n − 6)] + . . .
+ h(2){x(n − 2) + x(n − 5)} + h(3){x(n − 3) + x(n − 4)}
Z-1
( )
x n
Z-1
+
Z-1
Z-1
+
+
+
Z-1
Z-1
+ Z-1
+
+
( )
y n
h(1) h(2) h(3)
h(0)
( 1)
x n − ( 2)
x n − ( 3)
x n −
( 1)
x n −
( 6)
x n − ( 5)
x n − ( 4)
x n −
( 7)
x n −
October 18, 2016 21 / 151

For Odd M h(n) = h(M − 1 − n)
H(z) = h

M − 1
2

z
−

M−1
2

+
(M−3)/2
X
n=0
h(n)
h
z−n
+ z−(M−1−n)
i
H(z) =
Y (z)
X(z)
Y (z)
X(z)
= h

M − 1
2

z
−

M−1
2

+
(M−3)/2
X
n=0
h(n)
h
z−n
+ z−(M−1−n)
i
Y (z) = h

M − 1
2

z

− M−1
2

X(z) +
(M−3)/2
X
n=0
h(n)
h
z−n
+ z−(M−1−n)
i
X(z)
By expanding the summation
Y (z) = h

M − 1
2

z
−

M−1
2

X(z) + h(0)
h
1 + z−(M−1)
i
X(z)
= +h(1)
h
z−1
+ z−(M−2)
i
X(z) + . . .
October 18, 2016 22 / 151

y(n) = h

M − 1
2

x

n −
M − 1
2

+ h(0)[x(n) + x(n − (M − 1))] +
+ h(1)[x(n − 1) + x(n − (M − 2))] + . . .
+ h

M − 3
2

x

n −
M − 3
2

+ x

n −
M + 1
2

for M=9
y(n) = h(4)x(n − 4) + h(0)[x(n) + x(n − 8)] + h(1)[x(n − 1) + x(n − 7)] + . . .
+ h(2){x(n − 2) + x(n − 6)} + h(3){x(n − 3) + x(n − 5)}
Z-1
( )
x n
Z-1
+
Z-1
Z-1
+
+
+
Z-1
Z-1
+
Z-1
+
+
( )
y n
h(1) h(2) h(3)
h(0)
( 1)
x n − ( 2)
x n − ( 3)
x n −
( 1)
x n −
( 6)
x n − ( 5)
x n −
( 4)
x n −
( 7)
x n −
Z-1
h(4)
( 8)
x n −
Figure 7: Linear phase FIR structure
October 18, 2016 23 / 151

Realize a linear phase FIR filter with the following impulse response. Give necessary equations
h(n) = δ(n) + 1
2
δ(n − 1) − 1
4
δ(n − 2) + δ(n − 4) + 1
2
δ(n − 3)
Solution: h(n) = {1, 1
2
, −1/4, 1
2
, 1}. Here M=5 h(0) = h(4), h(1) = h(3)
h(n) = δ(n) +
1
2
δ(n − 1) −
1
4
δ(n − 2) +
1
2
δ(n − 3) + δ(n − 4)
H(z) = 1 +
1
2
z−1
−
1
4
z−2
+
1
2
z−3
+ z−4
Y (z) = X(z)H(z) =

1 +
1
2
z−1
−
1
4
z−2
+
1
2
z−3
+ z−4

X(z)
Y (z) = X(z) +
1
2
z−1
X(z) −
1
4
z−2
X(z) +
1
2
z−3
X(z) + z−4
X(z)
y(n) = x(n) +
1
2
x(n − 1) −
1
4
x(n − 2) +
1
2
x(n − 3) + x(n − 4)
y(n) = [x(n) + x(n − 4)] +
1
2
[x(n − 1) + x(n − 3)] −
1
4
x(n − 2)
Z-1
( )
x n
Z-1
+ +
+ +
( )
y n
h(1)=1/2 h(3)=-1/4
h(0)=1
( 1)
x n − ( 2)
x n −
( 3)
x n −
( 4)
x n −
Z-1
Z-1
October 18, 2016 24 / 151

DEC 2010,2011, 2012 Realize a linear phase FIR filter having the following impulse response
h(n) = δ(n) + 1
4
δ(n − 1) − 1
8
δ(n − 2) + 1
4
δ(n − 3) + δ(n − 4)
4
, −1/8, +1
4
, 1}. Here M=5 h(0) = h(4), h(1) = h(3)
h(n) = δ(n) +
1
4
δ(n − 1) −
1
8
δ(n − 2) +
1
4
δ(n − 3) + δ(n − 4)
H(z) = 1 +
1
4
z−1
−
1
8
z−2
+
1
4
z−3
+ z−4
Y (z) = X(z)H(z) =

1 +
1
4
z−1
−
1
8
z−2
+
1
4
z−3
+ z−4

X(z)
Y (z) = X(z) +
1
4
z−1
X(z) −
1
8
z−2
X(z) +
1
4
z−3
X(z) + z−4
X(z)
y(n) = x(n) +
1
4
x(n − 1) −
1
8
x(n − 2) +
1
4
x(n − 3) + x(n − 4)
y(n) = [x(n) + x(n − 4)] +
1
4
[x(n − 1) + x(n − 3)] −
1
8
x(n − 2)
Z-1
( )
x n
Z-1
+ +
+ +
( )
y n
1
( 1)
x n − ( 2)
x n −
( 3)
x n −
( 4)
x n −
Z-1
Z-1
1
8
−
1
4
October 18, 2016 25 / 151

May 2010 Realize a linear phase FIR filter having the following impulse response
h(n) = δ(n) − 1
4
δ(n − 1) + 1
2
δ(n − 2) + 1
2
δ(n − 3) − 1
4
δ(n − 4) + δ(n − 5)
Solution: h(n) = {1, −1
4
, 1
2
, 1
2
, −1
4
, 1}. Here M=6 h(0) = h(5), h(1) = h(4), h(2) = h(3)
h(n) = δ(n) −
1
4
δ(n − 1) +
1
2
δ(n − 2) +
1
2
δ(n − 3) −
1
4
δ(n − 4) + δ(n − 5)
H(z) = 1 −
1
4
z−1
+
1
2
z−2
+
1
2
z−3
−
1
4
z−4
+ z−5
Y (z) = X(z)H(z) =

1 −
1
4
z−1
+
1
2
z−2
+
1
2
z−3
−
1
4
z−4
+ z−5

X(z)
Y (z) = X(z) −
1
4
z−1
X(z) +
1
2
z−2
X(z) +
1
2
z−3
X(z) −
1
4
z−4
X(z) + z−5
X(z)
y(n) = x(n) −
1
4
x(n − 1) +
1
2
x(n − 2) +
1
2
x(n − 3) −
1
4
x(n − 4) + x(n − 5)
y(n) = [x(n) + x(n − 5)] −
1
4
[x(n − 1) + x(n − 4)] +
1
2
[x(n − 2) + x(n − 3)]
Z-1
( )
x n
Z-1
+
Z-1
Z-1
+
+
+ Z-1
+
+
( )
y n
1
( 1)
x n − ( 2)
x n −
( 3)
x n −
( 4)
x n −
( 5)
x n −
1
4
−
1
2
October 18, 2016 26 / 151

May 2010
Obtain the direct form Realization of linear phase FIR system given by
H(z) = 1 + 2
3
z−1 + 15
8
z−2 + 2
3
z−3 + z−4
3
, 15/8, 2
3
, 1}. Here M=5 h(0) = h(4), h(1) = h(3)
H(z) = 1 +
2
3
z−1
+
15
8
z−2
+
2
3
z−3
+ z−4
Y (z) = X(z)H(z) =

1 +
2
3
z−1
+
15
8
z−2
+
2
3
z−3
+ z−4

X(z)
Y (z) = X(z) +
2
3
z−1
X(z) +
15
8
z−2
X(z) +
2
3
z−3
X(z) + z−4
X(z)
y(n) = x(n) +
2
3
x(n − 1) +
15
8
x(n − 2) +
2
3
x(n − 3) + x(n − 4)
y(n) = [x(n) + x(n − 4)] +
2
3
[x(n − 1) + x(n − 3)] +
15
8
x(n − 2)
Z-1
( )
x n
Z-1
+ +
+ +
( )
y n
1
( 1)
x n − ( 2)
x n −
( 3)
x n −
( 4)
x n −
Z-1
Z-1
15
8
2
3
October 18, 2016 27 / 151

Realize the following system function by linear phase FIR structure
H(z) =
2
3
z + 1 +
2
3
z−1
Solution:
H(z) = 1 +
2
3
(z + z−1
)
Y (z) = X(z)H(z) =

1 +
2
3
(z + z−1
)

X(z)
Y (z) = X(z) +
2
3
(z + z−1
)X(z)
y(n) = x(n) +
2
3
[x(n − 1) + x(n + 1)]
( )
x n
Z-1
+
( 1)
x n −
( 1)
x n +
Z
+
2
3
( )
y n
Figure 8: Linear phase FIR structure
October 18, 2016 28 / 151

Realize the following system function by linear phase FIR structure
H(z) = 1 +
z−1
4
+
z−2
4
+ z−3
Solution:
H(z) = 1 +
1
4
(z−1
+ z−2
) + z−3
Y (z) = X(z)H(z) =

1 +
1
4
(z−1
+ z−2
) + z−3

X(z)
= X(z) +
1
4
(z−1
X(z) + z−2
X(z)) + z−3
X(z)
y(n) = [x(n) + x(n − 3)] +
1
4
[x(n − 1) + x(n − 2)]
Z-1
( )
x n
Z-1
+ +
+
Z-1
( )
y n
h(0)=1
( 1)
x n −
( 2)
x n −
( 3)
x n −
1
(1)
4
h =
October 18, 2016 29 / 151

Realize the following system function by linear phase FIR structure: h(n) = [1, 2, 3, 4, 3, 2, 1]
Solution:
H(z) = 1 + 2z−1
+ 3z−2
+ 4z−3
+ 3z−4
+ 2z−5
+ 1z−6
Y (z) = X(z)H(z) =

1 + 2z−1
+ 3z−2
+ 4z−3
+ 3z−4
+ 2z−5
+ 1z−6

X(z)
= X(z) + 2z−1
X(z) + 3z−2
X(z) + 4z−3
X(z) + 3z−4
X(z) + 2z−5
X(z) + 1z−6
X(z)
y(n) = x(n) + 2x(n − 1) + 3x(n − 2) + 4x(n − 3) + 3x(n − 4) + 2x(n − 5) + 1x(n − 6)
y(n) = 1[x(n) + x(n − 6)] + 2[x(n − 1) + x(n − 5)] + 3[x(n − 2) + x(n − 4)] + 4x(n − 3)
Z-1
( )
x n
Z-1
+
Z-1
Z-1
+
+
+
Z-1
Z-1
+
+
( )
y n
2 3 4
1
( 1)
x n − ( 2)
x n − ( 3)
x n −
( 5)
x n − ( 4)
x n −
( 6)
x n −
October 18, 2016 30 / 151

Realize a direct form for the following linear phase filters
h(n) = [1, 2, 3, 3, 2, 1]
Solution:
H(z) = 1 + 2z−1
+ 3z−2
+ 3z−3
+ 2z−4
+ 1z−5
]
Y (z) = X(z)H(z) =

1 + 2z−1
+ 3z−2
+ 3z−3
+ 2z−4
+ 1z−5

X(z)
= X(z) + 2z−1
X(z) + 3z−2
X(z) + 3z−3
X(z) + 2z−4
X(z) + 1z−5
X(z)
y(n) = x(n) + 2x(n − 1) + 3x(n − 2) + 3x(n − 3) + 2x(n − 4) + 1x(n − 5)
y(n) = 1[x(n) + x(n − 5)] + 2[x(n − 1) + x(n − 4)] + 3[x(n − 2) + x(n − 3)]
Z-1
( )
x n
Z-1
+
Z-1
Z-1
+
+
+ Z-1
+
+
( )
y n
1
( 1)
x n − ( 2)
x n −
( 3)
x n −
( 4)
x n −
( 5)
x n −
2 3
October 18, 2016 31 / 151

Realization of FIR system Frequency Sampling for FIR Systems
Frequency Sampling for FIR Systems
October 18, 2016 32 / 151

Frequency sampling realization is used when an FIR filter is to operate on some desired
frequency.
The desired frequency may be defined and this reduces the complexity of the system.
Consider a frequency ω
ωk =
2π
M
k k = 0, 1, . . . M − 1
where ωk is the frequency at discrete points.
Let the unit sample response of FIR system be h(n). The fourier transform of the h(n) is
defined as
H(ω) =
M−1
X
n=0
h(n)e−jωn
H(ω) at ω=ωk = 2π
M
k
H(ωk ) = H

2π
M
k

=
M−1
X
n=0
h(n)e−j2πkn/M
H(ωk ) is also written as H(k) and defined as
H(k) =
M−1
X
n=0
h(n)e−j2πkn/M
k = 0, 1, . . . M − 1
October 18, 2016 33 / 151

This equation represents the M point DFT of h(n) and it is defined as
h(n) =
1
M
M−1
X
n=0
H(k)ej2πkn/M
n = 0, 1, . . . M − 1
z transform is defined as
H(z) =
M−1
X
n=0
h(n)z−n
substituting the value of h(n)
H(z) =
M−1
X
n=0
1
M
M−1
X
n=0
H(k)ej2πkn/M
!
z−n
Interchanging the order of summations in the above equation
H(z) =
M−1
X
n=0
H(k)
1
M
M−1
X
n=0
ej2πk/M
z−1
!n
October 18, 2016 34 / 151

