Discrete time control systems

Library of Congress Cataloging-in-PublicationData
Ogata, Katsuhiko.
Discrete-time control systems / Katsuhiko Ogata. -2nd ed.
p. cm.
Hncludes bibliographical references and inciex.
HSBN 0-13-034281-5
1. Discrete-time systems. 2. Control theory. H. IEltle.
A402.04 1994 94-19896
629,8'Sdc20 CHP
Editoriallproduction supervision: Lynda GriEitithsiT
Cover design: Karen Salzbach
roduction coordinator: David Dickeyl
O 1995, 1987 by Prentics-Hall,Inc.
Upper SaddleRiver, New Jersey 07458
All rights reserved. No part of this book may be
reproduced,in any form or by any means,
without permission in writing from the pubiisher.
Frinted in the Wnited States of America
1 0 9 8 7
I S B N D-23-034281-5
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1-1 INTRODUCYION,1
9-2 DIGITAL CONTROL SY
1-3 QUANTIZING AND QUANTIZATIOII ERROR, 8
1-4 DATA ACQUISITION, CONVERSION, AND DISTRIBUTIONSYSTEMS, 1 1
1-5 CONCLUDING COMMENTS, 20
2-1 INTRODUCTION, 23
2-2 THE z TRANSFORM, 24
2-3 z TRANSFORMS OF ELEMENTARY FUNCTIONS, 25
2-4 IMPORTANT PROPERTIESAND THEOREMS OF THE z TRANSFORM, 31
2-5 THE INVERSEz TRANSFORM, 37
2-6 z TRANSFORM METHOD FOR SOLVING DIFFERENCEEQUATIONS, 52
2-7 CONCLUDING COMMENTS, 54
EXAMPLE PROBLEMS AND SOLUTIONS, 55
PROBLEMS, 70

ontents
3-2 IMPULSESAMPLINGAND DATA HOLD, 75
3-3 OBTAININGTHEzTRANSFORMBYTHECONVOLUTIONINTEGRALMETHOD,83
3-4 RECONSTRUCTINGORIGINAL SIGNALS FROM SAMPLED SIGNALS, 90
3-5 THE PULSE TRANSFER FUNCTION, 98
3-6 REALlZATlONOF DlGlTAL CONTROLLERS AND DlGlTAL FILTERS, 122
PROBLEMS, 166
4-2 MAPPING BETWEENTHE s PLANE AND THE z PLANE, 174
4-3 STABlLlTY ANALYSIS OF CLOSED-LOOPSYSTEMS IN THE z PLANE, 182
4-4 TRANSIENT AND STEADY-STATERESPONSE ANALYSIS, 193
4-5 DESIGN BASED ON THE ROOT-LOCUSMETHOD, 204
4-6 DESIGN BASED ON THE FREQUENCY-RESPONSEMETHOD, 225
4-7 ANALYTICAL DESIGN METHOD, 242
EXAMPLE PROBLEMSAND SOLUTIONS, 257
PROBLEMS, 288
INTRODUCTION, 293
STATE-SPACE REPRESENIATIONSOF DISCRETE-TIMESYSTEMS, 297
SOLVING DISCRETE-TIMESTATE-SPACE EQUATIONS, 302
PULSE-TRANSFER-FUNCTIONMATRIX, 310
DlSCRETlZATlON OF CONTINUOUS-TIMESTATE-SPACEEQUATIONS, 312
LIAPUNOV STABlLlTY ANALYSIS, 321
EXAMPLE PROBLEMSAND SOLUTIONS, 336
PROBLEMS, 370
6-2 CONTROLLABILITY,378
6-3 OBSERVABILITY, 388
6-4 USEFULTRANSFORMATIONS IN STATE-SPACEANALYSIS AND DESIGN, 396
6-5 DESIGNVIA POLE PLACEMENT, 402
6-6 STATE OBSERVERS,421
6-7 SERVO SYSTEMS, 460
PROBLEMS,510
7-2 DIOPHANTINE EQUATION, 518
7-3 ILLUSTRATIVE EXAMPLE, 522
7-4 POLYNOMIAL EQUATIONSAPPROACH TO CONTROL SYSTEMS DESIGN, 525
7-5 DESIGN OF MODEL MATCHING CONTROL SYSTEMS, 532
PROBLEMS,562
8-2 QUADRATIC OPTIMAL CONTROL, 569
8-3 STEADY-STATE QUADRATIC OPTIMAL CONTROL, 587
8-4 QUADRATIC OPTIMAL CONTROL OF A SERVO SYSTEM, 596
PROBLEMS, 629
A-1
A-2
A-3
A-4
A-5
A-6
A-7
A-U
DEFINITIONS,633
DETERMINANTS, 633
INVERSIONOF MATRICES, 635
RULES OF MATRIX OPERATIONS, 637
VECTORS AND VECTOR ANALYSIS, 643
EIGENVALUES, EIGENVECTORS,AND SlMlLARlTYTRANSFORMATION, 649
QUADRATIC FORMS, 659
PSEUDO1NVERSES, 663

8-1 INTRODUCTION,681
8-2 USEFULTHEOREMS OF THE z TRANSFORMTHEORY, 681
8-3 INVERSEz TRANSFORMATION AND INVERSIONINTEGRALMETHOD, 686
8-4 MODIFIEDzTRANSFORM METHOD, 691
C-9 INTRODUCTION, 304
C-2 PRELIMINARY DISCUSSIONS, 704
C-3 POLE PLACEMENT DESIGN, 707

X Preface Preface
publisher. This book can also serve as a self-study book for practicing engineers w
discrete-time control theory by themselves.
tive cornrnents absut the material ín this book.

recent years there has been a rapid increase in the gital controllers in
ontrol systems. Digital controls are used for achieving performance-f~r
example, in the form of maximum productivity, maximu t9minimum ~ 0 %or
minimum energy use.
recently, the application of cornputer control has made possible "intelli-
on in industrial robots, the optimization of fue1economy '
and refinements in the operation of household appliances and m
rnicrowave ovens and sewing machines, among others. Decision-m
and flexibilityin the controlprogram aremajor advantagesof digitalcontrolsysterns.
The current trend toward digital rather than analog ntrd of dynamicsystems
is mainly due to the availability of low-cost digital com ters and the advantages
found in working with digital signals rather than continuous-time signals.
continuous-time signal is a signal defined over a contin-
uous plitude may assume a continuous range of values or may
assume only a finite number of distinctvalues. The process o£representing a variable
by a set of distinct values is called quantization, and the resulting distinct values are
called quantized values. The quantized variable changes only by a set of distinct
steps.
n analog signal is a signal defined over a continuous range of time whose am-
plitude can assume a continuous range of values. Figure 1-l(a) shows a continuous-
time analog signal, and Figure 1-l(b) shows a continuous-time quantized signal
(quantized in amplitude only).

lntroduction to Bíscrete-Time Control Systems Chap. 1
Figwe 1-1 (a) Continuous-time analog
signal; (b) continuous-time quantized
signal; (c) sampled-data signal;
(d) digital signal.
Notice that the anaiog signal is a special case of the continuous-time signal. In
practice, however, we frequently use the terminology "continuous-time" in lieu of
"analog." Thus in the literature, including this book, the terms "continuous-time
signal" and "analog signal" are frequently interchanged, although strictly speaking
they are not quite synonymous.
A discrete-time signal is a signal defined only at discrete instants of time (that
is, one in which the independent variable t is quantized). n a diXrete-h-le signal9
if the amplitude can assume a continuous range of values, then the signal is called
a sampled-data sigizal. A sampled-data signal can be generated by sampling an
analog signal at discrete instants of time. It is an amplitude-modulated pulse signal.
Figure 1-1(c) shows a sarnpled-data signal.
A digital signal i:, a dlscrete-time signalwith quantized arnplitude. Sucha signal
can be represented by a sequence of nurnbers, for example, in t
ec. 1-1 introdlaction
numbers. (In practice, many digital signals are obtained by sampling analog signals
and then quantizing them; it is the quantization that allows these analog signals to
be read as finite binary words.) Figure 1-l(d) depicts a digital signal. Clearly, it is
a signal quantized both in amplitude and in time. The use of the
requires quantization of signals both in amplitude and in time.
The term "discrete-time signal" is broader than the term "digital signal" or the
term "sampled-data signal." 1can refer either to a digital
signal or to a sampled-data signal. "discrete time7'and
""dgital" are often interchan
used in theoretical study, wh
ware or software realizations.
In control engineering,
ysical plant or process or a nonphysical process such as an economic process.
t plants and processes involve continuous-time signals; therefore, if digital
controllers are involved in the control systems, signal conversions (analog to dibital
and digital to analog) become necess y. Standard techniques are available for such
signal conversions; we shall discuss em in Section 1-4.
Loosely speaking, terminologies such as discrete-time con
pled-data control systems, and digital control systems imply the
similar types of control systems. Preciselyspeaking, there are, of
in these systems. For example, in a sa -data control system both continuous-
time and discrete-time signals exist in the system; the discrete-time signals are
amplitude-modulated pulse signals. Digital control systemsmay include both contin-
us-time and discrete-time signals; here, the latter are in a numerically coded form.
th sampled-data control systems and digital control systems are discrete-time
any industrial control systemsinclude continuous-time signals, sampled-data
signals, and digital signals. Therefore, in this book we use the term "discrete-time
control systems" to describe the control systemsthat include sorneformsof sarnpled-
data signals (amplitude-modulated pulse signals) and/or digital signals (signals in
numerically coded form).
. The discrete-time control systems consid-
ered in this book are mostly linear and time invariant, although nonlinear and/or
time-varying systems are occasionallyincluded in discussions. A linear systemis one
in which the principie of superposition applies. Thus, if y, is the response of the
system to input xl and y2 the response to input x2, then the system is linear if and
only if, for every scalar a and P, the response to input al+ Px2 is cuyl + ,By2.
A linear systern may be described by linear differential or linear difference
equations. A time-invariant linear system is one in which the coefficients in the
differential equation or difference equation do not vary with time, that is, one in
which the properties of the system do not change with time.
Discrete-time control systemsare control systemsin which one or more variables can
change only at discrete instants of time. These instants, which we shall denote by
kTor tk(k = 091,2,. . . ),may specify the times at which some physicalrneasurement

lntroduction to Discrete-TimeControl Ystems Chap. 1
is performed or the times at which the memory of a digital computer is read out. The
time interval between two discrete instants is taken to be sufficiently short that the
data for the time between them can be approximated by simple interpolation.
Discrete-time control systems differ from continuous-time control systems in
that signals for a discrete-time control system are in sampled-data form or in digital
form. If a digital computer is involved in a control system as a digital controller, any
sampled data must be converted into digital data.
Continuous-time systems, whose signals are continuous in time, may be de-
scribed by differential equations. Discrete-time systems, which involve sample
data signals or digital signals and possibly continuous-time signals as well, may be
described by difference equations after the appropriate discretization of continuous-
time signals.
esses. The sarnpling of a continuous-time signal replaces the
origi -time signal by a sequence of values at discrete time points. A
sampling process is used whenever a control system involves a digital controller,
since a sampling operation and quantization are necessary to enter data into such
a controller. Also, a sampling process occurs whenever measurements necessary for
control are obtained in an intermittent fashion. For example, in a radar tracking
system, as the radar antenna rotates, information about azimuth and elevation is
obtained once for each revolutio of the antenna. Thus, the scanning operation of
the radar produces sampled data n another example, a sampling process is needed
whenever a large-scale controller or computer is time-shared by several plants in
order to save cost. Then a control signal is sent out to each plant only periodically
and thus the signal becomes a sampled-data signal.
The sampling process is usually followed by a quantization process. In the
quantiration process the sampled analog amplitude is replaced by a digital ampli-
tude (represented by a binary number). Then the digital signal is processed by the
computer. The output of the cornputer is sampled and fed to a hold circuit. The
output of the hold circuit is a continuous-time signal and is fed to the actuator.
shall present details of such signal-processing methods in the digital controlle
Section 1-4.
The term "discretization," rather than "sampling," is frequently used in the
analysis of multiple-input-multiple-output systems, although both mean basically
tant to note that occasionally the sampling operation or discretiza-
fictitious and has been introduced only to simplify the analysis of
control systems that actually contain only continuous-time signals. In fact, we often
use a suitable discrete-time model for a continuous-time system. An example is a
digital-computer simulation of a continuous-time system. Such a digital-computer-
simulated system can be analyzed to yield parameters that will optimize a given
performance index.
ost of the material presented in this book deals with control systemsthat can
be modeled as linear time-invariant discrete-time systems. It isim
that many digital control systems are based on continuous-time
Since a wealth of experience has been accumulated in the design of continuous-time
Sec. 1-2 Digital Control Cystems
controllers, a thorough knowledge of them is highly valuable in designing discrete-
time control systems.
Egure 1-2 depicts a block diagram of a
tion of the basic control scheme. The sy
feedforward control. In designing such
"goadness" of the control system depe
choose an appropriate performance index for a given case and design a controller
so as to optimize the chosen performance index.
. Figure 1-3 shows a block diagram
of the system are shownby the blocks.
ñhe controller operation is controlled by the clock. In such a digital control system,
somepoints of the system pass signalsof varyingamplitude in either continuous time
or discrete time, while other points pass signalsin numerical code, as de
of the plant is a continuous-time signal. The error signal is con-
1 form by the sample -hold circuit and the analog-to-digital
converter. The conversion is done at sampling time. The digital computer
l i
1
Clock
L _I
Digital controlier
Noise
Figure 1-2 Block diagram of a digital control system.

