The z-Transform

Why z-Transform?
• A generalization of Fourier transform
• Why generalize it?
– FT does not converge on all sequence
– Notation good for analysis
– Bring the power of complex variable theory deal
with the discrete-time signals and systems
 Definition
• The z-transform of sequence x(n) is defined by

X ( z)   x ( n) z
n  
n

Fourier
Transform
 Let z = ej.

X (e )  j
 x ( n )e
n 
 j n
 z-Plane


 x ( n) z
Im
n
X ( z) 
z = ej
n  

 Re
j
X (e )   x ( n )e
n 
 j n

Fourier Transform is to evaluate z-transform

on a unit circle.
 z-Plane
Im
X(z)
z = ej

Re

Im

Re
 Periodic Property of FT

X(ej)
X(z)

  

Im

Re Can you say why Fourier Transform is

a periodic function with period 2?
 Region of Convergence (RoC)

• Give a sequence, the set of values of z for

which the z-transform converges, i.e.,
|X(z)|<, is called the region of convergence.
 
| X ( z ) |  x (
n  
n ) z n
  | x
n  
( n ) || z | n


ROC is centered on origin and

consists of a set of rings.
Example: Region of Convergence

 
| X ( z ) |  x (
n  
n ) z n
  | x
n  
( n ) || z | n


Im
ROC is an annual ring centered
on the origin.
r
Re Rx  | z | Rx 

ROC  {z  re j | Rx  r  Rx }

 Stable Systems

• A stable system requires that its Fourier

transform is uniformly convergent.
Im  Fact: Fourier transform is to
evaluate z-transform on a unit
circle.
1
 A stable system requires the
Re ROC of z-transform to include
the unit circle.
Example: A right sided Sequence

x ( n)  a n u ( n)

x(n)

... n
-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10
Example: A right sided Sequence

For convergence of X(z), we

x ( n)  a u ( n)
n
require that


 |
 1
| az 1 | 1
X ( z)   a u(n)z
n  
n n | az
n 0

 | z || a |
  a n z n 
1 z
n 0 X ( z )   (az ) 
1 n
1

 n 0 1  az za
  (az 1 ) n
| z || a |
n 0
Example: A right sided Sequence ROC
for x(n)=anu(n)

z
X ( z)  , | z || a | Which one is stable?
za

Im Im

1 1
a a a a
Re Re
Example: A left sided Sequence

x(n)  a nu (n  1)

-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10
... n

x(n)
 Example: A left sided Sequence

For convergence of X(z), we

x(n)  a u (n  1)
n
require that
 
X ( z )    a u (n  1)z
 z|
1
n
| a 1 z | 1
n

n  
| a
1
n 0
   a n z n
n  
| z || a |

  a  n z n 
1 z
n 1 X ( z )  1   (a z )  1 
1 n
1

 n 0 1 a z z  a
 1   a n z n
n 0 | z || a |
Example: A left sided Sequence ROC
for x(n)=anu( n1)

z
X ( z)  , | z || a | Which one is stable?
za

Im Im

1 1
a a a a
Re Re
Represent z-transform as a Rational
Function

P( z )
X ( z) 
where P(z) and Q(z) are
polynomials in z.
Q( z )

Zeros: The values of z’s such that X(z) = 0

Poles: The values of z’s such that X(z) = 
Mr. GANESH V N , Dept. of E&C,
MITE, Moodabidri
Example: A right sided Sequence

z
x ( n)  a n u ( n) X ( z)  , | z || a |
za
Im

ROC is bounded by the

pole and is the exterior
a
Re of a circle.
Example: A left sided Sequence

z
x(n)  a nu (n  1) X ( z)  , | z || a |
za
Im

ROC is bounded by the

pole and is the interior
a
Re of a circle.
 Example: Sum of Two Right Sided Sequences

x(n)  ( 12 ) n u(n)  ( 13 ) n u(n)

z z 2 z ( z  121 )
X ( z)   
z2 z3
1 1
( z  12 )( z  13 )
Im
ROC is bounded by poles
and is the exterior of a circle.
1/12
1/3 1/2 Re

ROC does not include any pole.

Example: A Two Sided Sequence

x(n)  ( 13 ) n u(n)  ( 12 ) n u(n 1)

z z 2 z ( z  121 )
X ( z)   
z3 z2
1 1
( z  13 )( z  12 )
Im
ROC is bounded by poles
and is a ring.
1/12
1/3 1/2 Re

ROC does not include any pole.

 Example: A Finite Sequence

x ( n)  a n , 0  n  N 1
N 1 N 1
1  (az 1 ) N 1 zN  aN
X ( z)   a z n n
  ( az )
1 n
  N 1
n 0 n 0 1  az 1 z za
Im
N-1 zeros
ROC: 0 < z < 
N-1 poles ROC does not include any pole.
Re
Always Stable
 Properties of ROC
• A ring or disk in the z-plane centered at the origin.
• The Fourier Transform of x(n) is converge absolutely iff the ROC includes
the unit circle.
• The ROC cannot include any poles
• Finite Duration Sequences: The ROC is the entire z-plane except possibly
z=0 or z=.
• Right sided sequences: The ROC extends outward from the outermost finite
pole in X(z) to z=.
• Left sided sequences: The ROC extends inward from the innermost
nonzero pole in X(z) to z=0.
 More on Rational z-Transform

Consider the rational z-transform

with the pole pattern:
Im

Find the possible a b c

ROC’s Re
 More on Rational z-Transform

Consider the rational z-transform

with the pole pattern:
Im
Case 1: A right sided Sequence.