The geometric series formula is
N
X
n=0
an
=
1 − aN+1
1 − a
Using the formula where a = e2πkn/M z−1
H(z) =
M−1
X
k=0
H(k)
1
M
1 − ej2πk/M z−1
M
=
M−1
X
k=0
H(k)
1
M
1 − ej2πk z−M
ej2πk = cos(2πk) + jsin(2πk) = 1
October 18, 2016 35 / 151

H(z) =
M−1
X
k=0
H(k)
1
M
1 − z−M
H(z) =
1 − z−M
M
M−1
X
k=0
H(k)
The equation can be considered as multiplication of two systems and defined as
H(z) = H1(z).H2(z)
where
H1(z) =
1 − z−M
M
H2(z) =
M−1
X
k=0
H(k)
H1(z) and H2(z) are realized independently. H(z) is obtained by multiplication of H1(z)
and H2(z).
October 18, 2016 36 / 151

H(z) =
1 − z−M
M
M−1
X
k=0
H(k)
Z-M
+
( )
x n
Z-1
+ +
H(2)
4 /
j M
e π
Z-1
+ +
H(0)
Z-1
+ +
H(1)
2 /
j M
e π
Z-1
+ +
H(M-1)
2 ( 1)/
j M M
e π −
( )
y n
Realization
of H1
(z)
1
M
Realization
of H2
(z)
1
-
Figure 9: Frequency sampling FIR structure
October 18, 2016 37 / 151

Determine the transfer function H(z) of an FIR filter to implement
h(n) = δ(n) + 2δ(n − 1) + δ(n − 2). Using frequency sampling technique
Solution:
h(0) = 1, h(1) = 2, h(2) = 1 The DFT of h(n)
H(k) =
M−1
X
n=0
h(n)e−j2πkn/M
With M=3
H(k) =
2
X
n=0
h(n)e−j2πkn/3
H(k) =
2
X
n=0
h(n)e−j2πkn/3
= h(0) + h(1)e−j2πk/3
+ h(2)e−j4πk/3
k=0,1,2
H(0) = 1 + 2 + 1 = 4
H(1) = 1 + 2e−j2π/3
+ e−j4π/3
= −0.5 − j0.866 = e−j2π/3
H(2) = 1 + 2e−j4π/3
+ e−j8π/3
= −0.5 − j0.866 = e−j2π/3
October 18, 2016 38 / 151

The system function is
H(z) =
1 − z−M
M
M−1
X
k=0
H(k)
1 − e2πk/M z−1
For M=3
H(z) =
1 − z−3
3
2
X
k=0
H(k)
1 − e2πk/3z−1
=
1 − z−3
3

H(0)
1 − z−1
+
H(1)
1 − e−j2π/3z−1
+
H(2)
1 − e−j4π/3z−1

= H1(z) × H2(z)
where
H1(z) =
1 − z−3
3
H2(z) =

4
1 − z−1
+
e−j2π/3
1 − e−j2π/3z−1
+
e−j4π/3
1 − e−j4π/3z−1
#
October 18, 2016 39 / 151

Z-1
+ +
4 /3
j
e π
Z-1
+ +
4
1
Z-1
+ +
4 /3
j
e π
−
Z-3
+
( )
x n
1
3
2 /3
j
e π
−
2 /3
j
e π
( )
y n
Figure 10: Frequency sampling FIR structure
October 18, 2016 40 / 151

Realization of IIR system Realization of IIR system
Realization of IIR system
Direct form structure for IIR system
1 Direct form-I structure of IIR system
2 Direct form-II structure of IIR system
Cascade form structure for IIR system
Parallel form structure for IIR system
Lattice structure of IIR system
An Infinite Impulse Response (IIR) filters are digital filters with infinite impulse response.
Unlike FIR filters, they have the feedback (a recursive part of a filter) and are known as
recursive digital filters therefore.
IIR filters are computationally more efficient than FIR filters as they require fewer
coefficients due to the fact that they use feedback.
If the coefficients deviate from their true values then the feedback can make the filter
unstable.
October 18, 2016 41 / 151

Realization of IIR system Realization of IIR system
LTI systems are represented by the following
difference equation.
y(n) = −
N
X
k=1
ak y(n − k) +
M
X
k=0
bk x(n − k)
Y (z) = −
N
X
k=1
ak z−k
Y (z) +
M
X
k=0
bk z−k
X(z)
Y (z)

1 +
N
X
k=1
ak z−k
#
=
M
X
k=0
bk z−k
X(z)
The system function H(z) is defined as
H(z) =
Y (z)
X(z)
=
M
P
k=0
bk z−k
1 +
N
P
k=1
ak z−k
The general expression of an IIR
filter can be expressed as follows:
H(z) =
M
P
k=0
bk z−k
1 +
N
P
k=1
ak z−k
H1(z) =
M
X
k=0
bk z−k
H2(z) =
1
1 +
N
P
k=1
ak z−k
H(z) = H1(z).H2(z)
October 18, 2016 42 / 151

Realization of IIR system Direct form-I Structure of IIR System
Direct form Structure of IIR System
Structures in which the multiplier coefficients are precisely the coefficients of the transfer
function are called direct form structure.
The direct form structures for H1(z) H2(z) are
H1(z) =
M
X
k=0
bk z−k
= b0 + b1z−1
+ . . . bM z−M
H1(z) is defined as
H1(z) =
Y1(z)
X1(z)
Y1(z) = b0X1(z)+b1z−1
X1(z)+. . . bM z−M
X1(z)
Its inverse z transform is
y1(n) = b0x1(n)+b1x1(n−1)+. . . bM x1(n−M)
Z-1
+
b0
Z-1
+
1( )
y n
b1
+
b2
Z-1
( )
x n
bM
October 18, 2016 43 / 151

H2(z) is the all pole system and is given by
H2(z) =
1
1 +
N
P
k=1
ak z−k
H2(z) is also expressed in terms of system function
H2(z) =
Y2(z)
X2(z)
=
1
1 +
N
P
k=1
ak z−k
Y2(z)[1 +
N
X
k=1
ak z−k
] = X2(z)
Y2(z) = −
N
X
k=1
ak z−k
Y2(z) + X2(z)
Z-1
+
Z-1
+
-a1
+
-a2
Z-1
-aN
2 ( )
x n 2 ( )
y n
Expanding the above function
Y2(z) = −a1z−1
Y2(z) − a2z−2
Y2(z) − a3z−3
Y2(z) + . . . − aN z−N
Y2(z) + X2(z)
Its inverse z transform is
y2(n) = −a1y2(n − 1) − a2y2(n − 2) − a3y2(n − 3) − . . . − aN y2(n − N) + x2(n)
October 18, 2016 44 / 151

1
1
1
( )
( )
( )
Y z
H z
X z
= 2
2
2
( )
( )
( )
Y z
H z
X z
=
1
( ) ( )
x n x n
= 1 2
( ) ( )
y n x n
= 2 ( ) ( )
y n y n
=
Figure 11: Direct form I realization of IIR system
Z-1
+
Z-1
+
-a1
Z-1
+
-a2
Z-1
+
-a3
+
-aN-1
-aN
2 ( )
x n 2 ( ) ( )
y n y n
=
Z-1
+
b0
Z-1
+
1( )
y n
b1
Z-1
+
b2
Z-1
+
1
( ) ( )
x n x n
=
b3
+
bM-1
bM
All zero system All pole system
Figure 12: Direct form I realization of IIR system
October 18, 2016 45 / 151

Realization of IIR system Direct form II structure for IIR system
Direct form-II Structure of IIR System
Direct form II is also called as canonic form because the number delay elements is same as
the order of difference equation.
H(z) =
M
P
k=0
bk z−k
1 +
N
P
k=1
ak z−k
H(z) =
Y (z)
X(z)
=
Y (z)
X(z)
W (z)
W (z)
=
W (z)
X(z)
Y (z)
W (z)
= H1(z).H2(z)
H1(z) =
W (z)
X(z)
=
1
1 +
N
P
k=1
ak z−k
W (z)[1 +
N
X
k=1
ak z−k
] = X(z)
W (z) = X(z) −
N
X
k=1
ak z−k
W (z)
= X(z) − a1z−1
W (z) − a2z−2
W (z) − − . . . aN z−N
W (z)
w(n) = x(n) − a1w(n − 1) − a2w(n − 2) − . . . − aN w(n − N)
October 18, 2016 46 / 151

H2(z) =
Y (z)
W (z)
=
M
X
k=0
bk z−k
Y (z) =
M
X
k=0
bk z−k
W (z)
= b0W (z) + b1z−1
W (z) + b2z−2
W (z) + . . . + bM z−M
W (z)
By inverse z transform
y(n) = b0w(n) + b1w(n − 1) + b2w(n − 2) + . . . + bM w(n − M)
Z-1
+
Z-1
+
-a1
+
-a2
Z-1
-aN
( )
x n ( )
w n
Z-1
+
b0
Z-1
+
( )
y n
b1
+
b2
Z-1
( )
w n
bM
October 18, 2016 47 / 151

1
( )
( )
( )
W z
H z
X z
= 2
( )
( )
( )
Y z
H z
W z
=
( )
x n ( )
w n ( )
y n
Figure 13: Cascade connection of H1(z) H2(z)
( )
w n
( )
x n
All zero system All pole system
Z-1
+
Z-1
+
-a1
Z-1
+
-a2
Z-1
+
-a3
+
-aN-1
-aN
( )
y n
Z-1
+
b0
Z-1
+
b1
Z-1
+
b2
Z-1
+
b3
+
bM-1
bM
Figure 14: Direct form-II Structure
( )
w n
( )
x n
Z-1
+
Z-1
+
-a1
Z-1
+
-a2
Z-1
+
-a3
+
-aN-1
-aN
( )
y n
+
b0
+
b1
+
b2
+
b3
+
bM-1
bM
October 18, 2016 48 / 151

A difference equation describing a filter is given below
y(n) −
3
4
y(n − 1) +
1
8
y(n − 2) = x(n) +
1
2
x(n − 1)
Draw the direct form I and direct form II structures
Solution:
y(n) =
3
4
y(n − 1) −
1
8
y(n − 2) + x(n) +
1
2
x(n − 1)
( )
y n
Z-1
+
1
0.5
( )
x n
Z-1
+
Z-1
+
1
8
−
3
4
Figure 16: Direct form-I
( )
w n
( )
x n
Z-1
+
Z-1
+
( )
y n
+
1
+
3
4
1
8
−
1
2
Figure 17: Direct form-II
October 18, 2016 49 / 151

[July 2013]:
y(n) = −0.1y(n − 1) + 0.2y(n − 2) + 3x(n) + 3.6x(n − 1) + 0.6x(n − 2)obtain the the direct
form I and direct form II structures
Solution:
y(n) = −0.1y(n − 1) + 0.2y(n − 2) + 3x(n) + 3.6x(n − 1) + 0.6x(n − 2)
( )
y n
Z-1
+
3
3.6
( )
x n
Z-1
+
Z-1
+
0.2
0.1
−
Z-1
+
0.6
( )
w n
( )
x n
Z-1
+
Z-1
+
-0.1
0.2
( )
y n
+
3
+
3.6
0.6
October 18, 2016 50 / 151

June 2010 EC
Obtain direct form I and direct form II for the system described by
y(n) = −0.1y(n − 1) + 0.72y(n − 2) + 0.7x(n) − 0.252x(n − 2)
Solution:
y(n) = −0.1y(n − 1) + 0.72y(n − 2) + 0.7x(n) − 0.252x(n − 2)
( )
y n
Z-1
+
0.7
( )
x n
Z-1
+
Z-1
+
0.1
−
0.72
( 1)
y n −
( 2)
y n −
( 1)
x n −
Z-1
-0.252
( 2)
x n −
( )
w n
( )
x n
Z-1
+
Z-1
+
-0.1
0.72
( )
y n
+
0.7
-0.252
October 18, 2016 51 / 151

A system is represented by transfer function H(z) is given by
H(z) = 3 +
4z
z − 1
2
−
2
z − 1
4
(i) Does this H(z) represent a FIR or IIR filter ? Why ?
(ii) Give a difference equation realization of this system using direct form I
(iii) Draw the block diagram for the direct form II canonic realization, and give the
governing equations for implementation
Solution:
H(z) = 3 +
4z
z − 1
2
−
2
z − 1
4
= 3 +
4z
z − 0.5
−
2
z − 0.25
=
7z2 − 5.25z + 1.375
z2 − 0.75z + 0.125
(i): By observing the system function it has numerator of polynomial of order 2 as well as
denominator of polynomial of order 2. The system function has poles as well zeros, hence it
represents IIR filter.
October 18, 2016 52 / 151

(ii): Direct form I
H(z) =
7z2 − 5.25z + 1.375
z2 − 0.75z + 0.125
=
7 − 5.25z−1 + 1.375z−2
1 − 0.75z−1 + 0.125z−2
=
Y (Z)
X(Z)
Y (Z)[1 − 0.75z−1
+ 0.125z−2
] = 7X(Z) − 5.25z−1
X(Z) + 1.375z−2
X(Z)
y(n) − 0.75y(n − 1) + 0.125y(n − 2) = 7x(n) − 5.25x(n − 1) + 1.375x(n − 2)
y(n) = 0.75y(n − 1) − 0.125y(n − 2) + 7x(n) − 5.25x(n − 1) + 1.375x(n − 2)
( )
y n
Z-1
+
Z-1
+
7
-5.25
1.375
( )
x n
Z-1
+
Z-1
+
-1.25
0.75
Figure 22: Direct form-I Structure
October 18, 2016 53 / 151

