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SYSTEM DYNAMICS AN INTRODUCTION

DEREK ROWELL DAVID N. WORMLEY

·--,

System Dynamics: An Introduction Derek Rowell David N. Wormley

Prentice Hall, Upper Saddle River. New Jersey 07458

Ubrtlry of Congress Cataloging-in-Publication Data Rowell, Derek. System dynamics : an introduction I Derek Rowell, David N. Wormley

p.

em.

Includes bibliographical references and index.

ISBN

~13-210808-9

1. Systems engineering.

2. System analysis.

I. Wormley, D. N.

IL ntlc. TA168.R69 1997 620' .OOJ 'l--dc20

96-27622 CIP

Acquisitions editor: Bill Stenquist Production editor: bookworks Editorial production supervision: Sharyn Varano Editor-in-Chief: Marcia Horton Managing editor: Bayani MendoZil DeLeon Copy editor: Carol Dean Cover designer: Karen Salzbach Director of production and manufacturing: David W. Riccanli Manufacturing buyer: Julia Meehan Editorial assistant: Meg We~t Composition: PriljJ{, Inc.

The author and publisher of this book have used their best effons 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 .documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs.

«:> 1997 by Prentice-Hall, Inc. A Pearson Education Company Upper Saddle River, NJ 07458

All rights reserved. No pan of this book may be reproduced, in any form or by any means, without permission in writing from the publisher.

Printed in the United States of America

JO 9 8 7 6 5 4 3 2

ISBN

0-13-210808-9

Prentice-Hall International (UK) Limited,London Prentice-Hall of Australia Pty. Limited, Sydney Prentice-Hall Canada Inc., Toronto Prentice-Hall Hispanoamericana, S.A., Mexico Prentice-Hall of India Private Limited, New Delhi Prentice-Hall of Japan, Inc., Tokyo Pearson Education Asia Pte. Ltd., Singapore Editora Prentice-Hall do Brasil, Ltda., Rio de Janeiro

To our families, especially Marjorie and Shirley, for their continuing support and encouragement

Contents

Preface xiii

1

Introduction 1.1 1.2 1.3

System Dynamics 1 State-Determined Systems 5 Physical System Units 12 References 17

2

Energy and Power Flow in State-Determined Systems

2.1

Introduction 2.1.1 2.1.2

2.2

Definition of Power Flow Variables 30 Primitive Rotational Element Definitions 32

Electric System Elements 37 2.4.1 2.4.2

2.5

Definition of Power Flow Variables 22 Primitive Translational Element Definitions 23

Mechanical Rotational Systems 30 2.3.1 2.3.2

2.4

19

Energy Conservation in Physical Systems 19 Spatial Lumping in Physical Systems 20

Mechanical Translational System Elements 21 2.2.1 2.2.2

2.3

19

Definition of Power Flow Variables 37 Primitive Electric Element Definitions 38

Fluid System Elements 44 2.5.1 2.5.2

Definition of Power Flow Variables 44 Primitive Fluid Element Definitions 46

v

~

2.6

Conren~

Thermal System Elements 53 2.6. 1 2.6.2

Definition of Power Flow Variables 53 Primitive Thennal Element Definitions 54

References 65

66

3

Summary of One-Port Primitive Elements

3.1 3.2 3.3

Introduction 66 Generalized Through- and Across-Variables 67 Generalization of One-Port Elements 71 3.3.1 A-TYPe Energy Storage Elemen~ 71 3.3.2 3.3.3 3.3.4

T-'JYpe Energy Storage Elemen~ 76 D-Type Dissipative Elements 78 Ideal Sources 80

3.4 3.5

Causa1ity 82 Linearization of Nonlinear Elements 83 References 91

4

Formulation of System Models

4.1 4.2 4.3

Introduction to Linear Graph Models 92 Linear Graph Representation of One-Port Elements 93 Element Interconnection Laws 95 4.3.1 4.3.2

4.3.3

4.4 4.5

Compatibility 95 Continuity 97 Series and Parallel Connection of Elements

98

Sign Conventions on One-Port System Elements 98 Linear Graph Models of Systems of One-Port Elements 4.5.1 4.5.2 4.5.3 4.5.4 4.5.5

Mechanical Translational System Models Mechanical Rotational Systems 103 Linear Graph Models of Electric Sysrems Fluid Sysrem Models 105 Thermal System Models 107

4.6

Physical Source Modeling References 119

5

State Equation Formulation

5.1

State Variable System Representation 5.1.1 5.1.2

5.1.3 5.1.4

5.2

92

104

108

120 120

Definition of Sysrem State 120 The State Equations 121 Output Equations 123 State Equation-Based Modeling Procedure

Linear Graphs and System Structural Properties 5.2.1 5.2.2

101

101

Linear Graph Properties Graph Trees 127

124

124

124

vii

Contents 5.2.3 5.2.4

5.3 5.4

Input Derivative Form 145 Transformation to the Standard Form

145 147

State Equation Generation Using Linear Algebra Nonlinear Systems 152 5.6.1 5.6.2

Linearization of State Equations References 168

6

Energy-Transducing System Elements

6.1 6.2

Introduction 169 IdeaJ Energy Transduction 6.2.1 6.2.2

150

General Considerations 152 Examples of Nonlinear System Model Formulation

5.7

6.3 6.4

128

State Equation Formulation 135 Systems with Nonstandard State Equations 5.4.1 5.4.2

5.5 5.6

System Graph Structural Constraints The System Normal Tree 129

158

173

Transformer Models 175 Gyrator Models 180

Graph Trees for Systems of 1\vo-Port Transduction Element~ Specification of Causality for Two-Port Elements 188 Derivation of the Normal Tree 190 State Equation Generation 193

References

Operational Methods for Linear Systems

7.1 7.2

Introduction 205 Introduction to Linear Time Domain Operators 207 7.2.1 7.2.2 7.2.3 7.2.4

205

System Operators 207 Operational Block Diagrams 209 Primitive Linear System Operators 209 Superposition for Linear Operators 211

Representation of Linear Systems with Block Diagrams 7.3.1 7.3.2

188

204

7

7.4 7.5

169

Multipart Element Models 181 State Equation Formulation 187 6.4.1 6.4.2 6.4.3 6.4.4

7.3

153

212

Block Diagrams Based on the System Linear Graph 212 Block Diagrams Based on the State Equations 217

Input-Output Linear System Models Linear Operator Algebra 220

219

7 .5.1 7 .5.2 7.5.3

221

Interconnected Linear Operators Polynomial Operators 222 The Inverse Operator 223

viii

Contents

7.6 7.7 7.8 7.9

The System Transfer Operator 225 Transformation from State Space Equations to Classical Form 226 Transformation from Classical Fonn to State Space Representation 233 The Matrix Transfer Operator 237 References 243

8

System Properties and Solution Techniques

8.1 8.2

Introduction 244 System Input Function Characterization 245 8.2.1 8.2.2 8.2.3

8.3

246

Classical Solution of Linear Differential Equations 251 8.3. J 8.3.2 8.3.3

8.4

Singularity Input Functions SinusoidaJ Inputs 248 Exponential Inputs 248

Solution of the Homogeneous Differential Equation 252 Solution of the Nonhomogeneous Differential Equation 255 The Complete Solution 257

System Properties 259 8.4.1 8.4.2 8.4.3 8.4.4

System Stability 259 Time Invariance 261 Superposition for Linear Time-Invariant Systems 262 Differentiation and Integration Properties of LTI Systems

8.5

Convolution 264 References 275

9

First- and Second-Order System Response

9.1 9.2

Introduction 276 First-Order Linear System Transient Response 277 9.2.1 9.2.2 9.2.3 9.2.4

9.3

244

263

276

The Homogeneous Response and the F'JJ'St-Order TlDle Constant The Characteristic Response ofFJl'St-Order Systems 283 System Input-Output Transient Response 286 Summary of Singularity Function Responses 288

279

Second-Order System Transient Response 295 9.3.1 9.3.2 9.3.3

Solution of the Homogeneous Second-Order Equation 300 Characteristic Second-Order System Transient Response 308 Second-Order System Transient Response 315

References 330

10

General Solution of the Linear State Equations

10.1 10.2

Introduction 331 State Variable Response of Linear Systems 332 10.2.1 10.2.2

The Homogeneous State Response 332 The Forced State Response of Linear Systems

331

334

Contents

10.3 10.4

ix

The System Output Response 336 The State Transition Matrix 337 10.4.1 10.4.2 10.4.3 10.4.4 10.4.5 10.4.6 10.4. 7

10.5

Properties of the State Transition Matrix 337 System Eigenvalues and Eigenvectors 338 A Method for Determining the State Transition Matrix 344 Systems with Complex Eigenvalues 345 Systems with Repeated Eigenvalues 347 Stability of Linear Systems 347 Transfonnation of State Variables 349

The Response of Linear Systems to the Singularity Input Functions 10.5.1 10.5.2 10.5.3

References

365

11

Solution of System Response by Numerical Simulation

11.1

Introduction

11.2

Solution of State Equations by Numerical Integration 11.2.1 11.2.2 11.2.3 1J.2.4

11.3

394

The Transfer Function

12.1 12.2 12.3 12.4

Introduction 395 Single-Input Single-Output Systems 396 Relationship to the Transfer Operator 398 System Poles and Zeros 398 The Pole-Zero Plot 400 System Poles and the Homogeneous Response 400 System Stability 404

Geometric Evaluation of the Transfer Function 405 Transfer Functions of Interconnected Systems 407 State Space-Formulated Systems 408 12.7.1 12.7.2

384

Step-Invariant Simulation 384 Ramp-Invariant Simulation 386

12

12.5 12.6 12.7

367

Numerical Integration Techniques 367 Euler Integration of a First-Order State Equation 368 Euler Integration Methods for a System of Order n 374 Higher-Order Integration Techniques 375

References

12.4.1 12.4.2 12.4.3

366

366

Numerical Simulation Methods Based on the State Transition Matrix 11.3.1 11.3.2

353

The Impulse Response 354 The Step Response 356 The Ramp Response 359

Single-Input Single-Output Systems 408 Multiple-Input Multiple-Output Systems 412

References

420

395

Contents

X

422

13

Impedance-Based Modeling Methods

13.1

Introduction 422

13.2

Driving Point Impedances and Admittances · 422 13.2.1 The Impedance ofldeal Elements 424

13.3

The Impedance of Interconnected Elements 426 13.3.1 Series Connection of Elements 426 13.3.2 The Impedance of Parallel-Connected Elements 428 13.3.3 General Interconnected Impedances 430 13.3.4 Impedance Relationships for Two-Port Elements 431

13.4

Transfer Function Generation Using Impedances

13.5

Source Equivalent Models 441 13.5.1 Thevenin Equivalent System Model 441 13.5.2 Norton Equivalent System Model 443 References

434

452

453

14

Sinusoidal Frequency Response of Linear Systems

14.1

Introduction

14.2

The Steady-State Frequency Response

14.3

The Complex Frequency Response

14.4

The Sinusoidal Frequency Response

14.5

The Frequency Response of First- and Second-Order Systems 14.5.1 First-Order Systems 460 14.5.2 Second-Order Systems 463

14.6

Logarithmic (Bode) Frequency Response Plots 467 14.6.1 Logarithmic Amplitude and Frequency Scales 469 14.6.2 Asymptotic Bode Plots of Low-Order Transfer Functions 471 14.6.3 Bode Plots of Higher-Order Systems 478

14.7

Frequency Response and the Pole-Zero Plot 483 14.7.1 A Simple Method for Constructing the Magnitude Bode Plot Directly from the Pole-Zero Plot 486 References 499

15

Frequency Domain Methods

15 .I

Introduction

15.2

Fourier Analysis of Periodic Waveforms 502 15.2.1 Computation of the Fourier Coefficients 507 15.2.2 Properties of the Fourier Series 509

15.3

The Response of Linear Systems to Periodic Inputs

453 453

454 457 460

500

500

514

Contents

15.4

xi

Fourier Analysis of Transient Waveforms 519 15.4.1 15.4.2

15.5

Fourier Transform-Based Properties of Linear Systems 531 15.5.1 15.5.2 15.5.3 15.5.4 15.5.5

15.6

Response of Linear Systems to Aperiodic Inputs 531 The Frequency Response Defined Directly from the Fourier Transform 535 Relationship between the Frequency Response and the Impulse Response 535 The Convolution Property 537 The Frequency Response of Interconnected Systems 538

The Laplace Transfonn 15.6.1 15.6.2 15.6.3

15.7

Fourier Transform Examples 523 Properties of the Fourier Transform 528

539

Laplace Transform Examples 542 Properties of the Laplace Transform 545 Computation of the Inverse Laplace Transform 548

Laplace Transform Applications in Linear Systems 550 15. 7.1 Solution of Linear Differential Equations 550 15.7.2 15.7.3 15.7.4 15.7.5

Solution of State Equations 554 The Convolution Property 555 The Relationship between the Transfer Function and the Impulse Response 556 The Steady-State Response of a Linear System 556

References 563

A

Introduction to Matrix Algebra

A.l A.2 A.3 A.4

Definition 564 Elementary Matrix Arithmetic 565 Representing Systems of Equations in Matrix Form 567 Functions of a Matrix 568 A.4.1 A.4.2 A.4.3

The Transpose of a Matrix 568 The Determinant 569 The Matrix Inverse 570

A.5 A.6

Eigenvalues and Eigenvectors Cramer's Method 572

B

Complex Numbers

8.1 8.2 8.3 B.4

Introduction 574 Complex Number Arithmetic 575 Polar Representation of Complex Numbers 577 Euler's Theorem 579

c

Partial Fraction Expansion of Rational Functions

C.I C.2 C.3

Introduction 580 Expansion Using Linear Algebra 582 Direct Computation of the Coefficients 584

