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Slide 1 Click to edit Master subtitle style Nguyen Thanh Tuan, M Eng Department of Telecommunications (113B3) Ho Chi Minh City University of Technology Email nttbk97@yahoo com z Transform Chapter 5 Di[.]

Click to edit Master subtitle style

Nguyen Thanh Tuan, M.Eng

Department of Telecommunications (113B3)

Ho Chi Minh City University of Technology

z-Transform

Chapter 5

 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

Content

1 z-transform

2 Properties of the z-transform

3 Causality and Stability

4 Inverse z-transform

1 ( )

0 ( )

1 ( )

2 ( )

( )

X

n

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

} )

( )

(

| C

X z

ROC

 The z-transform of impulse response h(n) is called the transform

function of the filter:

Example 2

 Determine the z-transform of the signal

a) x(n)=(0.5)nu(n)

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

z-transform and ROC

 It is possible for two different signal x(n) to have the same

z-transform 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

ROC of an anticausal signal

is the interior of a circle

Example 3

 Determine the z-transform of the signal

) 1 (

) ( )

(n  a u n b u n

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

2 Properties of the z-transform

 Linearity:

1 1

1 (n) X (z) with ROC

2 2

2 1

2 Properties of the z-transform

 Time shifting:

) ( )

(n X z

) ( )

Example: Determine the z-transform of the signal x(n)=2nu(n-1)

 Scaling in the z-domain:

2

1 | | :

ROC )

( )

(n X z r z r

if

2 1

( )

(n X a z a r z a r x

then

for any constant a, real or complex

Example: Determine the z-transform of the signal x(n)=ancos(w0n)u(n)

2 Properties of the z-transform

 Time reversal:

if

then

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

 Convolution of two sequence:

2

1 | | :

ROC )

( )

(n X z r z r

if and

) ( )

( )

( )

( )

( )

(n x1 n x2 n X z X1 z X2 z

then

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] ?

1 2

|

| r

1 : ROC )

( )

(

r

z z

X n

) ( )

2 Properties of the z-transform

 Differentiation in the z-domain

if

then

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

) ( )

nx( )  z  ( )

the ROCs of both are the same

3 Causality and stability

will have z-transform

 A causal signal of the form







 ( ) ( ) )

(n A1p1u n A2 p2u n

|

| max

|

|

ROC 1

p

A z

p

A z

the ROC of causal signals are outside of the circle

 A anticausal signal of the form

(n A1p1u n A2p2u n

|

| min

|

|

ROC 1

p

A z

p

A z

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

4 Inverse z-transform

ROC

), ( )

(n z transform X z

) ( ROC

), (z inversez-transform x n

( )

(anticausa |

a

|

| z

| ROC if

) 1 (

signals) (causal

| a

|

| z

| ROC if

)

( )

(

n u a

n u

a n

n

1 -

az - 1

1 )

(z 

X

Partial fraction expression method

 In general, the z-transform is of the form

 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

) 1

( ) 1

)(

1 ( ) (z   p1z1  p2z1  p z1

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

where

M M

N N z a z

a

z b z

b b

z D

z

N z

1 1 0

1 )

(

)

( )

(

Example 4 d

 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

Example 5 od

Partial fraction expression method

 If N=M

Where and for i=1,…,M

 If N> M

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

Example 6 (cont.)

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

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

Example 8

 Determine the causal inverse z-transform of

Solution:

Example 8 (cont.)

Some common z-transform pairs

Review

 Định nghĩa biến đổi z

 Ý nghĩa miền hội tụ của biến đổi z

 Mối liên hệ giữa miền hội tụ với đặc tính nhân quả và ổn định của tín hiệu/hệ thống-LTI rời rạc

 Biến đổi z của một số tín hiệu cơ bản: (n), anu(n), anu(-n-1)

 Một số tính chất cơ bản (tuyến tính, trễ, tích chập) của biến đổi z

 Phân chia đa thức và biến đổi z ngược

Homework 1

Homework 2

Homework 3

Homework 4

Homework 5