lntroduction to Discrete-Time Control Cysterns Chap. ?
ipre 1-3 Block diagram of adigital control system showingsignalsinbinary or graphic form.
rocesses the sequences of numbers by means of an algorithm and produces new
quences of numbers. At every sampling instant a coded number (usually a binary
number consisting of eight or more binary d ) must be converted to a physical
control signal, which is usually a continuous e Or anal% signal- The digital-to-
analog converter and the hold circuit convert sevence ~f numbers in numerical
code into a piecewise continuous-time signal. The real-time clock in the cornputer
synchronizes the events. The output of the hold circuit, a continuous-time signal, is
fed to the plant, either directly or through the actuator, to control its
The operation that transforms continuous-time signals into discrete-time
data is called sampling or discretizaíion. The reverse operation, the operation that
transforms discrete-time data into a continuous-time signal, is called data-hold; it
amounts to a reconstruction of a continuous-time signal from the sequence of
discrete-time data. It isusuallydone using one of the many extrapolation techniques.
Pn many cases it is ne by keeping the signal constant between the successive
sampling instants. ( shall discuss such extrapolation techniques in Section 1-4.)
The sample-and-hold (SIH) circuit and analog-to-digital (AID) converter con-
vert the continuous-time signal into a sequence of numerically coded binary words.
Such an AID conversion process is called coding or encoding. The combination of
the SIH circuit and analog-to-digital converter may be visualized as a switch that
closesinstantaneously at every time interval Tand generates a sequence of numbers
in numerical code. The digitalcornputer operateson suchnumbers in numerical code
and generates a desired sequence of numbers in numerical code. The digital-to-
analog (DIA) conversion process is called decoding.
efore we discuss digital control systems in detail, we
need to define some of the terms that appear in the block diagram of Figure 1-3.
Sample-and-Hold (SIH). "Sarnple-and-hold" is a general term used for a
sarnple-and-hold amplifier. It describes a circuit that receives an analog input signal
and holds this signal at a constant value for a specified period of time. Usually the
signal is elebrical, but other forrns are possible, such as optical and mechanical.
igital Control Systerns 7
Analog-to-Digital Gonverter (AlD). An analog-to-digital converter, also
called an encoder, is a device that converts an analog signal into a digital signal,
usually a numerically coded signal. Such a converter is needed as an interface
between an analog component and a digital com onent. A sample-an
is often an integral part of a commercialliyavailable A/D converter. The conversion
of an analog signal into the corresponding digital signal (binary number) is an
ation, because the analog signal ke on an infinite number of values,
he variety of different nurnbers a finite set of digits
This approximation process is c ore on quantization
is presented in Section 1-3.)
Digital-to-Analog Converter (DIA). A digital-to-analog converter, also
called a decoder, is a device that converts a digital signal (numerically co
into an analog signal. Such a converter is needed as an interface between a digital
component and an analog component.
Plant or Process. A plant is any ical object to be controlled. Examples
are a furnace, a chemical reactor, and of machine parts functioning together
to perform a particular operation, such as a servo system or a spacecraft.
A process is generally defined as a progressive operation or develop
marked by a series of gradual changes that succeed one another in a relatively
fixed way and lead toward a particular result or end. In this book we cal1 any
operation to be controlled a process. Examples are chemical, economic, and biolog-
ical processes.
The most difficult part in the design of control systems may lie in the accurate
modeling of a physical plant or process. There are many approaches to the plant or
process model, but, even so, a difficulty may exist, mainly because of the absence
of precise process dynamics and the pr of poorly defined random parameters
in many physical plants or processes. ,in designing a digital controller, it is
necessary to recognize the fact that the mathematical model of a plant or process
in many cases is onIy an approximation of the physical one. Exceptions are found
in the modeling of electromechanical systems and hydraulic-mechanical systems,
since these may be modeled accurately. For example, the modeling of a robot arm
system may be accompIished with great accuracy.
Transducer. A transducer is a device that converts an input signal into an
output signal of another form, such as a device that converts a pressure signal into
a voltage output. The output signal, in general, depends on the past history of the
input.
Transducers may be classified as analog transducers, sampled-data transduc-
ers, or digital transducers. An analog transducer is a transducer in which the input
and output signalsare continuous functions of time. The magnitudes of these signals
may be any values within the physical limitations of the system. A sampled-data
transducer is one in which the input and output signalsoccur only at discrete instants
of time (usually periodic), but the magnitudes of the signals, as in the case of the
analog transducer, are unquantized. A igital transducer is one in which the input
and output signals occur only at discrete instants of time and the signal magnitudes
are quantized (that is, they can assume only certain discrete levels).

lntroductisn to Discrete-TimeControl
ns, As stated earlier, a si
discrete-time signal. A sa
in transforming a continuous-time g*al into a dkxete-
There are several different t es of sampling operations of practica1 impar-
tance:
n this case, the campling instants are equally spaced, or
tk = kT(k =O,1,2,... ). eriodic sampling is the II-lost conventional tYPe of
sampling operation.
Multiple-order sampling. The pattern of the tk's is repeated
is, tk+r- tk is constant for al1 k .
ltiple-ratesampling. In a control system havi?
stant involved in one loop may be quite
ence, it may be advisable to sample slow
time constant, while in a loop involvingonly small time constants the sampling
rate must be fast. Thus, a digital control system may have different sampling
riods in different feedback paths or may have multiple sam
ndorn samplirzg. In this case, the sampling instaiats are ra
random variable.
n this book we shall treat only the case where the sampling is periodic.
The main functions involvedin analog-to-digital conversion are sampling, amplitude
quantizing, and coding. hen the vahe of any sample falls between two adjacent
"'perrnitted" output stat it must be read as the permitted output state nearest the
actual value of the signal. The rOWss of representing a ~ontinuousor anal% sigrial
by a finite number of discrete states is called amplitude quantization. That is,
"quantizing" means transforming a continuous or analog signalinto a set of
states. (Note that quantizing occurs whenever a physical quantity is represented
numerically.)
The output state of each quantized sample is then described by a numerical
code. The process of representing a sample value by a numerical code (such as a
binary code) is called encoding or coding. Thus, encoding is a process of assigning
a digital word or code to each discrete state. The sampling period and quantizing
levels affect the performance of digital control systems. Sothey must be determined
carefully.
. The standard number system used for processing digital signals
is the binary number system. In this system the code group consists of n pulses each
indicating either "on" (1) or "off" (O). In the case of quantizing, n "on-off" pulses
can represent 2" amplitude levels or output states.
The quantization level Q is defined as the range between two adjacent decision
points and is given by
ec. 1-3 Quantizing and Quantization
where the FSR is the full-scalerange. Note that the st bit of the natural binary
as the most weight (one-half of the full scale) he most significant
SB). The rightmost bit has the least weight (112" t e full scale) and is
called the least significant bit (LS
The least significant bit is t e quantization leve1 Q.
Error. Sincethe nu igitalword is finite,AID
in a finite resolution. output can assume only
a finite number of levels, and therefore an analog number must be rounded off to
the nearest digital level. Menee, any AID conversion involves quantization error.
Such quantization error varies between O and t $ Q S error depends on the
fineness of the quantization level and can be made as s
quantization level smaller (that is, by increasing the
there is a maximum for the number of bits n, and so there is always some error due
to quantization. The uncertainty present in the quantization process is called quan-
tization noise.
To determinethe desired sizeof the quantization level (or thenumber of output
states) in a givendigital control system, the engineermust have a good understanding
of the relationship between the size of the quantization level and the resulting error.
Thevarianceof the quantization noise isan important measure of quantization error,
since the variance is proportional to the average power associated with the noise.
Figure 1-4(a) shows a block diagram of a quantizer together with its input-
output characteristics. For an analog input x(t), the outputy(t)takes on only a finite
number of levels, which are integral multiples of the quantization level Q .
In numerical analysis the error resulting from neglecting the remaining digits
is called the round-off error. Since the quantizing process is an approximating
process in that the analog quantity is approximated by a finite digital number, the
quantization error is a round-off error. Clearly, the finer the quantization level is,
the smaller the round-off error.
Figure 1-4(b) shows an analog input x(t) and the discrete output y(t), which
is in the form of a staircase function. The quantization error e(t) is the difference
between the input signal and the quantized output, or
Note that the magnitude of the quantized error is
For a small quantization leve1 ,the nature of the quantization error is similar
to that of random noise. nd, in effect, the quantization process acts as a source of
random noise. In what followswe shallobtain the variance of the qu
Scach variance can be obtained in terms of the quantization level

lntroduction to Diccrete-Time Ontrol Chala. 1
(a) Block diagram of a quantizer and its input-output characteristics;
(b) analog input x(t) and discrete output y(t); (c) probability distribution P(e) of
quantization error e(f).
Suppose that the quantization level Q is small and we assume that the quan-
tization error e(t) is distributed uniformly between -4 Se that this error
acts as a white noise. [This is obviously a rather rough assumption. However, since
the quantization error signal e(t) is o£a small amplitude, such an assumption may
be aiceptable as a first-order approximation.] The probability distribution P(e) of
Sec. 1-4 Dala Acquicition, Conversion, and
signar e(t) may be plotted as shown in Figure 1-4(c).The average value of e(t) iszem,
or e(t) = O. Then the variance u h f the quantization noise is
n level Q is small co pared with the average amplitude o£
variance of the quantization noise is seen to be one-twelfth
of the square of the quantization level.
ith the rapid growth in the use of digital computers to perform digital control
actions, both the data-ac uisition system and the distribution system have become
an important part of the entire control system.
The signalconversion that takes place in the digital control system involvest
following operations:
ultiplexing and demultiplexing
.Sample and hold
Analog-to-digital conversion (quantizing and encoding)
.Digital-to-analog conversion (decodlng)
Figure 1-5(a) shows a block diagram o£a acquisition system, and Figure 1-5(b)
shows a block diagram of a data-distrib
n the data-acquisitionsystem the input to the system is a physicalvariable such
asposition, velocity, acceleration, temperature, orpressure. uch a physicalvariable
is first converted into an electrical signal (a voltage or current signal) by a suitable
Physical To digital
variable controller
From digital
controller
To actuator
Figure 1-5 (a) Block diagram of a data-acquisition system; (b) block diagram of a data-
distribution system.

