discrete-time-systems and discetre time fourier

Important

1. UNIT - 8: Implementation of Discrete-time
Systems[ ?,?,?,?]
Dr. Manjunatha. P
[email protected]
Professor
Dept. of ECE
J.N.N. College of Engineering, Shimoga
October 18, 2016

2. Unit 8 SyllabusIntroduction
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 gures/content used in this material are redrawn, some
of the gures/pictures are downloaded from the Internet.
This material is not for
This material is prepared based on
ECE/TCE
syllabus (Karnataka State, India).
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 2 / 151

3. Unit 8 SyllabusIntroduction
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
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 3 / 151

4. Implementation of Discrete-time SystemsIntroduction
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, Ampliers.
Linearity:
If inputx1(n) produces responsey1(n)
and ifx2(n) produces responsey2(n) and
ax1(n) +bx2(n) =ay1(n) +by2(n) then the
system is calledLinear system.System
1
()xn 1
()yn
System
2
()xn 2
()yn
System
12
() ()ax n bx n+
12
() ()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 inputx(n) produces responsey(n) then ifx(nn0) produces responsey(nn0), then the
system is called as time invariant
Causality:A system is causal if the output depends only on, but not future
inputs. All memoryless systems are causal.
y[n] =x[n] +x[n1] +x[n2]: : :
Some of the examples for such analog systems are Oscillator, regulated power supply.,,
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 4 / 151

5. Realization of FIR systemRealization of FIR system
Digital lters are discrete
equations
The discrete time sytems can be of
response IIR
A FIR lter is a lter whose impulse response is of nite duration, because it settles to
zero in nite 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
−
()xn () ( 1)yn xn=−
Delay Element
a
()xn () ()yn axn=
Multiplier
+
2
()xn
1
()xn
12
() () ()yn x n x n=+
Adder
Figure 2: Basic Elements
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 5 / 151

6. Realization of FIR systemRealization of FIR system
Realization of FIR system
The dierence equation is a formula for computing an output sample at timenbased on past
and present input samples and past output samples in the time domain. The general, causal,
LTI dierence equation is as follows:
y(n) = b0x(n) +b1x(n1) +: : :bkx(nk)a1y(n1)a2y(n2): : :aky(nk)
=
M
X
k=0
bkx(nk)
N
X
k=1
aky(nk)
wherexis the input signal,yis the output signalakandbkare called the coecients.
The second term in this equation is usually termed as feedback for the system. This is the
equation used to represent Innite 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
bkx(nk)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 6 / 151

7. Realization of FIR systemRealization of FIR system
LTI systems are represented by the following dierence equation.
y(n) =
N
X
k=1
aky(nk) +
M
X
k=0
bkx(nk)
By taking z-transform on both sides
Y(z) =
N
X
k=1
akz
k
Y(z) +
M
X
k=0
bkz
k
X(z)
Y(z)
"
1 +
N
X
k=1
akz
k
#
=
M
X
k=0
bkz
k
X(z)
The system functionH(z) is dened as
H(z) =
Y(z)
X(z)
H(z) =
MP
k=0
bkz
k
1 +
NP
k=1
akz
k
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 7 / 151

8. Realization of FIR systemRealization of FIR system
An FIR system does not have feedback. Hencey(nk) term is absent in the system. FIR
output is expressed as
y(n) =
M
X
k=0
bkx(nk)
If there are M coecients then
y(n) =
M1
X
k=0
bkx(nk)
By taking z-transform on both sides
Y(z) =
M
X
k=0
bkz
k
X(z)
System functionH(z) is dened as
H(z) =
Y(z)
X(z)
=
M1
X
k=0
bkz
k
By taking inverse z transform
h(n) =

bnfor0nM1
0 otherwise
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 8 / 151

9. Realization of FIR systemDirect form Structure of FIR System
Direct form Structure of FIR System
Structures in which the multiplier coecients are precisely the coecients of the transfer
function are called direct form structure.
Sinceh(n) =bnthen y(n) is
y(n) =
M1
X
k=0
h(k)x(nk)
Expanding the summation
y(n) =h(0)x(n) +h(1)x(n1) +h(2)x(n2) +: : :h(M1)x(nM+ 1)Z
-1
h(0)
+
Z
-1
h(1)
+
Z
-1
h(M-2)
+
Z
-1
h(2)
+
h(M-1)
()xn
+
()(0)xnh
(1) ( 1)hxn− (2) ( 2)hxn−
(1)( 1)hM xn M− −+
()(0)
(1) ( 1)
xnh
hxn+−
()(0)
(1) ( 1)
(2) ( 2)
xnh
hxn
hxn
+−
+−
1
0
() ()( )
M
k
yn hkxn k
−
=
= −∑
Figure 3:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 9 / 151

10. Realization of FIR systemDirect form Structure of FIR System
Realize a direct form FIR lter for the following impulse response.
h(n) =(n) +
1
2
(n1)
1
4
(n2) +(n4) +
1
2
(n3)
Solution:h(n) =(n) +
1
2
(n1)
1
4
(n2) +
1
2
(n3) +(n4)
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(n1)
1
4
x(n2) +
1
2
x(n3) +x(n4)Z
-1
+
Z
-1
Z
-1
+
Z
-1
+
y(n)
()xn
+
(0) 1h= (4) 1h=
1
(1)
2
h=
1
(2)
4
h
−
=
1
(3)
2
h=
(1)xn−
(4)xn−(3)xn−(2)xn−
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 10 / 151

11. Realization of FIR systemDirect form Structure of FIR System
DEC-2010 EE
Realize the system functionH(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(n1) +
4
5
x(n2) +
5
9
x(n3) +
1
9
x(n4)Z
-1
+
Z
-1
+
Z
-1
()xn
(0) 1h=
3
(1)
2
h=
4
(2)
5
h=
Z
-1
++
5
(3)
9
h=
1
(4)
9
h=
y(n)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 11 / 151

12. Realization of FIR systemDirect form Structure of FIR System
June 2012 EC
A FIR lter is given byy(n) =x[n] +
2
5
x[n1] +
3
4
x[n2] +
1
3
x[n3] draw the direct form.
Solution:Z
-1
+
Z
-1
+
Z
-1
y(n)
()xn
+
(0) 1h=
2
(1)
5
h=
3
(2)
4
h=
1
(3)
3
h=
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 12 / 151

13. Realization of FIR systemDirect form Structure of FIR System
Determine a direct form realization for the following linear phase lters
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(n1) + 3x(n2) + 4x(n3) + 3x(n4) + 2x(n5) + 1x(n6)Z
-1
+
Z
-1
+
Z
-1
()xn
(0) 1h= (1) 2h= (2) 3h=
Z
-1
++
(3) 4h=
Z
-1
+
(5) 2h=(4) 3h=
Z
-1
+
y(n)
(6) 1h=
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 13 / 151

14. Realization of FIR systemDirect form Structure of FIR System
Determine a direct form realization for the following linear phase lters
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(n1) + 3x(n2) + 3x(n3) + 2x(n4) + 1x(n5)+Z
-1
+
Z
-1
+
Z
-1
()xn
(0) 1h= (1) 2h= (2) 3h=
Z
-1
++
(3) 3h=
Z
-1
+
(5) 1h=(4) 2h=
y(n)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 14 / 151

15. Realization of FIR systemDirect form Structure of FIR System
For the following FIR lter 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
()xn
(0) 1h= (1) 2.88h= (2) 3.4048h=
Z
-1
++
(3) 1.74h= (4) 0.4h=
y(n)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 15 / 151

16. Realization of FIR systemDirect form Structure of FIR System
Realize direct form FIR lter with impulse responseh(n) is given
h(n) = 4(n) + 5(n1) + 6(n2) + 7(n3). With inputx(n) = [1;2;3] calculate output
y(n)
Solution:
h(n) = 4(n) + 5(n1) + 6(n2) + 7(n3)
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(n1) + 6x(n2) + 7x(n3)0123
2
3
n
1
()xn
Figure 4: Input x(n) to the FIR lterZ
-1
+
Z
-1
+
Z
-1
()xn
(0) 4h= (1) 5h= (2) 6h=
+
(3) 7h=
y(n)
(1)xn− (3)xn−(2)xn−
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 16 / 151

17. Realization of FIR systemDirect form Structure of FIR System
y(n) = 4x(n) + 5x(n1) + 6x(n2) + 7x(n3)
y(0) = 4x(0) + 5x(01) + 6x(02) + 7x(03) = 41 = 4
y(1) = 4x(1) + 5x(0) = 42 + 51 = 13
y(2) = 4x(2) + 5x(1) + 6x(0) = 43 + 52 + 61 = 28
y(3) = 4x(3) + 5x(2) + 6x(1) + 7x(0) = 40 + 53 + 62 + 71 = 34
y(4) = 4x(4) + 5x(3) + 6x(2) + 7x(1) = 0 + 0 + 63 + 72 = 32
y(5) = 4x(5) + 5x(4) + 6x(3) + 7x(2) = 0 + 0 + 0 + 73 = 21
y(n) = [4;13;28;34;32;21]012345
n
()yn
4
13
21
34 32
28
Figure 5: Output y(n) of the FIR
lterZ
-1
+
Z
-1
+
Z
-1
()xn
(0) 4h= (1) 5h= (2) 6h=
+
(3) 7h=
y(n)
(1)xn− (3)xn−(2)xn−
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 17 / 151

18. Realization of FIR systemDirect form Structure of FIR System
June 2015Obtain 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
+
()xn
(0) 1h=
2
(1)
3
h=
15
(2)
8
h=
y(n)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 18 / 151

19. Realization of FIR systemLinear Phase FIR structure
Linear Phase FIR structure
Linear phase is a property of a
lter, where the phase response
of the lter 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 lters have a
symmetric impulse response.
The FIR lter has linear phase if
its unit sample response satises
the following condition:
h(n) =h(M1n) n= 0;1;2; : : : ;M10 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
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 19 / 151

20. Realization of FIR systemLinear Phase FIR structure
The Z transform of the unit sample response is expressed asH(z) =
M1P
n=0
h(n)z
n
For even Mh(n) =h(M1n)
H(z) =
M=21
X
n=0
h(n)
h
z
n
+z
(M1n)
i
The system expression is
H(z) =
Y(z)
X(z)
Y(z)
X(z)
=
M=21
X
n=0
h(n)
h
z
n
+z
(M1n)
i
Y(z) =
M=21
X
n=0
h(n)
h
z
n
+z
(M1n)
i
X(z)
By expanding the summation
Y(z) = h(0)
h
1 +z
(M1)
i
X(z) +h(1)
h
z
1
+z
(M2)
i
X(z) +: : :
+h(M=21)
h
z
(M=21)
+z
(M=2)
i
X(z)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 20 / 151

21. Realization of FIR systemLinear Phase FIR structure
By taking inverse z transform
y(n) = h(0)[x(n) +x(n(M1))] +h(1)[x(n1) +x(n(M2))] +: : :
+h(M=21)x[n(M=21)] +x[n(M=2)]
By considering M=8 theny(n) is
y(n) = h(0)[x(n) +x(n7)] +h(1)[x(n1) +x(n6)] +: : :
+h(2)fx(n2) +x(n5)g+h(3)fx(n3) +x(n4)gZ
-1
()xn
Z
-1
+
Z
-1
Z
-1
+
+
+
Z
-1
Z
-1
+
Z
-1
++
()yn
h(1) h(2) h(3)
h(0)
(1)xn
− (2)xn− (3)xn−
(1)xn−
(6)xn− (5)xn−
(4)xn−
(7)xn−
Figure 6:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 21 / 151

22. Realization of FIR systemLinear Phase FIR structure
For Odd Mh(n) =h(M1n)
H(z) =h

M1
2

z


M1
2

+
(M3)=2
X
n=0
h(n)
h
z
n
+z
(M1n)
i
H(z) =
Y(z)
X(z)
Y(z)
X(z)
=h

M1
2

z


M1
2

+
(M3)=2
X
n=0
h(n)
h
z
n
+z
(M1n)
i
Y(z) =h

M1
2

z


M1
2

X(z) +
(M3)=2
X
n=0
h(n)
h
z
n
+z
(M1n)
i
X(z)
By expanding the summation
Y(z) = h

M1
2

z


M1
2

X(z) +h(0)
h
1 +z
(M1)
i
X(z)
= +h(1)
h
z
1
+z
(M2)
i
X(z) +: : :
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 22 / 151

