on z-transform and its applications
TRANSCRIPT
An-Najah National University
Faculty of Graduate Studies
On Z-transform and Its Applications
By
Asma Belal Fadel
Supervisor
Dr. ''Mohammad Othman'' Omran
This thesis is submitted in Partial Fulfillment of the Requirements for
the Degree of Master of Mathematics , Faculty of Graduate Studies, An-
Najah National University, Nablus, Palestine.
2015
II
On Z-transform and Its Applications
By
Asma Belal Fadel
This thesis was defended successfully on 30/6 /2015 and approved by:
Defense Committee Members Signature
Dr. "Mohammad Othman" Omran / Supervisor โฆโฆ.โฆ.โฆโฆโฆ.
Dr. Saed Mallak /External Examiner .โฆ.โฆโฆ.โฆโฆโฆ
Dr. Mohammad Assa'd /Internal Examiner โฆ..โฆโฆ.โฆโฆโฆ
III
Dedication
To my parents and brothers.
IV
Acknowledgements
I would like to express my sincere gratitude to my Supervisor
Dr. ''Mohammad Othman'' Omran for the continuous
support, helpful suggestion, patience and encouragement
throughout the course of this thesis. His guidance helped me
at all times of research and writing of this thesis.
I would also like to thank my defense thesis committee: Dr.
Mohammad Assa'd and Dr. Saed Mallak.
I want to take this opportunity to express gratitude to all
my friends and to all people who help me do this work.
Last but not least, I wish to thank my parents, for the
unceasing encouragement, support and attention they
provide me throughout my life.
V
ูุฑุงุง ุงูุฅ
ุฃูุง ุงูู ููุฑุน ุฃุฏูุงู ู ูุฏู ุงูุงุณุงูุฉ ุงูุชู ุชุญู ู ุงูุนููุงู
On Z-transform and Its Applications
ุงูุฑุง ุจุฃู ู ุง ุงุดุชู ูุช ุนููู ูุฐู ุงูุงุณุงูุฉ ุฅูู ุง ูู ูุชุงุฌ ุฌูุฏู ุงูุฎุงุตุ ุจุงุณุชุซูุงุก ู ุง ุชู ุช ุงูุฅุดุง ุฉ
ุฅููู ุญูุซู ุง ู ุฏ ูุฃู ูุฐู ุงูุงุณุงูุฉ ููู ุฃู ุฃู ุฌุฒุก ู ููุง ูู ููุฏู ู ู ูุฑุจู ูููู ุฃูุฉ ุฏ ุฌุฉ ุนูู ูุฉ ุฃู
ุจุญุซ ุนูู ู ุฃู ุจุญุซู ูุฏู ุฃูุฉ ู ุคุณุณุฉ ุชุนููู ูุฉ ุฃู ุจุญุซูุฉ ุฃุฎุงู.
Declaration
The work provided in this thesis, unless otherwise referenced, is the
researcher's own work, and has not been submitted elsewhere for any other
degree or qualifications.
Student's name: :ุงุณู ุงูุทุงูุจ
Signature: :ุงูุชููุฑูุน
Date: :ุงูุชุง ูุฎ
VI
List of Contents No Subject page
Dedication iii Acknowledgments iv
Declaration v
List of Contents vi List of Figures viii Abstract ix Introduction 1
1 Chapter One Definitions and Concepts 3
2 Chapter Two The Z-transform 8
2.1 Definition of Z-transform 8
2.2 Properties of Z-transform. 15
2.3 Examples on the Properties of Z-transform. 21
2.4 Definition and Properties of the One-Sided Z-transform. 26
3 Chapter Three The Inverse Z-transform 34
3.1 The Inverse Z-transform 34
3.2 The Relation Between Z-transform and the Discrete
Fourier Transform.
48
3.3 The Relation Between Z-transform and Laplace Transform. 49
3.4 The Two-Dimensional Z-transform. 50
3.4.1 Definition of the Two-Dimensional Z-transform. 50
3.4.2 Properties of the Two-Dimensional Z-transform. 51
3.4.3 The Inverse of the Two-Dimensional Z-transform 53
4 Chapter Four Z-transform and Solution of Some
Difference Equations.
55
4.1 Linear Difference Equations with Constant Coefficients. 55
4.2 Volterra Difference Equations of Convolution Type. 59
5 Chapter Five Z-transform and Digital Signal
Processing.
61
5.1 Introduction. 61
5.2 Analysis of Linear Shift-Invariant (LSI) Systems and Z-
transform.
64
5.3 Realization of FIR Systems. 72
5.4 Realization of IIR Systems. 79
5.5 Design of IIR Filters From Analog Filters. 87
6 Chapter Six The Chirp Z-transform Algorithm and Its
Applications.
95
Conclusion 101
References 102
VII
Appendix A: Some Maple Commands on Z-transform: 105
Appendix B: Example on using Chirp Z-transform
Algorithm.
106
Appendix C: A Table of Properties of Z-transform. 109
Appendix D: A Table of Common Z-transform Pair. 110
ุจ ุงูู ูุฎุต
VIII
List of Figures
No Subject page
Figure (2.1) ๐ ๐๐ถ for Z-transform in Example 2.3 11
Figure (2.2) ๐ ๐๐ถ for Z-transform in Example 2.4 12
Figure (2.3) ๐ ๐๐ถ for Z-transform in Example 2.5 where
|๐| > |๐ผ| 13
Figure (5.1) Direct-form realization of FIR systems. 73
Figure (5.2) Cascade-form realization of FIR systems. 74
Figure (5.3) An (๐ โ 1)-stage lattice filter with typical stage. 76
Figure (5.4) Direct-form ฮ realization of IIR systems. 81
Figure (5.5) Direct-form ฮฮ realization of IIR systems. 81
Figure (5.6) Cascade-form realization of IIR systems. 82
Figure (5.7) Parallel -form realization of IIR systems. 83
Figure (5.8) Lattice-form realization of IIR systems. 87
IX
On Z-transform and Its Applications
by
Asma Belal Fadel
Supervisor
Dr. ''Mohammad Othman'' Omran
Abstract
In this thesis we study Z-transform (the two-sided Z-transform), the one-
sided Z-transform and the two-dimensional Z-transform with their
properties, their inverses and some examples on them. We also present the
relation between Z-transform and Laplace transform and between Z-
transform and Fourier transform. Some applications of Z-transform
including solutions of some kinds of linear difference equations, analysis of
linear shift-invariant systems, implementation of FIR and IIR filters and
design of IIR filters from analog filters are discussed. Chirp Z-transform
algorithm is also presented with two applications: enhancement in poles and
high resolution and narrow band frequency analysis.
1
Introduction
Transformation is a very powerful mathematical tool so using it in
mathematical treatment of problem is arising in many applications [1].
The idea of Z-transform back to 1730 when De Moivre introduced the
concept of โgenerating functionsโ to probability theory [8]. In 1947 a
transform of sampled signal or sequence defined by W. Hurewicz as a
tractable way to solve linear constant-coefficients difference equations. The
transformation named "Z-transform" by Ragazzini and Lotfi Zadeh in the
sampled-data control group at Columbia University in 1952 [21].
Z-transform is transformation for discrete data equivalent to the Laplace
transform of continuous data and its a generalization of discrete Fourier
transform [6].
Z-transform is used in many areas of applied mathematics as digital signal
processing, control theory, economics and some other fields [8].
In this thesis, we present Z-transform, the one-sided Z-transform and the two-
dimensional Z-transform with their properties, finding their inverse and some
examples on them. Many applications of Z-transform are discussed as solving
some kinds of linear difference equations, applications in digital signal
processing. Finally chirp Z-transform is represented.
In the first chapter, some basic definitions and concepts of sequences are
presented together with some theorems on integration in complex plane
[1,2,5,6,10,14,19].
2
In the second chapter, the definition of Z-transform and one-sided Z-
transform are discussed as well as some important properties and examples
of them [6,8,9,13,14].
In the third chapter, methods for determining the inverse of Z-transform are
represented, also we have discussed the relation between Z-transform and
Laplace transform and discrete Fourier transform. The chapter is closed by
describing the definition and properties of two-sided Z-transform in addition
to its inverse [5,9,12,13,14,20].
In the fourth chapter, Z-transform is used to solve some kind of linear
difference equations as linear difference equation of constant coefficient and
Volterra difference equations of convolution type [3,4,7,18].
In the fifth chapter, applications of Z-transform in digital signal processing
such as analysis of linear shift-invariant systems, implementation of finite-
duration impulse response (FIR) and infinite-duration impulse response (IIR)
systems and design of IIR filters from analog filters [1,6,9,11,14].
In the sixth chapter, the chirp Z-transform algorithm is studied with two
applications of it such as: enhancement of poles and high resolution and
narrow band frequency analysis [9,16,17,19].
3
Chapter One
Definitions and Concepts
In this chapter we give some basic definitions, concepts and theorems
important for our thesis.
Definition 1.1: [5] The complex sequence {๐๐} is called geometric
sequence if โ a constant ๐ โ โ s.t ๐๐+1๐๐
= ๐ , โ๐ โ โ (1.1)
In that case
๐๐ = ๐๐ ๐ (1.2)
A geometric series is of the form
โ๐๐ ๐โ
๐=0
= ๐ + ๐๐ + ๐๐ 2 +โฏ (1.3)
Note that a finite geometric series is summable with
โ๐๐ ๐๐
๐=0
= {๐(1 โ ๐ ๐+1)
1 โ ๐ , ๐ โ 1
๐(๐ + 1) , ๐ = 1
(1.4)
If |๐ | < 1, then
โ๐๐ ๐โ
๐=0
=๐
1 โ ๐ (1.5)
Definition 1.2: [10]A sequence ๐ฅ(๐) is called:
a) causal if:
๐ฅ(๐) = 0, for ๐ < 0 (1.6)
b) anticausal if:
๐ฅ(๐) = 0, for ๐ โฅ 0 (1.7)
4
Definition 1.3:[1,19]The unit step sequence or Heaviside step sequence is
defined as
๐ข(๐) = {1, ๐ โฅ 00, ๐ < 0
(1.8)
And for two-dimensional space it has the form
๐ข(๐,๐) = {1, ๐,๐ โฅ 0 0, ๐๐ก๐๐๐ค๐๐ ๐
(1.9)
Definition 1.4:[1,19] The unit impulse or unit sample sequence is defined
as
๐ฟ(๐) = {1, ๐ = 0 0, ๐๐กโ๐๐๐ค๐๐ ๐
(1.10)
And for two-dimensional space it has the form
๐ฟ(๐,๐) = {1, ๐ = ๐ = 0 0, ๐๐กโ๐๐๐ค๐๐ ๐
(1.11)
Definition 1.5: [6] The convolution between two infinite sequences ๐ฅ(๐)
and ๐ฆ(๐) is defined as
๐ฅ(๐) โ ๐ฆ(๐) = โ ๐ฅ(๐)๐ฆ(๐ โ ๐)
โ
๐=โโ
(1.12)
Example 1.1: Find the convolution of the two sequences
๐ฅ(๐) = ๐ฟ(๐) โ 5๐ฟ(๐ โ 1), ๐ฆ(๐) = 5๐๐ข(๐)
Solution:
๐ฅ(๐) โ ๐ฆ(๐) = โ ๐ฅ(๐)๐ฆ(๐ โ ๐)
โ
๐=โโ
= โ [๐ฟ(๐) โ 5๐ฟ(๐ โ 1)]5๐โ๐๐ข(๐ โ ๐)
โ
๐=โโ
For ๐ < ๐, ๐ข(๐ โ ๐) = 0,
For ๐ โ 0 and ๐ โ 1, ๐ฟ(๐) โ 5๐ฟ(๐ โ 1) = 0
So
๐ฅ(๐) โ ๐ฆ(๐) = 5๐ ๐ข(๐) โ 5. 5๐โ1๐ข(๐ โ 1)
5
= 5๐ (๐ข(๐) โ ๐ข(๐ โ 1))
Since ๐ฟ(๐) = 1 when ๐ = 1 and ๐ฟ(๐ โ 1) = 1 when ๐ = 1.
Then we have
๐ฅ(๐) โ ๐ฆ(๐) = {1, ๐ = 0 0, ๐๐กโ๐๐๐ค๐๐ ๐
Or it can be written as
๐ฅ(๐) โ ๐ฆ(๐) = ๐ฟ(๐)
Definition 1.6: [14] The correlation between two sequences ๐ฅ(๐) and ๐ฆ(๐)
is defined as
๐๐ฅ๐ฆ(๐) = โ ๐ฅ(๐)๐ฆ(๐ โ ๐)
โ
๐=โโ
= ๐ฅ(๐) โ ๐ฆ(โ๐) (1.13)
where ๐ is an integer.
Definition 1.7: [14] The autocorrelation ๐๐ฅ๐ฅ(๐) of a sequence ๐ฅ(๐) is the
correlation with itself.
Definition 1.8: A function ๐(๐ง) is analytic at a point ๐ง0 if it has a
derivative at each point in some neighborhood of ๐ง0.
Theorem 1.1:[2] If a function ๐(๐ง) is analytic at all points interior to and
on a simple contour ๐ถ then,
โฎ๐(๐ง)๐๐ง๐ถ
= 0 (1.14)
Theorem 1.2:[2] Laurentโs Theorem
Suppose that a function ๐(๐ง) is analytic throughout an annular domain ๐ <
|๐ง โ ๐ง0| < ๐ , centered at ๐ง0, and let ๐ถ be any positively oriented simple
closed contour a round ๐ง0 and lying in that domain. Then, at each point in
the domain, ๐(๐ง) has the series representation
6
๐(๐ง) = โ ๐๐
โ
๐=โโ
(๐ง โ ๐ง0)๐, ๐ < |๐ง โ ๐ง0| < ๐ (1.15)
where
๐๐ =1
2๐๐ โฎ
๐(๐ง)
(๐ง โ ๐ง0)๐+1
๐๐ง๐ถ
, ๐ โ โค (1.16)
The representation of ๐(๐ง) in Eq(1.15) is called a Laurent series.
Definition 1.9: [2] If ๐ง0 is an isolated singular point of a function ๐(๐ง), then
there is a positive number ๐ such that ๐(๐ง) is analytic at each point ๐ง for
which 0 < |๐ง โ ๐ง0| < ๐ . Consequently, ๐(๐ง) has a Laurent series
representation as in Eq(1.15). The complex number ๐โ1, which is the
coefficient of 1/(๐ง โ ๐ง0) in Eq(1.15) is called the residue of ๐(๐ง) at the
isolated singular point ๐ง0, and we shall often write
๐โ1 = ๐ ๐๐ [๐(๐ง), ๐ง0 ]
Theorem 1.3: [2] An isolated singular point ๐ง0 of a function ๐(๐ง) is a pole
of order ๐ if and only if ๐(๐ง) can be written in the form
๐(๐ง) =๐(๐ง)
(๐ง โ ๐ง0)๐
where ๐(๐ง) is analytic and nonzero at ๐ง0.
Moreover, if ๐ = 1 then,
๐ ๐๐ [๐(๐ง), ๐ง0 ] = ๐ ๐๐ [๐(๐ง)
๐ง โ ๐ง0, ๐ง0 ] = ๐(๐ง0) (1.17)
And if ๐ โฅ 2
๐ ๐๐ [๐(๐ง), ๐ง0 ] = ๐ ๐๐ [๐(๐ง)
(๐ง โ ๐ง0)๐ , ๐ง0 ]
=1
(๐ โ 1)!
๐๐ โ1
๐๐ง๐ โ1๐(๐ง)| .๐ง = ๐ง0
๐๐ด (1.18)
Theorem 1.4: [2] Residue Theorem or Cauchy's Residue Theorem.
7
If a function ๐(๐ง) is analytic inside and on a simple closed contour ๐ถ
(described in the positive sense) except for a finite number of singular
points ๐ง๐ , ๐ = 1, 2,โฆ , ๐ inside ๐ถ then,
1
2๐๐ โฎ๐(๐ง)๐๐ง๐ถ
= โ๐ ๐๐ [๐(๐ง), ๐ง๐ ]
๐
๐=1
(1.19)
Theorem 1.5:[2] Taylorโs Theorem
Suppose that a function ๐(๐ง) is analytic throughout a disk |๐ง โ ๐ง0| < ๐
centered at ๐ง0 and with radius ๐ . Then ๐(๐ง) has the power series
representation
๐(๐ง) = โ๐๐
โ
๐=0
(๐ง โ ๐ง0)๐, |๐ง โ ๐ง0| < ๐ (1.20)
where
๐๐ =๐(๐)(๐ง)
๐! , ๐ = 0, 1, 2, โฆ (1.21)
The series in Eq(1.20) is called the Taylor series of ๐(๐ง) about ๐ง = ๐ง0 and
its converges to ๐(๐ง) when ๐ง lies in the given open disk.
8
Chapter Two
The Z-transform
In this chapter we introduce the two-sided and the one-sided Z-transforms,
investigate their properties and give some examples on them.
Section 2.1: Definition of Z-transform.
Definition 2.1:[9] Given an infinite complex sequence ๐ฅ(๐), we define
its Z-transform ๐(๐ง) by the two sided infinite power series
๐(๐ง) = ๐ง[๐ฅ(๐)] = โ ๐ฅ(๐)๐งโ๐โ
๐=โโ
(2.1)
where ๐ง is a complex variable. This Z-transform is called a two-sided or a
bilateral Z-transform.
Note: Whenever we talk about Z-transform we mean the two-sided Z-
transform.
Z-transform exists only for those values of ๐ง for which the series in Eq(2.1)
converges. These values of ๐ง define the region of convergence (๐ ๐๐ถ) of
๐(๐ง). Thus, the ๐ ๐๐ถ is the domain of the Z-transform. So, whenever ๐(๐ง)
is found, its ๐ ๐๐ถ should be also defined.
The region of convergence of ๐(๐ง) is identical to the set of all values of ๐ง
which make the sequence in Eq(2.1) absolutely summable, i.e [9]
โ |๐ฅ(๐) ๐งโ๐| < โ
โ
๐=โโ
(2.2)
Example 2.1: Determine the Z-transform of the sequence
๐ฅ(๐) ={โฆ , 0, 1,0,โ2โ, 0,0,7,0,0, โฆ }= {
1 , ๐ = โ2 2 , ๐ = 0 7 , ๐ = 3 0 , ๐๐กโ๐๐๐ค๐๐ ๐
9
Note: The arrow in this sequence indicates the position of ๐ฅ(0).
Solution:
๐(๐ง) = ๐ง[๐ฅ(๐)] = โ ๐ฅ(๐)๐งโ๐โ
๐=โโ
= ๐ง2 + 2 + 7 ๐งโ3 ,
๐(๐ง) converges for the entire z-plane except ๐ง = 0 and ๐ง = โ, so
๐ ๐๐ถ: 0 < |๐ง| < โ
Example 2.2: Determine the Z-transform of the following sequences.
a) ๐ฅ(๐) = (1
2)๐
, where ๐ โฅ 0
b) (๐) = (1
2)๐
, where ๐ < 0
c) (๐) = (1
2)๐
, where ๐ โ โค
Solution:
a)
๐(๐ง) = โ ๐ฅ(๐)๐งโ๐โ
๐=โโ
=โ(1
2)๐
๐งโ๐โ
๐=0
= โ (1
2๐ง)๐
=
โ
๐=0
1
1 โ (2๐ง)โ1=
๐ง
๐ง โ12
The above series is an infinite geometric series that converges only if |1
2๐ง| < 1
or |z| >1
2. Thus, the ๐ ๐๐ถ is |z| >
1
2 , the exterior of a circle of radius
1
2
centered at the origin of the complex plane.
Note that: From Definition 1.2 (a) The sequence ๐ฅ(๐) is causal.
b)
๐(๐ง) = โ ๐ฆ(๐)๐งโ๐โ
๐=โโ
= โ (1
2)๐
๐งโ๐โ1
๐=โโ
= โ (2๐ง)โ๐
โ1
๐=โโ
Substitute ๐ = โ๐ we get
11
๐(๐ง) = โ(2๐ง)๐โ
๐=1
=1
1 โ 2๐งโ 1 =
๐ง
12โ ๐ง
Again, this is an infinite geometric series that converges only if |2z| < 1 or
|z| <1
2. Thus, the ๐ ๐๐ถ is |๐ง| <
1
2. In this case the ๐ ๐๐ถ is the interior of a
circle centered at the origin of the complex plane of radius 1
2.
Note that: From Definition 1.2 (b) The sequence ๐ฆ(๐) is anticausal.
c)
๐ (๐ง) = โ ๐(๐)๐งโ๐โ
๐=โโ
= โ (1
2)๐
๐งโ๐โ
๐=โโ
= โ (1
2)๐
๐งโ๐โ1
๐=โโ
+โ(1
2)๐
๐งโ๐โ
๐=0
= โ ๐ฆ(๐)๐งโ๐โ1
๐=โโ
+ โ๐ฅ(๐)๐งโ๐
โ
๐=1
But the first power series converges if |๐ง| < 1
2 where the second power
series converges if |๐ง| > 1
2, so there is no ๐ง in the region of convergence
for ๐ (๐ง).
โด ๐ ๐๐ถ = โ (the empty set).
Example 2.3: Determine the Z-transform of the sequence
๐ฅ(๐) = ๐ผ ๐ ๐ข(๐), ๐คโ๐๐๐ ๐ผ โ โ
Solution:
๐ฅ(๐) = ๐ผ ๐ ๐ข(๐) = { ๐ผ๐ , ๐ โฅ 00 , ๐ < 0
Note that this sequence is the same as the sequence in Example 2.2(a) with
๐ผ =1
2 so,
11
๐(๐ง) = โ ๐ฅ(๐)๐งโ๐โ
๐=โโ
=โ๐ผ ๐๐งโ๐โ
๐=0
=๐ง
๐ง โ ๐ผ
โด ๐ง[๐ผ ๐ ๐ข(๐)] =๐ง
๐ง โ ๐ผ (2.3)
๐ ๐๐ถ: |๐ง| > |๐ผ| the exterior of a circle centered at the origin of the complex
plane having radius |๐ผ|.
Example 2.4: Determine the Z-transform of the sequence
๐ฅ(๐) = โ๐ ๐ ๐ข(โ ๐ โ 1)
Solution:
๐ฅ(๐) = โ๐ ๐ ๐ข(โ ๐ โ 1) = {0 , ๐ โฅ 0
โ๐๐, ๐ < 0
Note that this sequence is the negative of the sequence in Example 2.2 (b)
so
๐(๐ง) = โ โ๐๐๐งโ๐โ1
๐=โโ
= โโ(๐โ1๐ง)๐
โ
๐=1
=๐ง
๐ง โ ๐
โด ๐ง[โ๐ ๐ ๐ข(โ ๐ โ 1)] =๐ง
๐ง โ ๐ (2.4)
๐ ๐๐ถ: |๐ง| < |๐| the interior of a circle centered at the origin of the complex-
plane having radius |๐|.
Figure 2.1 ๐ ๐๐ถ for Z-transform in Example
2.3
|๐ผ|
12
Note that: if ๐ = ๐ผ then the function ๐(๐ง) in Example 2.3 is identical to
that in Example 2.4 with different ๐ ๐๐ถ. This illustrates the important fact
that specifying Z-transform of a sequence requires not only the function ๐(๐ง)
but also its region of convergence.
