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  1. 1. System Dynamics Fourth Edition Katsuhiko Ogata University ofMinnesota PEARSON -------Pnmticc HidI Upper Saddle River, NJ 07458
  2. 2. Library of Congress Cataloging-in-PubUcation Data on me. Vice President and Editorial Director, ECS: Marcia J. Horton Acquisitions Editor: Laura Fischer Vice President and Director of Production and Manufacturing, ESM: David W. Riccardi Executive Managing Editor: Vince O'Brien Managing Editor: David A. George Production Editor: Scott Disanno Director of Creative Services: Paul Belfanti Creative Director: Jayne Conte Art Editor: Greg Dulles Manufacturing Manager: Trudy Pisciotti Manufacturing Buyer: Lisa McDowell Marketing Manager: Holly Stark © 2004, 1998, 1992, 1978 Pearson Education, Inc. Pearson Prentice Hall Pearson Education, Inc. Upper Saddle River, 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 publisher. Pearson Prentice Hall® is a trademark of Pearson Education, Inc. MATLAB is a registered trademark ofThe MathWorks, Inc., 3 Apple Hill Drive, Natick, MA 01760-2098. The author and publisher of this book have used their best efforts in preparing this book.These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind. expressed or implied, with regard to these programs or the documenta- tion contained in this book. The author and publisher shall not be liable in any event for incidental or consequen- tial damages in connection with, or arising out of. the furnishing, performance, or use of these programs. Printed in the United States ofAmerica 109 ISBN D-13-1424b2-9 Pearson Education Ltd., London Pearson Education Australia Pty. Ltd.• Sydney Pearson Education Singapore, Pte. Ltd. Pearson Education North Asia Ltd., Hong Kong Pearson Education Canada Inc., Toronto Pearson Educati6n de Mexico, S.A. de c.v. Pearson Education-Japan, Tokyo Pearson Education Malaysia, Pte. Ltd. Pearson Education, Inc., Upper Saddle River, New Jersey
  3. 3. Contents PREFACE 1 INTRODUCTION TO SYSTEM DYNAMICS 1-1 Introduction 1 1-2 Mathematical Modeling of Dynamic Systems 3 1-3 Analysis and Design of Dynamic Systems 5 1-4 Summary 6 2 THE LAPLACE TRANSFORM 2-1 Introduction 8 2-2 Complex Numbers, Complex Variables, and Complex Functions 8 2-3 Laplace Transformation 14 2-4 Inverse Laplace Transformation 29 2-5 Solving Linear,TIlDe-Invariant Differential Equations 34 Example Problems and Solutions 36 Problems 49 3 MECHANICAL SYSTEMS 3-1 Introduction 53 3-2 Mechanical Elements 57 3-3 Mathematical Modeling of Simple Mechanical Systems 61 3-4 Work, Energy, and Power 73 Example Problems and Solutions 81 Problems 100 vii 1 8 53 iii
  4. 4. iv Contents 4 TRANSFER-FUNCTION APPROACH TO MODELING DYNAMIC SYSTEMS 4-1 Introduction 106 4-2 Block Diagrams 109 4-3 Partial-Fraction Expansion with MATLAB 112 4-4 Transient-Response Analysis with MATLAB 119 Example Problems and Solutions 135 Problems 162 5 STATE-SPACE APPROACH TO MODELING DYNAMIC SYSTEMS 5-1 Introduction 169 5-2 Transient-Response Analysis of Systems in State-Space Form with MATLAB 174 5-3 State-Space Modeling of Systems with No Input Derivatives 181 5-4 State-Space Modeling of Systems with Input Derivatives 187 5-5 "fransformation of Mathematical Models with MATLAB 202 Example Problems and Solutions 209 Problems 239 6 ELECTRICAL SYSTEMS AND ELECTROMECHANICAL 106 169 SYSTEMS 251 6-1 Introduction 251 6-2 Fundamentals of Electrical Circuits 254 6-3 Mathematical Modeling of Electrical Systems 261 6-4 Analogous Systems 270 6-5 Mathematical Modeling of Electromechanical Systems 274 6-6 Mathematical Modeling of Operational-Amplifier Systems 281 Example Problems and Solutions 288 Problems 312 7 FLUID SYSTEMS AND THERMAL SYSTEMS 323 7-1 Introduction 323 7-2 Mathematical Modeling of Liquid-Level Systems 324 7-3 Mathematical Modeling of Pneumatic Systems 332 7-4 Linearization of Nonlinear Systems 337 7-5 Mathematical Modeling of Hydraulic Systems 340 7-6 Mathematical Modeling ofThermal Systems 348 Example Problems and Solutions 352 Problems 375
  5. 5. Contents v B TIME-DOMAIN ANALYSIS OF DYNAMIC SYSTEMS 383 8-1 Introduction 383 8-2 Transient-Response Analysis of First-Order Systems 384 8-3 Transient-Response Analysis of Second-Order Systems 388 8-4 Transient-Response Analysis of Higher Order Systems 399 8-5 Solution of the State Equation 400 Example Problems and Solutions 409 Problems 424 9 FREQUENCY-DOMAIN ANALYSIS OF DYNAMIC SYSTEMS 9-1 Introduction 431 9-2 Sinusoidal Transfer Function 432 9-3 Vibrations in Rotating Mechanical Systems 438 9-4 Vibration Isolation 441 9-5 Dynamic Vibration Absorbers 447 9-6 Free Vibrations in Multi-Degrees-of-Freedom Systems 453 Example Problems and Solutions 458 Problems 484 10 TIME-DOMAIN ANALYSIS AND DESIGN 431 OF CONTROL SYSTEMS 491 10-1 Introduction 491 10-2 Block Diagrams and Their Simplification 494 10-3 Automatic Controllers 501 10-4 Thansient-Response Analysis 506 10-5 Thansient-Response Specifications 513 10-6 Improving Transient-Response and Steady-State Characteristics 522 10-7 Stability Analysis 538 10-8 Root-Locus Analysis 545 10-9 Root-Locus Plots with MATLAB 562 10-10 Thning Rules for PID Controllers 566 Example Problems and Solutions 576 Problems 600 11 FREQUENCY-DOMAIN ANALYSIS AND DESIGN OF CONTROL SYSTEMS 60B 11-1 Introduction 608 11-2 Bode Diagram Representation of the Frequency Response 609 11-3 Plotting Bode Diagrams with MATLAB 629 11-4 Nyquist Plots and the Nyquist Stability Criterion 630
  6. 6. 11-5 Drawing Nyquist Plots with MATLAB 640 11-6 Design of Control Systems in the Frequency Domain 643 Example Problems and Solutions 668 Problems 690 APPENDIX A SYSTEMS OF UNITS APPENDIXB CONVERSION TABLES APPENDIXC VECTOR-MATRIX ALGEBRA APPENDIXD INTRODUCTION TO MATLAB REFERENCES INDEX 695 700 705 720 757 759
  7. 7. Preface A course in system dynamics that deals with mathematical modeling and response analyses of dynamic systems is required in most mechanical and other engineering curricula. This book is written as a textbook for such a course. It is written at the junior level and presents a comprehensive treatment of modeling and analyses of dynamic systems and an introduction to control systems. Prerequisites for studying this book are first courses in linear algebra, intro- ductory differential equations, introductory vector-matrix analysis, mechanics, cir- cuit analysis, and thermodynamics.Thermodynamics may be studied simultaneously. Main revisions made in this edition are to shift the state space approach to modeling dynamic systems to Chapter 5, right next to the transfer function approach to modeling dynamic systems, and to add numerous examples for modeling and response analyses of dynamic systems.All plottings of response curves are done with MATLAB. Detailed MATLAB programs are provided for MATLAB works pre- sented in this book. This text is organized into 11 chapters and four appendixes. Chapter 1 presents an introduction to system dynamics. Chapter 2 deals with Laplace transforms of commonly encountered time functions and some theorems on Laplace transform that are useful in analyzing dynamic systems. Chapter 3 discusses details of mechan- ical elements and simple mechanical systems.This chapter includes introductory dis- cussions of work, energy, and power. Chapter 4 discusses the transfer function approach to modeling dynamic sys- tems. 'lransient responses of various mechanical systems are studied and MATLAB is used to obtain response curves. Chapter 5 presents state space modeling of dynam- ic systems. Numerous examples are considered. Responses of systems in the state space form are discussed in detail and response curves are obtained with MATLAB. Chapter 6 treats electrical systems and electromechanical systems. Here we included mechanical-electrical analogies and operational amplifier systems. Chapter 7 vii
  8. 8. viii Preface deals with mathematical modeling of fluid systems (such as liquid-level systems, pneumatic systems, and hydraulic systems) and thermal systems. A linearization technique for nonlinear systems is presented in this chapter. Chapter 8 deals with the time-domain analysis of dynamic systems. Transient- response analysis of first-order systems, second-order systems, and higher order sys- tems is discussed in detail. This chapter includes analytical solutions of state-space equations. Chapter 9 treats the frequency-domain analysis of dynamic systems. We first present the sinusoidal transfer function, followed by vibration analysis of mechanical systems and discussions on dynamic vibration absorbers. Then we dis- cuss modes of vibration in two or more degrees-of-freedom systems. Chapter 10 presents the analysis and design of control systems in the time domain. After giving introductory materials on control systems, this chapter discusses transient-response analysis of control systems, followed by stability analysis, root-locus analysis, and design of control systems. Fmally, we conclude this chapter by giving tun- ing rules for PID controllers. Chapter 11 treats the analysis and design of control sys- tems in the frequency domain. Bode diagrams, Nyquist plots, and the Nyquist stability criterion are discussed in detail. Several design problems using Bode diagrams are treated in detail. MATLAB is used to obtain Bode diagrams and Nyquist plots. Appendix A summarizes systems of units used in engineering analyses. Appendix B provides useful conversion tables. Appendix C reviews briefly a basic vector-matrix algebra. Appendix D gives introductory materials on MATLAB. If the reader has no prior experience with MATLAB, it is recommended that he/she study Appendix D before attempting to write MATLAB programs. Throughout the book, examples are presented at strategic points so that the reader will have a better understanding of the subject matter discussed. In addition, a number of solved problems (A problems) are provided at the end of each chapter, except Chapter 1. These problems constitute an integral part of the text. It is sug- gested that the reader study all these problems carefully to obtain a deeper under- standing of the topics discussed. Many unsolved problems (B problems) are also provided for use as homework or quiz problems. An instructor using this text for hislher system dynamics course may obtain a complete solutions manual for B prob- lems from the publisher. Most of the materials presented in this book have been class tested in courses in the field of system dynamics and control systems in the Department of Mechani- cal Engineering, University of Minnesota over many years. If this book is used as a text for a quarter-length course (with approximately 30 lecture hours and 18 recitation hours), Chapters 1 through 7 may be covered. After studying these chapters, the student should be able to derive mathematical models for many dynamic systems with reasonable simplicity in the forms of transfer func- tion or state-space equation. Also, he/she will be able to obtain computer solutions of system responses with MATLAB. If the book is used as a text for a semester- length course (with approximately 40 lecture hours and 26 recitation hours), then the first nine chapters may be covered or, alternatively, the first seven chapters plus Chapters 10 and 11 may be covered. If the course devotes 50 to 60 hours to lectures, then the entire book may be covered in a semester.