23. Realization of FIR systemLinear Phase FIR structure
By taking inverse z transform
y(n) = h

M1
2

x

n
M1
2

+h(0)[x(n) +x(n(M1))] +
+h(1)[x(n1) +x(n(M2))] +: : :
+h

M3
2

x

n
M3
2

+x

n
M+ 1
2

for M=9
y(n) = h(4)x(n4) +h(0)[x(n) +x(n8)] +h(1)[x(n1) +x(n7)] +: : :
+h(2)fx(n2) +x(n6)g+h(3)fx(n3) +x(n5)gZ
-1
()xn
Z
-1
+
Z
-1
Z
-1
+
+
+
Z
-1
Z
-1
+
Z
-1
++
()yn
h(1) h(2) h(3)
h(0)
(1)xn− (2)xn
− (3)xn−
(1)xn−
(6)xn− (5)xn−
(4)xn−
(7)xn−
Z
-1
h(4)
(8)xn−
Figure 7:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 23 / 151

24. Realization of FIR systemLinear Phase FIR structure
Realize a linear phase FIR lter with the following impulse response. Give necessary equations
h(n) =(n) +
1
2
(n1)
1
4
(n2) +(n4) +
1
2
(n3)
Solution:h(n) =f1;
1
2
;1=4;
1
2
;1g. Here M=5h(0) =h(4);h(1) =h(3)
h(n) = (n) +
1
2
(n1)
1
4
(n2) +
1
2
(n3) +(n4)
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(n1)
1
4
x(n2) +
1
2
x(n3) +x(n4)
y(n) = [x(n) +x(n4)] +
1
2
[x(n1) +x(n3)]
1
4
x(n2)Z
-1
()xn
Z
-1
+ +
+ +
()yn
h(1)=1/2
h(3)=-1/4h(0)=1
(1)xn− (2)xn
−
(3)xn−(4)xn−
Z
-1
Z
-1
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 24 / 151

25. Realization of FIR systemLinear Phase FIR structure
DEC 2010,2011, 2012 Realize a linear phase FIR lter having the following impulse response
h(n) =(n) +
1
4
(n1)
1
8
(n2) +
1
4
(n3) +(n4)
Solution:h(n) =f1;
1
4
;1=8;+
1
4
;1g. Here M=5h(0) =h(4);h(1) =h(3)
h(n) = (n) +
1
4
(n1)
1
8
(n2) +
1
4
(n3) +(n4)
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(n1)
1
8
x(n2) +
1
4
x(n3) +x(n4)
y(n) = [x(n) +x(n4)] +
1
4
[x(n1) +x(n3)]
1
8
x(n2)Z
-1
()xn
Z
-1
+ +
+ +
()yn
1
(1)xn− (2)xn−
(3)xn−(4)xn−
Z
-1
Z
-1
1
8
−
1
4
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 25 / 151

26. Realization of FIR systemLinear Phase FIR structure
May 2010 Realize a linear phase FIR lter having the following impulse response
h(n) =(n)
1
4
(n1) +
1
2
(n2) +
1
2
(n3)
1
4
(n4) +(n5)
Solution:h(n) =f1;
1
4
;
1
2
;
1
2
;
1
4
;1g. Here M=6h(0) =h(5);h(1) =h(4);h(2) =h(3)
h(n) = (n)
1
4
(n1) +
1
2
(n2) +
1
2
(n3)
1
4
(n4) +(n5)
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(n1) +
1
2
x(n2) +
1
2
x(n3)
1
4
x(n4) +x(n5)
y(n) = [x(n) +x(n5)]
1
4
[x(n1) +x(n4)] +
1
2
[x(n2) +x(n3)]Z
-1
()xn
Z
-1
+
Z
-1
Z
-1
+
+
+
Z
-1
++
()yn
1
(1)xn− (2)xn−
(3)xn−(4)xn−(5)xn−
1
4
−
1 2
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 26 / 151

27. Realization of FIR systemLinear Phase FIR structure
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
Solution:h(n) =f1;
2
3
;15=8;
2
3
;1g. Here M=5h(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(n1) +
15
8
x(n2) +
2
3
x(n3) +x(n4)
y(n) = [x(n) +x(n4)] +
2
3
[x(n1) +x(n3)] +
15
8
x(n2)Z
-1
()xn
Z
-1
+ +
+ +
()yn
1
(1)xn− (2)xn−
(3)xn−(4)xn−
Z
-1
Z
-1
15
8
2 3
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 27 / 151

28. Realization of FIR systemLinear Phase FIR structure
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(n1) +x(n+ 1)]()xn
Z
-1
+
(1)xn−
(1)xn+
Z
+
2
3
()yn
Figure 8:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 28 / 151

29. Realization of FIR systemLinear Phase FIR structure
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(n3)] +
1
4
[x(n1) +x(n2)]Z
-1
()xn
Z
-1
+ +
+
Z
-1
()yn
h(0)=1
(1)xn−
(2)xn−
(3)xn−
1
(1)
4
h=
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 29 / 151

30. Realization of FIR systemLinear Phase FIR structure
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(n1) + 3x(n2) + 4x(n3) + 3x(n4) + 2x(n5) + 1x(n6)
y(n) = 1[x(n) +x(n6)] + 2[x(n1) +x(n5)] + 3[x(n2) +x(n4)] + 4x(n3)Z
-1
()xn
Z
-1
+
Z
-1
Z
-1
+
+
+
Z
-1
Z
-1
++
()yn
2 34
1
(1)xn− (2)xn
− (3)xn−
(5)xn− (4)xn−(6)xn−  
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 30 / 151

31. Realization of FIR systemLinear Phase FIR structure
Realize a direct form for the following linear phase lters
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(n1) + 3x(n2) + 3x(n3) + 2x(n4) + 1x(n5)
y(n) = 1[x(n) +x(n5)] + 2[x(n1) +x(n4)] + 3[x(n2) +x(n3)]Z
-1
()xn
Z
-1
+
Z
-1
Z
-1
+
+
+
Z
-1
++
()yn
1
(1)xn− (2)xn−
(3)xn−(4)xn−(5)xn−
2 3
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 31 / 151

32. Realization of FIR systemFrequency Sampling for FIR Systems
Frequency Sampling for FIR Systems
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 32 / 151

33. Realization of FIR systemFrequency Sampling for FIR Systems
Frequency sampling realization is used when an FIR lter is to operate on some desired
frequency.
The desired frequency may be dened and this reduces the complexity of the system.
Consider a frequency!
!k=
2
M
k k = 0;1; : : :M1
where!kis the frequency at discrete points.
Let the unit sample response of FIR system beh(n). The fourier transform of theh(n) is
dened as
H(!) =
M1
X
n=0
h(n)e
j!n
H(!) at!=!k=
2
M
k
H(!k) =H

2
M
k

=
M1
X
n=0
h(n)e
j2kn=M
H(!k) is also written as H(k) and dened as
H(k) =
M1
X
n=0
h(n)e
j2kn=M
k= 0;1; : : :M1
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 33 / 151

34. Realization of FIR systemFrequency Sampling for FIR Systems
This equation represents the M point DFT of h(n) and it is dened as
h(n) =
1
M
M1
X
n=0
H(k)e
j2kn=M
n= 0;1; : : :M1
z transform is dened as
H(z) =
M1
X
n=0
h(n)z
n
substituting the value of h(n)
H(z) =
M1
X
n=0

1
M
M1
X
n=0
H(k)e
j2kn=M
!
z
n
Interchanging the order of summations in the above equation
H(z) =
M1
X
n=0
H(k)

1
M
M1
X
n=0
e
j2k=M
z
1
!n
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 34 / 151

35. Realization of FIR systemFrequency Sampling for FIR Systems
The geometric series formula is
N
X
n=0
a
n
=
1a
N+1
1a
Using the formula wherea=e
2kn=M
z
1
H(z) =
M1
X
k=0
H(k)
1
M

1e
j2k=M
z
1

M
1e
j2k=M
z
1
=
M1
X
k=0
H(k)
1
M
1e
j2k
z
M
1e
j2k=M
z
1
e
j2k
=cos(2k) +jsin(2k) = 1
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 35 / 151

36. Realization of FIR systemFrequency Sampling for FIR Systems
H(z) =
M1
X
k=0
H(k)
1
M
1z
M
1e
j2k=M
z
1
H(z) =
1z
M
M
M1
X
k=0
H(k)
1e
j2k=M
z
1
The equation can be considered as multiplication of two systems and dened as
H(z) =H1(z):H2(z)
where
H1(z) =
1z
M
M
H2(z) =
M1
X
k=0
H(k)
1e
j2k=M
z
1
H1(z) andH2(z) are realized independently.H(z) is obtained by multiplication ofH1(z)
andH2(z).
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 36 / 151

37. Realization of FIR systemFrequency Sampling for FIR Systems
H(z) =
1z
M
M
M1
X
k=0
H(k)
1e
j2k=M
z
1Z
-M
+
()xn
Z
-1
+ +
H(2)
4/jM
e
π
Z
-1
+ +
H(0)
Z
-1
+ +
H(1)
2/jM
e
π
Z
-1
+ +
H(M-1)
2( 1)/jMM
e
π−
()yn
Realization
of H
1
(z)
1
M
Realization
of H
2
(z)
1
-
Figure 9:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 37 / 151

38. Realization of FIR systemFrequency Sampling for FIR Systems
Determine the transfer function H(z) of an FIR lter to implement
h(n) =(n) + 2(n1) +(n2). Using frequency sampling technique
Solution:
h(0) = 1,h(1) = 2,h(2) = 1 The DFT ofh(n)
H(k) =
M1
X
n=0
h(n)e
j2kn=M
With M=3
H(k) =
2
X
n=0
h(n)e
j2kn=3
H(k) =
2
X
n=0
h(n)e
j2kn=3
=h(0) +h(1)e
j2k=3
+h(2)e
j4k=3
k=0,1,2
H(0) = 1 + 2 + 1 = 4
H(1) = 1 + 2e
j2=3
+e
j4=3
=0:5j0:866 =e
j2=3
H(2) = 1 + 2e
j4=3
+e
j8=3
=0:5j0:866 =e
j2=3
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 38 / 151

39. Realization of FIR systemFrequency Sampling for FIR Systems
The system function is
H(z) =
1z
M
M
M1
X
k=0
H(k)
1e
2k=M
z
1
For M=3
H(z) =
1z
3
3
2
X
k=0
H(k)
1e
2k=3
z
1
=
1z
3
3

H(0)
1z
1
+
H(1)
1e
j2=3
z
1
+
H(2)
1e
j4=3
z
1

=H1(z)H2(z)
where
H1(z) =
1z
3
3
H2(z) =
"
4
1z
1
+
e
j2=3
1e
j2=3
z
1
+
e
j4=3
1e
j4=3
z
1
#
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 39 / 151

40. Realization of FIR systemFrequency Sampling for FIR SystemsZ
-1
+ +
4/3j
e
π
Z
-1
+ +
4
1
Z
-1
+ +
4/3j
e
π−
Z
-3
+
()xn
1
3
2/3j
e
π−
2/3j
e
π
()yn
Figure 10:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 40 / 151

41. Realization of IIR systemRealization 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 Innite Impulse Response (IIR) lters are digital lters with innite impulse response.
Unlike FIR lters, they have the feedback (a recursive part of a lter) and are known as
recursive digital lters therefore.
IIR lters are computationally more ecient than FIR lters as they require fewer
coecients due to the fact that they use feedback.
If the coecients deviate from their true values then the feedback can make the lter
unstable.
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 41 / 151

42. Realization of IIR systemRealization of IIR system
LTI systems are represented by the following
dierence equation.
y(n) =
N
X
k=1
aky(nk) +
M
X
k=0
bkx(nk)
By taking z-transform on both sides
Y(z) =
N
X
k=1
akz
k
Y(z) +
M
X
k=0
bkz
k
X(z)
Y(z)
"
1 +
N
X
k=1
akz
k
#
=
M
X
k=0
bkz
k
X(z)
The system functionH(z) is dened as
H(z) =
Y(z)
X(z)
=
MP
k=0
bkz
k
1 +
NP
k=1
akz
k
The general expression of an IIR
lter can be expressed as follows:
H(z) =
MP
k=0
bkz
k
1 +
NP
k=1
akz
k
H1(z) =
M
X
k=0
bkz
k
H2(z) =
1
1 +
NP
k=1
akz
k
H(z) =H1(z):H2(z)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 42 / 151