Example 2.5: Determine the Z-transform of the sequence
๐ฅ(๐) = ๐ผ๐ ๐ข(๐) โ๐๐ ๐ข(โ ๐ โ 1)
Solution:
๐ฅ(๐) = ๐ผ๐ ๐ข(๐) โ๐๐ ๐ข(โ ๐ โ 1) = { ๐ผ๐ , ๐ โฅ 0โ๐๐ , ๐ < 0
๐(๐ง) = โ ๐ฅ(๐)๐งโ๐โ
๐=โโ
= โ โ๐๐๐งโ๐โ1
๐=โโ
+โ๐ผ๐๐งโ๐โ
๐=0
= โ โ(๐โ1๐ง)๐
โ
๐=1
+โ(๐ผ ๐งโ1)๐โ
๐=0
The first power series converges if |๐โ1๐ง| < 1 or |z| < |๐| where the
second power series converges if |ฮฑ zโ1| < 1 or |z| > |๐ผ|, this gives us
two cases for ๐(๐ง):
Case 1: If |๐| โค |๐ผ| then, there is no common region of convergence, i.e
๐(๐ง) does not exist.
Figure 2.2 ๐ ๐๐ถ for Z-transform in Example
2.4
|๐|
13
Case 2: If |๐| > |๐ผ|, then there is a common region of convergence, which
is |๐ผ| < |z| < |๐|, and in this case ๐(๐ง) will be
๐(๐ง) = ๐ง
๐ง โ ๐ +
๐ง
๐ง โ ๐ผ=
๐ง(2๐ง โ ๐ผ โ ๐)
(๐ง โ ๐)(๐ง โ ๐ผ)
Note: For any sequence ๐ฅ(๐) with rational Z-transform the region of
convergence cannot contain any poles and is bounded by poles or by zero or
infinity.
A Note on the ๐น๐ถ๐ช of the Z-transform for two sided sequences [9].
Let ๐(๐ง) be the Z-transform of the sequence ๐ฅ(๐), then
๐(๐ง) = โ ๐ฅ(๐)
โ
๐=โโ
๐งโ๐
= โ ๐ฅ(๐)๐งโ๐ + โ ๐ฅ(๐) ๐งโ๐โ
๐=๐1
๐0
๐=โโ
(2.5)
For the series
โ ๐ฅ(๐) ๐งโ๐โ
๐=๐1
(2.6)
Suppose that Eq(2.6) is a absolutely convergent for ๐ง = ๐ง1, so
โ |๐ฅ(๐) ๐ง1โ๐| < โ
โ
๐=๐1
(2.7)
Figure 2.3 ๐ ๐๐ถ for Z-transform in Example 2.5 where |๐| > |๐ผ|
|๐| |๐ผ|
14
We have two cases for the value of ๐1 .
Case 1: If ๐1 โฅ 0, then for any ๐ง โ โ such that |๐ง| > |๐ง1| we get,
โ |๐ฅ(๐) ๐ง1โ๐|
โ
๐=๐1
< โ |๐ฅ(๐) ๐งโ๐| < โ
โ
๐=๐1
(2.8)
So the ๐ ๐๐ถ in this case is |๐ง| > ๐๐ฅ where ๐๐ฅ is the smallest value of |๐ง|
make Eq(2.6) convergent
Case 2: If ๐1 < 0, then we express the series in Eq(2.6) as
โ ๐ฅ(๐) ๐งโ๐โ
๐=๐1
= โ ๐ฅ(๐) ๐งโ๐โ1
๐=๐1
+โ๐ฅ(๐) ๐งโ๐โ
๐=0
(2.9)
The first series on the right-hand side of Eq(2.9) is finite for any finite
value of ๐ง so, its convergent for all values of ๐ง except for ๐ง = โ. Where the
second series by (case 1) convergent for |๐ง| > ๐๐ฅ. Thus, the series in
Eq(2.6) has a region of convergence that is the exterior of a circle centered
at the origin of complex-plane with radius ๐๐ฅ with the exception of ๐ง = โ
for ๐1 < 0.
For the series
โ ๐ฅ(๐)๐งโ๐
๐0
๐=โโ
(2.10)
If we change the index of summation through the substitution ๐ = โ๐, we
obtain the series
โ ๐ฅ(โ๐)๐ง๐โ
๐=โ๐0
(2.11)
Applying the result of Eq(2.6) on Eq(2.10) (with n replaced by โ๐ and ๐ง
by ๐งโ1) then, the region of convergence of Eq(2.10) is the interior of a
circle centered at the origin of complex-plane with radius ๐ ๐ฅ where ๐ ๐ฅis the
15
largest value of |๐ง| make Eq(2.10) convergent and not equal zero for ๐0 >
0.
So for Eq(2.5), The first series has a region of convergence |๐ง| < ๐ ๐ฅ while
the region of convergence for the second series is ๐๐ฅ < |๐ง|. If ๐๐ฅ < ๐ ๐ฅ then
the region of convergence for ๐(๐ง) is ๐๐ฅ < |๐ง| < ๐ ๐ฅ; otherwise there is no
region of convergence for ๐(๐ง) .
Section 2.2: Properties of Z-transform.
In studying discrete-time signals and systems, Z-transform is a very
powerful tool due to its properties [14]. In this section, we examine some of
these properties leaving the Examples to the next section.
Let ๐(๐ง) be the Z-transform of ๐ฅ(๐) with ๐ ๐๐ถ ๐๐ฅ < |๐ง| < ๐ ๐ฅ and let ๐(๐ง)
be the Z-transform of ๐ฆ(๐) with ๐๐ถ ๐๐ฆ < |๐ง| < ๐ ๐ฆ. Let ๐ = max(๐๐ฅ, ๐๐ฆ)
and ๐ = min(๐ ๐ฅ, ๐ ๐ฆ), where ๐ can be as small as 0 and ๐ can be as large as
โ.
1. Linearity.
For any two complex numbers ฮฑ, ฮฒ we have
๐[๐ผ ๐ฅ(๐) + ๐ฝ ๐ฆ(๐)] = ๐ผ ๐(๐ง) + ๐ฝ ๐(๐ง), ๐ < |๐ง| < ๐ (2.12)
Proof:
Let
๐ค(๐) = ๐ผ ๐ฅ(๐) + ๐ฝ ๐ฆ(๐)
then
๐(๐ง) = โ ๐ค(๐)๐งโ๐โ
๐=โโ
16
= โ [๐ผ ๐ฅ(๐) + ๐ฝ ๐ฆ(๐)]๐งโ๐โ
๐=โโ
= โ ๐ผ ๐ฅ(๐)๐งโ๐ + โ ๐ฝ ๐ฆ(๐) ๐งโ๐โ
๐=โโ
โ
๐=โโ
= ๐ผ โ ๐ฅ(๐)๐งโ๐ + ๐ฝ โ ๐ฆ(๐) ๐งโ๐โ
๐=โโ
โ
๐=โโ
= ๐ผ ๐(๐ง) + ๐ฝ ๐(๐ง)
๐ ๐๐ถ of ๐(๐ง) = ๐ผ ๐(๐ง) + ๐ฝ ๐(๐ง) contains all ๐ง such that
๐ง โ ๐ ๐๐ถ(๐(๐ง)) โฉ ๐ ๐๐ถ(๐(๐ง))
then
{๐๐ฅ < |๐ง| < ๐ ๐ฅ} โฉ {๐๐ฆ < |๐ง| < ๐ ๐ฆ}
โด max(๐๐ฅ, ๐๐ฆ) < |๐ง| < min(๐ ๐ฅ, ๐ ๐ฆ)
If the linear combination canceled some poles the region of convergence
may be larger. For example, the sequences ๐ ๐๐ข(๐) and ๐ ๐๐ข(๐ โ 1) both
have a region of convergence defined by |๐ง| > |๐|, but the sequence
corresponding to the difference [๐ ๐ ๐ข(๐) โ ๐ ๐ ๐ข( ๐ โ 1)] = ๐ฟ(๐) has the
entire z-plane as its ๐ ๐๐ถ.
2. Shifting
If ๐ is any integer then,
๐[๐ฅ(๐ + ๐)] = ๐ง๐ ๐(๐ง), ๐๐ฅ < |๐ง| < ๐ ๐ฅ (2.13)
Proof:
๐[๐ฅ(๐ + ๐)] = โ ๐ฅ(๐ + ๐)๐งโ๐โ
๐=โโ
(2.14)
Substitute ๐ = ๐ + ๐ in Eq(2.14) ,we get
๐[๐ฅ(๐ + ๐)] = โ ๐ฅ(๐)๐งโ(๐โ๐)โ
๐=โโ
17
= ๐ง๐ โ ๐ฅ(๐)๐งโ๐โ
๐=โโ
= ๐ง๐ ๐(๐ง)
The ๐ ๐๐ถ of ๐(๐ง), and ๐[๐ฅ(๐ + ๐)] are identical, with the possible
exception of ๐ง = 0 or ๐ง = โ.
3. Multiplication by Exponential.
If ๐ผ is any complex number then,
๐[๐ผ๐๐ฅ(๐)] = ๐(๐ผโ1๐ง), |๐ผ| ๐๐ฅ < |๐ง| < |๐ผ| ๐ ๐ฅ (2.15)
Proof:
๐[๐ผ๐๐ฅ(๐)] = โ ๐ผ๐๐ฅ(๐)๐งโ๐โ
๐=โโ
= โ ๐ฅ(๐)(๐ผโ1๐ง)โ๐โ
๐=โโ
= ๐(๐ผโ1๐ง)
where ๐๐ฅ < |๐ผโ1 ๐ง| < ๐ ๐ฅ or |๐ผ|๐๐ฅ < |๐ง| < |๐ผ|๐ ๐ฅ
4. Time Reversal.
In discrete-time signal the variable ๐ in the sequence ๐ฅ(๐) refer to time.
๐[๐ฅ(โ๐)] = ๐(๐งโ1), 1
๐ ๐ฅ< |๐ง| <
1
๐๐ฅ (2.16)
Proof:
๐[๐ฅ(โ๐)] = โ ๐ฅ(โ๐)๐งโ๐โ
๐=โโ
(2.17)
Let ๐ = โ๐, then Eq(2.17) become
โ ๐ฅ(๐)(๐ง)๐โ
๐=โโ
= โ ๐ฅ(๐)(๐งโ1)โ๐โ
๐=โโ
= ๐(๐งโ1)
๐ ๐๐ถ ๐๐ฅ < |๐งโ1| < ๐ ๐ฅ ๐๐
1
๐ ๐ฅ< |๐ง| <
1
๐๐ฅ
The above means that if ๐ง0 belongs to the ๐ ๐๐ถ of ๐(๐ง) then 1/๐ง0 is in the
๐ ๐๐ถ of ๐(๐งโ1).
18
5. Conjugation.
๐[๐ฅโ(๐)] = ๐โ(๐งโ), ๐๐ฅ < |๐ง| < ๐ ๐ฅ (2.18)
where ๐ฅโ(๐) is the complex conjugate of ๐ฅ(๐).
Proof:
๐[๐ฅโ(๐)] = โ ๐ฅโ(๐)๐งโ๐ = โ [๐ฅ(๐)(๐งโ)โ๐]โโ
๐=โโ
โ
๐=โโ
= [ โ ๐ฅ(๐)(๐งโ)โ๐โ
๐=โโ
]
โ
= ๐โ(๐งโ)
With ๐ ๐๐ถ identical to the ๐ ๐๐ถ of ๐(๐ง).
6. Multiplication by n or Differentiation of the Transform
๐[๐ ๐ฅ(๐)] = โ๐ง๐๐(๐ง)
๐๐ง, ๐๐ฅ < |๐ง| < ๐ ๐ฅ (2.19)
Proof:
From the definition of Z-transform
๐(๐ง) = โ ๐ฅ(๐)๐งโ๐โ
๐=โโ
Differentiate both sides of the previous equation with respect to z ,we obtain
๐๐(๐ง)
๐๐ง= โ ๐ฅ(๐) (โ ๐)๐งโ๐โ1
โ
๐=โโ
= โ๐งโ1 โ [๐๐ฅ(๐)] ๐งโ๐โ
๐=โโ
= โ๐งโ1๐[๐ ๐ฅ(๐)]
โด ๐[๐ ๐ฅ(๐)] = โ๐ง๐๐(๐ง)
๐๐ง
With the same ๐ ๐๐ถ of ๐(๐ง) which is ๐๐ฅ < |๐ง| < ๐ ๐ฅ
7. Convolution of Two Sequences.
The convolution property is one of the most powerful properties of Z-
transform.
19
๐[๐ฅ(๐) โ ๐ฆ(๐)] = ๐(๐ง)๐(๐ง) (2.20)
Where the ๐ ๐๐ถ is, at least, the intersection of ๐ ๐๐ถ of ๐(๐ง) and ๐ ๐๐ถ of
๐(๐ง). However, the ๐ ๐๐ถ may be larger if there is a pole-zero cancelation in
the product ๐(๐ง)๐(๐ง).
Proof:
Let ๐(๐) be the convolution of ๐ฅ(๐) and ๐ฆ(๐), then ๐(๐) is defined as
๐(๐) = ๐ฅ(๐) โ ๐ฆ(๐) = โ ๐ฅ(๐)๐ฆ(๐ โ ๐)
โ
๐=โโ
(2.21)
Taking the Z-transform of Eq(2.21), we obtain
๐ (๐ง) = ๐[๐(๐)] = โ ๐(๐) ๐งโ๐โ
๐=โโ
= โ [ โ ๐ฅ(๐)๐ฆ(๐ โ ๐)
โ
๐=โโ
]
โ
๐=โโ
๐งโ๐ (2.22)
By interchanging the order of the summations of Eq(2.22) and applying the
shifting property, we obtain
๐ (๐ง) = โ ๐ฅ(๐) [ โ ๐ฆ(๐ โ ๐)๐งโ๐โ
๐=โโ
]
โ
๐=โโ
= ๐(๐ง) โ ๐ฅ(๐)๐งโ๐โ
๐=โโ
= ๐(๐ง) ๐(๐ง)
The ๐ ๐๐ถ of the convolution is, at least, the intersection of ๐ ๐๐ถ of ๐(๐ง)
and ๐ ๐๐ถ of ๐(๐ง).
8. Correlation of Two Sequences [14].
๐[๐๐ฅ๐ฆ(๐)] = ๐ ๐ฅ๐ฆ(๐ง) = ๐(๐ง)๐(๐งโ1) (2.23)
The ๐ ๐๐ถ is, at least, the intersection of ๐ ๐๐ถ of ๐(๐ง) and ๐ ๐๐ถ of ๐(๐งโ1).
Proof:
We know from Definition 1.6 that
21
๐๐ฅ๐ฆ(๐) = ๐ฅ(๐) โ ๐ฆ(โ๐) (2.24)
Taking Z-transform of both sides of Eq(2.24), and use the convolution and
time reversal properties, we get
๐ ๐ฅ๐ฆ(๐ง) = ๐[๐ฅ(๐)]๐[๐ฆ(โ๐)] = ๐(๐ง)๐(๐งโ1).
The proof of the following property and theorem will be delayed till the
introduction of the inverse Z-transform in Chapter 3.
9. Multiplication of Two Sequences
๐(๐ฅ(๐)๐ฆ(๐)) =1
2๐๐ โฎ ๐(๐ฃ)๐ (
๐ง
๐ฃ)1
๐ฃ๐๐ฃ
๐ถ
, (2.25)
With ๐ ๐๐ถ: ๐๐ฅ๐๐ฆ < |๐ง| < ๐ ๐ฅ๐ ๐ฆ
where ๐ถ is a counterclockwise closed contour that encloses the origin and
lies within the common region of convergence of ๐(๐ฃ) and ๐ (๐ง
๐ฃ).
Parsevalโs Theorem
If
๐๐ฅ๐๐ฆ < |๐ง| = 1 < ๐ ๐ฅ๐ ๐ฆ
then we have
โ ๐ฅ(๐)๐ฆโ(๐)
โ
๐=โโ
= 1
2๐๐ โฎ ๐(๐ง)Yโ (
1
๐งโ)1
๐ง๐๐ง
๐ถ
(2.26)
where ๐ถ is a counterclockwise closed contour that encloses the origin and
lies within the common region of convergence of ๐(๐ง) and Yโ(1/zโ).
Note: If ๐ฆ(๐) is real sequences, then Parsevalโs theorem will be
โ ๐ฅ(๐)๐ฆ(๐)
โ
๐=โโ
= 1
2๐๐ โฎ ๐(๐ง)๐ (
1
๐ง)1
๐ง๐๐ง
๐ถ
(2.27)
A table of properties of Z-transform is given in Appendix C.
21
Section 2.3: Examples on the Properties of Z-transform.
Example 2.6: Determine the Z-transform of the sequence
๐ฅ(๐) = ๐๐๐ (๐๐ค) ๐ข(๐), ๐ค โ โ
Solution:
By using Euler's identity
๐๐๐ (๐๐ค) =๐๐๐ค๐ + ๐โ๐๐ค๐
2
๐(๐ง) will be
๐(๐ง) = ๐[๐๐๐ (๐๐ค)๐ข(๐)] = ๐ [๐๐๐ค๐ + ๐โ๐๐ค๐
2๐ข(๐) ]
By using linear property, we get
๐(๐ง) =1
2[๐[๐๐๐ค๐๐ข(๐)] + ๐[๐โ๐๐ค๐๐ข(๐)]]
By Example 2.3, with ๐ผ = ๐ยฑ๐ค๐ (|๐ผ| = |๐+๐ค๐| = 1), we get
๐(๐ง) =1
2[
๐ง
๐ง โ ๐๐ค๐+
๐ง
๐ง โ ๐โ๐ค๐]
=1
2[2๐ง2 โ (๐โ๐ค๐ + ๐๐ค๐)๐ง
(๐ง2 โ ๐ง(๐๐ค๐ + ๐โ๐ค๐) + 1]
=๐ง2 โ ๐ง๐๐๐ (๐ค)
๐ง2 โ 2๐ง๐๐๐ (๐ค) + 1
Since the ๐ ๐๐ถ of ๐[๐๐๐ค๐๐ข(๐)] is |๐ง| > 1and the ๐ ๐๐ถ of ๐[๐โ๐๐ค๐๐ข(๐)] is
also |๐ง| > 1 and there is no pole-zero cancelation, then the ๐ ๐๐ถ of ๐(๐ง) is
the intersection of the two regions, which is |๐ง| > 1.
Example 2.7: Determine the Z-transform of the sequence
๐ฅ(๐) = ๐3๐๐ข(โ๐)
Solution:
Using Example 2.3, with ๐ผ =1
3 (|๐ผ| = |
1
3| =
1
3) we get
22
๐ [1
3
๐
๐ข(๐)] =๐ง
๐ง โ13
, |๐ง| >1
3
Then by time reversal property, we obtain
๐[3๐๐ข(โ๐)] = ๐งโ1
๐งโ1 โ13
=3
3 โ ๐ง, |๐ง| < 3
Finally, using multiplication by n property, it follows that
๐[๐3 ๐๐ข(โ๐)] = โ๐ง๐
๐๐ง 3
3 โ ๐ง
= โ3๐ง
(3 โ ๐ง)2, |๐ง| < 3
Example 2.8: Determine the Z-transform of the convolution of
๐ฅ(๐) = ๐ฟ(๐) โ 5๐ฟ(๐ โ 1) and ๐ฆ(๐) = 5๐๐ข(๐)
Solution:
๐[๐ฅ(๐)] = ๐[๐ฟ(๐) ] โ 5๐[๐ฟ(๐ โ 1)]
Using shifting property
๐[๐ฅ(๐)] = ๐[๐ฟ(๐) ] โ 5๐งโ1๐[๐ฟ(๐)]
= 1 โ 5๐งโ1 โ 1 =๐ง โ 5
๐ง, |๐ง| > 0
๐[๐ฆ(๐)] =๐ง
๐ง โ 5, |๐ง| > 5
However, Z-transform of the convolution of ๐ฅ(๐) and ๐ฆ(๐) is
๐[๐ฅ(๐) โ ๐ฆ(๐)] = ๐(๐ง)๐(๐ง) =๐ง โ 5
๐ง.๐ง
๐ง โ 5= 1
Which due to the pole-zero cancelation (see property 7), has a region of
convergence that is the entire z-plane.
Note: From Example ๐. ๐ we find that
๐ฅ(๐) โ ๐ฆ(๐) = [๐ฟ(๐) โ 5๐ฟ(๐ โ 1)] โ 5๐๐ข(๐) = ๐ฟ(๐)
So
23
๐[๐ฅ(๐) โ ๐ฆ(๐)] = ๐[๐ฟ(๐)] = 1
๐ ๐๐ถ is the entire z-plane.
Example 2.9: Determine the Z-transform of the autocorrelation of the
sequence
๐ฅ(๐) = (0.1)๐ ๐ข(๐)
Solution:
Since the autocorrelation of the sequence is its correlation with itself, then
๐๐ฅ๐ฅ(๐) = ๐ฅ(๐) โ ๐ฅ(โ๐)
From correlation property
๐ ๐ฅ๐ฅ(๐ง) = ๐[๐๐ฅ๐ฅ(๐)] = ๐(๐ง)๐(๐งโ1)
๐(๐ง) =๐ง
๐ง โ 0.1 |๐ง| > 0.1
๐(๐งโ1) =1
1 โ 0.1๐ง |๐ง| < 10
๐ ๐ฅ๐ฅ(๐ง) =๐ง
๐ง โ 0.1โ
1
1 โ 0.1๐ง
=๐ง
โ0.1๐ง2 + 1.01๐ง โ 0.1
=โ10๐ง
๐ง2 โ 10.1๐ง + 1
๐ ๐๐ถ 0.1 < |๐ง| < 10
Example 2.10: Determine the Z-transform of the sequence ๐ค(๐) = ๐ฅ(๐) โ
๐ฆ(๐) where ๐ฅ(๐) = 2 ๐ ๐ข(๐) and ๐ฆ(๐) = 3 ๐ ๐ข(๐).
Solution:
๐(๐ง) =๐ง
๐ง โ 2, |๐ง| > 2
๐(๐ง) =๐ง
๐ง โ 3, |๐ง| > 3
Using multiplication of two sequences property
๐(๐ง) =1
2๐๐ โฎ ๐(๐ฃ)๐ (
๐ง
๐ฃ)1
๐ฃ๐๐ฃ
๐ถ
24
=1
2๐๐ โฎ
๐ฃ
๐ฃ โ 2
(๐ง โ ๐ฃ)
[(๐ง โ ๐ฃ) โ 3] 1
๐ฃ๐๐ฃ
๐ถ
=1
2๐๐ โฎ
1
๐ฃ โ 2
๐ง
(๐ง โ 3๐ฃ) ๐๐ฃ
๐ถ
The integral has two poles, one located at ๐ฃ = 2 and the second at
๐ฃ = ๐ง โ 3. The contour of integration must be within the region of
convergence of ๐(๐ฃ), and consequently will enclose the pole at ๐ฃ = 2. To
determine whether it encloses the pole at ๐ฃ = ๐ง โ 3, we consider that the Z-
transform ๐(๐ง) is only valid for |z| > 3. Therefore, the corresponding
expression for Y(z/v) is only valid for |๐ง/๐ฃ| > 3. Thus if
|๐ง
๐ฃ| > 3
Then
|๐ง
3| > |๐ฃ|
For the ๐ ๐๐ถ of ๐[๐ฅ(๐) โ ๐ฆ(๐)] we get
|๐ฃ| > 2 and |๐ง
3| > |๐ฃ|
So
|๐ง
3| > |๐ฃ| > 2
โด ๐ ๐๐ถ of ๐[๐ฅ(๐) โ ๐ฆ(๐)] is |๐ง| > 6
Consequently, the pole |๐ง
3| lie outside the contour of integration in ๐ฃ.