  9. 9. Preface ix Fmally, I wish to acknowledge deep appreciation to the following professors who reviewed the third edition of this book prior to the preparation of this new edi- tion: R. Gordon Kirk (Vrrginia Institute of Technology), Perry Y. Li (University of Minnesota), Sherif Noah (Texas A & M University), Mark L. Psiaki (Cornell Uni- versity), and William Singhose (Georgia Institute of Technology). Their candid, insightful, and constructive comments are reflected in this new edition. KATSUHIKO OGATA
  10. 10. Introduction to System Dynamics 1-1 INTRODUCTION System dynamics deals with the mathematical modeling of dynamic systems and response analyses of such systems with a view toward understanding the dynamic nature of each system and improving the system's performance. Response analyses are frequently made through computer simulations of dynamic systems. Because many physical systems involve various types of components, a wide variety of different types of dynamic systems will be examined in this book. The analysis and design methods presented can be applied to mechanical, electrical, pneumatic, and hydraulic systems, as well as nonengineering systems, such as eco- nomic systems and biological systems. It is important that the mechanical engineer- ing student be able to determine dynamic responses of such systems. We shall begin this chapter by defining several terms that must be understood in discussing system dynamics. Systems. A system is a combination of components acting together to per- form a specific objective.A component is a single functioning unit of a system. By no means limited to the realm of the physical phenomena, the concept of a system can be extended to abstract dynamic phenomena, such as those encountered in eco- nomics, transportation, population growth, and biology. 1
  11. 11. 2 Introduction to System Dynamics Chap. 1 A system is called dynamic if its present output depends on past input; if its current output depends only on current input, the system is known as static.The out- put of a static system remains constant if the input does not change. The output changes only when the input changes. In a dynamic system, the output changes with time if the system is not in a state of equilibrium. In this book, we are concerned mostly with dynamic systems. Mathematical models. Any attempt to design a system must begin with a prediction of its performance before the system itself can be designed in detail or ac- tually built. Such prediction is based on a mathematical description of the system's dynamic characteristics. This mathematical description is called a mathematical model. For many physical systems, useful mathematical models are described in terms of differential equations. Linear and nonlinear differential equations. Linear differential equations may be classified as linear, time-invariant differential equations and linear, time- varying differential equations. A linear, time-invariant differential equation is an equation in which a depen- dent variable and its derivatives appear as linear combinations. An example of such an equation is d2x dx - + 5- + lOx =0 dt2 dt Since the coefficients of all terms are constant, a linear, time-invariant differential equation is also called a linear, constant-coefficient differential equation. In the case of a linear, time-varying differential equation, the dependent vari- able and its derivatives appear as linear combinations, but a coefficient or coeffi- cients of terms may involve the independent variable. An example of this type of differential equation is d2 x - + (1 - cos 2t)x = 0 dt2 It is important to remember that, in order to be linear, the equation must con- tain no powers or other functions or products of the dependent variables or its derivatives. A differential equation is called nonlinear if it is not linear. Two examples of nonlinear differential equations are and
  12. 12. Sec. 1-2 Mathematical Modeling of Dynamic Systems 3 Linear systems and nonlinear systems. For linear systems, the equations that constitute the model are linear. In this book, we shall deal mostly with linear sys- tems that can be represented by linear, time-invariant ordinary differential equations. The most important property oflinear systems is that the principle ofsuperpo- sition is applicable.This principle states that the response produced by simultaneous applications of two different forcing functions or inputs is the sum of two individual responses. Consequently, for linear systems, the response to several inputs can be calculated by dealing with one input at a time and then adding the results. As a result of superposition, complicated solutions to linear differential equations can be derived as a sum of simple solutions. In an experimental investigation of a dynamic system, if cause and effect are proportional, thereby implying that the principle of superposition holds, the system can be considered linear. Although physical relationships are often represented by linear equations, in many instances the actual relationships may not be quite linear. In fact, a careful study of physical systems reveals that so-called linear systems are actually linear only within limited operating ranges. For instance, many hydraulic systems and pneumatic systems involve nonlinear relationships among their variables, but they are frequently represented by linear equations within limited operating ranges. For nonlinear systems, the most important characteristic is that the principle of superposition is not applicable. In general, procedures for finding the solutions of problems involving such systems are extremely complicated. Because of the mathe- matical difficulty involved, it is frequently necessary to linearize a nonlinear system near the operating condition. Once a nonlinear system is approximated by a linear mathematical model, a number of linear techniques may be used for analysis and design purposes. Continuous-time systems and discrete-time systems. Continuous-time systems are systems in which the signals involved are continuous in time. These sys- tems may be described by differential equations. Discrete-time systems are systems in which one or more variables can change only at discrete instants of time. (These instants may specify the times at which some physical measurement is performed or the times at which the memory of a digital computer is read out.) Discrete-time systems that involve digital signals and, possi- bly, continuous-time signals as well may be described by difference equations after the appropriate discretization of the continuous-time signals. The materials presented in this text apply to continuous-time systems; discrete- time systems are not discussed. 1-2 MATHEMATICAL MODELING OF DYNAMIC SYSTEMS Mathematical modeling. Mathematical modeling involves descriptions of important system characteristics by sets of equations. By applying physical laws to a specific system, it may be possible to develop a mathematical model that describes the dynamics of the system. Such a model may include unknown parameters, which
  13. 13. 4 Introduction to System Dynamics Chap. 1 must then be evaluated through actual tests. Sometimes, however, the physical laws governing the behavior of a system are not completely defined, and formulating a mathematical model may be impossible. Ifso, an experimental modeling process can be used. In this process, the system is subjected to a set of known inputs, and its out- puts are measured.Then a mathematical model is derived from the input-output re- lationships obtained. Simplicity of mathematical model versus accuracy of results of analysis. In attempting to build a mathematical model, a compromise must be made between the simplicity of the model and the accuracy of the results of the analysis. It is im- portant to note that the results obtained from the analysis are valid only to the ex- tent that the model approximates a given physical system. In determining a reasonably simplified model, we must decide which physical variables and relationships are negligible and which are crucial to the accuracy of the model. To obtain a model in the form of linear differential equations, any dis- tributed parameters and nonlinearities that may be present in the physical system must be ignored. If the effects that these ignored properties have on the response are small, then the results of the analysis of a mathematical model and the results of the experimental study of the physical system will be in good agreement. Whether any particular features are important may be obvious in some cases, but may, in other instances, require physical insight and intuition. Experience is an important factor in this connection. Usually, in solving a new problem, it is desirable first to build a simplified model to obtain a general idea about the solution.Afterward, a more detailed math- ematical model can be built and used for a more complete analysis. Remarks on mathematical models. The engineer must always keep in mind that the model he or she is analyzing is an approximate mathematical descrip- tion of the physical system; it is not the physical system itself In reality, no mathe- matical model can represent any physical component or system precisely. Approximations and assumptions are always involved. Such approximations and as- sumptions restrict the range of validity of the mathematical model. (The degree of approximation can be determined only by experiments.) So, in making a prediction about a system's performance, any approximations and assumptions involved in the model must be kept in mind. Mathematical modeling procedure. The procedure for obtaining a math- ematical model for a system can be summarized as follows: L Draw a schematic diagram of the system, and define variables. 2. Using physical laws, write equations for each component, combine them according to the system diagram, and obtain a mathematical model. 3. To verify the validity of the model, its predicted performance, obtained by solving the equations of the model, is compared with experimental results. (The question of the validity of any mathematical model can be answered only by experiment.) If the experimental results deviate from the prediction
  14. 14. Sec. 1-3 Analysis and Design of Dynamic Systems 5 to a great extent, the model must be modified. A new model is then derived and a new prediction compared with experimental results. The process is re- peated until satisfactory agreement is obtained between the predictions and the experimental results. 1-3 ANALYSIS AND DESIGN OF DYNAMIC SYSTEMS This section briefly explains what is involved in the analysis and design of dynamic systems. Analysis. System analysis means the investigation, under specified condi- tions, of the performance of a system whose mathematical model is known. The first step in analyzing a dynamic system is to derive its mathematical model. Since any system is made up of components, analysis must start by developing a mathematical model for each component and combining all the models in order to build a model of the complete system. Once the latter model is obtained, the analysis may be formulated in such a way that system parameters in the model are varied to produce a number of solutions. The engineer then compares these solutions and interprets and applies the results of his or her analysis to the basic task. H should always be remembered that deriving a reasonable model for the complete system is the most important part of the entire analysis. Once such a model is available, various analytical and computer techniques can be used to ana- lyze it. The manner in which analysis is carried out is independent of the type of physical system involved-mechanical, electrical, hydraulic, and so on. Design. System design refers to the process of finding a system that accom- plishes a given task. In general, the design procedure is not straightforward and will require trial and error. Synthesis. By synthesis, we mean the use of an explicit procedure to find a system that will perform in a specified way. Here the desired system characteristics are postulated at the outset, and then various mathematical techniques are used to synthesize a system having those characteristics. Generally, such a procedure is com- pletely mathematical from the start to the end of the design process. Basic approach to system design. The basic approach to the design of any dynamic system necessarily involves trial-and-error procedures. Theoretically, a synthesis of linear systems is possible, and the engineer can systematically deter- mine the components necessary to realize the system's objective. In practice, howev- er, the system may be subject to many constraints or may be nonlinear; in such cases, no synthesis methods are currently applicable. Moreover, the features of the com- ponents may not be precisely known. Thus, trial-and-error techniques are almost al- ways needed. Design procedures. Frequently, the design of a system proceeds as follows: The engineer begins the design procedure knowing the specifications to be met and
  15. 15. 6 Introduction to System Dynamics Chap. 1 the dynamics of the components, the latter of which involve design parameters.The specification may be given in terms of both precise numerical values and vague qualitative descriptions. (Engineering specifications normally include statements on such factors as cost, reliability, space, weight, and ease of maintenance.) It is impor- tant to note that the specifications may be changed as the design progresses, for de- tailed analysis may reveal that certain requirements are impossible to meet. Next, the engineer will apply any applicable synthesis techniques, as well as other meth- ods, to build a mathematical model of the system. Once the design problem is formulated in terms of a model, the engineer car- ries out a mathematical design that yields a solution to the mathematical version of the design problem. With the mathematical design completed, the engineer simu- lates the model on a computer to test the effects of various inputs and disturbances on the behavior of the resulting system. If the initial system configuration is not sat- isfactory, the system must be redesigned and the corresponding analysis completed. This process of design and analysis is repeated until a satisfactory system is found. Then a prototype physical system can be constructed. Note that the process of constructing a prototype is the reverse of mathemati- cal modeling. The prototype is a physical system that represents the mathematical model with reasonable accuracy. Once the prototype has been built, the engineer tests it to see whether it is satisfactory. If it is, the design of the prototype is com- plete. If not, the prototype must be modified and retested. The process continues until a satisfactory prototype is obtained. 1-4 SUMMARY From the point of view of analysis, a successful engineer must be able to obtain a mathematical model of a given system and predict its performance. (The validity of a prediction depends to a great extent on the validity of the mathematical model used in making the prediction.) From the design standpoint, the engineer must be able to carry out a thorough performance analysis of the system before a prototype is constructed. The objective of this book is to enable the reader (1) to build mathematical models that closely represent behaviors of physical systems and (2) to develop sys- tem responses to various inputs so that he or she can effectively analyze and design dynamic systems. Outline of the text. Chapter 1 has presented an introduction to system dy- namics. Chapter 2 treats Laplace transforms. We begin with Laplace transformation of simple time functions and then discuss inverse Laplace transformation. Several useful theorems are derived. Chapter 3 deals with basic accounts of mechanical sys- tems. Chapter 4 presents the transfer-function approach to modeling dynamic sys- tems. The chapter discusses various types of mechanical systems. Chapter 5 examines the state-space approach to modeling dynamic systems. Various types of mechanical systems are considered. Chapter 6 treats electrical systems and electromechanical systems, including operational-amplifier systems. Chapter 7 deals with fluid systems,
  16. 16. Sec. 1-4 Summary 7 such as liquid-level systems, pneumatic systems, and hydraulic systems, as well as thermal systems. A linearization technique for nonlinear systems is explored. Chapter 8 presents time-domain analyses of dynamic systems-specifically, transient-response analyses of dynamic systems. The chapter also presents the ana- lytical solution of the state equation. Chapter 9 treats frequency-domain analyses of dynamic systems. Among the topics discussed are vibrations of rotating mechanical systems and vibration isolation problems. Also discussed are vibrations in multi- degrees-of-freedom systems and modes of vibrations. Chapter 10 presents the basic theory of control systems, including transient- response analysis, stability analysis, and root-locus analysis and design. Also dis- cussed are tuning rules for PID controllers. Chapter 11 deals with the analysis and design of control systems in the frequency domain. The chapter begins with Bode diagrams and then presents the Nyquist stability criterion, followed by detailed design procedures for lead, lag, and lag-lead compensators. Appendix A treats systems of units, Appendix B summarizes conversion tables, and Appendix C gives a brief summary of vector-matrix algebra. Appendix D presents introductory materials for MATLAB. Throughout the book, MATLAB is used for the solution of most computa- tional problems. Readers who have no previous knowledge of MATLAB may read Appendix D before solving any MATLAB problems presented in this text.
