lec.3 Ztransform

8/8/2019 lec.3 Ztransform

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CISE318

Lecture 3: Z-transform

Dr. Amar KhoukhiTerm 092

CISE318_Topic 3 (c) Khoukhi 2010 1

Read Sections 2.1-2.4 of text book

Outlines

Series and their convergence.

Wh do we use Z-transform?

Definition of Z-transform

Z-Transform of simple functions

Properties of Z-transform

Inverse Z-transform

Long division


Partial fraction expansion

Solving Difference equations


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Objectives

Understand the definition of discrete time systems

Appreciate the use Z-transform to solve lineardifference equations

Understand the inverse Z-transform

Appreciate the use Z-transform to describe linear


time-invariant discrete-time systems To be able to obtain inverse Z-transform

To be able to solve difference equations

Series

For what values of a the series is defined?

What is the value of series?

........1

....1

2

2

N

N

aaaB

aaaA


........21

........

2

N

NaaaD

aaaC


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Answers

1 N

1,1

1........1

11

11....1

2

2

N

N

afordefineda

aaaB

aforN

aforaaaaA


?........21

1,11

1........

2

32

N

N

NaaaD

afordefinedaa

aaaC

What is Z-transform?

Z-transform is a tool that helps us to solve lineardifference e uations

Z-transform models such as pulse transfer

function are commonly used to describe:

linear time-invariant discrete-time systems

Difference Z-transform Algebraic


Equation Equation

Solution of theAlgebraicEquation

Inverse Z-transformSolution of

The DifferenceEquation


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Why do we use Z-transform

With Z-transform, it is easier to

ompu e e response o scre e- me sys em

Analyze discrete-time system

Stability

Performance

Steady state error


.

Design digital controllers

More useful for computer systems

Definition of Z-transform

,

Main tool is z-transform

0

)()()]([k

kzkfzFkfZ

The Z-transform of a sequence {f(k)} is defined as


f(k) F(z) , where z is complex

Analogous to Laplace transform for s-domain


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Series

0

1ifConverge

A?isWhatconverge?seriesthedoesWhen

....

a

aaaaAk


111

aA

Examples of Z-transformUnit Step

u(k)

Example 1

00

01)(

k

kkustepunit

. . .. . .

-2 -1 0 1 2 3 4

k


-

100 1

1)()}({

zzzkukuZ

k

k

k

k


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Examples of Z-transform Example 2


Examples of Z-transformUnit Ramp Example 3

00

0)(

k

kkTkrrampunit

The Z-transform of unit ramp r(k) is

Tzkk


200 1 zkk


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Example 4

>33210k

01-231f(k)

Find the Z-transform of the sequence


321

01231)()(

zzzzkfzF kk

Example 5


>33210k

11111f(k)

kk


1

321

00

1

1....1

zzzz

kk


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Example 6


>33210k

(0.5)k0.1250.250.51f(k)

)2(5.0)()(000

zzzkfzF kk

k

kk

k

k


5.05.01)2(1....8421 11

zzzzzz

Properties of Z-transformLinearity

)}({)( 11 kfZzFLet

)}({)( 22

zFakaZ

Then

numbersrealanyarebanda

kfZzF


)()()}()({

)()()}()({

2121

2121

zFbzFakfbkfaZ

zFzFkfkfZ


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Example 7

>33210k

11118f(k)

1 k

-


10 1 zk

Example 8


>33210k

22222f(k)

10

2)()(

zkfzF k

k



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zFkZLet

)()}({

dz

zdFzTkfkTZ


)()}({ zeFkfeZ aTakT


)()}({ zFkfZLet

stepunitshiftedtheis)(

)()}()({

nkuwhere

zFznkunkfZn



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Z-transform Properties mnfmfnfnf

m

2121

Convolution definition

zmnfmf

zmnfmf

mnfmfZnfnfZ

n

n

n

m

m

21

21

2121

Take z-transform

Z-transform definitionInterchange summation

Substitute r= nm


zFzF

zrfzmf

zrfmf

r

rm

m

m r

mr

m n

21

21

21

Z-transform definition

Example 9

Ratio of polynomial z-domain functions

13

12][

2

zzzX

Divide through by the highest

power of z

Factor denominator into first-order factors

22 zz

21

21

21

231

21][

zz

zzzX

2121][

zzzX


Use partial fraction decomposition

to get first-order terms

11 12

1

zz

1

2

1

10

1

2

11

][

z

A

z

ABzX


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Example 9 (cont)

