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for
Chapter 9 z-transforms and applications
Overview
The z-transform
is useful for the manipulation of discrete data sequences and has
acquired a new significance in the formulation and analysis of
discrete-time systems. It is used extensively today in the
areas of applied mathematics, digital signal processing, control
theory, population science, economics. These discrete
models are solved with difference equations in a manner that is
analogous to solving continuous models with differential
equations. The role played by the z-transform in the
solution of difference equations corresponds to that played by the
Laplace transforms in the solution of differential equations.
9.1 The z-transform
The function notation for sequences is
used in the study and application of
z-transforms. Consider a function
defined
for
that
is sampled at times
, where
is
the sampling period (or rate). We can write the
sample as a sequence using the notation
. Without
loss of generality we will set
and
consider real sequences such as,
. The
definition of the z-transform involves an infinite series of the
reciprocals
.
Definition 9.1 (z-transform) Given
the sequence
the
z-transform is defined as follows
(9-1)
,
which is a series involving powers of
.
Remark 9.1. The
z-transform is defined at points
where the Laurent series (9-1)
converges. The z-transform
region of convergence (ROC) for the Laurent series is chosen to
be
, where
.
Remark 9.2. The
sequence notation
is
used in mathematics to study difference equations and the function
notation
is
used by engineers for signal processing. It's a good idea
to know both notations.
Remark 9.3. In the
applications, the sequence
will
be used for inputs and the sequence
will
be used for outputs. We will also use the notations
, and
.
Theorem 9.1 (Inverse
z-transform) Let
be
the z-transform of the
sequence
defined
in the region
. Then
is given by the formula
(9-2)
,
where
is any positively oriented simple closed curve that lies in the
region
and
winds around the origin.
9.1.1 Admissible form of a
z-transform
Formulas for
do not arise in a vacuum. In an introductory course they
are expressed as linear combinations of z-transforms
corresponding to elementary functions such as
![]()
![]()
![]()
.
In Table 9.1, we will see that the z-transform
of each function in
is
a rational function of the complex variable
. It
can be shown that a linear combination of rational functions is a
rational function. Therefore, for the examples and
applications considered in this book we can restrict the z-transforms
to be rational functions. This restriction is emphasized
this in the following definition.
Definition 9.2 (Admissible
z-transform) Given the z-transform
we
say that
is an admissible z-transform,
provided that it is a rational function, that is
(9-3)
,
where
,
are polynomials of degree
,
respectively.
From our knowledge of rational functions, we
see that an admissible z-transform is
defined everywhere in the complex plane except at a finite number of
isolated singularities that are poles and occur at the points
where
. The
Laurent series expansion
in (9-1) can be obtained by a
partial fraction manipulation and followed by geometric series
expansions in powers of
. However,
the signal feature of formula (9-3) is
the calculation of the inverse z-transform
via residues.
Theorem 9.2 (Cauchy's
Residue
Theorem) Let D be
a simply connected domain, and let C be a
simple closed positively oriented contour that lies
in D. If f(z) is analytic
inside C and on C, except
at the points
that
lie inside C, then
.
Corollary 9.1 (Inverse
z-transform) Let
be
the z-transform of the sequence
. Then
is given by the formula
.
where
are
the poles of
.
Corollary 9.2 (Inverse
z-transform) Let
be
the z-transform of the
sequence. If
has
simple poles at the points
then
is given by the formula
.
Example 9.1. Find
the z-transform of
the unit pulse or impulse
sequence
.
Solution 9.1. This follows trivially from Equation
(9-1)
.
![[Graphics:Images/ZTransformIntroMod_gr_78.gif]](ztransform/ZTransformIntroMod/Images/ZTransformIntroMod_gr_78.gif)
Example 9.2. The
z-transform of the
unit-step sequence
is
.
Solution 9.2. From Equation
(9-1)
![[Graphics:Images/ZTransformIntroMod_gr_87.gif]](ztransform/ZTransformIntroMod/Images/ZTransformIntroMod_gr_87.gif)
Example 9.3. The
z-transform of the
sequence
is
.
Solution 9.3. From Definition 9.1
.
![[Graphics:Images/ZTransformIntroMod_gr_101.gif]](ztransform/ZTransformIntroMod/Images/ZTransformIntroMod_gr_101.gif)
Example 9.4. The
z-transform of the
exponential sequence
is
.
Solution 9.4. From Definition 9.1
![[Graphics:Images/ZTransformIntroMod_gr_118.gif]](ztransform/ZTransformIntroMod/Images/ZTransformIntroMod_gr_118.gif)
![[Graphics:Images/ZTransformIntroMod_gr_119.gif]](ztransform/ZTransformIntroMod/Images/ZTransformIntroMod_gr_119.gif)
9.1.2 Properties of the
z-transform
Given that
and
. We
have the following properties:
(i) Linearity.
.
(ii) Delay
Shift.
.
(iii) Advance
Shift.
, or
![]()
(iv) Multiplication
by
.
.
Example 9.5 (a).
The z-transform of the
sequence
is
.
Example 9.5 (b). The
z-transform of the sequence
is
.
Solution 9.5 (a).
Solution 9.5 (b). This is left as an exercise for the reader.
Remark 9.4. When
using the residue theorem to compute inverse z-transforms,
the complex form is preferred, i. e.

![[Graphics:Images/ZTransformIntroMod_gr_146.gif]](ztransform/ZTransformIntroMod/Images/ZTransformIntroMod_gr_146.gif)
9.1.3 Table of
z-transforms
We list the following table of z-transforms.
It can also be used to find the inverse
z-transform.