Index 587

564

57 J

574

580

Preface

The system dynamics approach provides important unifying concepts for the analysis of systems which span the traditional engineering disciplines, for example, electrical, mechanical, civil, and chemical engineering. The concepts embodied by system dynamics, while developed for engineering systems, have now been adapted and modified for use in many other types of systems, including economic, biological, and demographic. In this text the concepts of system dynamics are introduced, with the primary emphasis on applications to engineering systems. In particular, the basis for constructing a lumped-parameter system model from a set of primitive element~ in a systematic way and understanding the transient and frequency reponse performance of the model are developed in detail. Many of the concepts and techniques of system dynamics were developed independently within the individual engineering disciplines in the 1940s. The initial emphasis was on dynamic behavior of electric and electronic circuits and mechanical vibrational systems. The gradual realization that there was a set of unifying concepts, particularly related to energy and power flow, alJowed these techniques to be extended to other disciplines. In the 1950s, a new set of analysis techniques, the state space methods, were developed and are now commonly implemented in computer-assisted analyses. Today's practicing engineer must be equally conversant with both the classical input-output system representation and the modern state space approach. In this text, we develop modeling techniques using an energy-based state space fonnulation method and provide linkages to the classical system representations. This approach provides a basis for developing an understanding of basic system behavior and an ability to analyze more complex systems using computer-assisted techniques. This text is a direct outgrowth of our many years of teaching a sophomore level course in system dynamics within the Department of Mechanical Engineering at Massachusetts Institute of Technology. The course covers most of the material in the book but not at the depth indicated by some of the later chapters. In writing the text we have attempted to organize it for use in a range of courses throughout the undergraduate or early graduate levels. Each chapter is organized so that early chapter sections provide a basis for an xill

xiv

Preface

introductory level course, while later chapter sections are more advanced. The problems accompanying each chapter are similiarly organized. Thus, an instructor may select chapters and chapter sections for an introductory one-semester or one-quarter course, or may include more advanced materials for senior level or graduate students. Chapter sections suggested for an introductory level course, as weii as a more advanced cou!Se, are listed in Table 1. A suggested introductory level course includes model formulation techniques for mechanical, electric, and fluid and thermal systems in a systematic manner, resulting in a set of state equations, the use of operational techniques to transform between state space and classical system representations, input waveform characterization, and a detailed development of first- and second-order transient solution and sinusoidal frequency response techniques. The materials can be readily taught with the assistance of computer-based analysis packages by selection of appropriate homework problems. A more advanced course may include additional modeling materials on multienergy domain systems u~ing two-port primitive elements, advanced_solution techniques for statedetermined system models, nonlinear modeling and analysis techniques, impedance model formulation techniques, and advanced frequency domain analysis techniques. In developing this text we have given considerable thought to the choice of a graphical representation of system structure. Over the years we have taught modeling methods using the various graphical methods, for example, linear graphs, bond graphs, and block diagrams. On balance, we believe that the linear graph method has distinct advantages at the introductory level: It stresses the concepts of continuity and compatibility and is easy for students to grasp. In this book frequency domain methods and the Laplace transform have been delayed until the final chapters. Our decision to do this is based upon our desire to stress the time domain basis of the dynamic behavior of systems~ We have chosen to introduce and use linear operational algebra to manipulate the dynamic equations directly in the time domain. Fourier and Laplace methods are introduced toward the end of the book, as a method of waveform representation, and then used to defines-plane-based system properties. Derek Rowell David N. Wormley

Preface

XV

Table 1. Distribution of Course Material

TOPICS (COURSE MATERIALS) Introduction to System Dynamics Concepts (Ch. 1) Modeling (Cbs. 2 - 6) Energy Concepts (Ch. 2) Primitive Elements (Ch. 2) A Multi-Domain Unified Approach (Ch. 3) Linear Nonlinear Model Construction (Ch. 4) Equation Derivation (Cbs. 5 and 6) Single energy domain (Ch. 5) Multiple energy domains (Ch. 6) Operational Methods (Ch. 7) Block diagrams Equation manipulation System response Classical Time-Domain Solution Methods (Ch. 8) Low-Order SISO System Response (Ch. 9) FlfSt-Order Response Second-Order Response State-Space Time-Domain Solution Method~ (Ch. 10) Numerical Solution Techniques (Ch. 11) Transfer Functions (Ch. 12) Impedances (Ch. 13) Frequency Response Methods Sinusoidal Response (Ch. 14) Fourier and Laplace Transform Methods (Ch. 15) .

ThiTRODUCTORY

ADVANCED

1.1 - 1.3 2.1 2.1-2.6 3.1-3.4 3.5 4.1 -4.5

5.1-5.3

5.4-5.5 6.1-6.4

7.1-7.3 7.5-7.7

7.8-7.9

8.1-8.4

8.5

9.1 -9.2 9.3 10.1- 10.3 11.1-11.2 12.1- 12.6

10.4- 10.5 11.3 12.7 13.1 - 13.4

14.1- 14.6 15.1- 15.7

1

Introduction

1.1

SYSTEM DYNAMICS Engineering is a creative activity leading to the design, development, and manufacture of devices and systems to meet societal needs. Many of the engineering technologies and systems developed to address such issues as human health and safety, energy production and utilization, and the environment involve the integration of elements from diverse engineering, scientific, and social disciplines. Engineering systems, ranging from large structures such as tall bui1dings and bridges excited by wind and seismic forces, to ocean ships and platforms excited by wind and waves, to audio and video consumer products, represent systems in which the dynamic, or time-varying, response to external inputs is critical. In many cases the effectiveness of a system is intrinsically related to its dynamic behavior. For example, the design of the automobile air bag, developed to improve occupant safety, involves mechanical, electrical, and chemical engineering components as well as a detailed knowledge of human physiology. The dynamic response of the air bag, in terms of the requirement to detect a crash condition and deploy within a period of a few mi11iseconds, is critical to its effectiveness. Similarly, an electric power generation and distribution system utilizes technologies that span several engineering disciplines-fluidic, combustion, nuclear, electrical, and so on. The power system must respond in a timely manner to daily and seasonal changes in demand and react to sudden failures and faults in the system components without interruption of service. Inadequate responses to disturbances may result in blackouts, such as occurred in the northeastern United States in the fall of 1965. Historically engineers have developed specialized methods for analyzing the behavior of systems within their own discipline. For example, electrical engineers have developed and refined circuit analysis methods in order to determine the response of voltages and currents in electronic systems; and structural and mechanical engineers have developed methods of computing forces and displacements within systems assembled from mechanical components. The generalized discipline of system dynamics has been developed over the past five decades to provide a unified method of system representation and analysis that can 1

Introduction

2

Chap. 1

Gripper

Figure 1.1: A high-perfonnance industrial robot.

be applied across a· broad range of technologies. System dynamics concepts are now used in the analysis and design of many types of interconnected systems including mechanical, electric, thermal and fluid systems [1-9]. The general methodologies arising from this field have recently been extended to the analysis of many other types of systems including economics, biology, ecology, the social sciences, and medicine [10-12]. In dynamic systems, a number of questions related to system performance are often of interest. For example, consider robots, which are now commonly used in highly repetitive or hazardous tasks at automated manufacturing facilities. The dynamic behavior of a robot arm as it moves a payload from point tO point significantly affects its productivity. The design. of a high-performance robot, such as is shown in Fig. 1.1; must consider many factors. Of prime importance is the interaction between the robot, which may be defined as the system, and the forces and torques originating from its environment. Questions that may be addressed through dynamic analyses of the robot include the following:

• How fast can the robot perform its tasks, and what size· motors and actuators are needed to achieve the desired speed?

Sec. 1.1

System Dynamics

3

• What are the forces and torques on the various system components during the execution of a task? • Given a command to move a load held by the robot from one point to another, what is the trajectory of the arm? • How sensitive is the accuracy of the spatial position of the gripper to changes in the load? Such questions can often be answered by developing an analytical model of the robot and its environment and using the model to develop performance data as a function of system parameters such as the size of the electric motors selected to move the arm. Selection of a set of parameters that yield an acceptable performance from simulation studies forms the basis for the design and construction of a prototype. System dynamics is the study of the dynamic or time-varying behavior of a system and includes the following components: 1. Definition of the system, system boundaries, input variables, and output variables. 2. Formulation of a dynamic model of the physical system, usually in the form of mathematical or graphical relationships determined analytically or experimentally. 3. Determination of the dynamic behavior of the system model and the influence of system inputs on the system output variables of interest. 4. Formulation of recommendations or strategies to improve the system performance through modification of the system structure or parameter values. These implicit elements of dynamic analyses are commonly employed by engineers in the design and development of a wide range of complex systems including spacecraft, automobiles, energy production and distribution systems, computer and control systems, water distribution and purification systems, chemical processes, and manufacturing systems. The advent and proliferation of digital computing methods now allow analysis of proposed designs using dynamic simulation methods before expensive prototypes are constructed. The following example illustrates a number of elements of system dynamics. In the example engineers recognized the possibility of undesirable wind-induced motions in a structure and developed a novel solution using the principles of system dynamics. Example 1.1 Tall buildings. above 50 stories in height, often exhibit oscillatory motions when subjected to external forces and sway with a period of approximately one cycle every 6-10 s [13]. Under windy conditions the motion of the top floors may be several feet from side to side. The motion of such structures is important because when building motion is sufficiently large, people occupying the upper floors may become uncomfortable and even experience motion sickness. Figure 1.2 depicts a tall building responding to wind forces. The building structure is a dynamic system. with wind forces on the building acting as inputs and. the resulting motion as output A dynamic model of the system must include a description of the structural response to the forces generated on the building under various wind conditions. In the design of several tall buildings, it was recently anticipated that windinduced motion could be a problem. and systems known as tuned-mass dampers were included as an integral part of the building to modify the motion response characteristics. These units are dynamic systems consisting of a large mass sliding on lubricated bearing surfaces on an

Introduction

4

Chap. I

Mass velocity vm(l)

II II II

II II II

/

(a)

F..,(r)

vm(l)

0

"'~

.,c

.£

Building (unmodified)

~

System output

·g

II

u>

eo c

:§ ·;; ~

(b) F..,(T)

"" l

v \J

vu

vm(l)

System input

"'

~

.£

.,c ~

Building (modified)

System output

0 ·;:; 0

u>

eo c

:§

·;;

(c)

~

Reduction of building motion using a tuned-mass damper. (a) The structure of a tuned-ma~s damper showing the sliding mass coupled to the building through springs and dampers. (b) The building response to sinusoidal wind force s without the tuned-mass damper. (c) The reduction in building motion with the tuned-mass damper in action.

Figure 1.2:

upper floor of the building which are coupled to the building through a spring and a damper system. The system is designed so that as the building sways at its own natural frequency, the mass oscillates at the same frequency but in a direction opposite the building motion. The tuned-mass damper exerts a reaction force on the building structure as it moves, which can significantly reduce building motion. Building designers have successfully employed tunedmass dampers in the John Hancock Building in Boston, Massachusetts, and in the Citicorp Building in New York City. In these two cases the designers, after initially evaluating a system response and finding it unacceptable, were able to modi fy the original system by adding a set of dynamic mechanical elements to achieve an acceptable level of overall system performance.

Sec. 1.2

1.2

State-Determined Systems

s

STATE-DETERMINED SYSTEMS Fundamental to system dynamics is the interaction between a system and its environment. In the broadest context of system dynamics, a system and its environment are defined as abstract entities:

• System: A col1ection of matter, thoughts, or concepts contained within a real or imaginary boundary.

• Environment: All that is external to the system. The interaction between a system and its environment is characterized in terms of a set of system variables, as illustrated in Fig. 1.3, which in engineering systems may be timevarying physical quantities such as forces, voltages, or pressures or mathematical variables with no direct physical context. These variables may be internal to the system and reflect the state of an element, for example, the force acting on a spring, or they might express the time variation of some quantity at the interface between the system and its environment. It is useful to define two important classes of system variables:

• Inputs: An input is a system variable that is independently prescribed, or defined, by the system's environment The value of an input at any instant is independent of the system behavior or response. Inputs define the external excitation of the system and can be quantities such as the external wind force acting on a tall building system or the rainfall forming the input flow into a reservoir system. A system may have more than one input.

• Outputs: An output is defined as any system variable of interest. It may be a variable measured at the interface with the environment or a variable that is internal to the system and does not directly interact with the environment

Figure 1.3:

Schematic representation of a dynamic system.

The identification of a system and inputs and outputs may be illustrated by considering the design of an automobile suspension. The goal is to achieve both good handling characteristics, to ensure safe operation during cornering and driving maneuvers, and good ride comfort while traversing bumpy roads. Suspension design requires a trade-off in the selection of the stiffness of the springs and damping effects of the shock absorbers, to achieve the good handling (relatively stiff suspensions) associated with high-performance cars, and the good ride quality (relatively soft suspensions) associated with more conventional cars.

Introduction

6

Chap. 1

In this case the system may be defined as the automobile itself, with inputs from the environment defined as the road surface profile and the driver's steering actions. Output variables of interest are vehicle motions and accelerations and forces between the road and · the tires, which provide a measure of the handling and ride quality. Questions of interest to the designer include: Do all wheels remain in contact with the ground during emergency actions taken by the driver? What are the vertical accelerations under typical city and highway road conditions? To develop an initial suspension design from the system definition, a mathematical model, which can predict vehicle motions from the roadway and steering inputs of the system, is generated and implemented as a computer program. Representative road profiles and driver inputs are created and used as inputs to the simulation to determine the quality of the ride and the handling characteristics of the car. Through many simulations, based on different suspension parameters, the design trade-offs are identified. Finally a set of suspension parameters may be selected that are recommended for testing in a prototype vehicle. A central element in all dynamic analyses is the formulation of a mathematical model of the system. Many physical systems of interest to engineers may be represented by a set of mathematical equations that fonn a state-determined system model of the system. A state-determined system model has the characteristic that 1. a mathematical description of the system, 2. specification of a limited set of system variables at an initial time to, and 3. specification of the inputs to the system for all time t :::: to are necessary and sufficient conditions to determine the system behavior for all time t > to. The definition of a state-determined system model is developed from the concept of system state, which as described by Kalman [14] and others [15-17] is represented by the specification of a minimum set of variables, known as state variables, that uniquely define the system response at any given time. The mathematical model of the system is given by a set of state equations. This definition states that the response of a state-determined system model to any arbitrary input can be determined if the value of each of the state variables at time to (known as the initial conditions) and the time history of the inputs for time t :::: to are known. Thus, at any time to, the system state, defined by the values of the state variables, completely characterizes the present condition of the system and no information is required · concerning the past history of the system. Although the number of state variables required to represent a system is uniquely determined for a state-determined system model, the specific set of state variables is not unique. Many different sets of state variables may be used to describe a particular system, however, the set must be complete and ail the state variables in the set must be mathematically independent In this text we introduce and use a set of state variables that are both physically measurable and directly related to the energy stored in system elements. In the following example, the state variable concept is illustrated for a simple mechanical system.

Sec. 1.2

State-Determined Systems

7

Example 1.2 In Fig. I .4 an automobile is shown traveling along a straight road under the influence of a time-varying propulsive force F1,(r). We assume that the velocity v(r) of the car is of primary interest. A simple state-determined model of the car may be formulated by considering the system to consist of the car as a mass element m, acted o n by the sum of the propulsion force and forces resisting the motion (for example, wind drag and ground resistance). In this case the velocity, which determines the car's kinetic energy, is selected as the state variable and is also the output variable of interest. Velocity

I'---•~- v.,(r) Propulsive force Fp(l)

Figure 1.4:

Automobile traveling on a straight. flat road.

If at time to the velocity is u0 , the initial kinetic energy E:o of the vehit:le is (i)

For time r > r0 , the propulsion force F1,(t) from the engine is considered to be the input (independently specified by the driver), and if all resistance forces are neglected, the acceleration is determined by Newton's law: dv

mdt

= Fp(T)

(ii)

where v is the system state variable, Fp is the system input, and m is a system parameter. The velocity v(r) at any time may be determined by integrating Eq. (ii):

l

v(t) dv

vCto) v(r)

=

1t J

-

F1,(r) dr

(iii)

= -1 1t Fp(r) dt + v(to)

(iv)

to m

m

to

Equation (iv) shows that the velocity v(r) at any timet can be determined if 1. the initial velocity v(to) is known, and

2. the input F,(l) is known for timet

~ t0 •

The velocity thus satisfies the requirements of a state variable, and Eq. (ii) is a state-determined model of the car. If a constant propulsion force Fp(t) = F is applied to the mass and the initial time to = 0, the velocity fort > 0 is v(r)

= -IllF t + v(O)

(v)

8

Introduction

Chap. 1

The velocity v(t) therefore increases linearly with time when a constant propulsion force is . applied in the absence of any resistance forces. The resulting velocity v(t) for any other Fp(t) can also be found by simply substituting into Eq. {iv) and solving the integral. While velocity v is a logical and convenient choice for a state variable in this example, the choice of a state variable is not unique, and momentum p = mv could have equally well been chosen. Beeause velocity and momentum are directly related by the constant mass m, knowledge of either one determines the other variable and either one descnl>es the system state.