Using Cauchy residue theorem to evaluate ๐(๐ง), we obtain
๐(๐ง) = ๐ ๐๐ [1
๐ฃ โ 2
๐ง
(๐ง โ 3๐ฃ) ,2 ]
=๐ง
(๐ง โ 6), |๐ง| > 6
Example 2.11: Use Parseval's Theorem to find
25
โ ๐ฅ(๐)๐ฆ(๐)
โ
๐=โโ
where ๐ฅ(๐) = (1
2) ๐ ๐ข(๐) and ๐ฆ(๐) = (
1
3)๐ ๐ข(๐).
Solution:
From Parseval's theorem, since ๐ฆ(๐) is real we use the expression in
Eq(2.27) we obtain,
โ ๐ฅ(๐)๐ฆ(๐)
โ
๐=โโ
= 1
2๐๐ โฎ ๐(๐ง)๐ (
1
๐ง)1
๐ง๐๐ง
๐ถ
๐(๐ง) =๐ง
๐ง โ12
, |๐ง| >1
2
๐ (1
๐ง) =
3
3 โ ๐ง=
โ3
๐ง โ 3, |๐ง| < 3
So, the ๐ ๐๐ถ of ๐(๐ง) and ๐ (1
๐ง) is
1
2< |๐ง| < 3
โ ๐ฅ(๐)๐ฆ(๐)
โ
๐=โโ
= 1
2๐๐ โฎ
๐ง
๐ง โ12
โ3
๐ง โ 3
1
๐ง๐๐ง
๐ถ
= 1
2๐๐ โฎ
โ3
(๐ง โ12)(๐ง โ 3)
๐๐ง๐ถ
where ๐ถ is any circle with radius ๐, where 1
2< ๐ < 3.
โ ๐ฅ(๐)๐ฆ(๐)
โ
๐=โโ
= ๐ ๐๐ [โ3
(๐ง โ12)(๐ง โ 3)
,1
2 ] =
6
5
Example 2.12: If ๐(๐ง) = ๐[(1 + ๐)๐๐ข(๐)] =๐ง
๐งโ(1+๐), |๐ง| > โ2. Determine
the Z-transform of ๐ฅ(๐) = (1 โ ๐)๐๐ข(๐) .
Solution:
Note that ๐ฅ(๐) = [(1 + ๐)๐๐ข(๐)]โ
Using conjugation property
26
๐[๐ฅ(๐)] = ๐โ(๐งโ) = [๐งโ
๐งโ โ (1 + ๐)]โ
, |๐ง| > โ2
=๐ง
๐ง โ (1 โ ๐)ูซ |๐ง| > โ2
Example 2.13: Determine the Z-transform of the sequence
๐ฅ(๐) = (โ1
2)|๐|
Solution:
We can write ๐ฅ(๐) as
๐ฅ(๐) = (โ1
2)๐
๐ข(๐) + (โ1
2)โ๐
๐ข(โ๐) โ ๐ฟ(๐)
Using linearity and time reversal properties, we get
๐(๐ง) =๐ง
๐ง +12
+๐งโ1
๐งโ1 +12
โ 1, |โ1
2| < |๐ง| <
1
|โ12|
=3๐ง
(2๐ง + 1)(๐ง + 2) ,
1
2< |๐ง| < 2
2.4: Definition and Properties of the One-Sided Z-transform.
Definition 2.2:[6] The one-sided or unilateral Z-transform of a sequence is
๐+(๐ง) = ๐ง+[๐ฅ(๐)] = โ๐ฅ(๐)๐งโ๐โ
๐=0
(2.28)
where ๐ง is a complex variable.
The one-sided Z-transform differs from the two-sided Z- transform in the
lower limit of the summation, which is always zero whether or not the
sequence ๐ฅ(๐) is zero for ๐ < 0 (i.e., causal). So, it does not contain any
information about the sequence ๐ฅ(๐) for the negative values of ๐.
Note: The one-sided Z-transform ๐+(๐ง) of ๐ฅ(๐) is identical to the two-sided
Z- transform of the sequence ๐ฅ(๐)๐ข(๐). Since ๐ฅ(๐)๐ข(๐) is causal, the region
27
of convergence of its transform is always the exterior of a circle centered at
the origin of z-plane.
Example 2.14: Determine the one-sided Z-transform of the sequences.
a) ๐ฅ(๐) = {โ7โ, 3,0,1,2,6}
b) ๐ฆ(๐) = {2,โ3โ,โ4,0,5}
c) ๐(๐) = {7,3,2,โ3โ,โ4,0,5}
d) ๐ค(๐) = 4๐
e) โ(๐) = ๐ข(โ๐ โ 1)
Solution:
a)
๐+(๐ง) = โ๐ฅ(๐)๐งโ๐โ
๐=0
= 7 + 3 ๐งโ1 + 1 ๐งโ3 + 2 ๐งโ4 + 6๐งโ5
๐ ๐๐ถ: entire z-plane except ๐ง = 0
b)
๐+(๐ง) = 3 โ 4 ๐งโ1 + 5 ๐งโ3 , |๐ง| > 0
c)
๐ +(๐ง) = 3 โ 4 ๐งโ1 + 5 ๐งโ3 , |๐ง| > 0
d)
๐+(๐ง) =๐ง
๐ง โ 4 , |๐ง| > 4
e)
๐ป+(๐ง) = 0, ๐ ๐๐ถ is all z โ plane
28
Note: The one-sided Z-transform is not unique for noncausal sequence. for
example, in the previous example ๐+(๐ง) = ๐ +(๐ง) but ๐ฆ(๐) โ ๐(๐). Also
it's not unique for anticausal sequences it's always equal zero.
Most of the properties of the one-sided Z-transform are the same as those for
the two-sided Z-transform except the shifting property.
Shifting Property for the One-Sided Z-transform
If ๐+(๐ง) is the one-sided Z-transform of the sequence ๐ฅ(๐) with ๐ ๐๐ถ |๐ง| >
๐๐ฅ, and ๐ is any positive integer then,
a) ๐ง+[๐ฅ(๐ โ ๐)] = ๐งโ๐ [๐+(๐ง) +โ๐ฅ(โ๐)๐ง๐๐
๐=1
] , |๐ง| > ๐๐ฅ (2.29)
in case ๐ฅ(๐) is casual
๐ง+[๐ฅ(๐ โ ๐)] = ๐งโ๐๐+(๐ง)
b) ๐ง+[๐ฅ(๐ + ๐)] = ๐ง๐ [๐+(๐ง) โโ๐ฅ(๐)๐งโ๐๐โ1
๐=0
] , |๐ง| > ๐๐ฅ (2.30)
Proof:
a)
๐ง+[๐ฅ(๐ โ ๐)] =โ๐ฅ(๐ โ ๐)๐งโ๐โ
๐=0
(2.31)
Substitute ๐ = ๐ โ ๐ in Eq(2.31), we get
๐ง+[๐ฅ(๐ โ ๐)] = โ ๐ฅ(๐)๐งโ(๐+๐)โ
๐=โ๐
= ๐งโ๐ [โ ๐ฅ(๐)๐งโ๐ +โ๐ฅ(๐)๐งโ๐โ
๐=0
โ1
๐=โ๐
]
= ๐งโ๐ [โ ๐ฅ(๐)๐งโ๐ + ๐+(๐ง)
โ1
๐=โ๐
] (2.32)
Substitute ๐ = โ๐ in Eq(2.32), we obtain
29
๐ง+[๐ฅ(๐ โ ๐)] = ๐งโ๐ [๐+(๐ง) +โ๐ฅ(โ๐)๐ง๐๐
๐=1
]
b)
๐ง+[๐ฅ(๐ + ๐)] = โ๐ฅ(๐ + ๐)๐งโ๐โ
๐=0
(2.33)
Changing the index of summation of Eq(2.33) from n to ๐ = ๐ + ๐
๐ง+[๐ฅ(๐ + ๐)] = โ๐ฅ(๐)๐งโ(๐โ๐)โ
๐=๐
= ๐ง๐โ๐ฅ(๐)๐งโ๐)โ
๐=๐
= ๐ง๐ [โ ๐ฅ(๐)๐งโ๐โ
๐=0
โโ๐ฅ(๐)๐งโ๐๐โ1
๐=0
]
= ๐ง๐ [๐+(๐ง) โโ๐ฅ(๐)๐งโ๐๐โ1
๐=0
] (2.34)
Changing the index of summation of Eq(2.34) from m to ๐ = ๐
๐ง+[๐ฅ(๐ + ๐)] = ๐ง๐ [๐+(๐ง) โโ๐ฅ(๐)๐งโ๐๐โ1
๐=0
]
Example 2.15: Determine the one-sided Z-transform of the sequences.
a) ๐ฅ(๐) = 3๐, ๐ โ โค
b) ๐ฆ(๐) = ๐ฅ(๐ โ 2)
c) ๐ค(๐) = ๐ฅ(๐ + 3)
Solution:
a)
๐+(๐ง) = โ๐ฅ(๐)๐งโ๐โ
๐=0
= โ3๐๐งโ๐โ
๐=0
=๐ง
๐ง โ 3, |๐ง| > 3
31
b)
Using shifting property in Eq(2.29) for the one-sided Z-transform with
๐ = 2
๐ง+[๐ฅ(๐ โ 2)] = ๐งโ2 [๐+(๐ง) +โ๐ฅ(โ๐)๐ง๐2
๐=1
]
= ๐งโ2[๐+(๐ง) + ๐ฅ(โ1)๐ง1 + ๐ฅ(โ2)๐ง2]
= ๐งโ2๐+(๐ง) + ๐ฅ(โ1)๐งโ1 + ๐ฅ(โ2)
since ๐ฅ(โ1) = 3โ1 =1
3, ๐ฅ(โ2) = 3โ2 =
1
9, we obtain ,
๐ง+[๐ฅ(๐ โ 2)] =1
๐ง2 โ 3๐ง+1
3๐ง+1
9, |๐ง| > 3
c)
Using shifting property in Eq(2.30) for the one-sided Z-transform with
๐ = 3
๐ง+[๐ฅ(๐ + 3)] = ๐ง3 [๐+(๐ง) โโ๐ฅ(๐)๐งโ๐3โ1
๐=0
]
= ๐ง3[๐+(๐ง) โ (๐ฅ(0)๐ง0 + ๐ฅ(1)๐งโ1 + ๐ฅ(2)๐งโ2)]
= ๐ง3๐+(๐ง) โ ๐ฅ(0)๐ง3 โ ๐ฅ(1)๐ง2 + ๐ฅ(2)๐ง1
since ๐ฅ(0) = 30 = 1, ๐ฅ(1) = 31 = 3 and ๐ฅ(2) = 32 = 9, we obtain,
๐ง+[๐ฅ(๐ + 3)] = ๐ง3๐ง
๐ง โ 3โ ๐ง3 โ 3๐ง2 + 9๐ง
=๐ง4
๐ง โ 3โ ๐ง3 โ 3๐ง2 + 9๐ง, |๐ง| > 3
Now, we will talk about two important theorems in the analysis of sequences:
the Initial value theorem and Final value theorem. They are useful if we
are interested in asymptotic behavior of a sequences ๐ฅ(๐) and we know its
one-sided Z-transform but not the sequences.
31
Initial Value Theorem[6]
Let ๐ฅ(๐) be a sequence, the initial value, ๐ฅ(0), can be found from ๐+(๐ง) as
follows:
๐ฅ(0) = ๐๐๐๐งโโ
๐+(๐ง) (2.35)
Proof:
๐+(๐ง) = โ๐ฅ(๐) ๐งโ๐โ
๐=๐
= ๐ฅ(0) + ๐ฅ(1) ๐งโ1 + ๐ฅ(2) ๐งโ2 +โฏ
If we let ๐ง โ โ ,each term in ๐+(๐ง) goes to zero except the first one.
โด ๐๐๐๐งโโ
๐+(๐ง) = ๐ฅ(0)
Final Value Theorem[8]
If ๐+(๐ง) is the one-sided Z-transform of the sequence ๐ฅ(๐),then
๐๐๐๐โโ
๐ฅ(๐) = ๐๐๐๐งโ1
(๐ง โ 1)๐+(๐ง) , ๐๐ ๐๐๐๐โโ
๐ฅ(๐) exist (2.36)
Proof:
๐ง+[๐ฅ(๐ + 1) โ ๐ฅ(๐)] = ๐๐๐๐โโ
โ[๐ฅ(๐+1)โ๐ฅ(๐)]
๐
๐=0
๐งโ๐
๐ง๐+(๐ง) โ ๐ง๐ฅ(0) โ ๐+(๐ง) = ๐๐๐๐โโ
โ[๐ฅ(๐+1)โ๐ฅ(๐)]
๐
๐=0
๐งโ๐
(๐ง โ 1)๐+(๐ง) โ ๐ง๐ฅ(0) = ๐๐๐๐โโ
โ[๐ฅ(๐+1)โ๐ฅ(๐)]
๐
๐=0
๐งโ๐
By taking the limit as ๐ง โ 1 ,the above equation become
๐๐๐๐งโ1
(๐ง โ 1)๐+(๐ง) โ ๐ฅ(0)
= ๐๐๐๐โโ
โ[๐ฅ(๐+1)โ๐ฅ(๐)]
๐
๐=0
= ๐๐๐๐โโ
[๐ฅ(1) โ ๐ฅ(0) + ๐ฅ(2) โ ๐ฅ(1) + โฏ
๐ฅ(๐) โ ๐ฅ(๐ โ 1) + ๐ฅ(๐ + 1) โ ๐ฅ(๐)]
= ๐๐๐๐โโ
[โ๐ฅ(0) + ๐ฅ(๐ + 1)]
= โ๐ฅ(0) + ๐๐๐๐โโ
๐ฅ(๐)
โด ๐๐๐๐โโ
๐ฅ(๐) = ๐๐๐๐งโ1
(๐ง โ 1)๐+(๐ง)
32
Example 2. 16: If ๐+(๐ง) = ๐ง2 (๐ง โ 1)(๐ง โ ๐โ1)โ , |๐ง| > 1,is the one-sided
Z-transform of the sequence ๐ฅ(๐). Find the value of ๐ฅ(0) and ๐๐๐๐โโ
๐ฅ(๐).
Solution:
From initial value theorem
๐ฅ(0) = ๐๐๐๐งโโ
๐+(๐ง) = ๐๐๐๐งโโ
๐ง2
(๐ง โ 1)(๐ง โ ๐โ1)= 1
And from final value theorem
๐๐๐๐โโ
๐ฅ(๐) = ๐๐๐๐งโ1
(๐ง โ 1)๐+(๐ง)
= ๐๐๐๐งโ1
(๐ง โ 1)๐ง2
(๐ง โ 1)(๐ง โ ๐โ1)
=1
(1 โ ๐โ1)
Example 2.17: Generalize the Initial value theorem to find the value of
๐ฅ(1) for a causal sequence ๐ฅ(๐) and find it for
๐(๐ง) =4๐ง3 + 5๐ง2
8๐ง3 โ 2๐ง + 3
Solution:
If ๐ฅ(๐) is causal then
๐(๐ง) = ๐ฅ(0) + ๐ฅ(1)๐งโ1 + ๐ฅ(2)๐งโ2 +โฏ (2.37)
Subtract ๐ฅ(0) from both sides of Eq(2.37)
๐(๐ง) โ ๐ฅ(0) = ๐ฅ(1)๐งโ1 + ๐ฅ(2)๐งโ2 +โฏ (2.38)
Multiply both sides of Eq(2.38) with z
๐ง[๐(๐ง) โ ๐ฅ(0)] = ๐ง[๐ฅ(1)๐งโ1 + ๐ฅ(2)๐งโ2 +โฏ] (2.39)
If we take the limit as ๐ง โ โ for Eq(2.39)
๐๐๐๐งโโ
๐ง[๐(๐ง) โ ๐ฅ(0)] = ๐ฅ(1) (2.40)
Eq(2.40) is a generalize the Initial value theorem to find the value of ๐ฅ(1)
for a causal sequence ๐ฅ(๐).
33
For the given Z-transform
๐ฅ(0) = ๐๐๐๐งโโ
๐(๐ง) = 1
2
๐ฅ(1) = ๐๐๐๐งโโ
๐ง[๐(๐ง) โ ๐ฅ(0)]
๐ง[๐(๐ง) โ ๐ฅ(0)] = ๐ง [4๐ง3 + 5๐ง2
8๐ง3 โ 2๐ง + 3โ1
2] =
10๐ง3 + 2๐ง2 โ 3๐ง
16๐ง3 โ 4๐ง + 6
Therefore
๐ฅ(1) = ๐๐๐๐งโโ
10๐ง3 + 2๐ง2 โ 3๐ง
16๐ง3 โ 4๐ง + 6=5
8
Example 2.18: Let ๐ฅ(๐) be a left-sided sequence that is equal zero for ๐ >
0. If
๐(๐ง) =5๐ง + 4
9๐ง2 โ 3๐ง + 2
Find ๐ฅ(0).
Solution:
For a left-sided sequence that is equal zero for ๐ > 0, the Z-transform is
๐(๐ง) = ๐ฅ(0) + ๐ฅ(โ1)๐ง + ๐ฅ(โ2)๐ง2 +โฏ
If we take the limit as ๐ง โ 0, we get
๐๐๐๐งโ0
๐(๐ง) = ๐ฅ(0)
For our example
๐ฅ(0) = ๐๐๐๐งโ0
๐(๐ง) = ๐๐๐๐งโ0
5๐ง + 4
9๐ง2 โ 3๐ง + 2= 2
34
Chapter Three
The Inverse Z-transform
In this chapter we investigate methods for finding the inverse of the Z-
transform, clarify the relation between Z-transform and discrete Fourier-
transform and its relation with Laplace transform, then we introduce the
definition of two-dimensional Z-transform, investigate its properties and how
to find its inverse.
Section 3.1: The Inverse Z-transform
Just as important as technique for finding the Z-transform of a sequence are
methods that may be used to invert the Z-transform and recover the sequence
๐ฅ(๐) from ๐(๐ง). Three methods are often used for the evaluation of the
inverse of Z-transform.
1. Integration.
2. Power Series.
3. Partial-Fraction.
1. Integration Method.[9]
Integration method relies on Cauchy integral formula, which state that if ๐ถ is
a closed contour that encircles the origin in a counterclockwise direction then, 1
2๐๐ โฎ๐ง๐โ1๐๐ง๐ถ
= {1, ๐ = 00, ๐ โ 0
(3.1)
The Z-transform of a sequence ๐ฅ(๐) is given by
๐(๐ง) = โ ๐ฅ(๐)๐งโ๐โ
๐=โโ
(3.2)
35
Multiply both sides of Eq(3.2) by 1
2๐๐ ๐ง๐โ1 and integrating over a contour ๐ถ
that encloses the origin counterclockwise and lies entirely in the region of
convergence of ๐(๐ง), we obtain
1
2๐๐ โฎ๐(๐ง) ๐ง๐โ1๐๐ง๐ถ
=1
2๐๐ โฎ โ ๐ฅ(๐)
โ
๐=โโ
๐งโ๐+๐โ1๐๐ง๐ถ
(3.3)
Interchanging the order of integration and summation on the right-hand side
of Eq(3.3) (valid if the series is convergent) we get
1
2๐๐ โฎ๐(๐ง) ๐ง๐โ1๐๐ง๐ถ
= โ ๐ฅ(๐)
โ
๐=โโ
1
2๐๐ โฎ ๐งโ๐+๐โ1๐๐ง๐ถ
(3.4)
Applying Cauchy integral formula on the integral in the right hand side of
Eq(3.4), we get 1
2๐๐ โฎ ๐งโ๐+๐โ1๐๐ง๐ถ
= {1, ๐ = ๐0, ๐ โ ๐
(3.5)
So Eq(3.4) becomes 1
2๐๐ โฎ๐(๐ง) ๐ง๐โ1๐๐ง๐ถ
= ๐ฅ(๐) (3.6)
Therefore, the inverse of Z-transform is given by the integral
๐ฅ(๐) =1
2๐๐ โฎ๐(๐ง) ๐ง๐โ1๐๐ง๐ถ
(3.7)
where ๐ถ is a counterclockwise closed contour in the region of convergence
of ๐(๐ง) and encircling the origin of the z-plane and ๐ โ โค.
Note: For a rational Z-transform ๐(๐ง), ๐ฅ(๐) is often evaluated using the
residue theorem, i.e.,
๐ฅ(๐) = โ[๐๐๐ ๐๐๐ข๐ ๐๐ ๐(๐ง) ๐ง๐โ1๐๐ก ๐กโ๐ ๐๐๐๐๐ ๐๐๐ ๐๐๐ ๐ถ ] (3.8)
Example 3.1: Find the inverse Z-transform of
๐(๐ง) =๐ง
๐ง + 3 , |๐ง| > 3
36
Solution:
From Eq(3.7) ๐ฅ(๐), will be
๐ฅ(๐) =1
2๐๐ โฎ
๐ง
๐ง + 3 ๐ง๐โ1๐๐ง
๐ถ
=1
2๐๐ โฎ
๐ง๐
๐ง + 3๐๐ง
๐ถ
where the contour of integration, C is a circle of radius greater than 3.
For ๐ โฅ 0, the contour of integration encloses only one pole at ๐ง = โ3. So
๐ฅ(๐) = ๐ ๐๐ [ ๐ง๐
๐ง + 3,โ3 ] = (โ3)๐
For ๐ < 0, in an addition to the pole at ๐ง = โ3 there is a multiple-order
pole at ๐ง = 0 whose order depends on ๐.
For ๐ = โ1,
๐ฅ(โ1) =1
2๐๐ โฎ
1
๐ง(๐ง + 3) ๐๐ง
๐ถ
= ๐ ๐๐ [1
๐ง(๐ง + 3) , 0 ] + ๐ ๐๐ [
1
๐ง(๐ง + 3) , โ3 ] =
1
3 +โ1
3 = 0
For ๐ = โ2,
๐ฅ(โ2) =1
2๐๐ โฎ
1
๐ง2(๐ง + 3) ๐๐ง
๐ถ
= ๐ ๐๐ [1
๐ง2(๐ง + 3) , 0 ] + ๐ ๐๐ [
1
๐ง2(๐ง + 3) , โ3 ]
=1
(1)!
๐
๐๐ง
1
(๐ง + 3) | .๐ง = 0
๐๐ด +1
9 =โ1
9 +1
9 = 0
Continuing this procedure it can be verified that ๐ฅ(๐) = 0, ๐ < 0. So
๐ฅ(๐) = { (โ3)๐ , ๐ โฅ 00 , ๐ < 0
or
๐ฅ(๐) = (โ3)๐๐ข(๐)
Example 3.2: Find the inverse Z-transform of
๐(๐ง) = ๐ง2 + 6 + 7๐งโ3, 0 < |๐ง| < โ
37
Solution:
From Eq(3.7) ๐ฅ(๐), will be
๐ฅ(๐) =1
2๐๐ โฎ(๐ง2 + 6 + 7๐งโ3) ๐ง๐โ1๐๐ง๐ถ
where ๐ถ is the unit circle taken counterclockwise .