43. Realization of IIR systemDirect form-I Structure of IIR System
Direct form Structure of IIR System
Structures in which the multiplier coecients are precisely the coecients of the transfer
function are called direct form structure.
The direct form structures forH1(z)H2(z) are
H1(z) =
M
X
k=0
bkz
k
=b0+b1z
1
+: : :bMz
M
H1(z) is dened as
H1(z) =
Y1(z)
X1(z)
Y1(z) =b0X1(z)+b1z
1
X1(z)+: : :bMz
M
X1(z)
Its inverse z transform is
y1(n) =b0x1(n)+b1x1(n1)+: : :bMx1(nM)Z
-1
+
b
0
Z
-1
+
1
()yn
b
1
+
b
2
Z
-1
()xn
b
M
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 43 / 151

44. Realization of IIR systemDirect form-I Structure of IIR System
H2(z) is the all pole system and is given by
H2(z) =
1
1 +
NP
k=1
akz
k
H2(z) is also expressed in terms of system function
H2(z) =
Y2(z)
X2(z)
=
1
1 +
NP
k=1
akz
k
Y2(z)[1 +
N
X
k=1
akz
k
] = X2(z)
Y2(z) =
N
X
k=1
akz
k
Y2(z) +X2(z)Z
-1
+
Z
-1
+
-a
1
+
-a
2
Z
-1
-a
N
2
()xn
2
()yn
Expanding the above function
Y2(z) =a1z
1
Y2(z)a2z
2
Y2(z)a3z
3
Y2(z) +: : :aNz
N
Y2(z) +X2(z)
Its inverse z transform is
y2(n) =a1y2(n1)a2y2(n2)a3y2(n3): : :aNy2(nN) +x2(n)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 44 / 151

45. Realization of IIR systemDirect form-I Structure of IIR System1
1
1
()
()
()
Yz
Hz
Xz
=
2
2
2
()
()
()
Yz
Hz
Xz
=
1
() ()xnxn=
12
() ()yn xn=
2
() ()yn yn=
Figure 11: Direct form I realization of IIR systemZ
-1
+
Z
-1
+
-a
1
Z
-1
+
-a
2
Z
-1
+
-a
3
+
-a
N-1
-a
N
2
()xn 2
() ()yn yn=
Z
-1
+
b
0
Z
-1
+
1
()yn
b
1
Z
-1
+
b
2
Z
-1
+
1
() ()xnxn=
b
3
+
b
M-1
b
M
All zero system All pole system Figure 12: Direct form I realization of IIR system
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 45 / 151

46. Realization of IIR systemDirect form II structure for IIR system
Direct form-II Structure of IIR System
Direct form II is also called as
the order of dierence
H(z) =
MP
k=0
bkz
k
1 +
NP
k=1
akz
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 +
NP
k=1
akz
k
W(z)[1 +
N
X
k=1
akz
k
] = X(z)
W(z) = X(z)
N
X
k=1
akz
k
W(z)
=X(z)a1z
1
W(z)a2z
2
W(z) : : :aNz
N
W(z)
w(n) = x(n)a1w(n1)a2w(n2): : :aNw(nN)
By inverse z transform
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 46 / 151

47. Realization of IIR systemDirect form II structure for IIR system
H2(z) =
Y(z)
W(z)
=
M
X
k=0
bkz
k
Y(z) =
M
X
k=0
bkz
k
W(z)
=b0W(z) +b1z
1
W(z) +b2z
2
W(z) +: : :+bMz
M
W(z)
By inverse z transform
y(n) =b0w(n) +b1w(n1) +b2w(n2) +: : :+bMw(nM)Z
-1
+
Z
-1
+
-a
1
+
-a
2
Z
-1
-a
N
()xn ()wn Z
-1
+
b
0
Z
-1
+
()yn
b
1
+
b
2
Z
-1
()wn
b
M
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 47 / 151

48. Realization of IIR systemDirect form II structure for IIR system1
()
()
()
Wz
Hz
Xz
=
2
()
()
()
Yz
Hz
Wz
=
()xn ()wn ()yn
Figure 13: Cascade connection ofH1(z)H2(z)()wn()xn
All zero system All pole system
Z
-1
+
Z
-1
+
-a
1
Z
-1
+
-a
2
Z
-1
+
-a
3
+
-a
N-1
-a
N
()yn
Z
-1
+
b
0
Z
-1
+
b
1
Z
-1
+
b
2
Z
-1
+
b
3
+
b
M-1
b
M Figure 14: Direct form-II Structure()wn()xn
Z
-1
+
Z
-1
+
-a
1
Z
-1
+
-a
2
Z
-1
+
-a
3
+
-a
N-1
-a
N
()yn
+
b
0
+
b
1
+
b
2
+
b
3
+
b
M-1
b
M Figure 15: Direct form-II Structure
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 48 / 151

49. Realization of IIR systemDirect form II structure for IIR system
A dierence equation describing a lter is given below
y(n)
3
4
y(n1) +
1
8
y(n2) =x(n) +
1
2
x(n1)
Draw the direct form I and direct form II structures
Solution:
y(n) =
3
4
y(n1)
1
8
y(n2) +x(n) +
1
2
x(n1)()yn
Z
-1
+
1
0.5()
xn
Z
-1
+
Z
-1
+
1
8
−
3 4
Figure 16: Direct form-I()wn()xn
Z
-1
+
Z
-1
+
()yn
+
1
+
3
4
1 8
−
1
2 Figure 17: Direct form-II
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 49 / 151

50. Realization of IIR systemDirect form II structure for IIR system
[July 2013]:
y(n) =0:1y(n1) + 0:2y(n2) + 3x(n) + 3:6x(n1) + 0:6x(n2)obtain the the direct
form I and direct form II structures
Solution:
y(n) =0:1y(n1) + 0:2y(n2) + 3x(n) + 3:6x(n1) + 0:6x(n2)()yn
Z
-1
+
3
3.6()
xn
Z
-1
+
Z
-1
+
0.2
0.1−
Z
-1
+
0.6
Figure 18: Direct form-I()wn()xn
Z
-1
+
Z
-1
+
-0.1
0.2
()yn
+
3
+
3.6
0.6 Figure 19: Direct form-II
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 50 / 151

51. Realization of IIR systemDirect form II structure for IIR system
June 2010 EC
Obtain direct form I and direct form II for the system described by
y(n) =0:1y(n1) + 0:72y(n2) + 0:7x(n)0:252x(n2)
Solution:
y(n) =0:1y(n1) + 0:72y(n2) + 0:7x(n)0:252x(n2)()yn
Z
-1
+
0.7()xn
Z
-1
+
Z
-1
+
0.1−
0.72
(1)yn−
(2)yn−
(1)xn−
Z
-1
-0.252
(2)xn−
Figure 20: Direct form-I()wn()xn
Z
-1
+
Z
-1
+
-0.1
0.72
()yn
+
0.7
-0.252 Figure 21: Direct form-II
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 51 / 151

52. Realization of IIR systemDirect form II structure for IIR system
A system is represented by transfer functionH(z) is given by
H(z) = 3 +
4z
z
1
2

2
z
1
4
(i)
(ii)
(iii)
governing equations for implementation
Solution:
H(z) = 3 +
4z
z
1
2

2
z
1
4
= 3 +
4z
z0:5

2
z0:25
=
7z
2
5:25z+ 1:375
z
2
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 lter.
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 52 / 151

53. Realization of IIR systemDirect form II structure for IIR system
(ii): Direct form I
H(z) =
7z
2
5:25z+ 1:375
z
2
0:75z+ 0:125
=
75:25z
1
+ 1:375z
2
10:75z
1
+ 0:125z
2
=
Y(Z)
X(Z)
Y(Z)[10:75z
1
+ 0:125z
2
] = 7X(Z)5:25z
1
X(Z) + 1:375z
2
X(Z)
y(n)0:75y(n1) + 0:125y(n2) = 7x(n)5:25x(n1) + 1:375x(n2)
y(n) = 0:75y(n1)0:125y(n2) + 7x(n)5:25x(n1) + 1:375x(n2)()yn
Z
-1
+
Z
-1
+
7
-5.25
1.375()
xn
Z
-1
+
Z
-1
+
-1.25
0.75
Figure 22: Direct form-I Structure
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 53 / 151

54. Realization of IIR systemDirect form II structure for IIR system
(iii): Direct form II canonic realization
H(z) =
75:25z
1
+ 1:375z
2
10:75z
1
+ 0:125z
2
=
Y(z)
X(z)
H(z) =
Y(z)
W(z)
W(z)
X(z)
=
75:25z
1
+ 1:375z
2
10:75z
1
+ 0:125z
2
H(z) =H1(z):H2(z) =
1
10:75z
1
+ 0:125z
2
[75:25z
1
+ 1:375z
2
]
H1(z) =
W(z)
X(z)
=
1
10:75z
1
+ 0:125z
2
W(z)[10:75z
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(n1)0:125w(n2)()wn()xn
Z
-1
+
Z
-1
+
-1.25
0.75
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 54 / 151

55. Realization of IIR systemDirect form II structure for IIR system
H2(z) =
Y(z)
W(z)
= 75:25z
1
+ 1:375z
2
Y(z)
W(z)
= 75:25z
1
+ 1:375z
2
Y(z) = 7W(z)5:25z
1
W(z) + 1:375z
2
W(z)
Taking inverse z transform
y(n) = 7w(n)5:25w(n1) + 1:375w(n2)()wn ()yn
Z
-1
+
Z
-1
+
7
-5.25
1.375
Figure 23: Direct form ofH2(z)()wn()xn
Z
-1
+
Z
-1
+
-1.25
()yn
Z
-1
+
Z
-1
+
7
-5.25
0.75
1.375 Figure 24: Realization ofH(z) =H1(z):H2(z)()wn()xn
Z
-1
+
Z
-1
+
0.75
-0.125
()yn
+
7
+
-5.25
1.325 Figure 25: Direct form-II, canonic form
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 55 / 151

56. Realization of IIR systemDirect form II structure for IIR system
2012 DEC, 2012 June
Obtain the direct form II (canonic) realization for the following system.
H(z) =
(z1)(z
2
+ 5z+ 6)(z3)
(z
2
+ 6z+ 5)(z
2
6z+ 8)
Solution:
H(z) =
(z1)(z
2
+ 5z+ 6)(z3)
(z
2
+ 6z+ 5)(z
2
6z+ 8)
=
(z
2
4z+ 3)(z
2
+ 5z+ 6)
(z
2
+ 6z+ 5) + (z
2
6z+ 8)
=
z
4
+z
3
11z
2
9z+ 18
z
4
23z
2
+ 18z+ 40
=
1 +z
1
11z
2
9z
3
+ 18z
4
123z
2
+ 18z
3
+ 40z
4()wn()xn
Z
-1
+
Z
-1
()yn
+
-11
+
Z
-1
+
-9
+
1
Z
-1
+
40−
23
18−
+
18
Figure 26: Direct form-II
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 56 / 151

57. Realization of IIR systemDirect form II structure for IIR system
2011 July
Obtain the direct form II realizations of the following system.
H(z) =
(1 +z
1
)
(1
1
4
z
1
)(1z
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(n1)
3
4
w(n2) +
1
8
w(n3)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 57 / 151

58. Realization of IIR systemDirect form II structure for IIR system
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(n1)()wn()xn
Z
-1
+
Z
-1
+
()yn
Z
-1
+
1
-1
Z
-1
5
4
3
4
−
1 8
Figure 27:()wn()xn
Z
-1
+
Z
-1
+
()yn
+
1
-1
Z
-1
5
4
3
4
−
1 8 Figure 28:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 58 / 151

59. Realization of IIR systemDirect form II structure for IIR system
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
)()yn
Z
-1
+
1
5()
xn
Z
-1
+
Z
-1
+
1
4
−
Z
-1
+
5
24
1
12
−
(1)yn
−
(3)yn−
(2)yn−
(1)xn−
Figure 29: Direct form-I
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(n1) +
5
24
y(n2)
1
12
y(n3) =x(n) + 5x(n1)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 59 / 151