The concept of a state-detennined system model, while developed most fully for physical systems, has also been utilized in the study of social and economic systems. In the next example, a classic predator-prey population model is described following the presentation of Luenberger [10]. Example 1.3

On a remote islarid a colony of rabbits created serious ecological problems by depleting the ground cover vegetation and opening the soil to erosion by wind and rain. In an effort to control the rabbit population a number of foxes were brought to the island to reduce the rabbit population. After the foxes were introduced. a cyclic oscillation in the populations of the foxes and the rabbits was observed. A dynamic model of the population dynamics that explains the oscillatory behavior of the populations can be developed from generic population models [ 10]. Consider the island as defining the boundary of the system. Let the number of rabbits x1 {t) and the number of foxes x 2{t) be system variables. Before the introduction of the foxes. and in the presence of a plentiful supply of food. a simple model for the rabbit population assumes that the breeding rate of rabbits is constanL When the birth and death rates are taken into account. the rate of population growth is proportional to the size of the population iLc;elf, that is. (i)

where a is a positive constanL Equation (i) is a- dynamic model of unconstrained population growth. The system response may be found by reorganizing Eq. (i) and integrating:

dx1 -=adt Xt

1

XJ(f)

ZJ(O)

}

-dx1 =a Xt

1'

(ii)

dt

(iii)

0

with the result (iv) which indicates exponential growth. Clearly at some point additional factors not considered in the model represented by Eq. (i) must come into play because the finite food supply on the island cannot support an infinite rabbit population. Mer the predator is introduced. the growth rate is significantly altered. We may surmise that the death rate of rabbits due to foxes depends on both the number of rabbits and the number of foxes. and conjecture that the death rate is proportional to the product of the two populations [10]. The model in Eq. {i) is therefore modified by the inclusion of an extra tenn that accounts for the effect of the fox population: · dxt(t)

--;u- =

ax1 (t)·-

bx1 (t)x2(t)

{v)

Sec. 1.2

State-Determined Systems

9

where b is a positive constant. The presence of the foxes has a negative effect on the rabbit population growth rate. Equation (v) is not a complete model of the system because the fox population x 2 (t) is itself dependent on the rabbits. The fox population model is developed by considering two cases. First, because rabbits are the primary source of food for foxes, we assume that the fox population will die out in the absence of any rabbits on the island, giving a simple model for the fox population: (vi) where c is a positive constant The solution of this differential equation is (vii)

indicating an exponential decay in the number of foxes. Second, we assume that the growth rate of the fox population is affected by the product of the number of rabbits and foxes, with a resultant model of the fonn (viii) where d is a positive constant. The complete model of the predator-prey system is represented by Eqs. (v) and (viii) together:

(ix)

which are a pair of coupled nonlinear differential equations. This model is state-determined because, given the two equations with the values of the four system parameters a, b, c, and d and the initial values of the two state variables x 1(to) and x2(t0) at some time to, the response for all t ::::. to may be computed. This model has no input (forcing function), and the dynamic response of the two populations is determined only by the initial conditions. The response of the model equations may be determined by numerical integration on a digital computer using methods described in Chap. 11. A typical set of results showing the evolution of the populations can be seen in Fig. 1.5. In Fig. 1.5a the unconstrained exponential growth of the rabbit population is shown; this fonn of response is typical of an unstable system; in Fig. 1.5b a typical oscillatory response is shown where both populations undergo cyclic variations but the rabbit population is kept within acceptable bounds.

State-determined system models are the primary types of models utilized in engineering and science. However, for some systems it is not possible to formulate a state-determined model because the initial conditions cannot be completely determine~ the system description is inadequate, or the inputs cannot be completely specified. An example of such a system is the stock market, where investors are not able to formulate a state-determined model, that is, a system description relating stock prices to a finite set of variables, and cannot quantify the inputs that influence the system. · In this text the techniques of system dynamics are developed with primary application to engineering systems (including mechanical, electric, fluid, and thermal systems) and the· interactions occurring among these systems. An essential characteristic of these

10

Chap. 1

Introduction 1600

{a)

1400 1200 c

.i0

1000

-;

a.

8.

:E .c CIS ~

800 600 400 200 0 2500

2

Time (yr)

3

4

5

(b)

2000

= 1500

·~

:; Q, 0

Q.,

JOOO

500

0

5

10 Time(yr)

15

20

Figure 1.5: Typical responses of the population dynamics predicted by the predator-prey model with parameters a 1, b 0.002, c 0.5, and d = 0.001. (a) Predicted exponential growth of the rabbit population from an initial population of 10 rabbits in the absence of foxes. (b) Oscillatory population response caused by the introduction of 100 foxes into a population of 1000 rabbits.

=

=

=

physical systems is the association of their dynamic behavior with changes in the storage and dissipation of energy in the system· components. Thus, the central basis for formulating state-determined system models of these physical systems is the conservation and conversion of energy. It is often convenient to consider a system as an interconnection of several subsystems that are. wholly contained within th~ overall system. In this context th~ complete system is considered to be part of the environment for each of the subsystems. For example, an automated manufacturing machine might in~lude several electric subsystems (~otors and their control systems) and many mechanical subsyst~ms for parts handling and ~achining. This

Sec. 1.2

11

State-Determined Systems

process of subdividing a system m ay be continued unti l the subsystem is a single lumped

element that cannot be further divided. An element is the simplest prim itive unit f rom which all systems can be constructed. For the engineering systems discussed in this text primitive elements that supp ly, s tore, and dissipate energy in mechanical, electric, fluid, and thermal energy domains are de fin ed in Chap. 2. The formulation of state-determined system models of phys ical systems requires

1. selecti on of a set of primitive elements to represent the e nergy interaction within a system a nd between a system and its environment, and 2. definition of the system structure or the m anner in which the primitive e lem ents are connected.

Example 1.4 A simplified model of an automobile suspension system is required for some preliminary design studies. Only the vertical motions of the car need to be considered. The system is taken to be the car itself, and the input is defined by the profile of the road surface. The engineers decide that for this initial study it is adequate to model ( I) the car body as a single lumped-mass element, (2) the primary suspension as a spring representing the combined effects of all four springs, a dashpot representing the energy dissipation in the shock absorbers, and a mass element representing the axles, and (3) the tires as a spring element to model their compliance together with a dashpot to account for frictional losses as the tires are flexed. The structure of the model is shown schematically in Fig. 1.6. It consists of interconnected primitive mass, spring, and dashpot elements driven by a vertical velocity input source element. Although simplified, this model using seven lumped elements can give valuable insights into the dynamic response of automobiles as they travel down a rough road.

_____ .J vll) 7/7;~~7/7.:-rro~ ,:;~777./m.,_j v, (I) (a) An automobile Figure 1.6:

t

V,(l)

(b) Lumped parameter model

Simplified schematic representation of an automobile suspension system as an interconnection of seven primitive elements.

A technique based on the linear graph as a unifie d means of expressing the topology or structure of physical systems is developed in Chaps. 3 and 4 . Linear graphs are similar in form to electric circuit diagrams but prov ide a commo n fram ework for the representatio n of mechanical, electric, fluid, and thermal systems. In addition they prov ide a basis for the coupling of systems that involve more than one energy modality. The development of state-determined models from a linear graph representation is described in Chap. 5.

12 1.3

Introduction

Chap. 1

PHYSICAL SYSTEM UNITS In physical system models, the system parameters and variables are ~xpressed in terms of specific measures or units, some of which have evolved over a long period of time. In the United States, units in common use have been derived primarily from the English system. (Note, however, that England has now converted to the metric system and no longer uses the English system!) For example, the English unit of length is the foot (ft), which was originally defined as the length of the Icing's foot. As civilization has progressed, precise and repeatable standards have been developed to represent fundamental units of measurement The International System of Units, Systeme International d'Unites (known as SI units in all languages), is the system that has been adopted by the principal industrial nations of the world, including the United States. SI units are the primary system of units used throughout this book, however, because much engineering practice in the United States is based on the English unit system, the relationship between the SI and English unit systems is discussed below. Each physical quantity may be described in terms of a set of generalized dimensions such as length and time. For example, the dimension of the unit of meter or foot is length, and of area is length squared. In a consistent set of units, a minimal set of fundamental dimensions may·be defined in terms of particular units, and then all other quantities may be defined in terms of the units of these fundamental dimensions. SI units are based on a set of seven fiindaniental, or base, units as summarized in Table 1. 1. All other units of llieas.urement, known as derived units, may be expressed in terms of the base units. In this book the fundamental dimensions of time (t), l~ngth (L), mass (m), current (i), and temperature (T) are used primarily. TABLE 1.1:

Sl Physical System Units•

Quantity Time Length Mass Current Temperature Luminous intensity Amount of substance Area Volume Velocity Aeceleration Force Energy Power Voltage Pressure

Dimension

SI Unit

t

second meter kilogram ampere kelvin candela mole

L

m T

Symbol

s m

kg A K cd mol

m2 square meter m.l . cubic meter Lt-1 meter per second m/s Lt-2 meter per second squared mls 2 mLr-2 newton N (kg-rills 2) J (N-m) mL 2r 2 joule W (J/s) mL2 r 3 watt mL 2r- 3;-• V (W/A) volt mL -1,-2 pascal Pa (N/m 2 ) L2 L3

•The top section shows the seven primary quantities, while the bottom section shows a set of commonly used derived quantities.

Sec. 1.3

13

Physical System Units

The selection of a basic set of units is not unique; for example, either force or mass may be selected as a primary dimension. Newton's law provides the connection between the units of force and mass since a pure mass element m under the influence of an external force F experiences an acceleration a given as F=ma

(1.1)

If mass (m) is selected as a primary dimension, along with length (L) and time (t), then force may be defined in terms of mass (m) and acceleration (Lr- 2 ) as

F=ma

while if force (/) is selected as a primary dimension, then mass may be defined as F

m=a

In the SI system mass is chosen as the base quantity and force is the derived unit In the English system force is the base unit, expressed in pounds (lb), and the unit of mass, the slug, is defined as that mass that accelerates at I ft/s 2 when subjected to a constant force of lib. The mass of a quantity of material is related to its weight. Since weight is a measure of the gravitational attraction force on a particular object, although directly proportional to the mass, it is expressed in units of force. If the weight of an object is determined as W from a scale, then the mass is found by Newton's law F = ma; noting that F = Wand a = g, the acceleration due to gravity

w

m=g

(1.2)

At the surface of the earth g is 9.81 m/s 2 or 32.17 ft/s 2 , and so a mass of 1 slug has a weight of 32.17 lb, while in the SI system a mass of 1.0 kg requires a force of 9.81 N to support its weight The basic unit of mass is the kilogram or the slug. Energy is also identified in Table 1.1 with the basic dimension of mL 2 t- 2 , which is equivalent to force times length and in the SI system is expressed in newton-meters or joules (after James P. Joule who in the 1840s experimentally verified the first law of thermodynamics). Power is defined as the rate of flow of energy, or the energy flowing per unit time (mL 2r- 3), and thus has units ofN-rnls and is defined in watts (after James Watt who made significant advances in steam engine performance between 1760 and 1820). A power unit of 1 W is equivalent to 1 J/s. Conversion factors between SI and English units are summarized in Table 1.2. Also shown in the table are two common English measures of energy and power: the British thermal unit (Btu), which is a measure of energy, and the horsepower (hp), which is a measure o(power. In the SI system, pressure (force per unit area) is defined in pascals (after Jacques Pascal who made major contributions in the field of fluid dynamics) or N/m 2 • Example 1.5 How much electric power (W) is available from a 150-hp electric generator?

14

Introduction TABLE 1.2:

Chap. 1

Unit Conversion Factors for Common Sl and English Engineering Units

Quantity

From

Multiply by

To produce

Length

m

3.2808 0.3048 0.2248 4.4482 0.06852 14.594 0.7376 1.3558 0.02088 47.8806 9 5 s

ft m lb

ft Force

N

Mass

lb kg slug

Energy

J

Pressure

ft-lb Pa Jb/ft 2

Temperature

K

OR Other factors: Temperature

Energy Power Pressure Gravity

N

slug kg ft-lb J lb/ft2 Pa

OR K

9

°F = 0 R - 459.67 oc = K- 273.15 °C= ~(°F- 32)

1.0 Btu =778.16 ft-lb J.O hp =550 ft-lb/s 1.0 Jblin. 2 144 lb/ft2 1.0 Pa =1 N/m 2 g =9.81 m/s 2 = 32.17 ft/s 2

=

Solution From Table 1.2, the 150 hp may be converted to the fundamental English units of ft-lb/s by using the factor 550 ft-lb/slhp: (150 hp) (550 ft-lb/slhp) = 82,500 ft-lb/s This power level may be converted to SI unite; (W or J/s) by the factor of 1.3558 J/ft-lb in Table 1.2: · (82,500 ft-Jb/s) (1.3558 J/ft-lb)

=111,854 J/s

Noting that I W is equal to 1 J/s, approximately 111.854 kW is available from a 150-hp generator.

PROBLEMS 1.1. Consider a home heating system. The outside temperature and radiant heat from the sun influence the internal room temperature. The furnace and its thermostat control are used to maintain the house temperature at a desired level as the external weather conditions change. (a) Use an engineering sketch to describe the system of interest and its environment, identifying the system inputs and outputs. Identify the sources of heat flow between the system and the environment (i) when the room temperature is above the temperature set on the thennostat, and (ii) when the room temperature is below the set point. (b) Discuss how an increase in the effectiveness of wall insulation in the house walls influences the system behavior.

Chap. I

15

Problems

1.2. Wind energy is used in some areas as a renewable energy source. A wind turbine, illustrated in Fig. I. 7, is used to supply electric energy to a remote farm house. The rotational energy of the blades drives an electric generator through a mechanical gear train. The generator supplies electric energy to the house, where it is used directly or stored in baneries. The house draws water from an artesian well through an electric pump, and stores the water in a tank at the top of a tower.

Storage tank II II

,

·:

~·

II II

II II

II II

II II

L

w'"'

Wind

~:

j§

1

j Pump

Pu 1 I" Figure 1.7:

Electric power

A rural wind power generation system.

(a) Use a sketch to illustrate the energy flow in the wind to electric energy generation system, and identify the system inpuL~ and outputs. (b) Identify the major energy conversion and transmission elements that are important in the overall conversion of wind to electrical power.

(c) Would you expect the system to be 100% efficient? Is all of the available energy in the wind flow converted to electrical energy? Identify potential sites of energy dissipation. (d) Trace the flow and conversion of electrical energy through a typical appliance that rrtight be found in the farm house, such as a stove. Where does the energy ultimately end up?

(e) Discuss the storage and dissipation of energy associated with the water storage system.