(iii): Direct form II canonic realization
H(z) =
7 − 5.25z−1 + 1.375z−2
1 − 0.75z−1 + 0.125z−2
=
Y (z)
X(z)
H(z) =
Y (z)
W (z)
W (z)
X(z)
=
7 − 5.25z−1 + 1.375z−2
1 − 0.75z−1 + 0.125z−2
H(z) = H1(z).H2(z) =
1
1 − 0.75z−1 + 0.125z−2
[7 − 5.25z−1
+ 1.375z−2
]
H1(z) =
W (z)
X(z)
=
1
1 − 0.75z−1 + 0.125z−2
W (z)[1 − 0.75z−1
+ 0.125z−2
] = X(z)
W (z) = X(z)+0.75z−1
W (z)−0.125z−2
W (z)
Taking inverse z transform
w(n) = x(n) + 0.75w(n − 1) − 0.125w(n − 2)
( )
w n
( )
x n
Z-1
+
Z-1
+
-1.25
0.75
October 18, 2016 54 / 151

H2(z) =
Y (z)
W (z)
= 7 − 5.25z−1
+ 1.375z−2
Y (z)
W (z)
= 7 − 5.25z−1
+ 1.375z−2
Y (z) = 7W (z) − 5.25z−1
W (z) + 1.375z−2
W (z)
y(n) = 7w(n) − 5.25w(n − 1) + 1.375w(n − 2)
( )
w n ( )
y n
Z-1
+
Z-1
+
7
-5.25
1.375
Figure 23: Direct form of H2(z)
( )
w n
( )
x n
Z-1
+
Z-1
+
-1.25
( )
y n
Z-1
+
Z-1
+
7
-5.25
0.75
1.375
Figure 24: Realization of H(z) = H1(z).H2(z)
( )
w n
( )
x n
Z-1
+
Z-1
+
0.75
-0.125
( )
y n
+
7
+
-5.25
1.325
Figure 25: Direct form-II, canonic form
October 18, 2016 55 / 151

2012 DEC, 2012 June
Obtain the direct form II (canonic) realization for the following system.
H(z) =
(z − 1)(z2 + 5z + 6)(z − 3)
(z2 + 6z + 5)(z2 − 6z + 8)
Solution:
H(z) =
(z − 1)(z2 + 5z + 6)(z − 3)
(z2 + 6z + 5)(z2 − 6z + 8)
=
(z2 − 4z + 3)(z2 + 5z + 6)
(z2 + 6z + 5) + (z2 − 6z + 8)
=
z4 + z3 − 11z2 − 9z + 18
z4 − 23z2 + 18z + 40
=
1 + z−1 − 11z−2 − 9z−3 + 18z−4
1 − 23z−2 + 18z−3 + 40z−4
( )
w n
( )
x n
Z-1
+
Z-1
( )
y n
+
-11
+
Z-1
+
-9
+
1
Z-1
+
40
−
23
18
−
+
18
October 18, 2016 56 / 151

2011 July
Obtain the direct form II realizations of the following system.
H(z) =
(1 + z−1)
(1 − 1
4
z−1)(1 − z−1 + 1
2
z−2)
Solution:
H(z) =
(1 + z−1)
(1 − 5
4
z−1 + 3
4
z−2 − 1
8
z−3)
=
1
(1 − 5
4
z−1 + 3
4
z−2 − 1
8
z−3)
×
(1 + z−1)
1
= H1(z) × H2(z)
H1(z) =
W (z)
X(z)
=
1
(1 − 5
4
z−1 + 3
4
z−2 − 1
8
z−3)
W (z)[1 −
5
4
z−1
+
3
4
z−2
−
1
8
z−3
] = X(z)
W (z) = X(z) +
5
4
z−1
W (z) −
3
4
z−2
W (z) +
1
8
z−3
W (z)
w(n) = x(n) +
5
4
w(n − 1) −
3
4
w(n − 2) +
1
8
w(n − 3)
October 18, 2016 57 / 151

H2(z) = (1 + z−1)
H2(z) =
Y (z)
W (z)
= (1 + z−1
)
Y (z) = W (z) + z−1
W (z)
y(n) = w(n) + w(n − 1)
( )
w n
( )
x n
Z-1
+
Z-1
+
( )
y n
Z-1
+
1
-1
Z-1
5
4
3
4
−
1
8
( )
w n
( )
x n
Z-1
+
Z-1
+
( )
y n
+
1
-1
Z-1
5
4
3
4
−
1
8
October 18, 2016 58 / 151

June 2010 EE
Realize the following system function in direct form I
H(z) =
1 + 1
5
z−1
(1 − 1
2
z−1 + 1
3
z−2)(1 + 1
4
z−1)
Solution:
H(z) =
1 + 1
5
z−1
(1 − 1
2
z−1 + 1
3
z−2)(1 + 1
4
z−1)
H(z) =
Y (z)
X(z)
=
1 + 1
5
z−1
(1 − 1
4
z−1 + 5
24
z−2 + 1
12
z−3)
( )
y n
Z-1
+
1
5
( )
x n
Z-1
+
Z-1
+
1
4
−
Z-1
+
5
24
1
12
−
( 1)
y n −
( 3)
y n −
( 2)
y n −
( 1)
x n −
Y (z) −
1
4
z−1
Y (z) +
5
24
z−2
Y (z) −
1
12
z−3
Y (z) = X(z) + 5z−1
X(z)
By taking inverse Z transform on both sides
y(n) −
1
4
y(n − 1) +
5
24
y(n − 2) −
1
12
y(n − 3) = x(n) + 5x(n − 1)
October 18, 2016 59 / 151

Obtain direct form II for the system described by H(z) = 8Z3
−4Z2
+11Z−2
(Z− 1
4
)(Z2−Z+ 1
2
)
[December 2010 EC]
Solution:
H(z) =
8Z3 − 4Z2 + 11Z − 2
(Z − 1
4
)(Z2 − Z + 1
2
)
=
Y (z)
X(z)
=
8Z3 − 4Z2 + 11Z − 2
Z3 − 1.25Z2 + 0.75Z − 0.125
=
8 − 4Z−1 + 11Z−2 − 2Z−3
1 − 1.25Z−1 + 0.75Z−2 − 0.125Z−1
( )
w n
( )
x n
Z-1
+
Z-1
+
1.25
-0.75
( )
y n
+
8
11
+
Z-1
+
0.125 -2
+
-4
y(n)−1.25y(n −1)+0.75y(n −2)−0.125y(n −3) = 8x(n)−4x(n −1)+11x(n −2)−2x(n −3)
y(n) = 8x(n)−4x(n −1)+11x(n −2)−2x(n −3)+1.25y(n −1)−0.75y(n −2)+0.125y(n −3)
October 18, 2016 60 / 151

Obtain direct form II for the system described by
H(z) =
(z − 1)(z − 2)(z + 1)z
[z − ( 1
2
+ j 1
2
)][z − ( 1
2
− j 1
2
)](z − j
4
)(z + j
4
)
[December 2010 EC]
Solution:
H(z) =
Y (z)
X(z)
=
(Z2 − 1)(Z2 − 2Z)
(Z2 − Z + 1
2
)(Z2 + 1
16
)
=
Z4 − 2Z3 − Z2 + 2Z
Z4 − Z3 + 9
16
Z2 − 1
16
Z + 1
32
=
1 − 2Z−1 − Z−2 + 2Z−3
1 − Z−1 + 9
16
−2
− 1
16
Z−3 + 1
32
Z−4
y(n)−y(n−1)+
9
16
y(n−2)−
1
16
y(n−3)+
1
32
y(n−4) = x(n)−2x(n−1)−x(n−2)+2x(n−3)
y(n) = x(n)−2x(n−1)−x(n−2)+2x(n−3)+y(n−1)−
9
16
y(n−2)+
1
16
y(n−3)−
1
32
y(n−4)
October 18, 2016 61 / 151

( )
y n
Z-1
+
1
-2
( )
x n
Z-1
+
Z-1
+
1
Z-1
+
9
16
−
1
16
( 1)
y n −
( 3)
y n −
( 2)
y n −
( 1)
x n −
Z-1
-1
( 2)
x n −
Z-1
2
( 3)
x n −
+
+
Z-1
1
32
−
( 4)
y n −
+
( )
w n
( )
x n
Z-1
+
Z-1
+
1
( )
y n
+
1
-1
+
Z-1
+
2
+
-2
Z-1
+
1
32
−
9
16
−
1
16
October 18, 2016 62 / 151

Realize direct form-I and form -II for linear time invariant system which is described by the
following input output relation
2y(n) − y(n − 2) − 4y(n − 3) = 3x(n − 2)
Solution:
y(n) = 0.5y(n − 2) + 2y(n − 3) + 1.5x(n − 2)
( )
y n
Z-1
( )
x n
Z-1
+
Z-1
0.5
Z-1
1.5
Z-1
+
2
(a) Direct form-I
( )
w n
( )
x n
Z-1
+
Z-1
( )
y n
1.5
Z-1
+
0.5
2
(b) Direct form-II
Figure 33: Realization of IIR systems
October 18, 2016 63 / 151

Realize direct form-I and form -II for system given by
H(z) =
z−1 − 3z−2
(10 − z−1)(1 + 0.5z−1 + 0.5z−2)
Solution:
H(z) =
z−1 − 3z−2
(10 − z−1)(1 + 0.5z−1 + 0.5z−2)
=
z−1 − 3z−1
(10 + 4z−1 + 4.5z−2 − 0.5z−3)
Y (z)
X(z)
=
0.1z−1 − 0.3z−1
(1 + 0.4z−1 + 0.45z−2 − 0.05z−3)
By taking inverse z transform on both sides
y(n) + 0.4y(n − 1) + 0.45y(n − 2) − 0.05y(n − 3) = 0.1x(n − 1) − 0.3x(n − 2)
y(n) = −0.4y(n − 1) − 0.45y(n − 2) + 0.05y(n − 3) + 0.1x(n − 1) − 0.3x(n − 2)
( )
y n
Z-1
( )
x n
Z-1
+
Z-1
0.45
−
Z-1
-0.3
Z-1
+
0.05
+
0.1
+
0.4
−
(a) Direct form-I
( )
w n
( )
x n
Z-1
+
Z-1
( )
y n
-0.3
Z-1
+
0.45
−
0.05
+
0.4
−
+
0.1
(b) Direct form-II
October 18, 2016 64 / 151

Realization of IIR system Cascade Form Structure
Cascade Form Structure
H(z) =
M
P
k=0
bk z−k
1 +
N
P
k=1
ak z−k
=
b0 + b1z−1 + b2z−2 + bM z−M
1 + a1z−1 + a2z−2 + aN z−N
The system can be factored into a cascade of second order subsystems such that H(z) can be
expressed as
H(z) = H1(z) × H2(z) × H3(z) . . . × Hk (z) =
K
Y
k=1
Hk (z)
where K is the integer part of (N+1)/2 and Hk (z) has the general form
Hk (z) =
bk0 + bk1z−1 + bk2z−2
1 + ak1z−1 + ak2z−2
1
( ) ( )
x n x n
=
H1
(z)
1 2
( ) ( )
y n x n
=
H2
(z)
2 3
( ) ( )
y n x n
=
Hk
(z)
( ) ( )
k
y n y n
=
1( )
( )
k
k
y n
x n
− =
Figure 35: Cascade realization
October 18, 2016 65 / 151

In cascade each Hk (z) are represented as
Hk (z) =
Yk (z)
Xk (z)
=
Wk (z)
Xk (z)
Yk (z)
Wk (z)
= Hk1(z).Hk2(z)
where Hk1(z) and Hk2(z) defined as
Hk1(z) =
Wk (z)
Xk (z)
=
1
1 + ak1z−1 + ak2z−2
Hk2(z) = bk0 + bk1z−1
+ bk2z−2
( )
w n
1( ) ( )
k
y n x n
− =
Z-1
+
Z-1
+
+
+
2
k
a
−
1
k
a
−
2
k
b
1
k
b
0
k
b 1
( ) ( )
k k
y n x n
+
=
wk (n) = −ak1wk (n − 1) − ak2wk (n − 2) + −yk−1(n)
yk (n) = bk0wk (n) + bk1wk (n − 1) + bk2wk (n − 2)
yk−1(n = xk (n)
yk (n) = xk+1(n)
y0(n) = x(n)
y(n) = xk (n)
October 18, 2016 66 / 151

June 2010 EE
Realize the following system function in cascade form
H(z) =
1 + 1
5
z−1
(1 − 1
2
z−1 + 1
3
z−2)(1 + 1
4
z−1)
Solution:
H(z) = H1(z) × H2(z) =
1
(1 + 1
4
z−1)
.
1 + 1
5
z−1
(1 − 1
2
z−1 + 1
3
z−2)
( )
x n
Z-1
+
( )
y n
+
+
Z-1
Z-1
+
1
5
1
3
−
1
4
−
1
2
1
11
a
−
21
b
22
a
−
21
a
−
20
b
October 18, 2016 67 / 151