60. Realization of IIR systemDirect form II structure for IIR system
Obtain direct form II for the system described byH(z) =
8Z
3
4Z
2
+11Z2
(Z
1
4
)(Z
2
Z+
1
2
)
[December 2010 EC]
Solution:
H(z) =
8Z
3
4Z
2
+ 11Z2
(Z
1
4
)(Z
2
Z+
1
2
)
=
Y(z)
X(z)
=
8Z
3
4Z
2
+ 11Z2
Z
3
1:25Z
2
+ 0:75Z0:125
=
84Z
1
+ 11Z
2
2Z
3
11:25Z
1
+ 0:75Z
2
0:125Z
1()wn()xn
Z
-1
+
Z
-1
+
1.25
-0.75
()yn
+
8
11
+
Z
-1
+
0.125 -2
+
-4
Figure 30: Direct form-II
By taking inverse Z transform on both sides
y(n)1:25y(n1) + 0:75y(n2)0:125y(n3) = 8x(n)4x(n1) + 11x(n2)2x(n3)
y(n) = 8x(n)4x(n1) + 11x(n2)2x(n3) + 1:25y(n1)0:75y(n2) + 0:125y(n3)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 60 / 151

61. Realization of IIR systemDirect form II structure for IIR system
Obtain direct form II for the system described by
H(z) =
(z1)(z2)(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)
=
(Z
2
1)(Z
2
2Z)
(Z
2
Z+
1
2
)(Z
2
+
1
16
)
=
Z
4
2Z
3
Z
2
+ 2Z
Z
4
Z
3
+
9
16
Z
2

1
16
Z+
1
32
=
12Z
1
Z
2
+ 2Z
3
1Z
1
+
9
16
2

1
16
Z
3
+
1
32
Z
4
By taking inverse Z transform on both sides
y(n)y(n1) +
9
16
y(n2)
1
16
y(n3) +
1
32
y(n4) =x(n)2x(n1)x(n2) + 2x(n3)
y(n) =x(n)2x(n1)x(n2) + 2x(n3) +y(n1)
9
16
y(n2) +
1
16
y(n3)
1
32
y(n4)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 61 / 151

62. Realization of IIR systemDirect form II structure for IIR system()yn
Z
-1
+
1
-2()
xn
Z
-1
+
Z
-1
+
1
Z
-1
+
9
16
−
1
16
(1)yn
−
(3)yn−
(2)yn−
(1)xn−
Z
-1
-1
(2)xn−
Z
-1
2
(3)xn−
+
+
Z
-1
1
32
−
(4)yn
−
+
Figure 31: Direct form-I()wn()xn
Z
-1
+
Z
-1
+
1
()yn
+
1
-1
+
Z
-1
+
2
+
-2
Z
-1
+
1
32
−
9
16
−
1
16 Figure 32: Direct form-II
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63. Realization of IIR systemDirect form II structure for IIR system
Realize direct form-I and form -II for linear time invariant system which is described by the
following input output relation
2y(n)y(n2)4y(n3) = 3x(n2)
Solution:
y(n) = 0:5y(n2) + 2y(n3) + 1:5x(n2)()yn
Z
-1
()xn
Z
-1
+
Z
-1
0.5
Z
-1
1.5
Z
-1
+
2
(a) Direct form-I()wn()xn
Z
-1
+
Z
-1
()yn
1.5
Z
-1
+
0.5
2 (b) Direct form-II
Figure 33:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 63 / 151

64. Realization of IIR systemDirect form II structure for IIR system
Realize direct form-I and form -II for system given by
H(z) =
z
1
3z
2
(10z
1
)(1 + 0:5z
1
+ 0:5z
2
)
Solution:
H(z) =
z
1
3z
2
(10z
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(n1) + 0:45y(n2)0:05y(n3) = 0:1x(n1)0:3x(n2)
y(n) =0:4y(n1)0:45y(n2) + 0:05y(n3) + 0:1x(n1)0:3x(n2)()yn
Z
-1
()xn
Z
-1
+
Z
-1
0.45−
Z
-1
-0.3
Z
-1
+
0.05
+
0.1
+
0.4−
(a) Direct form-I()wn()xn
Z
-1
+
Z
-1
()yn
-0.3
Z
-1
+
0.45−
0.05
+
0.4−
+
0.1 (b) Direct form-II
Figure 34:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 64 / 151

65. Realization of IIR systemCascade Form Structure
Cascade Form Structure
H(z) =
MP
k=0
bkz
k
1 +
NP
k=1
akz
k
=
b0+b1z
1
+b2z
2
+bMz
M
1 +a1z
1
+a2z
2
+aNz
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 andHk(z) has the general form
Hk(z) =
bk0+bk1z
1
+bk2z
2
1 +ak1z
1
+ak2z
21
() ()xnxn=
H
1
(z)
12
() ()yn xn=
H
2
(z)
23
() ()yn xn=
H
k
(z)
() ()
k
yn yn=
1
()
()
k
k
yn
xn
−
=
Figure 35: Cascade realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 65 / 151

66. Realization of IIR systemCascade Form Structure
In cascade eachHk(z) are represented as
Hk(z) =
Yk(z)
Xk(z)
=
Wk(z)
Xk(z)
Yk(z)
Wk(z)
=Hk1(z):Hk2(z)
whereHk1(z) andHk2(z) dened as
Hk1(z) =
Wk(z)
Xk(z)
=
1
1 +ak1z
1
+ak2z
2
Hk2(z) =bk0+bk1z
1
+bk2z
2()wn
1
() ()
k
ynxn
−
=
Z
-1
+
Z
-1
+
+
+
2k
a−
1k
a−
2k
b
1k
b
0k
b
1
() ()
kk
yn x n
+
=
Figure 36: Cascade realization
wk(n) = ak1wk(n1)ak2wk(n2) +yk1(n)
yk(n) = bk0wk(n) +bk1wk(n1) +bk2wk(n2)
yk1(n=xk(n)
yk(n) = xk+1(n)
y0(n) = x(n)
y(n) = xk(n)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 66 / 151

67. Realization of IIR systemCascade Form Structure
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
)()xn
Z
-1
+
()yn
+
+
Z
-1
Z
-1
+
1
5
1
3
−
1
4
−
1
2
1
11
a−
21
b
22
a−
21
a−
20
b
Figure 37: Cascade realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 67 / 151

68. Realization of IIR systemCascade Form Structure
Realize the following system function in cascade form
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
)()xn
Z
-1
+
()yn
+
+
Z
-1
Z
-1
+
1
2
−
1
2
−
1
5
1
4
1
Z
-1
+
1
6
−
Figure 38: Cascade realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 68 / 151

69. Realization of IIR systemCascade Form Structure
Realize the following system function in cascade form
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
)()xn
Z
-1
+
()yn
+
+
Z
-1
Z
-1
+
1
3
1
8
−
1
5
3
4
1
Figure 39: Cascade realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 69 / 151

70. Realization of IIR systemCascade Form Structure
The transfer function of a discrete system is given as follows
H(z) =
1z
1
(10:2z
1
0:15z
2
)
Draw the cascade realization
Solution:
H(z) =H1(z)H2(z) = (1z
1
)
1
(10:2z
1
0:15z
2
)()xn
Z
-1
()yn
+
+
Z
-1
Z
-1
0.15
0.2
-1
1
+
1
()Hz
2
()Hz
Figure 40: Cascade realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 70 / 151

71. Realization of IIR systemCascade Form Structure
December 2010 EE
Draw the cascade realization for the following system
y(n) = 0:75y(n1)0:125y(n2) + 6x(n) + 7x(n1) +x(n2)
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
10:75z
1
+ 0:125z
2
H(z) =
Y(z)
X(z)
=
6 + 7z
1
+z
2
10:75z
1
+ 0:125z
2
=
6z
2
+ 7z+ 1
z
2
0:75z+ 0:125
=
(6z+ 1)(z+ 1)
(z0:5)(z0:25)
=
(6 +z
1
)(1 +z
1
)
(10:5z
1
)(10:25z
1
)
=H1(z)H2(z)
a= 1;b=0:75;c= 0:125
Roots of the quadratic equation for
ax
2
+bx+c= 0 are
b
q
b
2
4ac
2a
=
0:75
p
0:75
2
4(0:125)
2
=
0:75
p
0:0625
2
=
0:750::25
2
=)0:5;0:25
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 71 / 151

72. Realization of IIR systemCascade Form Structure
H1(z) =
6 +z
1
10:5z
1
H2(z) =
(1 +z
1
)
(10:25z
1
)()yn
2
()Hz
()xn
Z
-1
1
+
1
()Hz
+
0.5
6
Z
-1
1
++
0.25
1
Figure 41:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 72 / 151

73. Realization of IIR systemCascade Form Structure
A system function is specied by its transfer function H(z) given by
H(z) =
(z1)(z2)(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) =
(z1)(z2)(z+ 2)z
[z(
1
2
+j
1
2
)][z(
1
2
j
1
2
)](z
j
4
)(z+
j
4
)
=
z
2
3z+ 2
z
2
z+ 0:5

z
2
+ 2z
z
2
+ 0:0625
=
13z
1
+ 2z
2
1z
1
+ 0:5z
2

1 + 2z
1
1 + 0:0625z
2
=H1(z)H2(z)()xn
Z
-1
+
Z
-1
+
1
-0.5
+
1
+
-3
2
()yn
+
Z
-1
Z
-1
+
0.0625−
1
2
Figure 42: Cascade realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 73 / 151

74. Realization of IIR systemCascade Form Structure
Obtain the cascade realization of the following system. The system should have two biquadratic
sections. [EC December 2012]
H(z) =
(z1)(z
2
+ 5z+ 6)(z3)
(z
2
+ 6z+ 5)(z
2
6z+ 8)
Solution:
H(z) =
(z1)(z
2
+ 5z+ 6)(z3)
(z
2
+ 6z+ 5)(z
2
6z+ 8)
=
z
2
4z+ 3
z
2
+ 6z+ 5

z
2
+ 5z+ 6
z
2
6z+ 8
=
14z
1
+ 3z
2
1 + 6z
1
+ 5z
2

1 + 5z
1
+ 6z
2
16z
1
+ 8z
2()xn
Z
-1
+
Z
-1
+
-6
-5
+
1
+
-4
3
Z
-1
+
Z
-1
+
6
-8
+
1
+
5
6
()yn
Figure 43:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 74 / 151

75. Realization of IIR systemCascade Form Structure
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(z0:5)(z0:6667)(z+ 2)
(z0:75)(z0:125)[z(0:5 +j0:5)][z(0:5j0:5)]
=
10z(z0:5)
(z0:75)(z0:125)

(z0:6667)(z+ 2)
[z(0:5 +j0:5)][z(0:5j0:5)]
=
10z
2
5z
z
2
0:875z+ 0:0938

z
2
+ 1:333z1:333
z
2
z+ 0:5
=
105z
1
10:875z
1
+ 0:0938z
2

1 + 1:333z
1
1:333z
2
1z
1
+ 0:5z
2()xn
Z
-1
+
Z
-1
+
0.875
-0.0938
+
10
-5
Z
-1
+
Z
-1
+
1
-0.5
+
1
+
-1.333
()yn
1.333
Figure 44: Cascade realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 75 / 151

76. Realization of IIR systemCascade Form Structure
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()yn
Z
-1
+
()xn
Z
-1
+
Z
-1
+
1
4
−
1
2
−
+
1
4
1
2
−
Figure 45: Cascade realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 76 / 151

77. Realization of IIR systemCascade Form Structure
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
1z
1
+
1
2
z
2
=H1(z)H2(z)()yn
Z
-1
+
()xn
Z
-1
+
Z
-1
+
1
2
−
1
+
1
1 4
1
+
+
1
2
1
Figure 46: Cascade realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 77 / 151

78. Realization of IIR systemCascade Form Structure
EC 2011 July
Obtain the direct form II realizations of the following system.
H(z) =
(1 +z
1
)
(1
1
4
z
1
)(1z
1
+
1
2
z
2
)
H(z) =
(1 +z
1
)
(1
1
4
z
1
)(1z
1
+
1
2
z
2
)
=
1 +z
1
(1
1
4
z
1
)

1
1z
1
+z
2
=H1(z)H2(z)()yn
Z
-1
+
()xn
Z
-1
+
Z
-1
+
1−
1
+
1
1
4
Figure 47: Cascade realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 78 / 151