Introduction to Discrete-TimeControl Systems Chap. 1
transducer. Once the physical variable is converted into a voltage or current signal,
the rest of the data-acquisition process is done by electronic means.
In Figure 1-5(a) the amplifier (frequentl
Iows the transducer performs one or more of
the voltage output of the transducer; it converts a current signalinto a vo
or it buffers the signal. The low-pass filter that follows the amplifier attenuates the
high-frequency signalwmponents, such asnoise signals. (Note that
are random in nature and may be reduced by low-pass filters.
common electrical noises as power-line interference are generally p
be reduced by means of notch filters.) The output of the low-pass filter is an analog
signal. m i s signal is fed to the analog multiplexer. The output of the multiplexer is
fed to the sample-and-hold circuit, whose output is, in turn,
digital converter. The output of the converter is the signal in
to the digital controller.
The reverse of the data-acquisition process is the ata-distributionProcess “%+,
shown in Figure 1-5(b), a data-distribution system consists of registers, a demulti-
plexer, digital-to-analog converters, and hold circuits. It conver the signalin digital
form (binary numbers) into analog form. The output of the DI converter is fed tQ
the hold circuit. The output of the hold circuit is fed to the analog actuator, which,
in turn, directly control; the plant under consideration.
In the following, we shall discuss each in b.+hal ComPonent h ~ ~ l v e din the
signal-processing system.
lexer. An analog-to-digital converter is the most expensive
component in a data-acquisition system. The analog multiplexer is a device that
performs the function of time-sharing an AID converter among many analog chan-
nels. The processing of a number of channels with a digital controller is posible
because the width of each pulse representing the input signal is very narrow, so the
empty space during each sampling period rnay be used for other signals. If many
signalsareto be processed by a single digital controller, then these input signalsmust
be fed to the controller through a multiplexer.
Figure 1-6 shows a schematic diagram of an analog multiplexer. The analog
To sampler
Figure 1-6 Schematic diagram of an
analog multiplexer.
Data Acquisition, Conversion, and
multiplexer is a multiple switch (usually an electronic switc ) that sequentially
switches amongmany analoginput channelsin bed fashion. Thenumber
instant of time, only one
given input channel, the
for a specified period of
time. During the connection time the sampl ircuit samples the signal
voltage (analog signal) and holds its alue, while the analog-to-
converts the analog value into digital ta (binary numbers). Each channel is read
in a sequential order, and the corresponding values are converted into digital data
in tbe same sequence.
lexer. The demultiplexer, which is synchronized with the in
pling signal, separates the composite output digital data flrom the digital controller
into the original channels. Each channel is connected to a DIA converter to produce
the output analog signal for that channel.
er in a digital system convelrts an analog
pulses. The hold circuit holds the value
of the sampled pulse signal over a specified period of time. The sample-and-hold is
necessary in the AID converter to pro
input signal at the sampling instant.
available in a single unit, known as
however, the sampling operation and
(see Section 3-2). It is common practice to use a single analog-to-digital converter
and multiplex many sampled analog inputs into it.
practice, sarnplingduration isvery short compared with the sampling period
the sampling duration is negligible, the sampler may be considered an
mpler." An ideal sampler enables us to obtain a relatively simplemathemat-
ical model for a sample-and-hold. (Such a mathematical model will be discussed in
e 1-7 showsa simplifieddiagramfor the sample-and-hold.Th
g circuit (simply a voltage memory device) in which an inpu
acquired and then stored on a high-quality capacitor with low leakage and low
dielectric absorption characteristics.
In Figure 1-7 the electronic switch is connected to the hold capacitor. Opera-
tional amplifier 1 is an input r amplifier with a high input impedance. Op-
erational amplifier 2 is the o amplifier; it buffers the voltage on the hold
capacitor.
There are two modes of operation for a sample-and-hold circuit: the tracking
mode and the hold mode. n the switch is closed (that is, when the input signal
is connected), the operatin ing mode. The charge on the capacitor
in the circuit tracks the input voltage. the switch is open (the input signal is
disconnected), the operating mode is the hold mode and the capacitor voltage holds
constant for a specified time period. Figure 1-8 shows the tracking mode and the
hold mode.
Note that, practically speaking, switching from the tracking mode to the hold
mode is not instantaneous. If the hsld command is given whde the circuit is in the

Introduction to Discrete-TimeControl Cystems
Analog
input
Chap. 1
Arnp. 1 I 1 Arnp. 2I
Analog
output
Sarnple-and-hoid
cornrnand
Figure 1-7 Sample-and-hold circuit.
tracking mode, then the circuitwill stay in the tracking mode for a short whilebefore
reacting to the hold command. The time interval during which the switching takes
place (that is, the time interval when the measured amplitude is uncertain) is called
the aperture time.
The output voltage duringthe hold mode may decrease slightly.The hold mode
droop may be reduced by using a high-input-impedance output buffer amplifier.
Such an output buffer amplifier must have very low bias current.
The sample-and-hold operation is controlled by a periodic clock.
es oofAnalog-to-Digital earlier, the PrQcessby
which a sampled analog signal is quantized and converted to a binary number is
called analog-to-digital conversion. Thus, an AID converter transforms an analog
Inpuí Sarnple to Hold rnode
I signal hold offset droop
Tracking
mode 4 kp:;ye-
t
ttioíd command Figure 1-43 Traclcing mode and hold
is given here mode.
Sec. 1-4 Data Acquisition, Conversion, and istribution Systemc
signal (usually in the formof a voltage or current) into a
coded word. In practice, the logic is based on binary dig
and the representation has only a finite
performs the operations of sample-and-ho
in the digital system a pulse is supplied every sampling period T by a clock. %heAID
converter sends a digital signal (binary number) to the digital contiroller each time
a pulse arrives.
Among the many ID circuits available, the following types are used most
frequently:
ve-approximation type
Each of these four types has its own advantagesan
application, the conversion speed, accuracy, size
be considered in choosing the type of A/D convert
for example, the number of bits in the output signal must be increased.)
As will be seen, analog-to-digital converters art of thek feedback l o o p
digital-to-analogconverters. The simplestty verter is the counter type.
The basic principle on which it works is that clock pulses are applied to the digital
counter in such a way that the o
the feedback loop in the AID con
the output voltage
hen the output volt
the clock pulses are stopped. The counter output voltage is then the digital output.
The successive-approximation type of AID converter is much faster than the
counter type and is the one most frequently use Figure 1-9 shows a schematic
diagrarn of the successive-approximariontype of
D/A Analog
converter reference
Digital
output
Analog
input
Figure 1-9 Schematicdiagramof a successive-approximation-typeof BID converter.

lntroduction to Discrete-TimeControl
The principle of operation of this type of AID converter is as follows. T
successive-approximation register (SAR) first turns on the most significant bit (ha
the maximum) and compares it with the analog input. The com
whether to leave the bit on or turn it off. If the analog input voltage is larger, tfae
most significant bit is set on. The next step is to turn on bit 2 and t
analog input voltage with three-fourths of the maximum. After n
coqleted, the digital output of the successive-approximationreg
those bits that remain on and produces the desired digital code.
ID converter sets 1bit each clock cycle, and so it requires ody n clock cycles t s
generate n bits, where n is the resolution of the converter in bits. (The number n
of bits employed determines the accuracy of conversion.) The time required for the
conversion is approxirnately 2 psec or less for a 12-bit conversion.
ctual analog-to-digital signal converters differ
from the ideal signal converter in that the former always have some errors, such as
offset error, linearity error, and gain error, the characteristics o£which are shown
in Figure 1-10. Also, it is important to note that the input-output characteristics
change with time and temperature.
Finally, it is noted that commerciaI converters are specified for three basic
temperature ranges: cornmercial (0°C to 70°C), industrial (-25°C to 85"C), and
military (-55°C to 125°C).
onverters. At the output of the digital controller the
digital signal must be converted to an analog signal by the process called digital-to-
analog conversion.A DIA converter is a devicethat transforms a digital input (binary
numbers) to an analog output. The output, in most cases, is the voltage signal.
For the full range of the digital input, there are 2" corresponding different
analog values, induding O. Por the digital-to-analog conversion there is a one-to-one
~orrespondencebetween the digital input and the analog output.
Two methods are commonly used for digital-to-analog conversion: the method
usirig weighted resistors, and the one using the R-2R ladder network. The former
is simple in circuit configuration, but its accuracy may not be very good. The latter
is a little more complicated in configuration, but is more accurate,
Figure 1-13 shows a schematic diagram o£ a DIA converter using weighted
resistors. The input resistors of e operational amplifier havetheir resistance values
weighted in a binary fashion. hen the Iogic circuit recebes binary 1, itch
(actually an electronic gate) co ects the resistor to the referente voltage. the
logic circuit receives binary 0, the switch connects the resistor to ground. The
digital-to-analog converters used in common practice are of the parallel type: al1bits
act sirnultaneously upon application of a digital input (binary numbers).
The D/A converter thus generates the analog output voltage corresponding
to the given digital voltage. For the D/A converter shown in Figure 1-11, if the
binary number is b3b.,bl bo,where each of the b's can be either a O or a 1, then the
output is
Sec. 'l-4 Data Acquisition,Conversion, and
Errors in AID converters:
(a) offset error; (b) linearity error;
(c) gain error.
Notice that as the nurnber of bits is increased the range of resistor values becomes
large and consequently the accuracy becomes poor.
e 1-12 shows a schematic diagram of an n-bit DlA converter using an
-2 er circuit. Note that with the xception of the feedback resistor (which
is 3R) al1 resistors involved are either or 2R. This means that a high Ievel of
accuracy can be achieved. The output voltage in this case can be given by
Ciircuii. Shesarnplingoperation pro-
duces an amplitude-modulated pulse signal. The function of the hold operation is

lntroduction to Biscrete-Time Control
Figure 1-11 Schematic diagram of a D/A converter using weighted resistors.
to reconstruct the analog signalthat has been transmitted as a train of pulse sa
That is, the purpose o£the hold operation is to fill in the spaces between sampling
periods and thus roughly reconstruct the original analog input signal.
The hold circuitis designed to extrapolate the output signalbetween successive
points according to some prescribed manner. The staircase waveform of the output
shownin Figure 1-13 is the simplest way to reconstruct the original input signal. The
Id circuit that produces such a staircase waveform is called a zero-order hold.
cause of its simplicity, the zero-order hold is commonly used in digital control
systems.
Figure 1-12 n-Bit DIA converter using an R-2R ladder circuit
Output
Figure 1-13
hold.
Outpur from a zero-order
ore sophisticated hold circuits are available than the zero-orde
r-order hold circuits and include the first-order hold a
gher-order hold circuits wiIl generally reconstnict a signal more
a zero-order hold, but with some disadvantages, as explained next.
The first-order hold retains the val e of the previous sampk as we
present one, and predicts, by extrapolation, the next sam
generating an output slope equal to the slo
and present samples and projecting it from
in J?igure 1-14.
As can easilv be seen from the figure, if the slope of the original signal
not change much, the prediction is go
slope, then the prediction is wrong a
causing a large error for the sampli
An interpolativefirst-order ho kdd, reconstructsthe
nal signal much more accuratel generates a straight-line
ut whose slopeis equal to that joining the previous samplevalue and the present
samplevalue, but this time the projection ismade fromthe cuirent samplepoint with
Output