๐ฅ(๐) =1
2๐๐ [โฎ ๐ง๐+1๐๐ง
๐ถ
+ 6โฎ ๐ง๐โ1๐๐ง๐ถ
+ 7โฎ ๐ง๐โ4๐๐ง๐ถ
]
From Cauchy integral formula in Eq(3.1) we get,
๐ฅ(โ2) =1
2๐๐ [2๐๐ โ 1 + 0 + 0] = 1
๐ฅ(0) =1
2๐๐ [0 + 6 โ 2๐๐ + 0] = 6
๐ฅ(3) =1
2๐๐ [0 + 0 + 7 โ 2๐๐] = 7
For ๐ โ โ2, 0, 3
๐ฅ(๐) = 0
So
๐ฅ(๐) = {
1 , ๐ = โ2 6 , ๐ = 0 7 , ๐ = 3 0 , ๐๐กโ๐๐๐ค๐๐ ๐
Note: Integration method is particularly useful if only a specific values of
x(n) are needed.
We now prove the Multiplication of Two Sequences Property and
Parseval's Theorem.
Multiplication of Two Sequences Property:
If ๐(๐ง) is the Z-transform of ๐ฅ(๐) with ๐ ๐๐ถ ๐๐ฅ < |๐ง| < ๐ ๐ฅ and ๐(๐ง) is the
Z-transform of ๐ฆ(๐) with ๐ ๐๐ถ ๐๐ฆ < |๐ง| < ๐ ๐ฆ, then
38
๐(๐ฅ(๐)๐ฆ(๐)) =1
2๐๐ โฎ ๐(๐ฃ)๐ (
๐ง
๐ฃ)1
๐ฃ๐๐ฃ
๐ถ
, (3.9)
๐ ๐๐ถ: ๐๐ฅ๐๐ฆ < |๐ง| < ๐ ๐ฅ๐ ๐ฆ
where ๐ถ is a counterclockwise closed contour that encloses the origin and
lies within the common region of convergence of ๐(๐ฃ) and ๐ (๐ง
๐ฃ).
Proof: [14]
Let ๐ถ be as given in the above theorem and
๐ค(๐) = ๐ฅ(๐)๐ฆ(๐) (3.10)
Then, the Z-transform of ๐ค(๐) is
๐(๐ง) = โ ๐ฅ(๐)๐ฆ(๐)๐งโ๐โ
๐=โโ
(3.11)
But
๐ฅ(๐) =1
2๐๐ โฎ๐(๐ฃ) ๐ฃ๐โ1๐๐ฃ๐ถ
(3.12)
Substituting ๐ฅ(๐) in Eq(3.11) and interchanging the order of summation
and integration we obtain
๐(๐ง) =1
2๐๐ โฎ๐(๐ฃ)๐ถ
[ โ ๐ฆ(๐) (๐ง
๐ฃ)โ๐
โ
๐=โโ
]1
๐ฃ ๐๐ฃ (3.13)
The sum in the brackets of Eq(3.13) is simply the transform ๐(๐ง) evaluated
at ๐ง/๐ฃ. Therefore,
๐(๐ง) =1
2๐๐ โฎ ๐(๐ฃ)๐ (
๐ง
๐ฃ)1
๐ฃ๐๐ฃ
๐ถ
To obtain the ๐ ๐๐ถ of ๐(๐ง) we note that if ๐(๐ฃ) converges for ๐๐ฅ < |๐ฃ| <
๐ ๐ฅ and ๐(๐ง) converges for ๐๐ฆ < |๐ง| < ๐ ๐ฆ, then the ๐ ๐๐ถ of ๐(๐ง/๐ฃ) is
๐๐ฆ < |๐ง
๐ฃ| < ๐ ๐ฆ
Hence the ๐ ๐๐ถ for ๐(๐ง) is at least
๐๐ฅ๐๐ฆ < |๐ง| < ๐ ๐ฅ๐ ๐ฆ
39
Parsevalโs Theorem
Let ๐(๐ง) be the Z-transform of ๐ฅ(๐) with ๐ ๐๐ถ ๐๐ฅ < |๐ง| < ๐ ๐ฅ
and let ๐(๐ง) be the Z-transform of ๐ฆ(๐) with ๐ ๐๐ถ ๐๐ฆ < |๐ง| < ๐ ๐ฆ, with
๐๐ฅ๐๐ฆ < |๐ง| = 1 < ๐ ๐ฅ๐ ๐ฆ
then we have
โ ๐ฅ(๐)๐ฆโ(๐)
โ
๐=โโ
= 1
2๐๐ โฎ ๐(๐ง)Yโ (
1
๐งโ)1
๐ง๐๐ง
๐ถ
(3.14)
where ๐ถ is a counterclockwise closed contour that encloses the origin and
lies within the common region of convergence of ๐(๐ง) and Yโ(1/zโ).
Proof: [9]
Let
๐ค(๐) = ๐ฅ(๐)๐ฆโ(๐) (3.15)
Noting that
โ ๐ค(๐)
โ
๐=โโ
= ๐(๐ง)|๐ง=1 (3.16)
By the multiplication of two sequences and the conjugation of Z-transform
properties, Eq(3.16) becomes
โ ๐ฅ(๐)๐ฆโ(๐)
โ
๐=โโ
= 1
2๐๐ โฎ ๐(๐ฃ)๐โ (
1
๐ฃโ)1
๐ฃ๐๐ฃ
๐ถ
(3.17)
Replace the dummy variable ๐ฃ with ๐ง in Eq(3.17) we obtain
โ ๐ฅ(๐)๐ฆโ(๐)
โ
๐=โโ
= 1
2๐๐ โฎ ๐(๐ง)๐โ (
1
๐งโ)1
๐ง๐๐ง
๐ถ
where ๐ถ is a counterclockwise closed contour that encloses the origin and
lies within the common region of convergence of ๐(๐ง) and ๐โ(1/zโ).
41
2. Power Series Method
The idea of this method is to write ๐(๐ง) as a power series of the form
๐(๐ง) = โ ๐๐๐งโ๐
โ
๐=โโ
(3.18)
which convergence in the ๐ ๐๐ถ. Then by uniqueness of ๐(๐ง) we can say
that ๐ฅ(๐) = ๐๐ for all ๐.
Example 3.3: Find the inverse Z-transform of
๐(๐ง) = ๐ง2 + 6 + 7๐งโ3, 0 < |๐ง| < โ
Solution:
Since ๐(๐ง) is a finite-order integer power function, ๐ฅ(๐) is a finite-length
sequence. Therefore, ๐ฅ(๐) is the coefficient that multiplies ๐งโ๐ in ๐(๐ง).
Thus, ๐ฅ(3) = 7, ๐ฅ(0) = 6 and ๐ฅ(โ2) = 1.
โด ๐ฅ(๐) = {
1 , ๐ = โ2 6 , ๐ = 0 7 , ๐ = 3 0 , ๐๐กโ๐๐๐ค๐๐ ๐
Note: Comparing Example 3.3with Example 3.2 we note that the power
series method is easier than integration method in such cases.
Example 3.4: Find the inverse Z-transform of
๐(๐ง) = ๐ ๐๐ ๐ง , |๐ง| โฅ 0
Solution:
Using Taylor series for ๐ ๐๐ ๐ง about ๐ง = 0 we get,
๐(๐ง) = โ(โ1)๐๐ง2๐+1
(2๐ + 1)!
โ
๐=0
41
But
๐(๐ง) = โ ๐ฅ(๐)๐งโ๐โ
๐=โโ
So
โ ๐ฅ(๐)๐งโ๐โ
๐=โโ
= โ(โ1)๐๐ง2๐+1
(2๐ + 1)!
โ
๐=0
Comparing the powers of ๐ง on the both sides of the previous equation we
find
โ ๐ = 2๐ + 1,๐ โ โ
So โ ๐ is odd integer,
๐ = โ1,โ3,โ5,โฆ
Thus ๐ฅ(๐) becomes
๐ฅ(๐) =(โ1)
โ(๐+12)
(โ๐)! , ๐ = โ1,โ3,โ5,โฆ
Example 3.5: Find the inverse Z-transform of
a) ๐(๐ง) =๐ง
๐ง + 3 , |๐ง| > 3
b) ๐(๐ง) =๐ง
๐ง + 3 , |๐ง| < 3
Solution:
a) Since the region of convergence is the exterior of a circle, the sequence is
a right-sided sequence. Furthermore, since lim๐งโโ
๐(๐ง) = 1 = constant, it's a
causal sequence. By long division to obtain power series in ๐งโ1.
42
โด ๐(๐ง) = 1 โ 3๐งโ1 + 9๐งโ2 +โฏ =โ(โ3)๐โ
๐=0
๐งโ๐
So that
๐ฅ(๐) = (โ3)๐๐ข(๐)
b) Since the region of convergence is the interior of a circle, the sequence is
a left-sided sequence and since ๐(0) = 0, the sequence is anticausal. Thus
we divide to obtain a series in powers of z.
๐(๐ง) =1
3๐ง โ
1
9๐ง2 +
1
27๐ง3 +โฏ = โ โ(โ3)๐๐งโ๐
โ1
๐=โโ
32
4
43
3
32
2
2
27
1
9
1
3
1
27
1-
27
1
9
1
9
1
9
1
3
1
3
1-
3
1
3
zzz
z
zz
z
zz
z
zz
zz
21
3
32
2
21
1
1
1
931
72-
279
9
93
3
31
131
zz
z
zz
z
zz
z
z
z
43
โด ๐ฅ(๐) = โ(โ3)๐๐ข(โ๐ โ 1)
3. Partial-Fraction Method.
If ๐(๐ง) is rational function then the partial fraction method is often useful
method to find its inverse [9]. The idea of this method is to write ๐(๐ง) as
๐(๐ง) = ๐ผ1๐1(๐ง) + ๐ผ2๐2(๐ง) + โฏ+ ๐ผ๐๐๐(๐ง) (3.19)
where each ๐๐(๐ง) has inverse Z-transform ๐ฅ๐(๐) and ๐ผ๐ โ โ for ๐ =
1, 2,โฆ , ๐.
If ๐(๐ง) is rational function of ๐ง then ๐(๐ง) can be expressed as
๐(๐ง) =๐ด(๐ง)
๐ต(๐ง)=๐0 + ๐1๐ง + ๐2๐ง
2 + โฏ+ ๐๐๐ง๐
๐0 + ๐1๐ง + ๐2๐ง2 + โฏ+ ๐๐๐ง
๐ (3.20)
where ๐,๐ โ โ.
If ๐ โ 0 and ๐ < ๐ then ๐(๐ง) is called proper rational function, otherwise
๐(๐ง) is improper, and it can always be written as a sum of polynomial and
proper rational function. i.e.
๐(๐ง) = ๐0 + ๐1๐ง + ๐2๐ง2 + โฏ+ ๐๐๐ง
๐ +๐ด1(๐ง)
๐ต(๐ง) (3.21)
The inverse of the polynomial can be easily found but to find the inverse of
the proper rational function we need to write it as a sum of simple functions,
for this purpose we factorized the denominator into factor of poles ๐0, ๐1, โฆ
, ๐๐of ๐(๐ง).
Remark: [5] Sometimes it's better to expand ๐(๐ง)/๐ง rather than ๐(๐ง)
because most Z-transforms have the term z in their numerator.
We have two cases for the poles[5].
Case 1: Simple poles.
44
If all poles of ๐(๐ง) are simple then
๐(๐ง) =โ๐ด๐
๐ง โ ๐๐
๐
๐=1
(3.22)
where
๐ด๐ = (๐ง โ ๐๐)๐(๐ง)| .๐ง = ๐๐๐๐ด (3.23)
Then
๐โ1 [๐ง
๐ง โ ๐๐] = {
(๐๐)๐๐ข(๐) ๐๐ ๐ ๐๐ถ |๐ง| > |๐๐|
โ(๐๐)๐๐ข(โ๐ โ 1) ๐๐ ๐ ๐๐ถ |๐ง| < |๐๐|
(3.24)
If ๐ฅ(๐) is causal and some poles of ๐(๐ง) are complex then if ๐ is a pole
then ๐โ is also a pole and in this case
๐ฅ(๐) = [๐ด(๐)๐ + ๐ดโ(๐โ)๐]๐ข(๐) (3.25)
In polar form Eq(3.25) become
๐ฅ(๐) = 2|๐ด||๐|๐ ๐๐๐ (๐๐ + ๐)๐ข(๐) (3.26)
where ๐ and ๐ are the argument of the pole ๐ and the argument of the
partial fraction coefficient ๐ด, respectively.
Case 2: Multiple poles.
For a function ๐(๐ง) with a repeated pole of multiplicity ๐, ๐ partial fraction
coefficients are associated with this repeated pole. The partial fraction
expansion of ๐(๐ง) will be of the form
๐(๐ง) = โ๐ด1๐
(๐ง โ ๐1)๐+1โ๐
๐
๐=1
+ โ๐ด๐
๐ง โ ๐๐
๐
๐=๐+1
(3.27)
where
๐ด1๐ =1
(๐ โ 1)!
๐๐โ1
๐๐ง๐โ1(๐ง โ ๐1)
๐๐(๐ง)| .๐ง = ๐1๐๐ด , ๐ = 1, 2,โฆ , ๐ (3.28)
Example 3.6: Find the inverse Z-transform of
45
๐(๐ง) =๐ง2 + 3๐ง
๐ง2 โ 3๐ง + 2 if ๐ ๐๐ถ is
a) |๐ง| > 1
b) |๐ง| < 2
c) 1 < |๐ง| < 2
Solution:
First we write ๐(๐ง)/๐ง as a partial fraction ๐(๐ง)
๐ง=
๐ง + 3
๐ง2 โ 3๐ง + 2 =
๐ง + 3
(๐ง โ 2)(๐ง โ 1)
=๐ด
๐ง โ 2 +
๐ต
๐ง โ 1
๐ด = (๐ง โ 2)๐(๐ง)
๐ง| .๐ง = 2
๐๐ด = 5
๐ต = (๐ง โ 1)๐(๐ง)
๐ง| .๐ง = 1
๐๐ด = โ4
So
๐(๐ง) = 5๐ง
๐ง โ 2 โ 4
๐ง
๐ง โ 1
For (a):
Since the ๐ ๐๐ถ of ๐(๐ง) is |๐ง| > 1, the sequence ๐ฅ(๐) is causal sequence so
we obtain,
๐ฅ(๐) = (5 (2)๐ โ 4)๐ข(๐)
b) The ๐ ๐๐ถ of ๐(๐ง) is |๐ง| < 2 so the sequence ๐ฅ(๐) is anticausal so,
๐ฅ(๐) = (โ5 (2)๐ + 4)๐ข(โ๐ โ 1)
c) The last ๐ ๐๐ถ 1 < |๐ง| < 2 of ๐(๐ง) is annular, so the sequence ๐ฅ(๐) is two-
sided. Thus one of the terms is causal and the other is anticausal. The ๐ ๐๐ถ is
overlapping |๐ง| > 1and |๐ง| < 2 so the pole ๐1 = 1 provides the causal
sequence and the pole ๐2 = 2 provides the anticausal sequence.
Thus
46
๐ฅ(๐) = โ4๐ข(๐) โ 5 (2)๐๐ข(โ๐ โ 1)
Example 3.7: Find the inverse Z-transform of
๐(๐ง) =๐ง + 1
๐ง2 โ 2๐ง + 2 ๐ ๐๐ถ|๐ง| > โ2
Solution:
We write ๐(๐ง) = X(z)/z as a partial fraction
๐(๐ง) =๐(๐ง)
๐ง=
๐ง + 1
๐ง(๐ง2 โ 2๐ง + 2) =๐ด
๐ง+
๐ต
๐ง โ (1 + ๐) +
๐ถ
๐ง โ (1 โ ๐)
๐ด = ๐ง๐(๐ง)| .๐ง = 0๐๐ด =
1
2
๐ต = [๐ง โ (1 + ๐)]๐(๐ง) | .๐ง = 1 + ๐๐๐ด =
2 + ๐
2๐ โ 2= โ(
1 + 3๐
4)
๐ถ = [๐ง โ (1 โ ๐)]๐(๐ง) | .๐ง = 1 โ ๐๐๐ด =
2 โ ๐
โ(2 + 2๐)= โ(
1 โ 3๐
4)
๐(๐ง) =1
2โ (
1 + 3๐
4)
๐ง
๐ง โ (1 + ๐) โ (
1 โ 3๐
4)
๐ง
๐ง โ (1 โ ๐)
โด ๐ฅ(๐) =1
2๐ฟ(๐) โ (
1 + 3๐
4) (1 + ๐)๐๐ข(๐) โ (
1 โ 3๐
4) (1 โ ๐)๐๐ข(๐)
Note that:
๐ฅ(0) =1
2โ (
1 + 3๐
4) โ (
1 โ 3๐
4) = 0
So,
๐ฅ(๐) = {โ(
1 + 3๐
4) (1 + ๐)๐ โ (
1 โ 3๐
4) (1 โ ๐)๐ , ๐ โฅ 1
0 , ๐๐กโ๐๐๐ค๐๐ ๐
For ๐ โฅ 1, ๐ฅ(๐) can be written in polar form as:
๐ฅ(๐) = (โ10
2) (โ2)
๐ ๐๐๐ (
๐
4๐ + tanโ1 3)
= โ5(โ2)๐โ1
๐๐๐ (๐
4๐ + tanโ1 3)
Example 3.8: Find the inverse Z-transform of
๐(๐ง) =๐ง3
๐ง3 โ ๐ง2 โ 5๐ง โ 3 ๐ ๐๐ถ|๐ง| > 1
47
Solution:
We write ๐(๐ง) = ๐(๐ง)/๐ง as a partial fraction
๐(๐ง) =๐(๐ง)
๐ง=
๐ง2
๐ง3 โ ๐ง2 โ 5๐ง โ 3 =
๐ง2
(๐ง โ 3)(๐ง + 1)2
=๐ด
๐ง โ 3 +
๐ต
๐ง + 1 +
๐ถ
(๐ง + 1)2
๐ด = (๐ง โ 3)๐(๐ง)| .๐ง = 3๐๐ด =
9
16
๐ต =1
(2 โ 1)!
๐
๐๐ง(๐ง + 1)2๐(๐ง)| .๐ง = โ1
๐๐ด =7
16
๐ถ = (๐ง + 1)2๐(๐ง)| .๐ง = โ1๐๐ด =
โ1
4
๐(๐ง) =9
16
๐ง
๐ง โ 3 +7
16
๐ง
๐ง + 1 โ1
4
๐ง
(๐ง + 1)2
โด ๐ฅ(๐) = [9
16(3)๐ +
7
16(โ1)๐ +
1
4๐(โ1)๐] ๐ข(๐)
๐ฅ(๐) =1
16[(3)๐+2 + (โ1)๐(7 + 4๐)]๐ข(๐)
Example 3.9: Find the inverse Z-transform of
๐(๐ง) =๐ง5 + 6๐ง3 + 7
๐ง3, 0 < |๐ง| < โ
Solution:
๐(๐ง) is an improper rational function so it can be written as
๐(๐ง) = ๐ง2 + 6 +7
๐ง3
So,
๐ฅ(๐) = ๐ฟ(๐ + 2) + 6๐ฟ(๐) + 7๐ฟ(๐ โ 3)
or
๐ฅ(๐) = {
1 , ๐ = โ2 6 , ๐ = 0 7 , ๐ = 3 0 , ๐๐กโ๐๐๐ค๐๐ ๐
48
Section 3.2: The Relation Between Z-transform and the Discrete Fourier
Transform.
Let ๐ฅ(๐) be a sequence, then the discrete Fourier transform (DFT) of ๐ฅ(๐)
is [5],
๐(๐๐๐) = โ ๐ฅ(๐)๐โ๐๐๐โ
๐=โโ
(3.29)
where ๐ is real.
The Z-transform of ๐ฅ(๐) is
๐(๐ง) = โ ๐ฅ(๐)๐งโ๐โ
๐=โโ
(3.30)
where ๐ง is a complex variable.
In polar form ๐ง is written as
๐ง = ๐๐๐๐ (3.31)
where ๐ = |๐ง| and ๐ = ๐๐๐ (๐ง).
Substituting Eq(3.31) in Eq(3.30) we obtain
๐(๐ง) = โ ๐ฅ(๐)(๐๐๐๐)โ๐
โ
๐=โโ
= โ ๐ฅ(๐)๐โ๐(๐๐๐)โ๐โ
๐=โโ
(3.32)
which is the discrete Fourier-transform for [๐ฅ(๐)๐โ๐]. If ๐ = 1, then ๐ง =
๐๐๐and the Z-transform of ๐ฅ(๐) becomes the discrete Fourier-transform.
i.e
๐(๐ง) | .๐ง = ๐๐๐
๐๐ด = ๐(๐๐๐)
So, Z-transform is generalization of discrete Fourier-transform.
49
Section 3.3: The Relation Between Z-transform and Laplace Transform.
Let ๐(๐ก) be a continuous function, then we can take a sampled discrete
function ๐๐ (๐ก) which can be written as [13]
๐๐ (๐ก) = โ ๐(๐ก)๐ฟ(๐ โ ๐ก)
โ
๐=โโ
= โ ๐(๐)๐ฟ(๐ โ ๐ก)
โ
๐=โโ
(3.33)
The Laplace transform of sampled function ๐๐ (๐ก) is:
๐(๐ ) = โ[๐๐ (๐ก)] = โซ [ โ ๐(๐)๐ฟ(๐ก โ ๐)
โ
๐=โโ
] ๐โ๐ ๐ก๐๐ก
โ
โโ
(3.34)
where ๐ = ๐ + ๐๐, ๐ and ๐ are real variables, by interchanging the order of
the summation and the integration of Eq(3.34) we get
๐(๐ ) = โ ๐(๐)
โ
๐=โโ
[ โซ ๐ฟ(๐ก โ ๐)๐โ๐ ๐ก๐๐ก
โ
โโ
] = โ ๐(๐)
โ
๐=โโ
๐โ๐๐
= โ ๐(๐)
โ
๐=โโ
(๐๐ )โ๐ = ๐(๐ง) | .๐ง = ๐๐ ๐๐ด (3.35)
So the relation between Laplace-transform and Z-transform is
๐(๐ ) = ๐(๐ง) | .๐ง = ๐๐ ๐๐ด (3.36)
or
๐(๐ง) = ๐(๐ ) | .๐ = ๐๐ ๐ง๐๐ด (3.37)
The most important correspondence between the s-plane and z-plane are:
1. The points on the jฯ-axis in the s-plane mapped onto the unit circle in
the z-plane.
2. The points in the right half of the s-plane mapped outside the unit circle
in the z-plane where the points in the left half mapped inside the unit
circle.
51
3. The lines ๐ = ๐ parallel to the jฯ-axis in the s-plane mapped into
circles with radius|๐ง| = ๐๐in z-plane where the lines ๐ = ๐๐ข parallel to
the ๐-axis mapped into rays of the form ๐๐๐ ๐ง = ๐ข radians from ๐ง = 0.
