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Chapter 5
z-Transform
Nguyen Thanh Tuan, Click
M.Eng.
to edit Master subtitle style
Department of Telecommunications (113B3)
Ho Chi Minh City University of Technology
Email: nttbk97@yahoo.com


 The z-transform is a tool for analysis, design and implementation of
discrete-time signals and LTI systems.
 Convolution in time-domain  multiplication in the z-domain

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z-Transform



Content

1. z-transform

2. Properties of the z-transform
3. Causality and Stability
4. Inverse z-transform

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1. The z-transform
 The z-transform of a discrete-time signal x(n) is defined as the power
series:
X ( z) 





x (n ) z  n 

x( 2) z 2  x ( 1) z  x (0)  x(1) z 1  x(2) z 2 

n 

 The region of convergence (ROC) of X(z) is the set of all values of
z for which X(z) attains a finite value.
ROC  {z  C | X ( z ) 



 x ( n) z

n


 }

n  

 The z-transform of impulse response h(n) is called the transform
function of the filter:
H ( z) 



n
h
(
n
)
z


n  
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z-Transform


Example 1
 Determine the z-transform of the following finite-duration signals
a) x1(n)=[1, 2, 5, 7, 0, 1]

b) x2(n)=x1(n-2)
c) x3(n)=x1(n+2)
d) x4(n)=(n)
e) x5(n)=(n-k), k>0
f) x6(n)=(n+k), k>0

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z-Transform


Example 2
 Determine the z-transform of the signal

a) x(n)=(0.5)nu(n)
b) x(n)=-(0.5)nu(-n-1)

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z-transform and ROC
 It is possible for two different signal x(n) to have the same ztransform. Such signals can be distinguished in the z-domain by their
region of convergence.

 z-transforms:

and their ROCs:

ROC of a causal signal is the
exterior of a circle.
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ROC of an anticausal signal
is the interior of a circle.
z-Transform


Example 3
 Determine the z-transform of the signal
x(n)  a nu(n)  b nu(n  1)

 The ROC of two-sided signal is a ring (annular region).
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z-Transform


2. Properties of the z-transform
 Linearity:
if
and

z
x1 (n) 

X 1 ( z ) with ROC1
z
x2 (n) 

X 2 ( z ) with ROC2

then
z
x(n)  x1 (n)  x2 (n) 

X ( z)  X1 ( z)  X 2 ( z) with ROC  ROC1  ROC2

 Example: Determine the z-transform and ROC of the signals

a) x(n)=[3(2)n-4(3)n]u(n)
b) x(n)=cos(w0 n)u(n)
c) x(n)=sin(w0 n)u(n)

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z-Transform


2. Properties of the z-transform
 Time shifting:
if
then

z
x(n) 

X ( z)
z
x(n  D) 

z  D X ( z)

 The ROC of z  D X (z ) is the same as that of X(z) except for z=0 if
D>0 and z= if D<0.
Example: Determine the z-transform of the signal x(n)=2nu(n-1).
 Convolution of two sequence:
if

z
x1 (n) 

X1 ( z)

z
x2 (n) 

X 2 ( z)

and

z
then x(n)  x1 (n)  x2 (n) 

X ( z)  X1 ( z) X 2 ( z)
the ROC is, at least, the intersection of that for X1(z) and X2(z).

Example: Compute the convolution of x=[1 1 3 0 2 1] and h=[1, -2, 1] ?
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z-Transform


2. Properties of the z-transform
 Time reversal:
z
x(n) 

X ( z ) ROC : r1 | z | r2
if
1
1
z
then
x (  n) 

X ( z 1 ) ROC : | z |
r2

r1

Example: Determine the z-transform of the signal x(n)=u(-n).

 Scaling in the z-domain:
z
x(n) 

X ( z ) ROC : r1 | z | r2

if
then a n x(n) 
z

X (a 1 z ) ROC : | a | r1 | z || a | r2
for any constant a, real or complex
Example: Determine the z-transform of the signal x(n)=ancos(w0n)u(n).
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z-Transform


3. Causality and stability
 A causal signal of the form
x(n)  A1 p1nu(n)  A2 p2nu(n)  

will have z-transform
A1
A2
X ( z) 

  ROC| z | max | pi |
1
1
i
1  p1 z
1  p2 z

the ROC of causal signals are outside of the circle.

 A anticausal signal of the form
x(n)   A1 p1nu(n  1)  A2 p2nu(n  1)  

X ( z) 

A1
A2

  ROC| z | min | pi |
1
1
i
1  p1 z
1  p2 z

the ROC of causal signals are inside of the circle.
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z-Transform


3. Causality and stability
 Mixed signals have ROCs that are the annular region between two
circles.

 It can be shown that a necessary and sufficient condition for the
stability of a signal x(n) is that its ROC contains the unit circle.

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z-Transform


4. Inverse z-transform
 transform
x(n) z
 X ( z ), ROC
transform
X ( z), ROC inverse
z-

 x(n)
z
x(n) 

X ( z ), ROC

 In inverting a z-transform, it is convenient to break it into its partial
fraction (PF) expression form, i.e., into a sum of individual pole
terms whose inverse z transforms are known.
1
 Note that with X ( z ) 
-1 we have
1 - az

a nu (n) if ROC | z || a | (causal signals)
x ( n)   n
 a u (n  1) if ROC | z || a | (anticausal signals)
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Partial fraction expression method
 In general, the z-transform is of the form
N ( z ) b0  b1 z 1    bN z  N
X ( z) 

D( z )
1  a0 z 1   aM z  M

 The poles are defined as the solutions of D(z)=0. There will be M
poles, say at p1, p2,…,pM . Then, we can write
D( z )  (1  p1 z 1 )(1  p2 z 1 )(1  pM z 1 )

 If N < M and all M poles are single poles.

where

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Example 4d
 Compute all possible inverse z-transform of

Solution:
- Find the poles:

1-0.25z-2 =0  p1=0.5, p2=-0.5

- We have N=1 and M=2, i.e., N < M. Thus, we can write

where

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Example 5od

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Partial fraction expression method
 If N=M

Where

and

for i=1,…,M

 If N> M

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Example 6
 Compute all possible inverse z-transform of

Solution:
- Find the poles:

1-0.25z-2 =0  p1=0.5, p2=-0.5

- We have N=2 and M=2, i.e., N = M. Thus, we can write

where

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Example 6 (cont.)

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Example 7 (cont.)
 Determine the causal inverse z-transform of

Solution:
- We have N=5 and M=2, i.e., N > M. Thus, we have to divide the
denominator into the numerator, giving

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z-Transform


Partial fraction expression method
 Complex-valued poles: since D(z) have real-valued coefficients, the
complex-valued poles of X(z) must come in complex-conjugate pairs

Considering the causal case, we have

Writing A1 and p1 in their polar form, say,
with B1 and R1 > 0, and thus, we have
As a result, the signal in time-domain is

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z-Transform


Example 8
 Determine the causal inverse z-transform of

Solution:

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Example 8 (cont.)

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Some common z-transform pairs

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