Output
Figure 1-15 Output from an inter-
polative first-order hold (polygonal
hold).
the amplltude of the prevlous sample. ence7the accuracy in reconstructing the
original signal is better than for other hold circuits, but there is a one-sarnpling-
period delay, as shown in Figure 1-15. effect,the better aCcuracY is achieved at
the expense of a delay of one sampling period. From the viewpoint of the stability
of closed-loop systems, such a delay is not desirable, and so the interpolative
first-order hold (polygonal hold) is not used in control system applications.
In concluding this chapterwe shall compare digitalcontrollers and analogcontrollers
used in industrial control systems and review digital control of processes. Then we
shall present an outline of the book.
rs. Digital controllers operate only
on numbers. Decision making is one of their important Tunctions. They are often
used to solve problems involved in the optimal overall operation o£industrial plants.
Digital controllers are extremely versatile, They can handle nonlinear control
equations involving complicated computations or logic operations. A very much
wider class of control laws can be used in digital controllers than in analog con-
trollers. Also, in the digital controller, by merely issuing a new program the oper-
ations being performed can be changed completely. Shis feature is particularly
important if the control system is to receive operating information or instructions
from some computing center where econornic analysis and optirnization studies are
made.
Digital controllers are capable of performing complex computations with
constant accuracy at high speed and can have almost any desired degree of compu-
tational accuracy at relatively little increase in cost.
Originally, digital controllers were used as components only in large-scale
control systems. At present, however, thanks to the availability of inexpensive
microcornputers, digital controllers are being used in many large- and small-scale
control systemc. hn fact, digital controllers are replacing the analog controllers that
Sec. 7-5 Concluding Cornrtents
have been used in many srnall-scale control systems. Digital controllers ase often
superior in performance and lower in price than their analog counterparts.
Analog controllers represent the variables in an equation by continuous
ical quantities. The easily be designed to serve satisfactorily as non-decision-
e cost of analog computers or analog controllers increases
of the computations increases, if constant accuracy is to
There are additional advantages of digital controllers over analog controllers.
Egital components,such assample-and-hold circuits, AID and DlA converters, and
highly reliable, and often compact
ts have high sensitivity, are often
cheaper than their analog counterparts, and are less sensitive to noise signals. And,
as mentioned earlier, digital controllers are flexible in allowing programming
changes.
Digihl Ccpnkr~b industrial process control systems, it is gen-
erally not practica1t ry long time at steady state, because certain
changes rnay occur in production requirements, raw materials, economic factoss,
and processing equipments and techniques. Thus, the transient behavlor of indus-
trial processes must always be taken into consideration. Sincethere are interactions
amsng process variables, using only one p ariable for each control agent is
not suitable for really complete control. use of a digital controller, it ks
possible to take into account al1 process v together with econornic factors,
ents, equipment performance, and al1other needs, and thereby
al control of industrial processes.
Note that a system capable of controlling a process as coqlete%yas
will have to solve complex equations. The more complete the control, t
important it is that the correct relations between operating variables be known and
used. The system must be capable of accepting instructions from such varied sources
as computers and human operators and must also be capable of changing its control
subsystem completely in a short time. Digital controllers are most suitable in such
situations. %nfact, an advantage of the digital controller is flexibility, that is, ease
of changing control schemes by reprogramming.
In the digital control of a complex process, the designer must have a good
knowledge of the process to be controlled and must be able to obtain its mathemat-
ical model. (The mathematical model rnay be obtained in terms of differential
equations or difference equations, or in some other form.) The designer must be
familiar with the measurement technology associated with the output of the process
and other variables involved in the process. e or she must have a good working
knowledge of digital computers as well as modern control theory. Iíf the process is
complicated, the designer must investigate severa1different approaches to the design
of the control system. In this respect, a good knowledge of simulation techniques
is helpful.
The objective of this book is to present a detailed
acco~intof the control theory that is relevant to the analysis an design of discrete-
time control systems. Our enphasis is on understanding the basicconceptc involved.

lntroduction to Discrete-TimeControl Ystems ChaP '1
In this book, digital controllers are often designed in the form of pulse transfer
functionsor equivalent difference equations,which can be easilyimplemented in the
form of computer programs.
The outline of the book is as follows. Chapter 1has presented introductory ma-
terial. Chapter 2presents the z transform theory. ter includes z transforms
of elementary functions, important properties an
inverse z transform, and the solution of differe
method. Chapter 3 treats background materials
systems. This chapter includes discussions of impulse sampling and reconstruction
of original signalsfrom sampled signals, pulse transfer functions, and realization of
digital controllers and digital filters.
Chapter 4 first presents mapping between the S plane and the z plane and then
discusses stability analysis of closed-loop systems in the z plane, followed by tran-
sient and steady-state response analyses, design by the root-locus and frequency-
response methods, and an analytical design method. Chapter 5 gives state-
space representation of discrete-time systems, the solution of discrete-time state-
space equations, and the pulse transfer function matrix. Then, discretization of
continuous-time state-space equations and Liapunov stability analysis are treated.
Chapter 6 presents control systems design in the state s
chapter with a detailed presentation of controllability and obs
present design techniques based on pole placernent, follow
full-order state observen and minimum-order state observe
chapter with the design of servo systems. Chapter 7 treats the
approach to the design of control systems. e b%in the cha~terwith diX~ssionsof
Diophantine equations. Then we present the design of regulator systems and control
systems using the solution of Diophantine equations. The approach here is an
alternative to the pole-placement approach ombined with minimum-order observ-
ers. The design of model-matching control systems is included in this chapter.
Finally, Chapter 8 treats quadratic optimal control problems in detail.
The state-space analysisand design of discrete-time control systems, presented
in Chapters 5,6, and 8,make extensive use of vectors and rices-In studying h % e
chapters the reader may, as need arises, refer to App which ~ummarizecthe
basic materials of vector-matrix analysis. Appendix ts materiak in z tEUls-
form theory not included in Chapter 2. Appendix C treats pole-placement design
problems when the control is a vector quantity.
In each chapter, except Chapter 1,the main text isfollowedby solvedproblems
and unsolved problems. The reader should study al1 solved probl
Solved problems are an integral part of the text. Appendixes A,
followedby solvedproblems. The reader who studiesthese solvedproblems willhave
an increased understanding of the material presented.
atical tool commonly used for the analysis an esis of discrete-time
systems is similar to that of the Lapla form in continuous-time systems.
tions to linear difference equations become algebraic in nature. (Just as the Laplace
transformation transforrns linear time-invariant differential equations into algebraic
equations in S , the z transformation transforms linear time-invariant difference
equations into algebraic equations in z .)
The main objective of this chapter is to resent definitionsof the z transform,
basic theorems associated with the z transform, and methods for finding the inverse
z transform. Solving difference equations by the z transform method is also
cussed.
Signak, Diserete-t signals arise if the system involves a
sampling of continuous-time als. The samplled signal is x(O),x(T),
x(2T), ...,where T is the sampling period. Such a sequence of values arising from
the sampling operation is usually written asx(kT). If the system involvesan iterative
process carried out by a digital computer, the signal involved is a nu
x(O),x(l), x(2). . .. The sequence of numbers is usualhy written as x(k), where tbe
argument k indicates the order in which the number occurs in the sequence, for
example, x(O),x(l),42). .. . Although x(k) is a number sequence, it can be con-
sidered as a sampled signal of x(t) when t e sampling period T is 1sec.

The z Transform
The 2 transform applies to the continuous-time signal x(t), sampled signal
x(kT), and the number sequence x(k). dealing with the z transform~if no
confusion occurs in the discussion, we occasionallyuse x(kT) and x(k) interchange-
ably. [That is, to simplify the presentation, we occasionallydrop the explicit appear-
ance of T and write x(kT) as x(k).]
ection 2-1. has presented introductory remarks.
ection 2-2 presents the definition of the z transform and associated subjects.
Section 2-3 gives z transforms of elementary functions. mportant properties
theorems of the z transform are presented in Section 2-4.
computational methods for finding the inverse z transform are discussed in
2-5. Section 2-6 presents the solution of difference equations by the z transform
method. Finally, Section 2-7 gives concluding commeats.
The z transform method is an operational method that is very
working with discrete-time systems. In what followswe shall define
of a time function or a nurnber sequence.
In considering the z transform of a time function x(t), we consider
sampled values o£x(t), that is,x(O),x(T), x(2T), ...,where Tisthe samplin
The z transform of a time function x(t),where t isnonnegative, os o£a
of valuesx(kT), where k takes zero or positive integers and Tis the sampli
is defined by the following equation:
m
For a sequence of numbers x(k), the z transform is defined by
m
The z transforrn defined by Equation (2-1) or (2-2) is referred to as the one-sided
z transform.
The symbol7 denotes "the z transform of." In the one-sided z transform,
we assumer(t) = O fort < O orx(k) = Ofork < O. Note that z is a complexvariable.
Note that, when dealing with a time sequence x(kT) obtained by sampling a
time signal x(t), the z transform X(z) involves T explicitly. Owevery £m a Imnber
sequence x(k), the z transform X(z) does not involve T explicitly.
The z transform of x(t), where -.a < t < .a, or of x(k), where k takes integer
values (k = 0, +-1, t 2 , ), is defined by
m
X(z) = 27 [x(t)] = Z [x(kT)] =
k=-m
Sec. 2-3 z Transformc of Elementary Functions
The z transform defined by Equation (2-3) or (2-4) is referred to as the two-sided
z transform. In the two-sided z transfor e function x(t) is assumed to be
nonzero for t < O and the sequence x(k) is considered
h the one-sided and two-sided z transforms
involvesboth positive and negative powers of z
one-sided z transform is considered in detail.
For most engineering applications the one-sided z transform
venient closed-form solution in its r
an infinite series in z-', converges
radius of absolute convergente, in u ansform method for
time problems it is not necessary each cify the values of z over whichX(z)
ks convergent.
Notice that expansion of the right-hand side of Equation (2-1) gives
any continuous-timefunction x(t) may
zmkin this series indicates the position
inversisn integral method (see Section 2-5 for details.)
n the following we shall present z transfor S of several elementary functions. It is
noted that in one-sided z transform theory, in sampling a discontinuous function x(t),
we assume that the function is continuous from the right; that is, if discontinuity
occurs at t = O, then we assume that x(O) is equal to x(O+) rather than to the average
at the discontinuity, [x(O-) + x(O+)]/2.
Unit-StepFunc~on. ket us find the z transforrn of the unit-step function
S just noted, in sampling a unit-step function we assume that this function is
continuous from the right; that is, l(0) = l . Then, referring to Equation (2-l), we
have

The z Transform Chap. 2
Notice that the series converges if lzl > 1. fina"ir%the transform, the Variable
ator. It is not necessary to specify the region of z over which
suffices to know that such a region exists. The z transform
n ~ ( t )obtained in this way is valid throughout the z plane
except at poles of X(z).
t is noted that 1(k)as defined by
k = 0,1,2,.-.
is commonly called a unit-step seqraence.
on. Consider the unit-ramp function
Notice that
x(kT) = k T , k = 0,1,2,.
Figure 2-1 depicts the signal. The magnitudes of the sam
values are proportional iod T .The z transforrn of the unit-r
function can be written as
m m m
O T 2T 3T 4T t Figure 2-1 Sampled unit-ramp signai.
Sec. 2-3 z Transforms of Elementary Fcdnctions
Note that it is a function of the sampling period T
on ak. Let us obtain the z transform of x(k) as
where a is a constant. Referring to the definition of t e z mnsform given by
Equation (2-2), we obtain
m w
- z--
z - a
on. Let us find the z transform of
e-"', O S t
x(t) =
Since
x(kT) = e-"kT, k = 0,1,2,. . .
on. Consider the sinusoidal function
sin ot, O 5 t
x(t) =
10, t<O
Noting that
e;"' = cos ot +j sin ot
e-;"' = cos ot - j sin ot

Since the z transform of the exponential function is
7[e-"'] =
1
&
1- e-aTZ-l
we bave
- 1
X ( z ) = ;S[sinwt] = ;S -(elw' - eiw')]
l'j
- z-l sin wT
1- 2z-l cos oT + z - ~
-- z sinoT
z 2 - 2 z coswT+ 1
Obtain the z transform of the cosine function
coswt, O C l t
t < O
If we proceed in a manner similar to the way we treated the z transform o£the
sine function, we have
x ( ~ )= Y wt] = $Y[e1." + e-]" 1
- 1 - z-' COS wT
1 - 2.2-' cos wT + z-'
Obtain the z transform of
Whenever a function in S is given, one approach for finding the corresponding z
transform is to convert X(s) into x(t) and then find the z transform of x(t). Another
approach is to expand X(s) into partial fractions and use a z transform table to find the
z transforms of the expanded terms. Still other approaches will be discussed in
Cec. 2-3 z Transforms of Elementary Functions
The inverse Laplace transform of X e ) is
( t ) = 1- e O 5 t
Hence,
1 1
X(Z)= Z[l - e-'] = --
1- Z - l 1 - e-'Z-l
en&. Just as in working with t lace transformatio*, a table
transforrns of commonly encountered functions is very useful for solving problerns
in the field of discrete-time systems. Table 2-1 is such a table.