1.3. Solar heating systems use incident energy from the sun's radiation. In a typical solar heating system a fluid is circulated from an insulated storage tank through solar collectors, where heat is transferred to the fluid and returned to tbe tank. A second circulation system is used to pump the heated fluid from the tank through radiators in the building. (a) A variable speed pump, which is controlled by the temperature of the fluid leaving the solar collectors, sets the circulation rate of the input flow. Construct an engineering diagram showing the important elements of the heat collection system. Identify the system inputs and outputs. What variable characterizes the thermal energy stored in the insulated tank? (b) In the heat distribution system a fixed speed pump is turned on or off by a thermostat in the building and is used to force the hot water through the radiators. Augment the sketch in part (a) to include this second circulation loop and identify any additional system inputs and outputs. Describe how the two circulation systems are influenced by the total thermal energy stored in the tank.

(c) Comment on the design factors that you think would affect the efficiency of the solar collector panels.

16

Introduction

Chap. 1

1.4. Many urban areas receive their water supply from reservoirs .located in nearby mountains. Rain and melting snow in the catchment area flows through streams into the reservoir. Water is drawn from the reservoir by pumps for supply to the urban area. In addition, water evaporates into the atmosphere . from the surface of the reservoir, and water seeps into the ground around the reservoir. We are interested in the variation of the total volume of water stored in the reservoir. (a) Describe the system of interest using a sketch and identify the system inputs and the output of interest. (b) The net volume flow rate of water into the reservoir varies with the time of year, as does the net flow leaving the reservoir. The net input and output volume flow rates are shown in Fig. 1.8. If the net flow into the reservoir is Qin and the net flow out is Qou1• the resultant total reservoir net Qin- Qout determines the change in the total volume of water in the reservoir at flow Qnet any given time. Determine and plot the total reservoir net flow as a function of time.

=

f

Q(t)

2.5

e

2

-e·

1.5

~ B

0

>

Qjp(l)

I I I I I I I

Qout (t)

I I I I

r-------~

---

~

c

e=

1------.r 1

;;;-

I

0

:

·------------

0.5

2

4

6

8

10

12

Time (months) Figure 1.8:

Reservoir volume flow rates.

(c) The volume of water V(t) in the reservoir at any time, t, is equal to the initial volume V0 at time t = 0 plus the integral of the total reservoir net flow over time

V (t)

= Vo +

1'

Qnctdl

Determine the total volume of the water in the reservoir as a function of time if at time t = 0, V0 = 4 x 108 m 3 • Plot the volume as a function of time. When is the volume a minimum? In many urban areas, a water emergency is declared if a reservoir reaches a sufficiently low level, which for the example is 2.5 x 1011 m 3 • Does the volume of water in the reservoir ever decrease to an emergency level during the year'?

1.5. A large truck pulls into a highway weighing station to ensure that it meets state weight limits. The truck's weight is 24,000 lb. (a) What is the mass of the truck (i) in SI units and (ii) in English uiuts. (b) Use Ne\Vlon's laws of motion to determine the propulsive force that the truck's engine must develop to accelerate the truck at 2.5 ftls 2 in (i) Sl units and (ii) in English units. (c) If the truck accelerates from rest with a constant force of I 0,000 ,lb, how far has it traveled after 10 seconds? What is its speed at this time?

Chap.

t

17

References

1.6. One of the proposals to reduce pollution in congested areas is to encourage the use of electric battery-powered automobiles. In these vehicJes an electronic module controls the energy ftow from a battery to an electric motor, which drives the wheels of the car through a gear train. The electric current delivered to the motor is established by the driver through adjustment of the controller. We wish to formulate a model for the electric car that is appropriate for operation on a nominally straight, level road. (a) Using an engineering sketch, define the car and propulsion system including inputs and any outputs of interest (b) In a test drive on an electric vehicle it was found that an average of 20 horsepower was used over the trip duration of two hours. How many joules of electrical energy are used from the battery during the trip? If the car uses 24-volt batteries, what was the average current drawn during the trip (assume that the battery voltage remains constant).

1.7. A robotic machine is used to move a mass m in a cyclic motion along a straight line. For high speed motions the dominant force which the robot must provide is the inertial force associated with the mass. Assume that the prescribed cyclic motion is x = xo sin wt. where x is the displacement in meters and w is the angular frequency in radls, and xo is the amplitude of the motion. (a) Determine expressions for the mass velocity and acceleration as a function of time. If the displacement xo is 2.0 em, at what cyclic frequency is the maximum acceleration of the mass equal to 5 times the nominal acceleration of gravity. (b) The mass inertial force is equal to the product of the mass and the acceleration. For a frequency of (J) =50 radls what is the largest mass that can be moved with an amplitude of x 0 = 2.0 em if the robotic machine can provide a peak force of 10 N?

REFERENCES [1) Paynter, H. M., Analysis and Design of Engineering Systems, MIT Press, Cambridge, MA, 1961. [2] Koenig, H. E., Tokad, Y., Kesavan, H. K., and Hedges, H. G., Analysis of Discrete Physical Systems, McGraw-Hi11, New York, 1967. [3] Shearer, J. L., Murphy, A. T., and Richardson, H. H., Introduction to System Dynamics. AddisonWesley, Reading, MA, 1967. [4] Blackwell, W. A., Mathematical Modeling of Physical Networks, Macmillan, New

Yor~

1967.

[5] Crandall, S. H., Karnopp, D. C., Kurtz, E. F., Jr., and Predmore-Brown, D. D., Dynamics of Mechanical and Electromechanical Systems, McGraw-Hill, New York, 1968. [6] Doebelin, E. 0., System Dynamics Modeling and Response, Charles E. Merrill, Columbus, OH, 1972. [7] Ogata, K., System Dynamics, Prentice Hall, Englewood Cliffs, NJ, 1978. [8] Karnopp, D. C., Margolis, D. L., and Rosenberg, R. C., System Dynamics: A Unified Approach (2nd Ed.), John Wiley, New York, 1990. [9] Shearer, J. L., and Kulakowski, B. T., Dynamic Modeling and Control of Engineering Systems, Macmillan, New York, 1990. [10] Luenberger, D. G., Introduction to Dynamic Systems, Theory, Models, and Applications, John Wiley, New York, 1979. [11] Riggs, D. S., Control Theory and Physiological Feedback Mechanisms, Williams and Wilkins, Baltimore, MD, 1970. [12] Forrester, J. W., Principles of Systems, MIT Press, Cambridge, MA, 1968. [13] Wiesner, K. B., ..Taming Lively Buildings," Civil Engineering, 56(6), 54-57, June 1986.

18

Introduction

Chap. 1

[14] Kabnan, R. E., ••0n the General Theory of Control Systems:• Proceedings of the First IFAC Congress, 481-493, Butterworth, London, 1960.

[ 15] Athans, M., and Falb, P. L., Optimal Control: An Introduction to the Theory and Its Applications, McGraw-Hill, New York. 1968. [16] Schultz, D. G., and Melsa, J. L., State Functions and Linear Control Systems, McGraw-Hill, New York, 1967. [17] Timothy, L. K., and Bona, B. E., State Space Analysis: An Introduction, McGraw-Hill, New York, 1969.

2

Energy and Power Flow in State-Determined Systems

2.1

INTRODUCTION 2.1.1 Energy Conservation in Physical Systems

A set of primitive elements that form the basis for construction of dynamic models of a physical system may be defined from the energy flows within the system and between .the system and its environment. In this chapter we define such primitive elements, which characterize the generation, storage, and dissipation of energy in four energy domains: mechanical, electric, fluid, and thermal. In Chap. 6 an additional set of elements representing energy flows between energy domains are defined. The principle of energy conservation provides a fundamental basis for characterizing and defining the primitive elements. An idealization is adopted in which it is assumed that a system model exchanges energy with its environment through a finite set of energy or power ports [1] as shown in Fig. 2.1. For systems defined by such a boundary, the law of energy conservation may be written as d£ dt

P(t) = -

(2.1)

where tis time, £(t) is the instantaneous stored energy within the system boundary, and 'P(t) is the instantaneous net power flow across the system boundary (where power flow is defined as positive into the system and negative out of the system). Equation (2.1) states that the rate of change of stored energy in the system is equal to the net power flow 'P(t) across the system boundary. It is assumed that the system itself contains no sources of energy. All sources are external to the system and influence the dynamic behavior through the power flows across the system boundary. 19

20

Energy and Power Flow in State-Determined Systems

Figure 2.1:

Chap.2

Power flows into a dynamic system.

For the systems considered in this chapter, the power flow across the system boundary over an incremental time period dt may be written as 'Pdt

= L\W + L\H

(2.2)

where L\ W is the increment of work performed on the system by the external sources over the period dt and L\H is the increment of heat energy transferred to the system over the same period. Positive work and heat flow are defined as increasing the system total energy. The work done on the system and heat flow crossing the system boundary result in a change in the total energy level of the system as expressed by Eq. (2.1 ), which when integrated with respect to time and combined with Eq. (2.2) is a form of the first law of thermOdynamics [2]. 2.1.2 Spatial Lumping in Physical Systems In the formulation of system models. it is convenient to consider power flows across the

system boundary to be localized at a set of discrete locations on the system boundary as shown in Fig; 2.1. H the power flows are localized at n sites, the net power flow is the sum of the power. flows at each location: 'P(t)

= 'Pt (t) + 'P2(t) + ·· · + 'Pn(t) n

= L'P;

(2.3)

i=l

where 'P; (t) is the power flow at location i. The power flows include power delivered {or extracted) by energy sources in the environment as well as power flows from the system due to energy dissipation in system elements. The power flows in Eq. (2.3) do not include the flow of power between system elements contained within the boundary. .In a similar manner the total system energy· at any instant may be expressed as the sum of energies stored at m discrete locations within the system: e(t) = Et (t)

+ E2(t) + · ·· + Em(t) (2.4)

Sec. 2.2

Mechanical Translational System Elements

21

The m energies are associated with a set of m energy storage elements in the system model. The energy conservation law may then be expressed in terms of local power flow sites and energy storage elements as n

L:P;(t) i=l

m

d£·

i=J

dt

= L:-'

(2.5)

which states that the total power flow across the boundary is distributed among the m energy storage elements. Equation (2.5) may be applied directly to systems consisting of lumped-parameter elements where elements considered to be "lumped" in space have locally uniform system variables and parameters. In a lumped-parameter model, the variables at a given spatial location are used to represent the variables of regions in the near vicinity of the point In Example 1.2, which examines an automobile traveling along a road under the action of propulsion forces, a single velocity is used to represent the motion of the car as a whole; in effect the entire car mass m is considered to be located at a single position. All points on the body are represented by a single forward velocity. Differences in the velocity of different parts of the car, for example, resulting from vibration in the car structural members, are neglected. Similarly, in the automobile suspension examined in Example 1.4, the whole car is lumped into a small number of discrete, interconnected elements. In lumped-parameter dynamic models the variables at discrete points in the system are functions of time and are described by ordinary differential equations. They are not considered continuous functions of both time and position as is characteristic of spatially continuous or distributed models described by partial differential equations [3]. In the following sections, variables are defined for power and energy flow at discrete locations in mechanical, electric, fluid, and thermal systems. These variables provide the basis for defining a set of lumped-parameter elements representing energy sources, energy storage, energy dissipation, and energy transformation. Each-element is described in terms of a constitutive equation expressing the functional relationship for the element in terms of material properties and geometry.

~.2

MECHANICAL TRANSLATIONAL SYSTEM ELEMENTS

Mechanical translational systems are characterized by straight-line or linear motion of physical elements. The dynamics of these systems are governed by the laws of mechanical energy conservation and are described by Newton's laws of motion. The power flow to and from translational systems is through mechanical work supplied by external sources, and energy is dissipated within the system and transferred to the environment through conversion to heat by mechanical friction. There are two mechanisms for energy storage within a mechanical system: 1. As kinetic energy associated with moving elements of finite mass

2. As potential energy stored through elastic deformation of springlike elements.

Energy and Power Flow in State-Determined Systems

22

Chap.2

Two energy-conserving elements, based on these storage mechanisms, together with a third dissipative element representing frictional losses, are used as the basis for lumpedparameter modeling of translational systems. The functional definition of these elements may be developed by considering the mechanical translational power flow into a system. 2.2.1 Definition of Power Flow Variables

Figure 2.2 shows ·~m energy port into a mechanical system through which power is transferred by translational motion along a prescribed direction. For any such mechanical system the power flow 'P(t) (W) at any instant is the product of the velocity v(t) (rnls) and the collinear force F(t) (N): 'P(t)

= F(t)v(t)

(2.6)

The increment in energy expended or absorbed in the form of mechanical work flowing through a power port in an elemental time period dt is defined as ~W

= 'P(t) dt = F(t)v(t) dt

~W

(2.7)

By convention the external source is said to perform work on the system when the power is positive, that is, when F(t) and v(t) have the same sign (or act in the same direction), as shown in Fig. 2.2. When 'P(t) > 0, the external source supplies energy to the system. This energy may be stored within the system and later recovered, or it may be dissipated as heat and thus rendered unavailable to the system. If F(t) and v(t) act in opposite directions, the ·power flow is negative and the system performs work on its environment. The two variables F(t) and v(t) are the primary variables used to describe the dynamics of mechanical translational systems throughout this book. Reference

r--------. Power into port @l(t)

Reference direction

=F(t)v(t)

Figure 2.2: Mechanical system with a single power port.

~

Equation (2.7) may be integrated to determine the total mechanical work W transferred through the port in a time period 0 ~ t ~ T: W

=iT

'P(t)dt

=iT

(2.8)

Fvdt

It is useful to introduce two additional variables; the linear displacement x(t) (m), which is the integral of the velocity: x(t) =

fo' v(t)dt +x(O)

or

dx

= vdt

(2.9)

Sec. 2.2

Mechanical Translatianal System Elements

23

and the linear momentum p(t) (N-s), which is defined as the integral of the force: p(t)

= Ia' F(t) dt + p(O)

or

dp = Fdt

(2.10)

The work performed across a system boundary in time dt may be expressed in terms of the power variables and the integrated power variables in the following three forms: LlW LlW LlW

=

= =

F(vdt) v(Fdt) (Fv)dt

= = =

Fdx vdp (vF)dt

= =

dt:'kinetic

=

d£'dissipated

dEpotential

(2.11)

These three forms of power flow illustrate the origins of the definitions of the three lumped-parameter elements used in mechanical system models - the spring, mass, and damper elements. The first expression states that the mechanical work may result in a change in the system stored energy through a change in displacement dx, which is usuaiJy associated with potential energy storage in a springlike element. The second expression states that the work may result in a change in momentum dp, which is usually associated with energy storage in a masslike element. The third expression indicates the work is converted into heat, as occurs with friction in a mechanical system, and is no longer available in mechanical form. The energy is then considered to have been dissipated and transferred to the environment. Lumped-parameter models of physical systems may be defined in terms of these three elements. Often considerable engineering judgment is required in deciding how to represent physical components in terms of primitive elements. These decisions require knowledge of the function of the component within the. system. For example, a large coil spring that does not undergo any deflection but has a uniform velocity might be represented as a simple mass because the only energy stored is kinetic. A coil spring in an automobile suspension might be represented as a pure spring for slowly varying applied forces since the energy stored within it is based on the spring deflection. For very rapid1y varying deflections of the coil, however, a significant amount of kinetic energy might be associated with the rapid1y moving mass of the coil spring. In high-performance mechanical systems a model of a physical spring may need to include both primitive spring and mass effects. 2.2.2 Primitive Translational Element Definitions Energy Storage Elements Energy storage within a mechanical translational system may be described in terms of two lumped-parameter primitive elements defined in terms of the power and integrated power variables.

