signalsandsystemsztransformEEEmaterial.pdf

z transform

1. Z Transform
Dr.R.Helen, APEEE, TCE
9/14/2020

2. Z-Transform
Syllabus
•Definitions and Properties of z-transform
•Rational z-transforms
•Inverse z-transform
•One sided z-transform
•Analysis of LTI systems in z-domain
e-
2
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3. Z-Transform definition

3
Z-transform is mainly used for analysis of discrete signal and discrete
LTI system. Z.T of discrete time single x (n) is defined by the following
expression.
X (z)   x(n)z
n

n
where, X(z) z-transform of x(n)
zcomplex variable = re
jw
From the above definition of Z.T. it is clear that ZT is power series & it
exist for only for those values of z for which X(z) attains finite value (
convergence) ,which is defined by Region of convergence. (ROC)

Region of Convergence: (ROC)
Region of Convergence is set of those values of z for which power
series x (z) converges. OR for which power series, x (z) attains finite value.
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4. Z-Transform of Finite duration signal
Find the Z - Transform and mention the Region of Convergence
(ROC) for the following discrete time sequences.

1. x (n) = { 2 1 2 3}

2. x (n) = { 2, 1, 2 3 }

3. x (n) = { 1 2 1 -2 3 1}
1
st
2
nd
3
rd
example Is of causal signal
example Is of anti-causal signal
example Is of non-causal signal
4
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5. Solution(1)

n
X (z)   x(n)z
n
x (n) = { 2 1 2 3}

Z.T. is defined as
In this case x(z) is finite for all values of z, except |z| = 0.
Because at z = 0, x(z) = ∞.
Thus ROC is entire z-plane except |z| = 0.
ROC
X (z)  x(0)  x(1)z
1
 x(2)z
2
 x(3)z
3
X (z)  2 1z
1
 2z
2
 3z
3
X (z)  2  z
1
 2z
2
 3z
3
ROC is a set of those values of z for which x (z) is not infinite
Z=0
5
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6. Solution(2)

X (z)   x(n)z
n
n
X (z)  x(0)  x(1)z
1
 x(2)z
2
 x(3)z
3
x (n) = { 2 1 2 3}

Z.T. is defined as
In this case X(z) is finite for all values of z, except |z| = ∞.
Because at z = ∞, X(z) = ∞.
Thus ROC is entire z-plane except |z| = ∞.
ROC
X (z)  3  2z
1
1z
2
 2z
3
X (z)  3  2z  z
2
 2z
3
ROC is a set of those values of z for which x (z) is not infinite
Z= ∞
6
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7. Solution(3)

X (z)   x(n)z
n
x (n) = { 1 2 1 -2 3 1}
Z.T. is defined as
ROC is a set of those values of z for which x (z) is not infinite
In this case X(z) is finite for all values of z, except |z| = ∞.
Because at z = ∞, X(z) = ∞.
Thus ROC is entire z-plane except |z| = 0 &|z| = ∞.
ROC
Z= ∞
n

X (z)  x(2)z
2
 x(1)z
1
 x(0)  x(1)z
1
 x(2)z
2
 x(3)z
3
X (z)  1z
2
 2z
1
1 2z
1
 3z
2
1z
3
 z
3
X (z)  z
2
 2z
1
1 2z
1
 3z
2
Z=0
7
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8. Quiz
Find the Z - Transform and mention the Region of Convergence
(ROC) for the following discrete time sequences.
1. x (n) =

2. x (n) =
 (n  2)
 (n)
z
-2
Ans (1) ROC- entire z-plane except |z|= 0
Ans (2) 1 ROC- entire z-plane
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9. z-Transform of infinite duration signal
Find the z-transform for following discrete time sequences. Also mention
ROC for all the cases.
xn  a
n
U n
xn  a
n
U  n 1
x(n)  a
n
U (n)  b
n
U (n 1)
1
2
3
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10. Solution(1)
Sequence is causal xn  a
n
U n

X (z)   x(n)z
n

n
Z.T. for the given sequence x (n) is defined as

  a
n
U (n)z
n
n

X (z)   a
n
z
n
n0


1
n0
n
az  X (z) 
U(n)=0 for
n<0

n0
 a
n
 a
0
 a
1
 a
2
 ....... We know that
Series converges iff |a|<1

1
n0 1  a
 a
n
 We also know a  1 for
Thus, x (z) converges when | a z
–1
| < 1
1
1
1 az
X z 
z
z  a
X z  
–1
for | a z | < 1
for | a /z | < 1
for | z | > |a|
ROC is outside the
circle |z|=|a|
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11. Solution(2)
Sequence is anti-causal xn  a
n
U  n 1