Find B0 by polynomial2

1212

3

2

1 1212 zzzz

division

Express in terms of B0

15

23

1

12

z

zz

11

1

12

11

512][

zz

zzX

21


Solve for A1 and A2

8

2

1

121

2

11

21

9211

1

1

21

2

2

11

1

1

z

z

z

zzA

zzzA

Example 9 (cont)

Express X[z] in terms of B0, A1, and A2

Use table to obtain inverse z-transform

11 1

2

11

2][

zz

zX

nununnx

n

81

92


2

With the unilateral z-transform, or the bilateral z-transform withregion of convergence, the inverse z-transform is unique


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Properties of Z-transformInitial and final value theorems

)()}({ zFkfZLet

)(lim)0(

TheoremvalueFinal

zFf

TheoremvalueInitial

z


1.zatpolesingleapossiblyexcept

circleunitinsideareF(z)ofpolesallprovided

1

z

Example 10

What is f(0) and f() if6.0

)(2

zzF

2

0.6(0) lim ( ) 0 ' '

0.5 0.06( )

z z

z f F z L Hopital sRule

z zTo find f we need to check for applicability of

final value Theorem

..


21

( ) 0.3, 0.2

( 1)( 0.6)( ) lim( 1) ( )

0.5 0.0z

Poles of F z are we can use the Theorem

z z f z F z

z z

10

6 z


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Example 11

What is f(0) and f() if23

7.0)(

2

zzF

ofityapplicabilforchecktoneedweffindTo

zz

zzFf

zz

)(

023

7.0)(lim)0(

2


unboundedisf

TheoremtheuseNOTcanwearezFofPoleseoremva uena

)(

2,1)(

Example 12

?}{ kTeZisWhat akT

2)1(}{

:

zzTkTZ

Solution

aTakT


2)1(}{

ze

zeTkTeZ

aT

aTakT


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Example 13

6g n n u n u n

2 26 6

2 2 55

6 6 6 6

1 16

11 1

1 6

11 1

n u n n u n u n

z z zG z

z z zz z

z

z z z z z


5

2 25 55

6

25

11 6

1 11 1

6 5

1

zz z

z z z z z z z

z z

z z

Find the z-transform of the unit pulse or impulse sequence

Example 14

otherwise0

nxn

00

101][)(n

n

n

n

nn

nzzxxzzx

This follows trivially from Equation


1...001)}({ 21 zzkZ


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Example 15Find the z-transform of the unit step sequence

0

1

00

)()(n

n

n

n

n

n

n zzzxzx

0for0

nnuxn

From the definition


11

11

z

z

z

Exercise 1


>2210k

1100f(k)

?)( zF



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Exercise 2


>2210k

0-111f(k)

?)( zF


Inverse Z-transform

Several methods are available to obtaininverse Z-transform

Partial fraction expansion

Power series method

Inversion Formula

CISE318_Lesson5 (c) Khoukhi 2009 34


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z-Transforms of Common Functions

Impulse

Step

Ram

1

1z

z

2

z

1)( tf

ttf )(

1

s

1

2

1

00

01)(

t

ttf


Exponential

Sine

z

aez

z

1)(Cos2

Sin2 zaz

az

atetf )(

)sin()( ttf

s

as

1

22

1

s

z-transform Table (2)

numnnnn n

)1)...(2)(1(

No. zX nx

1 mz

10

m !

nunrn

)cos(

nnun cos

nnun

sin

22)cos2(

)cos(

zz

zz

22 )cos2(

)]cos(cos[

zz

zrz

22)cos2(

)sin(

zz

zz

11a

12a

12b


nunrn

)cos( je

nunr n )cos(

2

)5.0()5.0(

z

zre

z

zrejj

a 1cos

22 2

)(

azz

BAzz

22

222 )2

a

AaBBzAr

22

1tan

aA

BAa

12b

12c


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Partial Fraction Expansion

In Partial fraction ex ansion method F z is ex ressed as the sum of

In Inverse Z-transform, we expand F(z)/z thenmultiply by z to get the expansion of F(z)

simple terms and the Z-Transform Table is used to invert each one of them


s s use o ensure a e o a neexpansion of F(z) contains the factor z innumerator


-their Z-transform

Expand X(z) as the weighted sum of terms

available in the TableBecause most of term in the Z-transform table has

z in numerator, we will factor X(z)/z then


mu t p y y z to get z


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Partial Fraction Expansion, Example 16