79. Realization of IIR systemCascade Form Structure
June 2010 EC
Obtain cascade form for the system described by
y(n) =0:1y(n1) + 0:72y(n2) + 0:7x(n)0:252x(n2)
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)
=
0:70:252z
2
1 + 0:1z
1
0:72z
2
= (0:70:252z
2
)
1
1 + 0:1z
1
0:72z
2()xn
Z
-1
()yn
+
+
Z
-1
Z
-1
0.72
0.1−
-0.252
1
+
1
()Hz
2
()Hz
Z
-1
0.7
Figure 48:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 79 / 151

80. Realization of IIR systemCascade Form Structure
June 2015 EC
Find the transfer function and dierence equation realization shown in Figure()xn
Z
-1
+
Z
-1
()yn
+
4
+
2−
3
()wn
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 dierence equation is
y(n) =2y(n2) + 4x(n) + 3x(n1)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 80 / 151

81. Realization of IIR StructuresParallel Form Structure for IIR System
Parallel Form Structure for IIR System
The system function for IIR sytem is
H(z) =
MP
k=0
bkz
k
1 +
NP
k=1
akz
k
=
b0+b1z
1
+b2z
2
+bMz
M
1 +a1z
1
+a2z
2
+aNz
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()xn
()yn
2
()Hz
()
k
Hz
1
()Hz +
+
+
C
Figure 49: Parallel form realization for
IIR systems
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 81 / 151

82. Realization of IIR StructuresParallel Form Structure for IIR System
Hk(z) =
bk0+bk1z
1
1 +ak1z
1
+ak2z
2
wk(n) = ak1wk(n1)ak2wk(n2) +x(n)
yk(n) = bk0wk(n)bk1wk(n1)
y(n) = C x(n) +
K
X
k=1
yk(n)()
k
wn()
k
xn
Z
-1
+
Z
-1
+
()
k
yn
+
+
1k
a−
2k
a−
1k
b
0k
b
Figure 50: Direct form II of second order
subsystem
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 82 / 151

83. Realization of IIR StructuresParallel Form Structure for IIR System
Realize the following System in parallel form:
H(z) = 3 +
4z
z
1
2
+
2
z
1
4
Solution:
H(z) = 3 +
4
10:5z
1
+
2z
1
1
1
4
z
1()yn
+
Z
-1
+
0.25
Z
-1
+
0.5
()xn
3
4
2
Figure 51: Parallel realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 83 / 151

84. Realization of IIR StructuresParallel Form Structure for IIR System
Realize the following System in parallel form:
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)(z0:25)(z+ 0:125)
H(z) =
z(z+ 1)(z+ 2)
(z+ 0:5)(z0:25)(z+ 0:125)
H(z)
z
=
(z+ 1)(z+ 2)
(z+ 0:5)(z0:25)(z+ 0:125)
H(z)
z
=
A
(z+ 0:5)
+
B
(z0:25)
+
C
(z+ 0:125)
(z+ 1)(z+ 2)
(z+ 0:5)(z0:25)(z+ 0:125)
=
A
(z+ 0:5)
+
B
(z0:25)
+
C
(z+ 0:125)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 84 / 151

85. Realization of IIR StructuresParallel Form Structure for IIR System
Coecients A B and C are determined
A=
H(z)
z
[z+ 0:5]jz=0:5=
(z+ 1)(z+ 2)
(z0:25)(z+ 0:125)
=
(0:5 + 1)(0:5 + 2)
(0:50:25)(0:5 + 0:125)
= 2:66
B=
H(z)
z
[z0:25]jz=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]jz=0:125=
(z+ 1)(z+ 2)
(z+ 0:5)(z0:25)
=
(0:125 + 1)(0:125 + 2)
(0:125 + 0:5)(0:1250:25)
=11:66
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 85 / 151

86. Realization of IIR StructuresParallel Form Structure for IIR System
H(z)
z
=
2:66
(z+ 0:5)
+
10
(z0:25)
+
11:66
(z+ 0:125)
H(z) =
2:66
(1 + 0:5z
1
)
+
10
(10:25z
1
)
+
11:66
(1 + 0:125z
1
)()yn
+
Z
-1
+
0.25
Z
-1
+
0.5−
10
2.66
()xn
Z
-1
+
0.125−
11.66−
Figure 52: Parallel realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 86 / 151

87. Realization of IIR StructuresParallel Form Structure for IIR System
Realize the following System in parallel form:
H(z) =
1z
1
(10:2z
1
0:15z
2
)
Solution
H(z) =
1z
1
(10:2z
1
0:15z
2
)
H(z) =
z
2
z
(z
2
0:2z0:15)
a= 1;b=0:2;c=0:15
Roots of the equationax
2
+bx+c= 0 are =
b
p
b
2
4ac
2a
=
0:2
p
0:2
2
4(0:15)
2
=
0:2
p
0:64
2
=
0:20:8
2
)0:5;0:3
H(z)
z
=
z1
(z0:5)(z+ 0:3)
H(z)
z
=
A
(z0:5)
+
B
(z+ 0:3)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 87 / 151

88. Realization of IIR StructuresParallel Form Structure for IIR System
The values of A and B are determined by the following procedure.
A=
H(z)
z
(z0:5)jz=0:5=
z1
(z+ 0:3)
=
0:51
(0:5 + 0:3)
=0:625
B=
H(z)
z
(z+ 0:3)jz=0:3=
z1
(z0:5)
=
0:31
(0:30:5)
= 1:625
H(z)
z
=
0:625
z0:5
+
1:625
z+ 0:3
H(z) =
0:625
10:5z
1
+
1:625
1 + 0:3z
1
=H1(z) +H2(z)()yn
+
Z
-1
+
0.3−
Z
-1
+
0.5
1.625
0.625−
()xn
Figure 53: Parallel realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 88 / 151

89. Realization of IIR StructuresParallel Form Structure for IIR System
Realize the following System in parallel form:
H(z) =
1 + 0:33z
1
(10:75z
1
+ 0:125z
2
)
Solution
H(z) =
1 + 0:33z
1
(10:75z
1
+ 0:125z
2
)
=
z
1
(z+ 0:33)
z
2
(z
2
0:75z
1
+ 0:125z
2
)
H(z)
z
=
z+:33
(z
2
0:75z+ 0:125)
a= 1;b=0:75;c= 0:125
Roots of the equation are
=
0:75
p
0:75
2
4(0:125)
2
=
0:75
p
0:0625
2
=
0:750:25
2
)0:5;0:25
H(z)
z
=
z+ 0:33
(z0:5)(z0:25)
=
A
(z0:5)
+
B
(z0:25)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 89 / 151

90. Realization of IIR StructuresParallel Form Structure for IIR System
The values of A and B are determined by the following procedure.
A=
H(z)
z
(z0:5)jz=0:5=
z+ 0:33
(z0:25)
=
0:25 + 0:33
(0:50:25)
= 3:33
B=
H(z)
z
(z0:25)jz=0:25=
z+ 0:33
(z0:5)
=
0:25 + 0:33
(0:250:5)
=2:33
H(z)
z
=
3:33
(z0:5)
+
2:33
(z0:25)
H(z) =
3:33
10:5z
1
+
2:33
10:25z
1
=H1(z) +H2(z)()yn
+
Z
-1
+
0.25
Z
-1
+
0.5
2.33−
3.33
()xn
Figure 54: Parallel realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 90 / 151

91. Realization of IIR StructuresParallel Form Structure for IIR System
Realize the following System in parallel form:
y(n) = 0:75y(n1)0:125y(n2) + 6x(n) + 7x(n1) +x(n2)
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
10:75z
1
+ 0:125z
2
H(z) =
z
2
[6z
2
+ 7z+ 1]
z
2
[z
2
0:75z+ 0:125]
H(z) =
z
2
[6z
2
+ 7z+ 1]
z
2
[z
2
0:75z+ 0:125]
=
[6z
2
+ 7z+ 1]
[z
2
0:75z+ 0:125]
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 91 / 151

92. Realization of IIR StructuresParallel Form Structure for IIR System
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
p
0:0625
2
=
0:750:25
2
0:5;0:25
H(z) =
6z
2
+ 7z+ 1
(z0:5)(z0:25)
H(z)
z
=
6z
2
+ 7z+ 1
z(z0:5)(z0:25)
H(z)
z
=
A
z
+
B
(z0:5)
+
C
(z0:25)
A=H(z)zjz=0=
6z
2
+ 7z+ 1
z(z0:5)(z0:25)
z=
1
(0:5)(0:25)
= 8
B=H(z)(z0:5)jz=0:5=
6z
2
+ 7z+ 1
z(z0:5)(z0:25)
(z0:5z) =
6(0:5)
2
+ 7(0:5) + 1
0:5[0:50:25]
= 48
C=H(z)(z0:25z)jz=0:25=
6z
2
+ 7z+ 1
z(z0:5)(z0:25)
(z0:25)
=
6(0:25)
2
+ 7(0:25) + 1
0:25[0:250:5]
=50
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 92 / 151

93. Realization of IIR StructuresParallel Form Structure for IIR System
H(z) = 8 +
48
(10:5z
1
)
+
50
(10:25z
1
)
=H1(z) +H2(z) +H3(z)()yn
+
Z
-1
+
0.25
Z
-1
+
0.5
()xn
8
48
50−
Figure 55:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 93 / 151

94. Realization of IIR StructuresParallel Form Structure for IIR System
Realize the following System in parallel form:
H(z) =
8z
3
4z
2
+ 11z2
[z
1
4
][z
2
z+
1
2
]
H(z)
z
=
8z
3
4z
2
+ 11z2
z[z
1
4
][z
2
z+
1
2
]
H(z)
z
=
8z
3
4z
2
+ 11z2
z[z
1
4
][z
2
z+
1
2
]
H(z)
z
=
A
z
+
B
z0:25
+
Cz+D
z
2
z+:5
A=
H(z)
z
zjz= 0 =
8z
3
4z
2
+ 11z2
[z0:25][z
2
z+ 0:5]
=
2
[0:25][0:5]
= 16
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 94 / 151

95. Realization of IIR StructuresParallel Form Structure for IIR System
B=
H(z)
z
[z0:5]jz=0:5=
8z
3
4z
2
+ 11z2
z[z
2
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
=
8z
3
4z
2
+ 11z2
z[z0:25][z
2
z+ 0:5]
8z
3
4z
2
+ 11z2
z[z0:25][z
2
z+ 0:5]
=
A
z
+
B
z0:25
+
Cz+D
z
2
z+:25
8z
3
4z
2
+ 11z2 = A[z0:25][z
2
z+ 0:5] +
+Bz[z
2
z+:25] + (Cz+D)z[z0:25]
8z
3
4z
2
+ 11z2z=z
3
[A+B+C] +z
2
[1:25AB0:25C+D] +
+z[0:5A+:25B:25D]0:0625A
8 = A+B+C= 16 + 8 +C
C=16
4 = [1:25AB0:25C+D] = [208 + 4 +D]
D= 20
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 95 / 151

96. Realization of IIR StructuresParallel Form Structure for IIR System
H(z)
z
=
A
z
+
B
z0:25
+
Cz+D
z
2
z+:5
H(z)
z
=
16
z
+
8
z0:25
+
16z+ 20
z
2
z+:5
H(z) = 16 +
8
10:25z
1
+
16 + 20z
1
1z
1
+:5z
2()yn
+
Z
-1
+
1
Z
-1
+
0.25
()xn
16
8
-16
+
Z
-1
0.5−
+
20
Figure 56: Parallel realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 96 / 151

97. Realization of IIR StructuresParallel Form Structure for IIR System
EE 2010 May
Realize the following System in parallel 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) =
z
1
(z0:2)
z
2
[z
2
0:5z+ 0:33]z
1
[z+ 0:25]
H(z) =
z
2
(z0:2)
[z
2
0:5z+ 0:33z][z+ 0:25]
H(z)
z
=
z(z0:2)
[z
2
0:5z+ 0:33][z+ 0:25]
H(z)
z
=
Az+B
[z
2
0:5z+ 0:333]
+
C
[z+ 0:25]
z(z0:2)
[z
2
0:5z+ 0:33][z+ 0:25]
=
Az+B
[z
2
0:5z+ 0:333]
+
C
[z+ 0:25]
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 97 / 151