x(t) = 0, for t < 0.
x(1cT) = x(k) = 0, for lc < 0.
Unless otherwise noted, 1: = 0,1,2,3,. . . .
Cec. 2-4 lmportant Properties and Theorerns of the z Transform
The use of the z transform method in the analysis of discrete-time control systems
may be facilitated if theorems of the z transform are referred to. In this section we
present importantproperties and useful theor
the time function x is z-transformable al
where a is a constant.
To prove this, note thatm by definition
m
. The z transform possesses an Pmportant prop-
erty: linearity. This means that, if f(k) and g(k) are z-transformable and a and P
are scalars, then x(k) formed by a linear combination
x(k) = af(W + 13gW
where F(z) and G(z) are the z tr
The linearity property can be
X(z) = Z[x(k)] = Z[af(k) + pg(k)]
This can be proved as follows:m
m
. The shifting theorern presented here is also ieferred to as
on theorem. If x(t) = O for t < O and x(t) as Wz),

and
where n is zero or a positive integer.
To prove Equation (2-7), note that
m
By defining m = k - n, Equation (2-9) can be written as follows:
m
Sincex(mT) = O for m < O, we may change the lower limit of the summation from
m = -n to m = 0.
Thus, multiplication o£ a z transform by z-" has t e 0f delaying the time
function x(t) by time nT. (That is, move the funct lo the right by time nT.)
To p;ove-~quation(2-8)? we note that
-
For the number sequence x(k), Equation (2-8) can be written as follows:
From this last equation, we obtain
7[ ~ ( k+ l)]= zX(z) - zx(Q) (2-1 1)
Z[r(k + 211 = z l[x(k + 111 - zx(1) = z2X(Z) - z2x(0) - zx(1) (2-12)
lrnportant Properties and Theorems of the z Trancform
Similarly,
Z[x(k + n)]
= znX(z) - znx(0) - zn-'~(1)- zn-24 2 ) - .- - - zx(n-1) (2-13)
where n is a positive integer.
emember that multiplication of the z transform
advancing the signal x(kT) by one step (1sarnpiing per
of the z transform X(z) by z-' has the effect of delaying the signalx(kT) by one step
(1sampling period).
Find the z transforms of unit-step functions that are delayed by 1 sampling period
and 4 sampling periods, respectively, as shown in Figure 2-2(a) and (b).
Using the shifting theorem given by Equation (2-7), we have
1 - z-'
r[i(t - TI] = Z-~Z[I(~)]= Z-l- 1 - *-1 - -1 - z-l
(Note that z-' represents a delay of 1sampling period T, regardless of the value of T.)
o 2T 3T 4T 5T 6T 7T 8T ' Faginre2-2 (a) UnitSstepfunction
delayed by 1sampling period; (b) unit-
step function delayed by 4 sampling
(b) periods.

The z Transform
Weferring to Equation (2-7), we have
Z [ x ( k - l)]= z-%(z)
The z transform of ak is
and so
1 - z-'
z [ f ( a ) ]= Z[ak-']= z-'- - -
1-az-' 1-az-'
where k = 1,2,3, .. . .
Consider the function y(k),which is a sum of functions x(h),where h = 0,1,2,
such that
k
where y ( k ) = O for k < O. Obtain the z transform of y (k).
First note that
Therefore,
Z [ y ( k )- y(k - l ) ]= Z [ x ( k ) ]
which yields
Chap. 2 Cec. 2-4 lmportant Properties and Theorems of the z Transform
where X ( z ) = 2' [x(k)].
If x(t) has the z transform X(z), then the z
transform of e-"'x(t)can be given by X(zeuT).This is known as the complex trans-
lation theorern.
Noting that
Z-' sin o T
Z [sin wt] =
1 - 22-' coswT +
we substitute zeaTfor z to obtain the z transform of e-"' sin wt, as follows:
e-aT -1
z sinwT
Z [e-"' sin wt] =
1 - 2e-aT~-'cos o T + e-2nT~-2
Similarly, for the cosine function, we have
1 - Z-' cos wT
2' [cos wt] =
1 - 22-' coswT + zW2
y substituting zeaTfor z in the z transform of cos wt, we obtain
1 - e-"Tz-' cos wT
Z [e-"' cos ot] =
1 - 2e-aT~-'C O S ~ T+ e-2aTz-2
Obtain the z transform of te-"'.
Notice that
Thus,
. If x(t)has the z transfor X ( z ) and if limX ( z )exists,
then the initial value x(O) of x(t) or x(k) is given by z-t m
x (O) = limX ( z )
2 4 m
is theorem, note that
ketting z --+m in this last equation, we obtain Equation (2-15). The be
signal in the neighborhood of t = O or k = O can thus be determined by the behavior
of X ( z ) at z = m.
The initial value theorem is convenient for checking z transform calculations
for possible errors. ince x(0) is usually known, a check of the initiaI value by
limX(z) can easily spot errors in X(z), if any exist.
z-tm
To prove this theorern, note that
x
"[e-"'x(t)] = 2~ ( k T ) e - " ~ ' z - ~= x(kT)(~e"')-~= X(zeaT) (2-14)
k=O k=O
Thus, we cee that replacing z in X ( z ) by zeaTgives the z transform of e-"'x(t).
Determine the initial value x(0) if the z transform of x(t) is given by
Examplie 2-6 By using the initial value theorem, we find
Given the z transforms of sin wt and cos wt, obtain the z transforms of e-"' sin wt and
e- "' cos wt , respectively, by using the complex translation theorem.
x(0) = iim
(1 - e-')z-'
z-m(l - z-l)(l - CTz-')= O

The z Transforrn Chap. 2
Referring to Example 2-2, notice that this X ( z ) was the z transform of
x(t) = 1 - e-'
and thus x(0) = 0, which agrees with the result obtained earlier.
ence,
Taking the limit as z approaches unity, we have
m 1
ecause of the assumed stability condition and t e condition that x(k) = O for
k < 0, the left-hand side of this iast equation becomes
m
[x(k) - x(k - l)] = [x(O) - x(- l)] + [X (1)- x(O)]
k=O
t-[x(2) - x(l)l + - e * = X ( W ) = limx(k)
k-+ m
ence,
limx(k) = lim [(l- z-1)X(z)]
k-*m 1-2 1
which is Equation (2-16). The finalvalue theorem is very useful in determining the
behavior of x(k) as k - + frorn its z transform X(z).
Determine the final value x(m) of
1
X ( z ) = ----- -
1
1 - Z-1 1 - e - a T Z - 1
a > O
by using the final value theorem.
By applying the final value theorem to the given X(z), we obtain
ec. 2-5 The Inverse z Transforrn
Pt is noted that the given X ( z ) is actually the z transform of
y substituting t = m in this equation, we have
X ( W ) = lim (1 - e-"') = 1
f-m
As a matter of course, the two resdts agree.
. lin this section we have resented important roperties and theo-
rems of the z transforrn that will e to be useful in solving many z transforrn
problerns. For the purpose of conv nt referente, these important properties and
theorems are sumrnarized in Table 2-2. y of the theoserns presented in this
table are discussed in this section. cussed here but included in the table
are derived or proved in Ap
The z transforrnation serves the same role for discrete-time control systemsthat the
Laplace transformation serves for continuous-ti control systems. For the z trans-
form to be useful, we must be familiar with thods for finding the inverse z
transform.
The notation for the inverse z transform is Z-l. The inverse z transform of
X(z) yields the corresponding time sequence x(k).
Irt should be noted that only the time sequence at the sarnpling instants is
obtained frorn the inverse z transform. Thus, the inverse z transform of X(z) yields
a unique x(k), but does not yield a unique x(t). This rneans that the inverse z
transform yields a time sequence that specifies the values of x(t) only at discrete
instants of time, t = O, T,2T,. . . ,and says not ing about the values of x(t) at al1
other times. That is, many different ti e functions x(t) can have the sarne x(kT).
X(z), the z transform of x(kT) or x(k), is given, the operation that
onding x(kT) or x(k) is called the inverse z transformation.
n obvious method for finding the inverse z transform is to refer to a z transform
extensive table of z tra
a sum of simpler z tr
presented in this section.)
Other than referring to z transform fables, four methods for obfaining the
inverse z transform are commonly available:

The z Transform ec. 2-5 The Inverse z Transform
Figure 2-3 Two different continuous-time functions, xl(t) and xZ(t),that have the
same values at t = O, T, 2T,. ...
Direct division method
Gomputational method
. Partial-fraction-expansion method
nversion integral rnethod
In obtaining the inverse z transform, we assume, as usual, that the time
sequence x(kT) or x(k) is zero for k < 0.
efore we present the four methods, however, a few comments on poles and
zeros of the pulse transfer function are in order.
h e . Hn engineering appiications of the z transform
method, X(z) may have the form
where the p,'s (i = 1,2,.. . ,n) are the poles of X(z) and the z,'s ( j = 1,2, .. . ,m)
the zeros of X(z).
The locations of the poles and zeros of X(z) determine the characteristics of
x(k), the sequence of values or numbers. S in the case of the plane analysis of
linear continuous-time control systems, we often use a graphical display in the z
plane of the locations of the poles and zeros of X(z).
Note that in control engineering and signal processing X(z) is frequently
expressed as a ratio of polynomials in z-', as follows:

where z-' is interpreted as the unit del
properties and theorems of the z tra
expressed in terms of powers of z, as given ,or in terms of powers
of z-', as given by Equation (2-18),
e poles and zeros of
als in z . For example,
as poles at z = -1 and z = -2 and zeros at z = O and z = -0.5.
If X(z) is written as a ratio of polynomials in z-', however, the prece
be written as
Although poles at z = -1and z = -2 and a zero at z = -0.5 are clearly seen from
e expression, a zero at z = O is not explicitlyshown, and so the beginner may fail
see the existence of a zero at z = O. Sherefore, in dealing with the pol
of X(z),it is preferable to express X(z) as a ratio of polynomials in z,
polynomials in z-l. In addition, in obtaining the inverse z transform by use o
inversion integral method, it is desirable t Tres§ as a ratio of ~ o l ~ n o m i a l ~
in z, rather than polynomials in z-', to av anY possible erKJrs determininb the
number of poles at the origin of function X(z)zk-l.
. In the direet division rnethod we obtain the inverse
z transform by expanding X(z) into an infinite power series in z-l. This method is
useful when it is difficult to obtain the closed-form expression for the inverse z
transform or it is desired to find only the first several terms of x(k).
The direct division rnethod stems from the fact that if X(z) is expanded into
a power series in z-', that is, if
then x(kT) or x(k) is the coefficient of the zwkterm. ence7 the vahes of 4 k T ) 01
x(k) for k = 0,1,2, . . . can be determined by inspectlon.
If X(z) is givenin the form of a rational function,the expansion into an infinite
power series in increasing powers of z-' can be accomplished by simply dividing the
numerator by the demminator, where both the numerator and denominato
are wrihten in increasing pdswers of z-l. lif the resulting series is conver
Sec. 2-5 The lnverse z Transform
coefficients of the zWkterm in the series are the values x(kT) of the time sequence
or the values of x(k) of the number sequence.
h the present rnethod
,x(l),x(2), ... in a se
an expression for the general term from a set of values of x(kT) or x(k).
Find x(k) for k = 0,1,2,3,4when X(z) is given by
First, rewrite X(z) as a ratio of polynomials in z-', as follows:
Dividing the numerator by the denominator, we have
18.68~-~- 22.416~-~+ 3 . 7 3 6 ~ ~ ~
Thus,
By comparing this infinite series expansion of X(z) with X(z) =
we obtain
x(0) = o
As seen from this example,the direct division method may be carried out by hand
calculations if only the first several terms of the sequence are desired. In general, the
method does not yield a closed-forrn expression for x(k), except in special cases.
Find x(k) when X(z) is given by
1 - z-lX(z) = -- -
z + l 1 + z - l
By dividing the numerator by the denominator, we obtain