  17. 17. The Laplace Transform 2-1 INTRODUCTION The Laplace transform is one of the most important mathematical tools available for modeling and analyzing linear systems. Since the Laplace transform method must be studied in any system dynamics course, we present the subject at the begin- ning of this text so that the student can use the method throughout his or her study of system dynamics. The remaining sections of this chapter are outlined as follows: Section 2-2 reviews complex numbers, complex variables, and complex functions. Section 2-3 defines the Laplace transformation and gives Laplace transforms of several com- mon functions of time. Also examined are some of the most important Laplace transform theorems that apply to linear systems analysis. Section 2-4 deals with the inverse Laplace transformation. Finally, Section 2-5 presents the Laplace transform approach to the solution of the linear, time-invariant differential equation. 2-2 COMPLEX NUMBERS, COMPLEX VARIABLES, AND COMPLEX FUNCTIONS This section reviews complex numbers, complex algebra, complex variables, and complex functions. Since most of the material covered is generally included in the basic mathematics courses required of engineering students, the section can be omitted entirely or used simply for personal reference. S
  18. 18. Sec. 2-2 Complex Numbers, Complex Variables, and Complex Functions 1m o Re Figure 2-1 Complex plane representa- tion of a complex number z. 9 Complex numbers. Using the notation j = v=I, we can express all num- bers in engineering calculations as z = x + jy where z is called a complex number and x and jy are its real and imaginary parts, respectively. Note that both x and y are real and that j is the only imaginary quanti- ty in the expression. The complex plane representation of z is shown in Figure 2-1. (Note also that the real axis and the imaginary axis define the complex plane and that the combination of a real number and an imaginary number defines a point in that plane.) A complex number z can be considered a point in the complex plane or a directed line segment to the point; both interpretations are useful. The magnitude, or absolute value, of z is defined as the length of the directed line segment shown in Figure 2-1. The angle of z is the angle that the directed line segment makes with the positive real axis. A counterclockwise rotation is defined as the positive direction for the measurement of angles. Mathematically, magnitude of z = Izl = Vx2 + j, angle of z = 9 = tan-1l:'. x A complex number can be written in rectangular form or in polar form as follows: z = x + jy z = Izl(cos 9 + j sin 9) z= Izl~ z = Izl eifJ }rectangular fonns }polar forms In converting complex numbers to polar form from rectangular, we use Izl = Vx2 + y2, 8 = tan-II x To convert complex numbers to rectangular form from polar, we employ x = IzIcos 8, y = Izl sin 8 Complex conjugate. The complex conjugate of z = x + j y is defined as Z = x - jy
  19. 19. 10 The Laplace Transform Chap. 2 Figure 2-2 Complex number zand its complex conjugate Z, 1m o Re The complex conjugate ofzthus has the same real part as Z and an imaginary part that is the negative of the imaginary part of z. Figure 2-2 shows both zand Z. Note that z = x + jy = Izl il = Izi (cos 8 + j sin 8) z = x - jy = Izi /-8 = Izi (cas 8 - jsin8) Euler's theorem. The power series expansions of cos 8 and sin 8 are, respectively, and Thus, Since fi2 £t 86 cos 8 = 1 - - + - - - + 2! 4! 6! . 83 85 87 sm8 = 8 - - + - - - + 3! 5! 7! ('8)2 ('8)3 ('8)4 8 + ' , 8 = 1 + ('8) + -'- + -'- + -'- +cos ,sm , , 3' 4'2. . . it follows that cos 8 + j sin 8 = ejB This is known as Euler's theorem. Using Euler's theorem, we can express the sine and cosine in complex form. Noting that e-jB is the complex conjugate of ei6 and that eiB = cos 8 + j sin 8 e-jB = cos 8 - j sin 8 I
  20. 20. Sec. 2-2 Complex Numbers, Complex Variables, and Complex Functions 11 we find that ej8 + e-j8 cos 8 = 2 . ej8 - e-j8 sm8 = 2j Complex algebra. If the complex numbers are written in a suitable form, op- erations like addition, subtraction, multiplication, and division can be performed easily. Equality ofcomplex numbers. 1vo complex numbers z and ware said to be equal if and only if their real parts are equal and their imaginary parts are equal. So if two complex numbers are written z = x + jy, w = u + jv then z = w if and only if x = u and y = v. Addition. 1vo complex numbers in rectangular form can be added by adding the real parts and the imaginary parts separately: z + w = (x + jy) + (u + jv) = (x + u) + j(y + v) Subtraction. Subtracting one complex number from another can be consid- ered as adding the negative of the former: z - w = (x + jy) - (u + jv) = (x - u) + j(y - v) Note that addition and subtraction can be done easily on the rectangular plane. Multiplication. If a complex number is multiplied by a real number, the re- sult is a complex number whose real and imaginary parts are multiplied by that real number: az = a(x + jy) = ax + jay (a = real number) Iftwo complex numbers appear in rectangular form and we want the product in rec- tangular form, multiplication is accomplished by using the fact that P = -1. Thus, if two complex numbers are written z = x + jy, w = u + jv then zw = (x + j y)(u + jv) = xu + j yu + jxv + lyv = (xu - yv) + j(xv + yu) In polar form, multiplication of two complex numbers can be done easily. The mag- nitude of the product is the product of the two magnitudes, and the angle of the product is the sum of the two angles. So if two complex numbers are written z = Izl~, w = Iwl~ then zw = Izllwl/8 + cP
  21. 21. 12 The Laplace Transform Chap. 2 Multiplication byJ. It is important to note that multiplication by j is equiva- lent to counterclockwise rotation by 90°. For example, if z = x + jy then jz = j(x + jy) = jx + py = -y + jx or, noting that j = 1/90°, if z = Izl il then jz = 1/90° Izl il = Izl/8 + 90° Figure 2-3 illustrates the multiplication of a complex number z by j. Division. H a complex number z = IzIil is divided by another complex number w = IwIil., then z Izi il Izi w = Iwl L.!2. = M 18 - ~ That is, the result consists of the quotient of the magnitudes and the difference of the angles. Division in rectangular form is inconvenient, but can be done by mUltiplying the denominator and numerator by the complex conjugate of the denominator.This procedure converts the denominator to a real number and thus simplifies division. For instance, z x + jy (x + jy)(u - jv) (xu + yv) + j(yu - xv) - = - - = = ~--~--.;...~--..;... w u + jv (u + jv) (u - jv) u2 + v2 xu + yv + .yu - xv = u 2 + v 2 J u2 + v 2 o Figure 2-3 Multiplication of a complex number z by j. Re 1m o Re Figure 2-4 Division of a complex number z by j.
  22. 22. Sec. 2-2 Complex Numbers, Complex Variables, and Complex Functions 13 Division by j. Note that division by j is equivalent to clockwise rotation by 90°. For example, if z = x + jy, then z x + jy (x + jy)j jx - y . -=--= =--=y-]X j j jj -1 or z Izl L.P. j = 1 /90° = Izl /8 - 90° Figure 2-4 illustrates the division of a complex number z by j. Powers and roots. Multiplying z by itself n times, we obtain zn = (Izl L.P.)n = Izln / n8 Extracting the nth root of a complex number is equivalent to raising the number to the 1/nth power: For instance, and (8.66 - j5)3 = (10 /-30°)3 = 1000 /-90° = 0 - j 1000 = -j 1000 (2.12 - j2.12)112 = (9 / -45°)112 = 3 / -22.5° Comments. It is important to note that Izwl = Izllwl Iz + wi #: Izi + Iwl Complex variable. A complex number has a real part and an imaginary part, both of which are constant. If the real part or the imaginary part (or both) are variables, the complex number is called a complex variable. In the Laplace transfor- mation, we use the notation s to denote a complex variable; that is, s = u + jw where u is the real part and jw is the imaginary part. (Note that both u and ware real.) Complex function. A complex function F(s), a function of s, has a real part and an imaginary part, or F(s) = Fx + jFy where Fx and Fy are real quantities. The magnitude of F(s) is VFi + F~, and the angle 8 of F(s) is tan-1 (FylFx )' The angle is measured counterclockwise from the positive real axis.The complex conjugate of F(s) is pes) = Fx - jFy- Complex functions commonly encountered in linear systems analysis are single- valued functions of s and are uniquely determined for a given value of s.1}rpically,
  23. 23. 14 The Laplace Transform Chap. 2 such functions have the form K(s + ZI)(S + Z2) ... (s + Zm) F(s) - ---:...-..---.;...~----- - (s + PI)(S + P2) ... (s + Pn) Points at which F(s) equals zero are called zeros. That is, s = -Zh S = -Z2, ... , s = -Zm are zeros of F(s). [Note that F(s) may have additional zeros at infinity; see the illustrative example that follows.] Points at which F(s) equals infinity are called poles. That is, s = -PI, S = - P2, ... , s = - Pn are poles of F(s). If the denominator of F(s) involves k-multiple factors (s + Pl, then s = -pis called a multiplepole of order k or repeated pole of order k.1f k = 1, the pole is called a simple pole. As an illustrative example, consider the complex function JC(s + 2)(s + 10) G(s) - -------~ - s(s + l)(s + 5)(s + 15)2 G(s) has zeros at s = -2 and s = -10, simple poles at s = 0, s = -1, and s = -5, and a double pole (multiple pole of order 2) at s = -15. Note that G(s) becomes zero at s = 00. Since, for large values ofs, K G(s) * 3s it follows that G(s) possesses a triple zero (multiple zero of order 3) at s = 00. If points at infinity are included, G(s) has the same number of poles as zeros. To sum- marize, G(s) has five zeros (s = -2, s = -10, s = 00, s = 00, s = 00) and five poles (s = 0, s = -1, s = -5, s = -15, s = -15). 2-3 LAPLACE TRANSFORMATION The Laplace transform method is an operational method that can be used advanta- geously in solving linear, time-invariant differential equations. Its main advantage is that differentiation of the time function corresponds to multiplication of the trans- form by a complex variable s, and thus the differential equations in time become algebraic equations in s.The solution of the differential equation can then be found by using a Laplace transform table or the partial-fraction expansion technique. Another advantage of the Laplace transform method is that, in solving the differen- tial equation, the initial conditions are automatically taken care of, and both the par- ticular solution and the complementary solution can be obtained simultaneously. Laplace transformation. Let us define /(t) = a time function such that /(t) =°for t < 0 s = a complex variable 9!, = an operational symbol indicating that the quantity upon which it operates is to be transformed by the Laplace integral100 e-st dt F(s) = Laplace transform off(t)