98. Realization of IIR StructuresParallel Form Structure for IIR System
Coecients A B and C are determined
C=
H(z)
z
[z+ 0:25]jz=0:25=
(z0:2)
[z
2
0:5z+ 0:333]
=
0:25(0:250:2)
[(0:25)
2
(0:5)(0:25) + 0:333]
= 0:217
z(z0:2)
[z
2
0:5z+ 0:33][z+ 0:25]
=
Az+B
[z
2
0:5z+ 0:333]
+
C
[z+ 0:25]
z(z0:2) = (Az+B)[z+ 0:25] +C[z
2
0:5z+ 0:33]
z
2
0:2z=Az
2
+ 0:25Az+Bz+ 0:25B+Cz
2
0:5Cz+ 0:33C]
z
2
0:2z=z
2
(A+C) +z(0:25A+BC) + 0:25B+ 0:33C
Equating coecients
A+C= 1
A= 1C= 10:217 = 0:783
0:25B+ 0:33C= 0
B=
0:33C
0:25
=0:2864
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 98 / 151

99. Realization of IIR StructuresParallel Form Structure for IIR System
H(z)
z
=
0:78z+0:2864
[z
2
0:5z+ 0:333]
+
0:216
[z+ 0:25]
H(z)
z
=
z(0:780:2864z
1
)
z
2
(10:5z
1
+ 0:333z
2
)
+
0:216
z[1 + 0:25z
1
]
H(z) =
(0:780:2864z
1
)
(10:5z
1
+ 0:333z
2
)
+
0:216
(1 + 0:25z
1
)()yn
0.216
Z
-1
+
0.25
()xn
Z
-1
+
Z
-1
+
0.5
-0.333
+
0.78
-0.2864
+
Figure 57: Parallel realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 99 / 151

100. Realization of IIR StructuresParallel Form Structure for IIR System
EC 2012 December
Realize the following System in parallel form:
H(z) =
(z1)(z2)(z+ 2)z
[z(
1
2
+j
1
2
)][z(
1
2
j
1
2
)](z
j
4
)(z+
j
4
)
Solution:
H(z) =
(z1)(z2)(z+ 2)z
[z(
1
2
+j
1
2
)][z(
1
2
j
1
2
)](z
j
4
)(z+
j
4
)
H(z)
z
=
(z1)(z2)(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(0:5j0:5)
+
C
zj0:25
+
D
z+j0:25)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 100 / 151

101. Realization of IIR StructuresParallel Form Structure for IIR System
H(z)
z
=
A
z(0:5 +j0:5)
+
B
z(0:5j0:5)
+
C
zj0:25
+
D
z+j0:25)
A=
H(z)
z
[z(0:5 +j0:5)]jz=0:5+j0:5=
(z
2
1)(z2)
[z(0:5j0:25)][zj0:25][z+j0:25]
=
[(0:5 +j0:5)
2
1][(0:5 +j0:52]
[0:5 +j0:5(0:5j0:5)][z
2
+ 0:0625]
=
[0:25 +j0:50:251][(0:5 +j0:52]
[j1][z
2
+ 0:0625]
=
[1 +j0:5][(1:5 +j0:5]
[j1][j0:5 + 0:0625]
=
[1:5j0:5j0:75 + 0:25]
0:5 +j0:0625
=
1:25j1:25
0:5 +j0:0625
=
1:767\45
0:503\172
=
1:767\45
0:503\172
= 3:513\217 =2:8 +j2:1
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 101 / 151

102. Realization of IIR StructuresParallel Form Structure for IIR System
Similarly B, C and D are estimated.
H(z)
z
=
A
z(0:5 +j0:5)
+
B
z(0:5j0:5)
+
C
zj0:25
+
D
z+j0:25)
=
2:8 +j2:1
z(0:5 +j0:5)
+
2:7683j2:1517
z(0:5j0:5)
+
3:268j7:837
zj0:25
+
3:268 +j7:837
z+j0:25)
We have to design second order system, hence combine rst two terms and last two terms.
H(z)
z
=
5:536z+ 0:612
z
2
z+ 0:5
+
6:5366z3:918
z
2
+ 0:0625)
H(z)
z
=
z(5:536 + 0:612z
1
)
z
2
(1z
1
+ 0:5z
2
)
+
z(6:53663:918z
1
)
z
2
(1 + 0:0625z
2
)
H(z) =
(5:536 + 0:612z
1
)
(1z
1
+ 0:5z
2
)
+
(6:53663:918z
1
)
(1 + 0:0625z
2
)
=H1(z) +H2(z)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 102 / 151

103. Realization of IIR StructuresParallel Form Structure for IIR System
H(z) =
(5:536 + 0:612z
1
)
(1z
1
+ 0:5z
2
)
+
(6:53663:918z
1
)
(1 + 0:0625z
2
)
=H1(z) +H2(z)()yn
6.5366
Z
-1
+
()xn
Z
-1
+
Z
-1
+
1
-0.5
+
-5.536
0.6112
+
Z
-1
+
3.9184
-0.0625
Figure 58: Parallel realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 103 / 151

104. Realization of IIR StructuresParallel Form Structure for IIR System
EE 2010 May
Realize the following System in parallel form:
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
(z0:25)
z
2
[z
2
0:5z+ 0:25]z
1
[z+ 0:5]
H(z) =
z
2
(z+ 0:25)
[z
2
+ 0:5z+ 0:25][z+ 0:5]
H(z)
z
=
z(z0:25)
[z
2
+ 0:5z+ 0:25][z+ 0:25]
H(z)
z
=
Az+B
[z
2
+ 0:5z+ 0:25]
+
C
[z+ 0:5]
z(z0:25)
[z
2
+ 0:5z+ 0:25][z+ 0:5]
=
Az+B
[z
2
+ 0:5z+ 0:25]
+
C
[z+ 0:5]
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 104 / 151

105. Realization of IIR StructuresParallel Form Structure for IIR System
Coecients A B and C are determined
C=
H(z)
z
[z+ 0:5]jz=0:5=
z(z0:2)
[z
2
0:5z+ 0:333]
=
0:5(0:50:25)
[(0:5)
2
(0:5)(0:5) + 0:25]
= 0:5
z(z+ 0:25)
[z
2
+ 0:5z+ 0:25][z+ 0:5]
=
Az+B
[z
2
+ 0:5z+ 0:25]
+
C
[z+ 0:5]
z(z+ 0:25) = (Az+B)[z+ 0:5] +C[z
2
+ 0:5z+ 0:25]
z
2
+ 0:25z=Az
2
+ 0:5Az+Bz+ 0:5B+Cz
2
+ 0:5Cz+ 0:25C]
z
2
+ 0:25z=z
2
(A+C) +z(0:5A+B+ 0:5C) + 0:5B+ 0:25C
Equating coecients
A+C= 1
A= 1C= 10:5 = 0:5
0:25 = 0:5A+B+ 0:5C= 0:50:5 +B+ 0:50:5
B=0:25
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 105 / 151

106. Realization of IIR StructuresParallel Form Structure for IIR System
H(z)
z
=
0:5z0:25
[z
2
+ 0:5z+ 0:25]
+
0:5
[z+ 0:5]
H(z)
z
=
z(0:50:25z
1
)
z
2
(1 + 0:5z
1
+ 0:25z
2
)
+
0:5
z[1 + 0:5z
1
]
H(z) =
(0:50:25z
1
)
(1 + 0:5z
1
+ 0:25z
2
)
+
0:5
(1 + 0:5z
1
)()yn
0.5
Z
-1
+
-0.5
()xn
Z
-1
+
Z
-1
+
-0.5
-0.25
+
0.5
-0.25
+
Figure 59: Parallel realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 106 / 151

107. Realization of IIR StructuresParallel Form Structure for IIR System
Realize the following System in parallel form:
H(z) =
1
1
2
z
1
(1
1
3
z
1
)(1
1
4
z
1
)
Solution:
H(z) =
z
1
(z0:5)
z
1
(z0:33)z
1
(z0:25)
H(z) =
z(z0:5)
(z0:33)(z0:25)
H(z)
z
=
(z0:33)
(z0:33)(z0:25)
H(z)
z
=
A
z0:33
+
B
z0:25
(z0:5)
(z0:33)z
1
(z0:25)
=
A
z0:33
+
B
z0:25
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 107 / 151

108. Realization of IIR StructuresParallel Form Structure for IIR System
Coecients A and B are determined
A=
H(z)
z
[z0:33]jz=0:33=
z0:5
z0:25
=
0:330:5
0:330:25
=2
B=
H(z)
z
[z0:25]jz=0:25=
z0:5
z0:33
=
0:250:5
0:250:33
= 3()yn
+
Z
-1
+
0.25
Z
-1
+
1
3
3
2−
()xn
Figure 60: Parallel realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 108 / 151

109. Realization of IIR StructuresParallel Form Structure for IIR System
Realize the following System in parallel form:
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
(10:5z
1
)(10:25z
1
)
H(z) =
z
1
(z+ 0:33)
z
1
(z0:5)z
1
(z0:25)
H(z) =
z(z+ 0:33)
(z0:5)(z0:25)
H(z)
z
=
(z+ 0:33)
(z0:5)(z0:25)
H(z)
z
=
A
z0:5
+
B
z0:25
(z+ 0:33)
(z0:5)(z0:25)
=
A
z0:33
+
B
z0:25
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 109 / 151

110. Realization of IIR StructuresParallel Form Structure for IIR System
Coecients A and B are determined
A=
H(z)
z
[z0:5]jz=0:5=
z+ 0:33
z0:25
=
0:5 + 0:33
0:50:25
= 3:33
B=
H(z)
z
[z0:25]jz=0:25=
z+ 0:33
z0:5
=
0:25 + 0:33
0:250:5
=2:33
H(z)
z
=
3:33
z0:5
+
2:33
z0:25
H(z) =
3:33
z0:5
+
2:33
z0:25()yn
+
Z
-1
+
0.25
Z
-1
+
0.5
2.33−
3.333
()xn
Figure 61: Parallel realizationDr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 110 / 151

111. Realization of IIR StructuresParallel Form Structure for IIR System
EE 2010 May
Realize the following System in parallel form:
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(10:5z
1
)(1 + 2z
1
)(10:66z
1
)
(10:75z
1
)(10:125z
1
)[1(0:5 +j0:5)z
1
)][1(0:5j0:5)z
1
)]
H(z) =
10(10:5z
1
)(1 + 2z
1
)(10:66z
1
)
(10:75z
1
)(10:125z
1
)[1z
1
+ 0:5z
2
)]
H(z) =
10z
3
(z0:5)(z+ 2)(z0:66)
z
4
(z0:75)(z0:125)[z
2
z+ 0:5]
H(z)
z
=
10(z0:5)(z+ 2)(z0:66)
(z0:75)(z0:125)[z
2
z+ 0:5]
H(z)
z
=
A
[z0:75]
+
B
[z0:125]
+
Cz+D
[z
2
z+ 0:5]
10(z0:5)(z+ 2)(z0:66)
(z0:75)(z0:125)[z
2
z+ 0:5]
=
A
[z0:75]
+
B
[z0:125]
+
Cz+D
[z
2
z+ 0:5]
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 111 / 151

112. Realization of IIR StructuresParallel Form Structure for IIR System
Coecients A B and C are determined
A=
H(z)
z
[z0:75]jz=0:75=
10(z0:5)(z+ 2)(z0:66)
(z0:125)[z
2
z+ 0:5]
=
10(0:750:5)(0:75 + 2)(0:750:66)
(0:750:125)[0:75
2
0:75 + 0:5]
= 3:16
B=
H(z)
z
[z0:125]jz=0:125=
10(z0:5)(z+ 2)(z0:66)
(z0:75)[z
2
z+ 0:5]
=
10(0:1250:5)(0:125 + 2)(0:1250:66)
(0:1250:75)[0:125
2
0:125 + 0:5]
=17:46
z(z+ 0:25)
[z
2
+ 0:5z+ 0:25][z+ 0:5]
=
Az+B
[z
2
+ 0:5z+ 0:25]
+
C
[z+ 0:5]
z(z+ 0:25) = (Az+B)[z+ 0:5] +C[z
2
+ 0:5z+ 0:25]
z
2
+ 0:25z=Az
2
+ 0:5Az+Bz+ 0:5B+Cz
2
+ 0:5Cz+ 0:25C]
z
2
+ 0:25z=z
2
(A+C) +z(0:5A+B+ 0:5C) + 0:5B+ 0:25C
Equating coecients
A+C= 1
A= 1C= 10:5 = 0:5
0:25 = 0:5A+B+ 0:5C= 0:50:5 +B+ 0:50:5
B=0:25
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 112 / 151