H(z) =
1 − 1
2
z−1
(1 − 1
4
z−1 + 1
2
z−2)(1 − 1
5
z−1 + 1
6
z−2)
Solution:
H(z) = H1(z) × H2(z) =
1
(1 − 1
5
z−1 + 1
6
z−2)
.
1 − 1
2
z−1
(1 − 1
4
z−1 + 1
2
z−2)
( )
x n
Z-1
+
( )
y n
+
+
Z-1
Z-1
+
1
2
−
1
2
−
1
5
1
4
1
Z-1
+
1
6
−
October 18, 2016 68 / 151

H(z) =
1 + 1
3
z−1
(1 − 1
5
z−1)(1 − 3
4
z−1 + 1
8
z−2)
Solution:
H(z) = H1(z) × H2(z) =
1
(1 − 1
5
z−1)
×
1 + 1
3
z−1
(1 − 3
4
z−1 + 1
8
z−2)
( )
x n
Z-1
+
( )
y n
+
+
Z-1
Z-1
+
1
3
1
8
−
1
5
3
4
1
October 18, 2016 69 / 151

The transfer function of a discrete system is given as follows
H(z) =
1 − z−1
(1 − 0.2z−1 − 0.15z−2)
Draw the cascade realization
Solution:
H(z) = H1(z) × H2(z) = (1 − z−1
)
1
(1 − 0.2z−1 − 0.15z−2)
( )
x n
Z-1
( )
y n
+
+
Z-1
Z-1
0.15
0.2
-1
1
+
1( )
H z
2 ( )
H z
October 18, 2016 70 / 151

December 2010 EE
Draw the cascade realization for the following system
y(n) = 0.75y(n − 1) − 0.125y(n − 2) + 6x(n) + 7x(n − 1) + x(n − 2)
Solution:
Y (z) = 0.75z−1
Y (z) − 0.125z−2
Y (z) + 6X(z) + 7z−1
X(z) + z−2
X(z)
H(z) =
Y (z)
X(z)
=
6 + 7z−1 + z−2
1 − 0.75z−1 + 0.125z−2
H(z) =
Y (z)
X(z)
=
6 + 7z−1 + z−2
1 − 0.75z−1 + 0.125z−2
=
6z2 + 7z + 1
z2 − 0.75z + 0.125
=
(6z + 1)(z + 1)
(z − 0.5)(z − 0.25)
=
(6 + z−1)(1 + z−1)
(1 − 0.5z−1)(1 − 0.25z−1)
= H1(z) × H2(z)
a = 1, b = −0.75, c = 0.125
Roots of the quadratic equation for
ax2
+ bx + c = 0 are
−b±
q
b2−4ac
2a
=
0.75 ±
p
0.752 − 4(0.125)
2
=
0.75 ±
√
0.0625
2
=
0.75 ± 0..25
2
= ⇒ 0.5, 0.25
October 18, 2016 71 / 151

H1(z) =
6 + z−1
1 − 0.5z−1
H2(z) =
(1 + z−1)
(1 − 0.25z−1)
( )
y n
2 ( )
H z
( )
x n
Z-1
1
+
1( )
H z
+
0.5
6
Z-1
1
+
+
0.25
1
October 18, 2016 72 / 151

A system function is specified by its transfer function H(z) given by
H(z) =
(z − 1)(z − 2)(z + 2)z
[z − ( 1
2
+ j 1
2
)][z − ( 1
2
− j 1
2
)](z − j
4
)(z + j
4
)
Realize the cascade of two biquadratic sections.
Solution:
H(z) =
(z − 1)(z − 2)(z + 2)z
[z − ( 1
2
+ j 1
2
)][z − ( 1
2
− j 1
2
)](z − j
4
)(z + j
4
)
=
z2 − 3z + 2
z2 − z + 0.5
×
z2 + 2z
z2 + 0.0625
=
1 − 3z−1 + 2z−2
1 − z−1 + 0.5z−2
×
1 + 2z−1
1 + 0.0625z−2
= H1(z) × H2(z)
( )
x n
Z-1
+
Z-1
+
1
-0.5
+
1
+
-3
2
( )
y n
+
Z-1
Z-1
+
0.0625
−
1
2
October 18, 2016 73 / 151

Obtain the cascade realization of the following system. The system should have two biquadratic
sections. [EC December 2012]
H(z) =
(z − 1)(z2 + 5z + 6)(z − 3)
(z2 + 6z + 5)(z2 − 6z + 8)
Solution:
H(z) =
(z − 1)(z2 + 5z + 6)(z − 3)
(z2 + 6z + 5)(z2 − 6z + 8)
=
z2 − 4z + 3
z2 + 6z + 5
×
z2 + 5z + 6
z2 − 6z + 8
=
1 − 4z−1 + 3z−2
1 + 6z−1 + 5z−2
×
1 + 5z−1 + 6z−2
1 − 6z−1 + 8z−2
( )
x n
Z-1
+
Z-1
+
-6
-5
+
1
+
-4
3
Z-1
+
Z-1
+
6
-8
+
1
+
5
6
( )
y n
October 18, 2016 74 / 151

Obtain the cascade realization using second order sections.
H(z) =
10(1 − 1
2
z−1)(1 − 2
3
z−1)(1 + 2z−1)
(1 − 3
4
z−1)(1 − 1
8
z−1)[1 − ( 1
2
+ j 1
2
)z−1][1 − (1
2
− j 1
2
)z−1]
Solution:
H(z) =
10z(z − 0.5)(z − 0.6667)(z + 2)
(z − 0.75)(z − 0.125)[z − (0.5 + j0.5)][z − (0.5 − j0.5)]
=
10z(z − 0.5)
(z − 0.75)(z − 0.125)
×
(z − 0.6667)(z + 2)
[z − (0.5 + j0.5)][z − (0.5 − j0.5)]
=
10z2 − 5z
z2 − 0.875z + 0.0938
×
z2 + 1.333z − 1.333
z2 − z + 0.5
=
10 − 5z−1
1 − 0.875z−1 + 0.0938z−2
×
1 + 1.333z−1 − 1.333z−2
1 − z−1 + 0.5z−2
( )
x n
Z-1
+
Z-1
+
0.875
-0.0938
+
10
-5
Z-1
+
Z-1
+
1
-0.5
+
1
+
-1.333
( )
y n
1.333
October 18, 2016 75 / 151

EC December 2011
Obtain the cascade realization for the following system.
H(z) =
1 + 1
4
z−1
(1 + 1
2
z−1)(1 + 1
2
z−1 + 1
2
z−2)
Solution:
H(z) =
1 + 1
4
z−1
(1 + 1
2
z−1)(1 + 1
2
z−1 + 1
2
z−2)
=
1 + 1
4
z−1
1 + 1
2
z−1
×
1
1 + 1
2
z−1 + 1
2
z−2
( )
y n
Z-1
+
( )
x n
Z-1
+
Z-1
+
1
4
−
1
2
−
+
1
4
1
2
−
October 18, 2016 76 / 151

Obtain the cascade realization for the following system.
H(z) =
(1 + z−1)3
(1 − 1
4
z−1)(1 − 1
2
z−1 + 1
2
z−2)
Solution:
H(z) =
(1 + z−1)3
(1 − 1
4
z−1)(1 − 1
2
z−1 + 1
2
z−2)
=
1 + z−1
(1 − 1
4
z−1)
×
1 + 2z−1 + z−2
1 − z−1 + 1
2
z−2
= H1(z) × H2(z)
( )
y n
Z-1
+
( )
x n
Z-1
+
Z-1
+
1
2
−
1
+
1
1
4
1
+
+
1
2
1
October 18, 2016 77 / 151

EC 2011 July
Obtain the direct form II realizations of the following system.
H(z) =
(1 + z−1)
(1 − 1
4
z−1)(1 − z−1 + 1
2
z−2)
H(z) =
(1 + z−1)
(1 − 1
4
z−1)(1 − z−1 + 1
2
z−2)
=
1 + z−1
(1 − 1
4
z−1)
×
1
1 − z−1 + z−2
= H1(z) × H2(z)
( )
y n
Z-1
+
( )
x n
Z-1
+
Z-1
+
1
−
1
+
1
1
4
October 18, 2016 78 / 151

June 2010 EC
Obtain cascade form for the system described by
y(n) = −0.1y(n − 1) + 0.72y(n − 2) + 0.7x(n) − 0.252x(n − 2)
Solution:
Y (z) = −0.1z−1
Y (z) + 0.72z−2
Y (z) + 0.7X(z) − 0.252z−2
X(z) =
H(z) =
Y (z)
X(z)
=
0.7 − 0.252z−2
1 + 0.1z−1 − 0.72z−2
= (0.7 − 0.252z−2
) ×
1
1 + 0.1z−1 − 0.72z−2
( )
x n
Z-1
( )
y n
+
+
Z-1
Z-1
0.72
0.1
−
-0.252
1
+
1( )
H z
2 ( )
H z
Z-1
0.7
October 18, 2016 79 / 151

June 2015 EC
Find the transfer function and difference equation realization shown in Figure
( )
x n
Z-1
+
Z-1
( )
y n
+
4
+
2
−
3
( )
w n
Solution:
H(z) =
Y (z)
X(z)
=
4 + 3z−1
1 + 2z−2
Y (z) = −2z−2
Y (z) + 4X(z) + 3z−1
X(z)
The difference equation is
y(n) = −2y(n − 2) + 4x(n) + 3x(n − 1)
October 18, 2016 80 / 151

Realization of IIR Structures Parallel Form Structure for IIR System
Parallel Form Structure for IIR System
The system function for IIR sytem is
H(z) =
M
P
k=0
bk z−k
1 +
N
P
k=1
ak z−k
=
b0 + b1z−1 + b2z−2 + bM z−M
1 + a1z−1 + a2z−2 + aN z−N
The function can be expanded in partial
fractions as follows:
H(z) = C + H1(z) + H2(z) . . . + Hk (z)
C is constant and each
H1(z) + H2(z) . . . + Hk (z) are the second
order system which is as shown in Figure
50
( )
x n ( )
y n
2 ( )
H z
( )
k
H z
1( )
H z +
+
+
C
Figure 49: Parallel form realization for
IIR systems
October 18, 2016 81 / 151

Hk (z) =
bk0 + bk1z−1
1 + ak1z−1 + ak2z−2
wk (n) = ak1wk (n − 1) − ak2wk (n − 2) + x(n)
yk (n) = bk0wk (n) − bk1wk (n − 1)
y(n) = C x(n) +
K
X
k=1
yk (n)
( )
k
w n
( )
k
x n
Z-1
+
Z-1
+
( )
k
y n
+
+
1
k
a
−
2
k
a
−
1
k
b
0
k
b
Figure 50: Direct form II of second order
subsystem
October 18, 2016 82 / 151

Realize the following System in parallel form:
H(z) = 3 +
4z
z − 1
2
+
2
z − 1
4
Solution:
H(z) = 3 +
4
1 − 0.5z−1
+
2z−1
1 − 1
4
z−1
( )
y n
+
Z-1
+
0.25
Z-1
+
0.5
( )
x n
3
4
2
Figure 51: Parallel realization
October 18, 2016 83 / 151

H(z) =
(1 + z−1)(1 + 2z−1)
(1 + 1
2
z−1)(1 − 1
4
z−1)(1 + 1
8
z−1)
Solution:
H(z) =
z−2(z + 1)(z + 2)
z−3(z + 0.5)(z − 0.25)(z + 0.125)
H(z) =
z(z + 1)(z + 2)
(z + 0.5)(z − 0.25)(z + 0.125)
H(z)
z
=
(z + 1)(z + 2)
(z + 0.5)(z − 0.25)(z + 0.125)
H(z)
z
=
A
(z + 0.5)
+
B
(z − 0.25)
+
C
(z + 0.125)
(z + 1)(z + 2)
(z + 0.5)(z − 0.25)(z + 0.125)
=
A
(z + 0.5)
+
B
(z − 0.25)
+
C
(z + 0.125)
October 18, 2016 84 / 151

Coefficients A B and C are determined
A =
H(z)
z
[z + 0.5]|z=−0.5 =
(z + 1)(z + 2)
(z − 0.25)(z + 0.125)
=
(−0.5 + 1)(−0.5 + 2)
(−0.5 − 0.25)(−0.5 + 0.125)
= 2.66
B =
H(z)
z
[z − 0.25]|z=0.25 =
(z + 1)(z + 2)
(z + 0.5)(z + 0.125)
=
(0.25 + 1)(0.25 + 2)
(0.25 + 0.5)(0.25 + 0.125)
= 10
C =
H(z)
z
[z + 0.125]|z=−0.125 =
(z + 1)(z + 2)
(z + 0.5)(z − 0.25)
=
(−0.125 + 1)(−0.125 + 2)
(−0.125 + 0.5)(−0.125 − 0.25)
= −11.66
October 18, 2016 85 / 151

H(z)
z
=
2.66
(z + 0.5)
+
10
(z − 0.25)
+
−11.66
(z + 0.125)
H(z) =
2.66
(1 + 0.5z−1)
+
10
(1 − 0.25z−1)
+
−11.66
(1 + 0.125z−1)
( )
y n
+
Z-1
+
0.25
Z-1
+
0.5
−
10
2.66
( )
x n
Z-1
+
0.125
−
11.66
−
October 18, 2016 86 / 151