  24. 24. Sec. 2-3 Laplace Transformation 15 Then the Laplace transform off(t) is given by ~[f(t)] = F(s) = l"'e-n dt[f(t)] = l"'f(t)e-n dt The reverse process of finding the time function f(t) from the Laplace transform F(s) is called inverse Laplace trans/ormation.The notation for inverse Laplace trans- formation is ;r1. Thus, ;rl[F(s)] = /(t) Existence of Laplace transform. The Laplace transform of a function f(t) exists if the Laplace integral converges. The integral will converge iff(t) is piecewise continuous in every finite interval in the range t > 0 and if I(t) is of exponential order as t approaches infinity. A function f(t) is said to be of exponential order if a real, positive constant u exists such that the function e-atl/(t)I approaches zero as t approaches infinity. If the limit of the function e-utl/(t)I approaches zero for u greater than uc and the limit approaches infinity for u less than uC' the value uc is called the abscissa 0/convergence. It can be seen that, for such functions as t, sin wt, and t sin wt, the abscissa of convergence is equal to zero. For functions like e-ct , te-ct , and e-ct sin wt, the abscis- sa of convergence is equal to -c. In the case of functions that increase faster than the exponential function, it is impossible to find suitable values of the abscissa of convergence. Consequently, such functions as il and ter do not possess Laplace transforms. Nevertheless, it should be noted that, although er for 0 s t S 00 does not possess a Laplace transform, the time function defined by /(t) = er =0 for 0 :s; t :s; T < 00 for t < 0, T < t does, since / (t) = er for only a limited time interval 0 S t !5 T and not for oS t S 00. Such a signal can be physically generated. Note that the signals that can be physically generated always have corresponding Laplace transforms. If functions 11(t) and h(t) are both Laplace transformable, then the Laplace transform of11(t) + h(t) is given by ;e[fl(t) + h(t)] = ;e[f1(t)] + ;e[f2(t)] Exponential function. Consider the exponential function / (t) = 0 for t < 0 =Ae-at for t ~ 0 where A and a are constants. The Laplace transform of this exponential function can be obtained as follows: 1 00 100 A;e[Ae-at] = Ae-ate-st dt = A e-(a+s)t dt = - - o 0 s + a
  25. 25. 16 The Laplace Transform Chap. 2 In performing this integration, we assume that the real part of s is greater than -a (the abscissa of convergence), so that the integral converges. The Laplace trans- form F(s} of any Laplace transformable functionf(t) obtained in this way is valid throughout the entire s plane, except at the poles of F(s). (Although we do not pre- sent a proof of this statement, it can be proved by use of the theory of complex variables.) Step function. Consider the step function f (t) = 0 for t < 0 = A fort> 0 where A is a constant. Note that this is a special case of the exponential function Ae-at, where a = O. The step function is undefined at t = O. Its Laplace transform is given by 1 00 A ;£[A] = Ae-st dt = - o s The step function whose height is unity is called a unit-step function. The unit- step function that occurs at t = to is frequently written l(t - to), a notation that will be used in this book.The preceding step function whose height is A can thus be writ- tenA1(t}. The Laplace transform of the unit-step function that is defined by is lis, or l(t) = 0 =1 for t < 0 for t > 0 ~[l(t)] = !s Physically, a step function occurring at t =to corresponds to a constant signal suddenly applied to the system at time t equals to. Ramp function. Consider the ramp function f(t) = 0 = At for t < 0 for t ~ 0 where A is a constant.The Laplace transform of this ramp function is !'eIAt] = A ["'te-Sf dt To evaluate the integral, we use the formula for integration by parts: [budv =Uvl: - [bVdU
  26. 26. Sec. 2-3 Laplace Transformation In this case, U =t and dv =e-st dt. [Note that v = e-SII( -s).] Hence, 100 ( -sl I00 100 -Sf ) ;e[At] = A te-sr dt = A t ~ - ~dt o -s 0 o-s = A fooe-sr dt = A s 10 s2 Sinusoidal function. The Laplace transform of the sinusoidal function J(t) = 0 for t < 0 = A sin wt for t ~ 0 where A and ware constants, is obtained as follows: Noting that ejw1 = cos wt + j sin wt and e-jwt = cos wt - j sin wt we can write Hence, AlOO • •~[A sin wt] = --: (eJwt - e-Jwt)e-st dt 2J 0 Al A1 Aw = - - - - - - - = 2j s - jw 2j s + jw s2 + w2 Similarly, the Laplace transform of A cos wt can be derived as follows: As ;erA cos wt] = 2 2 S + w 17 Comments. The Laplace transform of any Laplace transformable function f(t) can be found by multiplying f(t) by e-st and then integrating the product from t = 0 to t = 00. Once we know the method of obtaining the Laplace transform, how- ever, it is not necessary to derive the Laplace transform of I(t) each time. Laplace transform tables can conveniently be used to find the transform of a given function f(t). Table 2-1 shows Laplace transforms of time functions that will frequently appear in linear systems analysis. In Table 2-2, the properties ofLaplace transforms are given. Translated function. Let us obtain the Laplace transform of the translated function f(t - a)l(t - a), where a ~ O. This function is zero for t < a. The func- tionsf(t)l(t) and f(t - a)1(t - a) are shown in Figure 2-5. By definition, the Laplace transform of J(t - a)l(t - a) is ~1f(1 - a)t(t - a)] = [,oJ(1 - a)t(1 - aV" dl
  27. 27. 18 The Laplace Transform Chap. 2 TABLE 2-1 Laplace Transform Pairs f(t) F(s) 1 Unit impulse cS{t) 1 2 Unit step 1(t) 1 - s 1 3 t s2 4 tn- 1 (n=1,2,3, ... ) 1 - (n - 1)! sn 5 tn (n=1,2,3, ... ) n! -sn+l 6 e-at 1 --s+a 1 7 te-at (s + a)2 1 n-l -at (n = 1, 2, 3, ... ) 1 8 (n - 1)! t e (s + a)n 9 tne-at (n=1,2,3, ... ) n! (s + a)n+l w 10 sin wt $2 + w2 S 11 coswt s2 + w2 w 12 sinh wt S2 - (J)2 $ 13 cosh wt ; - w2 14 !(1 - e-at ) 1 a s(s + a) 15 _1_(e-at _ e-bt) 1 b-a (s + a)(s + b) _1_(be-bt - ae-at ) s 16 (s + a)(s + b)b-a ~[1 + _1_(be-at - ae-bt)] 1 17 s(s + a)(s + b)ab a - b
  28. 28. TABLE 2-1 (continued) r I 1-- f(t) I lpes) - 18 1 2 (1 e 01 a ate 01) 1 s(s + a)2 - 19 1 '--- 2 (at 1 + e-al) 1 a s2(s + a) -w 20 e 01 sin wt - (s + a)2 + ;;;. ~ 21 l.- e 01 cos wt s+a - (s + a)2 +-;;; - 22 Wn_ e(wnt· ~~ smwn 1-,2 t w2 n 2 s + 2,wns + w2 n - 1- =e (Wnl' ( ~ 23 ~ smwn 1-,2 / -cJ» cJ> = tan-1YI -,2 _s 2 S + 2,wns + w2 , - n - 1 1_ _e(wnt· ( ~ 24 ~ smwn 1-,2 / +cJ» cJ> = tan-l !l -,2 w2 n S(S2 + 2,wns + w1) ' - - w2 25 1 cos wt -S(S2 + w2) ' - - w3 26 wt - sin wt - s2(s2 + ( 2) - 2w3 27 sin wt - wt cos wt - (s2 + w2)2 - 28 1 2wI sin wI s - (s2 + w2)2 - s2 - w2 29 t cos wt - (s2 + ( 2)2 - - I _ 1 30 ~ _:I(cos WIt cos i»2t) (WI ¢~) S (s2 + wI)(s2 + ~) - 31 1 2w (sin wt + wt cos wt) - S2 (s2 + (2)2 L.. 19
  29. 29. TABLE 2-2 Properties of Laplace Transforms 1 ~[Af(t)] :::: AF(s) 2 ~[fl(t) ± 12(t)] :::: P1(s) ± F2(S) 3 ~±[:tf(t)] = sF(s) - f(O±) 4 !:e±[:~f(t)] = s2p(s) - sf(O±) - i(O±) [ dn] n (k-l) ~± dtnf(t) = snp(s) - ~sn-k f(O±) 5 (k-l) dk- 1 where f(t) :::: dtk-1f(t) 6 [J ] F(s) If[(I) dll/eD< ~± f(t)dt = - + s s [f! ] F(s) If[(I) dll/eD< Iff[(I) dl dll/eD< 7 ~± f(t) dt dt = - 2 + 2 + S S s 8 ~±[/···1f(t)(dt)n] = F(:) + ±n_ 1 k+l [/···1f(t)(dt)k ] s k=l S I=O± 9 f£[!o'[(I) dl] = F~S) 10 ["'[(I) dl = lim F(s) if1.""[(I) dl exists o s-O 11 ~[e-a'f(t)] = F(s + a) 12 !:e[f(t - a)l(t - a)] = e-asF(s) a~O 13 dP(s) !:e[tf(t)] = - - - ds 14 d2 ~[t2f(t)] = -2 F(s) ds dn 15 ~[tnf(t)] = (-l)n-F(s) n = 1,2,3, ... dsn 16 f£[7[(I)] = 1""F(S) ds if lim.!.f(t) exists I-ot 17 ~[f(~)] = aF(as) 20
  30. 30. Sec. 2-3 Laplace Transformation f(t) l(t) f(1 - DC) l(t - DC) o o DC Figure 2-S Functionf(t)l(t) and translated function f(t - a)I(1 - a). By changing the independent variable from t to 7, where 7 = t - a, we obtain foof(t - £1')l(t - £1')e-st dt = l°Of (7)1(T)e-S(T+a) dT 10 -a 21 Noting that f(T)1(7) = 0 for T < 0, we can change the lower limit of integration from -a to O.Thus, l°O f (T)l(T)e-S(T+a> dT = f OO f (7)1(T)e-S(T+a) dT -a 10 =["'j(T)e-STe-as dT =e-as100 j(T)e-ST dT =e-asp(s) where P(s) = ~(f(t)] = 1°Oj (tv" dl Hence, ;Eff(t - a)l(t - a)] = e-aSF(s) a~O This last equation states that the translation of the time functionf(t)l(t) by a (where a ~ 0) corresponds to the multiplication of the transform F(s) bye-as. Pulse function. Consider the pulse function shown in Figure 2-6, namely, A f(t) =- to =0 where A and to are constants. for 0 < t < to for t < 0, to < t The pulse function here may be considered a step function of height Alto that begins at t = 0 and that is superimposed by a negative step function of height Alto
  31. 31. 22 The Laplace Transform Chap. 2 1(1) A to o 4_ 00 to Figure 2-6 Pulse function. Figure 2-7 Impulse function. beginning at t = to; that is, A A f(t) = -1(t) - -1(t - to) to to Then the Laplace transform off(t) is obtained as ~[J(t)l = ~[~1(t)] - ~[~1(t - to) ] A A -SI = - - -e 0 tos tos A _ = -(1 - e stO) tos (2-1) Impulse function. The impulse function is a special limiting case of the pulse function. Consider the impulse function f (t) = lim A for 0 < t < to to-O to = 0 for t < 0, to < t Figure 2-7 depicts the impulse function defined here. It is a limiting case ofthe pulse function shown in Figure 2-6 as to approaches zero. Since the height of the impulse function is Alto and the duration is to, the area under the impulse is equal to A. As the duration to approaches zero, the height Alto approaches infinity, but the area under the impulse remains equal to A. Note that the magnitude of the impulse is measured by its area. From Equation (2-1), the Laplace transform of this impulse function is shown to be :Eff(t)] = lim [~(1 -e-sto)] 10-0 tos ~[A(l - e-S1o )] . dto As =hm =-=A 10-0 d () s - tos dto
  32. 32. Sec. 2-3 Laplace Transformation 23 Thus, the Laplace transform of the impulse function is equal to the area under the impulse. The impulse function whose area is equal to unity is called the unit-impulse function or the Dirac delta function. The unit-impulse function occurring at t = to is usually denoted by eS(t - to), which satisfies the following conditions: eS(t - to) = 0 eS(t - to) = 00 J~6(1 - '0) dl = 1 for t *to for t = to An impulse that has an infinite magnitude and zero duration is mathematical fiction and does not occur in physical systems. If, however, the magnitude of a pulse input to a system is very large and its duration very short compared with the system time constants, then we can approximate the pulse input by an impulse function. For instance, if a force or torque input f(t) is applied to a system for a very short time duration 0 < t < to, where the magnitude of f(t) is sufficiently large so that J;o f(t) dt is not negligible, then this input can be considered an impulse input. (Note that, when we describe the impulse input, the area or magnitude of the impulse is most important, but the exact shape of the impulse is usually immaterial.) The impulse input supplies energy to the system in an infInitesimal time. The concept of the impulse function is highly useful in differentiating discon- tinuous-time functions. The unit-impulse function eS(t - to) can be considered the derivative of the unit-step function l(t - to) at the point of discontinuity t = to, or d eS(t - to) = dt l(t - to) Conversely, if the unit-impulse function eS(t - to) is integrated, the result is the unit- step function l(t - to). With the concept of the impulse function, we can differenti- ate a function containing discontinuities, giving impulses, the magnitudes of which are equal to the magnitude of each corresponding discontinuity. Multiplication of fIt) bye-at. If f(t) is Laplace transformable and its Laplace transform is F(s), then the Laplace transform of e-at f(t) is obtained as ~[e"""f(l)l = [X>e-alf(t)e-st dl = F(s + a) (2-2) We see that the multiplication of f(t) bye-at has the effect of replacing s by (s + a) in the Laplace transform. Conversely, changing s to (s + a) is equivalent to mUltiplyingf(t) bye-at. (Note that a may be real or complex.) The relationship given by Equation (2-2) is useful in finding the Laplace transforms of such functions as e-at sin wt and e-at cos wt. For instance, since ~[sin wt] = 2 W 2 = F(s) s + w and s ~[cos wt] = 2 2 = G(s) s + w