4. The origin of the s-plane mapped to the point z = 1 in the z-plane.
Section 3.4: The Two-Dimensional Z-transform.
3.4.1: Definition of the Two-Dimensional Z-transform.
Definition 3.1:[20] The two-dimensional Z-transform ๐(๐ง1, ๐ง2) of a
sequence ๐ฅ(๐,๐) is defined as
๐(๐ง1, ๐ง2) = โ โ ๐ฅ(๐,๐)๐ง1โ๐
โ
๐=โโ
โ
๐=โโ
๐ง2โ๐ (3.38)
where (๐ง1, ๐ง2) โ โ2 and the ๐ ๐๐ถ is the set of all (๐ง1, ๐ง2) points for which
โโ|๐ฅ(๐,๐)||๐ง1โ๐||๐ง2
โ๐|
๐
< โ
๐
Example 3.10: Find the two-dimensional Z-transform of
๐ฅ(๐,๐) = {
3, ๐ = 0,๐ = โ21, ๐ = 0,๐ = 04, ๐ = 5,๐ = 30, ๐๐กโ๐๐๐ค๐๐ ๐
Solution:
๐(๐ง1, ๐ง2) = โ โ ๐ฅ(๐,๐)๐ง1โ๐
โ
๐=โโ
โ
๐=โโ
๐ง2โ๐
= 3๐ง22 + 1 + 4๐ง1
โ5๐ง2โ3
๐ ๐๐ถ all โ2 except ๐ง1 = 0 or ๐ง2 = 0,โ.
Example 3.11: Find the two-dimensional Z-transform of
๐ฅ(๐,๐) = 2๐๐ฟ(๐ โ๐)๐ข(๐,๐)
Solution:
51
๐(๐ง1, ๐ง2) = โ โ ๐ฅ(๐,๐)๐ง1โ๐
โ
๐=โโ
โ
๐=โโ
๐ง2โ๐
= โ โ 2๐๐ฟ(๐ โ๐)๐ข(๐,๐)๐ง1โ๐
โ
๐=โโ
โ
๐=โโ
๐ง2โ๐
= โ โ 2๐๐ฟ(๐ โ๐)๐ข(๐)๐ข(๐)๐ง1โ๐
โ
๐=โโ
โ
๐=โโ
๐ง2โ๐
= โ โ 2๐๐ฟ(๐ โ๐)๐ง1โ๐
โ
๐=0
โ
๐=0
๐ง2โ๐
= โ2๐๐ง1โ๐
โ
๐=0
๐ง2โ๐ =
๐ง1๐ง2๐ง1๐ง2 โ 2
, |๐ง1||๐ง2| > 2
3.4.2: Properties of the Two-Dimensional Z-transform.
The properties of the two-dimensional Z-transform are the same as the
properties of Z-transform and their proofs are parallel with an additional
important property, the separable of sequences propriety.
Let ๐(๐ง1, ๐ง2) be the two-dimensional Z-transform of ๐ฅ(๐,๐) with ๐ ๐๐ถ ๏ฟฝฬ๏ฟฝ๐ฅ
and ๐(๐ง1, ๐ง2) the Z-transform of ๐ฆ(๐,๐) with ๐ถ ๏ฟฝฬ๏ฟฝ๐ฆ. Then the properties are
1. Linearity.
For any complex numbers ฮฑ, ฮฒ we have
๐[๐ผ๐ฅ(๐,๐) + ๐ฝ๐ฆ(๐,๐)] = ๐ผ๐(๐ง1, ๐ง2) + ๐ฝ๐(๐ง1, ๐ง2) (3.39)
๐ ๐๐ถ: at least ๏ฟฝฬ๏ฟฝ๐ฅ โฉ ๏ฟฝฬ๏ฟฝ๐ฆ
2. Shifting.
For ๐1, ๐1 โ โค
๐[๐ฅ(๐ + ๐1, ๐ +๐1)] = ๐ง1๐1 ๐ง2
๐1๐(๐ง1, ๐ง2) (3.40)
๐ ๐๐ถ: ๏ฟฝฬ๏ฟฝ๐ฅwith possible exceptions |๐ง1| = 0,โ and |๐ง2| = 0,โ
3. Multiplication by Exponential.
52
For ๐, ๐ โ โ
๐[๐๐๐๐๐ฅ(๐,๐)] = ๐(๐โ1๐ง1, ๐โ1๐ง2) (3.41)
๐ ๐๐ถ: ๏ฟฝฬ๏ฟฝ๐ฅscaled by |a| in the ๐ง1variable and by |๐| in the ๐ง2 variable.
4. Time Reversal.
๐[๐ฅ(โ๐,โ๐) ] = ๐(๐ง1โ1, ๐ง2
โ1) (3.42)
๐ ๐๐ถ: ๐ง1โ1, ๐ง2
โ1 ๐๐ ๏ฟฝฬ๏ฟฝ๐ฅ
5. Conjugation.
๐[๐ฅ โ(๐,๐)] = ๐โ(๐ง1โ, ๐ง2
โ) (3.43)
๐ ๐๐ถ: ๏ฟฝฬ๏ฟฝ๐ฅ
6. Multiplication by n or Differentiation of the Transform
๐[๐ ๐ ๐ฅ(๐,๐) ] = ๐ง1๐ง2๐2๐(๐ง1, ๐ง2)
๐๐ง1๐๐ง2 (3.44)
๐ ๐๐ถ: ๏ฟฝฬ๏ฟฝ๐ฅ
7. Convolution of Two Sequences.
๐[๐ฅ(๐,๐) โ ๐ฆ(๐,๐)] = ๐(๐ง1, ๐ง2) ๐(๐ง1, ๐ง2) (3.45)
๐ ๐๐ถ: at least ๏ฟฝฬ๏ฟฝ๐ฅ โฉ ๏ฟฝฬ๏ฟฝ๐ฆ
8. Multiplication of Two Sequences
๐[๐ฅ(๐,๐))๐ฆ(๐,๐)] = (1
2๐๐)2
โฎ โฎ ๐(๐ง1, ๐ง2)๐ (๐ง1
๐ฃ1,๐ง2
๐ฃ2)1
๐ฃ1
1
๐ฃ2๐๐ฃ1๐๐ฃ2๐ถ1๐ถ2
(3.46)
9. Separable of Sequences Property
If ๐1(๐ง1), ๐2(๐ง2) are the Z-transform of ๐ฅ1(๐), ๐ฅ2(๐) respectively and
If ๐ฅ(๐,๐) = ๐ฅ1(๐) โ ๐ฅ2(๐) then
๐(๐ง1, ๐ง2) = ๐1(๐ง1) โ ๐2(๐ง2) (3.47)
with ๐ ๐๐ถ: ๐ง1 โ ๐ ๐๐ถ ๐1(๐ง1) and ๐ง2 โ ๐ ๐๐ถ ๐2(๐ง2)
Example 3.11: Find the two-dimensional Z-transform of
53
๐ฅ(๐,๐) = 3๐โ๐
๐!๐ข(๐,๐)
Solution:
๐ฅ(๐,๐) = 3๐โ๐
๐!๐ข(๐,๐) =
3๐
๐!๐ข(๐) โ
1
3๐๐ข(๐) = ๐ฅ1(๐) โ ๐ฅ2(๐)
Using separable of sequences propriety we get,
๐(๐ง1, ๐ง2) = ๐1(๐ง1) โ ๐2(๐ง2)
= ๐3 ๐ง1โ โ๐ง2
๐ง2 โ13
๐ ๐๐ถ |๐ง1| > 0, |๐ง2| >1
3
3.4.3: The Inverse of the Two-Dimensional Z-transform
Definition 3.2:[12] The inverse of the two-dimensional Z-transform
๐(๐ง1, ๐ง2) is
๐ฅ(๐,๐) = (1
2๐๐)2
โฎ โฎ ๐(๐ง1, ๐ง2) ๐ง1๐โ1 ๐ง2
๐โ1๐๐ง1 ๐๐ง2
๐ถ1๐ถ2
(3.48)
where C1and ๐ถ2 are a counterclockwise closed contours encircling the
origin and within the ๐ ๐๐ถ of ๐(๐ง1, ๐ง2).
Example 3.12: Find the inverse of the two-dimensional Z-transform
๐(๐ง) =3๐ง1
3๐ง1 โ ๐ง2, |๐ง1| >
|๐ง2|
3
Solution:
๐ฅ(๐,๐) = (1
2๐๐)2
โฎ โฎ ๐(๐ง1, ๐ง2) ๐ง1๐โ1 ๐ง2
๐โ1๐๐ง1 ๐๐ง2
๐ถ1๐ถ2
= (1
2๐๐)2
โฎ โฎ3๐ง1
3๐ง1 โ ๐ง2 ๐ง1
๐โ1 ๐ง2๐โ1
๐๐ง1 ๐๐ง2 ๐ถ1๐ถ2
where C1and ๐ถ2 are a counterclockwise closed contours encircling the
origin and within the ๐ ๐๐ถ of ๐(๐ง1, ๐ง2).
54
=1
2๐๐ โฎ ๐ง2
๐โ1 [1
2๐๐โฎ
๐ง1๐
๐ง1 โ๐ง23
๐๐ง1 ๐ถ1
] ๐๐ง2๐ถ2
By using residue theorem and since |๐ง1| >|๐ง2|
3
1
2๐๐โฎ
๐ง1๐
๐ง1 โ๐ง23
๐๐ง1๐ถ1
= ๐ ๐๐ [ ๐ง1
๐
๐ง1 โ๐ง23
,๐ง23 ] = (
๐ง23)๐
๐ข(๐)
So
๐ฅ(๐,๐) =1
2๐๐ โฎ ๐ง2
๐โ1 (๐ง23)๐
๐ข(๐)๐๐ง2๐ถ2
=1
2๐๐ โฎ ๐ง2
๐+๐โ1 (1
3)๐
๐ข(๐)๐๐ง2๐ถ2
= (1
3)๐
๐ฟ(๐ +๐)
55
Chapter Four
Z-transform and Solution of Some Difference Equations
One of the most important applications of Z-transform is solving some linear
difference equations. Z-transform is also one of the most effective methods
for solving Volterra difference equations of convolution type[4].
Section 4.1: Linear Difference Equations with Constant Coefficients.
Definition 4.1: [3] A linear difference equation with constant coefficients
has the form,
๐ฆ(๐ + ๐) + ๐1๐ฆ(๐ + ๐ โ 1) + ๐2๐ฆ(๐ + ๐ โ 2) +โฏ+ ๐๐๐ฆ(๐) = ๐ฅ(๐) (4.1)
where ๐๐ 's are real or complex constants for all ๐ = 1, 2,โฆ , ๐. If ๐ฅ(๐) = 0
then the equation is called homogeneous otherwise its non homogeneous.
The order of linear difference equation is the difference between the largest
and smallest indices of the unknown sequence ๐ฆ(๐).
If the initial values ๐ฆ(0), ๐ฆ(1), ๐ฆ(2), โฆ , ๐ฆ(๐ โ 1) are all given, then Eq(4.1)
is called initial value problem (IVP).
Theorem 4.1: [7] On the Existence of a Unique Solution
Consider the general ๐th IVP,
๐ฆ(๐ + ๐) = ๐(๐, ๐ฆ(๐), ๐ฆ(๐ + 1),โฆ , ๐ฆ(๐ + ๐ โ 1), ๐ = 0, 1, 2, โฆ (4.2)
where ๐ฆ(0) = ๐ฆ0, ๐ฆ(1) = ๐ฆ1, ๐ฆ(2) = ๐ฆ2, โฆ , ๐ฆ(๐ โ 1) = ๐ฆ๐โ1, are given
and the function g is defined for all its arguments ๐, ๐ฆ(๐), ๐ฆ(๐ + 1),
๐ฆ(๐ + 2),โฆ , ๐ฆ(๐ + ๐ โ 1). Then the IVP has a unique solution
56
corresponding to each arbitrary set of the ๐ initial values
๐ฆ(0), ๐ฆ(1), ๐ฆ(2),โฆ , ๐ฆ(๐ โ 1).
Proof: see [7].
Method of Solution:
For solving linear difference equation by using Z-transform method we take
Z-transform of the difference equation which transforms the unknown
sequence ๐ฆ(๐) into an a algebraic equation on Z-transform ๐(๐ง) then we
obtain the sequence ๐ฆ(๐) from ๐(๐ง) by taking the inverse Z-transform of
๐(๐ง).[4]
In order to solve linear difference equations with constant coefficients with
non zero initial conditions we use the one sided Z-transform.
Example 4.1: Use the Z-transform method to solve the linear difference
equation
๐ฆ(๐ + 2) = ๐ฆ(๐ + 1) + ๐ฆ(๐)
where ๐ฆ(0) = 0 and ๐ฆ(1) = 1.
Solution:
Take the one-sided Z-transform for both sides of the linear difference
equation , we get,
๐+[๐ฆ(๐ + 2)] = ๐+[๐ฆ(๐ + 1)] + ๐+[๐ฆ(๐)]
๐ง2๐+(๐ง) โ ๐ง2๐ฆ(0) โ ๐ง ๐ฆ(1) = ๐ง๐+(๐ง) โ ๐ง๐ฆ(0) + ๐+(๐ง)
๐+(๐ง)[๐ง2 โ ๐ง โ 1] = ๐ง2๐ฆ(0) โ ๐ง๐ฆ(0) + ๐ง ๐ฆ(1)
Substitute ๐ฆ(0) = 0 and ๐ฆ(1) = 1 in the previous equation we get,
๐+(๐ง)[๐ง2 โ ๐ง โ 1] = ๐ง
So, let
57
๐(๐ง) =๐+(๐ง)
๐ง=
1
๐ง2 โ ๐ง โ 1=
1
[๐ง โ (1 + โ52 )] [๐ง + (
1 โ โ52 )]
Using partial fraction method we get,
๐(๐ง) = ๐ด
๐ง โ (1 + โ52 )
+๐ต
๐ง + (1 โ โ52 )
=1
โ5
1
๐ง โ (1 + โ52 )
โ1
โ5
1
๐ง + (1 โ โ52 )
So
๐+(๐ง) =1
โ5
๐ง
๐ง โ (1 + โ52 )
โ1
โ5
๐ง
๐ง + (1 โ โ52 )
Taking the inverse Z-transform we get,
๐ฆ(๐) =1
โ5 [(1 + โ5
2)
๐
โ (1 โ โ5
2)
๐
]
Using binomial theorem we get,
๐ฆ(๐) =1
2๐ โ5 [โ(
๐๐)
๐
๐=0
(โ5)๐โโ(
๐๐)
๐
๐=0
(โโ5)๐]
=2
2๐ โ5 [โ(
๐2๐ + 1
)
๐ฟ
๐=0
(โ5)2๐+1
] , ๐ฟ = โ๐ โ 1
2โ
=1
2๐โ1 [โ(
๐2๐ + 1
)
๐ฟ
๐=0
5๐] , ๐ > 0
Remark: The sequence in our example is called Fibonacci sequence.
Example 4.2: Solve the linear difference equation
๐ฆ(๐) โ1
3๐ฆ(๐ โ 1) = ๐ข(๐), ๐ฆ(โ1) = 2
58
Solution:
We take the one sided Z-transform of the linear difference equation, to get
๐+(๐ง) โ1
3[๐งโ1๐+(๐ง) + ๐ฆ(โ1)] =
๐ง
๐ง โ 1
Substitute ๐ฆ(โ1) = 2 in the previous equation we get,
๐+(๐ง) [1 โ1
3๐งโ1] โ
2
3=
๐ง
๐ง โ 1
By some mathematical operation we obtain,
๐+(๐ง) =๐ง(5๐ง โ 2)
(3๐ง โ 1)(๐ง โ 1)
Then, write as ๐+(๐ง)/z as a partial fraction
๐+(๐ง)
๐ง=
(5๐ง โ 2)
(3๐ง โ 1)(๐ง โ 1)=
๐ด
3๐ง โ 1+
๐ต
๐ง โ 1
Using partial fraction method we get,
๐ด =๐+(๐ง)
๐ง(3๐ง โ 1) | .
๐ง =13
๐๐ด =1
2
๐ต =๐+(๐ง)
๐ง(๐ง โ 1)| .๐ง = 1
๐๐ด =3
2
Then
๐+(๐ง) =1
2
๐ง
3๐ง โ 1+3
2
๐ง
๐ง โ 1
=1
6
๐ง
๐ง โ13
+3
2
๐ง
๐ง โ 1
Take the inverse Z-transform we obtain
๐ฆ(๐) = [1
6(1
3)๐
+3
2] ๐ข(๐)
=1
2[(1
3)๐+1
+ 3]๐ข(๐)
59
Section 4.2: Volterra Difference Equations of Convolution Type.
In this section we will use Z-transform to solve Volterra difference equations
of convolution type.
Definition 4.2: [18] Volterra difference equation of convolution type of the
first kind is of the form
๐ฅ(๐) = โ ๐พ(๐ โ๐)๐ฆ(๐)
๐
๐=0
(4.3)
where ๐ = 0, 1, 2, โฆ ๐พ(๐) and ๐ฆ(๐) are sequences and ๐พ(๐) is called the
kernel.
For solving Eq(4.3), take the Z-transform for both sides of Eq(4.3) we get,
๐(๐ง) = ๐พ(๐ง)๐(๐ง)
So
๐(๐ง) =๐(๐ง)
๐พ(๐ง)
Thus
๐ฆ(๐) = ๐โ1 [๐(๐ง)
๐พ(๐ง)]
Definition 4.3: [18] Volterra difference equation of convolution type of the
second kind is of the form
๐ฆ(๐) = ๐(๐) + โ ๐พ(๐ โ๐)๐ฆ(๐)
๐
๐=0
(4.4)
where ๐ = 0, 1, 2, โฆ. ๐(๐), ๐พ(๐), ๐ฆ(๐) are sequences and ๐พ(๐) is called the
kernel.
For solving Eq(4.4), take the Z-transform for both sides of Eq(4.4) we get,
61
๐(๐ง) = ๐น(๐ง) + ๐พ(๐ง)๐(๐ง)
So
๐(๐ง) =๐น(๐ง)
1 โ ๐พ(๐ง)
Thus
๐ฆ(๐) = ๐โ1 [๐น(๐ง)
1 โ ๐พ(๐ง)]
Example 4.3: Solve Volterra difference equation
๐2 = โ(๐ โ๐)๐ข(๐ โ๐)๐ฆ(๐)
๐
๐=0
Solution:
We can write the above equation as:
๐2 = ๐๐ข(๐) โ ๐ฆ(๐)
Take the Z-transform for both sides we get,
๐[๐2] = ๐[๐๐ข(๐) โ ๐ฆ(๐) ] ๐ง(๐ง + 1)
(๐ง โ 1)3=
๐ง
(๐ง โ 1)2๐(๐ง)
Simplify, we get
๐(๐ง) =๐ง + 1
๐ง โ 1
=๐ง
๐ง โ 1+
1
๐ง โ 1
=๐ง
๐ง โ 1+ ๐งโ1
๐ง
๐ง โ 1
Taking the inverse Z-transform we get
๐ฆ(๐) = ๐ข(๐) + ๐ข(๐ โ 1) = 2๐ข(๐) โ ๐ฟ(๐)
61
Chapter Five
Z-transform and Digital Signal Processing
In this chapter we study some applications of Z-transform in digital signal
processing such as analysis of linear shift invariant systems, realization of
finite-duration impulse response (FIR) and infinite-duration impulse
response (IIR) systems and design of IIR filters from analog filters.
Section 5.1: Introduction
In this section we introduce some necessary definitions and theorems in
digital signal processing.
A signal is any physical quantity that varies with time, space or any
independent variable or variables. Mathematically, we describe a signal as a
function of one or more independent variables. In this chapter, time will be
the independent variable whether it's continuous or discrete. If it's continuous
then the signal is called continuous-time signal and it's represented by a
function -of continuous variable ๐(๐ก). But, if the time is in discrete form the
signal is called discrete-time signal and is represented by a sequence of
numbers {๐๐} where โ โค [9,14].
Also, the range of the signal (amplitude) can be continuous or discrete.
Analog time signals are signals with continuous time and amplitude, where
digital signals are signals with discrete time and amplitude and they are
represented by a function of an integer independent variable ๐ and its values
are taken from a finite set [9,14].
62
Definition 5.1:[14] A system is a physical device that performs an operation
on a signal. These operations are usually referred to as signal processing.
Continuous-time systems are systems for which the input and output are
continuous-time signals, where the discrete-time systems are systems with
discrete-time input and output signals, and its represented by the notation ๐[. ]
which transforms the input signal ๐ฅ(๐) into the output signal ๐ฆ(๐). We write
๐[๐ฅ(๐)] = ๐ฆ(๐)
Note that any sequence ๐ฅ(๐) can be expressed as a sum of scaled and delayed
unit samples. i.e
๐ฅ(๐) = โฏ+ ๐ฅ(โ1)๐ฟ(๐ + 1) + ๐ฅ(0)๐ฟ(๐) + ๐ฅ(1)๐ฟ(๐ โ 1) + โฏ (5.1)
Eq(5.1) can be written as
๐ฅ(๐) = โ ๐ฅ(๐)๐ฟ(๐ โ ๐)
โ
๐=โโ
(5.2)
In discrete-time system, the output of ๐ฅ(๐) is
๐[๐ฅ(๐)] = ๐ฆ(๐) = ๐ [ โ ๐ฅ(๐)๐ฟ(๐ โ ๐)
โ
๐=โโ
] (5.3)
Definition 5.2:[6] A system is said to be linear if
๐[๐๐ฅ1(๐) + ๐๐ฅ2(๐)] = ๐๐[๐ฅ1(๐)] + ๐๐[๐ฅ2(๐)]
for any two inputs ๐ฅ1(๐) and ๐ฅ2(๐) and for any complex constants a and b.
If the system is linear, then Eq(5.3) become
๐ฆ(๐) = โ ๐
โ
๐=โโ
[๐ฅ(๐)๐ฟ(๐ โ ๐)] (5.4)
But ๐ฅ(๐) is constant with respect to ๐, so
๐ฆ(๐) = โ ๐ฅ(๐)๐[๐ฟ(๐ โ ๐)]
โ
๐=โโ
(5.5)
63
Let โ๐(๐) be the response of the system to at unit sample at ๐ = ๐, then
Eq(5.5) become
๐ฆ(๐) = โ ๐ฅ(๐) โ๐(๐)
โ
๐=โโ
(5.6)
Definition 5.3:[6] Let ๐ฆ(๐) be the response of a system to an arbitrary input
๐ฅ(๐). The system is said to be shift-invariant if, for any delay ๐0, the response
to ๐ฅ(๐ โ ๐0) is ๐ฆ(๐ โ ๐0). Otherwise, the system is shift-varying.
Definition 5.4: [6] A system is called a linear shift-invariant (LSI) system if
it's linear and shift-invariant.
If we have LSI system Eq(5.6) become
๐ฆ(๐) = โ ๐ฅ(๐)
โ
๐=โโ
โ(๐ โ ๐) (5.7)
= ๐ฅ(๐) โ โ(๐) (5.8)
where โ(๐) is the unit sample response (response of ๐ฟ(๐) ).
