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
CISE318_Topic 3 (c) Khoukhi 2010 2
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
CISE318_Topic 3 (c) Khoukhi 2010 3
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
CISE318_Topic 3 (c) Khoukhi 2010 4
........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
CISE318_Topic 3 (c) Khoukhi 2010 5
?........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
CISE318_Topic 3 (c) Khoukhi 2010 6
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
CISE318_Topic 3 (c) Khoukhi 2010 7
.
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
CISE318_Topic 3 (c) Khoukhi 2010 8
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
CISE318_Topic 3 (c) Khoukhi 2010 9
111
aA
Examples of Z-transformUnit Step
u(k)
Example 1
00
01)(
k
kkustepunit
. . .. . .
-2 -1 0 1 2 3 4
k
CISE318_Topic 3 (c) Khoukhi 2010 10
-
100 1
1)()}({
zzzkukuZ
k
k
k
k
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Examples of Z-transform Example 2
CISE318_Topic 3 (c) Khoukhi 2010 11
Examples of Z-transformUnit Ramp Example 3
00
0)(
k
kkTkrrampunit
The Z-transform of unit ramp r(k) is
Tzkk
CISE318_Topic 3 (c) Khoukhi 2010 12
200 1 zkk
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Example 4
>33210k
01-231f(k)
Find the Z-transform of the sequence
CISE318_Topic 3 (c) Khoukhi 2010 13
321
01231)()(
zzzzkfzF kk
Example 5
Find the Z-transform of the sequence
>33210k
11111f(k)
kk
CISE318_Topic 3 (c) Khoukhi 2010 14
1
321
00
1
1....1
zzzz
kk
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Example 6
Find the Z-transform of the sequence
>33210k
(0.5)k0.1250.250.51f(k)
)2(5.0)()(000
zzzkfzF kk
k
kk
k
k
CISE318_Topic 3 (c) Khoukhi 2010 15
5.05.01)2(1....8421 11
zzzzzz
Properties of Z-transformLinearity
)}({)( 11 kfZzFLet
)}({)( 22
zFakaZ
Then
numbersrealanyarebanda
kfZzF
CISE318_Topic 3 (c) Khoukhi 2010 16
)()()}()({
)()()}()({
2121
2121
zFbzFakfbkfaZ
zFzFkfkfZ
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Example 7
>33210k
11118f(k)
1 k
-
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10 1 zk
Example 8
Find the Z-transform of the sequence
>33210k
22222f(k)
10
2)()(
zkfzF k
k
CISE318_Topic 3 (c) Khoukhi 2010 18
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Properties of Z-transform
zFkZLet
)()}({
dz
zdFzTkfkTZ
CISE318_Topic 3 (c) Khoukhi 2010 19
)()}({ zeFkfeZ aTakT
Properties of Z-transform
)()}({ zFkfZLet
stepunitshiftedtheis)(
)()}()({
nkuwhere
zFznkunkfZn
CISE318_Topic 3 (c) Khoukhi 2010 20
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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
CISE318_Topic 3 (c) Khoukhi 2010 21
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
CISE318_Topic 3 (c) Khoukhi 2010 22
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
CISE318_Topic 3 (c) Khoukhi 2010 23
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
CISE318_Topic 3 (c) Khoukhi 2010 24
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
CISE318_Topic 3 (c) Khoukhi 2010 25
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
..
CISE318_Topic 3 (c) Khoukhi 2010 26
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
CISE318_Topic 3 (c) Khoukhi 2010 27
unboundedisf
TheoremtheuseNOTcanwearezFofPoleseoremva uena
)(
2,1)(
Example 12
?}{ kTeZisWhat akT
2)1(}{
:
zzTkTZ
Solution
aTakT
CISE318_Topic 3 (c) Khoukhi 2010 28
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
CISE318_Topic 3 (c) Khoukhi 2010 29
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
CISE318_Topic 3 (c) Khoukhi 2010 30
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
CISE318_Topic 3 (c) Khoukhi 2010 31
11
11
z
z
z
Exercise 1
Find the Z-transform of the sequence
>2210k
1100f(k)
?)( zF
CISE318_Topic 3 (c) Khoukhi 2010 32
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Exercise 2
Find the Z-transform of the sequence
>2210k
0-111f(k)
?)( zF
CISE318_Topic 3 (c) Khoukhi 2010 33
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
CISE318_Lesson5 (c) Khoukhi 2009 35
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
CISE318_Lesson5 (c) Khoukhi 2009 36
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
CISE318_Lesson5 (c) Khoukhi 2009 37
s s use o ensure a e o a neexpansion of F(z) contains the factor z innumerator
Partial Fraction Expansion
-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
CISE318_Lesson5 (c) Khoukhi 2009 38
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
CISE318_Lesson5 (c) Khoukhi 2009 39
122)2(2)1(2)(2
2
1
2)(
2)2)(1(
2)2(
2
2
kkk
z
kfz
z
z
zzF
zzza
Partial Fraction Expansion
, Example 17 Simple pole case
-
pz
ii
n
n
z
zFpza
pz
a
pz
a
pz
a
z
zFExpand
i
)()(
)(
2
2
1
1
CISE318_Lesson5 (c) Khoukhi 2009 40
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
CISE318_Lesson5 (c) Khoukhi 2009 41
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
CISE318_Lesson5 (c) Khoukhi 2009 42
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
CISE318_Lesson5 (c) Khoukhi 2009 43
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:
CISE318_Lesson5 (c) Khoukhi 2009 44
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
CISE318_Lesson5 (c) Khoukhi 2009 45
)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
CISE318_Lesson5 (c) Khoukhi 2009 46
][)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
CISE318_Lesson5 (c) Khoukhi 2009 49
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
CISE318_Topic 3 (c) Khoukhi 2010 50
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