  33. 33. 24 The Laplace Transform Chap. 2 it follows from Equation (2-2) that the Laplace transforms of e-at sin wt and e-at cos wt are given, respectively, by and w !e[e-al sin wt] = F(s + a) = - - - - - (s + a)2 + w2 s+a !e[e-al cos wt] = G(s + a) = -----::- (s + a)2 + w2 Comments on the lower limit of the Laplace integral. In some cases,Jtt) possesses an impulse function at t = O. Then the lower limit of the Laplace integral must be clearly specified as to whether it is 0- or 0+, since the Laplace transforms ofJtt) differ for these two lower limits. If such a distinction of the lower limit of the Laplace integral is necessary, we use the notations and .:£_[f(t)] = ["'[(t)e-n dt = .:£+[f(t)] + [0+[(t)e-st dt IfJtt) involves an impulse function at t = 0, then since [[(t)e-st dt .. 0 for such a case. Obviously, ifJtt) does not possess an impulse function at t = 0 (i.e., if the function to be transformed is finite between t = 0- and t = 0+), then !e+[f(t)] = !e_[f(t)] Differentiation theorem. The Laplace transform of the derivative of a function/(t) is given by .:£[:/(t)] = sF(s) - [(0) (2-3) where .f(0) is the initial value of.f(t), evaluated at t = O. Equation (2-3) is called the differentiation theorem. For a given function .f(t), the values of /(0+) and 1(0-) may be the same or different, as illustrated in Figure 2-8. The distinction between 1(0+) and 1(0-) is important when.f(t) has a discontinuity at t = 0, because, in such a case, dJtt)/dt will
  34. 34. Sec. 2-3 Laplace Transformation /(/) /(0 +) /(/) Flgure 2-8 Step function and sine function indicating initial values at 1 = 0- and 1 = 0+. 25 involve an impulse function at t = O. If 1(0+) ::F 1(0-), Equation (2-3) must be modified to ~+[:,t(t)] = sF(s) - [(0+) ~-[:,t(t)]= sF(s) - /(0-) To prove the differentiation theorem, we proceed as follows: Integrating the Laplace integral by parts gives roo -st I00 (:JO[ d ] -sf 10 I(t)e- st dt = I(t) ~s 0- 10 dtl(t) ~s dt Hence, 1(0) 1 [d ]F(s) = - + -;£ -/(t) s s dt It follows that ~[:t[(t)]= sF(s) - [(0) Similarly, for the second derivative ofI(t), we obtain the relationship ~[;:[(t)]= h(s) - s/(O) - j(O) where 1(0) is the value of dl(t)ldt evaluated at t = O. To derive this equation, define d dtl(t) = g(t)
  35. 35. 26 Then The Laplace Transform Chap. 2 !'£[:I:/(/)] =!'£[:lg(/)] = s!'£[g(t)] - g(O) = S!'£[:,t(/) ] - j(O) =l-P(s) - sl(O) - i(O) Similarly, for the nth derivative of f(t), we obtain !'£[:/:t</)] = s"F(s) - r'/(O) - s"-2j(0) - ... ~fW) • (n-l) 1 where f(O), f(O), ... , f(O) represent the values of f(t), dl(t)ldt, ... , dn- f(t)1 dtn- 1, respectively, evaluated at t = O. If the distinction between :£+ and ;£_ is necessary, we substitute t = 0+ or t = 0- into I(t), df(t)ldt, ... , dn-1f(t)ldtn-1, depending on whether we take ;£+ or ;E_. Note that, for Laplace transforms of derivatives of f(t) to exist, dnf(t)ldrn (n = 1, 2, 3, ... ) must be Laplace transformable. Note also that, if all the initial values of f(t) and its derivatives are equal to zero, then the Laplace transform of the nth derivative off(t) is given by snF(s). Final-value theorem. The final-value theorem relates the steady-state behav- ior off(t) to the behavior of sF(s) in the neighborhood of s = O. The theorem, howev- er, applies if and only if lim/-+oof(t) exists [which means thatf(t) settles down to a definite value as t --+ 00].Ifall poles ofsF(s) lie in the left halfs plane, then lim,_oo f(t) exists, but if sF(s) has poles on the imaginary axis or in the right half s plane,f(t) will contain oscillating or exponentially increasing time functions, respectively, and lim,-+oo I (t) will not exist.The final-value theorem does not apply to such cases. For in- stance, if f(t) is a sinusoidal function sin wt, then sF(s) has poles at s = ±jw, and lim/-+oo f(t) does not exist.Therefore, the theorem is not applicable to such a function. The final-value theorem may be stated as follows: If f(t) and df(t)ldt are Laplace transformable, if P(s) is the Laplace transform of f(t), and if lim,_oof(t) exists, then lim f(t) = lim sF(s),_00 s-o To prove the theorem, we let s approach zero in the equation for the Laplace trans- form of the derivative off(t), or lim (oo[dd f(t)]e-s,dt = lim [sP(s) - 1(0)] s-o}o t s-o Since lims_oe-s, = 1, if lim,_oo f(t) exists, then we obtain ['[:,1(/)]dl = 1(/)r= I( 00) - 1(0) = lim sP(s) - 1(0) s-o
  36. 36. Sec. 2-3 Laplace Transformation 27 from which it follows that /(00) = lim /(t) = lim sF(s) 1_00 s-o Initial-value theorem. The initial-value theorem is the counterpart of the final-value theorem. Using the initial-value theorem, we are able to find the value of f(t) at t = 0+ directly from the Laplace transform of/(t). The theorem does not give the value ofI(t) at exactly t = 0, but rather gives it at a time slightly greater than zero. The initial-value theorem may be stated as follows: Iff(t) and df(t)/dt are both Laplace transformable and if lims_ oo sF(s) exists, then /(0+) = lim sF(s) s-OO To prove this theorem, we use the equation for the ~+ transform of df(t)ldt: ~+[:,t(t)]= sF(s) - /(0+) For the time interval 0+ ~ t ~ 00, as s approaches infinity, e-SI approaches zero. (Note that we must use ~+ rather than ~_ for this condition.) Hence, lim [OO[dd /(t)]e-sl dt = lim [sF(s) - /(0+)] = 0 s_oo Jo+ t. s-OO or /(0+) = lim sF(s)s_oo In applying the initial-value theorem, we are not limited as to the locations of the poles of sF(s).Thus, the theorem is valid for the sinusoidal function. Note that the initial-value theorem and the final-value theorem provide a con- venient check on the solution, since they enable us to predict the system behavior in the time domain without actually transforming functions in s back to time functions. Integration theorem. Iff(t) is of exponential order, then the Laplace trans- form of J/(t) dt exists and is given by ~[fJet) dt] = F~S) + r~(o) (2-4) where F(s) = ~[f(t)] and /-1(0) = J/(t) dt, evaluated at t = o. Equation (2-4) is called the integration theorem. The integration theorem can be proven as follows: Integration by parts yields ~[JJet) dt] = f'[fJet) dt]e-"dt = [JJet) dt] e~;I~ -100 f(t) ~; dt
  37. 37. 28 The Laplace Transform Chap. 2 =!jl(t) dti + ! rooI(t)e-st dt s 1;:0 S Jo 1-1(0) F(s) =--+-- s s and the theorem is proven. Note that, if f(t) involves an impulse function at t = 0, then rl(O+)~ /-1(0-). So if f(t) involves an impulse function at t = 0, we must modify Equation (2-4) as follows: ~+[Jf(t) dt] = F~S) + rl~o+) ~-[Jf(t) dt] = F~S) + rl~O-) We see that integration in the time domain is converted into division in the s domain. If the initial value of the integral is zero, the Laplace transform of the inte- gral off(t) is given by F(s)ls. The integration theorem can be modified slightly to deal with the definite inte- gral off!..t). Iff(t) is of exponential order, the Laplace transform of the definite inte- gral Jof(t) dt can be given by ~[[f(t) dt] = F~S) (2-5) To prove Equation (2-5), first note that [f(t) dt = j f(t) dt - rl(O) where rl(O) is equal to Jf(t) dt, evaluated at t = 0, and is a constant. Hence, ~[[f(t) dt] =~[Jf(t)dt - rl(O) ] =~[Jf(t) dt] - ~[rl(O)l Referring to Equation (2-4) and noting that [-1(0) is a constant, so that ;£ff-l(O)] = 1-1(0) S we obtain
  38. 38. Sec. 2-4 Inverse Laplace Transformation 29 Note that, ifJtt) involves an impulse function at t = 0, then J/ f(t) dt ¢ J/ f(t) dt, and the following distinction must be observed: 0+ 0- ~{£ f(t) dt] = ~+~(t)] ~-[.Lf(t) dt] = ~-~(t)l 2-4 INVERSE LAPLACE TRANSFORMATION The inverse Laplace transformation refers to the process of finding the time func- tionJtt) from the corresponding Laplace transform F(s). Several methods are avail- able for finding inverse Laplace transforms.The simplest of these methods are (1) to use tables of Laplace transforms to find the time function Jtt) corresponding to a given Laplace transform F(s) and (2) to use the partial-fraction expansion method. In this section, we present the latter technique. [Note that MATLAB is quite useful in obtaining the partial-fraction expansion of the ratio of two polynomials, B(s)IA(s). We shall discuss the MATLAB approach to the partial-fraction expan- sion in Chapter 4.] Partial-fraction expansion method for finding inverse Laplace transforms. HF(s), the Laplace transform ofJtt),is broken up into components, or F(s) = Ft(s) + F2(S) + ... + Fn(s) and if the inverse Laplace transforms of Ft(s), F2(S), ... , Fn(s) are readily avail- able, then ,rl[F(s)] = ,rt[Fl(S)] + ,rt[F2(S)] + ... + ;e--l[Fn(s)] = ft(t) + h(t) + ... + fn(t) where ft(t),h(t), ... ,fn(t) are the inverse Laplace transforms of F1(s), F2(s),"', Fn(s), respectively. The inverse Laplace transform of F(s) thus obtained is unique, except possibly at points where the time function is discontinuous. Whenever the time function is continuous, the time function f(t) and its Laplace transform F(s) have a one-to-one correspondence. For problems in systems analysis, F(s) frequently occurs in the form B(s) F(s) = A(s) where A(s) and B(s) are polynomials in s and the degree of B(s) is not higher than that of A (s). The advantage of the partial-fraction expansion approach is that the individ- ual terms of F(s) resulting from the expansion into partial-fraction form are very simple functions of s; consequently, it is not necessary to refer to a Laplace trans- form table if we memorize several simple Laplace transform pairs. Note, however, that in applying the partial-fraction expansion technique in the search for the
  39. 39. 30 The Laplace Transform Chap. 2 inverse Laplace transform of F(s) = B(s)/A(s), the roots of the denominator poly- nomial A(s) must be known in advance.That is, this method does not apply until the denominator polynomial has been factored. Consider F(s) written in the factored form B(s) K(s + Zt)(s + zz)'" (s + Zm) F(s) = A(s) = (s + Pt)(s + P2)'" (s + Pn) where Pt, P2, ... , Pn and ZI, Z2, ... , Zm are either real or complex quantities, but for each complex Pi or Zi, there will occur the complex conjugate of Pi or Zi, respective- ly. Here, the highest power ofsin A(s) is assumed to be higher than that in B(s). In the expansion of B(s)/A(s) into partial-fraction form, it is important that the highest power ofs in A(s) be greater than the highest power ofs in B(s) because ifthat is not the case, then the numerator B(s) must be divided by the denominator A(s) in order to produce a polynomial in s plus a remainder (a ratio of polynomials in s whose numerator is of lower degree than the denominator). (For details, see Example 2-2.) Partial-fraction expansion when F(s) involves distinct poles only. In this case, F(s) can always be expanded into a sum of simple partial fractions; that is, B(s) at a2 an F(s) =-=--+--+ ... +-- (2-6) A(s) s + PI S + P2 s + Pn where ak(k = 1, 2, ... , n) are constants. The coefficient ak is called the residue at the pole at s = - Pk' The value of ak can be found by multiplying both sides of Equation (2-6) by (s + Pk) and letting s = - Pk, giving [(S + Pk) B(S)] = [_a_t_(S + Pk) + ~(s + Pk) + ... A(s) s=-p" s + PI S + P2 ak a n ]+ --(s + Pk) + ... + --(s + Pk) s + Pk S + Pn S=-Pk = ak We see that all the expanded terms drop out, with the exception of ak' Thus, the residue ak is found from [ B(s) 1ak = (s + Pk) A() _ S S--Pk (2-7) Note that since f(t) is a real function of time, if PI and P2 are complex conjugates, then the residues al and a2 are also complex conjugates. Only one of the conjugates, al or a2, need be evaluated, because the other is known automatically. Since f(t) is obtained as f(t) = ~-I[F(s)] = ale-PIt + a2e-P2t + ... + ane-Pllt t ::: 0