An LSI system are classified as finite-duration impulse response (FIR) and
infinite-duration impulse response (IIR) according to whether โ(๐) has finite
or infinite duration, respectively.
Definition 5.5: [11]: A system is said to be causal if the output of the system
at ๐๐ depends on the input at ๐ โค ๐๐.
Theorem 5.1:[14] A LSI system is causal if and only if the unit sample
response โ(๐) equals zero for ๐ < 0.
Proof: see [14]
Definition 5.6:[11] A system is said to be bounded inputโbounded output
(BIBO) stable if the response of any bounded input remains bounded.
64
Theorem 5.2:[9] A LSI system is (BIBO) stable if and only if the unit
sample response โ(๐) is absolutely summable.
โ |โ(๐)| < โ
โ
๐=โโ
Proof: see [9]
Definition 5.7:[6]: A system whose output ๐ฆ(๐) depends on any number of
past outputs is called recursive. In contrast, if the output of the system
depends only on the present and past inputs is called non recursive.
Section 5.2: Analysis of Linear Shift-Invariant (LSI) Systems and Z-
transform.
If we have an LSI system with input ๐ฅ(๐) and output ๐ฆ(๐), then as in Eq(5.8)
๐ฆ(๐) = ๐ฅ(๐) โ โ(๐)
where โ(๐) is the sample response.
Taking Z-transform for both sides of Eq(5.8) we get,
๐(๐ง) = ๐(๐ง)๐ป(๐ง) (5.9)
where ๐ป(๐ง) is the Z-transform of โ(๐) and is called the transfer function.
Theorem 5.3:[9] The LSI system is causal if and only if the ๐ ๐๐ถ of transfer
function is the exterior of a circle centered at the origin.
Proof:
The LSI system is causal if and only if the unit sample response is causal if
and only if the ๐ ๐๐ถ of the Z-transform of the sample response (transfer
function) is the exterior of a circle centered at the origin.
65
Theorem 5.4:[14] The LSI system is BIBO stable if and only if the ๐ ๐๐ถ of
transfer function contains the unit circle.
Proof: a)
Suppose the LSI system is BIBO stable
The transfer function of LSI system is
๐ป(๐ง) = โ โ(๐)
โ
๐=โโ
๐งโ๐
Then
|๐ป(๐ง)| = | โ โ(๐)
โ
๐=โโ
๐งโ๐| โค โ |โ(๐)๐งโ๐|
โ
๐=โโ
= โ |โ(๐)|
โ
๐=โโ
|๐งโ๐|
If |๐ง| = 1, then
|๐ป(๐ง)| โค โ |โ(๐)|
โ
๐=โโ
But the LSI system is BIBO stable so from Theorem (5.2)
โ |โ(๐)|
โ
๐=โโ
< โ
Therefore
|๐ป(๐ง)| โค โ |โ(๐)|
โ
๐=โโ
< โ
So โ(๐)๐งโ๐ is summable at each value of ๐ง of magnitude 1, so
{|๐ง| = 1} โ ๐ ๐๐ถ of ๐ป(๐ง)
b) Suppose {|๐ง| = 1} โ ๐ ๐๐ถ of ๐ป(๐ง)
Since {|๐ง| = 1} โ ๐ ๐๐ถ, then โ(๐)๐งโ๐ is summable for each value of ๐ง of
magnitude 1, which gives
โ |โ(๐)|
โ
๐=โโ
< โ
66
So, from Theorem (5.3) the system is BIBO stable.
There are many methods for analyzing the behavior or response of linear
systems. One of these methods is to solve the input-output equation of the
system which is for an LSI system with input ๐ฅ(๐) and output ๐ฆ(๐) is a linear
difference equation with constant-coefficient of the form
โ๐(๐)๐ฆ(๐ โ ๐)
๐
๐=0
=โ๐(๐)๐ฅ(๐ โ ๐)
๐
๐=๐
(5.10)
where ๐,๐ are integers and ๐ is the order of the difference equation.
Example 5.1: An LSI system is represented by the linear difference equation
๐ฆ(๐) โ ๐ฆ(๐ โ 1) +1
4๐ฆ(๐ โ 2) = ๐ฅ(๐) โ
1
4๐ฅ(๐ โ 1)
a) Find the unit sample response of the system.
b) Is the system stable?
c) Find the response of the input signal ๐ฅ(๐) = (1
4)๐๐ข(๐).
Solution:
a) We take the Z-transform of the linear difference equation and get
๐(๐ง) โ ๐งโ1๐(๐ง) +1
4๐งโ2๐(๐ง) = ๐(๐ง) โ
1
4๐งโ1๐(๐ง)
๐(๐ง) (1 โ ๐งโ1 +1
4๐งโ2) = ๐(๐ง) (1 โ
1
4๐งโ1)
So the transfer function is,
๐ป(๐ง) =๐(๐ง)
๐(๐ง)=
1 โ14๐งโ1
1 โ ๐งโ1 +14๐งโ2
=๐ง (๐ง โ
14)
๐ง2 โ ๐ง +14
โด๐ป(๐ง)
๐ง=
๐ง โ14
๐ง2 โ ๐ง +14
=๐ง โ
14
(๐ง โ12)
2 =๐ด
๐ง โ12
+๐ต
(๐ง โ12)
2
๐ด (๐ง โ1
2) + ๐ต = ๐ง โ
1
4 (5.11)
67
If ๐ง =1
2 then, ๐ต =
1
4
Differentiate both sides of Eq(5.11) with respect to ๐ง we get ๐ด = 1, then the
transfer function can be written as,
๐ป(๐ง) =๐ง
๐ง โ12
+1
4
๐ง
(๐ง โ12)
2
Since the linear difference equation is causal, the ๐ ๐๐ถ of ๐ป(๐ง) is |๐ง| >1
2 , so
the inverse of the transfer function (sample response of the system) is,
โ(๐) = (1
2)๐
๐ข(๐) +1
2๐ (1
2)๐
๐ข(๐)
= [1 +๐
2] (1
2)๐
๐ข(๐)
b) Since the ๐ ๐๐ถ of the transfer function contains the unit circle the system
is stable.
c) To find the response ๐ฆ(๐) of the input signal ๐ฅ(๐)we first find the Z-
transform of ๐ฆ(๐)
๐(๐ง) = ๐(๐ง)๐ป(๐ง) =๐ง
๐ง โ14
๐ง (๐ง โ
14)
๐ง2 โ ๐ง +14
=๐ง2
๐ง2 โ ๐ง +14
=๐ง2
(๐ง โ12)
2
= ๐ง
๐ง โ12
+1
2
๐ง
(๐ง โ12)
2
The ๐ ๐๐ถ of ๐ป(๐ง) can be either |๐ง| >1
2 or |๐ง| <
1
2 but since the ๐ ๐๐ถ of ๐(๐ง)
is |๐ง| >1
4 , and by convolution property of Z-transform the ๐ ๐๐ถ of ๐(๐ง) is
at least the intersection of ๐ ๐๐ถ of ๐(๐ง) and ๐ ๐๐ถ of ๐ป(๐ง), the
๐ ๐๐ถ of ๐ป(๐ง) is |๐ง| >1
2 so the response of ๐ฆ(๐) is,
๐ฆ(๐) = (1
2)๐
๐ข(๐) + ๐ (1
2)๐
๐ข(๐)
68
= [1 + ๐] (1
2)๐
๐ข(๐).
Consider Eq(5.10) again with ๐(0) = 1 we get,
๐ฆ(๐) = โโ๐(๐)๐ฆ(๐ โ ๐)
๐
๐=1
+โ๐(๐)๐ฅ(๐ โ ๐)
๐
๐=๐
(5.12)
To find the transfer function for the LSI system we take Z-transform of both
sides of Eq(5.12), we get
๐(๐ง) +โ๐(๐)๐งโ๐๐(๐ง)
๐
๐=1
=โ๐(๐)๐งโ๐๐(๐ง)
๐
๐=๐
(5.13)
So the transfer function is of the form
๐ป(๐ง) =๐(๐ง)
๐(๐ง)=
โ ๐(๐)๐งโ๐๐๐=๐
1 + โ ๐(๐)๐งโ๐๐๐=1
(5.14)
So, an LSI system represented by a linear difference equation with constant
coefficient has a rational transfer function as in Eq(5.14). There are many
cases for the values of ๐(๐) and ๐(๐).
Case 1: ๐(๐) = 0 for 1 โค ๐ โค ๐ then, Eq(5.14) becomes,
๐ป(๐ง) =โ๐(๐)๐งโ๐๐
๐=๐
=1
๐ง๐โ๐(๐)๐ง๐โ๐๐
๐=๐
(5.15)
From Eq(5.15) there is a multiple-pole of order ๐ at ๐ง = 0 and ๐ zeros for
๐ป(๐ง). If the transfer function in Eq(5.15) contains ๐ nontrivial zeros then
the system is called all-zero system, and has finite-duration impulse response
(FIR).
Case 2: if ๐(๐) = 0 for 1 โค ๐ โค ๐ then, Eq(5.14) become,
๐ป(๐ง) =๐(0)
1 + โ ๐(๐)๐งโ๐๐๐=1
(5.16)
=๐(0)๐ง๐
โ ๐(๐)๐ง๐โ๐๐๐=0
, ๐(0) = 1 (5.17)
69
In Eq(5.17) ๐ป(๐ง) has ๐ poles and a multiple-zero of order ๐ at ๐ง = 0. If the
transfer function in Eq(5.17) has ๐ nontrivial poles then the system is called
all-pole system. Due to the presence of poles, the impulse response of such
systems is infinite in duration, and hence it is an infinite-duration impulse
response (IIR). But the general case of ๐ป(๐ง) is to have both zeros and poles,
such system is called a pole-zero system and due to the presence of poles it
is an IIR system.
Schรผr-Cohn Stability Test [14].
Schรผr-Cohn stability test is a test used to determine if the causal linear shift
invariant system is stable or not depends on the theorem which say a linear
shift invariant system is causal and stable if and only if all its poles lies inside
the unit circle.
Before set Schรผr-Cohn stability test we need to define some important
notations.
We know that the transfer function of LSI system is rational, i.e.
๐ป(๐ง) =๐ด(๐ง)
๐ต(๐ง)
Where
๐ต(๐ง) = 1 + ๐(1)๐งโ1 + ๐(2)๐งโ2 +โฏ+ ๐(๐)๐งโ๐ =โ๐(๐)๐งโ๐๐
๐=0
Let ๐ต๐(๐ง) be a polynomial in ๐งโ1 of degree ๐ of the form,
๐ต๐(๐ง) = โ๐๐(๐)๐งโ๐
๐
๐=0
, ๐๐(0) = 1
where ๐ = 0, 1,โฆ ,๐
And ๐ถ๐(๐ง) is the reverse polynomial in ๐งโ1 of degree ๐ defined as,
71
๐ถ๐(๐ง) = ๐งโ๐๐ต๐(๐ง
โ1) = ๐งโ๐โ๐๐(๐)๐ง๐
๐
๐=0
= โ๐๐(๐ โ โ)๐งโโ๐
โ=0
, โ = ๐ โ ๐
We define the reflection coefficients ๐พ1, ๐พ2, โฆ, ๐พ๐ taken from ๐ต๐(๐ง). First
note that
๐ต(๐ง) = ๐ต๐(๐ง)
Let
๐พ๐ = ๐๐(๐)
For computing a lower degree of ๐ต๐(๐ง), ๐ = ๐,๐ โ 1,๐ โ 2,โฆ , 1 we use
the recursive equation,
๐ต๐โ1(๐ง) =๐ต๐(๐ง) โ ๐พ๐๐ถ๐(๐ง)
1 โ ๐พ๐2 (5.18)
where ๐พ๐is defined as,
๐พ๐ = ๐๐(๐)
From Schรผr-Cohn stability test the causal LSI system will be stable if and
only if |๐พ๐| < 1, for all ๐ = 1, 2,โฆ , ๐.
Example 5.2: A LSI system has the transfer function
๐ป(๐ง) =โ๐ง
12๐ง2 โ 7๐ง + 3
a) Find the difference equation that represented the system.
b) Is the system stable?
Solution:
a) We write ๐ป(๐ง) in term of ๐งโ1
๐ป(๐ง) =โ๐งโ1
12 โ 7๐งโ1 + 3๐งโ2
So
71
๐(๐ง)
๐(๐ง)=
โ๐งโ1
12 โ 7๐งโ1 + 3๐งโ2 (5.19)
By cross multiplication of Eq(5.19) we obtain,
12๐(๐ง) โ 7๐งโ1๐(๐ง) + 3๐งโ2๐(๐ง) = โ๐งโ1๐(๐ง) (5.20)
We take the inverse Z-transform for both sides of Eq(5.20) and do some
operations we get the difference equation which represented the system,
๐ฆ(๐) =7
12๐ฆ(๐ โ 1) โ
3
12๐ฆ(๐ โ 2) โ
1
12๐ฅ(๐ โ 1)
b) Since the system is causal LSI system we can use Schรผr-Cohn stability test
to determine the stability of the system.
First we write ๐ป(๐ง) as,
๐ป(๐ง) =โ112๐งโ1
1 โ712๐งโ1 +
312๐งโ2
where
๐ต2(๐ง) = 1 โ7
12๐งโ1 +
3
12๐งโ2
Hence
๐พ2 = ๐2(2) =3
12
|๐พ2| = |3
12| =
3
12< 1
Now
๐ถ2(๐ง) = ๐งโ2๐ต2(๐ง
โ1) =3
12โ7
12๐งโ1 + ๐งโ2
And
๐ต1(๐ง) =๐ต2(๐ง) โ ๐พ2๐ถ2(๐ง)
1 โ ๐พ22 = 1 โ
63
135๐งโ1
So
๐พ1 = ๐1(1) =โ63
135,
|๐พ1| = |โ36
135| =
63
135< 1
72
Since |๐พ๐| < 1, for ๐ = 1, 2, the system is stable.
Example 5.3: Is the LSI system represented by the following linear
difference equation stable?
๐ฆ(๐) = 2.5๐ฆ(๐ โ 1) โ ๐ฆ(๐ โ 2) + ๐ฅ(๐) โ 5๐ฅ(๐ โ 1) โ 6๐ฅ(๐ โ 2)
Solution:
Since the system is a causal LSI system we can use Schรผr-Cohn stability test.
First, we take Z-transform for both sides of the linear difference equation we
get,
๐(๐ง) = 2.5๐งโ1๐(๐ง) โ ๐งโ2๐(๐ง) + ๐(๐ง) โ 5๐งโ1๐(๐ง) โ 6๐งโ2๐(๐ง)
Then, the transfer function is
๐ป(๐ง) =๐(๐ง)
๐(๐ง)=1 โ 5๐งโ1 โ 6๐งโ2
1 โ 2.5๐งโ1 + ๐งโ2
So
๐ต2(๐ง) = 1 โ 2.5๐งโ1 + ๐งโ2
From ๐ต2(๐ง) we get ๐พ2 = 1
Since |๐พ2| = 1 โฎ 1, then by Schรผr-Cohn stability test the system is unstable.
Section 5.3: Realization of FIR Systems.
There are several methods for implementing an FIR system. In this section
we will present some of these methods like direct-form, cascade-form, and
lattice realization for an FIR system.
In general an FIR system is described by the linear difference equation
๐ฆ(๐) = โ ๐(๐)๐ฅ(๐ โ ๐)
๐โ1
๐=0
(5.21)
or by the transfer function
73
๐ป(๐ง) = โ ๐(๐)๐งโ๐๐โ1
๐=0
(5.22)
where the unit sample response of FIR system is,
โ(๐) = {๐(๐), 0 โค ๐ โค ๐ โ 1 0 , ๐๐กโ๐๐๐ค๐๐ ๐
1. Direct-Form Realization.
Direct-form realization is the most common way to implement the FIR
systems. It follows directly from the non recursive difference equation given
by Eq(5.21) or equivalently by,
๐ฆ(๐) = โ โ(๐)๐ฅ(๐ โ ๐)
๐โ1
๐=0
(5.23)
2. Cascade-Form Realization.
Cascade form is alternative to the direct form by factoring the transfer
function into second-order FIR systems as,
๐ป(๐ง) = ๐บโ(1 + ๐๐(1)๐งโ1 + ๐๐(2)๐ง
โ2)
๐พ
๐=1
(5.24)
where ๐พ is the integer part of (๐ + 1) 2โ and ๐บ is called the gain parameter.
Figure (๐. ๐): Direct-form realization of FIR systems. [14]
74
3. Lattice Realization.
Let us start by defining a sequence of FIR filters with transfer function
๐ป๐(๐ง) = ๐ด๐(๐ง) (5.25)
where ๐ด๐(๐ง) is a polynomial of degree ๐ defined by,
๐ด๐(๐ง) = 1 +โ๐๐(๐)๐งโ๐
๐
๐=1
, ๐ โฅ 1 (5.26)
and ๐ด0(๐ง) = 1.
So, the unit sample response โ๐(๐) is
โ๐(๐) = { 1 , ๐ = 0๐๐(๐), 1 โค ๐ โค ๐
Lattice realization or FIR lattice filter is a cascade of two-port networks
each one of them has two inputs ๐๐โ1(๐) and ๐๐โ1(๐) related to the two
outputs ๐๐(๐), ๐๐(๐) by the two difference equations give us a relation
between ๐-order direct-form FIR filter and ๐-stage lattice filter,
๐๐(๐) = ๐๐โ1(๐) + ๐พ๐๐๐โ1(๐ โ 1), ๐ = 1, 2,โฆ ,๐ โ 1 (5.27)
๐๐(๐) = ๐พ๐๐๐โ1(๐) + ๐๐โ1(๐ โ 1), ๐ = 1, 2,โฆ ,๐ โ 1 (5.28)
For ๐ = 0
๐0(๐) = ๐0(๐) = ๐ฅ(๐)
Figure (๐. ๐): Cascade-form realization of FIR systems.
[14]
75
And the output of (๐ โ 1)-stage is
๐ฆ(๐) = ๐๐โ1(๐)
Where ๐พ๐ is the reflection coefficient which is the same reflection coefficient
of Schรผr-Cohn stability test.
For each ๐ = 0, 1,โฆ ,๐ โ 1, we will define the transfer function as,
the transfer function in Eq(5.25), the relation between the input ๐ฅ(๐) and the
intermediate output ๐๐(๐) is given by,
๐๐(๐) = โ๐๐(๐)๐ฅ(๐ โ ๐)
๐
๐=0
(5.29)
where ๐๐(0) = 1.
By taking the Z-transform of both sides of Eq(5.29) we get,
๐น๐(๐ง) = ๐ด๐(๐ง)๐(๐ง) (5.30)
From Eq(5.27) and Eq(5.28) we can see that for each ๐ = 1, 2,โฆ ,๐ โ 1 the
coefficients of ๐๐(๐) are the same as those of ๐๐(๐) but in reverse order, i.e
๐๐(๐) = โ๐๐(๐ โ ๐)๐ฅ(๐ โ ๐)
๐
๐=0
= โ๐๐(๐)๐ฅ(๐ โ ๐),
๐
๐=0
๐๐(๐) = ๐๐(๐ โ ๐)
By taking the Z-transform the previous equation we get,
๐บ๐(๐ง) = ๐ต๐(๐ง)๐(๐ง)
where
๐ต๐(๐ง) = โ๐๐(๐)๐งโ๐
๐
๐=0
=โ๐๐(๐ โ ๐)๐งโ๐๐
๐=0
=โ๐๐(๐)๐ง๐โ๐
๐
๐=0
, ๐ = ๐ โ ๐
76
= โ๐๐(๐)๐ง๐โ๐
๐
๐=0
= ๐งโ๐โ๐๐(๐)๐ง๐
๐
๐=0
= ๐งโ๐๐ด๐(๐งโ1) (5.31)
So
๐บ๐(๐ง) = ๐งโ๐๐ด๐(๐ง
โ1)๐(๐ง) (5.31)
Taking the Z-transform for both sides of Eq(5.27) and Eq(5.28) and divide
them with ๐(๐ง) then solve the result equations with Eq(5.30) and Eq(5.31)
to get the recurrence relation
๐ด๐(๐ง) = ๐ด๐โ1(๐ง) + ๐พ๐๐งโ๐๐ด๐โ1(๐ง
โ1), ๐ = 1,2,โฆ ,๐ โ 1 (5.32)
and ๐ด0(๐ง) = 1.
This recurrence relation called step-up recurrence and it used to converse the
lattice reflection coefficient to direct-form filter coefficient.
Example 5.4: Given a three-stage FIR lattice filter with reflection
coefficients ๐พ1 =1
2, ๐พ2 =
โ1
3, ๐พ3 =
1
4, determine the FIR filter coefficients for
the direct-form structure.
Solution:
Figure (5.3):An (๐ โ 1)-stage lattice filter with typical stage
[14].
77
First,
๐ด0(๐ง) = ๐ด0(๐งโ1) = 1
From the step-up recurrence in Eq(5.32) with ๐ = 1 we get the single-stage
lattice
๐ด1(๐ง) = ๐ด0(๐ง) + ๐พ1๐งโ1๐ด0(๐ง
โ1)
= 1 +1
2๐งโ1
So, the coefficients of an FIR filter corresponding to single-stage lattice are
๐1(0) = 1, ๐1(1) =1
2.
Next, for ๐ = 2 we get the second-stage of lattice filter,
๐ด2(๐ง) = ๐ด1(๐ง) + ๐พ2๐งโ2๐ด1(๐ง
โ1)
= 1 +1
2๐งโ1 +
โ1
3 ๐งโ2 (1 +
1
2๐ง)
= 1 +1
3๐งโ1 โ
1
3๐งโ2
Hence, the coefficients of an FIR filter corresponding to second-stage lattice
are ๐2(0) = 1, ๐2(1) =1
3. ๐2(2) =
โ1
3.
Finally, for ๐ = 3 we get the result of the third-stage of lattice filter,
๐ด3(๐ง) = ๐ด2(๐ง) + ๐พ3๐งโ3๐ด2(๐ง
โ1)
= 1 +1
3๐งโ1 โ
1
3๐งโ2 +
1
4 ๐งโ3 (1 +
1
3๐ง โ
1
3๐ง2 )
= 1 +1
4๐งโ1 โ
1
4๐งโ2 +
1
4๐งโ3
Hence, direct-form of an FIR filter is characterized the coefficients,
๐3(0) = 1, ๐3(1) =1
4, ๐3(2) =
โ1
4, ๐3(3) =
1
4.