X (z)   x(n)z
n

n
Z.T. for the given sequence x (n) is defined as

   a
n
U (n 1)z
n
n

1
1
n
n
az  X (z)  
U(-n-1)=0 for
n>-1

n0
 a
n
 a
0
 a
1
 a
2
 ....... We know that
Series converges iff |a|<1

1
n0 1  a
 a
n
 We also know a  1 for
Thus, x (z) converges when | a
–1
z | < 1
a z
1
1 a
1
z
X z  
z
z  a
X z  
–1
for | a z | < 1
for | z /a | < 1
for | z | < |a|
ROC is inside the
circle |z|=|a|
Put n=-m
 
1
1
1
1
m
m
m
m
a z  az    X (z)  
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U(-n-1)=1 n less than 0

13. Problems
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14. Solution(1)
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15. Solution(2)
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16. Solution(3)
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17. Problem:
1
3 (1
1 z
1
)(1 2z
1
)
X (z) 
Show all possible ROC’s and pole-zero
diagram z-transform given below
1/3
2 1
|z|=1
1/3
2 1
|z|=1
1/3 1
|z|=1
If x[n] is left sided signal
i.e. anti-causal signal If x[n] is right sided signal
i.e. causal signal
If x[n] is double sided
signal
i.e. non-causal signal
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18. Inverse z-transform
•Synthetic Division Method


•Partial Fraction Method


•Cauchy’s Integration Method
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19. Synthetic division Example(1)
1
1 az
1
X (z) 
1 az
1 1
1 az
1
1
az
1
az
1
 a
2
z
2
az
1
a
2
z
2
a
2
z
2
 a
3
z
3
a
3
z
3
•Since ROC is |z| >a, x[n] is causal
sequence or right sided sequence.
•Quotient series should have –ve
powers of z as z-transform of causal
sequence has –ve powers of z
 a
2
z
2
a
4
.....} a
2
a
3
x[n]  {1 a

| z || a |
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20. Synthetic division Example(2)
1
1 az
1
X (z) 
 az
1
1 1
1 a
1
z
a
1
z
a
1
z  a
2
z
2
a
1
z
a
2
z
2
a
2
z
2
 a
3
z
3
a
3
z
3
•Since ROC is |z| >a, x[n] is anti-causal
sequence or left sided sequence.
•Quotient series should have +ve powers
of z , as z-transform of anti-causal
sequence has +ve powers of z
0 .......}

 a
1
 a
2
x[n]  {....... a
3
| z || a |
 a
2
z
2
 a
3
z
3
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21. Problem
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22. Solution (a)
2 2 1
3 z
1

1 z
2
X (z) 
1
ROC :| z | 1
As |z|>1, x[n] is causal sequence
1 2
1
3
2 z 
1
2 z
1
1
1 2
1
3
2 z 
1
2 z
2

3 z
1
4

7 z
2
8

15 z
3
16

31 z
4
3 z
1

1 z
2
2 2
3 z
1

9 z
2

3 z
3
2 4 4
7 z
2

3 z
3
4 4
4 14 8
7 
21 
7 z
2
z
3
z
4
......}
31
16
15
8
7
4
3
2
15 z
3

7 z
4
8 8

x[n]  {1

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23. Solution (b)
2 2 1
3 z
1

1 z
2
X (z) 
1
2 ROC :| z |
1
As |z|<1/2, x[n] is anti-causal sequence
1

3
2 z 1
1 z
2
2 1
1 3z  2z
2
3z  2z
2
 2z
2
 6z
3
14z
4
62 30 14 6 2 0 0}

x[n]  {......
 30z
5  62z
6
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24. Z-transform Pairs
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25. Z-transform Pairs
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26. Z-transform Pairs
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27. Partial Fraction Method
2 1
1 6z  8z
 4  8z
1
H (z) 