The r Transform Chap. 2
By comparíng this infinite series expansion of X ( r ) with X ( z ) =
obtain
x(O) = O
x(1) = 1
x(2) = -1
4 3 ) = 3
n(4) = -1
This is an alternating signal of 1and -1, which starts from k = 1. Figure 2-4 shows
a plot of this signal.
Obtain the inverse z transform of
X ( z ) = 14- 22-1 + 3ze2+ 4t-3
The transform X ( z ) is aiready in the form of a power series in z
has a finite number of terrns, it corresponds to a signal of finite length.
we find
x(0) = 1
x(1) -2
4 2 ) = 3
x(3) = 4
Al1 other x(k) vaiues are zero.
En what follows, we present two corn
proaches to obtain the inverse z transform.
Difference equation approach
Consider a system G(z) defined by
n finding the inverse z transform, we utilize the Kronecker delta function So(kT),
wbere
-1 0 4 Figure 2-4 Alternating signal of 1and
-1 starting from k = 1.
Sec. 2-5 The Inverse z Transform
= O, for k f O
Assume that x(k), the input to t e system G(z), is the Kronecker delta
input, or
x(k) = 1, for k = O
= 0, for k f O
The z transform of the Kronecker delta input is
X(z) = 1
Using the Kronecker delta input, Equation (2-19) can be rewritten as
Approach. MATLA can be used for finding t
transforrn. Weferring to -20), the input X(z) is the z transform of the
Kronecker delta input. the Kronecker delta input is given by
where N corresponds to the end of the discrete-time duration of the process consid-
ered.
Since the z transforrn of the Kronecker elta input X(z) is equali to unity, the
response of the system to this input is
ce the inverse z transform of G(z) is given by y(O),y (l),y(2), .. ..Let us obtain
up to k = 40.
To obtain the inverse z transform of G(z) with
follows: Enter the numerator and denominator as follows:
num = [O
den = [l
Enter the Kronecker delta input.
X =
Then enter the command
Y =
to obtaán the response y(k) from k = O tcs k = 40.

e inverse z transforrn or the
response to the Kroneck rograrn 2-1.
If this program is executed, the screen will show the output y(k) from k = O to 40
as follows:
Y =
Columns 1 through 7
O 0.4673 0.3769 0.2690 0.1632 0.0725 0.0032
Columns 8 through 14
-0.0429 -0.0679 -0.0758 -0.0712 -0.0591 -0.0436 -0.0277
-0.01 37 -0.0027 0.0050 0.0094 0.01 11 0.01O8 0.0092
0.0070 0.0046 0.0025 0.0007 -0.0005 -0.001 3 -0.001 6
-0.001 6 -0.001 4 -0.001 1 -0.0008 -0.0004 -0.0002 0.0000
0.0002 0.0002 0.0002 0.0002 0.0002 0.0001
computations begin from column 1and end at column 41,
rather than from column O to column 40.) These values give the inverse z transforrn
of G(z). That is,
Y (0) = 0
y(1) = 0.4673
y(2) = 0.3769
-5 The lnverse z Transform
33 plot the values of t e inverse z transfor
given in the folliowing.
onding plot is skown in Figure 2-5.
Response to Kronecker Delta Input
re 2-5 Response of the system defined by Equation (2-20) to the Kronecker
delta input.

1f oints (open circles, o)
need to plot(k,y,'ol)to plot(k,y,'o',k,y,'-').
we can convert this equation into t
k) = O for k f O,
(O) and y(l) can be
k = -2 into Eqaiation (2-21), we fin$
Finding the inverse z transfor of Y(z) IXxV b3=2o
following difference equation for y(k):
the initial data y (O) = ) = O for k # O.
ation (2-22) can be solv ,FBRTRAN, or
other.
presented here, which is paralle
Laplace transformation, is widely used in
The method requires that all terms in
recognizable in the table sf z tr
To find the invesse z eran
Sec. 2-5 The Inverse r Transforrn
efore we discuss the partial-fraction-expansion method, we shall review the shifting
theorem. Consider fhe following X(z):
By writing zX(z) as Y ( z ) ,we obfain
1
zX(z) = Y ( z )= -1 - az-'
Referring to Table 2-1, the inverse z transforrn of Y ( z )can be obtained as follows:
T 1 [ Y ( z ) ]= y(k) = ak
Hence, the inverse z transform of X ( z ) = z-' Y ( z )is given by
2-'[X(z)] = x(k) = y(k - 1)
Since y(k) is assumed to be zero for al1 k < O, we have
Consider X(z) as given by
To expand X(z) into partial fractions, we first factor t
of X(z) and find the poles of X(z):
then expand X(z)/z into partial fractions so t at each term is easily recognizable
in a table o£ z transforms. If the shifting theorem is utilized in
transforms, however, X(z), instead of X(z)/z, may be expanded
tions. The inverse z transform of X(z) is obtained as the sum
transforms of the partial fractions.
A comrnonlyused procedure for the cas here allthepoles are o£simpleorder
and there is at least one zero at the origin ( t is, bm = O) is to divide both sides
o£X(z) by z and then expand X(z)/z into partial fractions. OnceX(z)/zis expanded,
it will be of the form
The coefficient aican be determined by multiplying both sides of this last equation
by z - p, and setting z = pi.Thiswill result in zero for allthe terms on the right-hand
side except ehe ai term, in which the multiplicative factor z - pihas been canceled
by the denominator.

The r Transform Chap. 2
Note that such determination of a, is valid only for sim
olves a multiple pole
n X(z)/z will have t
The coefficients cl and c2are determined from
Tt is noted that if X(z)/zinvolves a triple
must include a term (z +pJ(z - ~ 7 ~ ) ~ .(
Given the z transform
where a is a constant and T is the sampling period, determine the inverse z transform
x(kT) by use of the partial-fraction-expansion method.
The partial fraction expansion of X(z)/z is found to be
1X o = L - -
z z - 1 z - e-aT
Thus,
From Sable 2-1 we find
r 1
Hence, the inverse z transform of X ( z ) is
x(kT) = 1 - k = 0,1,2,. ..
Let us obtain the inverse z transform of
by use of the partial-fraction-expansion method.
We may expand X ( z ) into partial fractions as follows:
Sec. 2-5 The lnverse z Transform
Noting that the two poles involved in the quadratic term of this last equation are
complex conjugates, we rewrite X ( z ) as follows:
Since
1 - e-aTz-' cos wT
2' [e-"" TOS wkT] = 1 - 2eWaTz-'cos o T + e-2aTz-2
e - a T z - l
sin wTZ [e-"" sin wkT] = 1 - 2e-aT~-'cos o T + e-2aTz-2
by identifying e-"T = 1 and cos oT = $ in this case, we have wT = ~ 1 3and
sin wT = 1/3/2. Hence, we obtain
- -
and
Thus, we have
( k - 1 ) ~ 1 (k - 1 ) ~
x ( k ) = 4(lkP1)- 3(lk-') c o s + (1.-') sin--
3
Wewriting, we have
The first severa1values of x(k) are given by
Note that the inverse z transform of X ( z ) can also be obtained as follows:
Since

and
y-1
I z-l
krr
m 1 - z - 1 + z - 2 ]= ;sí(1.) sin-3
krr 4 ( k - l ) r r
- 2 ~ ?sin- + -sin-
3 v'3 3 '
k = l , 2 ,3,. .
x ( k ) =
k 6 0
Although this solution may look difierent from the one obtained earlier, both solutions
are correct and yield the sarne values for x(k).
Pn is a useful techniq
verse z al for the ztra~sfo
its center at the srigin of t
of complex variables. It c
m
[residue o£x(z)z"-' at pole z = ziaf (2-24)
i=l
(2-25)
Ef X(z)zk-' contains a mult
Sec. 2-5 The Inverse z Transform
partial-fraction-expansion method rnay prove lo be simpler to apply. On the other
hand, in certain problems the partial-fraction-expansion approach may become
laborious. Then, the inversion integral method proves to be very convenient.
Qbtain x(kT) by using the inversion integral method when X ( z ) is given by
Note that
For k = O, 1,2,. . .,X(z)zk-' has two simple poles, z = zl = 1 and z = z2 = e-OT.
Hence, from Equation (2-24), we have
x(k) =
--
Kl =
--
K2 =
--
Hence,
[residue of (1 - e-OT)zk at pole z = z,
1 = 1 ( Z - 1)(z - ePaT) 1
where
[residue at simple pole z = 11
[residue at simple pole z = e-OT]
Qbtain the inverse z transform of
by using the inversion integral rnethod.
Notice that
For k = 0,1,2, .. . ,X(z)zk-' has a simple pole at z = zl = e-"Tand a double pole at
z = z2 = 1. Hence, from Equation (2-24), we obtain
Z k t l
x ( k ) = [residue of at pole z = z,
I = I ( z - I ) ~ ( z- e-OT)
= Ki + KZ
1

where
The z Transform
M, = [residue at simple pole z = e-."?
.k+l
]=
,-a(k+l)T
(Z - ~ ) ~ ( z- e-"') (1 - e-aT)2
K2 = [residue at double pole z = 11
(k + l)zk(z - e-aT)- zk+l
= lim
z-. I ( z - ePaT)'
Difference equations can be solved easily by use o£a
nurnerical values of al1coefficients and parameters are
expressions for x(k) cannot be obtained fr com~uters0lLhX-l7 e*cePt forvepY
special cases. The usefulness of the z "can ethod 6t h t it enables tQ
the clased-form expressisn for x(k).
Consider the linear time-invariant discrete-time system characterized by the
following linear difference equation:
where u(k) and x(k) are the systern's input and output, respedively, at the kth
iteration. In describing such a difference equation in the z plane, we take the z
transform of each term in the equation.
Eet us define
q x(k)]= X(z)
Thenx(k + l),x(k + 2),x(k + 3),.. . andx(k - l),x(k - 2),x(k - 3),...can be
expressed in terms of X(z) and the initial conditions. Their exact z transforms were
derived in Section 2-4 and are surnrnarized in Table 2-3 for convenient referente.
Next we present two exarnple problems for solvingdifference equations by t
z transfcwm method.
-6 P Transform Method for
P TRANSFORWLS OF x(k + m) AND x(k - m)
Discrete function z Pansform
1
Soive the following difference equation by use of the z transform method:
x ( k + 2 ) + 3 x ( k + I ) + & ( k ) = O , x(O)=O, x ( l ) = l
First note that the z transforms of x(k + 2), x(k + l j , and x ( k ) are given,
respectively, by
Z [ x ( k + 2)] = z 2 X ( z )- z2x(0)- zx(1)
Z [ x ( k + 1)]= z X ( z ) - zx(0)
Z[x(k)]= X ( z )
Taking the z transforms of both sides of the given difference equation, we obtain
z2X(zj - z2x(0)- zx(1) + 3zX(z) - 3zx(O) + 2%(z) = O
Substituting the initial data and simplifying gives
1 1=---
1 + z-' 1 + 22-'
Noting that
Obtain the solution of the following difference equation in terms of x(0) and x(1):
x(k + 2) + (a + b)x(k + 1) + abx(k) = O
where a and b are constants and k = 0,1,2,. . . .