Translational Spring: A pure mechanical translational spring element is defined as an element in which the displacement x across the element is a single-valued monotonic function of the force· F: X= :F(F) (2.12) where FO is a single-valued monotonic function, as shown in Fig. 2.3. Equation (2.12) is known as the constitutive equation for a spring. The displacement x is the net spring

Energy and Power Flow in State-Determined Systems

24

Chap. 2

deflection, as shown in Fig. 2.3, expressed in tenns·of the difference of two measurements in the fixed coordinate system, less the spring rest (natural) length lo, that is, x = x2-Xt -lo. v = v2 -v1 VJ(t) /.

ttK

X

v2(t)

2~

cu

F(t) . ~~~ F(t)

e

x,~,_J

-5.8 xo "' Q F0

F

Force Figure 2.3: Definition of the pure translational spring element.

From Eq. (2.11) the energy stored in a spring with displacement x is &=

fo" Fdx

(2.13)

and is illustrated as the shaded area in Fig. 2.3. The energy stored in the spring is a direct function of x and is zero when x = 0. A pure spring is any element described by Eq. (2.12) and may be represented by a nonlinear relationship between x and F. The restriction that the relationship is single-valued and monotonic ensures that a unique relationship exists between X and F so that given a displacement x, the force may be uniquely determined, or given F, the displacement may be uniquely determined. An ideal or linear spring is defined as a pure spring in which the relationship between displacement and force is linear so that Eq. (2.12) becomes x·=CF

(2.14)

where the constant of proportionality C is defined to be the spring compliance (miN). In engineering practice linear springs are usually described by the reciprocal of the compliance, and the linear relationship ofEq. (2.14) is written F=Kx

. (2.15)

where K = 1I C (N/m) is defined to be the spring constant or stiffness. Equation (2.15) is Hooke's Jaw for a linear spring [4]. For an ideal spring with a displacement x from the rest position and an applied force F K x, the energy stored is

=

E=

XFdx= LX Kxdx=-Kx I 1 2 = - F2 . o 2 2K Lo

(2.16)

The stored energy is always a positive quantity since it depends on the square of the displacement or force.

Sec. 2.2

Mechanical Translational System Elements

25

The relationships between force and displacement for springs depend on geometry and material properties. Tabulations of spring constants and derivations of constitutive equations for simple coil and beamlike springs are contained in several references [4, 5]. Figure 2.4 shows several configurations for mechanical springs, including a coil spring, a deflecting beam, and a compressible material. The definition of a pure or ideal spring and its stored energy, given in Eq. (2.16), requires that the displacements be defined as shown in Fig. 2.4, and so at x = 0 the force and energy are both zero and any finite displacement x is associated with energy storage. F(t)

t

(a) Coil spring

(b) Cantilevered beam

(c) Compressible material

Figure 2.4: Examples of translational springs.

An equation for the ideal spring, expressed directly in terms of the power variables force and velocity, may be derived by differentiating the ideal constitutive equation Eq. (2.15): dF (2.17) -=Kv dt Equation (2.17) is called an elemental equation because it expresses the element characteristic in terms of power variables. Mass Element: A pure translational mass element m is defined as an element in which the linear momentum pis a single-valued monotonic function of the velocity v, that is

P = :F(v)

(2.18)

In general for very high velocities, the constitutive relationship for a pure mass is given by the nonlinear relativistic relationship [6]

mv

p= -;:::=== 2 (vfc)

Jt-

(2.19)

where m is the mass at rest (kg), and c is the velocity of light (m/s). This constitutive relationship is shown in Fig. 2.5. The energy of the mass is

E=

1P vdp

and is indicated by the shaded area in the figure.

(2.20)

Energy and Power Flow in State-Determined Systems

26

Chap. 2

Constant velocity reference v1

v0 Velocity Figure 2.5:

v

Definition of the pure translational mass element

At velocities much less than the speed of light, v equation for an idea] mass element:

0. A tank operator HT (} may be defined, and the relationship between Pc(t) and Q(t) written in operational form: Pc(t)

= HT {Q(t)} s -1 c

f.' 0

Qdt

(7.6)

The output Pc(t) is interpreted as the result of HT {} acting on Q(t). A cause and effect is implicit in this operational statement of the system response; the output is the result of an action taken by the system on the input. In the three examples shown in Fig. 7.2 the required mathematical actions are one or more of multiplication by a constant (scaling}, differentiation, and integration. An operator such as HT {} represents a mathematical transformation on a system function. The actions defined by an operator may be complex, for example, representing the overall input-output behavior of a high-order system, or may be as simple as the algebraic scaling operation of the voltage divider in the example in Fig. 7.2b. An operator is similar

208

Operational Methods for Linear Systems

Chap. 7

Output vK(t)

Input ~ F K(t} ~ O:::::::::::J---1

(a} Massless spring driven by a force source

v,.(r)

--8--

•.

(b) Electric voltage divider

Q ( t ) - - & P.,(t) Input Q(t)

__

~

P.,(t) =

~f.'o_t!t+P.,(O)

_..._,_.. (c) Vertical walled fluid tank capacitance

Figure 7.2: Three simple linear systems showing a physical system representation. a block diagram implying a relationship between input and output variables. and the mathematical operation relating output to input

to an algebraic function, such as a sine or exponential function, in the sense that both transform (or map) an input within their domain to an output Unlike a function, which maps instantaneous values of the input to the output, an operator maps an input function to an output function. At any instant the output of an operator may depend not only on the current value of the input but also on its complete time history. For example, the operator Hr {}defined for the tank computes the integral of the input from timet = 0 to the present time. Such operators are defined to be dynamic operators. The voltage divider in Fig. 7 .2b contains no energy storage elements, and its output depends on only the current value of the input; it is an algebraic or static operator. Algebraic functions, such as the square root, sine, and exponential may be considered nonlinear static operators. The use of operational methods for manipulating and solving linear differential equations in engineering systems was developed by the English electrical engineer Oliver Heaviside in the late 1800s. His work was originally criticized for its apparent lack of rigor, and it was not until many years after his death that its importance was realized. The basis for the operational algebra and calculus based on Heaviside's work has since been developed in detail [9- 1 1] but is beyond the scope of this text. In this chapter we draw on some of the results of functional methods to develop alternative descriptions and methods for manipulating linear systems.

Sec. 7.2

Introduction to Linear TliDe Domain Operators

209

7.2.2 Operational Block Diagrams

Functional operators describing a system are often depicted graphically in an operational block diagram. The system behavior defined by the input-output transformation is shown graphically as a processing block with a single input and a single output An arrow, pointing into the block, defines the input and therefore the causality. Each block in the diagram specifies the operational relationship between the input function and the output as indicated in Fig. 7.1. Block diagrams can be drawn at many levels of detail, from a complete description of the input-output relationship of a system in a single block to a detailed interconnection diagram showing the primitive mathematical operations implied by each system element. Each block is labeled with its appropriate operator function. The block diagram structure for any system is not unique; many equivalent block diagrams may be constructed for a given system. While transfonnations on single variables are described by operators, two or more system variables are combined through the arithmetic operations of addition, subtraction, multiplication, and division shown in block diagram fonn in Fig. 7.3. Although all four arithmetic operations may be needed to describe complex nonlinear systems, only two (addition and subtraction) are needed in the description of linear systems. (Strictly only the process of addition is required because the negation of a variable may be defined through a scaling operator before summation.) A "signed" addition element, as shown in Fig. 7.3, is used in block diagrams to allow either addition or subtraction at any of its inputs. u 1 (t)~

Ut(t)Yy(t)

u2(t)

= u 1(t) + u2(t)

+

~ y{t)

= u 1(t)

X u2(t)

u2(r)

(c) Multiplication

(a) Addition

Ut(l)~ ~ y(t):::u 1(t)/u2(t)

u2(t) (b) Subuaction (signed addition)

Figure 7.3:

(d) Division

Arithmetic operations to combine system variables.

7.2.3 Primitive Linear System Operators

All linear time-invariant systems may be represented by an interconnection of three primitive operators: 1. The constant scaling operator: The scaling operator multiplies the input function by a constant factor. It is denoted by the value of the constant, either numerically or symbolically, for example, 2.0 {} , 1/m {}, RI/(Rl + R2) {},or a{}.

210

Operational Methods for Linear Systems

Chap. 7

2. The differential operator: If a function f(t) is differentiable and has the property that f(t) = 0 at timet= 0, then the differential operator, designated S {},generates the time derivative of the input f(t) for time t > 0: 1 y(t) = s {f(t)}

=dtd {/(t)}

fort> 0

(7.7)

Notice that the output of the operator is defined only in the period t 2: 0.

3. The integral operator: The integral operator, written s- 1 {} (or sometimes as 1/S), is defined as

y(t) =

s- 1 lfl

"'fo' J

dt

(7.8)

where it is assumed that at timet = 0, the output y(O) = 0. If an initial condition y(O) is specified, y(t) =

s-I {f(t)} + y(O)

(7.9)

and a separate summing block is included at the output of the integrator to account for the initial condition. Figure 7.4 is the block diagram representation of the primitive operators. In addition, two other useful operators may be defined:

4. The identity operator: The identity operator I {} leaves the value of the input unchanged, that is, y(t) =I (J(t)}

= f(t)

(7.10)

The action of the identity operator is therefore equivalent to scaling the input by unity.

5. The null operator: The null operator N (} identically produces an output of zero for any input; that is, y(t)

= N (f(t)} = 0

(7.11)

for any f (t). The action of the null operator is equivalent to scaling the input by zero.

The operational calculus, based upon generalized functions [9-11 ], provides a definition of the differential operator that aJlows for nonzero values of the function f (t) at the time origin t = 0. In particular the full definition is y(t)

= s {/(t)} =dtd {/(t)} + /(0) &(t)

where «S(t) is the Dirac delta function introduced in Chap. 8. This definition is not pursued further in this text; we simply state here that results derived in this chapter generalize to situations involving nonzero initial conditions.

Sec. 7.2

211

Introduction to Linear Time Domain Operators (a) Scaling

u(l)

(b)Diffetendation

u(r)

(c) Integration

u(t)

-8---------0-

y(r)

y(t)

y(t)

y{O)

Figure 7.4:

The three primitive operational elements required for linear time-invariant system analysis.

7.2.4 Superposition for Linear Operators

A linear operator, written £ {}, is defined as one that satisfies the principle of superposition. When the input f(t) to a linear operator C {}is the sum of two component variables, that is, f (t) = ft (t) + fz (t), the principle of superposition requires that C{ft(t)

+ /2(t)} =

£{/I(t)}

+ C{/2(t)}

(7.12)

The result of any linear operator C {} acting on the combined variable [/I (t) + /2(t)] is the same as the sum of the effects of that operator acting on the two variables independently. Figure 7.5 illustrates this required operational equivalence in block diagram form. All the three primitive system operators defined above are linear since a {/t (t) + /2(t)} =aft (t) + a/2(t) d d S {/I (t) + /2(t)} = dt /1 (t) + dt f2(t)

s-• lf•J =

£

hOO

fl(t)dt

+

(7.13) (7.14)

£

(7.15)

Jz(t)dt

hOO y(t)

fl(t)

-

y(t)

fl(t)

Figure 7.5:

Block diagram equivalent structures for linear operators.

212 7.3

Operational Methods for Unear Systems

Chap. 7

REPRESENTATION OF LINEAR SYSTEMS WITH BLOCK DIAGRAMS

7.3.1 Block Diagrams Based on the System Linear Graph Linear state-determined systems may be represented directly by block diagrams based on the three primitive linear operators and the summation operation. The elemental equations for ideal A-type, T-type, and D-type elements described in Chap. 3 may be expressed directly by operational blocks in terms of the through- and across-variables. For the energy storage elements a causality must be assigned; for example, the elemental equation for a mass element may be written in two forms:

dvm _.!_F. dt - m m

or

F, m

=m dvm dt

(7.16)

The output variable is implicitly on the left-hand side; in the first case integration of the input (Fm) is required:

defining an integral causality, while in the second case derivative causality is implied:

Fm

= m dvm dt

or operationally

Fm

= m {S {vm}}

(7.18)

Figure 7.6 shows block diagram representations for the elemental equations of the three types of lumped system elements, indicating both integral and derivative causality for the two energy storage elements. Integral causality is preferred in any system representation because it implicitly incorporates the initial conditions associated with any energy storage element into the system structure and provides a direct form for computer-based solution techniques. The steps in constructing a system operational block diagram using integral causality parallel the procedure described in Chap. 5 for deriving state equations from the linear graph and the normal tree. The nonnal tree is used to define a set of independent energy storage elements together with a set of primary and secondary variables. One or more blocks are drawn for each element, with causality chosen such that the output is a primary variable and the input is a secondary variable. A compatibility or a continuity constraint equation is associated with each element Each such equation expresses a single secondary variable as a sum of primary variables. Each constraint equation is therefore represented on the block diagram as a summation element that generates a secondary variable. Each elemental block requires a summer at its input.

Sec. 7.3

Representation of Linear Systems with Block Diagrams f(t)

v(t)

v(t)

213

~f(t) Derivative causality

(a) A-type elements

V(l)

f(t)

f(t) - - 8 - D - - v ( t )

Derivative causality (b) T-type elements

f(t)-0--

v(t)

v ( t > - - 0 - f(t)

Algebraic causality

Algebraic causality (c) D-type elements

Rgure 7.6: Operational block diagram representation of ideal elements in integral and differential causality.

The steps for constructing a block diagram from a linear graph model are as follows. Step 1: Construct operational block representations for each passive system element as in Fig. 7 .6, using the secondary variable associated with that element as the input and the primary variable as the output. Step 2: Use the set of independent continuity and compatibility constraint equations to express the secondary variable at each input in tenns of primary variables. A summer at each elemental input represents the constraint equation with the inputs to each summer as either system inputs or primary variables and the output as the secondary variable. Step 3: Complete the diagram by connecting the summer inputs to the primary variables at the outputs of the appropriate elemental blocks. Example7.1 Draw an operational block diagram for the system shown in Fig. 7. 7 from its linear graph and normal tree. R

C

v.CJ

R

c

R

c

Vrcr=O

{a) Circuit diagram Rgure 7.7:

(b) System graph

(c) Normal tree

A series RLC circuit, its system graph, and a nonnal tree.

214

Operational Methods for Linear Systems

Chap. 7

Solution The normal tree results in the elemental equations

d;~

Primary variables

!

diL

dt

VR

=

= =

~ic I

_!_VL

(i)

Secondary variables

L RiR

with the continuity and compatibility equations ic

=h (ii)

The three steps in constructing the block diagram are illustrated in Fig. 7 .Sa-c. Three elemental subdiagrams appear in Fig. 7.8a, representing each of the elemental equations. In Fig. 7.8b summer blocks are added to each input to determine the secondary variables, and in Fig. 7.8c the interconnections are made. Summers with a single input have been removed.