H(z) =
1 − z−1
(1 − 0.2z−1 − 0.15z−2)
Solution
H(z) =
1 − z−1
(1 − 0.2z−1 − 0.15z−2)
H(z) =
z2 − z
(z2 − 0.2z − 0.15)
a = 1, b = −0.2, c = −0.15
Roots of the equation ax2 + bx + c = 0 are =
−b±
√
b2−4ac
2a
=
0.2 ±
p
0.22 − 4(−0.15)
2
=
0.2 ±
√
0.64
2
=
0.2 ± 0.8
2
⇒ 0.5, − 0.3
H(z)
z
=
z − 1
(z − 0.5)(z + 0.3)
H(z)
z
=
A
(z − 0.5)
+
B
(z + 0.3)
October 18, 2016 87 / 151

The values of A and B are determined by the following procedure.
A =
H(z)
z
(z − 0.5)|z=0.5 =
z − 1
(z + 0.3)
=
0.5 − 1
(0.5 + 0.3)
= −0.625
B =
H(z)
z
(z + 0.3)|z=−0.3 =
z − 1
(z − 0.5)
=
−0.3 − 1
(−0.3 − 0.5)
= 1.625
H(z)
z
=
−0.625
z − 0.5
+
1.625
z + 0.3
H(z) =
−0.625
1 − 0.5z−1
+
1.625
1 + 0.3z−1
= H1(z) + H2(z)
( )
y n
+
Z-1
+
0.3
−
Z-1
+
0.5
1.625
0.625
−
( )
x n
October 18, 2016 88 / 151

H(z) =
1 + 0.33z−1
(1 − 0.75z−1 + 0.125z−2)
Solution
H(z) =
1 + 0.33z−1
(1 − 0.75z−1 + 0.125z−2)
=
z−1(z + 0.33)
z−2(z2 − 0.75z−1 + 0.125z−2)
H(z)
z
=
z + .33
(z2 − 0.75z + 0.125)
a = 1, b = −0.75, c = 0.125
Roots of the equation are
=
0.75 ±
p
0.752 − 4(0.125)
2
=
0.75 ±
√
0.0625
2
=
0.75 ± 0.25
2
⇒ 0.5, 0.25
H(z)
z
=
z + 0.33
(z − 0.5)(z − 0.25)
=
A
(z − 0.5)
+
B
(z − 0.25)
October 18, 2016 89 / 151

The values of A and B are determined by the following procedure.
A =
H(z)
z
(z − 0.5)|z=0.5 =
z + 0.33
(z − 0.25)
=
0.25 + 0.33
(0.5 − 0.25)
= 3.33
B =
H(z)
z
(z − 0.25)|z=0.25 =
z + 0.33
(z − 0.5)
=
0.25 + 0.33
(0.25 − 0.5)
= −2.33
H(z)
z
=
3.33
(z − 0.5)
+
−2.33
(z − 0.25)
H(z) =
3.33
1 − 0.5z−1
+
−2.33
1 − 0.25z−1
= H1(z) + H2(z)
( )
y n
+
Z-1
+
0.25
Z-1
+
0.5
2.33
−
3.33
( )
x n
October 18, 2016 90 / 151

y(n) = 0.75y(n − 1) − 0.125y(n − 2) + 6x(n) + 7x(n − 1) + x(n − 2)
Solution
Y (z) = 0.75z−1
Y (z) − 0.125z−2
Y (z) + 6X(z) + 7z−1
X(z) + z−2
X(z)
H(z) =
Y (z)
X(z)
=
6 + 7z−1 + z−2
1 − 0.75z−1 + 0.125z−2
H(z) =
z−2[6z2 + 7z + 1]
z−2[z2 − 0.75z + 0.125]
H(z) =
z−2[6z2 + 7z + 1]
z−2[z2 − 0.75z + 0.125]
=
[6z2 + 7z + 1]
[z2 − 0.75z + 0.125]
October 18, 2016 91 / 151

a = 1, b = −0.75, c = 0.125
Roots of the equation are
=
0.75 ±
q
(−0.75)2
− 4(0.125)
2
=
0.75 ±
√
0.0625
2
=
0.75 ± 0.25
2
0.5, 0.25
H(z) =
6z2 + 7z + 1
(z − 0.5)(z − 0.25)
H(z)
z
=
6z2 + 7z + 1
z(z − 0.5)(z − 0.25)
H(z)
z
=
A
z
+
B
(z − 0.5)
+
C
(z − 0.25)
A = H(z)z|z=0 =
6z2 + 7z + 1
z(z − 0.5)(z − 0.25)
z =
1
(−0.5)(−0.25)
= 8
B = H(z)(z − 0.5)|z=0.5 =
6z2 + 7z + 1
z(z − 0.5)(z − 0.25)
(z − 0.5z) =
6(0.5)2 + 7(0.5) + 1
0.5[0.5 − 0.25]
= 48
C = H(z)(z − 0.25z)|z=0.25 =
6z2 + 7z + 1
z(z − 0.5)(z − 0.25)
(z − 0.25)
=
6(0.25)2 + 7(0.25) + 1
0.25[0.25 − 0.5]
= −50
October 18, 2016 92 / 151

H(z) = 8 +
48
(1 − 0.5z1)
+
−50
(1 − 0.25z1)
= H1(z) + H2(z) + H3(z)
( )
y n
+
Z-1
+
0.25
Z-1
+
0.5
( )
x n
8
48
50
−
October 18, 2016 93 / 151

H(z) =
8z3 − 4z2 + 11z − 2
[z − 1
4
][z2 − z + 1
2
]
H(z)
z
=
8z3 − 4z2 + 11z − 2
z[z − 1
4
][z2 − z + 1
2
]
H(z)
z
=
8z3 − 4z2 + 11z − 2
z[z − 1
4
][z2 − z + 1
2
]
H(z)
z
=
A
z
+
B
z − 0.25
+
Cz + D
z2 − z + .5
A =
H(z)
z
z|z = 0 =
8z3 − 4z2 + 11z − 2
[z − 0.25][z2 − z + 0.5]
=
−2
[−0.25][0.5]
= 16
October 18, 2016 94 / 151

B =
H(z)
z
[z − 0.5]|z=0.5 =
8z3 − 4z2 + 11z − 2
z[z2 − z + 0.5]
=
8(0.5)3 − 4(0.5)2 + 11(0.5) − 2
(0.5)[(0.5)2 − (0.5) + 0.5]
= 8
H(z)
z
=
8z3 − 4z2 + 11z − 2
z[z − 0.25][z2 − z + 0.5]
8z3 − 4z2 + 11z − 2
z[z − 0.25][z2 − z + 0.5]
=
A
z
+
B
z − 0.25
+
Cz + D
z2 − z + .25
8z3
− 4z2
+ 11z − 2 = A[z − 0.25][z2
− z + 0.5] +
+Bz[z2
− z + .25] + (Cz + D)z[z − 0.25]
8z3
− 4z2
+ 11z − 2z = z3
[A + B + C] + z2
[−1.25A − B − 0.25C + D] +
+z[0.5A + .25B − .25D] − 0.0625A
8 = A + B + C = 16 + 8 + C
C = −16
−4 = [−1.25A − B − 0.25C + D] = [−20 − 8 + 4 + D]
D = 20
October 18, 2016 95 / 151

H(z)
z
=
A
z
+
B
z − 0.25
+
Cz + D
z2 − z + .5
H(z)
z
=
16
z
+
8
z − 0.25
+
−16z + 20
z2 − z + .5
H(z) = 16 +
8
1 − 0.25z−1
+
−16 + 20z−1
1 − z−1 + .5z−2
( )
y n
+
Z-1
+
1
Z-1
+
0.25
( )
x n
16
8
-16
+
Z-1
0.5
−
+
20
October 18, 2016 96 / 151

EE 2010 May
H(z) =
1 − 1
5
z−1
[1 − 1
2
z−1 + 1
3
z−2][1 + 1
4
z−1]
Solution:
H(z) =
z−1(z − 0.2)
z−2[z2 − 0.5z + 0.33]z−1[z + 0.25]
H(z) =
z2(z − 0.2)
[z2 − 0.5z + 0.33z][z + 0.25]
H(z)
z
=
z(z − 0.2)
[z2 − 0.5z + 0.33][z + 0.25]
H(z)
z
=
Az + B
[z2 − 0.5z + 0.333]
+
C
[z + 0.25]
z(z − 0.2)
[z2 − 0.5z + 0.33][z + 0.25]
=
Az + B
[z2 − 0.5z + 0.333]
+
C
[z + 0.25]
October 18, 2016 97 / 151

C =
H(z)
z
[z + 0.25]|z=−0.25 =
(z − 0.2)
[z2 − 0.5z + 0.333]
=
−0.25(−0.25 − 0.2)
[(−0.25)2 − (0.5)(−0.25) + 0.333]
= 0.217
z(z − 0.2)
[z2 − 0.5z + 0.33][z + 0.25]
=
Az + B
[z2 − 0.5z + 0.333]
+
C
[z + 0.25]
z(z − 0.2) = (Az + B)[z + 0.25] + C[z2
− 0.5z + 0.33]
z2
− 0.2z = Az2
+ 0.25Az + Bz + 0.25B + Cz2
− 0.5Cz + 0.33C]
z2
− 0.2z = z2
(A + C) + z(0.25A + B − C) + 0.25B + 0.33C
Equating coefficients
A + C = 1
A = 1 − C = 1 − 0.217 = 0.783
0.25B + 0.33C = 0
B =
−0.33C
0.25
= −0.2864
October 18, 2016 98 / 151

H(z)
z
=
0.78z + −0.2864
[z2 − 0.5z + 0.333]
+
0.216
[z + 0.25]
H(z)
z
=
z(0.78 − 0.2864z−1)
z2(1 − 0.5z−1 + 0.333z−2)
+
0.216
z[1 + 0.25z−1]
H(z) =
(0.78 − 0.2864z−1)
(1 − 0.5z−1 + 0.333z−2)
+
0.216
(1 + 0.25z−1)
( )
y n
0.216
Z-1
+
0.25
( )
x n
Z-1
+
Z-1
+
0.5
-0.333
+
0.78
-0.2864
+
October 18, 2016 99 / 151

EC 2012 December
H(z) =
(z − 1)(z − 2)(z + 2)z
[z − ( 1
2
+ j 1
2
)][z − ( 1
2
− j 1
2
)](z − j
4
)(z + j
4
)
Solution:
H(z) =
(z − 1)(z − 2)(z + 2)z
[z − ( 1
2
+ j 1
2
)][z − ( 1
2
− j 1
2
)](z − j
4
)(z + j
4
)
H(z)
z
=
(z − 1)(z − 2)(z + 2)z
[z − ( 1
2
+ j 1
2
)][z − ( 1
2
− j 1
2
)](z − j
4
)(z + j
4
)
H(z)
z
=
A
z − (0.5 + j0.5)
+
B
z − (0.5 − j0.5)
+
C
z − j0.25
+
D
z + j0.25)
October 18, 2016 100 / 151

H(z)
z
=
A
z − (0.5 + j0.5)
+
B
z − (0.5 − j0.5)
+
C
z − j0.25
+
D
z + j0.25)
A =
H(z)
z
[z − (0.5 + j0.5)]|z=0.5+j0.5 =
(z2 − 1)(z − 2)
[z − (0.5 − j0.25)][z − j0.25][z + j0.25]
=
[(0.5 + j0.5)2 − 1][(0.5 + j0.5 − 2]
[0.5 + j0.5 − (0.5 − j0.5)][z2 + 0.0625]
=
[0.25 + j0.5 − 0.25 − 1][(0.5 + j0.5 − 2]
[j1][z2 + 0.0625]
=
[−1 + j0.5][(−1.5 + j0.5]
[j1][j0.5 + 0.0625]
=
[1.5 − j0.5 − j0.75 + 0.25]
−0.5 + j0.0625
=
1.25 − j1.25
−0.5 + j0.0625
=
1.767∠ − 45
0.503∠172
=
1.767∠ − 45
0.503∠172
= 3.513∠ − 217 = −2.8 + j2.1
October 18, 2016 101 / 151

Similarly B, C and D are estimated.
H(z)
z
=
A
z − (0.5 + j0.5)
+
B
z − (0.5 − j0.5)
+
C
z − j0.25
+
D
z + j0.25)
=
−2.8 + j2.1
z − (0.5 + j0.5)
+
−2.7683 − j2.1517
z − (0.5 − j0.5)
+
3.268 − j7.837
z − j0.25
+
3.268 + j7.837
z + j0.25)
We have to design second order system, hence combine first two terms and last two terms.
H(z)
z
=
−5.536z + 0.612
z2 − z + 0.5
+
6.5366z − 3.918
z2 + 0.0625)
H(z)
z
=
z(−5.536 + 0.612z−1)
z2(1 − z−1 + 0.5z−2)
+
z(6.5366 − 3.918z−1)
z2(1 + 0.0625z−2)
H(z) =
(−5.536 + 0.612z−1)
(1 − z−1 + 0.5z−2)
+
(6.5366 − 3.918z−1)
(1 + 0.0625z−2)
= H1(z) + H2(z)
October 18, 2016 102 / 151

H(z) =
(−5.536 + 0.612z−1)
(1 − z−1 + 0.5z−2)
+
(6.5366 − 3.918z−1)
(1 + 0.0625z−2)
= H1(z) + H2(z)
( )
y n
6.5366
Z-1
+
( )
x n
Z-1
+
Z-1
+
1
-0.5
+
-5.536
0.6112
+
Z-1
+
3.9184
-0.0625
October 18, 2016 103 / 151