  40. 40. Sec. 2-4 Inverse laplace Transformation 31 Example 2-1 Fmd the inverse Laplace transform of F(s) _ s + 3 - (s + l)(s + 2) The partial-fraction expansion of F(s) is F( ) s + 3 al a2 s = (s + l)(s + 2) = s + 1 + s + 2 where al and a2 are found by using Equation (2-7): a1 =[(s+1) s+3 ] =[S+3] =2 (s + l)(s + 2) s=-1 s + 2 s=-1 a2 = [(S + 2) s + 3 ] = [~] = -1 (s + l)(s + 2) s=-2 s + 1 s=-2 Thus, f(t) = ~1[F(s)] =~1[s !1] + Tl[S~12] = 2e-t - e-2t t ;;:: 0 Example 2-2 Obtain the inverse Laplace transform of G S3 + 5s2 + 9s + 7 (s) = (s + l)(s + 2) Here, since the degree of the numerator polynomial is higher than that of the denominator polynomial, we must divide the numerator by the denominator: s+3 G(s) = s + 2 + (s + l)(s + 2) Note that the Laplace transform of the unit-impulse function ~(t) is unity and that the Laplace transform of dS(t)ldt is s. The third term on the right-hand side of this last equation is F(s) in Example 2-1. So the inverse Laplace transform of G(s) is given as get) = ..!!.-S(t) + 2~(t) + 2e-t - e-2t dt t ;;:: 0- Comment. Consider a function F(s) that involves a quadratic factor s2 + as + b in the denominator. If this quadratic expression has a pair of complex- conjugate roots, then it is better not to factor the quadratic, in order to avoid com- plex numbers. For example, if F(s) is given as F(s) = pes) s(s2 + as + b)
  41. 41. 32 The Laplace Transform Chap. 2 where a ~ 0 and b > 0, and if s2 + as + b = 0 has a pair of complex-conjugate roots, then expand F(s) into the following partial-fraction expansion form: c ds + e F(s) = -; + -s2-+- a - s -+-b (See Example 2-3 and Problems A-2-15, A-2-16, and A-2-19.) ExampJe2-3 Find the inverse Laplace transform of F(s) = 2s + 12 s2 + 2s + 5 Notice that the denominator polynomial can be factored as s2 + 2s + S = (s + 1 + j2)(s + 1 - j2) The two roots of the denominator are complex conjugates. Hence, we expand F(s) into the sum of a damped sine and a damped cosine function. Noting that; + 2s + S = (s + 1)2 + 22 and referring to the Laplace trans- forms of e-al sin wt and e-al cos Cdt, rewritten as and _ s+a .'i[e at cos Cdt] = 2 2 (s + a) + w we can write the given F(s) as a sum of a damped sine and a damped cosine function: 2s + 12 10 + 2(s + 1) F(s) - - - - - - s2 + 2s + S (s + 1)2 + 22 =S 2 +2 s+1 (s + 1)2 + 22 (s + 1)2 + 22 It follows that f(t) = .'i-I[F(s)] = SX- 1 [(S + 1~2 + 22] + 2!C 1 [(S +S1;21+ 22] = Se-1 sin 2t + 2e-1 cos 2t t ~ 0 Partial-fraction expansion when F(s) involves multiple poles. Instead of discussing the general case, we shall use an example to show how to obtain the partial-fraction expansion of F(s). (See also Problems A-2-17 and A-2-19.) Consider F (s) = s2 + 2s + 3 (s + 1)3
  42. 42. Sec. 2-4 Inverse Laplace Transformation 33 The partial-fraction expansion of this F(s) involves three terms: F( B(s) ~ ~ b1 s) = A(s) = (s + 1? + (s + 1)2 + S + 1 where b3, b2, and b1 are determined as follows: Multiplying both sides of this last equation by (s + 1?, we have B(s) (s + 1)3 A(s) = ~ + b,,(s + 1) + b1(s + 1)2 (2-8) Then, letting s = -1, we find that Equation (2-8) gives [ (s + 1)3 B (S)] = ~ A(s) s=-1 Also, differentiating both sides of Equation (2-8) with respect to s yields :s [(s + 1)3 !~:~] = ~ + 2bt(s + 1) (2-9) H we let s = -1 in Equation (2-9), then !£[(S + 1?B(S)] = ~ ds A(s) s=-1 Differentiating both sides of Equation (2-9) with respect to s, we obtain ~[(S + 1)3 B (S)] = 2b1ds2 A(s) From the preceding analysis, it can be seen that the values of ~, ~, and b1 are found systematically as follows: ~ = [(s + 1)3!~:~L-I = (s2 + 2s + 3)S=-1 =2 ~ = {:s [(s + 1)3 !~:~]L-I = [~(S2 + 2s + 3)]ds s=-1 = (2s + 2)s=-1 =0 b1= ~ {if.[(S + 1)3 B(S)]} 2! ds2 A(s) s=-1 = J,[d 2 2(s2 + 2s + 3)]2. ds s=-1 =1.(2) = 1 2
  43. 43. 34 The Laplace Transform Chap. 2 We thus obtain f(t) = ~-l[F(s)] _;e-l[ 2 ] + :e-1[ 0 ] + ;e-l[_1_]- (s + 1)3 (s + 1)2 S + 1 = t2e-t + 0 + e-t = (t2 + 1)e-t t ~ 0 2-5 SOLVING LINEAR, TIME-INVARIANT DIFFERENTIAL EQUATIONS In this section, we are concerned with the use of the Laplace transform method in solving linear, time-invariant differential equations. The Laplace transform method yields the complete solution (complementary solution and particular solution) of linear, time-invariant differential equations. Classical methods for finding the complete solution of a differential equation require the evaluation of the integration constants from the initial conditions. In the case of the Laplace transform method, however, this requirement is unnecessary because the initial conditions are automatically included in the Laplace transform of the differential equation. If all initial conditions are zero, then the Laplace transform of the differential equation is obtained simply by replacing dldt with s, d21dt2 with s2, and so on. In solving linear, time-invariant differential equations by the Laplace trans- form method, two steps are followed: 1. By taking the Laplace transform of each term in the given differential equa- tion, convert the differential equation into an algebraic equation in s and ob- tain the expression for the Laplace transform of the dependent variable by rearranging the algebraic equation. 2. The time solution of the differential equation is obtained by rmding the in- verse Laplace transform of the dependent variable. In the discussion that follows, two examples are used to demonstrate the solu- tion of linear, time-invariant differential equations by the Laplace transform method. Example 2-4 Find the solution x(t) of the differential equation :i + 3x + 2x = 0, x(O) = a, x(O) = b where a and b are constants. Writing the Laplace transform ofx(t) as Xes), or ;e[x(t)] = Xes) we obtain !e[x] = sX(s) - x(O) !e[:i] = s2Xes) - sx(O) - x(O)
  44. 44. Sec. 2-5 Solving Linear, lime-Invariant Differential Equations The Laplace transform of the given differential equation becomes [s2X(s) - sx(O) - x(O)] + 3[sX(s) - x(O)] + 2X(s) = 0 Substituting the given initial conditions into the preceding equation yields [s2X(s) - as - b] + 3[sX(s) - a] + 2X(s) = 0 or (S2 + 3s + 2)X(s) = as + b + 3a Solving this last equation for X(s), we have X (s) = as + b + 3a = as + b + 3a = _2o_+_b _ _a _+_b s2 + 3s + 2 (s + 1)(s + 2) s + 1 s + 2 The inverse Laplace transform of X(s) produces x(t) = ~-l[X(S)] = ~-1[2a + b]_ ;£-I[a + b] s+1 s+2 = (20 + b)e-I - (a + b)e-2t t ~ 0 35 which is the solution of the given differential equation. Notice that the initial condi- tions a and b appear in the solution. Thus,x(t) has no undetermined constants. Example 2-5 Find the solution x(t) of the differential equation x + 2X + 5x = 3, x(O) = 0, x(O) = 0 Noting that ;£[3] = 3/s, x(O) = 0, and X(O) = 0, we see that the Laplace trans- form of the differential equation becomes 3 S2X(s) + 2sX(s) + 5X(s) = - s Solving this equation for X(s), we obtain 3 X(s) = s(S2 + 2s + 5) 31 3 s+2 =-- - 5 s 5 s2 + 2s + 5 31 3 2 5s Hence, the inverse Laplace transform becomes x(t) = ;£-l[X(s)) 3 s + 1 5(s+I)2+22 -~;£-1[!]_ ~~-1[ 2 ]_ ~;£-1[ S + 1 ] - 5 s 10 (s + 1)2 + 22 5 (s + 1)2 + 22 3 3 -I • 2 3 -I 2= - - - e SID t - - e cos t 5 10 5 t~O which is the solution of the given differential equation.
  45. 45. 36 The Laplace Transform Chap. 2 EXAMPLE PROBLEMS AND SOLUTIONS Problem A-2-1 Obtain the real and imaginary parts of 2 + j1 3 + j4 Also, obtain the magnitude and angle of this complex quantity. Solution 2 + j1 (2 + j1)(3 - j4) - - = 3 + j4 (3 + j4)(3 - j4) 2 .1 = 5" - '5" Hence, 6+j3-j8+4 9 + 16 10 - j5 25 2 real part = - 5' . . .1 nnagmary part = -,-5 The magnitude and angle of this complex quantity are obtained as follows: magnitude = ~n)' + (~1)' = &= ~ = 0.447 -115 -1 angle = tan-1- - = tan-1- = -26.565° 215 2 Problem A-2-2 Find the Laplace transform of Solution Since f(l) = 0 1<0 = te-3r t ~ 0 1 ~[t] = G(s) =: - s2 referring to Equation (2-2), we obtain F(s) = ;e[te-3r] = G(s + 3) = 1 (s + 3)2 Problem A-2-3 What is the Laplace transform of where 8 is a constant? Solution Noting that f(t) = 0 1<0 = sin(wl + 8) I == 0 sin(wt + 8) = sin wI cos 8 + cos wI sin 8
  46. 46. Example Problems and Solutions 37 we have !£[sin(wt + 8)] = cos 8 ~[sin wt] + sin 8 !e[cos wt] w . s = cos 8-2 - - 2 + sm 8-2 - - 2 s+w s+w w cos 8 + s sin 8 s2 + w2 Problem A-2-4 Find the Laplace transform F(s) of the functionj(t) shown in Figure 2-9. Also, find the limiting value of F(s) as a approaches zero. Solution The functionj(t) can be written Then 1 2 1 [(t) = -1(t) - -1(t - a) + -1(t - 2a) a2 a2 a2 F(s) = !eff(t)) 1 2 1 = - !e[1(t)) - - !e[l(t - a)] + - !e[l(t - 2a)] a2 a2 a2 1 1 2 1 -as 1 1 -2as=-----e +--e a2 s a2 s a2 s = ...!....(1 - 2e-as + e-2aS) a2 s As a approaches zero, we have ~(1 - 2e-as + e-2as) 1 - 2e-as + -2as d limF(s) = lim e = Iim-a------- a-O a--O Qls a--O d Ql da ( s) 2se-as _ 2se-2a.f -as -2as = lim = lime -e a-O 2as a-O a f(/} 1 -;; t---""'I 2a o a 1 I-;;Z Figure 2-9 Functionf(/}.
  47. 47. 38 The Laplace Transform Chap. 2 d ( -as -2os)-e -e 20 = lim _d_a____ = lim -se-os + 2se- S -0 d 0-0 1 o -(a) da =-s + 2s = s Problem A-2-S Obtain the Laplace transform of the functionj(t) shown in Figure 2-10. Solution The given functionj(t) can be defined as follows: 1(1) = 0 b = -t a =0 O<tSa a<t Notice that j(t) can be considered a sum of the three functions 'I(t), /2(t), and /3(t) shown in Figure 2-11. Hence,j(I) can be written as 1(1) = fl(t) + /2(1) + /3(t) b b =-t ·1(t) - -(I - a) '1(1 - a) - b'l(t - a) a a Figure 2-10 Functionf(t). Figure 2-11 Functions II(t), 12(t), and !J(t). f(t) b I(t) b o a b II(t) = Ii t· l(t) o~__~~_____________ 12(t) = - E.. (t-a) • l(t - a) a -b "13(t) =-b· l(t - a)
  48. 48. Example Problems and Solutions 39 Then the Laplace transform off(1) becomes F(s) = ~.!. - ~ .!.e-as - b!e-as ai as2 s = ~(1 - e-as) - £e-as as2 s The same F(s) can, of course, be obtained by performing the following Laplace inte- gration: Problem A-2-6 J. ab 100~[f(I)] = -Ie-st dl + Oe-st dl o a a b e-st la J.ab e-sl =-1- - --dl a -s 0 0 a -s e-as b e-sll a =b-+-- -s as -s 0 -as b =b_ e - - _(e-as - 1) -s as2 = ~(1 - e-as) - £e-as as2 s Prove that if the Laplace transform off(l) is F(s), then, except at poles of F(s), d ~[lf(I)] = - ds F(s) d2 ~[12f(I)] = ds2F(s) and in general, n = 1,2,3, ... Solution {OO (OO d ~[lf(l)] = Jo If(l)e-SI dl = - Jo f(l) ds (e-Sl ) dl d J.OO d= - - f(t)e-Sl dl = - - F(s) ds 0 ds Similarly, by defining If(l) = g(I), the result is ~[12f(I)] = ~[lg(I)] = - :sG(s) = - :s[- :sF(s)1 d2 d2 = (-1)2 ds2F(S) = ds2F(S) Repeating the same process, we obtain n = 1,2,3, ...