21

z

ZFind

2)2)(1(

2)1(

21)2)(1(

2)(

1

1

21

zzz

za

z

a

z

a

zzz

zFExpand


122)2(2)1(2)(2

2

1

2)(

2)2)(1(

2)2(

2

2

kkk

z

kfz

z

z

zzF

zzza


, Example 17 Simple pole case

-

pz

ii

n

n

z

zFpza

pz

a

pz

a

pz

a

z

zFExpand

i

)()(

)(

2

2

1

1


k

nn

kk

n

n

papapakf

pz

za

pz

za

pz

zazF

)(...)()()(

)(

2211

2

2

1

1


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Partial Fraction ExpansionExample 18

21

z

ZFind

2)2)(1(

2)1(

21)2)(1(

2)(

1

1

21

zzz

za

z

a

z

a

zzz

zFExpand


122)2(2)1(2)(2

2

1

2)(

2)2)(1(

2)2(

2

2

kkk

z

kfz

z

z

zzF

zzza

Inversion Formula

kzzEofresidueskx

1)()(

az

k

az

k

zzXof

polesall

mtymultipliciwithpolesfor

zzEazzzEofresidues

polessimplefor

k

11

)(

)()()(

1


General form is obtained

az

km

m

m

az

kzzEaz

dz

d

mzzEofresidues

1

1

11 )()(

)!1(

1)(


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Example 19Inversion Formula

)(1

)(1

1

zzzXzX

kk

)()(

2,10

2,1,00

)(

1

1

zzEofresidueskx

arepoleskfor

arepoleskfor

zzXof

polesall

k

k


General form is obtained

)2)(1()(

1

zzz

zzkxk

k

More Examples

)3)(2(

198][

zz

zzX

Find inverse z-transform-real unique poles

Find the inverse z-transform of :

)3)(2(

198][

zzz

z

z

zX

3

)3/5(

2

)2/3()6/19(][

zzzz

zX

5319 zz

Step 1: Divide both sides by z :

Step 2: Perform partial fraction:


33226 zz

Step 4: Obtain inverse z-transform of each term from table (#1 & #6) :

][33

52

2

3][

6

19][ nunnx

nn


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Find inverse z-transform repeat real poles

Find the inverse z-transform of:3

2

)2)(1(

)12112(][

zz

zzzzX

Example 20

v e o s es y z an expan :

)2()2()2(1)2)(1(

12112][ 22

1

3

0

3

2

z

a

z

a

z

a

z

k

zz

zz

z

zX

3)2)(1(

12112

1

3

2

zzz

zzk 2

)2)(1(

12112

2

3

2

0

zzz

zza

Use covering method to find k and :0a


)2()2()2(

2

1

3

)2)(1(

12112][ 22

1

33

2

z

a

z

a

zzzz

zz

z

zX

To find , multiply both sides by z and let :

30030 22 aa2a z

e ge :

Find inverse z-transform repeat real poles

To find let z = 0 : 12

3

44

13

8

121

1 aa

1a

Example 20, Cont.

2

3

)2(

1

)2(

2

1

3][23

zzzzz

zX

23

)2()2(2

13][

23

z

z

z

z

z

z

z

zzX

Therefore, we find :

Use pairs #6 & #10


][)2(3)2(2

)2(8

)1(23][ nu

nnnnx nnn

][2)12(4

13

2 nunn n


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Inverse Z-Transform by Power Series Expansion

nznxzX

Example 22

The z-transform is power seriesn

2112 z2xz1x0xz1xz2xzX

12

1112

11

z1z1z2

11zzX

2n1

In expanded from

Z-transforms of this form


z2

z2

z

1n2

1n1n

2

12nnx

2n0

1n2

1

0n1

1n2nx

can generally be inversed

Especially useful for finite-length

Example

Homework problems

Check the course webCT for the Homework


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