EE 2010 May
H(z) =
1 + 1
4
z−1
[1 + 1
2
z−1 + 1
4
z−2][1 + 1
2
z−1]
Solution:
H(z) =
z−1(z − 0.25)
z−2[z2 − 0.5z + 0.25]z−1[z + 0.5]
H(z) =
z2(z + 0.25)
[z2 + 0.5z + 0.25][z + 0.5]
H(z)
z
=
z(z − 0.25)
[z2 + 0.5z + 0.25][z + 0.25]
H(z)
z
=
Az + B
[z2 + 0.5z + 0.25]
+
C
[z + 0.5]
z(z − 0.25)
[z2 + 0.5z + 0.25][z + 0.5]
=
Az + B
[z2 + 0.5z + 0.25]
+
C
[z + 0.5]
October 18, 2016 104 / 151

C =
H(z)
z
[z + 0.5]|z=−0.5 =
z(z − 0.2)
[z2 − 0.5z + 0.333]
=
−0.5(−0.5 − 0.25)
[(−0.5)2 − (0.5)(−0.5) + 0.25]
= 0.5
z(z + 0.25)
[z2 + 0.5z + 0.25][z + 0.5]
=
Az + B
[z2 + 0.5z + 0.25]
+
C
[z + 0.5]
z(z + 0.25) = (Az + B)[z + 0.5] + C[z2
+ 0.5z + 0.25]
z2
+ 0.25z = Az2
+ 0.5Az + Bz + 0.5B + Cz2
+ 0.5Cz + 0.25C]
z2
+ 0.25z = z2
(A + C) + z(0.5A + B + 0.5C) + 0.5B + 0.25C
A + C = 1
A = 1 − C = 1 − 0.5 = 0.5
0.25 = 0.5A + B + 0.5C = 0.5 × 0.5 + B + 0.5 × 0.5
B = −0.25
October 18, 2016 105 / 151

H(z)
z
=
0.5z − 0.25
[z2 + 0.5z + 0.25]
+
0.5
[z + 0.5]
H(z)
z
=
z(0.5 − 0.25z−1)
z2(1 + 0.5z−1 + 0.25z−2)
+
0.5
z[1 + 0.5z−1]
H(z) =
(0.5 − 0.25z−1)
(1 + 0.5z−1 + 0.25z−2)
+
0.5
(1 + 0.5z−1)
( )
y n
0.5
Z-1
+
-0.5
( )
x n
Z-1
+
Z-1
+
-0.5
-0.25
+
0.5
-0.25
+
October 18, 2016 106 / 151

H(z) =
1 − 1
2
z−1
(1 − 1
3
z−1)(1 − 1
4
z−1)
Solution:
H(z) =
z−1(z − 0.5)
z−1(z − 0.33)z−1(z − 0.25)
H(z) =
z(z − 0.5)
(z − 0.33)(z − 0.25)
H(z)
z
=
(z − 0.33)
(z − 0.33)(z − 0.25)
H(z)
z
=
A
z − 0.33
+
B
z − 0.25
(z − 0.5)
(z − 0.33)z−1(z − 0.25)
=
A
z − 0.33
+
B
z − 0.25
October 18, 2016 107 / 151

Coefficients A and B are determined
A =
H(z)
z
[z − 0.33]|z=0.33 =
z − 0.5
z − 0.25
=
0.33 − 0.5
0.33 − 0.25
= −2
B =
H(z)
z
[z − 0.25]|z=0.25 =
z − 0.5
z − 0.33
=
0.25 − 0.5
0.25 − 0.33
= 3
( )
y n
+
Z-1
+
0.25
Z-1
+
1
3
3
2
−
( )
x n
October 18, 2016 108 / 151

H(z) =
1 + 1
3
z−1
1 − 3
4
z−1 + 1
8
z−2
Solution:
H(z) =
1 + 1
3
z−1
1 − 3
4
z−1 + 1
8
z−2
H(z) =
1 + 0.33z−1
(1 − 0.5z−1)(1 − 0.25z−1)
H(z) =
z−1(z + 0.33)
z−1(z − 0.5)z−1(z − 0.25)
H(z) =
z(z + 0.33)
(z − 0.5)(z − 0.25)
H(z)
z
=
(z + 0.33)
(z − 0.5)(z − 0.25)
H(z)
z
=
A
z − 0.5
+
B
z − 0.25
(z + 0.33)
(z − 0.5)(z − 0.25)
=
A
z − 0.33
+
B
z − 0.25
October 18, 2016 109 / 151

Coefficients A and B are determined
A =
H(z)
z
[z − 0.5]|z=0.5 =
z + 0.33
z − 0.25
=
0.5 + 0.33
0.5 − 0.25
= 3.33
B =
H(z)
z
[z − 0.25]|z=0.25 =
z + 0.33
z − 0.5
=
0.25 + 0.33
0.25 − 0.5
= −2.33
H(z)
z
=
3.33
z − 0.5
+
−2.33
z − 0.25
H(z) =
3.33
z − 0.5
+
−2.33
z − 0.25
( )
y n
+
Z-1
+
0.25
Z-1
+
0.5
2.33
−
3.333
( )
x n
October 18, 2016 110 / 151

EE 2010 May
H(z) =
10(1 − 1
2
z−1)(1 + 2z−1)(1 − 2
3
z−1)
(1 − 3
4
z−1)(1 − 1
8
z−1)[1 − ( 1
2
+ j 1
2
)z−1)][1 − (1
2
− j 1
2
)z−1)]
Solution:
H(z) =
10(1 − 0.5z−1)(1 + 2z−1)(1 − 0.66z−1)
(1 − 0.75z−1)(1 − 0.125z−1)[1 − (0.5 + j0.5)z−1)][1 − (0.5 − j0.5)z−1)]
H(z) =
10(1 − 0.5z−1)(1 + 2z−1)(1 − 0.66z−1)
(1 − 0.75z−1)(1 − 0.125z−1)[1 − z−1 + 0.5z−2)]
H(z) =
10z−3(z − 0.5)(z + 2)(z − 0.66)
z−4(z − 0.75)(z − 0.125)[z2 − z + 0.5]
H(z)
z
=
10(z − 0.5)(z + 2)(z − 0.66)
(z − 0.75)(z − 0.125)[z2 − z + 0.5]
H(z)
z
=
A
[z − 0.75]
+
B
[z − 0.125]
+
Cz + D
[z2 − z + 0.5]
10(z − 0.5)(z + 2)(z − 0.66)
(z − 0.75)(z − 0.125)[z2 − z + 0.5]
=
A
[z − 0.75]
+
B
[z − 0.125]
+
Cz + D
[z2 − z + 0.5]
October 18, 2016 111 / 151

A =
H(z)
z
[z − 0.75]|z=0.75 =
10(z − 0.5)(z + 2)(z − 0.66)
(z − 0.125)[z2 − z + 0.5]
=
10(0.75 − 0.5)(0.75 + 2)(0.75 − 0.66)
(0.75 − 0.125)[0.752 − 0.75 + 0.5]
= 3.16
B =
H(z)
z
[z − 0.125]|z=0.125 =
10(z − 0.5)(z + 2)(z − 0.66)
(z − 0.75)[z2 − z + 0.5]
=
10(0.125 − 0.5)(0.125 + 2)(0.125 − 0.66)
(0.125 − 0.75)[0.1252 − 0.125 + 0.5]
= −17.46
z(z + 0.25)
[z2 + 0.5z + 0.25][z + 0.5]
=
Az + B
[z2 + 0.5z + 0.25]
+
C
[z + 0.5]
z(z + 0.25) = (Az + B)[z + 0.5] + C[z2
+ 0.5z + 0.25]
z2
+ 0.25z = Az2
+ 0.5Az + Bz + 0.5B + Cz2
+ 0.5Cz + 0.25C]
z2
+ 0.25z = z2
(A + C) + z(0.5A + B + 0.5C) + 0.5B + 0.25C
A + C = 1
A = 1 − C = 1 − 0.5 = 0.5
0.25 = 0.5A + B + 0.5C = 0.5 × 0.5 + B + 0.5 × 0.5
October 18, 2016 112 / 151

H(z)
z
=
0.5z − 0.25
[z2 + 0.5z + 0.25]
+
0.5
[z + 0.5]
H(z)
z
=
z(0.5 − 0.25z−1)
z2(1 + 0.5z−1 + 0.25z−2)
+
0.5
z[1 + 0.5z−1]
H(z) =
(0.5 − 0.25z−1)
(1 + 0.5z−1 + 0.25z−2)
+
0.5
(1 + 0.5z−1)
( )
y n
0.5
Z-1
+
-0.5
( )
x n
Z-1
+
Z-1
+
-0.5
-0.25
+
0.5
-0.25
+
October 18, 2016 113 / 151

Realization of FIR system Lattice structure for FIR Systems
Lattice structure for FIR Systems
The system function of mth order FIR filter is represented by the polynomial Am(z)
Hm(z) = Am(z)
where Am(z) is the polynomial and is defined as
Am(z) = 1 +
m
X
i=1
am(i)z−i
A0(z) = 1 The system function becomes
Hm(z) = 1 +
m
X
i=1
am(i)z−i
H0(z) = 1 the z transform the input and output are related as
Hm(z) =
Y (z)
X(z)
Y (z)
X(z)
= 1 +
m
X
i=1
am(i)z−i
October 18, 2016 114 / 151

Y (z) = X(z) +
m
X
i=1
am(i)z−i
X(z)
y(n) = x(n) +
m
X
i=1
am(i)x(n − i)
Consider the order of the filter m=1
y(n) = x(n) + a1(1)x(n − 1)
From the figure y(n) is
y(n) = f0(n) + k1g0(n − 1)
f0(n) = x(n) and g0(n − 1) = x(n − 1)
y(n) = x(n) + k1x(n − 1)
+
+
Z-1 1
K
1
K
0 ( )
f n
0 ( )
g n 0 ( 1)
g n −
1( ) ( )
f n y n
=
1( )
g n
( )
x n
Figure 63: Single stage Lattice filter
October 18, 2016 115 / 151

Consider the filter of order of two.
y(n) = x(n) +
2
X
i=1
am(i)x(n − i)
= x(n) + a2(1)x(n − 1) + a2(2)x(n − 2)
The first stage output is
f1(n) = x(n) + K1x(n − 1)
g1(n) = K1x(n) + x(n − 1)
The second stage output is
f2(n) = f1(n) + K2g1(n − 1)
g2(n) = K2f1(n) + g1(n − 1)
The output y(n) is
y(n) = f1(n) + K2g1(n − 1)
y(n) = x(n) + K1x(n − 1) + K2[K1x(n − 1) + x(n − 2)]
= x(n) + K1(1 + K2)x(n − 1) + K2x(n − 2)
+
+
Z-1 1
K
1
K
0 ( )
f n
0 ( )
g n
1( )
f n
1( )
g n
( )
x n
+
+
Z-1 2
K
2
K
2 ( ) ( )
f n y n
=
2 ( )
g n
Figure 64: Two stage Lattice filter
October 18, 2016 116 / 151

y(n) = x(n) + K1(1 + K2)x(n − 1) + K2x(n − 2)
This equation represents the second order FIR system if:
a2(1) = K1(1 + K2) and a2(2) = K2
K1 =
a2(1)
(1 + K2)
and K2 = a2(2)
The lattice structure for mth order is obtained from direct form coefficients by the following
recursive equations.
Km = am(m)
am−1(i) =
am(i) − am(m)am(m − i)
1 − K2
m
i = 1, 2, . . . m − 1
October 18, 2016 117 / 151

Draw the lattice structure for the following FIR filter function
H(z) = 1 + 2z−1
+
1
3
z−2
Solution:
H(z) = 1 + 2z−1
+
1
3
z−2
a2(1) = 2, a2(2) = 1
3
Km = am(m)
for m=2
K2 = a2(2) = 1
3
am−1(i) =
1 − K2
m
a1(i) =
a2(i) − a2(2)a2(2 − i)
1 − K2
2
a1(1) =
a2(1) − a2(2)a2(1)
1 − K2
2
=
2 − (0.333)(2)
1 − (0.333)2
= 1.5
+
+
Z-1
1 1.5
K =
( )
x n
( )
y n
2 ( )
g n
1 1.5
K =
+
+
Z-1
2 0.333
K =
2 0.333
K =
Figure 65: Lattice filter
October 18, 2016 118 / 151

A FIR filter is given by
y(n) = x(n) +
2
5
x(n − 1) +
3
4
x(n − 2) +
1
3
x(n − 3)
Draw the lattice structure
Solution:
y(n) = x(n) +
2
5
x(n − 1) +
3
4
x(n − 2) +
1
3
x(n − 3)
By taking z transform
Y (z) = X(z) +
2
5
z−1
X(z) +
3
4
z−2
X(z) +
1
3
z−3
X(z)
H(z) =
Y (z)
X(z)
= 1 +
2
5
z−1
+
3
4
z−2
+
1
3
z−3
= 1 + 0.4z−1
+ 0.75z−2
+ 0.333z−3
a3(1) = 0.4, a3(2) = 0.75, a3(3) = 0.333
October 18, 2016 119 / 151

for m=3, K3 = a3(3) = 0.333
am−1(i) =
1 − K2
m
a2(i) =
a3(i) − a3(3)a3(3 − i)
1 − K2
3
a2(1) =
a3(1) − a3(3)a3(2)
1 − K2
3
=
0.4 − (0.333)(0.75)
1 − (0.333)2
= 0.169
a2(2) =
a3(2) − a3(3)a3(1)
1 − K2
3
=
0.75 − (0.333)(0.4)
1 − (0.333)2
= 0.693
for m=2, K2 = a2(2) = 0.693
a1(i) =
a2(i) − a2(2)a2(2 − i)
1 − K2
2
a1(1) =
a2(1) − a2(2)a2(1)
1 − K2
2
=
0.169 − (0.693)(0.169)
1 − (0.693)2
= 0.0998
for m=1, K1 = a1(1) = 0.0998
+
+
Z-1
1 0.0998
K =
( )
x n
( )
y n
3 ( )
g n
1 0.0998
K =
+
+
Z-1
2 0.693
K =
2 0.693
K =
+
+
Z-1
3 0.333
K =
3 0.333
K =
October 18, 2016 120 / 151