  49. 49. 40 The Laplace Transform Chap. 2 Problem A-~7 Find the Laplace transform of Solution Since f(t) = 0 t < 0 = t2 sin wt t20 ~[sin wt] = 2 W 2 S + Cd referring to Problem A-2-6, we have d 2 [ w 1 -2w 3 + 6ws 2 :£[f(t)] = :£[t2 sin wt] = -2 -2--2 = (2 2)3 ds S + w s + w Problem A-2-8 Prove that if the Laplace transform off(t) is F(s), then ~[f(;)1= aF(as) a > 0 Solution Ifwe define tla = 'T and as = SJ, then ~[/(;)1= [I(;)e-" dt = [1(T)e-na dT = a1.00/(T)e-'" dT = aF(s,) = aF(as) Problem A-~9 Prove that if f(t) is of exponential order and if fooof(/) dl exists [which means that 1000f(t) dl assumes a definite value], then where F(s) = !e[f(/)]. Solution Note that roof(t) dl = lim F(s) Jo s-O roof(t) dt = lim tf(/) dl Jo ,-ooJo Referring to Equation (2-5), we have ~[[I(t) dt1= F~S) Since ftf(t) dt exists, by applying the final-value theorem to this case, we obtain l' F(s) lim f(t) dt = lim s - ,_00 0 s-O S or roof(t) dt = lim F(s) Jo s-O
  50. 50. Example Problems and Solutions Problem A-2-10 The convolution of two time functions is defmed by [fI(T)[,(1 - T) dT A commonly used notation for the convolution is 11(1)*/2(1), which is defined as f,(I).h(l) = [fl(T)h(1 - T) dT = [11(1 - T)h(T) dT Show that if11(t) and 12(1) are both Laplace transformable, then :£[/"/l(T)h(1 - T) dT] = Fl(S)F2(S) where FI(S) = :e[ft(t)] and F2(S) = :£[f2(1)]. Solution Noting that 1(t - T) = 0 for t < T, we have !e[[fl(T)"(1 - T) dT] = !e[/.oofl(T)"(1 - T)l(1 - T) dT] = /.ooe-'r[/.oofl(T),,(1 - T)l(1 - T) dT] dl 41 = /.oof,(T) dT /.00['(1 - T)l(1 - T)e-" dl Changing the order of integration is valid here, since II(t) and h(I) are both Laplace transformable, giving convergent integrals. If we substitute A = 1 - T into this last equation, the result is or !e[[fl(T)"(1 - T) dT] = /.00f,(TV" dT/.00"p.V'A d)' = FI(S)F2(S) :£[f.(I)*!2(t)] = Fl(S)F2(S) Thus, the Laplace transform of the convolution of two time functions is the product of their Laplace transforms. Problem A-2-11 Determine the Laplace transform of 11(1)*12(1), where Solution Note that II(t) =h(t) = 0 1.(1) = t h(t) = 1 - e-' for t < 0 for 1 ~ 0 for 1 ~ 0
  51. 51. 42 The Laplace Transform The Laplace transform of the convolution integral is given by !£[fl(t)*!2(t)] = Fl(S)F2(S) = :2(.; - s!1) 1 1 1 1 1 1 =-- =---+---- s3 s2(s + 1) s3 s2 S S + 1 Chap. 2 To verify that the expression after the rightmost equal sign is indeed the Laplace trans- form of the convolution integral, let us frrst integrate the convolution integral and then take the Laplace transform of the result. We have fl(t)*!2(t) = 1'''[1 - e-('-T)] d" = [(I - T)(l - e~) dT = 1'(1-T - le~ + Te~)dT Noting that we have Thus, Problem A-2-12 Prove that if/(t) is a periodic function with period T, then Solution 1T f(t)e-" dl ;£[f(I)] = 0 1 _ e-Ts 10 00 00 l<n+l )T ;e[J(t)] = f(t)e-sr dt = ~ f(t)e-sr dt o n=O nT By changing the independent variable from t to" = t - nT, we obtain 00 1T;£[f(t)] = ~e-nTs f(" + nT)e-ST d" n=O 0
  52. 52. Example Problems and Solutions Since f(t) is a periodic function with period T, f(l' + nT) = f(1'). Hence, Noting that we obtain It follows that Problem A-2-13 CXl J.T~[f(t)] = ~e-nTs 0 f( 1')e-SI d1' CXl Le-nTs = 1 + e-Ts + e-2Ts + n=O = 1 + e-TS(1 + e-Ts + e-2Ts + ... ) = 1 + e-TS( ~e-nTs) n=O CXl 1 Le-nTs = ---=- n=O 1 - e-Ts J.Tf(t)e-" dt ~[f(t)] = 0 1 -Ts -e What is the Laplace transform of the periodic function shown in Figure 2-12? Solution Note that J. Tf(t)e-SI dt = J.Tf2e-sl dt + [T(-1 )e-SI dt o 0 JTf2 f(t) 1 o u-1 = e- U I Tf2 _ e-SII T -s 0 -s Tf2 e-(lf2)Ts _ 1 e-Ts _ e-(lf2)Ts ----+------s s = ! [e-Ts - 2e-(1f2)Ts + 1] s = ! [1 - e-(1I2)Tsj2 s Figure 2-12 Periodic function (square wave). 43
  53. 53. 44 The Laplace Transform Chap. 2 Consequently, J.Tf{t )e-" dt (lis)[1 _ e-(ll2)T'l' F(s) = 1 -Ts 1 -Ts -e -e 1 - e-(II2)Ts 1 Ts - - - - - = - tanh- s[1 + e-(ll2)T'1 s 4 Problem A-2-14 Find the initial value of df(I)/dl, where the Laplace transform off(l) is given by 2s + 1 F(s) = ~[f(I)] = s2 + s + 1 Solution Using the initial-value theorem, we obtain s(2s + 1) lim /(1) = lim sF(s) = lim 2 = 2 ,-0+ $-00 s-oo S + s + 1 Since the ~+ transform of d/(I)ldl = g(l) is given by ~+[g(I)] = sF(s) - /(0+) = s(2s + 1) _ 2 = -s - 2 s2 + s + 1 S2 + s + 1 the initial value of df(t)/dt is obtained as d/(I) lim -d- = g(O+) = lims[sF(s) - /(0+)] 1-0+ t s-OO To verify this result, notice that 2(s + 0.5) F(s) = = ~[2e-o.s1 cos 0.866t] (s + 0.5)2 + (0.866)2 Hence, /(1) = 2e-o.51 cos 0.866t and i(t) = _e-o·51 cos 0.8661 + 2e-o·5IO.866 sin 0.8661 Thus, i(o) = -1 + 0 = -1 Problem A-2-lS Obtain the inverse Laplace transform of F(s) = cs + d (s2 + 2as + a2) + b2 where a, b, c, and d are real and a is positive.
  54. 54. Example Problems and Solutions Solution Since F(s) can be written as we obtain Problem A-2-16 F(s) _ c(s + a) + d - ca - (s + a)2 + b2 c(s + a) d - ca b --~-"'7 + - - --~-"'7 (s + a)2 + b2 b (s + a)2 + b2 I(t) = ce-at cos bl + d - ca e-al sin bt b Fmd the inverse Laplace transform of 1 F(s) = -s(-s2=--+-2-s-+-2-) Solution Since S2 + 2s + 2 = (s + 1 + jl)(s + 1 - jl) 45 it follows that F(s) involves a pair of complex-conjugate poles, so we expand F(s) into the form where ah a2, and a3 are determined from 1 = at(s2 + 2s + 2) + (a2s + a3)s By comparing corresponding coefficients of the S2, s, and sO terms on both sides of this last equation respectively, we obtain from which it follows that Therefore, 11 1 s+2 F(s) ::: 2: -; - 2: s2 + 2s + 2 11 1 1 1 s+1 ::: 2: -; - 2(s + 1)2 + 12 - 2(s + 1)2 + 12 The inverse Laplace transform of F(s) is I() 1 1 -I • 1 -I I ::: 2- 2e sm I - 2e cos I t~O
  55. 55. 46 The Laplace Transform Chap. 2 Problem A-2-17 Derive the inverse Laplace transform of 5(s + 2) F(s) = -S2-(s--=-+-I-)(-s":""'+-3-) Solution 5(s + 2) ~ bl at a2 F(s) = s2(s + l)(s + 3) = s2 + -;- + s + 1 + s + 3 where 5(s + 2) I 5 at = s2(s + 3) s=-t = 2: 5(s + 2) I 5 a2 = S2(S + 1) s=-3 = 18 5(s + 2) I 10 ~ = (s + 1)(s + 3) s=O =3 d [ 5(s + 2) 1 bl = ds (s + l)(s + 3) s=O = 5(s + l)(s + 3) - 5(s + 2)(2s + 4) I = _ 25 (s + 1)2(s + 3)2 s=O 9 Thus, 10 1 25 1 5 1 5 1 F(s) = 3 s2 - 9" -; + 2: s + 1 + 18 s + 3 The inverse Laplace transform of F(s) is f(t) = 10t - 25 + ~e-t + 2-e- 3t t ~ 0 3 9 2 18 Problem A-2-18 Fmd the inverse Laplace transform of F(s) = S4 + 2s 3 + 3s 2 + 4s + 5 s(s + 1) Solution Since the numerator polynomial is of higher degree than the denominator polynomial, by dividing the numerator by the denominator until the remainder is a fraction, we obtain 2 2s + 5 at a2 F(s) = s + s + 2 + = S2 + s + 2 + - + - - s(s + 1) s s + 1 where 2s + 51at = - - = 5 s + 1 s=o a2 = 2s + 51 = -3 S s..-I
  56. 56. Example Problems and Solutions It follows that 5 3 F(s) = s2 + s + 2 + - --- s s + 1 The inverse Laplace transform of F(s) is d2 d f(t) = ~-t[F(s)] = -2eS(t) + -deS(t) + 2eS(t) + 5 - 3e-1 dt t Problem A-2-19 Obtain the inverse Laplace transform of F(s) = 2s2 + 4s + 6 s2(s2 + 2s + 10) 47 t ~ 0- (2-10) Solution Since the quadratic term in the denominator involves a pair of complex- conjugate roots, we expand F(s) into the following partial-fraction form: F() at a2 bs + c s :=::-+-+---- s2 S S2 + 2s + 10 The coefficient al can be obtained as _ 2s2 + 4s + 61 - 0 6al - - s2 + 2s + 10 s=O • Hence, we obtain F() 0.6 a2 bs + c s =-+-+---- s2 s s2 + 2s + 10 (a2 + b )s3 + (0.6 + 202 + c)s2 + (1.2 + 10a2)s + 6 s2(s2 + 2s + 10) (2-11) By equating corresponding coefficients in the numerators of Equations (2-10) and (2-11), respectively, we obtain a2 + b = 0 0.6 + 2a2 + c = 2 1.2 + 10a2 :=:: 4 from which we get a2 = 0.28, b = -0.28, c = 0.84 Hence, F() 0.6 0.28 -O.28s + 0.84 s = - + - + ---:----- S2 S s2 + 2s + 10 0.6 0.28 -0.28(s + 1) + (1.12/3) x 3 = - + - + ---------- s2 S (s + 1)2 + 32 The inverse Laplace transform of F(s) gives f(t) = 0.6t + 0.28 - 0.28e-1 cos 3t + 1~2 e-1 sin 3t
  57. 57. 48 The Laplace Transform Chap. 2 Problem A-2-20 Derive the inverse Laplace transform of 1 F(s) = -s(-s2-+-(J)2-) Solution 1 1(1 s)F(s) = = - - - - -S(s2 + c.i) (J)2 s s2 + (J)2 1 l I s = (J)2 ~ - w2 s2 + (J)2 Thus, the inverse Laplace transform of F(s) is obtained as f(t) = ~-l[F(s)] = ~(1 - cos wt) t ~ 0 (J) Problem A-2-21 Obtain the solution of the differential equation x+ ax = A sin (J)t, x(O) = b Solution Laplace transforming both sides of this differential equation, we have or (J) [sX(s) - x(O)] + aX(s) = A-2- -2s + (J) A(J) (s + a)X(s) =-2--2 + b s + (J) Solving this last equation for X(s), we obtain A(J) b X(s) = +-- (s + a)(s2 + ( 2) S + a Aw (1 s - a ) b = a2 + (J)2 s + a - s2 + w2 + S + a ( b AW) 1 Aa w Aw s = + a2 + w2 s + a + a2 + w2 s2 + w2 - a2 + w2 s2 + (Ii The inverse Laplace transform of X(s) then gives x(t) = ~-l[X(S)] (b AW) -aJ Aa. Aw = + 2 2 e + 2 2 sm (J)t - 2 2 cos wt a +(J) a +w a +w t~O
  58. 58. Problems 49 PROBLEMS Problem 8-2-1 Derive the Laplace transform of the function f(t) = 0 t < 0 = te-21 t 2 0 Problem 8-2-2 Fmd the Laplace transforms of the following functions: (8) ft(t) = 0 t < 0 = 3sin(51 + 45°) t 2 0 (b) h(t) = 0 t < 0 = 0.03(1 - cos 2t) 1 2 0 Problem 8-2-3 Obtain the Laplace transform of the function defined by f(t) = 0 1 < 0 = t 2e-or 1 ~ 0 Problem B-2-4 Obtain the Laplace transform of the function f(t) = 0 1 < 0 = cos 2w1 • cos 3(1)1 t 2 0 Problem 8-2-5 What is the Laplace transform of the function f(1) shown in Figure 2-131 Problem 8-2-6 Obtain the Laplace transform of the pulse functionf(t) shown in Figure 2-14. f(t) f(1) c ----- bl------~ o a a+b o a b Figure 2-13 Function f(1). Figure 2-14 Pulse function.