Theorem 9.3 (Residues at
Poles)
(i) If
has a simple pole at
, then
the residue is
.
(ii) If
has a pole of order
at
, then
the residue is
.
(iii) If
has a pole of order
at
, then
the residue is
.
Subroutines for finding the inverse z-transform
Example 9.6. Find
the inverse z-transform
. Use
(a) series, (b) table of z-transforms, (c) residues.
The following two theorems about z-transforms are useful in finding the solution to a difference equation.
Theorem 9.4 (Shifted Sequences &
Initial Conditions) Define the sequence
and let
be
its z-transform. Then
(i)
(ii)
(iii)
Theorem 9.5
(Convolution) Let
and
be
sequences with z-transforms
,
respectively. Then
where the operation
is
defined as the convolution sum
.
9.1.4 Properties of the
z-transform
The following properties of z-transforms
listed in Table 9.2 are well known in the field of digital signal
analysis.
The reader will be asked to prove some of these properties in the
exercises.

Example
9.7. Given
. Use
convolution to show that the z-transform is
.
Solution 9.7.
Let both
be
the unit step sequence, and both
and
. Then
,
so that
is
given by the convolution
.
9.1.5 Application to signal
processing
Digital signal processing often involves the
design of finite impulse response (FIR) filters. A simple
3-point FIR filter can be described as
(9-4)
.
Here, we choose real coefficients
so that the homogeneous difference equation
(9-5)
has solutions
. That
is, if the linear combination
is
input on the right side of the FIR filter equation, the output
on the left side of the equation will be zero.
Applying the time delay property to the
z-transforms of each term in
(9-4), we obtain
. Factoring,
we get
(9-6)
, where
(9-7) ![]()
represents the filter transfer function. Now, in order for
the filter to suppress the inputs
,
we must have
and an easy calculation reveals that
, and
.
A complete discussion of this process is given in Section
9.3 of this chapter.
Example
9.8. (FIR filter
design) Use
residues to find the inverse z-transform
of
.
Then, write down the FIR filter equation that suppresses
.
![[Graphics:Images/ZTransformIntroMod_gr_341.gif]](ztransform/ZTransformIntroMod/Images/ZTransformIntroMod_gr_341.gif)
![[Graphics:Images/ZTransformIntroMod_gr_342.gif]](ztransform/ZTransformIntroMod/Images/ZTransformIntroMod_gr_342.gif)
9.1.6 First
Order Difference Equations
The solution of difference equations
is analogous to the solution of differential
equations. Consider the first
order homogeneous equation
where
is a constant. The following
method is often used.
Trial solution
method.
Use the trial
solution
, and
substitute it into the above equation and
get
. Then
divide through by ![]()
and simplify to obtain
. The
general solution to the difference equation is
.
Familiar
models of difference equations are given in the table
below.

9.1.7 Methods
for Solving First Order Difference Equations
Consider the
first order linear constant coefficient difference equation
(LCCDE)
with
the initial condition
.
Trial solution method.
First, solve the homogeneous
equation
and
get
. Then
use a trial solution that is appropriate for the sequence
on the right side of the equation and solve to obtain a particular
solution
. Then
the general solution is
.
The shortcoming of this method is that an extensive list of
appropriate trial solutions must be available. Details can
be found in difference equations textbooks. We will
emphasize techniques that use the z-transform.
z-transform method.
(i) Use the time
forward property
and
take the z-transform of each term and
get
![]()
(ii) Solve the
equation in (i) for
.
(iii) Use partial
fractions to expand
in
a sum of terms, and look up the inverse z-transform(s)
using Table 1, to get
![]()
Residue method.
Perform steps (i) and (ii) of the above
z-transform method. Then
find the solution using the formula
(iii)
.
where
are
the poles of
.
Convolution method.
(i) Solve the
homogeneous equation
and
get
.
(ii) Use the
transfer function
and construct the unit-sample
response
.
(iii) Construct
the particular solution
,
in convolution
form
.
(iv) The general
solution to the nonhomogeneous difference equation is
.
(v) The
constant
will
produce the proper initial condition
. Therefore,
.
Remark 9.6. The
particular solution
obtained
by using convolution has the initial condition
Example 9.9. Solve
the difference equation
with
initial condition
.
9 (a). Use the
z-transform and Tables 9.1 - 9.2 to
find the solution.
9 (b). Use residues to
find the solution.
Example 9.10. Solve
the difference equation
with
initial condition
.
9.10 (a). Use the
z-transform and Tables 9.1 - 9.2 to
find the solution.
9.10 (b). Use residues
to find the solution.
Example 9.11. Given
the repeated dosage drug level model
with the initial condition
.
9.11 (a). Use the trial solution
method.
9.11 (b). Use
z-transforms to find the
solution.
9.11 (c). Use residues
to find the solution.
9.11 (d). Use
convolution to find the solution.
An illustration of the dosage model using
the parameters
and initial condition
is shown in Figure 1 below.
![[Graphics:Images/ZTransformIntroMod_gr_574.gif]](ztransform/ZTransformIntroMod/Images/ZTransformIntroMod_gr_574.gif)
Figure
9.1. The solution to
with
.
Exercises for Section 9.1. The z-Transform
The Next Module for Z-Transforms is
Homogeneous Difference Equations
Return to the Complex Analysis Modules
Return to the Complex Analysis Project
This material is coordinated with our book Complex Analysis for Mathematics and Engineering.
(c) 2012 John H. Mathews, Russell W. Howell