To find the lattice reflection coefficients from direct-form FIR filter
coefficients we use the step-down recurrence relation determined from step-
up recurrence in Eq(5.32),
78
๐ด๐(๐ง) = ๐ด๐โ1(๐ง) + ๐พ๐๐งโ๐๐ด๐โ1(๐ง
โ1), ๐ = 1,2,โฆ ,๐ โ 1
If we substitute ๐งโ1 instead of ๐ง in Eq(5.32) we get,
๐ด๐(๐งโ1) = ๐ด๐โ1(๐ง
โ1) + ๐พ๐๐ง๐๐ด๐โ1(๐ง), ๐ = 1,2,โฆ ,๐ โ 1 (5.33)
If we solve Eq(5.33) for ๐ด๐โ1(๐งโ1) we obtain,
๐ด๐โ1(๐งโ1) = ๐ด๐(๐ง
โ1) โ ๐พ๐๐ง๐๐ด๐โ1(๐ง), ๐ = 1,2,โฆ ,๐ โ 1 (5.34)
Substitute Eq(5.34) in Eq(5.32),
๐ด๐(๐ง) = ๐ด๐โ1(๐ง) + ๐พ๐๐งโ๐(๐ด๐(๐ง
โ1) โ ๐พ๐๐ง๐๐ด๐โ1(๐ง))
๐ = 1,2,โฆ ,๐ โ 1 (5.35)
If we solve Eq(5.35) for ๐ด๐โ1(๐ง) we get,
๐ด๐โ1(๐ง) =1
1 โ ๐พ๐2 (๐ด๐(๐ง) โ ๐พ๐๐ง
โ๐๐ด๐(๐งโ1)) (5.36)
where ๐ = ๐ โ 1,๐ โ 2,โฆ , 1
Observe that the step-down recurrence works if |๐พ๐| โ 1, ๐ =
1,2,โฆ ,๐ โ 1
Example 5.5: Determine the lattice coefficients corresponding to the third-
order FIR filter with transfer function.
๐ป(๐ง) = ๐ด3(๐ง) = 1 +1
4๐งโ1 โ
1
4๐งโ2 +
1
4๐งโ3
Solution:
๐พ3 = ๐3(3) =1
4
From the step-down relation Eq(5.36) with ๐ = 3 we get,
๐ด2(๐ง) =1
1 โ ๐พ32 (๐ด3(๐ง) โ ๐พ3๐ง
โ3๐ด3(๐งโ1))
=1
1 โ (14)
2 [1 +1
4(๐งโ1 โ ๐งโ2 + ๐งโ3) โ
1
4๐งโ3 (1 +
1
4(๐ง โ ๐ง2 + ๐ง3 ))]
=16
15(15
16+5
16๐งโ1 โ
5
16๐งโ2)
79
= 1 +1
3๐งโ1 โ
1
3๐งโ2
So,
๐พ2 = ๐2(2) =โ1
3
Now, by repeating the step-down relation in Eq(5.36) with ๐ = 1 we get,
๐ด1(๐ง) =1
1 โ ๐พ22 (๐ด2(๐ง) โ ๐พ2๐ง
โ2๐ด2(๐งโ1))
=1
1 โ (โ13 )
2 [1 +1
3๐งโ1 โ
1
3๐งโ2 โ
โ1
3๐งโ2 (1 +
1
3๐ง โ
1
3๐ง2 )]
=9
8(8
9+4
9๐งโ1)
= 1 +1
2๐งโ1
Hence,
๐พ1 = ๐1(1) =1
2
Section 5.4: Realization of IIR Systems.
As in FIR systems, IIR systems have many types of realization, including
direct form, cascade form, parallel form and lattice form.
IIR systems are described by the difference equation in Eq(5.12) or by the
transfer function in Eq(5.14) for ๐ โ 0.
1. Direct-Form Realization.
In direct-form realization of IIR systems the transfer function in Eq(5.14) can
be written as a cascade of two transfer functions, as
๐ป(๐ง) = ๐ป1(๐ง)๐ป2(๐ง) (5.37)
where
๐ป1(๐ง) =โ๐(๐)๐งโ๐๐
๐=๐
(5.38)
81
and
๐ป2(๐ง) =1
1 + โ ๐(๐)๐งโ๐๐๐=1
(5.39)
๐ป1(๐ง) consists of the zeros of ๐ป(๐ง) where ๐ป2(๐ง) consists of the poles of
๐ป(๐ง). There are two direct forms to realize IIR systems depending whether
๐ป1(๐ง) precedes ๐ป2(๐ง) or vice versa. If all-zero filter ๐ป1(๐ง) placed before all-
pole filter ๐ป2(๐ง), then we get direct-form ฮ realization which need ๐ +๐ +
1 multiplications, ๐ +๐ additions and ๐ +๐ delays and is depicted in
Fig(5.4). But if the first filter is ๐ป2(๐ง), then we obtain direct-form ฮฮ and the
system becomes a cascade of the recursive system
๐ค(๐) = โโ๐(๐)๐ค(๐ โ ๐)
๐
๐=1
+ ๐ฅ(๐) (5.40)
with input ๐ฅ(๐) and output ๐ค(๐), followed by the non recursive system
๐ฆ(๐) = โโ๐(๐)๐ค(๐ โ ๐)
๐
๐=1
(5.41)
which has ๐ค(๐) as input and ๐ฆ(๐) output.
Direct-form ฮฮ require ๐ +๐ + 1 multiplications, ๐ +๐ additions but
๐๐๐ฅ{๐,๐} delays and its depicted in Fig(5.5). Since ๐ป2(๐ง) need less
number of delays to realize ๐ป(๐ง) it's called canonic.
81
Figure (๐. ๐): Direct-form ฮ realization of IIR systems [14].
Figure (๐. ๐): Direct-form ฮฮ realization of IIR systems for ๐ = ๐
[14].
82
2. Cascade-Form Realization.
Cascade-form realization is derived by factoring the transfer function in
Eq(5.14) into a cascade of second-order sections. We start by considering
๐ โฅ ๐ then ๐ป(๐ง) factored as
๐ป(๐ง) = ๐บโ๐ป๐(๐ง)
๐พ
๐=1
(5.42)
where is ๐พ the integer part of (๐ + 1) 2โ and ๐ป๐(๐ง) has the general form
๐ป๐(๐ง) =1 + ๐๐(1)๐ง
โ1 + ๐๐(2)๐งโ2
1 + ๐๐(1)๐งโ1 + ๐๐(2)๐ง
โ2 (5.43)
Where ๐บ is the gain parameter and equal ๐(0), the coefficients ๐๐(๐), ๐๐(๐)
are real coefficients for ๐ = 1,2 and ๐ = 1 , โฆ , ๐พ.The chosen of the
numerator and denominator of ๐ป๐(๐ง) are arbitrary. If ๐ < ๐, then some
coefficients of the denominator of the second-order sections will equal zero.
3. Parallel-Form Realization.
Parallel-form realization can be obtained by expanding the transfer function
๐ป(๐ง) in Eq(5.14) using partial fraction method. Without loss of generality,
we assume that the poles of ๐ป(๐ง) are distinct. If ๐ > ๐, then ๐ป(๐ง) expressed
as
๐ป(๐ง) = โ๐ด๐
1 โ ๐๐๐งโ1
๐
๐=1
(5.44)
Figure (๐. ๐): Cascade-form realization of IIR system [14].
83
where ๐ด๐ 's are the coefficients of ๐ป(๐ง) and ๐๐ 's are the poles.
In general, the poles of ๐ป(๐ง) are complex so there corresponding coefficients
are complex. To avoid multiplication of complex numbers we combine the
complex-conjugate poles to have second-order section ๐ป๐(๐ง) of the form
๐ป๐(๐ง) =๐๐(0) + ๐๐(1)๐ง
โ1
1 + ๐๐(1)๐งโ1 + ๐๐(2)๐ง
โ2 (5.45)
where the coefficients are real, then ๐ป(๐ง) become
๐ป(๐ง) = โ๐ป๐(๐ง)
๐พ
๐=1
(5.46)
where ๐พ is the integer part of (๐ + 1) 2โ .
If ๐ โค ๐, then the partial fraction expansion will contain a term of the form
๐ถ = ๐0 + ๐1๐งโ1 +โฏ+ ๐๐โ๐๐ง
โ(๐โ๐)
which is FIR filter placed in parallel with the other terms of expansion of
๐ป(๐ง).
Example 5.6: Determine the transfer function for cascade and parallel
realizations for the system described by the transfer function
Figure (๐. ๐): Parallel -form realization of IIR systems. [14]
84
๐ป(๐ง) =(1 โ
12๐งโ1) (1 +
34๐งโ1)
(1 โ (12+12๐) ๐งโ1) (1 โ (
12โ12๐) ๐งโ1) (1 โ
14๐งโ1)
Solution:
Since ๐ = 3 > ๐ = 2 then ๐พ = integer part of (3 + 1) 2โ = 2
For cascade realization,
๐ป(๐ง) =โ๐ป๐(๐ง)
2
๐=1
A possible paring for poles and zeros is
๐ป1(๐ง) =1 โ
12๐งโ1
(1 โ (12+12๐) ๐งโ1) (1 โ (
12โ12๐) ๐งโ1)
=1 โ
12๐งโ1
1 โ ๐งโ1 +12๐งโ2
๐ป2(๐ง) =1 +
34๐งโ1
1 โ14๐งโ1
So
๐ป(๐ง) =1 +
34๐งโ1
1 โ14๐งโ1
1 โ
12๐งโ1
1 โ ๐งโ1 +12๐งโ2
For parallel-form realization we need to expand ๐ป(๐ง) using partial fraction
method,
๐ป(๐ง) = โ๐ด๐
1 โ ๐๐๐งโ1
3
๐=1
=๐ด1
1 โ14๐งโ1
+๐ด2
1 โ (12+12๐) ๐งโ1
+๐ด3
1 โ (12โ12๐) ๐งโ1
Where the coefficients ๐ด1, ๐ด2, ๐ด3 are
85
๐ด1 = ๐ป(๐ง) (1 โ1
4๐งโ1) | .
๐งโ1 = 4๐๐ด =
โ4
5
๐ด2 = ๐ป(๐ง) [1 โ (1
2+1
2๐) ๐งโ1] | .
๐งโ1 = 1 โ ๐๐๐ด =
9 โ 8๐
10
๐ด3 = ๐ป(๐ง) [1 โ (1
2โ1
2๐) ๐งโ1] | .
๐งโ1 = 1 + ๐๐๐ด =
9 + 8๐
10
So, the transfer function become
๐ป(๐ง) =โ4
5
1
1 โ14๐งโ1
+1
10
9 โ 8๐
1 โ (12+12๐) ๐งโ1
+1
10
9 + 8๐
1 โ (12โ12๐) ๐งโ1
or
๐ป(๐ง) =โ4
5
1
1 โ14๐งโ1
+1
10
18 โ ๐งโ1
1 โ ๐งโ1 +12๐งโ2
4. Lattice-Form Realization.
In this section we will develop a lattice filter structure equivalent to IIR
systems, and our notations is the same as in the Lattice realization of FIR
filters.
Let us start with all pole system which has the transfer function
๐ป(๐ง) =1
1 + โ ๐๐(๐)๐งโ๐๐
๐=1
=1
๐ด๐(๐ง) (5.47)
The direct form realization of the previous system represented by the
difference equation
๐ฆ(๐) = โโ๐๐(๐)๐ฆ(๐ โ ๐)
๐
๐=1
+ ๐ฅ(๐) (5.48)
If we interchange the input signal ๐ฅ(๐) with the output signal ๐ฆ(๐) and the
output ๐ฆ(๐) with the input ๐ฅ(๐) we get
๐ฅ(๐) = โโ๐๐(๐)๐ฅ(๐ โ ๐)
๐
๐=1
+ ๐ฆ(๐) (5.49)
Which can be written as
86
๐ฆ(๐) = ๐ฅ(๐) +โ๐๐(๐)๐ฅ(๐ โ ๐)
๐
๐=1
(5.50)
Which is the linear difference equation describe an FIR system having the
transfer function ๐ป(๐ง) = ๐ด๐(๐ง).
So, we can obtain the IIR lattice filter from an FIR by interchanging the input
and the output. Therefore, the definition of IIR lattice filter is the opposite of
FIR lattice filter.
By solving Eq(5.27) for ๐๐โ1(๐) and let Eq(5.28) stay the same for ๐๐(๐)
we get the following rules for IIR lattice filter
๐๐โ1(๐) = ๐๐(๐) โ ๐พ๐๐๐โ1(๐ โ 1), ๐ = ๐,๐ โ 1,โฆ , 1 (5.51)
๐๐(๐) = ๐พ๐๐๐โ1(๐) + ๐๐โ1(๐ โ 1), ๐ = ๐,๐ โ 1,โฆ , 1 (5.52)
with
๐0(๐) = ๐0(๐) = ๐ฆ(๐)
๐๐(๐) = ๐ฅ(๐)
In general, the transfer function for all-pole IIR system is
๐ป๐(๐ง) =๐(๐ง)
๐(๐ง)=๐น0(๐ง)
๐น๐(๐ง)=
1
๐ด๐(๐ง) (5.53)
And for all-zero FIR system is
๐ป๐(๐ง) =๐บ๐(๐ง)
๐(๐ง)=๐บ๐(๐ง)
๐บ0(๐ง)= ๐งโ๐๐ด๐(๐ง
โ1) (5.54)
So, for the IIR system contains both poles and zeros has the transfer function
๐ป(๐ง) =โ ๐๐(๐)๐ง
โ๐๐๐=1
1 + โ ๐๐(๐)๐งโ๐๐
๐=1
=๐ถ๐(๐ง)
๐ด๐(๐ง) (5.55)
Without loss of generality, if ๐ โค ๐, then ๐ป(๐ง) consists of two components
the first is an all-pole lattice with parameters ๐พ๐, 1 โค ๐ โค ๐, and the other
is a tapped delay line with coefficients ๐๐(๐).
If ๐ = ๐ then
87
๐ถ๐(๐ง) = ๐ถ๐โ1(๐ง) + ๐๐(๐)๐งโ๐๐ด๐(๐ง
โ1),๐ = 1,2,โฆ ,๐ (5.56)
Section 5.5: Design of IIR Filters From Analog Filters.
Before talking about designing IIR filter from analog filter we need to sample
the analog signal ๐ฅ๐(๐ก) by using uniform sample described by the relation
๐ฅ(๐) = ๐ฅ๐(๐๐) (5.57)
where ๐ is the sample period and ๐ฅ(๐) is the discrete signal obtained by
sampling the analog signal ๐ฅ๐(๐ก).
Note: The subscript ๐ will denote analog signal.
An analog filter may be described by the transfer function
๐ป๐(๐ ) =๐ต(๐ )
๐ด(๐ )=โ ๐(๐)๐ ๐๐๐=0
โ ๐(๐)๐ ๐๐๐=0
(5.58)
where ๐(๐)'s, ๐(๐)'s are the filter coefficients.
Or by its impulse response related to ๐ป๐(๐ ) by Laplace transform
๐ป๐(๐ ) = โซ โ(๐ก)๐โ๐ ๐ก๐๐ก
โ
โโ
(5.59)
Figure (๐. ๐): Lattice-form realization of IIR systems
[14].
๐1(1)
๐๐โ1(๐ โ 1) ๐๐(๐)
๐0(0)
๐2(2)
88
Also, such filters can be described by the differential equation
โ๐(๐)๐๐๐ฆ(๐ก)
๐๐ก๐
๐
๐=0
=โ๐(๐)
๐
๐=0
๐๐๐ฅ(๐ก)
๐๐ก๐ (5.60)
where ๐ฅ(๐ก) is the input signal and ๐ฆ(๐ก) is the output signal.
If all poles of the transfer function are in the left half of s-plane then the
system is stable. For effective conversion from analog signal to digital signal
the mapping which we use should map the points on the ๐ฮฉ-axis in the s-plane
into the unit circle in the z-plane and the points in the right half of the s-plane
into outside the unit circle in the z-plane where the points in the left half
mapped inside the unit circle.
In this section we will design IIR filter from analog filter by using two
methods impulse invariant and bilinear transform.
1. Impulse Invariant Method.
In impulse invariant method the digital IIR filter is designed by sampling the
impulse response of the analog filter.
โ(๐) = โ๐(๐๐) (5.61)
where T is the sampling interval.
From sampling theorem, if we have a continuous time signal ๐ฅ๐(๐ก) with
spectrum ๐๐(๐น) sampled at a rate ๐น๐ = 1 ๐โ sample per second, the spectrum
of the sampled signal ๐(๐) is the periodic repetition of the scaled spectrum
๐น๐ ๐๐(๐น) with period ๐น๐ . Specially the relation is
๐(๐) = ๐น๐ โ ๐๐[(๐ โ ๐)๐น๐ ]
โ
๐=โโ
(5.62)
where ๐ = ๐น ๐น๐ โ is the normalized frequency.
89
If we apply Eq(5.62) for impulse response of an analog filter with frequency
response ๐ป๐(๐น), then the unit sample response โ(๐) = โ๐(๐๐) for digital
filter has the frequency response.
๐ป(๐) = ๐น๐ โ ๐ป๐[(๐ โ ๐)๐น๐ ]
โ
๐=โโ
(5.63)
or, equivalent to
๐ป(๐) = ๐น๐ โ ๐ป๐[(๐ โ 2๐๐)๐น๐ ]
โ
๐=โโ
(5.64)
or
๐ป(ฮฉ๐) =1
๐โ ๐ป๐ (ฮฉ โ
2๐๐
๐)
โ
๐=โโ
(5.65)
To obtain the mapping between the z-plane and the s-plane implied by
sampling process, with a generalization of Eq(5.65) which relates the Z-
transform of โ(๐) to Laplace transform of โ๐(๐๐) by the relation
๐ป(z) | .๐ง = ๐๐ ๐
๐๐ด =1
๐โ ๐ป๐ (s โ ๐
2๐๐
๐)
โ
๐=โโ
(5.66)
where
๐ป(z) = โโ(๐)๐งโ๐โ
๐=0
So
๐ป(z) | .๐ง = ๐๐ ๐
๐๐ด =โโ(๐)๐โ๐ ๐โ
๐=0
(5.67)
The general characteristic of our mapping is
๐ง = ๐๐ ๐ (5.68)
where
๐ = ๐ + ฮฉ๐ (5.69)
91
Substitute Eq(5.69) in Eq(5.68), then express the result equation in polar
form we get
๐๐๐๐ = ๐๐๐๐ฮฉT๐
Clearly
๐ = ๐๐๐
๐ = ฮฉT
To see how poles and zeros of the analog filter mapped using impulse
invariant method we express the transfer function of the analog filter in partial
fraction form. With assumption that the poles of the transfer function of the
analog filter are distinct we get
๐ป๐(๐ ) = โ๐๐
๐ โ ๐๐
๐
๐=1
(5.70)
where ๐ is a positive integer, ๐๐ , ๐ = 1, 2,โฆ ,๐ are the poles of the transfer
function of the analog filter and ๐๐, ๐ = 1, 2,โฆ , ๐ are the coefficients of the
partial fraction expansion. So, the unit sample response is
โ๐(๐ก) = โ๐๐๐๐๐t
๐
๐=1
, ๐ก โฅ 0 (5.71)
If we sampled โ๐(๐ก) periodically at ๐ก = ๐๐ we have
โ(๐) = โ๐(๐๐) = โ๐๐๐๐๐Tn
๐
๐=1
(5.72)
Now, substitute the unit sample response in Eq(5.72) in the transfer function
of digital IIR filter we get
๐ป(z) = โโ(๐)๐งโ๐โ
๐=0
91
= โ(โ๐๐๐๐๐Tn
๐
๐=1
)
โ
๐=0
๐งโ๐
=โ๐๐โ[๐๐๐T๐งโ1]๐โ
๐=0
๐
๐=1
(5.73)
The inner summation of Eq(5.73) is converges because ๐๐ < 0, so
๐ป(z) = โ๐๐1
1 โ ๐๐๐T๐งโ1
๐
๐=1
(5.74)
Therefore, the transfer function of the digital IIR filter has poles at
๐ง = ๐๐๐T, ๐ = 1, 2,โฆ ,๐ (5.75)
If some poles are complex we can combined them to form two-pole filter
section.
The impulse invariant method is appropriate only for low pass filters and a
limited class of band pass filters.
Example 5.7: Convert the analog filter with transfer function
๐ป๐(๐ ) =๐ + 1
(๐ + 1)2 + 9
into a digital IIR filter by impulse invariant method.
Solution:
๐ป๐(๐ ) has two poles at ๐ = โ1 + 3๐,โ1 โ 3๐, so the transfer function can
be written as
๐ป๐(๐ ) =๐ + 1
[๐ โ (โ1 + 3๐)][๐ โ (โ1 โ 3๐)]
By partial fraction expansion
๐ป๐(๐ ) =1
2
1
[๐ โ (โ1 + 3๐)]+1
2
1
[๐ โ (โ1 โ 3๐)]
Then
92
๐ป(๐ง) =1
2
1
[1 โ ๐(โ1+3๐)๐๐งโ1]+1
2
1
[1 โ ๐(โ1โ3๐)๐๐งโ1]
=1 โ ๐โ๐ cos 3๐ ๐งโ1
1 โ 2๐โ๐ cos 3๐ ๐งโ1 + ๐โ2๐๐งโ2
2. Bilinear Transformation Method.
The bilinear transform overcome the limitations of impulse invariant. The
bilinear transformation transform the ๐ฮฉ-axis in the s-plane into the unit circle
in the z-plane only once. The bilinear transformation linked to the trapezoidal
formula for numerical integration. For example, consider an analog filter has
the transfer function.
๐ป(๐ ) =๐
๐ + ๐ (5.76)
The transfer function in Eq(5.76) can be represented by the differential
equation ๐๐ฆ(๐ก)
๐๐ก+ ๐๐ฆ(๐ก) = ๐๐ฅ(๐ก) (5.77)
Integrate the derivative in Eq(5.77)
๐ฆ(๐ก) = โซ๐ฆโฒ(๐)๐๐
๐ก
๐ก0
+ ๐ฆ(๐ก0) (5.78)
where ๐ฆโฒ(๐ก) is the derivative of y(t).
By using the trapezoidal formula at ๐ก = ๐๐ and ๐ก0 = ๐๐ โ ๐ to approximate
the integral in Eq(5.78), we get
๐ฆ(๐๐) =๐
2[๐ฆโฒ(๐๐) + ๐ฆโฒ(๐๐ โ ๐)] + ๐ฆ(๐๐ โ ๐) (5.79)
The derivatives of ๐ฆ(๐ก) at ๐ก = ๐๐ and ๐ก = ๐๐ โ ๐ according to Eq(5.77) are
๐ฆโฒ(๐๐) = โ๐๐ฆ(๐๐) + ๐๐ฅ(๐๐) (5.80)
๐ฆโฒ(๐๐ โ ๐) = โ๐๐ฆ(๐๐ โ ๐) + ๐๐ฅ(๐๐ โ ๐) (5.81)
93
Substitute Eq(5.80) and Eq(5.81) in Eq(5.79), we obtain the linear
difference equation. With ๐ฆ(๐) = ๐ฆ(๐๐) and ๐ฅ(๐) = ๐ฅ(๐๐) we get
(1 +๐๐
2)๐ฆ(๐) โ (1 โ
๐๐
2)๐ฆ(๐ โ 1) =
๐๐
2[๐ฅ(๐) + ๐ฅ(๐ โ 1)] (5.82)
Take the Z-transform of the difference equation in Eq(5.82), we get
(1 +๐๐
2)๐(๐ง) โ (1 โ
๐๐
2) ๐งโ1๐(๐ง) =
๐๐
2(1 + ๐งโ1)๐(๐ง) (5.83)
So, the transfer function is
๐ป(๐ง) =๐(๐ง)
๐(๐ง)=
(๐๐ โ 2)(1 + ๐งโ1)
1 + ๐๐ 2โ โ (1 โ๐๐2 )
๐งโ1 (5.84)
or, equivalent to
๐ป(๐ง) =๐
2๐ (1 โ ๐งโ1
1 + ๐งโ1) + ๐
(5.85)
Therefore, the mapping from the s-plane to the z-plane is
๐ =2
๐(1 โ ๐งโ1
1 + ๐งโ1) (5.86)
Which called the bilinear transformation.