(1 4z
1
)(1 2z
1
)
 4  8z
1

1 2z
1
A
2
1 4z
1
A
1
H (z) 
4
1
z
1

1
1 2z
 4  8z
1
A
1 
2
 
8
1  8
z
1

1
1 4z
1
 4  8z
1
 
6
(1/ 2)  12 A
2 
8

12
1 2z
1
1 4z
1
 H (z) 
Taking IZT ,we get h[n]  [12(4)
n
 8(2)
n
]u(n)
Partial Fraction
Expansion
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28. Different Pole’s Cases
a)Distinct Real Poles
b)Complex Conjugate and Distinct Poles
c)Repeated Poles
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29. Distinct Real Poles
D(z)
H (z) 
N (z) Where N(z) and D(z) are polynomials of order m
and n respectively
If m<n,
N (z)
(z  p
1 )(z  p
2 )(z  p
3 ) .......(z  p
n )
H (z) 
A
n A
3 A
1 A
2
(z  p
n )
   .... 
(z  p
1 ) (z  p
2 ) (z  p
3 )
H (z) 
where A
k  (z  p
k )H (z)
z  p
k
If m>=n, use division
D(z)
H (z)  Q 
N (z)
such that m’<n, if not repeat division
Q- Quotient
N(z)- remainder
D(z)- Divisor
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30. Problem
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31. Contd..
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32. Complex-Conjugate & Distinct Poles
(z  2)(z
2
 2z  5)
z(z
2
 2z)
X (z) 

(z  2)(z
2
 2z  5)
2z z
2
z
X (z) z
2
2z

(z  2)(z  2  j1)(z  2  j1)
A
3 A
1 A
2
z
 
X (z)

z2
(z  2  j1)(z  2  j1)
2z
(z  2) (z  2  j1)
z
2
A
1 
(z  2  j1)
 0
z
2
z (2 j1)
(z  2)(z  2  j1)
2z
A
2 
j
2

1
3
z
2
z(2 j1)
(z  2)(z  2  j1)
2z
A 
j
2

1
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33. Contd..
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34. Repeated Poles
  
r
k
q
k
A B
k 1 z  p
k k 1 z  p
k
H (z) 
where q  no. of distinct poles and not repetitive
r  no. of repetition of repetitive pole

A
k  is calculated by method as described earlier
B
k  is calculated by following equation
r
i k z  p
j
d
r k
B  (z  p ) F (z)
(r  k )! dz
r k
1
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35. Problem
(z 1)(z  2)
3
z(z
2
 9)
H (z) 
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36. Cauchy’s Integration (Residue) Method
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37. Problem
(z 1)(z  2)
10z
X (z) 
(z 1)(z  2) (z 1)(z  2)
10z 10z
n
z
n1
 G(z)  X (z)z
n1

G(z) has two poles ,z =1 & z=2
x(n) = Residue of G(z) at z=1 + Residue of G(z) at z=2
 10
1 2

10.1
n
z 1
(z 1)(z  2)
10z
n
R
z 1  (z 1)
1
10.(2)
n
10z
n
 10.(2)
n

(z 1)(z  2)
R  (z  2)
z 2
z 2
x(n)  10 10(2)
n
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38. Problem
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39. Problem
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40. Quiz
Find IZT for following z-transforms
1
(z 1)(z  0.5)
X (z) 
(1 e
a
)z
X (z) 
(z 1)(z  e
a
)
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41. z-transform properties
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42. Contd..
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43. Contd..
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44. Problem on Properties 1-8
a
n
u(n) 1) Find z-transform of
We know
1
1 z
1
u(n) 
z

a
n
u(n) 
z
U (
z )
a
and Property 3
1
)
1
a 1 (
z
a
n
u(n) 
z

1

1 az
1
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45. Contd..
2) Find IZT of X (z)  log(1 az
1
)
|z|>|a|
We know
2

x
3

x
4
3 4
log(1 x)  x 
x
2

 (1)  ..... 
n1
n1
n
x
n


 (1)
1 n
n1 (az )
n
X (z)  log(1 az
1
)

n 
a
n
 
n1 
 (1)
n1
z
n
a
n
n
n1





n 


 u(n 1) z (1)
n1
n
Comparing above equation with z-transform definition,

n
X (z)   x(n)z
n
we get
a
n
x(n)  (1)
n1
u(n 1)
n



 0
n x ( n ) 
(  1)
a
n
n 1
n  1
otherwise
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46. Contd..
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47. Contd..
We know , property of differentiation in z domain
ROC  R
nx[n]
z
Z
d
( X (z))
dz
If x[n]
z
 X (z) with then
Thus we have
1
1 az
1
a
n
u(n) 
z

1
dz 1 az
1
na
n
u(n) 
z
 z
d
1
(a.(1).z
2
)
(1 az
1
)
2
 z(1)
1 2
(1 az )
az
2
 z
1 2