A FIR filter is given by
y(n) = x(n) + 3.1x(n − 1) + 5.5x(n − 2) + 4.2x(n − 3) + 2.3x(n − 4)
Draw the lattice structure
Solution:
y(n) = x(n) + 3.1x(n − 1) + 5.5x(n − 2) + 4.2x(n − 3) + 2.3x(n − 4)
Y (z) = X(z) + 3.1z−1
X(z) + 5.5z−2
X(z) + 4.2z−3
X(z) + 2.3z−4
X(z)
H(z) =
Y (z)
X(z)
= 1 + 3.1z−1
+ 5.5z−2
+ 4.2z−3
+ 2.3z−4
a4(1) = 3.1, a4(2) = 5.5, a4(3) = 4.2 a4(4) = 2.3
October 18, 2016 121 / 151

For m=4 K4 = a4(4) = 2.3
am−1(i) =
1 − K2
m
a3(i) =
a4(i) − a4(4)a4(4 − i)
1 − K2
4
a3(1) =
a4(1) − a4(4)a4(3)
1 − K2
4
=
3.1 − (2.3)(4.2)
1 − (2.3)2
= 1.529
a3(2) =
a4(2) − a4(4)a4(2)
1 − K2
4
=
5.5 − (2.3)(5.5)
1 − (2.3)2
= 1.6667
a3(3) =
a4(3) − a4(4)a4(1)
1 − K2
4
=
4.5 − (2.3)(3.1)
1 − (2.3)2
= 0.683
For m=3 K3 = a3(3) = 0.683
am−1(i) =
1 − K2
m
a2(i) =
a3(i) − a3(3)a3(3 − i)
1 − K2
3
a2(1) =
a3(1) − a3(3)a3(3)
1 − K2
3
=
1.529 − (0.683)(1.667)
1 − (0.683)2
= 0.732
a2(2) =
a3(2) − a3(3)a3(1)
1 − K2
3
=
1.667 − (0.683)(1.529)
1 − (0.683)2
= 1.167
For m=2, K2 = a2(2) = 1.167
October 18, 2016 122 / 151

For m=2, K2 = a2(2) = 1.167
For m=1
k1 = a1(1) =
1 − K2
m
k1 = a1(1) =
a2(1) − a2(2)a2(1)
1 − K2
2
=
0.732 − (1.167)(0.732)
1 − (1.167)2
= 0.338
+
+
Z-1
1 0.338
K =
( )
x n
( )
y n
4 ( )
g n
1 0.338
K =
+
+
Z-1
2 1.167
K =
2 1.167
K =
+
+
Z-1
3 0.683
K =
3 0.683
K =
+
+
Z-1
4 2.3
K =
4 2.3
K =
1( )
f n 3 ( )
f n
2 ( )
f n
3 ( )
g n
2 ( )
g n
1( )
g n
October 18, 2016 123 / 151

Determine all the FIR filters which are specified by the lattice parameters K1 = 0.1, K2 = 0.2,
and K3 = 0.3 and draw the structure
Solution:
am(0) = 1
am(m) = Km
am(i) = am−1(i) + am(m)am−1(m − i)
For m=1
a1(0) = 1
a1(1) = K1 = 0.1
For m=2
a2(0) = 1
a2(2) = K2 = 0.2
a2(i) = a1(i) + a2(2)a1(2 − i)
a2(1) = 0.1 + (0.2)0.1 = 0.12
For m=3
a3(0) = 1
a3(3) = K3 = 0.3
a3(i) = a2(i) + a3(3)a2(3 − i)
a3(1) = a2(1) + a3(3)a2(2)
= 0.12 + (0.3)(0.2) = 0.18
a3(2) = a2(2) + a3(3)a2(1)
= 0.2 + (0.3)(0.12) = 0.236
October 18, 2016 124 / 151

H(z) = 1 +
m
X
i=1
am(i)z−i
= 1 +
3
X
i=1
a3(i)z−i
= 1 + a3(1)z−1
+ a3(2)z−2
+ a3(3)z−3
= 1 + 0.18z−1
+ 0.236z−2
+ 0.3z−3
Z-1
+
Z-1
+
Z-1
( )
x n
1 0.18 0.236
+
0.3
y(n)
October 18, 2016 125 / 151

Determine all the FIR filters which are specified by the lattice parameters K1 = 0.5, K2 = 0.333,
and K3 = 0.25 and draw the structure
Solution:
am(0) = 1
am(m) = Km
am(i) = am−1(i) + am(m)am−1(m − i)
For m=1
a1(0) = 1
a1(1) = K1 = 0.5
For m=2
a2(0) = 1
a2(2) = K2 = 0.333
a2(i) = a1(i) + a2(2)a1(2 − i)
a2(1) = 0.5 + (0.333)0.5 = 0.665
For m=3
a3(0) = 1
a3(3) = K3 = 0.25
a3(i) = a2(i) + a3(3)a2(3 − i)
a3(1) = a2(1) + a3(3)a2(2)
= 0.665 + (0.25)(0.333) = 0.75
a3(2) = a2(2) + a3(3)a2(1)
= 0.333 + (0.667)(0.12) = 0.5
a3(3) = k3 = 0.25
a3(0) = 1, a3(1) = 0.75, a3(2) = 0.5, and a3(3) = 0.25
October 18, 2016 126 / 151

a3(0) = 1, a3(1) = 0.75, a3(2) = 0.5, and a3(3) = 0.25
H(z) = 1 +
3
X
i=1
a3(i)z−i
= K3 = 1 + a3(1)z−1
+ a3(2)z−2
+ a3(3)z−3
= 1 + 0.75z−1
+ 0.5z−2
+ 0.25z−3
Z-1
+
Z-1
+
Z-1
( )
x n
1 0.75 0.5
+
0.25
y(n)
October 18, 2016 127 / 151

Determine the impulse response of a FIR filter with reflection coefficients K1 = 0.6, K2 = 0.3,
K3 = 0.5 and K4 = 0.9 Also draw the direct form structure
Solution:
am(0) = 1
am(m) = Km
am(i) = am−1(i) + am(m)am−1(m − i)
For m=1
a1(0) = 1
a1(1) = K1 = 0.5
For m=2
a2(0) = 1
a2(2) = K2 = 0.3
a2(i) = a1(i) + a2(2)a1(2 − i)
a2(1) = 0.6 + (0.3)0.6 = 0.78
For m=3
a3(0) = 1
a3(3) = K3 = 0.5
a3(i) = a2(i) + a3(3)a2(3 − i)
a3(1) = a2(1) + a3(3)a2(2)
= 0.78 + (0.5)(0.3) = 0.93
a3(2) = a2(2) + a3(3)a2(1)
= 0.3 + (0.5)(0.78) = 0.69
a3(3) = k3 = 0.25
October 18, 2016 128 / 151

For m=4
a4(0) = 1
a4(i) = a3(i) + a4(4)a3(4 − i)
a4(1) = a3(1) + a4(4)a3(3) = 0.93 + (0.9)(0.5) = 1.38
a4(2) = a3(2) + a4(3)a3(2) = 0.69 + (0.9)(0.69) = 1.311
a4(3) = a3(3) + a4(3)a3(1) = 0.5 + (0.9)(0.93) = 1.337
a4(4) = K4 = 0.9
a4(0) = 1, a4(1) = 1.38, a4(2) = 1.311, and a4(3) = 1.337 and a4(4) = 0.9
H(z) = 1 +
3
X
i=1
a3(i)z−i
= 1 + a4(1)z−1
+ a4(2)z−2
+ a4(3)z−3
+ a4(4)z−3
= 1 + 1.38z−1
+ 1.311z−2
+ 1.337z−3
+ 0.9z−4
Z-1
+
Z-1
+
Z-1
( )
x n
1 1.38 1.311
+
1.337
y(n)
Z-1
+
0.9
y(n)
October 18, 2016 129 / 151

Realization of IIR system Lattice structure for IIR Systems
Lattice structure for IIR Systems
October 18, 2016 130 / 151

The system function of mth order IIR filter is represented by
H(z) =
M
P
k=0
bk z−k
1 +
N
P
k=1
ak z−k
Let us consider an all pole system with system function
H1(z) =
1
1 +
N
P
k=1
aN (k)z−k
=
1
AN (z)
But the system function H1(z) =
Y (z)
X(z)
Y (z)
X(z)
=
1
1 +
N
P
k=1
aN (k)z−k
Y (z) +
N
X
k=1
aN (k)z−k
Y (z) = X(z)
October 18, 2016 131 / 151

Y (z) +
N
X
k=1
aN (k)z−k
Y (z) = X(z)
y(n) = −
N
X
k=1
aN (k)y(n − k) + x(n)
Interchange the input and output
x(n) = −
N
X
k=1
aN (k)x(n − k) + y(n)
Rearranging the above equation
y(n) = x(n) +
N
X
k=1
aN (k)x(n − k)
The equation is similar to the FIR
Consider the order of the filter m=1
y(n) = a1(1)y(n − 1) + x(n)
From the above equation
K1 = a1(1)
+
+
Z-1
1
K 1( ) ( )
f n x n
=
1( )
g n
0 ( ) ( )
f n y n
=
1
K
Figure 71: Single stage Lattice
filter
+
+ Z-1
f0
(n)=y(n)
g1
(n)
f1
(n)=x(n)
K1
K1
-
Figure 72: Single stage Lattice
October 18, 2016 132 / 151

Consider the filter of order of two.
f2(n) = x(n)
f1(n) = f2(n) − K2g1(n − 1)
g2(n) = K2f1(n) + g1(n − 1)
f0(n) = f1(n) − K1g0(n − 1)
g1(n) = K1f0(n) + g0(n − 1)
y(n) = f0(n) = g0(n)
y(n) = −K1(1 + K2)y(n − 1) − K2y(n − 2) + x(n)
g2(n) = K2y(n) + K1(1 + K2)y(n − 1) − y(n − 2)
+
+ Z-1
f1
(n)
g2
(n)
f2
(n)=x(n)
K2
K2
+
+ Z-1
f0
(n)=y(n)
g1
(n)
K1
K1
- -
g0
(n)
Figure 73: Single stage Lattice filter
October 18, 2016 133 / 151

y(n) = −
N
X
k=1
aN (k)y(n − k) + x(n)
y(n) = −
2
X
k=1
a2(k)y(n − k) + x(n)
y(n) = −
2
X
k=1
a2(k)y(n − k) + x(n)
= −a2(1)y(n − 1) − a2(2)y(n − 2) + x(n)
y(n) = x(n) + K1(1 + K2)x(n − 1) + K2x(n − 2)
This equation represents the second order FIR system if:
a2(1) = K1(1 + K2) and a2(2) = K2
K1 =
a2(1)
(1 + a2(2))
and K2 = a2(2)
October 18, 2016 134 / 151

The lattice structure for mth order is obtained by the following recursive equations.
Km = am(m)
am−1(i) =
1 − K2
m
=
am(i) − Kmam(m − i)
1 − K2
m
i = 1, 2, . . . m − 1
The IIR system function is
Hsystem(z) =
M
P
k=0
bk z−k
1 +
N
P
k=1
ak z−k
=
BM (z)
AN (z)
βi = bi −
M
X
m=i+1
βmam(m − i) i = M, M − 1, M − 2, . . . 1, 0
+
+ Z-1
fN
(n)=x(n)
KN
-
+
N
β
+
+ Z-1
f0
(n)
K1
-
+
1
β
+
+ Z-1
f2
(n)
K2
-
+
2
β
fN-1(n)
KN
f1
(n)
gN-1
(n)
+
1
N
β −
g1(n)
gN
(n) g2
(n)
0
β
g0
(n)
0
β
( )
y n
Figure 74: Lattice ladder structure
October 18, 2016 135 / 151

Realize the lattice structure for all pole filter given by H(z) = 1
1+ 2
5
z−1+ 3
4
z−2+ 1
3
z−3
Solution:
m=3 a3(0) = 1, a3(1) = 2
5
, a3(2) = 3
4
, a3(3) = 1
3
am−1(i) =
am(i)−am(m)am(m−i)
1−K2
m
for m=3
a2(i) =
a3(i) − a3(3)a3(3 − i)
1 − K2
3
a2(1) =
a3(1) − a3(3)a3(2)
1 − K2
3
a2(1) =
0.4 − (0.333)(0.75)
1 − (0.33)2
= 0.16875
a2(2) =
a3(2) − a3(3)a3(1)
1 − K2
3
a2(2) =
0.75 − (0.333)(0.4)
1 − (0.33)2
= 0.6937
for m=2
a1(1) =
a2(1) − a2(2)a2(1)
1 − K2
2
=
0.1687 − (0.6973)(0.1687)
1 − (0.697)2
= 0.0996
K3 = a3(3) = 1
3
,
K2 = a2(2) = 0.6937
K1 = a1(1) = 0.0996
+
+ Z-1
g3
(n)
f3
(n)=x(n)
0.333
+
+ Z-1
f0(n)=y(n)
g2
(n)
- -
g1
(n)
+
+ Z-1
g0
(n)
0.333
0.6973 0.0996
0.6973 0.0996
f2
(n) f1(n)
October 18, 2016 136 / 151