Our previous work was for first-order differential equation. In general for ๐th
order differential equation let.
๐ง = ๐๐๐๐
๐ = ๐ + ฮฉ๐
Then Eq(5.86) can be expressed as
๐ =2
๐
๐ง โ 1
๐ง + 1
=2
๐
๐๐๐๐ โ 1
๐๐๐๐ + 1
=2
๐[
๐2 โ 1
1 + ๐2 + 2๐ cos๐+ ๐
2๐ sin๐
1 + ๐2 + 2๐ cos๐] (5.87)
94
So
๐ =2
๐
๐2 โ 1
1 + ๐2 + 2๐ cos๐ (5.88)
ฮฉ =2
๐
2๐ sin๐
1 + ๐2 + 2๐ cos๐ (5.89)
Note, if ๐ < 1, then ๐ < 0 and if ๐ > 1, then ๐ > 0. But, if ๐ = 1 then ๐ = 0
and
ฮฉ =2
๐
sin๐
1 + cos๐
=2
๐tan
๐
2 (5.90)
or
๐ = 2 tanโ1ฮฉ๐
2 (5.91)
95
Chapter Six
The Chirp Z-transform Algorithm and Its Applications
The chirp Z-transform algorithm is a computational algorithm for numerical
evaluation of Z-transform for a sequence of ๐ points. By using it we can
easily find Z-transform of ๐ points in the z-plane lying on circular or spiral
contour beginning at any an arbitrary point in the z-plane. The algorithm
based on the fact that the value of Z-transform on a circular or spiral contour
can be expressed as a discrete convolution [16].
We will restrict our work to the Z-transform of sequences of a finite number
๐ of non zero points. Therefore Z-transform of these sequences can be
written without loss of generality as,
๐(๐ง) = โ ๐ฅ(๐)๐งโ๐๐โ1
๐=0
(6.1)
Note that ๐(๐ง) converges for all ๐ง except ๐ง = 0.
If we have a train of impulses with equally space ๐ of magnitude ๐ฅ(๐), it will
be of the form
โ๐ฅ(๐)
๐
๐ฟ(๐ก โ ๐๐)
Its Laplace transform will be
โ๐ฅ(๐)
๐
๐โ๐ ๐๐
which is the same of Eq(6.1) if we set
๐ง = ๐๐ ๐ (6.2)
Since we can compute Z-transform for a finite set of samples, we can evaluate
it for a finite number of points ๐ง๐where 0 โค ๐ โค ๐ โ 1 in z-plane, so Z-
transform of ๐ฅ(๐) evaluated at ๐ง๐ will be written as,
96
๐(๐) = ๐(๐ง๐) = โ ๐ฅ(๐)[๐ง๐]โ๐
๐โ1
๐=0
(6.3)
We are interested in the set of points equally spaced around the unit circle,
these points are of the form
๐ง๐ = ๐๐ฅ๐ (๐2๐
๐๐) , k = 0, 1,โฆ , N โ 1 (6.4)
If we substitute Eq(6.4) in Eq (6.3) we get
๐(๐) = โ ๐ฅ(๐)๐๐ฅ๐ (โ๐2๐
๐๐๐)
๐โ1
๐=0
, ๐ = 0, 1,โฆ , ๐ โ 1 (6.5)
Eq(6.5) is the discrete Fourier transform with (๐ =2๐๐
๐) which has many
applications but finding it requires ๐2additions and multiplications so we use
fast Fourier transform to compute it which requires ๐ log2๐ operation if ๐
is a power of 2 and ๐โ ๐๐๐ operation if ๐ is not a power of 2 where ๐๐ is
the prime factored of ๐.
Because fast Fourier transform has many limitations we use chirp Z-
transform which eliminates some of these limitations, for example the
number of samples of the sequence ๐ฅ(๐) need not to be equal to the number
of samples in the z-plane. Neither ๐ nor ๐ need be composite integers, they
can be primes. The angular spacing of ๐ง๐ is arbitrary and the contour need
not to be a circle, it can be a spiral in or out with respect to the origin [9,17].
Let's compute Z-transform on the more general contour of the form
๐ง๐ = ๐ด๐โ๐ , ๐ = 0, 1,โฆ ,๐ โ 1 (6.6)
where ๐ is arbitrary integer and ๐ด, ๐ are arbitrary complex numbers of the
form
๐ด = ๐ด0๐๐ฅ๐ (๐2๐๐0)
97
and
๐ = ๐0๐๐ฅ๐ (๐2๐๐0)
The case where ๐ด = 1, ๐ = ๐ and ๐ = ๐๐ฅ๐ (โ๐2๐ ๐โ ) correspond to the
discrete Fourier transform.
The equivalent s-plane contour of Eq(6.6) begins with the point
๐ 0 = ๐0 + ๐๐ค0 =1
๐ln๐ด (6.7)
And the form of general contour on the s-plane is
๐ ๐ = ๐ 0 + ๐(โ๐ + ๐โ๐ค)
=1
๐(ln๐ด โ ๐ ln๐), ๐ = 0, 1,โฆ ,๐ โ 1 (6.8)
Note that the z-plane contour maps into arbitrary finite length straight line in
the s-plane.
To compute Z-transform a long the contour described by Eq(6.6) we need
๐๐ multiplications and additions.
Now we will derive the chirp Z-transform algorithm [16,19].
If we find the Z-transform for the point along the contour represented in
Eq(6.6) we get,
๐(๐) = โ ๐ฅ(๐)๐ดโ๐๐๐๐
๐โ1
๐=0
, ๐ = 0, 1,โฆ ,๐ โ 1 (6.9)
If we use the Bluestein's ingenious substitution
๐๐ =๐2 + ๐2 โ (๐ โ ๐)2
2 (6.10)
in Eq(6.9) we get
๐(๐) = โ ๐ฅ(๐)๐ดโ๐๐(๐2 2โ )๐(๐2 2โ )๐โ(๐โ๐)2 2โ
๐โ1
๐=0
๐ = 0, 1,โฆ ,๐ โ 1 (6.11)
98
Eq(6.11) can be divided into three steps,
1. Forming a sequence ๐ฆ(๐) of the form
๐ฆ(๐) = ๐ฅ(๐)๐ดโ๐๐(๐2 2โ ), ๐ = 0, 1,โฆ ,๐ โ 1 (6.12)
2. Define the sequence ๐ฃ(๐) as
๐ฃ(๐) = ๐โ๐2 2โ (6.13)
To get the sequence ๐(๐)
๐(๐) = โ ๐ฆ(๐)๐ฃ(๐ โ ๐)
๐โ1
๐=0
, ๐ = 0, 1,โฆ ,๐ โ 1 (6.14)
which is a convolution of ๐ฆ(๐) and ๐ฃ(๐) so it can be expressed as
๐(๐) = ๐ฆ(๐) โ ๐ฃ(๐) (6.15)
3. Multiply ๐(๐) with ๐(๐2 2โ ) to give ๐(๐)
๐(๐) = ๐(๐)๐(๐2 2โ ), ๐ = 0, 1,โฆ ,๐ โ 1 (6.16)
The steps 1 and 3 need ๐, ๐ multiplications respectively where step 2
which is the main step in computation needs the most time.
In summary the chirp Z-transform algorithm consist of the following steps
1. Choose ๐ฟ to be the smallest integer greater than or equal to ๐ +
๐ โ 1 to be compatible with discrete Fourier transform which
mean for most users that ๐ฟ is a power of 2.
2. Form the sequence y(n) of L points by the equation
๐ฆ(๐) = {๐ฅ(๐)๐ดโ๐๐(๐2 2โ ) , ๐ = 0, 1,โฆ ,๐ โ 1
0 , ๐ = ๐,๐ + 1,โฆ , ๐ฟ โ 1 (6.17)
3. Use the fast Fourier transform to compute the discrete Fourier
transform of ๐ฆ(๐) and call them ๐(๐) , ๐ = 0, 1,โฆ , ๐ฟ โ 1.
4. Define the sequence ๐ฃ(๐) as
99
๐ฃ(๐) = {๐โ๐2 2โ , 0 โค ๐ โค ๐ โ 1 0 ,๐ โ 1 < ๐ < ๐ฟ โ ๐ + 1 ๐๐ ๐ฟ > ๐ +๐ โ 1
๐โ(๐ฟโ๐)2 2โ , ๐ฟ โ ๐ + 1 โค ๐ < ๐ฟ
(6.18)
Note that if ๐ฟ = ๐ +๐ โ 1, then there is no value of ๐ฃ(๐) equal zero.
5. Compute the ๐ฟ point discrete Fourier transform of ๐ฃ(๐) and call
them ๐(๐) , ๐ = 0, 1,โฆ , ๐ฟ โ 1.
6. Multiply ๐(๐) and ๐(๐) to get ๐บ(๐)
๐บ(๐) = ๐(๐)๐(๐) , ๐ = 0, 1,โฆ , ๐ฟ โ 1
7. Compute the ๐ฟ point of ๐(๐) by taking the inverse discrete Fourier
transform of ๐บ(๐).
8. Multiply ๐(๐) and ๐(๐2 2โ ) to get ๐(๐)
๐(๐) = ๐(๐)๐(๐2 2โ ), ๐ = 0, 1,โฆ ,๐ โ 1
where ๐(๐) for ๐ โฅ ๐ are discarded.
The computational time for the chirp Z-transform is proportional to ๐ฟ log2 ๐ฟ.
Therefore the direct method for computing ๐(๐) is most efficient for small
values of ๐, ๐ where chirp Z-transform will be efficient for large values or
even for relatively modest values of ๐ and ๐ of the order of 50 [9,16].
Chirp Z-transform algorithm has many applications as:
1. Enhancement of Poles.
Evaluating Z-transform at points outside and inside the unit circle is one of
the advantages of Z-transform over fast Fourier transform, this advantage
help us to find the poles and zeros for systems whose transfer function is of
polynomial form by making the contour which we use closer to the poles and
zeros.
111
2. High Resolution, Narrow Band Frequency Analysis.
The ability to evaluate high resolution, narrow band frequency is an important
application of chirp Z-transform algorithm because it allows us to chose the
initial frequency and frequency space independent of the number of time
sample. Where by using the fast Fourier transform, if the frequency resolution
โค โ๐น and sampling frequency 1 ๐โ , then we require
๐ โฅ 1 (๐โ๐น)โ samples which will be very large for small values of โ๐น.
111
Conclusion
For finding the inverse of Z-transform three methods are used: integration
method is useful for finding a few values of ๐ฅ(๐), where the power series
method is efficient when we find the inverse of finite-order integer power
function with partial fraction method is suitable for finding the inverse of
rational Z-transform. Z-transform is a transform for discrete data equivalent
to Laplace transform for continuous data and it's a generalization of discrete
Fourier transform. It's an efficient method for solving linear difference
equations with constant coefficients and Volterra difference equations of
convolution type. Also, it has many important applications in digital signal
processing as analysis of linear shift-invariant systems, implementation of
FIR and IIR filters and design of IIR filters from analog filters. Chirp Z-
transform algorithm is an important algorithm that overcomes the limits of
fast Fourier transform and has many applications such as enhancement of
poles and high resolution, narrow band frequency analysis.
112
References
[1] El Attar, Refaat: Lecture Notes on Z-Transform, Refaat El Attar,
2007.
[2] Brown, James Ward Churchill, Ruel V.: Complex Variables and
Applications, Eighth Edition, The McGraw-Hill Companies, Inc, 2009.
[3] Debashis, Datta: Textbook Of Engineering Mathematics, Second
Edition, New Age International Publishers, 2005.
[4] Elaydi, Saber: An Introduction to Difference Equations, Third
Edition, Springer Science+Business Media, Inc, 2005
[5] Fadali, M. Sam: Digital Control Engineering Analysis and Design,
Elsevier Inc, 2009.
[6 ] Hayes, Monson H.: Schaum's Outline of Theory and Problems of
Digital Signal Processing, Schaum's Outline Series, The McGraw-Hill
Companies, Inc. 1999.
[7] Jerri, Abdul J.: Linear Difference Equations with Discrete
Transform Methods, Kluwer Academic Publishers, 1996.
[8] Jury, E. I.: Theory and Application of the Z-transform Method,
Robert E. Krieger Publishing Co. Huntington, New York ,1964.
[9] Oppenheim, Alan V., Schafer, Ronald W.: Digital Signal Processing,
Prentice-Hall, Inc, Englewood Cliffs, New Jersey, 1975.
[10] Oppenheim, Alan V., Willsky, Alan S. with Nawab, S. Hamid:
Signals & Systems, Second Edition, Prentice-Hall International, Inc,
1992 .
113
[11 ] Orfanidis, Sophocles J.: Introduction to Signal Processing,
Sophocles J. Orfanidis, 2010.
[12] Poularikas, Alexander D.: The Handbook of Formulas and Tables
for Signal Processing, Boca Raton: CRC Press LLC, 1999.
[13] Poularikas, Alexander D.: The Transforms and Applications
Handbook, Second Edition, Boca Raton: CRC Press LLC,2000.
[14] Proakis, John G., Manolakis, Dimitris G.: Introduction to Digital
Signal Processing, Macmillan Publishing Company, a division of
Macmillan, Inc, 1988.
[15] Smith, Steven W.: The Scientist and Engineer's Guide to Digital
Signal Processing, Second edition, California Technical Publishing.
1997-1999.
[16] Rabiner, Lawrenge R., Schafer, Ronald W. And Rader, Charles M.:
The Chirp Z-transform Algorithm and Its Application, Manuscript
received November 21,1968, The Bell System Technical Journal,
May-June 1969, pages 1949-1992.
[17] Rabiner, L. R., Schafer, R. W., Rader, C. M.: The Chirp Z-
transform Algorithm, June 1969, IEEE Transactions On Audio And
Electroacoustics Vol. Au-17, No. 2.
[18] Salih, Raghad Kadhim; Z-Transformation for Solving Volterra
Integral Equations of Convolution Type, AL-fateh, Journal,
No26,2006.
[19] Shilling, Steve Alan: A Study Of The Chirp Z-transform And Its
Application, Manhattan, Kansas, Kansas State University, 1970, p57.
114
[20]http://www.dip.ee.uct.ac.za/~nicolls/lectures/eee401f/01_mdsp_slides.p
df
[21] Wikipedia.org, Z-transform, Internet:
http://en.wikipedia.org/wiki/Z-transform
115
Appendix A
Some Maple Commands on Z-transform.
Example 2.3:
๐ง๐ก๐๐๐๐ (๐๐, ๐, ๐ง)
Example 2.4:
๐ ๐ข๐(โ(๐ โ ๐งโ1)๐, ๐ = โโ. . โ1)
Example 2.14:
๐ง๐ก๐๐๐๐ (7๐โ๐๐๐๐๐[0](๐) + 3๐โ๐๐๐๐๐[1](๐) + 0๐โ๐๐๐๐๐[2](๐)
+1๐โ๐๐๐๐๐[3](๐) + 2๐โ๐๐๐๐๐[4](๐) + 6๐โ๐๐๐๐๐[5](๐), ๐, ๐ง)
Example 3.6: (a)
๐๐๐ฃ๐ง๐ก๐๐๐๐ (๐ง2 + 3๐ง
๐ง2 โ 3๐ง + 2 , ๐ง, ๐)
Example 3.8:
๐๐๐ฃ๐ง๐ก๐๐๐๐ (๐ง3
๐ง3 โ ๐ง2 โ 5๐ง โ 3 , ๐ง, ๐)
Example 4.1:
๐๐ ๐๐๐ฃ๐({๐ฆ(0) = 0, ๐ฆ(1) = 1, ๐ฆ(๐ + 2) = ๐ฆ(๐ + 1) + ๐ฆ(๐)}, ๐ฆ)
Example 4.2:
๐๐ ๐๐๐ฃ๐ ({๐ฆ(โ1) = 2, ๐ฆ(๐) =1
3๐ฆ(๐ โ 1) + 1} , ๐ฆ)
116
Appendix B
Example on using Chirp Z-transform Algorithm.
Example: Use chirp Z-transform algorithm to find the Z-transform of the
sequence
๐ฅ(๐) = {1 , ๐ = 0 2 , ๐ = 1 0 , ๐๐กโ๐๐๐ค๐๐ ๐
at the given points
๐ง๐ = ๐ด๐โ๐ , ๐ = 0, 1, 2
where
๐ด = 1 and ๐ =1
2โ ๐
โ๐๐2
Solution:
1. For choosing ๐ฟ
๐ +๐ โ 1 = 2 + 3 โ 1 = 4
So ๐ฟ = 4 because 4 is a power of 2.
2.
๐ฆ(๐) = {๐ฅ(๐)1โ๐ [
1
2โ ๐
โ๐๐2 ]
(๐2 2โ )
, ๐ = 0, 1
0 , ๐ = 2, 3
= {1 , ๐ = 0 1 โ ๐ , ๐ = 1 0 , ๐ = 2, 3
3.
๐(๐) = โ๐ฆ(๐)๐โ2๐๐๐๐
๐ฟ
๐ฟโ1
๐=0
, ๐ = 0, 1,โฆ , ๐ฟ โ 1
๐(๐) = โ๐ฆ(๐)๐โ๐๐๐๐2
3
๐=0
, ๐ = 0, 1, 2,3
117
= {
2 โ ๐ , ๐ = 0 โ๐ , ๐ = 1 ๐ , ๐ = 2 2 + ๐ , ๐ = 3
4.
๐ฃ(๐) =
{
[1
2โ ๐
โ๐๐2 ]
โ๐2 2โ
, 0 โค ๐ โค 2
[1
2โ ๐
โ๐๐2 ]
โ(4โ๐)2 2โ
, 3 โค ๐ < 4
= {
1 , ๐ = 0 1 + ๐ , ๐ = 1 โ4 , ๐ = 2 1 + ๐ , ๐ = 3
5.
๐(๐) = โ๐ฃ(๐)๐โ๐๐๐๐2
3
๐=0
, ๐ = 0, 1, 2,3
= {
โ1 + 2๐ , ๐ = 0 5 , ๐ = 1 โ5 โ 2๐ , ๐ = 2 5 , ๐ = 3
6.
๐บ(๐) = ๐(๐)๐(๐) , ๐ = 0, 1, 2 , 3
= {
5๐ , ๐ = 0 โ5๐ , ๐ = 1 2 โ 5๐ , ๐ = 2 10 + 5๐ , ๐ = 3
7.
๐(๐) =1
๐ฟโ๐บ(๐)๐
2๐๐๐๐๐ฟ
๐ฟโ1
๐=0
, ๐ = 0, 1,โฆ , ๐ฟ โ 1
๐(๐) =1
4โ๐บ(๐)๐
๐๐๐๐2
3
๐=0
, ๐ = 0, 1, 2,3
118
= {
3 , ๐ = 0 2 , ๐ = 1 โ2 , ๐ = 2 โ3 + 5๐ , ๐ = 3
8.
๐(๐) = ๐(๐) [1
2โ ๐
โ๐๐2 ]
(๐2 2โ )
, ๐ = 0, 1, 2
= {
3 , ๐ = 0 1 โ ๐ , ๐ = 1 1
2 , ๐ = 2
119
Appendix C
A Table of Properties of Z-transform
Property Sequence Z-transform ROC
Linearity ๐ผ๐ฅ(๐) + ๐ฝ๐ฆ(๐) ๐ผ ๐(๐ง) + ๐ฝ ๐(๐ง) ๐ < |๐ง| < ๐
Shifting ๐ฅ(๐ + ๐) ๐ง๐ ๐(๐ง) ๐๐ฅ < |๐ง| < ๐ ๐ฅ
Multiplication
by Exponential ๐ผ๐๐ฅ(๐) ๐(๐ผโ1๐ง)
|๐ผ|๐๐ฅ < |๐ง|< |๐ผ|๐ ๐ฅ
Time Reversal ๐ฅ(โ๐) ๐(๐งโ1) 1
๐ ๐ฅ< |๐ง| <
1
๐๐ฅ
Conjugation ๐ฅโ(๐) ๐โ(๐งโ) ๐๐ฅ < |๐ง| < ๐ ๐ฅ
Multiplication
by n ๐ ๐ฅ(๐) โ๐ง
๐๐(๐ง)
๐๐ง ๐๐ฅ < |๐ง| < ๐ ๐ฅ
Convolution of
Two Sequences ๐ฅ(๐) โ ๐ฆ(๐) ๐(๐ง)๐(๐ง)
at least, ๐ ๐๐ถ of
๐(๐ง) โฉ ๐ ๐๐ถ of
๐(๐ง)
Correlation of
Two Sequences ๐๐ฅ๐ฆ(๐) ๐(๐ง)๐(๐งโ1)
at least, ๐ ๐๐ถ of
๐(๐ง) โฉ ๐ ๐๐ถ of
๐(๐งโ1)
Multiplication
of Two
Sequences ๐ฅ(๐)๐ฆ(๐)
1
2๐๐โฎ ๐(๐ฃ)๐ (
๐ง
๐ฃ)1
๐ฃ๐๐ฃ
๐ถ
๐๐ฅ๐๐ฆ < |๐ง|
< ๐ ๐ฅ๐ ๐ฆ
Note: ๐(๐ง) is the Z-transform of ๐ฅ(๐) with ๐ ๐๐ถ ๐๐ฅ < |๐ง| < ๐ ๐ฅ and ๐(๐ง) is the Z-
transform of ๐ฆ(๐) with ๐ถ ๐๐ฆ < |๐ง| < ๐ ๐ฆ, ๐ = max(๐๐ฅ, ๐๐ฆ) and ๐ = min(๐ ๐ฅ, ๐ ๐ฆ).
111
Appendix D
A Table of Common Z-transform Pair.
Sequence Z-transform ROC
๐ฟ(๐) 1 all ๐ง
๐ข(๐) ๐ง
๐ง โ 1 |๐ง| > 1
๐ ๐ข(๐) ๐ง
(๐ง โ 1)2 |๐ง| > 1
๐2 ๐ข(๐) ๐ง(๐ง + 1)
(๐ง โ 1 )3 |๐ง| > 1
๐ผ๐ ๐ข(๐) ๐ง
๐ง โ ๐ผ |๐ง| > |๐ผ|
๐๐ผ๐ ๐ข(๐) ๐ผ๐ง
(๐ง โ ๐ผ )2 |๐ง| > |๐ผ|
โ๐๐ ๐ข(โ ๐ โ 1) ๐ง
๐ง โ ๐ |๐ง| < |๐|
โ๐ ๐๐ ๐ข(โ ๐ โ 1) ๐ผ๐ง
(๐ง โ ๐ผ )2 |๐ง| < |๐|
๐๐๐ (๐๐ผ)๐ข(๐) ๐ง2 โ ๐ง๐๐๐ (๐ผ)
๐ง2 โ 2๐ง๐๐๐ (๐ผ) + 1 |๐ง| > 1
๐ ๐๐(๐๐ผ)๐ข(๐) ๐ง๐ ๐๐(๐ผ)
๐ง2 โ 2๐ง๐๐๐ (๐ค) + 1 |๐ง| > 1
5102
ุจ