VL~iL it(O)

··--0-i·

(a) Draw elemental blocks

~~

vc(O)

VR~iL (b) Add compatibility and continuity summers

vc(O)

(c) Complete the block diagram Figure 7.8:

Construction of an operational block diagram for the RLC circuit

Similar procedures may be used to construct block diagrams for systems that include twoport elements. Each two-port element is defined by a pair of elemental equations, thus requiring two functional blocks.

Representation of Linear Systems with Block Diagrams

Sec. 7.3

215

The state variables for block diagrams formed from the system linear graph are the outputs of the integral causality blocks. The derivatives of the state variables are therefore defined as the inputs to the integrators in the diagram. The state equations may often be derived by inspection from the block diagram by writing an expression for the variable at the input to each integrator in terms of other state variables and inputs. Similarly, the output equations may be easily determined because all primary and secondary system variables are on the block diagram. Example7.2 Verify that the block diagram in Fig. 7.9 describes the electromechanical system analyzed in Example 6.4. Derive the state equations and output equations for VL and T8 from the block diagram by inspection.

Figure 7.9: Block diagram for electromechanical system analyzed in Example 6.4.

Solution Verification of the block diagram is left to the reader. Notice that two blocks are used to represent the de electric motor two-port element and that the state variables iL and 0 1 are the outputs of the integrator blocks. The state equations may be found by observing that the derivatives of the state variables exist at the inputs of the two integrator blocks. Then, working backward through each summer, (i)

(ii) and (iii)

(iv) The output equations may be determined by inspection in a similar manner. (v)

(vi)

216

Operational Methods for Linear Systems

Chap. 7

For systems containing dependent energy storage elements, direct application of the method described generates derivative causality blocks in the diagram. These dependent elements may often be eliminated by redefinition of the system variables as described in Sec. 5.4. If the derivative block cannot be eliminated, the state and output equations can still be derived from the block diagram. In the examples presented above, linear systems are represented by operational block diagrams. In general, nonlinear and time-varying operations may also be represented within a block diagram. For example, nonlinear blocks may be used to characterize phenomena such as aerodynamic drag (which is a function of the square of the velocity), flow through an orifice (which is a function of the square root of the pressure drop across the orifice), and Coulomb friction (in which the force is a nonlinear function of velocity). Example7.3 Draw the block diagram and write the state equations forthe nonlinear system shown in Fig. 7.I 0 where an automobile of mass m has a net propulsion force Fp which is used to accelerate the mass and overcome aerodynamic drag Fv and tire-road resistance F,.

-f.--- Fv Aerodynamic drag

Figure 7.10:

Nonlinear model of an automobile with its linear graph.

I

Solution In the model, the automobiJe is represented by the followin g elemental equations: dvm

Primary variables

I

dt F,

;Bl,v, Fm

Fv

Rv lvvl vv

Secondary variables

(i)

The continuity equation is ( ii)

and compatibility shows that Vm

= Vr = VD

(iii)

The nonlinear block diagram may be constructed as shown in Fig. 7 . I I. The state equation for the system is derived by noting that the derivative of the mass velocity may be expressed in terms of the state variable vm and inputs as (iv)

Sec. 7.3

Representation of Linear Systems with Block Diagrams

217

Fo~•o Nonlinear (a) Lumped-parameter elements

Nonlinear (b) Complete block diagram

Figure 7.11: Block diagram for a nonlinear system. (a) Representation by lumped elements. and (b) the complete diagram derived by using continuity and compatibility.

7.3.2 Block Diagrams Based on the State Equations

The state equations express the derivatives of the state variables explicitly in terms of the states themselves and the inputs. In this form the state vector is expressed as the direct result of vector integration. The block diagram representation is shown in Fig. 7.12. This general block diagram shows the matrix operations from input to output in terms of the A, B, C, and D matrices but does not show the path of individual variables.

Figure 7.12:

Vector block diagram for a linear system described by state space system dynamics.

218

Operational Methods for Linear Systems

Chap. 7

In state-determined systems, the state variables may always- be taken as the outputs of integrator blocks. A system of order n has n integrators in its block diagram. The derivatives of the state variables are the inputs to the integrator blocks, and each state equation expresses a derivative as a sum of weighted state variables and inputs. A detailed block diagram representing a system of order n may be constructed directly from the state and output equations as follows: Step 1: Drawn integrator (S- 1) blocks and assign a state variable to the output of each block. Step 2: At the input to each block (which represents the derivative of its state variable) draw a summing element Step 3: Use the state equations to connect the state variables and inputs to the summing elements through scaling operator blocks. Step 4: Expand the output equations and sum the state variables and inputs through a set of scaling operators to fonn the components of the output. Example7.4 Draw a block diagram for the general second-order, single-input single-output system

[~~] = [::: ::] [;~] + [!~] u(t} y(t}

= (Ct

C2]

(i}

[;~] + du(t}

Solution The block diagram shown in Fig. 7.13 was drawn using the four steps described above.

u(t)

y(t)

Figure 7.13:

Block diagram for a state equation-based second-order system.

Sec. 7.4

7.4

219

Input-Output Linear System Models

INPUT-OUTPUT LINEAR SYSTEM MODELS

In analyzing linear time-invariant system models, we often need to consider the relationship between a single output variable and the system inputs. For systems that have only one input, it is frequently convenient to work with the classical system input-output description in Eq. (7 .I), consisting of a single nth-order differential equation relating the output to the system input: (7.19)

where y(t) is the system variable of interest and u(t) is the single system input The constant coefficients a; and b; are defined by the system parameters and for physical systems m ::5 n. The classical form of the system description requires n initial conditions to be specified in order for the system to be completely described. It is usual to specify the output y(t) at timet = 0 and the first n - I derivatives of y(t) evaluated at t = 0 as the initial conditions. The classical system representation is unique; there is only one differential equation that relates a given output variable to the input. The classical nth-order differential equation may be derived directly from the system state equations by combining the state and output equations in such a way as to eliminate all variables except the output variable y(t) and the input u(t). · Example7.5 Derive the classical second-order differential equation that relates the position y(t) of the mass m to the input force FAt) from the state equation model of the system shown in Fig. 7.14. The state equations are (i) (ii)

and the position of the mass is directly related to the force in the spring: 1

y= -FK K

(iii)

Also, specify a set of initial conditions for the classical equation that corresponds to the initial states Vm(O) = 0 and FK(O) = 0.

=

Solution The system is second-order, n 2, and therefore a second-order differential equation in the displacement y is required. We begin by noting that

dy dt =

Vm,

and

(iv)

220

Operational Methods for Linear Systems

Chap. 7

Output y(t)

Rolling resistance 8 Figure 7.14:

A second-order mechanicaJ system.

and then from Eq. (i),

(v)

Substituting for the state variables from Eqs. (iii) and (iv),

2

B dy K I -ddt2y = ----y+ -F (t) m dt m m 5

(vi)

and rearranging the terms gives the required second-order differential equation

d 2y

B dy

K

I

-dt2 + - + -y = -Fs(t) m dt m m

(vii)

The second-order equation requires two initial conditions, the output variable and its derivative at time t o:·

=

and

-dy' dt

r=O

= Vm(O) = 0

In following sections, we develop fonnal methods based on the system transfer operator to relate the classical input-output fonn to the state equations.

7.5

UNEAR OPERATOR ALGEBRA In this section we develop a set of relationships that allow us to transform and manipulate linear differential equations as if they were algebraic equations [9, 10]. We start by considering some general properties of scalar linear operators and then apply these properties to operational representations of linear differential equations.

Sec. 7.5

221

Linear Operator Algebra

7.5.1 Interconnected Linear Operators If two linear operators £ 1 {} and £2 {} are applied in series, or cascade, so that £2 {} acts on the output of £ 1 {} as shown in Fig. 7.15, that is, if z(t) .C1 {x(t)} is the variable generated by the first operator and this becomes the input to the second, then

=

(7.20)

This relationship defines the cascade, or sequential, connection of the two operators. The notation adopted in writing cascaded operators is to retain the braces around the original input quantity only; for example, .C1 {£2 {x(t)}} is written £1£2 {x(t)}. As an example we can write an elemental equation for a mass element as F(t) = mS {v(t)}

(7.21)

as a pair of cascaded scaling and differential operators.

u(t)

-GJ----W-- ~(:£ 1 tu(t)J}==

u(t)

-§-;e,_:£1(u(t)}

(a) Cascade application of two operators

u(t)

-0--0-----

:if(:iftu(t)} J:= u(t)

-0--

!fl{u(t)}

(b) Repetitive application of the same operator

Figure 7.15: Equivalence of two cascaded operators and a single operator.

If the same operator .C {} is applied in succession n times, as in Fig. 7.15, it is written with an exponent

.c,n {x(t)} := £ {.C {.C ••• {£ {.C {.C {x(t)}}}} ... }}

(7.22)

This notation is not unusual; for example, the high-order derivatives of a variable are conventionally written d { dt d { ... dt d {x(t)}... S n {x(t)} =_ dt

II =

dn dtnx(t)

(7.23)

where the interpretation is that of repeated application of the derivative operator S to a single variable. Similarly, an expression such as a 3 {},while conventionally interpreted as multiplication by a factor a 3 , may also be interpreted as three successive scaling operations by the same constant a, that is, a {a {a{}}). Operational expressions such as £1£2 {} and £ 2 (} do not denote algebraic products. They define the cascaded application of one or more operators in which the output from one operator is passed sequentially to the next. If £ 1 {} and £2 {} are both linear operators, then the combined action .C 1.C2 {} is also a linear operator.

222

Operational Methods for Linear Systems

Chap. 7

For cascaded scalar linear operators, the output is independent of the order in which they are appJied, that is, y(t)

= .Ct (£2 {x(t)}} = £2 {.Ct {x(t)}}

(7.24)

This general commutative property of linear operators states that the result of an operation on a variable by a series of linear scalar operators is independent of the order in which they are applied. Thus, Sa {x(t)} is equivalent to aS {x(t)}. In the block diagram in Fig. 7.16 a system variable y(t) is shown as the summation of the outputs of two linear operator blocks, .Ct and .C2, each acting on the same input variable. This is defined to be a parallel combination of operators and is written y(t) = £1 (x(t)}

= [.Ct

+ .C2 {x(t)}

+ .C2]{x(t)}

If .Ct {}and .C2 {}are both linear operators, then the parallel operator [.Ct linear operator.

(7.25)

+ .C2] (} is also a

u(t)

Figure 7.16:

Parallel application of linear operators.

Care must be taken with the interpretation, because the plus sign in the parallel operator [.Ct + £2] does not imply addition of the operators in the usual arithmetic sense; it denotes the addition of the two results of the application of the two component operators. An operational expression such as y = [S + a]{x(t)} implies that

dx

y(t) = -

dt

+ ax(t)

(7.26)

7.5.2 Polynomial Operators A linear operator of the form (7.27) . is defined to be a polynomial operator in the operator .C {}. While clearly not an algebraic polynomial, because it denotes a set of cascaded and parallel operators, an operator of this

Sec. 7.5

223

Linear Operator Algebra

fonn has many properties that allow it to be manipulated using the rules of polynomial arithmetic. For example, the expansion of two cascaded parallel operators each containing a common operator .Ct {} y(t)

= [.Ct + .C2] [.Ct + .C3] {x(t)} = [.Ct + .C2l {.Ct {x(t)) + .C3 {x(t))) = [.cr + to. 1\vo broad classes of methods are commonly used to determine the response of a state-detennined system model to its initial conditions and inputs: 1. Analytical solution techniques, in which closed-form expressions for the output variables are derived for a specified system input and set of initial conditions 2. Numerical solution techniques, in which the system state equations are integrated using approximate numerical algorithms to determine the response in a numerical format Both analytical and numerical solution methods are important in engineering work. The solution methodology that is most appropriate for a given study depends on the system order and complexity, and in particular on whether the models are linear or nonlinear [1-3]. Analytical solution methods generate closed-fonn expressions for the system response in tenns of the system par~meters, which can lead to an understanding of the 244

Sec. 8.2

System Input Function Characterization

245

influence of elements on system behavior. In practice these methods are generally applied to low-order linear systems and are usually restricted to limited classes of input functions for which solution methods exist. Numerical solution methods, implemented in the form of computer-based simulation software packages, are applicable to a broader class of systems, both linear and nonlinear. They generate tabulated or plotted output representing the system response to a specified (or tabulated) input function at discrete times. The output generated by the computer program is purely numerical (even in plotted form), and the simulation must be repeated if a system parameter is changed or if a new set of initial conditions or inputs is applied. Numerical simulation methods are described in Chap. 11. In this chapter we define a set of families of input functions ~sed as typical system inputs in dynamic analysis and introduce classical analytical solution techniques for linear, state-determined systems. 8.2

SYSTEM INPUT FUNCTION CHARACTERIZATION

The inputs to physical systems are prescribed variables with a known form. In mechanical systems inputs are forces and velocities, in electric systems voltages and currents, in fluid systems pressures and flows, and in thermal systems temperature and heat flows. In general, input functions u(t) may be classified in two groups: 1. Deterministic inputs in which the input is a well-defined prescribed function of time, for example, u(t) = sin(wt), and 2. Random or stochastic inputs in which the input function cannot be described as any specific function and only its statistical properties, such as the mean value, variance, and average frequency content may be specified. Although many naturally occurring phenomena, such as wind- and wave-generated forces on structures, are random in nature, the response of systems to this class of inputs is beyond the scope of this book [4]. Deterministic functions may be further divided into two classes: 1. Aperiodic (or transient functions), which are not repetitive 2. Periodic functions, which repeat at regular intervals T, that is, f(t) = f(t + nT) for all n = 1, 2, 3, .... Many physical events are transient in nature, for example, an electric current surge into a capacitor caused by the closing of a switch, the force on an automobile during a collision, and the torque on the shaft of a lathe as the tool enters the workpiece. Many other physical phenomena may be modeled by periodic input functions. System inputs from sources such as a 60-Hz electric power distribution system, structural vibrations induced by rotating machinery, and acoustic waves may often be approximated by periodic functions. Engineers frequently analyze systems by studying the response to a small set of representative functions known to elicit important system response characteristics. The suitability of a system to its intended task is then inferred from its response to these test inputs. These inputs often consist of members of the family of singularity functions and periodic functions.

System Properties and Solution Techniques

Chap. 8

8.2.1 Singularity Input Functions The singularity functions are a family of transient waveforms frequently used to characterize the response of systems to discontinuous inputs. The singularity functions are either discontinuous, or have discontinuous derivatives, at time t = 0 and are defined to be zero for all time t < 0.

The Unit Step Function The unit step function Us(t) is widely used to study the way a system responds to discontinuous, or sudden, changes in its input. It is defined as

u

_ {

s (1) -

0 for t ::; 0 1 for t > 0

(8.1)

and is shown in Fig. 8.1. Step changes of other amplitudes can be formed by multiplying the unit step by a constant.

Us(t)

I

1.0 _~------------

0

Tune

Rgure 8.1: The unit step function.