(1 az )
az
1
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48. Contd..
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49. Contd..
(1 az
1
)
3
2a
2
z
1
n(n 1)a
n1
u(n 1) 
z

2
1
a
2
z
1
n(n 1)a
n1
u(n 1) 
z

2
1
(1 az
1
)
3
(1 az
1
)
3
z
1
n(n 1)a
n1
u(n 1) 
z

Multiply by 1/2
Divide by a
2
If a=0.5, we get
1
)
3
(1 0.5z
z
1
1 n(n 1)0.5
n1
u(n 1) 
z

2
 X
1 (z)
------------(2)
Multiply by 3z
-1
in eq. (2) and putting a=0.5, we get
2
3
(1 az
1
)
3
3z
2
(n 1)(n)0.5
n2
u(n) 
z
  X
2 (z)
x
1 (n) 
x
2 (n) 
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50. Contd..
x(n)  x
1 (n)  x
2 (n)

 x(n) 
1 n(n 1)0.5
n1
u(n 1) 
3 (n 1)(n)0.5
n2
u(n)
2 2
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51. Problem on Initial & Final value theorem
Find the initial and final value of x(n) ,if
(z 1)(z  0.5)
z(z  2)
X (z) 
z(z  2)
x[0]  lim
z (z 1)(z  0.5)
We know initial value theorem
x[0]  lim X (z)
z
 lim
z
2
(1 2z
1
)
z z
2
(1 z
1
)(1 0.5z
1
)
 1
(1 0)(1 0)
1 0
x[0] 
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52. Contd..
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53. Contd..
Find the initial and final value of following functions a) u(n) b) r(n)
Solution (a): we know
1
1 z
1
u(n) 
z

u[0]  limU (z)
z
1 1
 1
1 0
  lim
z 1 z
1
1
1 z
1
u()  lim(1 z
1
)
z1
 1
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54. Contd..
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55. Problems on convolution
Find x(n) using convolution theorem if
(z 1)
2
z
X (z) 
X (z)  
(z 1)
2
(z 1) (z 1)
z z 1
X (z)  X
1 (z).X
2 (z)
X (z) 
z

1
1
(z 1) (1 z
1
)
Taking IZT ,we get
1
x (n)  u(n)
1
(z 1)
X
2 (z) 
1
1 z
1
u(n) 
z

We know
1
1 z
1
u(n 1) 
z
 z
1
1
z 1
 u(n 1) 
z
 x
2 (n)  X
2 (z)
2
x (n)  u(n 1)
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56. Contd..
Now we know convolution property x (n)* x (n) 
z
 X (z).X (z)
1 2 1 2

 x(n)  x
1 (n)* x
2 (n)
 u(n)*u(n 1)
 u(k)u(n  k 1)
k 
u(k)
u(-k-1)
n<0,x(n)=0
n=0 ,x(n)=0
n=1 ,x(n)=1
u(1-k-1)
n=2 ,x(n)=2
u(2-k-1)
 x(n)  nu(n)
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57. Contd..
Find out convolution of two sequences given below

a)x(n)  {2 1 1 0 3}& h(n)  {1 2 1}
 


b)x(n)  {1 3 2 1}& h(n)  {1 1}
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58. Contd..

h(n)  {1
x(n)  {2 1 1 0 3}
2 1}

Solution (a)
 z
3
 z
5
 2z
1
 z
2
X (z)  2  z
1
 z
2
 0  3z
4
H (z)  z  2  z
1
Using convolution property
x(n)* h(n)  Z
1
{X (z).H (z)}
X (z).H (z)  (2  z
1
 z
2
 0  3z
4
)(z  2  z
1
)
 2z 1 z
1
 3z
3
 4  2z
1
 2z
2
 6z
4
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59. Contd..
X (z).H (z)  2z  5  z
1
 3z
2
 4z
3
 6z
4
 3z
5
x(n)* h(n)  Z
1
{X (z).H (z)}
x(n)* h(n)  {2 5 1  3 4 6  3}

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60. Contd..
Solution (b)
x(n)  {1 3 2 1}
h(n)  {1 1}
 z
4
 2z
3
 z
3
X (z)  1 3z
1
 2z
2
 3z
3
H (z)  1 z
1
X (z).H (z)  (1 3z
1
 2z
2
 3z
3
)(1 z
1
)
 1 3z
1
 2z
2
 z
1
 3z
2
 1 4z
1
 5z
2
 z
3
 z
4

1} x(n)* h(n)  {1 4 5 1

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