  59. 59. 50 The Laplace Transform Chap. 2 f(t) f(t) 10 ;; 12 ;; o 0 a a 2.5 5 II 12--:; -;;a" Figure 2-15 Function f(1). Figure 2-16 Function f(t). Problem 8-2-7 What is the Laplace transform of the functionf(t) shown in Figure 2-15? Also, what is the limiting value of ;e(f(t)] as a approaches zero? Problem 8-2-8 Find the Laplace transform of the functionf(t) shown in Figure 2-16.Also, find the lim- iting value of ;e[f(t)] as a approaches zero. Problem 8-2-9 Given 5($ + 2) F(s) = s(s + 1) obtain f( 00). Use the final-value theorem. Problem 8-2-10 Given 2(s + 2) F(s)------ - s(s + 1)(s + 3) obtain f(O+). Use the initial-value theorem. Problem 8-2-11 Consider a function x(t). Show that x(O+) = lim [s2X(s) - sx(O+)] s.....oo
  60. 60. Problems Problem B-2-12 Derive the Laplace transform of the third derivative of f(t). Problem B-2-13 What are the inverse Laplace transforms of the following functions? (a) F (s) _ s + 5 1 - (s + l)(s + 3) (b) 3(s + 4) F2(s) = -s(-s-+":-I-)(-s-=-+-2-) Problem B-2-14 Fmd the inverse Laplace transforms of the following functions: (0) (b) Problem B-2-15 r'()_6s+3qS - - - s2 F s _ 5s + 2 2( ) - (s + l)(s + 2)2 Find the inverse Laplace transform of Problem B-2-16 F(s) = 2s2 + 4s + 5 s(s + 1) Obtain the inverse Laplace transform of Problem B-2-17 F(s) = s2 + 2s + 4 s2 Obtain the inverse Laplace transform of F(s) _ s - s2 + 2s + 10 Problem B-2-18 Obtain the inverse Laplace transform of Problem B-2-19 F(s) = s2 + 2s + 5 S2(S + 1) Obtain the inverse Laplace transform of F(s) _ 2s + 10 - (s + 1)2(s + 4) 51
  61. 61. 52 The Laplace Transform Chap. 2 Problem 8-2-20 Derive the inverse Laplace transform of Problem 8-2-21 Obtain the inverse Laplace transform of c _ b _ F(s) = -(1 - e as) - -e as s2 S where a> O. Problem 8-2-22 Find the solution x(t) of the differential equation x + 4x = 0, x(O) = 5, x(O) = 0 Problem 8-2-23 Obtain the solution x(t) of the differential equation x+ w~x = t, x(O) = 0, x(O) = 0 Problem 8-2-24 Determine the solution x(t) of the differential equation 2x + 2x + x = 1, x(O) = 0, x(O) =2 Problem 8-2-25 Obtain the solution x(t) of the differential equation x+ x = sin 3t, x(0) = 0, x(O) = 0
  62. 62. Mechanical Systems 3-1 INTRODUCTION This chapter is an introductory account ofmechanical systems. Details ofmathematical modeling and response analyses ofvarious mechanical systems are given in Chapters 4, 5, 7, 8, and 9. We begin with a review of systems of units; a clear understanding of which is necessary for the quantitative study of system dynamics. Systems of units. Most engineering calculations in the United States are based on the International System (abbreviated SI)1 of units and the British engi- neering system (BES) of measurement. The International System is a modified met- ric system, and, as such, it differs from conventional metric absolute or metric gravitational systems of units. Table 3-1 lists some units of measure from each of the International System, conventional metric systems, and British systems of units. (The table presents only those units necessary to describe the behavior of mechani- cal systems. Units used in describing the behaviors of electrical systems are given in Chapter 6. For additional details on systems of units, refer to Appendix A.) The chief difference between "absolute" systems of units and "gravitational" systems of units lies in the choice of mass or force as a primary dimension. In the IThis "backward" abbreviation is for the French Systeme International. 53
  63. 63. 54 Mechanical Systems Chap. 3 TABLE 3-1 Systems of Units ~ Absolute systems Gravitational systemsof units Metric Metric British Quantity SI mks British engineering engineeringcgs Length m m em ft m it Mass kg kg g lb kgr s2 Ibr s2 - - slug =-- m ft TIme s s s s s s N N dyn poundal Force kg-m kg-m g-cm lb-ft kg[ lb[ =-- =-- =-- =-- s2 s2 s2 S2 Energy J = N-m J = N-m erg ft-poundal kgrm ft-Ib[ = dyn-cm or Btu N-m N-m dyn-cm ft-poundal kgrm ft-lb[ Power w=- w=- -- s s s s s s orhp absolute systems (SI and the metric and British absolute systems), mass is chosen as a primary dimension and force is a derived quantity. Conversely,in gravitational sys- tems (metric engineering and British engineering systems) of units, force is a prima- ry dimension and mass is a derived quantity. In gravitational systems, the mass of a body is defined as the ratio of the magnitude of the force to that of acceleration. (Thus, the dimension of mass is force/acceleration.) Mass. The mass of a body is the quantity of matter in it, which is assumed to be constant. Physically, mass is the property of a body that gives it inertia, that is, re- sistance to starting and stopping. A body is attracted by the earth, and the magni- tude of the force that the earth exerts on the body is called its weight. In practical situations, we know the weight w of a body, but not the mass m.We calculate mass m from w m=- g where g is the gravitational acceleration constant. The value ofg varies slightly from point to point on the earth's surface. As a result, the weight of a body varies slightly at different points on the earth's surface, but its mass remains constant. For engi- neering purposes, g = 9.807 m1s2 = 980.7 cm/s2 = 32.174 ft/S2 = 386.1 in.ls2 Far out in space, a body becomes weightless. Yet its mass remains constant, so the body possesses inertia.
  64. 64. Sec. 3-1 Introduction 55 The units of mass are kg, g, lb, kgr s2/m, and slug, as shown in Table 3-1. If mass is expressed in units of kilograms (or pounds), we call it kilogram mass (or pound mass) to distinguish it from the unit of force, which is termed kilogram force (or pound force). In this book, kg is used to denote a kilogram mass and kg, a kilogram force. Similarly, lb denotes a pound mass and lb, a pound force. A slug is a unit of mass such that, when acted on by a I-pound force, a I-slug mass accelerates at 1 ftls2 (slug = Ibrs2 /ft). In other words, if a mass of 1 slug is acted on by a 32.174-pound force, it accelerates at 32.174 ftls2 (= g). Hence, the mass of a body weighing 32.174Ib, at the earth's surface is 1 slug, or m = w = 32.174 lbr = 1 slu g 32.174 ftls2 g Force. Force can be defined as the cause which tends to produce a change in motion ofa body on which it acts.To move a body, force must be applied to it.1vo types of forces are capable of acting on a body: contact forces and field forces. Contact forces are those which come into direct contact with a body, whereas field forces, such as grav- itational force and magnetic force, act on a body, but do not come into contact with it. The units of force are the newton (N), dyne (dyn), poundal, kg" and lb,. In SI units and the mks system (a metric absolute system) of units, the force unit is the newton. One newton is the force that will give a 1-kg mass an acceleration of 1 m/s2, or I N = 1 kg-m/s2 This implies that 9.807 N will give a 1-kg mass an acceleration of 9.807 m/s2 • Since the gravitational acceleration constant is g = 9.807 m/s2 , a mass of 1 kg will produce a force of 9.807 N on its support. The force unit in the cgs system (a metric absolute system) is the dyne, which will give a 1-g mass an acceleration of 1 cm/s2, or 1 dyn = I g-Cm/S2 The force unit in the metric engineering (gravitational) system is kg" which is a primary dimension in the system. Similarly, in the British engineering system, the force unit is lb" a primary dimension in this system of units. Comments. The SI units of force, mass, and length are the newton (N), kilo- gram mass (kg), and meter (m). The mks units of force, mass, and length are the same as the SI units. The cgs units for force, mass, and length are the dyne (dyn), gram (g), and centimeter (em), and those for the BES units are the pound force (lb,), slug, and foot (ft). Each system of units is consistent in that the unit of force accelerates the unit of mass 1 unit of length per second per second. A special effort has been made in this book to familiarize the reader with the various systems of measurement. In examples and problems, for instance, calcula- tions are often made in SI units, conventional metric units, and BES units, in order to illustrate how to convert from one system to another. Table 3-2 shows some conve- nient conversion factors among different systems of units. (Other detailed conver- sion tables are given in Appendix B.)
  65. 65. 56 Mechanical Systems Chap. 3 TABLE 3-2 Conversion Table 1 1 m = 100cm Length 2 1 ft = 12 in. 1 in. = 2.54 em 3 1 m = 3.281 ft 1 ft = 0.3048 m 4 1 kg = 2.2046 lb 1 lb = 0.4536 kg 5 1 kg = 0.10197 kgrs2 /m 1 kgrs2 /m = 9.807 kg Mass 6 1 slug = 14.594 kg 1 kg = 0.06852 slug 7 1 slug = 32.1741b 1 lb = 0.03108 slug 8 1 slug = 1.488 kgrs2 /m 1 kgrs2 /m = 0.6720 slug 9 1 slug-ft2 = 1.356 kg-m2 1 kg-m2 = 0.7376 slug-ft2 Moment of inertia 10 1 slug-ft2 = 0.1383 kgr s2 -m 1 kgr s2 -m = 7.233 s}ug-ft2 11 1 slug-ft2 = 32.174Ib-tt2 11b-ft2 = 0.03108 slug-tt2 12 1 N = 105 dyn 13 1 N = 0.10197 kg, 1 kg, = 9.807 N 14 1 N = 7.233 poundals 1 poundal = 0.1383 N Force 15 1 N = 0.2248 lb, lib, = 4.4482 N 16 1 kg, = 2.2046Ib, lib, = 0.4536 kg, 17 1lbf = 32.174 poundals 1 poundal =0.031081bf 18 1 N-m = 1 J = 1 W-s 1 J = 0.10197 kgrm 19 1 dyn-em = 1 erg = 10-7 J 1 kgrm = 9.807 N-m Energy 20 1 N-m = 0.7376 ft-Ibf 1 ft-lbf = 1.3557 N-m 21 1 J = 2.389 X 10-4 keal 1 keal = 4186 J 22 1 Btu = 778 ft-Ibf 1 ft-lbf = 1.285 X 10-3 Btu 23 1 W = 1 J/s Power 24 1 hp = 550 ft-lbfls 1 ft-Ibtls = 1.818 X 10-3 hp 25 1 hp = 745.7W 1 W = 1.341 X 10-3 hp Outline of the chapter. Section 3-1 has presented a review of systems of units necessary in the discussions of dynamics of mechanical systems. Section 3-2 treats mechanical elements. Section 3-3 discusses mathematical modeling of me- chanical systems and analyzes simple mechanical systems. Section 3-4 reviews the concept of work, energy, and power and then presents energy methods for deriving mathematical models of conservative systems (systems that do not dissipate energy).
  66. 66. Sec. 3-2 Mechanical Elements 57 3-2 MECHANICAL ELEMENTS Any mechanical system consists of mechanical elements. There are three types of basic elements in mechanical systems: inertia elements, spring elements, and damper elements. Inertia elements. By inertia elements, we mean masses and moments ofinertia. Inertia may be defined as the change in force (torque) required to make a unit change in acceleration (angular acceleration). That is, . . change in force N mertla (mass) = h . I ' --2 or kg c ange m acce eratIOn mls change in torque N-m inertia (moment of inertia) = . . --or kg-m2 change m angular acceleratlon rad/s2 Spring Elements. A linear spring is a mechanical element that can be de- formed by an external force or torque such that the deformation is directly propor- tional to the force or torque applied to the element. Consider the spring shown in Figure 3-1(a). Here, we consider translational motion only. Suppose that the natural length of the spring is X, the spring is fixed at one end, and the other end is free. Then, when a force fis applied at the free end, the spring is stretched. The elongation of the spring is x. The force that arises in the spring is proportional to x and is given by F = kx (3-1) where k is a proportionality constant called the spring constant. The dimension of the spring constant k is force/displacement. At point P, this spring force F acts oppo- site to the direction of the force f applied at point P. Figure 3-1(b) shows the case where both ends (denoted by points P and Q) of the spring are deflected due to the forces f applied at each end. (The forces at each /J---x-j p ..r---_ _ 1 /A-----x+x~ (a) X+X1JX2~Q P 1- - I /---x+X, - X'-! (b) Figure 3-1 (a) One end of the spring is deflected; (b) both ends of the spring are deflected. (X is the natural length of the spring.)
  67. 67. 58 Mechanical Systems Chap. 3 end of the spring are on the same line and are equal in magnitude but opposite in direction.) The natural length of the spring is X. The net elongation of the spring is xI - X2. The force acting in the spring is then (3-2) At point P, the spring force F acts to the left. At point Q, F acts to the right. (Note that the displacements X + Xl and X2 of the ends of the spring are measured rela- tive to the same frame of reference.) Next, consider the torsional spring shown in Figure 3-2(a), where one end is fixed and a torque T is applied to the other end. The angular displacement of the free end is 8. Then the torque Tthat arises in the torsional spring is T = k8 (3-3) At the free end, this torque acts in the torsional spring in the direction opposite that of the applied torque T. For the torsional spring shown in Figure 3-2(b), torques equal in magnitude, but opposite in direction, are applied to the ends of the spring. In this case, the torque T acting in the torsional spring is (3-4) At each end, the spring torque acts in the direction opposite that of the applied torque at that end. The dimension of the torsional spring constant k is torque/angu- lar displacement, where angular displacement is measured in radians. When a linear spring is stretched, a point is reached in which the force per unit displacement begins to change and the spring becomes a nonlinear spring. If the spring is stretched farther, a point is reached at which the material will either break or yield. For practical springs, therefore, the assumption of linearity may be good only for relatively small net displacements. Figure 3-3 shows the force-displace- ment characteristic curves for linear and nonlinear springs. For linear springs, the spring constant k may be defined as follows: spring constant k (for translational spring) change in force N = change in displacement of spring m e ~~--+-r-r;. C~// C9 T (a) (b) Flgure 3-2 (a) A torque T is applied at one end of torsional spring. and the other end is ftxed; (b) a torque T is applied at one end, and a torque T, in the opposite direction, is applied at the other end.
  68. 68. Sec. 3-2 F o Mechanical Elements Nonlinear spring x Figure 3-3 Force-displacement charac- teristic curves for linear and nonlinear springs. spring constant k (for torsional spring) change in torque =------------~----~--------- N-m change in angular displacement of spring rad 59 Spring constants indicate stiffness; a large value of k corresponds to a hard spring, a small value of k to a soft spring. The reciprocal of the spring constant k is called compliance or mechanical capacitance C. Thus, C = 11k. Compliance or mechanical capacitance indicates the softness of a spring. Practical spring versus ideal spring. All practical springs have inertia and damping. In our analysis in this book, however, we assume that the effect of the mass of a spring is negligibly small; that is, the inertia force due to acceleration of the spring is negligibly small compared with the spring force. Also, we assume that the damping effect of the spring is negligibly small. An ideal linear spring, in comparison to a practical spring, will have neither mass nor damping and will obey the linear force-displacement law as given by Equations (3-1) and (3-2) or the linear torque-angular displacement law as given by Equations (3-3) and (3-4). Damper elements. A damper is a mechanical element that dissipates ener- gy in the form of heat instead of storing it. Figure 3-4(a) shows a schematic diagram of a translational damper, or dashpot. It consists of a piston and an oil-filled cylin- der. Any relative motion between the piston rod and the cylinder is resisted by oil (a) (b) Figure 3-4 (a) Translational damper; (b) torsional (or rotational) damper.

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