The Unit Impulse Function The unit impulse tS(t) is used to detennine system response to short-duration transient inputs. Figure 8.2 shows a unit pulse function tSr(t), that is, a brief rectangular pulse function of duration T defined to have a constant amplitude 1IT over its duration, and so the area T x I IT under the pulse is unity: 0 tST(t)

=

l/ T { 0

fort.::: 0 }

0 0

(8.5)

00

and we therefore assert: 8(t)

dus dt

=-

(8.6)

The Unit Ramp Function The unit ramp function u,(t) is defined to be a linearly increasing function of time with a slope of unity:

u t _ {0 '()t

for t .:5 0 for t > 0

(8.7)

as shown in Fig. 8.3. The ramp is the integral of the unit step

u,(t) =

L:

u,(t) dt

and is used to study the response of systems to constantly changing inputs.

(8.8)

System Properties and Solution Techniques

248

Chap. 8

u,.(t)

1.0

0

1.0 Time

The unit ramp function.

Figure 8.3:

Relationships Among Singular Functions The ramp, step, and impulse functions represent a family of functions which, as shown in Fig. 8.4 are related by successive integrations.

l

u,(t)

u,.(t)

l.Ot-----

0

1.0

Integration

Integration

Differentiation

Differentiation --- 0

t

Time

Time Figure 8.4:

Time

The re1ationsbip between singu1arity functions.

Time Shifting of Singularity Functions The singularity functions may be used to describe transient inputs that take place at a time other than t = 0. The discontinuity associated with each function accurs when the function argument is zero; therefore, a step that occurs at time to may be written as Us (t - to) since t - to 0 at t to. This property may be used to synthesize a transient function from a sum of singularity functions; for example, Fig. 8.5 shows the function u(t) Us(t)- 2us(t- 1) + Ur(t- 2)- u,(t- 3).

=

=

8.2.2 Sinusoidal Inputs

=

=

Sinusoidal input functions such as u(t) A sin (wt + t/J) and u(t) A cos (wt + t/J), shown in Fig. 8.6, are periodic with period T = 2n'I w seconds. These functions are described by three parameters: w, the angular frequency, (rad/s); t/J, the phase (rad); and A, the amplitude of the waveform. The frequency f of a periodic waveform is defined directly from the period f = I IT (Hz or cycles/second). The frequency of a sinusoid is related to the angular frequency w = 21rf = 2n IT. It is also common practice to express the phase in degrees instead of radians, with 36()0 2n' rad. Sinusoidal waveforms are used to represent many naturally occurring periodic phenomena. Furthermore, they are used as the basis for representing other periodic and transient waveforms through the process of Fourier synthesis, as described in Chap. 15.

=

8.2.3 Exponential Inputs Another class of theoretically and practically important input functions includes exponential

Sec. 8.2

249

System Input Function Characterization

u(l)

u(r)

2

2 u,(r)

0

0 Time -I

-I

-2u,(t - I)

-2

-2

(a) Component step and ramp functions.

(b) Resultant function formed

by summing components. Figure 8.5:

A transient functi on u(r) = u,(r)- 2u,(r- I)+ u,(r- 2) -u,(r- 3) synthesized from unit singularity function s.

inputs u(t) = est , where the exponent s may be either real or complex. Exponential wavefonns occur naturally as the response of linear systems and also provide a means for expressing sinusoidal functions in a compact fonn. For real exponents the value of the function increases without bound if s > 0, or decays exponentially to zero if s < 0, as shown in Fig. 8.7. The exponents defines the rate of decay or growth of the wavefonn. Table 8.1lists some properties of the exponential wavefonn that are of importance in system dynamics. When the exponent s is imaginary, that is, s = jw , where j = R. est is complex and the Euler fonnulas (App. B) define the real and imaginary parts: ejwt e- jwt

+j

sin (wt)

(8.9)

= cos (wt) - j sin (wt)

(8.1 0)

= cos (wt)

A sin(wr + )

Amplitude A

r

I

Period 2'TT-T=w

Figure 8.6:

A sinusoidal function.

2SO

System Properties and Solution Techniques

Chap. 8

exp (-at) l.Or----

0

0

Time

Time

Figure 8.7: Real exponential waveforms.

The exponential eiOJt is therefore periodic with angular frequency w and period T = 231'1w. The two functions, Eqs. (8.9) and (8.1 0), may be added and subtracted to give cos (wt)

. ) = -2] (e 1. + e-JOJt

sin (wt)

=

1

0)

1 ( . . j eJOJt - e-1"'')

2

(8.11)

(8.12)

The real periodic waveforms sin (wt) and cos (cut) may therefore be considered to consist of two complex exponential components, one with a positive frequency w and the second with a negative frequency -w. This complex representation of real waveforms is examined in detail in Chaps. 14 and 15. TABLE 8.1:

Elementary Properties ofthe Exponential Function es'

e

'50 >

3

U)

2

-

~

U)

!

-2

N

N

.!:!

I I I I

U)

2

-2

-5 -5

0 State variable 1

5

-2

0 State variable 1

2

State space ttajectories of stable and unstable second-order systems.

8.4.2 Time lnvariance Consider a system, initially at rest, that has an output response y(t) to an arbitrary input u(t). The system is defined to be time-invariant if its response to u(t- T), that is, the same input delayed by a finite time T, is y(t - T). A system described by Eqs. (8.29) is time-invariant All the elements of the matrices A, B, C, and D are constants, and the form of the responses x(t) and y(t) is independent of the time of occurrence of the input In a more general case, the elements of the matrices in Eqs. (8.29) may be explicit functions of time, and the matrices can be written A(t), B(t), C(t), and D(t). In this case the system is time-varying, and the response to a given input depends on the time of occurrence. Time-varying systems may result from natural component aging, for example, the drying out of the lubricant in a bearing, causing a gradual change in the damping coefficient; from changes in operating conditions such as are experienced by a robot arm as it moves loads of differing mass; or from variation in internal system parameters, such as the loss of mass in a rocket as fuel is burned and expelled. In many situations a quasi-time-invariant analysis, in which it is assumed that the system is time-invariant for the duration of the analysis, may be used to describe slowly changing operating conditions. nme-varying

System Properties and Solution Techniques

262

Chap. 8

systems are not discussed in this book, however, the numerical methods introduced in Chap. 11 may be used directly or with slight modification to study the response of timevarying systems [2-3].

8.4.3 Superposition for Linear Time-Invariant Systems Linear time-invariant systems, such as described by Eqs. (8.29) with A, B, C, and D constant, obey the principle of superposition, which in a general form states: The response of an LTI system to a set of given initial conditions and an input consisting of several components, U(t) = UJ (1)

+ U2(t) + ··· + Dt(t)

(8.30)

may be found by determining the response to (1) the initial conditions and (2) each of the k individual input components and then summing all component responses to determine the total response. The principle of superposition has several important consequences for linear systems:

1. The response of the output vector y(t) [and by inference the state vector x(t)] of a linear time-invariant system may be partitioned into (a) an unforced response component Yic(t) due to the system initial conditions with the assumption of zero input u(t) 0, and (b) the forced response y1 (t) due to the input u(t) with assumed zero initial conditions x(O) 0. The total response is the superposition, or sum, of the two components:

=

=

y(t)

= Yic(t) + YJ(t)

(8.31)

Figure 8.10 shows schematically how the two response components may be analyzed independently and combined to fonn the total response. Many engineering studies concentrate on determining the forced response YJ(t) because the separability of the components allows the total response with arbitrary initial conditions to be found by computing the initial condition response separately and forming the sum. 2. If the input n(t) is the weighted sum of several components, the forced response yf(t) is simply the superposition, or weighted sum, of the responses to each individual component. If the forced component of the response of a linear system to an input u 1(t) is y/• (t) and the component due to a second input u2 (t) is y12 (t), then the forced response to any other input u (t) that is a linear combination of u 1 (t) and u2 (t), such as u(t) k1u1 (t) + k2u2(t), where kt and k2 are arbitrary scalar constants, is

=

(8.32) 3. If a set of initia1 conditions x(O) is a weighted sum of several components, the unforced response Yu (t) is the superposition of the individual unforced responses. If the initial condition response of a linear system to initial conditions XI (0) is Yic 1 (t), and to a second set of initial conditions x2 (0) is Yic2 (t), then the unforced response to a linear combination of Xt (0) and x2 (0), such as x(O) = k1x 1(0) + k2x2 (0), where k1 and k2 are arbitrary sca1ar constants, is

(8.33)

Sec. 8.4

System Properties

263

Forced response

u(t)~ ~

°

x(O)= 0

Initial condition response

u(t)

~

TlDle

y(t) + y-=(t)

TotaJ response ) -

Yic(t)

=~ x(O)

·..___ - _____

r--"'lll~..._,--_..

0

0 ..........

Time

=.xo

Figure 8.10:

Superposition of forced and unforced components of the response of a linear time-invariant system.

8.4.4 Differentiation and Integration Properties of LTI Systems

Two further properties of linear systems that allow the forced response to a given input to be generalized to other input functions are

1. If the forced component of the response of a linear system to an input u(t) for which u(t) = 0 fort < 0 is YJ(I), then the forced component of the response of the system to a new input u 1 (t) that is the derivative of u(t), that is, Ut (t)

d = dt u(t)

(8.34)

is the derivative of the original output YJ(t), d

Y!• (t) = dt YJ(t)

(8.35)

2. Similarly, if the forced component of the response of a linear system to an input u(t) for which u(t) = 0 fort < 0 is y(t), then the forced response of the system to a new input u 1 (t) that is the integral of the input, that is, U1 (I)=

L'

U(l) dt

(8.36)

is the integral of the original output function y1 (t), Y!o (1)

=

L'

YJ(t) dt

(8.37)

These properties are useful in extending the range of known system responses because they state that if the forced response to a given waveform is known, then so is the response to the successive derivatives and integrals of that waveform.

264

8.5

System Properties and Solution Techniques

Chap.8

CONVOLUnON

In this section we derive a computational fonn of the system transfer operator H {u(t)}, defined in Chap. 7, that is based on a system's response to an impulsive inpuL We assume that the system is initially at rest, that is, all initial conditions are zero at time t = 0, and examine the time domain forced response y(t) to a continuous input waveform u(t). In Fig. 8.11 an arbitrary continuous input function u(t) has been approximated by a staircase function ur(t) ~ u(t), consisting of a series of piecewise constant sections each of an arbitrary fixed duration T, where ur(t) = u(nT) for nT ~ t < (n

+ l)T

(8.38)

for ann. It can be seen from Fig. 8.11 that as the interval Tis reduced, the approximation becomes more exact, and in the limit u(t) = lim ur(t)

r-o

The staircase approximation ur (t) may be considered to be a sum of nonoverlapping delayed pulses Pn(t), each with duration T but with a different amplitude u(nT): 00

ur(t)

=

E

(8.39)

Pn(t)

n=-oo

where (t) = { u(nT)

0

Pn

nT ~ t < (n

+ l)T

otherwise

(8.40)

Each component pulse Pn (t) may be written in tenns of a delayed unit pulse Br (t) as defined in Sec. 8.2.1, that is, (8.41) Pn(t) = u(nT) Br(t- nT)T and so Eq. (8.39) may be written ur(t)

=

00

L

u(nT) Br(t- nT)T

n=-oo

u(t)

u(t)

i

i

.5

.5 u(3T)

e

!

~ >.

V,)

V,)

0

Figure 8.11:

Staircase approximation to a continuous input function u(t).

(8.42)

Convolution

Sec. 8.5

265

We now assume that the system response to aT (t) is a known function and is designated hT (t) as shown in Fig. 8.12. Then if the system is linear and time-invariant, the response to a delayed unit pulse, occurring at time n T, is simply a delayed version of the pulse response:

= hT(I- nT)

Yn(t)

8r(t-nT)

Yr(t)l

. ._. . . .,-'-. L.S~r. ,. .l -

liT '--'__,_.......

0

(8.43)

nT)

-8----

nT (n + l)T

Yr(t)

I

0

~

I

nT (n + l)T

.. t

Rgure 8.12: System response to a unit pulse of duration T.

The principle of superposition allows the total system response to uT(t) to be written as the sum of the responses to all the component weighted pu1ses in Eq. (8.42): 00

YT(t) =

L u(nT)hr(t- nT)T n=-oo

(8.44)

as shown in Fig. 8.13. For physica1 systems the pulse response hr(t) is zero for timet < 0, and future components of the input do not contribute to the sum. So the upper limit of the summation may be rewritten as N

Yr(t)

=L

u(nT)hT(t- nT)T

for NT ~ t < (N + l)T

(8.45)

n=-oo

Equation (8.45) expresses the system response to the staircase approximation of the input in terms of the system pulse response hT(t). If we now let the pulse width T become very small, write nT = T, T = dT, and note that limr-o 8r(t) = 8(t), the summation becomes an integral: N

y(t)

L

= T-o lim

n=-oo

=

U(T)h(t- T) dT

1:

u(nT)hT(t- nT)T

(8.46)

(8.47)

00

where h(t) is defined to be the system impulse response h(t)

= r-o lim hr(t)

(8.48)

266

System Properties and Solution Techniques

Chap. 8

yr(t)

:l y,{t)

c

I

~ 0 1-LL..L...L~::::;:~:;;,;.L..-___.....,. Figure 8.13:

0

System response to individual pulses in the staircase approximation to u(t).

Equation (8.48) is an important integral in the study of linear systems and is known as the convolution or superposition integral. It states that the system is entirely characterized by its response to an impulse function ~(t) in the sense that the forced response to any arbitrary input u (t) may be computed from knowledge of the impulse response alone. The convolution operation is often written using the symbol 1r:

y(t)

=u(t) * h(t) = ["" u(~)h(t - ~) d~

(8.49)

Equation (8.49) is in the form of a linear operator as defined in Chap. 7 in that it transforms, or maps, an input function to an output function through a linear operation. It is a direct computational form of the system transfer operator H (u(t)} defined in Sec. 7.2, that is, y(t)

= H (u(t)} = u(t) * h(t)

The form of the integral in Eq. (8.48) is difficult to interpret because it contains the term

h (t - 't') in which the variable of integration has been negated. The steps implicitly involved in computing the convolution integral may be demonstrated graphically as in Fig. 8.14, in which the impulse response h('t') is reflected about the origin to create h( -'t') and then shifted to the right by t to form h(t- 't'). The product u(t)h(t- 't') is then evaluated and integrated to find the response. This graphical representation is useful for defining the limits necessary in the integration. For example, since for a physical system the impulse response h(t) is zero for all t < 0, the reflected and shifted impulse response h(t- 't') will be zero for all time 't' > t. The upper limit in the integral is then at most t. If in addition the input u(t) is time-limited, that is, u(t) = 0 fort < t1 and t > t2, the limits are

['

YJ(t)

={

l\ It

u(~)h(t- ~)d~

fort< t2

u('t')h(t- 't') d't'

fort

(8.50) ~

t2

Sec. 8.5

Convolution System impulse response

h(l)~0

-Jf~>

Time reversal

v

II

I

~

nshifting

System input

h(tJ-T)

II

T

T

Multiplication

0

T

\

n

lnte~OD

\

\ \

t

Response at time t 1 is defined by the area under the curve. I

I

System response

I

y(t)

-------

....,,'/

'

I

I

I

'

I

0 Rgure 8.14:

T

ll

~·>~ 0

II

Graphical demonstration of the convolution integral.