?'he z transform of this difference equation can be given by
[z2X(z)- z2r(0)- rx(l)]+ (a + b)[zX(z)- zx(O)] + abX(z) = O
Solving this last equation for X(z) gives
[z2+ (a + b)z]x(O)+ zx(1)
X ( z ) =
z2 + (a + b)z + ab
Notice that constants a and b are the negatives of the two roots of the characteristic
shall now consider separately two cases: (a) a # b and (
(a) For the case where a # b , expanding.X(z)/zPto partial fractions, we obtain
from which we get
The inverse z transform of X ( z ) @ves
bx(0) +x(1) ax(0) + x(1)
x ( k ) = (-a)" + (-b)k, a # b
b - a a - b
where k = 0,1,2, .. . .
) For the case where a = b , the z transform X(z) becomes
The inverse z transforrn of X ( z ) gives
x(k) = ~ ( 0 ) ( - a ) ~+ jax(0) +~ ( l ) ] k ( - a ) ~ - l , a = b
where k = 0,1,2,. . ..
In this chapter the basic theory of the z transform rnet een presenteda T'he
z transform serves the same purpose for linear time-i Saek-tirne sY5&Xls
asthe Laplace transform provides for linear time-invariant continuous-time systems.
method of analyzing data in discrete time results in difference
z transforrn rnethod, linear time-invariant difference equations
into algebraic equations. This facilitates the transient response
analysis of the digital control system. Also, the z transform e t k d all0VJsL E t* We
Chap. 2 Example Problemc and Solutions
conventional analysis and design techniques available to analog (continuous-time)
control systems, such as the root-locus technique. Frequenc
design can be carrie rting the z plane into
z-transforrned char
as the Sury stability
3 and 4.
Obtain the z transform of Gk,where G is an n x n constant matrix.
y definition, the z transform of G ~ S
m
h[ G ~ ]= E~ ~ z - ~
k=O
= 1 + GZ-' + G ~ ~ - " + 3 z - 3 + ...
Note that G~can be obtained by taking the inverse z transform of (
or (21 - G)-' Z . That is,
Obtain the z transform of k2
y definition, the t transform of k2is
m
Z [ k 2 ]= 2 k2z-k = Z-' + 4z-2 + 9z-3 + l6zp4+ e
k=O
= z-'(1 + z-l)(l + 32-' + 6z-2 + IOZ-~+ 1 5 ~ - ~+ )
ere we have used the closed-form ex
involved in the problem. (See Appendix
Obtain the z transforrn of kak-' by two methods.
Method l. By definition, the z transform of kak-' is given by
m

ethod 2. The summation expression for the z transform of kak-' can also be written
as follows:
-Ir
)7[kak-ll = 5kak-lz-k = a-1 k a k z k = 2 (a)k=O k=O ak=O
o~blemA-2-
Show that
and
Also show that
where 15 i 5 k - 1.
Solintaon Define
k
y(k) = 2x(h), k = O, 1,2,. . .
h =O
Then, clearly
Y ( k ) - Y (k - 1) = x ( k )
y writing the z transforms of x ( k ) and y(k) as X ( z ) and Y ( z ) ,respectively, and by
taking the z transform of this last equation, we have
Y ( z ) - z-1 Y ( z ) = X ( z )
Chap. 2 Exarnple Problerns and Solutions
By using the final value theorem, we find
k- m
2 ~ ( h )= Ex ( k ) = limX ( z )
h=O k=O 2-1
Next, to prove Equation (2-29), first define
k
j ( k ) = h=iEx(h) = x(i) + x(i i- 1) + + x ( k )
where 1 r i r k - 1. Define also
X ( z ) = x(i)zWi+ x(i + 1)z-'"" + .- + x(k)z7*+ - -
Then, noting that
we obtain
1-1
Since
the z transform of this last equation becomes
Y ( Z ) - z-' P(z) = Z ( z )
[Note that the z transform of x(k),which begins with k = i, is X(z),not X(z).]Thus,
1
B ( z ) = -1 - 2-1 [,(Z) - h = ~i ( h ) z h1-2-5
Obtain the z transform of the curve x(t)shawn in Figure 2-6. Assume that the sampling
period T is 1 sec.
ution From Figure 2-6 we obtain
x(0) = O
x(1) = 0.25

Then the z transform of x(k) can be given by
Notice that the curve x(t) can be written as
~ ( t )= $ t - g t - 4j1~t- 4)
where I(t - 4) is the unit-step function occurring at f = 4. Cince the sampling period
T = 1sec, the z transform of x(t) can also be obtained as follows:
X(z) = Z [x(t)] = Z [$f ] -7[a(t - 4)l(t - 4)]
Consider X(z), where
Obtain the inverse z transform of X(z).
Sgalntigan We shall expand X(z)/z into partial fractions as follows:
ap. 2 Example
Then
The inverse z transforms- of the individual- terms give
and therefore
~ ( k ) = 9 k ( 2 ~ - ' ) - 2 ~ + 3 , k = 0 , 1 , 2,...
olution Expanding X(z) into partiaf fractions, we obtain
(Note that in this example X(z) involves a double pole at z = O. Hence the partial
fraction expansion must include the terms ll(z2) and llz.] By referring to Table 2-1,
we find the inverse z transform of each term of this last equation. That is,
ence, the inverse z transform of X(z) can be given by
o - o - o = o , k = O
1 - o - 1 = 0 , k = 1
x(k) =
2 - 1 - 0 = 1 , k = 2
2k-'-O-O=2k-19 k = 3 , 4 , 5 , ...
Rewriting, we have
k = 0 , 1
/O9 k = 2x(k)= 1,
2k-1, k = 3 , 4 , 5,...
Toverify this result, the direct division method rnay be applied to this problem. Noting
that

The z Transforrn Chap. 2
we find
z-2
>
X(z) = (1 - z-1 >3
utionr The inverse z transform of ~ - ~ / ( 1- z-')~is not available from most z
transform tables. It ispossible, however, to write the givenX(z) as a sum of z transforms
that are commonly available in z transform tables. Since the denominator of X(z) is
(1 - z-')~and the z transform of k2is z-l(l + z-')!(1 - z-')~,let us rewrite X(z) as
from which we obtain the following partial fraction expansion:
The z transforms of the two terms on the rigkt-hand side of this last equation can be
found from Table 2-1. Thus,
It is noted that if the given X(z) is expanded into other partial fractions then the
inverse z transform rnay not be obtained.
As an aiternative approach, the inverse z transform of X(z) may be obtained by
use of the inversion integral method. First, note that
Hence, for k = O, 1,2, .. .,X(z)zk-' has a triple pole at z = 1.Referring to Equation
(2-24), we have
! zk
x(k) = residue of ----at triple pole z = 1
(Z - 113
Chap. 2 Exarnple Probierns and
=- l lirn? (z - 1)3-
(3 - q !z - ~ dzd2 ' (Z -zk113I
Using the inversion integral method, obtain the inverse z transform of
Sola~tion Note that
For k = 0, notice that X(z)zk-' becomes
Hence, for k = O, X(z)zk-' has t h ~ e esimple poles, z = zl = 1, z = z2 = 2, and z =
z3 = O. For k = 1,2,3, . .. ,however, X(z)zbl has only two simple poles, z = zl = 1
and z = z2= 2. Therefore, we must consider x(0) and x(k) (where k = 1,2,3,...)
separately.
For k = O. For this case, referring to Equation (2-24), we have
lo
at pole z = zi
(Z - 1)(z - 2)z 1
where
K1 = [residue at simple pole z = 11
= -10
(2 - 1)(z - 2)z
K2= [residue at simple pole z = 21
z-2 (z - 1)(z - 2)z
K3 = [residue at simple pole z = O]

The r Transform
ence,
x(0) = K1 + K2 + K3 = -10 + 5 + 5 = O
For k = 1,2,3, . . . . Eor this case, Equation (2-24) becomes
10zk--'
at pole z = zi
( z - 1)(2 - 2)
= Ki + K2
where
K1 = [residue at simple pole z = 11
K2 = [residue at simple poIe z = 21
Thus,
x(k)=K1+K2=-10+10(2k-1)=10(2k-1-1), k = 1 , 2 9 3 9
Mence, the inverse z transform of the given X ( z ) can be written
An alternative way to write x(k) for k 2 O is
x(k) = 560(k) + 10(2~-'- l ) , k = 0,1,2,...
where &(k) is the Kronecker delta function and is given by
for k = 0
"(w=jt: t o r k i o
'4-2-1
Obtain the inverce z transform of
Chap. 2
(2-30)
by use of the four methods presented in Section 2-5.
utiora
Method 1: Direcr division method. We first rewrite X ( z ) as a ratio of two polynomiak
in z-':
Dividing the numerator by the denominator, we get
X ( Z )= 1 + 4 ~ 4+ 7 ~ - 2+ ~ o z - ~+- . e .
ence,
x(0) = 1
Chap. 2 Example roblemc and Solutions
Method 2: Compuiational method (MATLAB approach). %(z) can be written as
Hence, the inverse z transform of X ( z ) can be obtained with
Define
num = [l 2 O]
d e n = [ l -2 11
If the values of x(k) for k = O,1,2, . . . ,30 are desired, then enter the Kronecker delta
input as follows:
Then enter the command
Program 2-3. [The screen will show the output x(k) from k = 0 té,
k = 30.1 (MATLAB computations begin from column 1 and end at column 31, rather
37 40 43 46 49 52 55 58 61 64 67 70

The z Transform Ckiap. 2
than from column O to column 30.) Thevalues x(k)givethe inverse z transform of X(z).
That is,
x(0) = 1
x(1) = 4
x(2) = 7
Method 3: Partial-fraction-expansion method. expand X ( z ) into the foliowing
partial fractions:
Then, noting that
we obtain
x(0) = 1
x ( k ) = 3 k + l , k = 1 , 2 , 3 ,...
which can be combined into one equation as follows:
x ( k ) = 3 k + l , k = 0 , 1 , 2,...
Note that if we expand X ( z ) into the following partia1 fractions
then the inverse z transform of X ( z ) becomes
x(0) = 1
which is the same as the result obtained by expanding X ( z ) into the other partial
fractions. [Remember that X ( z ) can be expanded into different partial fractions, but
the final result for the inverse z transform is the same.]
Method 4: Inversion integral rnethod. First, note that
Chap. 2 Exarnple Problerns and
For k = O, 1,2, ... ,X(z)zk-' has a double pole at z = P.
(2-24), we have
( 2 + 2)zk
residue of -----
(2 - 1)2
at double pole z = 1
Thus,
- -
x ( k ) = -----i i m - [ ( z - I ) 2 W ]
(2 - l ) !Z-i dz
d
= lim-[(zZ-+I dz + 2)zk]
= 3 k + 1 , k = 0 , 1 , 2,...
Solve the following difference equation:
2x(k) - 2x(k - 1) + x(k - 2) = u(k)
where x(k) = O for k < O and
y taking the z transform of the given difference equation,
1
2X(z) - 22-l X ( z ) + Z - ~ X ( Z )= -1 - z-'
Solving this last equation for X(z), we obtain
Expanding X ( z ) into partial fractions, we get
Notice that the two poles involved in the quadraticterm in this last equation are compiex
conjugates. Hence, we rewrite X ( z ) as follows:
y referring to the formulas for the z transforms of damped cosine and damped sine
functions, we identify e-2aT= 0.5 and cos oT = 11V5 for this problem. Hence, we get
wT = ~ 1 4 ,sin wT = l i d , and e-"= = l i d . Then the inverse z transform of X ( z )can
be written as
x ( k ) = 1 - Se-akTcos wkT + $ewakTsin wkT
from which we obtain

The z TTansform Chap. 2
Consider the difference equation
x(k + 2) - 1.3679x(k + 1) + 0.3679x(k) = 0.3679u(k + 1) + 0.2642u(k)
where x(k) is the output and x ( k ) = O for k 5 O and where u ( k ) is the input and is
given by
u ( k ) = O , k < O
u(0) = 1
u ( l ) = 0.2142
u(2) = -0.2142
u ( k ) = O , k = 3 , 4 , 5 ,...
Determine the output x(k).
olintlon Taking the z transform of the given difference equation, we obtain
[ z 2 X ( z )- z2x(0)- zx(1)l - 1.3679[zX(z)- zx(O)]+ 0.3679X(z)
= 0.3679[zU(z)- zu(0)l + 0.2642U(z) (2-31)
By substituting k = -1 into the given difference equation, we find
x(1) - 1.3679x(O) + 0.3673x(-1) = 0.3679u(O) + 0.2642u(-1)
Since x(0) = x(- 1) = O and since u(- 1) = O and u(0) = 1, we obtain
x(1) = 0.3679u(O) = 0.3679
By substituting the initial data
x(0) = O, x(1) = 0.3679, u(0) = 1
into Equation (2-31), we get
~ " ( 2 )- 0.36792 - 1.3679zX(z) + 0.3679X(z)
= 0.3679zU(z) - 0.36792 + 0.2642U(z)
Solving for X(z), we find
The z transforrn of the input u ( k ) is
U ( Z )= Z [ u ( k ) ]= 1 + 0.21422-' - 0 . 2 1 4 2 ~ - ~
Chap. 2 Example roblems and Solutions
ence,
Thus, the inverse z transform of X ( z ) gives
where x(0) = O and x(1) = 1. Note that x(2) = 1,x(3) = 2,x(4) = 3, .... The series
U, 1,1,2,3,5,8,13, ... is known as the Fibonacci series. Obtain the general solution
x ( k ) in a closed form. Show that the limiting value of x(k + l)/x(k)as k approaches
infinity is (1 + 1/5)/2, or approximately 1.6180.
olution By taking the z transform of this difference equation, we obtain
olving for X ( z ) gives
By substituting the initial data x(0) = O and x(1) = 1into this last equation, we have
The inverse 2 transform of X ( z ) is
Note that although this lact equation involves '~6the square roots in the right-hand side
of this last equation cancel out, and the values of x ( k ) for k = 0,1,2, . .. turn out to
be positive integers.