113. Realization of IIR StructuresParallel Form Structure for IIR System
H(z)
z
=
0:5z0:25
[z
2
+ 0:5z+ 0:25]
+
0:5
[z+ 0:5]
H(z)
z
=
z(0:50:25z
1
)
z
2
(1 + 0:5z
1
+ 0:25z
2
)
+
0:5
z[1 + 0:5z
1
]
H(z) =
(0:50:25z
1
)
(1 + 0:5z
1
+ 0:25z
2
)
+
0:5
(1 + 0:5z
1
)()yn
0.5
Z
-1
+
-0.5
()xn
Z
-1
+
Z
-1
+
-0.5
-0.25
+
0.5
-0.25
+
Figure 62: Parallel realization
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 113 / 151

114. Realization of FIR systemLattice structure for FIR Systems
Lattice structure for FIR Systems
The system function of mth order FIR lter is represented by the polynomialAm(z)
Hm(z) =Am(z)
whereAm(z) is the polynomial and is dened 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
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 114 / 151

115. Realization of FIR systemLattice structure for FIR Systems
Y(z) =X(z) +
m
X
i=1
am(i)z
i
X(z)
Taking inverse z transform
y(n) =x(n) +
m
X
i=1
am(i)x(ni)
Consider the order of the lter m=1
y(n) =x(n) +a1(1)x(n1)
From the gure y(n) is
y(n) =f0(n) +k1g0(n1)
f0(n) =x(n) andg0(n1) =x(n1)
y(n) =x(n) +k1x(n1)+
+Z
-1
1K
1
K
0
()fn
0
()gn
0
(1)gn−
1
() ()fn yn=
1
()gn
()xn
Figure 63: Single stage Lattice lter
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 115 / 151

116. Realization of FIR systemLattice structure for FIR Systems
Consider the lter of order of two.
y(n) = x(n) +
2
X
i=1
am(i)x(ni)
=x(n) +a2(1)x(n1) +a2(2)x(n2)
The rst stage output is
f1(n) = x(n) +K1x(n1)
g1(n) = K1x(n) +x(n1)
The second stage output is
f2(n) = f1(n) +K2g1(n1)
g2(n) = K2f1(n) +g1(n1)
The outputy(n) is
y(n) = f1(n) +K2g1(n1)
y(n) = x(n) +K1x(n1) +K2[K1x(n1) +x(n2)]
=x(n) +K1(1 +K2)x(n1) +K2x(n2)+
+Z
-1
1K
1
K
0
()fn
0
()gn
1
()fn
1
()gn
()xn
+
+Z
-1 2
K
2
K
2
() ()fn yn=
2
()gn
Figure 64: Two stage Lattice lter
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 116 / 151

117. Realization of FIR systemLattice structure for FIR Systems
y(n) = x(n) +K1(1 +K2)x(n1) +K2x(n2)
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 coecients by the following
recursive equations.
Km=am(m)
am1(i) =
am(i)am(m)am(mi)
1K
2
m
i= 1;2; : : :m1
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 117 / 151

118. Realization of FIR systemLattice structure for FIR Systems
Draw the lattice structure for the following FIR lter 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
am1(i) =
am(i)am(m)am(mi)
1K
2
m
a1(i) =
a2(i)a2(2)a2(2i)
1K
2
2
a1(1) =
a2(1)a2(2)a2(1)
1K
2
2
=
2(0:333)(2)
1(0:333)
2
= 1:5
K1=a1(1) = 1:5+
+Z
-1
1
1.5K=
()xn
()yn
2
()gn
1
1.5K=
+
+Z
-1
2
0.333K=
2
0.333K= Figure 65:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 118 / 151

119. Realization of FIR systemLattice structure for FIR Systems
A FIR lter is given by
y(n) =x(n) +
2
5
x(n1) +
3
4
x(n2) +
1
3
x(n3)
Draw the lattice structure
Solution:
y(n) =x(n) +
2
5
x(n1) +
3
4
x(n2) +
1
3
x(n3)
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
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 119 / 151

120. Realization of FIR systemLattice structure for FIR Systems
for m=3,K3=a3(3) = 0:333
am1(i) =
am(i)am(m)am(mi)
1K
2
m
a2(i) =
a3(i)a3(3)a3(3i)
1K
2
3
a2(1) =
a3(1)a3(3)a3(2)
1K
2
3
=
0:4(0:333)(0:75)
1(0:333)
2
= 0:169
a2(2) =
a3(2)a3(3)a3(1)
1K
2
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(2i)
1K
2
2
a1(1) =
a2(1)a2(2)a2(1)
1K
2
2
=
0:169(0:693)(0:169)
1(0:693)
2
= 0:0998
for m=1,K1=a1(1) = 0:0998+
+Z
-1
10.0998K=
()xn
()yn
3
()gn
10.0998K=
+
+Z
-1
2
0.693K=
2
0.693K=
+
+Z
-1
3
0.333K=
3
0.333K=
Figure 66:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 120 / 151

121. Realization of FIR systemLattice structure for FIR Systems
A FIR lter is given by
y(n) =x(n) + 3:1x(n1) + 5:5x(n2) + 4:2x(n3) + 2:3x(n4)
Draw the lattice structure
Solution:
y(n) =x(n) + 3:1x(n1) + 5:5x(n2) + 4:2x(n3) + 2:3x(n4)
By taking z transform
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) = 4:2a4(4) = 2:3
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 121 / 151

122. Realization of FIR systemLattice structure for FIR Systems
For m=4K4=a4(4) = 2:3
am1(i) =
am(i)am(m)am(mi)
1K
2
m
a3(i) =
a4(i)a4(4)a4(4i)
1K
2
4
a3(1) =
a4(1)a4(4)a4(3)
1K
2
4
=
3:1(2:3)(4:2)
1(2:3)
2
= 1:529
a3(2) =
a4(2)a4(4)a4(2)
1K
2
4
=
5:5(2:3)(5:5)
1(2:3)
2
= 1:6667
a3(3) =
a4(3)a4(4)a4(1)
1K
2
4
=
4:5(2:3)(3:1)
1(2:3)
2
= 0:683
For m=3K3=a3(3) = 0:683
am1(i) =
am(i)am(m)am(mi)
1K
2
m
a2(i) =
a3(i)a3(3)a3(3i)
1K
2
3
a2(1) =
a3(1)a3(3)a3(3)
1K
2
3
=
1:529(0:683)(1:667)
1(0:683)
2
= 0:732
a2(2) =
a3(2)a3(3)a3(1)
1K
2
3
=
1:667(0:683)(1:529)
1(0:683)
2
= 1:167
For m=2,K2=a2(2) = 1:167
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 122 / 151

123. Realization of FIR systemLattice structure for FIR Systems
For m=2,K2=a2(2) = 1:167
For m=1
k1=a1(1) =
am(i)am(m)am(mi)
1K
2
m
k1=a1(1) =
a2(1)a2(2)a2(1)
1K
2
2
=
0:732(1:167)(0:732)
1(1:167)
2
= 0:338+
+Z
-1
1
0.338K=
()xn
()yn
4
()gn
1
0.338K=
+
+Z
-1
2
1.167K=
2
1.167K=
+
+Z
-1
3
0.683K=
3
0.683K=
+
+Z
-1
4
2.3K=
4
2.3K=
1
()fn
3
()fn
2
()fn
3
()gn
2
()gn
1
()gn
Figure 67:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 123 / 151

124. Realization of FIR systemLattice structure for FIR Systems
Determine all the FIR lters which are specied by the lattice parametersK1= 0:1,K2= 0:2,
andK3= 0:3 and draw the structure
Solution:
am(0) = 1
am(m) = Km
am(i) = am1(i) +am(m)am1(mi)
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(2i)
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(3i)
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
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 124 / 151

125. Realization of FIR systemLattice structure for FIR Systems
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
3Z
-1
+
Z
-1
+
Z
-1
()xn
1
0.18 0.236
+
0.3
y(n)
Figure 68:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 125 / 151

126. Realization of FIR systemLattice structure for FIR Systems
Determine all the FIR lters which are specied by the lattice parametersK1= 0:5,K2= 0:333,
andK3= 0:25 and draw the structure
Solution:
am(0) = 1
am(m) = Km
am(i) = am1(i) +am(m)am1(mi)
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(2i)
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(3i)
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
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 126 / 151

127. Realization of FIR systemLattice structure for FIR Systems
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
3Z
-1
+
Z
-1
+
Z
-1
()xn
1
0.75 0.5
+
0.25
y(n)
Figure 69:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 127 / 151

128. Realization of FIR systemLattice structure for FIR Systems
Determine the impulse response of a FIR lter with reection coecientsK1= 0:6,K2= 0:3,
K3= 0:5 andK4= 0:9 Also draw the direct form structure
Solution:
am(0) = 1
am(m) = Km
am(i) = am1(i) +am(m)am1(mi)
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(2i)
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(3i)
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
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 128 / 151

129. Realization of FIR systemLattice structure for FIR Systems
For m=4
a4(0) = 1
a4(i) = a3(i) +a4(4)a3(4i)
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) = 1:337and 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
4Z
-1
+
Z
-1
+
Z
-1
()xn
1
1.38 1.311
+
1.337
y(n)
Z
-1
+
0.9
y(n)
Figure 70:
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 129 / 151

130. Realization of IIR systemLattice structure for IIR Systems
Lattice structure for IIR Systems
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 130 / 151

131. Realization of IIR systemLattice structure for IIR Systems
The system function of mth order IIR lter is represented by
H(z) =
MP
k=0
bkz
k
1 +
NP
k=1
akz
k
Let us consider an all pole system with system function
H1(z) =
1
1 +
NP
k=1
aN(k)z
k
=
1
AN(z)
But the system functionH1(z) =
Y(z)
X(z)
Y(z)
X(z)
=
1
1 +
NP
k=1
aN(k)z
k
Y(z) +
N
X
k=1
aN(k)z
k
Y(z) =X(z)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 131 / 151

132. Realization of IIR systemLattice structure for IIR Systems
Y(z) +
N
X
k=1
aN(k)z
k
Y(z) =X(z)
Taking inverse z transform
y(n) =
N
X
k=1
aN(k)y(nk) +x(n)
Interchange the input and output
x(n) =
N
X
k=1
aN(k)x(nk) +y(n)
Rearranging the above equation
y(n) =x(n) +
N
X
k=1
aN(k)x(nk)
The equation is similar to the FIR
Lattice system.
Consider the order of the lter m=1
y(n) =a1(1)y(n1) +x(n)
From the above equation
K1=a1(1)+
+Z
-1
1
K 1() ()fn xn=
1
()gn
0
() ()fn yn=
1
K
Figure 71: Single stage Lattice
lter+
+ Z
-1
f
0
(n)=y(n)
g
1
(n)
f
1
(n)=x(n)
K
1
K
1-
Figure 72: Single stage Lattice
lter
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 132 / 151

133. Realization of IIR systemLattice structure for IIR Systems
Consider the lter of order of two.
f2(n) = x(n)
f1(n) = f2(n)K2g1(n1)
g2(n) = K2f1(n) +g1(n1)
f0(n) = f1(n)K1g0(n1)
g1(n) = K1f0(n) +g0(n1)
y(n) = f0(n) =g0(n)
y(n) = K1(1 +K2)y(n1)K2y(n2) +x(n)
g2(n) = K2y(n) +K1(1 +K2)y(n1)y(n2)+
+ Z
-1
f
1
(n)
g
2
(n)
f
2
(n)=x(n)
K
2
K
2
+
+ Z
-1
f
0
(n)=y(n)
g
1
(n)
K
1
K
1-
-
g
0
(n)
Figure 73: Single stage Lattice lter
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 133 / 151

134. Realization of IIR systemLattice structure for IIR Systems
y(n) =
N
X
k=1
aN(k)y(nk) +x(n)
y(n) =
2
X
k=1
a2(k)y(nk) +x(n)
y(n) =
2
X
k=1
a2(k)y(nk) +x(n)
=a2(1)y(n1)a2(2)y(n2) +x(n)
y(n) = x(n) +K1(1 +K2)x(n1) +K2x(n2)
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)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 134 / 151

