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- 1. 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 Prentice-HallInternationaI (UK) Limited, London Prentice-Hall of Australia Pty. Limited, Sydney Prentice-Hall CanadaInc., Toronto Prentice-HallHispanoamericana,S.A., Mexico Prentice-Hall of India Private Limited, New Delhi Prentice-Hall of Japan, linc., Tokyo Prentbce-HallAsia Re. Ltd., Siizgapore Editora Prentice-Ealldo Brasil, Etda., Rio de Sanerio 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
- 2. ontents 3-1 INTRODUCTION, 74 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 EXAMPLE PROBLEMS AND SOLUTIONS, 138 PROBLEMS, 166 4-1 INTRODUCTION, 173 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 6-1 INTRODUCTION, 377 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 EXAMPLE PROBLEMS AND SOLUTIONS, 474 PROBLEMS,510 7-1 INTRODUCTION, 517 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 EXAMPLE PROBLEMS AND SOLUTIONS, 540 PROBLEMS,562 8-1 INTRODUCTION, 566 8-2 QUADRATIC OPTIMAL CONTROL, 569 8-3 STEADY-STATE QUADRATIC OPTIMAL CONTROL, 587 8-4 QUADRATIC OPTIMAL CONTROL OF A SERVO SYSTEM, 596 EXAMPLE PROBLEMS AND SOLUTIONS, 609 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 EXAMPLE PROBLEMS AND SOLUTIONS, 666
- 3. 8-1 INTRODUCTION,681 8-2 USEFULTHEOREMS OF THE z TRANSFORMTHEORY, 681 8-3 INVERSEz TRANSFORMATION AND INVERSIONINTEGRALMETHOD, 686 8-4 MODIFIEDzTRANSFORM METHOD, 691 EXAMPLE PROBLEMS AND SOLUTIONS, 697 C-9 INTRODUCTION, 304 C-2 PRELIMINARY DISCUSSIONS, 704 C-3 POLE PLACEMENT DESIGN, 707 EXAMPLE PROBLEMS AND SOLUTIONS, 718
- 4. 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.
- 5. 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).
- 6. 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
- 7. 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.
- 8. 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).
- 9. 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
- 10. 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.
- 11. 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
- 12. 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.
- 13. 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
- 14. 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
- 15. 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.
- 16. 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.
- 17. 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
- 18. 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
- 19. The z Transform Chap. 2 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.
- 20. The z Transform Chap. 2 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),
- 21. and The z Transform Chap. 2 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.) Obtain the z transform of 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.
- 22. 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
- 23. 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:
- 24. 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:
- 25. The z Transform Chap. 2 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
- 26. 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.
- 27. The z Transform Chap. 2 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 Columns 15 through 21 -0.01 37 -0.0027 0.0050 0.0094 0.01 11 0.01O8 0.0092 Columns 22 through 28 0.0070 0.0046 0.0025 0.0007 -0.0005 -0.001 3 -0.001 6 Columns 29 through 35 -0.001 6 -0.001 4 -0.001 1 -0.0008 -0.0004 -0.0002 0.0000 Columns 36 through 41 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.
- 28. 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.
- 29. 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
- 30. 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
- 31. 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,. . . .
- 32. The z Transform Chap. 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
- 33. 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
- 34. The z Transform Chap. 2 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,... Obtain the inverse z transform of 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
- 35. The z Transforrn Chap. 2 we find Obtain the inverse z transform of 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]
- 36. 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 Columns 13 through 24 37 40 43 46 49 52 55 58 61 64 67 70 Columns 25 through 31
- 37. 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
- 38. 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 Consider the difference equation 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.
- 39. 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 Columns 7 through 12 8 13 21 34 55 89 Columns 13 through 18 144 233 377 61O 987 1597 Columns 19 through 24 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 Consider the difference equation x(k + 2) + m ( k + 1) + ,4?x(k)= O (2-32)
- 40. The z Transform Chap. 2 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. Obtain the z transform of where a is a constant. -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 where a is a constant. -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
- 41. The z Transform Chap. 2 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 Obtain the inverse z transform of Use (1)the invewion integral method and (2)the -12 Obtain the inverse z transform of 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. Obtain the inverse z transform of 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.
- 42. 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.
- 43. 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