Realize the lattice structure for all pole filter given by H(z) = 1
1−0.9z−1+0.64z−2−0.576z−3
Solution:
m=3 a3(0) = 1, a3(1) = −0.9 a3(2) = 0.64 a3(3) = −0.576
for m=3, i=1,2
a2(i) =
a3(i) − a3(3)a3(3 − i)
1 − K2
3
a2(1) =
a3(1) − a3(3)a3(2)
1 − K2
3
a2(1) =
−0.9 − (−0.576)(0.64)
1 − (−0.5763)2
= −0.795
a2(2) =
a3(2) − a3(3)a3(1)
1 − K2
3
a2(2) =
0.64 − (−0.576)(−0.9)
1 − (−0.5763)2
= 0.1819
for m=2
a1(1) =
a2(1) − a2(2)a2(1)
1 − K2
2
=
−0.795 − (0.1819)(−0.795)
1 − (0.1819)2
= −0.6726
K3 = a3(3) = −0.576,
K2 = a2(2) = 0.1819
K1 = a1(1) = −0.6726
+
+ Z-1
g3
(n)
f3
(n)=x(n)
-0.576
+
+ Z-1
f0(n)=y(n)
g2
(n)
- -
g1
(n)
+
+ Z-1
g0
(n)
-0.576
0.1819 -0.6726
0.1819 -0.6726
f2
(n) f1(n)
October 18, 2016 137 / 151

Develop the lattice ladder structure for the filter with difference equation
y(n) +
3
4
y(n − 1) +
1
4
y(n − 2) = x(n) + 2x(n − 1)
Solution:
y(n) +
3
4
y(n − 1) +
1
4
y(n − 2) = x(n) + 2x(n − 1)
Y (z) +
3
4
z−1
Y (z) +
1
4
z−2
Y (z) = X(z) + 2z−1
X(z)
H(z) =
Y (z)
X(z)
=
1 + 2z−1
1 + 3
4
z−1 + 1
4
z−2
B(z) = 1 + 2z−1
A(z) = 1 + 3
4
z−1 + 1
4
z−2
a2(1) = 0.75, a2(2) = 0.25
October 18, 2016 138 / 151

for m=2
K2 = a2(2) = 0.25
am−1(i) =
1 − K2
m
a1(i) =
a2(i) − a2(2)a2(2 − i)
1 − K2
2
a1(1) =
a2(1) − a2(2)a2(1)
1 − K2
2
=
0.75 − (0.25)(0.75)
1 − (0.25)2
= 0.6
for m=1
K1 = a1(1) = 0.6
Ladder coefficients are
B(z) = 1 + 2z−1 M=1
b0 = 1, b1 = β1 = 2
βi = bi −
M
X
m=i+1
βmam(m − i)
β0 = b0 −
1
X
m=1
βmam(m)
= b0 − β1a1(1) = 1 − 2(0.6) = −0.2
x(n)
+
+ Z-1
f0
(n)
-
+
1 2
β =
+
+ Z-1
K2=0.25
-
+
f1
(n)
g1(n)
0 0.2
β = −
g0
(n)
( )
y n
K1
=0.6
October 18, 2016 139 / 151

A linear time invariant system is defined by
H(z) =
0.129 + 0.38687z−1 + 0.3869z−2 + 0.129z−3
1 − 0.2971z−1 + 0.3564z−2 − 0.0276z−3
Realize the IIR transfer function using lattice ladder structure
Solution:
H(z) =
1
A(z)
B(z)
where
A(z) = 1 − 0.2971z−1
+ 0.3564z−2
− 0.0276z−3
B(z) = 0.129 + 0.38687z−1
+ 0.3869z−2
+ 0.129z−3
a3(3) = −0.0276, a3(2) = 0.3564, a3(1) = −0.2971
October 18, 2016 140 / 151

for m=3
K3 = a3(3) = −0.0276
am−1(i) =
1 − K2
m
a2(i) =
a3(i) − a3(3)a3(3 − i)
1 − K2
3
a2(1) =
a3(1) − a3(3)a3(2)
1 − K2
3
=
−0.2971 − (−0.0276)(0.3564)
1 − (−0.2971)2
= −0.2875
a2(2) =
a3(2) − a3(3)a3(1)
1 − K2
3
=
0.3564 − (−0.0276)(−0.2971)
1 − (−0.0.0276)2
= 0.3485
for m=2
K2 = a2(2) = 0.3485
a1(i) =
a2(i) − a2(2)a2(2 − i)
1 − K2
2
a1(1) =
a2(1) − a2(2)a2(1)
1 − K2
2
=
0.2875 − (0.3485)(−0.2875)
1 − (0.3485)2
= −0.2132
for m=1
K1 = a1(1) = −0.2132
October 18, 2016 141 / 151

Ladder coefficients are: B(z) = 0.129 + 0.38687z−1 + 0.3869z−2 + 0.129z−3
βi = bi −
M
X
m=i+1
βmam(m − i)
b0 = 0.129, b1 = 0.38687, b2 = 0.3869, b3 = 0.129 = β3
i = 2
β2 = b2 −
3
X
m=3
βmam(m − 2) = b2 − β3a3(1) = 0.3869 − (0.129)(−0.2971) = 0.4252
i = 1
β1 = b1 −
3
X
m=2
βmam(m − 1)
= b1 − β2a2(1) − β3a3(2) = 0.3867 − (0.4252)(−0.2875) − (0.129)(0.3564) = 0.4630
i = 0
β0 = b0 −
3
X
m=1
βmam(m)
= b0 − β1a1(1) − β2a2(2) − β3a3(3)
= 0.129 − (0.4630)(−0.2132) − (0.4252)(0.3485) − (0.129)(−0.0276) = 0.0831
October 18, 2016 142 / 151

+
+ Z-1
fN
(n)=x(n)
K3=-0.0276
-
+
3 0.129
β =
+
+ Z-1
f0
(n)
-
+
1 0.4630
β =
+
+ Z-1
f2
(n)
-
+
fN-1(n)
K3
f1
(n)
gN-1
(n)
2 0.4252
β =
g1(n)
gN
(n) g2
(n)
0 0.0831
β =
g0
(n)
( )
y n
K1
=-0.2132
K2=-0.3485
K2
=-0.3485
K1=-0.2132
October 18, 2016 143 / 151

H(z) =
1 + z−1 + z−2
(1 + 0.5z−1)(1 + 0.3z−1)(1 + 0.4z−1)
Realize the IIR transfer function using lattice ladder structure
Solution:
H(z) =
1 + z−1 + z−2
(1 + 0.5z−1)(1 + 0.3z−1)(1 + 0.4z−1)
=
1 + z−1 + z−2
1 + 1.2z−1 + 0.47z−2 + 0.06z−3
H(z) =
1
A(z)
B(z)
where
A(z) = 1 + 1.2z−1
+ 0.47z−2
+ 0.06z−3
B(z) = 1 + z−1
+ z−2
a3(3) = 0.06, a3(2) = 0.47, a3(1) = 1.2
October 18, 2016 144 / 151

for m=3
K3 = a3(3) = 0.06
am−1(i) =
1 − K2
m
a2(i) =
a3(i) − a3(3)a3(3 − i)
1 − K2
3
a2(1) =
a3(1) − a3(3)a3(2)
1 − K2
3
=
1.2 − (−0.06)(0.47)
1 − (−0.06)2
= 1.176
a2(2) =
a3(2) − a3(3)a3(1)
1 − K2
3
=
0.47 − (−0.06)(1.2)
1 − (0.06)2
= 0.4
for m=2
K2 = a2(2) = 0.4
a1(i) =
a2(i) − a2(2)a2(2 − i)
1 − K2
2
a1(1) =
a2(1) − a2(2)a2(1)
1 − K2
2
=
1.176 − (0.4)(1.176)
1 − (0.4)2
= 0.84
for m=1
K1 = a1(1) = 0.84
October 18, 2016 145 / 151

Ladder coefficients are
B(z) = 1 + z−1 + z−2 M=1
b0 = 1, b1 = 1, b2 = 1 = β2
βi = bi −
M
X
m=i+1
βmam(m − i)
β2 = b2 = 1
i = 1
β1 = b1 −
3
X
m=2
βmam(m − 1)
= b1 − β2a2(1)
= 1 − 1.176 = −0.176
i = 0
β0 = b0 −
1
X
m=1
βmam(m)
= b0 − β1a1(1) − β2a2(2)
= 1 + .84(0.176) − 0.4 = 0.748
+
+ Z-1
fN
(n)=x(n)
K3=-0.06
-
+
+ Z-1
f0
(n)
-
+
1 0.176
β = −
+
+ Z-1
f2
(n)
-
+
fN-1(n)
K3
=0.06
f1
(n)
gN-1
(n)
2 1
β =
g1(n)
gN
(n) g2
(n)
0 0.748
β =
g0
(n)
( )
y n
K1
=-0.84
K2=-0.4
K2
=-0.4
K1=-0.84
October 18, 2016 146 / 151

Realization of IIR system Signal Flow Graph
Signal Flow Graph
Signal flow graph (SFG) is graphical representation of block diagram structure.
The basic elements of SFG are branches and nodes. The signal out of branch is equal to
the branch gain.
Adders and pick-off points are replaced by nodes. Multipliers and delay elements are
indicated by the the branch transmittance.
The input to the system originates at source node and output signal is extracted from a
sink node.
The importance of SFG in FIR or IIR structure is transposition or flow graph reversal
theorem.
Theorem states that “if we reverse the directions of all branch transmittances and
interchange the input and output in the flow graph, the system function remains
unchanged”. The resulting structure is called a transposed structure or a transposed
form.
October 18, 2016 147 / 151

( )
x n
Z-1
+
Z-1
+
+
+
2
a
−
1
a
−
2
b
1
b
0
b ( )
y n
Figure 80: Direct Form II
( )
x n
2
a
−
1
a
−
2
b
1
b
0
b ( )
y n
1
Z−
1
Z−
Source
Node
1
5
4
3
2 Sink Node
Figure 81: Signal Flow graph
( )
y n
2
a
−
1
a
−
2
b
1
b
0
b ( )
x n
1
Z−
1
Z−
1
5
4
3
2
Figure 82: Transposed Structure SFG
( )
x n
2
a
−
1
a
−
2
b
1
b
0
b
+
+
+
Z-1
Z-1
( )
y n
Figure 83: Direct form-II realization
Figure 80 shows the direct form II structure while Figure 81 shows its signal flow graph
Figure 82 shows the Transposed Structure while Figure 83 shows its realization.
October 18, 2016 148 / 151

H(z) =
1 − 1
2
z−1 + 1
4
z−2
1 + 1
8
z−1 − 1
5
z−2 + 1
6
z−3
Draw the signal flow graph, transposed signal flow graph and transposed direct form-II structure
( )
x n
1
6
−
( )
y n
1
Z−
1
Z−
Source
Node
1
5
4
3
2 Sink Node
1
4
1
2
−
1
8
−
1
5
Figure 84: Signal flow graph
( )
y n
1
6
−
( )
x n
1
Z−
1
Z−
Source
Node
1
5
4
3
2
1
4
1
2
−
1
8
−
1
5
Figure 85: Transposed Structure signal flow
graph
( )
x n
1
5
1
8
−
1
4
1
2
−
+
+
+
Z-1
Z-1
( )
y n
Z-1
1
6
−
October 18, 2016 149 / 151

For the flow graph write difference equations and system function
( )
x n ( )
y n
1
Z−
1
Z−
1
5
1
12
−
5
24
−
1
4
1
Z−
1
Z−
Figure 87: Direct Form I
Solution:
y(n) =
1
4
y(n − 1) −
5
24
y(n − 2) −
1
12
y(n − 3) + x(n) +
1
5
x(n − 1)
H(z) =
1 + 1
5
z−1
1 − 1
4
z−1 + 5
24
z−2 + 1
12
z−3
October 18, 2016 150 / 151

For the flow graph write difference equations and system function
( )
x n ( )
y n
3
2
1
1
1
Z−
1
Z−
1
Z−
1
Z−
( )
w n
Figure 88: Direct Form II Cascade
Solution:
w(n) = x(n) + 3w(n − 1) + w(n − 2)
W (z)
X(z)
=
1
1 − 3z−1 − z−2
y(n) = w(n) + y(n − 1) + 2y(n − 2)
Y (z)
W (z)
=
1
1 − z−1 − 2z−2
H(z) =
Y (z)
X(z)
=
Y (z)
W (z)
W (z)
X(z)
=
1
(1 − z−1 − 2z−2)(1 − 3z−1 − z−2)
=
1
1 − 4z−1 + 7z−3 + 2z−4
Y (z)[1 − 4z−1
+ 7z−3
+ 2z−4
] = X(z)
Y (z) = X(z)+4z−1
Y (z)−7z−3
Y (z)−2z−4
Y (z)
y(n) = x(n)+4y(n−1)−7y(n−3)−2y(n−4)
October 18, 2016 151 / 151

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