a b c
Re
 More on Rational z-Transform

Consider the rational z-transform

with the pole pattern:
Im
Case 2: A left sided Sequence.

a b c
Re
 More on Rational z-Transform

Consider the rational z-transform

with the pole pattern:
Im
Case 3: A two sided Sequence.

a b c
Re
 More on Rational z-Transform

Consider the rational z-transform

with the pole pattern:
Im
Case 4: Another two sided Sequence.

a b c
Re
 Z-Transform Pairs

Sequence z-Transform ROC

(n) 1 All z
All z except 0 (if m>0)
(n  m) z m
or  (if m<0)
1
| z | 1
u (n) 1  z 1
1
 u(n  1) | z | 1
1  z 1

1
n | z || a |
a u (n) 1  az 1
1
 a nu (n  1) | z || a |
1  az 1
 Z-Transform Pairs

Sequence z-Transform ROC

1  [cos 0 ]z 1
[cos 0 n]u (n) | z | 1
1  [2 cos 0 ]z 1  z  2

[sin 0 ]z 1
[sin 0 n]u (n) | z | 1
1  [2 cos 0 ]z 1  z  2

1  [r cos 0 ]z 1
[r n cos 0 n]u(n) | z | r
1  [2r cos 0 ]z 1  r 2 z  2

[r sin 0 ]z 1
[r n sin 0 n]u(n) | z | r
1  [2r cos 0 ]z 1  r 2 z  2

a n 0  n  N 1 1 a N zN
 | z | 0
0 otherwise 1  az 1
Properties of z-Transform
 Linearity

Z[ x(n)]  X ( z ), z  Rx
Z[ y(n)]  Y ( z), z  Ry

Z[ax(n)  by(n)]  aX ( z)  bY ( z), z  Rx  Ry

Overlay of
the above two
ROC’s
 Time Shift

Z[ x(n)]  X ( z ), z  Rx

Z[ x(n  n0 )]  z X ( z)
n0
z  Rx
Multiplication by an Exponential Sequence

Z[ x(n)]  X ( z ), Rx- | z | Rx 

1
Z[a x(n)]  X (a z)
n
z | a | Rx
Differentiation of X(z)

Z[ x(n)]  X ( z ), z  Rx

dX ( z )
Z [nx(n)]   z z  Rx
dz
 Time Reversal

Z[ x(n)]  X ( z ), z  Rx

1
Z[ x(n)]  X ( z ) z 1 / Rx
Initial Value Theorem

x(n)  0, for n  0

x(0)  lim X ( z )
z 
Convolution of Sequences

Z[ x(n)]  X ( z ), z  Rx
Z[ y(n)]  Y ( z), z  Ry

Z[ x(n) * y(n)]  X ( z)Y ( z) z  Rx  Ry

 Convolution of Sequences


x ( n) * y ( n)   x(k ) y (n  k )
k  

 

 n
Z[ x(n) * y (n)]     x(k ) y (n  k )  z
n    k   
   
  x(k )  y(n  k )z n
 
k  
x(k ) z  k  y (
n  
n )z n

k   n  

 X ( z)Y ( z )
 Inverse z-Transform: Examples

Example Find the inverse z-transform of

Solution We get,
Using table,

Example Find the inverse z-transform of

Solution
We get,

Using table,
 Inverse z-Transform: Examples
Example Find the inverse z-transform of

Solution Since,
By coefficient matching,

Therefore,

Example Find the inverse z-transform of

Solution

Using Table
Inverse z-Transform: Using Partial
Fraction
Problem: Find the inverse z-transform of

Solution: First eliminate the negative power of z.

Dividing both sides by z,

Finding the
constants:

Therefore, inverse z-transform is:

Problem:

Solution:
Dividing Y(z) by z,

Applying the partial

fraction expansion,
We first find B:

Next find A:
 Using the polar form,

Now we have:

Therefore, the inverse z-transform is:

Problem:

Solution:
Dividing both sides by z:

Where,
Using the formulas for mth-order,

m=2, p=0.5
 Then,

From Table,

Finally we get,
Partial Fraction Expansion Using MATLAB
Problem: Find the partial expansion of

Solution:
The denominator polynomial
can be found using MATLAB:

Therefore,

and

The solution is:

residues
poles direct term
Problem:

Solution:
Problem:

Solution:

Then
Transform domain Analysis
 Shift-Invariant System

x(n) y(n)=x(n)*h(n)
h(n)

X(z) H(z) Y(z)=X(z)H(z)

 Shift-Invariant System

X(z) Y(z)
H(z)
Y ( z)
H ( z) 
X ( z)
 Nth-Order Difference Equation

N M

a
k 0
k y (n  k )   br x(n  r )
r 0

N M
Y ( z ) ak z k  X ( z ) br z r
k 0 r 0

M N
r k
H ( z )   br z  ak z
r 0 k 0
Representation in Factored Form

Contributes poles at 0 and zeros at cr

M
A (1  cr z 1 )
H ( z)  N
r 1

 r )
(1
k 1
 d z 1

Contributes zeros at 0 and poles at dr

 Stable and Causal Systems

Causal Systems : ROC extends outward from the outermost pole.

Im
M
A (1  cr z 1 )
H ( z)  N
r 1
Re
 r )
(1 
k 1
d z 1
 Stable and Causal Systems

Stable Systems : ROC includes the unit circle.

Im
M
A (1  cr z 1 ) 1
H ( z)  N
r 1
Re
 r )
(1
k 1
 d z 1
 Example

Consider the causal system characterized by

y(n)  ay(n  1)  x(n) Im

1 1
H ( z) 
1  az 1 a Re

h( n)  a n u ( n)