The z Trancform Chap. 2
The limitingvalue of x(k + l)lx(k)as k approachesinfinity isobtained asfollows:
Since 1(1 -
x(k i1)
lim -
k-+m ~ ( k )
Hence,
( 1 + v5""
x(k + ') 1 -
- 1.6180lim -= lim
k - w x(k) 2
Referring to Problem A-2-13, write a ATLAB program to generate the Fibonacci
series. Carry out the Fibonacci series to k = 30.
Solistiora The z transform of the difference equation
is given by
Solving this equation for X ( z ) and substituting the initial data x(0) = O and x(1) = 1,
we get
The inverse ztransform of X ( z ) will give the Fibonacci series.
To get the inverse z transform of X(z),obtain the response of this system to the
Kronecker delta input. MATLAB Program 2-4 wilf.yield the Fibonacci series.
% ***** The Fibona
% response of X(z) t
O/O x(Z) = 2/(2"2 - z - 1) *****
num = [O 1 01;
den = [l -1 -41;
u = [l zeros(1,3O)l;
x = fiIter(num,den,d
Chap. 2 Example Problemc and Solutionc
The filtered output y shown next gives the Fibonacci series.
X =
Columns 1 through 6
o 1 1 2 3 5
8 13 21 34 55 89
144 233 377 61O 987 1597
2584 4181 6765 10946 17711 28657
4301umns 25 through 30
46368 75025 121393 196418 317811 514229
, Column 31
1 832040
Note that column 1 corresponds to k = O and column 31 corresponds to k = 30. The
Fibonacci series is given by
x(0) = 0
x(29) = 514,229
x(30) = 832,040
x(k + 2) + m ( k + 1) + ,4?x(k)= O (2-32)

Figure 2-7 Region in the cup plane in
which the solution series of Equation
(2-32), subjected to initial conditions, is
finite,
Find the conditions on a and P for which the solution series x ( k ) for k = 0,1,2, ...,
subjected to initial conditions, is finite.
Then, referring to Example 2-19, the solution x ( k )for k = 0,1,2, ... can be given by
The solution series x(k) for k = 0,1,2,. . . , subjected to initial conditions x(O) and
x(l),is finite if the absolute values o£a and b are less than unity. Thus, on the ap plane,
three critica1points can be located:
The interior of the region bounded by lines connecting these points satisfies the con-
dition /a/< 1,lb1 < 1. The boundary lines can be given by ,6 = 1, a - P = 1, and
a! -t- p = -1. See Figure 2-7. If point ( a ,p) lies inside the shaded triangular region,
then the solution seriesx ( k )for k = 0,1,2, ...,subjected to initial conditions x(0) and
x(l), is finite.
-2
Obtain the z transform of k3.
Obtain the z transform of t2e-"'.
Obtain the z transform of the following x(k):
x ( k ) = 9k(2k-1)- 2k + 3, k = 0,1,2,. .
Assume that x ( k ) = O for k < 0.
-5
Find the z transform of
-2-
Show that
-2-7
Qbtain the z transform of the curve x(t) shown in Figure 2-8.
-2-
Obtaln the inverse z transform of

Find the inverse z transfornl of
Use (1)the partial-fraction-expansion method and (2)the
a MATLAB program for finding x(k), the inverse z transform of X(z).
Given the z transform
z-'
X ( z ) =
(1 - z-')(l + 1.32-' + O.4zw2)
determine the initial and final values of x(k). Also find x(k),the inverse z transform
of X(z), in a closed form.
1
Use (1)the invewion integral method and (2)the
-12
z - ~
X ( z ) =
(1 - 2-l)(1 - 0.22-l)
in a closed form.
By using the inversion integral method, obtain the inverse z transform of
-2-1
Find the inverse z transform of
Use (1)the direct division method and (2) the MATLAB method.
by use of the inversion integral method.
Chap. roblerns
-1
Find the solution of the following difference equation:
x(k + 2) - 1.3x(k + 1) + 0.4x(k) = u(k)
where x(0) = x ( l ) = O and x ( k ) = O for k < O. For the input function u(k),consider
the following two cases:
and
u(0) = 1
u(k) = O, k # O
Solve this problem both analytically and computationally with MATLA
-2-17
Solve the following difference equation:
x(k + 2) - x ( k + 1) + 0.25x(k) = u(k + 2)
where x(0) = 1and x ( l ) = 2. The input function u(k) ic given by
u ( k ) = 1 , k = 0 , 1 , 2,...
Solve thic problem both anaiytically and computationally with
Consider the difference equation:
x(k + 2) - 1.3679x(k + 1) + 0.3679x(k) = 0.3679u(k + 1) + 0.2642u(k)
where x ( k ) = O for k 5 O. The input u(k) is given by
u ( k ) = O , k < 0
u(0) = 1.5820
u(1) = -0.5820
u(k)=O, k = 2 , 3 , 4,...
Determine the output x(k). Solve this problem both analytically and computationally
with MATLAB.

The z transform method is particularly useful for an
input-single-output lineartime-invariantdiscrete-time
presents background material necessary for the analysis and desig
control systemsin the zplane. The main advantage of the z transf
it enables the engineer to apply conventional continuous-time design methods to
discrete-time systems that may be partly discrete time and partl
Throughout this book we assume that the sampling operation is uniform; that
is, only one sampling rate exists in the system and the sampling period is constant.
If a discrete-time control system involves two or more samplers in the system, we
assume that al1 samplers are synchronized and have the same sarnpling rate or
sarnpling frequency.
er. The outline of this chapter is as follows. Section 3-1
gives int ks. Section 3-2 presents a rnethod to treat the sampling
operation as a mathernatical representation of the operation of taking samplesx(kT)
from a continuous-time signal x(t) by impulse modulation. This section includes
derivations of the transfer functions of the zero-order hold and first-order hold.
Section 3-3 deals with the convolution integral method for obtaining the z
transform. Reconstructing the original continu time signal from the sampled
signal is the main subject matter of Sectlon 3-4. ed on the fact that the Laplace
transform of the sampled slgnal is periodic, resent the sampling the~rem.
ection 3-5 discusses the pulse transfer functio thematical modeling of digital
controllers in terms of pulse transfer functions is discussed. Section 3-6 treats the
realization of digital controllers and digital filters.
ec. 3-2 impulse ampling and Data
Discrete-time control systems may opera in discrete time and partly in
continuous time. Thus, in such control syst signals appear as discr
functions (often in the form of a sequence o or a nurnerical code) a
signals as continuous-time functions. In analyzing discrete-ti
plays an important role. To see why the z transform method is
shall consider a fictitious S er cornmonlycalled an
impu f this sarnpler is consider
that begins with t = O, with the sampling period equal to Tan
pulse equal to the sampled value of the continuous-ti
pictorial diagrarn of the i
sented by an arrow with an
Tke impulse-sampled output is a sequence of impulses, with the st~engthof
ulse equal to the magnitude of x(t) at the ding instant of time.
kT, the impulse is x(kT)6(t - e that 6(t - kT) = O
hall use the notation x"(t) to r
signalx*(t), a train of impulses,
infinite summation
m
x*(t) = x(kT)S(t - kT)
k=O
or
shall define a train of unit impulses as ST(t),or
k=O
The sampler output is equal to the product of t e continuous-time input x(t) and the
train of unit impulses OT(t).Consequently, the sampler may be considered a rnodu-
lator with the input x(t) as the modulating signal and the train of unit impulses &(t)
as the carrier, as shown in Figure 3-2.
x * ( t )
X(S) 6r x"(s) Figure 3-1 Impulse sampler.

z-Plane Analycis of Discrete-TimeControl Cysterns Chap. 3
Carrier
signal 1
1
Figure 3-2 Impulse sampler as a modulator.
Next, consider the Eaplace transform of Equation (3-1):
X"(s) = 2[x"'(t)] = x(O)%[6(i)] + x(T)Ce[G(t - T)]
+x(2T>ce[8(t- 2T)] +
= x(O) + x ( T ) ~ - ~ '+ ~ ( 2 T ) e - ~ ~ '+ .
m
= ~ ( k T ) e - ~ "
k=O
Notice that if we define
eTs -- z
then Equation (3-2) becornes
I m
The right-hand side of Equation (3-3) ic exactly the same as the right-hand side of
t is the z transform o£the seque e x(O),x(T),x(2T), . .. ,gener-
t = kT, where k = 0,1,2,. . .. ence we may write
and Equation (3-3) becomes
Note that the variable z is a cornplexvariable and Tis the sarnplingperiod. [ht should
be stressed that the notation X(z) does not signify X(s) with s replaced by z, but
ratber X' (S = T.-' Ira z).]
Cec. 3-2 Impulse Sarnpling and Data Hold
. Let us summarizewhat we have just
signal x(t) is impulse sampled in a periodic manner,
signal may be represented by
m
x*(t) = 2x(t)s(t - kT)
k=O
converts a continuous-ti S occurring at the sampling
instants t = 0, T,2T,. .., od. (Note that between any
no information. Two signals
sampled signal.)
Data-hold is a process of generating a continuous-time signal h(t) from a
discrete-time sequence x(kT). A hold circuit converts the sampled cignal into a
continuous-time signal, which approximately reproduces the signaf applied to the
sampler. %hesignalh(t) during the time interval kT 5 t < (k + 1)Tmaybe approx-
imated by a polynomial in r as follows:
where O 5 T < T. Note that signal h(kT) must equal x(kT), or
ence, Equation (3-5) can be written as follows:
Hf the data-hold circuit is an nth-order polynomial extrapolator, it is called an
nth-order hold. Thus, if n = 1, it is called a first-order hold. [The nth-order hold
uses the past n + 1discrete data x((k - n)T),x((k - n + 1)T),. ..,x(kT) to gen-
erate a signal h(kT t r).]
ecause a higher-order hold uses past sarnples to extrapolate a continuous-
nal between the present sampling instant and the next sampling instant, the
accuracy of approximating the ntinuous-time signal improves as the number of
past samples used is increased. wever, this better accuracy is obtained at the cost
of a greater time delay. Hn closed-loop cont ystems, any added time delay in the
loopwill decrease the stability of the system in some casesrnay even cause system
instability.
The simplest data-hold is obtained when n = O in Equation (3-6), that is, when

Discrete time control systems

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