135. Realization of IIR systemLattice structure for IIR Systems
The lattice structure for mth order is obtained by the following recursive equations.
Km=am(m)
am1(i) =
am(i)am(m)am(mi)
1K
2
m
=
am(i)Kmam(mi)
1K
2
m
i= 1;2; : : :m1
The IIR system function is
Hsystem(z) =
MP
k=0
bkz
k
1 +
NP
k=1
akz
k
=
BM(z)
AN(z)
i=bi
M
X
m=i+1
mam(mi) i=M;M1;M2; : : :1;0+
+ Z
-1
f
N
(n)=x(n)
K
N
-
+
N
β
+
+ Z
-1
f
0
(n)
K1-
+
1
β
+
+ Z
-1
f
2
(n)
K
2
-
+
2
β
f
N-1
(n)
K
N
f
1
(n)
g
N-1
(n)
+
1N
β
−
g
1
(n)g
N
(n) g
2
(n)
0
β
g
0
(n)
0
β
()yn
Figure 74: Lattice ladder structure
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 135 / 151

136. Realization of IIR systemLattice structure for IIR Systems
Realize the lattice structure for all pole lter given byH(z) =
1
1+
2
5
z
1
+
3
4
z
2
+
1
3
z
3
Solution:
m=3a3(0) = 1;a3(1) =
2
5
;a3(2) =
3
4
;a3(3) =
1
3
am1(i) =
am(i)am(m)am(mi)
1K
2
m
for m=3
a2(i) =
a3(i)a3(3)a3(3i)
1K
2
3
a2(1) =
a3(1)a3(3)a3(2)
1K
2
3
a2(1) =
0:4(0:333)(0:75)
1(0:33)
2
= 0:16875
a2(2) =
a3(2)a3(3)a3(1)
1K
2
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)
1K
2
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
g
3
(n)
f
3
(n)=x(n)
0.333
+
+ Z
-1
f
0
(n)=y(n)
g
2
(n)
-
-
g
1
(n)
+
+ Z
-1
g
0
(n)
0.333
0.6973 0.0996
0.6973 0.0996
f
2
(n)
f
1
(n)
Figure 75: Lattice lterDr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 136 / 151

137. Realization of IIR systemLattice structure for IIR Systems
Realize the lattice structure for all pole lter given byH(z) =
1
10:9z
1
+0:64z
2
0:576z
3
Solution:
m=3a3(0) = 1;a3(1) =0:9a3(2) = 0:64a3(3) =0:576
for m=3, i=1,2
a2(i) =
a3(i)a3(3)a3(3i)
1K
2
3
a2(1) =
a3(1)a3(3)a3(2)
1K
2
3
a2(1) =
0:9(0:576)(0:64)
1(0:5763)
2
=0:795
a2(2) =
a3(2)a3(3)a3(1)
1K
2
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)
1K
2
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
g
3
(n)
f
3
(n)=x(n)
-0.576
+
+ Z
-1
f
0
(n)=y(n)
g
2
(n)
-
-
g
1
(n)
+
+ Z
-1
g
0
(n)
-0.576
0.1819 -0.6726
0.1819 -0.6726
f
2
(n)
f
1
(n)
Figure 76: Lattice lter
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 137 / 151

138. Realization of IIR systemLattice structure for IIR Systems
Develop the lattice ladder structure for the lter with dierence equation
y(n) +
3
4
y(n1) +
1
4
y(n2) =x(n) + 2x(n1)
Solution:
y(n) +
3
4
y(n1) +
1
4
y(n2) =x(n) + 2x(n1)
By taking z transform
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
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 138 / 151

139. Realization of IIR systemLattice structure for IIR Systems
for m=2
K2=a2(2) = 0:25
am1(i) =
am(i)am(m)am(mi)
1K
2
m
a1(i) =
a2(i)a2(2)a2(2i)
1K
2
2
a1(1) =
a2(1)a2(2)a2(1)
1K
2
2
=
0:75(0:25)(0:75)
1(0:25)
2
= 0:6
for m=1
K1=a1(1) = 0:6
Ladder coecients are
B(z) = 1 + 2z
1
M=1
b0= 1;b1=1= 2
i=bi
M
X
m=i+1
mam(mi)
0=b0
1
X
m=1
mam(m)
=b01a1(1) = 12(0:6) =0:2x(n)
+
+ Z
-1
f
0
(n)
-
+
1
2β=
+
+ Z
-1
K
2
=0.25-
+
f
1
(n)
g
1
(n)
0
0.2β
=−
g
0
(n)
()yn
K
1
=0.6
Figure 77: Lattice lter
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 139 / 151

140. Realization of IIR systemLattice structure for IIR Systems
A linear time invariant system is dened by
H(z) =
0:129 + 0:38687z
1
+ 0:3869z
2
+ 0:129z
3
10:2971z
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
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 140 / 151

141. Realization of IIR systemLattice structure for IIR Systems
for m=3
K3=a3(3) =0:0276
am1(i) =
am(i)am(m)am(mi)
1K
2
m
a2(i) =
a3(i)a3(3)a3(3i)
1K
2
3
a2(1) =
a3(1)a3(3)a3(2)
1K
2
3
=
0:2971(0:0276)(0:3564)
1(0:2971)
2
=0:2875
a2(2) =
a3(2)a3(3)a3(1)
1K
2
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(2i)
1K
2
2
a1(1) =
a2(1)a2(2)a2(1)
1K
2
2
=
0:2875(0:3485)(0:2875)
1(0:3485)
2
=0:2132
for m=1
K1=a1(1) =0:2132
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 141 / 151

142. Realization of IIR systemLattice structure for IIR Systems
Ladder coecients are:B(z) = 0:129 + 0:38687z
1
+ 0:3869z
2
+ 0:129z
3
i=bi
M
X
m=i+1
mam(mi)
b0= 0:129;b1= 0:38687;b2= 0:3869;b3= 0:129 =3
i= 2
2=b2
3
X
m=3
mam(m2) =b23a3(1) = 0:3869(0:129)(0:2971) = 0:4252
i= 1
1=b1
3
X
m=2
mam(m1)
=b12a2(1)3a3(2) = 0:3867(0:4252)(0:2875)(0:129)(0:3564) = 0:4630
i= 0
0=b0
3
X
m=1
mam(m)
=b01a1(1)2a2(2)3a3(3)
= 0:129(0:4630)(0:2132)(0:4252)(0:3485)(0:129)(0:0276) = 0:0831
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 142 / 151

143. Realization of IIR systemLattice structure for IIR Systems+
+ Z
-1
f
N
(n)=x(n)
K
3
=-0.0276
-
+
3
0.129β=
+
+ Z
-1
f
0
(n)
-
+
1
0.4630β=
+
+ Z
-1
f
2
(n)
-
+
f
N-1
(n)
K
3
f
1
(n)
g
N-1
(n)
2
0.4252β=
g
1
(n)g
N
(n) g
2
(n)
0
0.0831β=
g
0
(n)
()yn
K
1
=-0.2132K
2
=-0.3485
K
2
=-0.3485
K
1
=-0.2132
Figure 78: Lattice lter
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 143 / 151

144. Realization of IIR systemLattice structure for IIR Systems
A linear time invariant system is dened by
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
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 144 / 151

145. Realization of IIR systemLattice structure for IIR Systems
for m=3
K3=a3(3) = 0:06
am1(i) =
am(i)am(m)am(mi)
1K
2
m
a2(i) =
a3(i)a3(3)a3(3i)
1K
2
3
a2(1) =
a3(1)a3(3)a3(2)
1K
2
3
=
1:2(0:06)(0:47)
1(0:06)
2
= 1:176
a2(2) =
a3(2)a3(3)a3(1)
1K
2
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(2i)
1K
2
2
a1(1) =
a2(1)a2(2)a2(1)
1K
2
2
=
1:176(0:4)(1:176)
1(0:4)
2
= 0:84
for m=1
K1=a1(1) = 0:84
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 145 / 151

146. Realization of IIR systemLattice structure for IIR Systems
Ladder coecients are
B(z) = 1 +z
1
+z
2
M=1
b0= 1;b1= 1;b2= 1 =2
i=bi
M
X
m=i+1
mam(mi)
2=b2= 1
i= 1
1=b1
3
X
m=2
mam(m1)
=b12a2(1)
= 11:176 =0:176
i= 0
0=b0
1
X
m=1
mam(m)
=b01a1(1)2a2(2)
= 1 +:84(0:176)0:4 = 0:748+
+ Z
-1
f
N
(n)=x(n)
K
3
=-0.06
-
+
+ Z
-1
f
0
(n)
-
+
1
0.176β=−
+
+ Z
-1
f
2
(n)
-
+
f
N-1
(n)
K
3
=0.06
f
1
(n)
g
N-1
(n)
2
1β=
g
1
(n)g
N
(n) g
2
(n)
0
0.748β=
g
0
(n) ()yn
K
1
=-0.84K
2
=-0.4
K
2
=-0.4
K
1
=-0.84
Figure 79: Lattice lter
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 146 / 151

147. Realization of IIR systemSignal Flow Graph
Signal Flow Graph
Signal ow graph (SFG) isgraphical representationof block diagram structure.
The basic elements of SFG arebranches and nodes. The signal out of branch is equal to
the branch gain.
Addersandpick-o pointsare replaced bynodes.Multipliersanddelay elementsare
indicated by the the branch transmittance.
Theinputto the system originates atsource nodeandoutput signalis extracted from a
sink node.
The importance of SFG in FIR or IIR structure istransposition or ow graph reversal
theorem.
Theorem states that \if we reverse the directions of all branch transmittances and
interchange the input and output in the ow graph, the system function remains
unchanged". The resulting structure is called atransposed structure or a transposed
form.
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 147 / 151

148. Realization of IIR systemSignal Flow Graph()xn
Z
-1
+
Z
-1
+
+
+
2
a−
1
a−
2
b
1
b
0
b ()yn
Figure 80: Direct Form II()xn
2
a−
1
a−
2
b
1
b
0
b ()yn
1
Z
−
1
Z
−
Source
Node
1
5
4
3
2
Sink Node Figure 81: Signal Flow graph()yn
2
a−
1
a−
2
b
1
b
0
b ()xn
1
Z
−
1
Z
−
1
5
4
3
2 Figure 82: Transposed Structure SFG()xn
2
a−
1
a−
2
b
1
b
0
b
+
+
+
Z
-1
Z
-1
()yn Figure 83: Direct form-II realization
Figure 80 shows the direct form II structure while Figure 81 shows its signal ow graph
Figure 82 shows the Transposed Structure while Figure 83 shows its realization.
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 148 / 151

149. Realization of IIR systemSignal Flow Graph
A linear time invariant system is dened by
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 ow graph, transposed signal ow graph and transposed direct form-II structure()xn
1
6
−
()yn
1
Z
−
1
Z
−
Source
Node
1
5
4
3
2
Sink Node
1
4
1
2
−
1 8
−
1
5
Figure 84: Signal ow graph()yn
1
6
−
()
xn
1
Z
−
1
Z
−
Source
Node
1
5
4
3
2
1
4
1
2
−
1 8
−
1
5 Figure 85: Transposed Structure signal ow
graph()xn
1
5
1
8
−
1
4
1
2
−
+
+
+
Z
-1
Z
-1
()yn
Z
-1
1
6
−
Figure 86: Direct form-II Structure
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 149 / 151

150. Realization of IIR systemSignal Flow Graph
For the ow graph write dierence equations and system function()xn ()yn
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(n1)
5
24
y(n2)
1
12
y(n3) +x(n) +
1
5
x(n1)
H(z) =
1 +
1
5
z
1
1
1
4
z
1
+
5
24
z
2
+
1
12
z
3
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 150 / 151

151. Realization of IIR systemSignal Flow Graph
For the ow graph write dierence equations and system function()xn ()yn
3
2
1
1
1
Z
−
1
Z
−
1
Z
−
1
Z
−
()wn
Figure 88: Direct Form II Cascade
Solution:
w(n) =x(n) + 3w(n1) +w(n2)
W(z)
X(z)
=
1
13z
1
z
2
y(n) =w(n) +y(n1) + 2y(n2)
Y(z)
W(z)
=
1
1z
1
2z
2
H(z) =
Y(z)
X(z)
=
Y(z)
W(z)
W(z)
X(z)
=
1
(1z
1
2z
2
)(13z
1
z
2
)
=
1
14z
1
+ 7z
3
+ 2z
4
Y(z)[14z
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(n1)7y(n3)2y(n4)
Dr. Manjunatha. P (JNNCE) UNIT - 8: Implementation of Discrete-time Systems[ ?,?,?,?]October 18, 2016 151 / 151