System Properties and Solution Techniques

268

Chap. 8

Example8.8

A mass element, shown in Fig. 8.15, at rest on a viscous plane is subjected to a very short unit impulsive force of duration 0.()() 1 s and magnitude 1000 N and observed to respond with a velocity vm(t) = e-31 • Fmd the response of the same mass element to a ramp in applied force F(t) t fort > 0.

=

Vm(l)

F(t)

1000

~ vm(t)

0.001 s F(l)

-1

m

MH~±'

0

0.5

]

1

0

Viscous friction

Time

2

t

Tune (s)

Figure 8.15: A sliding mass element and its impulse response.

Solution The product of the impulsive force and its duration is unity, and because of its brief duration, the pulse may be considered to approximate an impulse. The measured response may then be taken as the system impulse response h(t), and we assume that h(t)

= e-31

(i)

The response to a ramp in input force, F(t) = t fort > 0, may be found by direct substitution into the convolution integral using the assumed impulse response: v(t)

=

1'

-re-JCt-t) d-r

= e-Jt

11

reJt

(ii)

dr

(iii)

where the limits have been chosen because the system is causal and the input is identically zero for all t < 0. Integration by parts gives the solution v(t)

= 31r - '91 + 91e-31

(iv)

Convolution is a linear operation and is commutative, associative, and distributive, that is, u(t) * h(t) u(t) *(hi (t) * h2(t)] u(t) * [ht {t) + h2(t)]

=

= =

h(t) * u(t) (u(t) * ht (t)] * h2(t) [u(t) * ht (t)] [u(t)

+

* h2{t)]

(commutative) {associative) {distributive)

{8.51)

The associative property may be interpreted as an expression for the response on two systems in cascade or series and indicates that the impulse response of two systems is h1 (t) •h2(t), as shown in Fig. 8.16. Similarly, the distributive property may be interpreted as the impulse response of two systems connected in parallel and to mean that the equivalent impulse response is ht (t) + h2(t).

Chap. 8

269

Problems Parallel sy~tems: u(t)

Cascade systems: [u(t) ·h 1(t)] •h2(t)

u(t)

u(t) • [h 1(t) • h2(r)]

Equivalent system u(l)

-8-Figure 8.16:

Equivalent system u(t) • [h 1(t) • h,(t))

u(t)

-8--

u(t) • [h 1(t) + h2(t))

Impulse response of series and parallel connected systems.

PROBLEMS 8.1. A signal synthesizing module consisting of differentiator,

gai~

and integrator blocks is shown

in Fig. 8.17. The module has three adjustable coefficients which generate a signal of period T from a unit amplitude square wave generator.

dO a-

J.-r~

1.0

dt

b

t

j{t)

-1.0 cf()dt (a)

j{t)

2

(b)

Figure 8.17:

(a) A wavefonn synthesizer module and (b) a sample generated signal.

=

=

=

(a) Sketch the signal generator output for (i) a= l, b = 0.5, c 0; (il) a= 0, b 0, c 1; and (iii) a 0, b 2, c -0.5 assuming T 2 s. (b) Determine the values of a, b, and c required to synthesize the signal illustrated in Fig. 8.17(b).

=

=

=

=

270

System Properties and Solution Techniques

Chap. 8

8.2. The family of singularity functions provide a basis for the representation of many types of transient waveforms. Write an expression for each of the signals shown in Fig. 8.18 in tenns of weighted and delayed singularity functions. Dlustrate how the terms in each of the expressions yield the desired function.

I

0

1

2

3

4

t

-1 r-

0

2

3

4

-1 (b)

(a)

2

0 -1

-2 (d)

(c)

Figure 8.18:

Four waveforms synthesized from singularity functions.

8.3. The impulse function ~(t) is used to simulate short-duration, high-amplitude phenomena. A useful property of ~(t) is the sifting property that states.

i:

f(t)~(t -

-r)dt

= /(-r)

that is, the integral takes the value of the function f(t) when the impulse occurs. (a) Prove the sifting property by considering the function f(t) to be constant over the duration of the impulse. (b) Evaluate the integral /_: (t 2 +

5)e'- 1 ~(t- 1)dt

8.4. Sinusoidal and exponential signals occur naturally in the response of linear systems. For each of the waveforms shown in Fig. 8.19, partially express the signal in tenns of sinusoidal and exponential terms and estimate the amplitude, frequency and phase for the sinusoidal components, and the amplitude and exponent for the exponential components.

271

Problems

Chap. 8 j(t)

j(t)

10

100

80 0

60 ~~~+-~~~-+~~~¥-~-

2

6

4

8

10 t (s)

40

20 -10

0

2

3

4

5

6 t (s)

(b)

(a) j(t)

1000

500

(c)

figure 8.19: Three exponential waveforms.

8.5. Write the waveform y(t) = 3 sin (lOr+ 0.5) + 2cos (lOt- 2) as (i) a single real sinusoid and (ii) as a pair of complex exponentials.

8.6. 1\vo sinusoidal waveforms ft (t) =At sin (w 1t + tPt) and /2(t) = A2 sin (Wlt + t/>2) are multiplied together in an electronic multiplier. Express the output y(t) = f 1 (t) · f2(t) as the sum of two sinusoidal waveforms. What are the frequencies of the two sinusoids?

8.7. The waveform f (t) = sin t +sin 1OOOt is passed through (i) a differentiator and (ii) an integrator. Derive expressions for the output waveform in each case, and comment on the relative amplitudes of the components in the output. 8.8. For each of the homogeneous differential equations given below, determine the order, the system characteristic equation, and its roots. Write the homogeneous response of each equation in terms of arbitrary constants. dy

(a) dt

+ay = 0

d 2y (b) dt2

dy

+ 7 dt + 12y = 0

272

System Properties and Solution Techniques d 2y

dy

(c) dt2

+ 6 dt + 9y = 0

d 2y (d) dt2

+ dt + 9y = 0

d 2y (e) dt2

d 3y (f) dt 3

Chap. 8

dy

dy

+ 12 dt + 9y = 0 d 2y

dy

+ 3 dt2 + 2 dt

=0

8.9. Derive the solutions to the following homogeneous differential equations: (a)

i, + d 2y

(b) dt 2

d 2y

ay = 0,

given y(O)

dy

+ 3 dt + 2y = 0 dy

(c) dt 2

+ 6 dt + 9y = 0

d 2y (d) dt 2

+ dt + 9y = 0

dy

= 1.0.

given y(O) = 1.0, and dyjdt given y(O)

= 0 at t = 0.

= 1.0, and dyjdt = 0 at t = 0.

given y(O) = 1.0, and dyjdt

= 0 at t = 0.

· 8.10. Show that if Yp(l) is a particular solution to a linear differential equation for an input u(t), then so is Yp(t) + Ce'J, where A. is any root of the characteristic equation and Cis an arbitrary constant. Show, therefore, that there is no unique particular solution to a given differential equation. 8.11. Write the input function u(t) = e'l' cos (et>t) as a pair of complex exponentials. Use this form to find the particular solution of the differential equation dy dt

+ y = 6e- 21 cos (4t)

8.12. Consider the differential equation

Find the particular integral corresponding to inputs (a) u(t) = 4us(l) (b) u(t)

= 3t

(c) u(t) = 5 sin 2t (d) u(t) = 2e- 3'

(e) u(t) = 4e 5'

8.13. Derive the solutions to the following differential equations. (a)

i, +

ay

= 2u.r(t), given y(O) = 0.

d 2y (b) dt 2

+ 3 dt + 2y = 2u.r(t), giVen y(O) = 0, and dy/dt = 0 at t = 0.

d 2y (c) dtl

+ 6 dt + 9y = u.r(t) gtven y(O) = 0, and dyjdt = 0 at t = 0.

dy

dy

.

.

Chap. 8

273

Problems

8.14. The principle of superposition is a useful technique in developing solutions to differential equations after an initial solution is derived. In developing such solutions it is often convenient to consider separately the response of a system to initial conditions with all forcing function inputs equal to zero, and, secondly, the response to the forcing function with all initial conditions equal to zero. Consider the following differential equation:

d; + ay = with y(O)

bu1 (t)

=c.

(a) Determine the solution y; (t) to the equation for nonzero, finite values of a and c, and for b

= 0.

(b) Determine the solution Yu(t) of the equation for nonzero, finite values of a and b, and for c

= 0.

(c) Let the solution resulting from adding the solutions from (a) and (b) be designated y1 (t) = Y;(t) + Yu(t).

(d) Determine the general solution to the equation for the case in which a, b, and care all finite. Let this solution be designated as Yr(t).

(e) Is Ys(t) equal to Yr(t)? (f) What is the solution to the equation for the two following cases: (i) a

a

= 1, b = 2, c = 1?

= b = c = 1 and (ii)

8.15. Consider the following first-order differential equation

dy +ay = bu(t) dt

-

(a) Detennine the total solution when the input is a unit step function, that is, u(t)

y(O)

= 0.

=u

1

(t) with

(b) Using the derivative property detennine the solution when the input is a unit impulse u(t)

with y(O) = 0.

= cS(t)

(c) Use the principle of superposition and the results of parts (a) and (b) to determine the solution

of the following equation

dy dt with y(O)

+ ay = 2u

1

(t)

+ 4&(1)

= 0.

8.16. Consider the first-order differential equation

dy dt with initial condition y(O)

+ 2y = 2u(t)

= 0.

(a) Find the solution of this equation when u(t)

= u (t), and when u(t) = t (the unit ramp). 1

(b) Express the two waveforms in Fig. 8.20 in terms of weighted and shifted singularity functions.

(c) Use the superposition principle to derive the solutions of the differential equation to the inputs in Fig. 8.20, using the solution components found in part (a).

274

System Properties and Solution Techniques u(t)

Chap. 8

u(t)

4 t----.,

0

4

1

4

2

0

t(s)

(a)

t(s)

(b) Rgure 8.20:

1\vo transient functions.

=

8.17. A linear system exhibits a unit step respon~ Ystep(t) (1- e-21 ). Find the response of this to an input u(t) U(t) + Su6 {t) - r{t) where 8{t) is the unit impulse, u 6 {t) is the unit step, and r{t) is the unit ramp. Assume that the system is initially at rest

=

8.18. For each of the following differential equations, obtain the solution and state whether the solution represents a stable, neutrally stable, or an unstable response.

dy

(a) dt +ay

d2y

(b) dtl

d2y

{c) dtl

+

=0

dy dt - ay

+ 9y =

0

for (i) a = 2 and (ii) a = -2.

=0

with y(O)

with y(O)

= 1, and dyjdt = 0 at t = 0.

= 0, and dyjdt = 0 at t = 0.

8.19. An electronic circuit is excited with a voltage pulse of amplitude 10 V and duration 10 psec. The response is monitored on an oscilloscope and found to be closely approximated as y(t) o.se- 100t.

=

(a) Determine (approximately) the system impulse response.

(b) Use the convolution integral to compute the unit step response of the system. (c) Compute the response of the system to an input u(t) that the system is at rest at t 0.

=

= 3e-20t fort > 0, with the assumption

8.20. Prove the commutative and associative properties of the convolution integral as stated in Eqs. {8.5 1), that is,

= h(~ * u{t) u(t) * [h1 (t) * h2{t)] = [u(t) * ht (t)] * h2(t)

(a) u(t) * h{t)

(b)

8.21. Consider a linear system described by the differential equation

d 2y dy -+2-+Sy=O dt2 dt

{a) Detennine the system response to an input u (t)

= 3e-3' by solving the equation directly.

(b) Determine the system step response by solving the equation directly, and find the impulse re-

sponse.

{c) Use the convolution integral to find the response to the input given in part (a) and compare the results.

Chap. 8

275

References

REFERENCES [l] Slotine, J. I., and Li, W., Applied Nonlinear Control, Prentice Hall, Englewood Cliffs, NJ, 1991. [2] Chen. C. T., Unear System Theory and Design, Holt. Rinehart and Winston, New York, 1984. [3] Luenberger, D. 0.,/ntroduction to Dynamic Systems, John Wiley, New York, 1979. [4] Lanning, J. H., and Battin, R. H., Random Processes in Automatic Control, McGraw-Hill, New York. 1956. [5] Edwards, C. H., Jr., and Penney, D. E., Elementary Differential Equations with Boundary Value Problems (2nd ed.), Prentice Hall, Englewood Cliffs, NJ, 1989. [6] Simmons, G. F., Differential Equations, McGraw-Hill, New York, 1972.

9

First- and Second-Order System Response

9.1

INTRODUCTION An understanding of the dynamic responses of simple first- and second-order linear timeinvariant systems is fundamental to understanding the dynamics of higher-order linear systems since the responses of complex linear systems contain components of first- and second-order system responses. An engineer who thoroughly understands the dynamics of first- and second-order systems is able to generalize this knowledge to higher-order models. A thorough knowledge of simple dynamic system responses to initial conditions and external inputs is important in all engineering disciplines [1-4]. The treatment in this chapter is based on the classical input-output differential equation formulation and the direct solution of linear differential equations as described in Chap. 8. Although the earlier chapters of this book stressed modeling procedures based on the state equation format, much practical engineering analysis is based on a set of descriptive system parameters such as the time constant T for a first-order system and the undamped natural frequency Wn and damping ratio ~ for a second-order system, all of which have their roots in classical input-output models. It is assumed in this chapter that a state equationbased model can be transformed to a classical model using the techniques described in Chap. 7. In Chap. 10 analysis techniques based directly on the state equation formulation are introduced. These two chapters are complementary; they both provide analytical tools for the analysis of systems. In this chapter the dynamic behavior is described in tenns of parameters directly related to the system response characteristics, and in the next chapter the response is characterized by properties of the A, B, C, and D matrices in the state space formulation. 276

Sec. 9.2

.2

277

First-Order Linear System Transient Response

FIRST-ORDER LINEAR SYSTEM TRANSIENT RESPONSE The dynamics of many systems of interest to engineers may be represented by a simple model containing one independent energy storage element. For example, the braking of an automobile, the discharge of an electronic camera flash, the flow of fluid from a tank, and the cooling of a cup of coffee may all be approximated by a first-order differential equation which may be written in a standard form as

i

dy dt

+ y(t) = f(t)

(9.1)

where the system is defined by the single parameter i, the system time constant, and f(t) is a forcing function. For example, if the system is described by a linear first-order state equation and an associated output equation

i =ax+bu

(9.2)

y =cx+du

(9.3)

and the selected output variable is the state variable, that is, y(t) = x(t), Eq. (9.3) may be rearranged as

dy - -ay=bu dt

(9.4)

and rewritten in the standard form (in terms of a time constant through by -a:

1 dy a dt

i

b a

-- - + y(t) = --u(t)

= -1 I a) by dividing (9.5)

where the forcing function is f(t) = (-bja)u(t). If the chosen output variable y(t) is not the state variable, Eqs. (9.2) and (9.3) may be combined to form an input-output differential equation in the variable y(t) by using the operational methods described in Chap. 7. In Example 7.7 the result is shown to be v(t)

·

= cb + [S- a]d {u(t)} S-a

(9.6)

which may be reorganized and written as a differential equation in the output variable y(t):

dy du - - ay = d dt dt

+ (be -

ad) u

(9.7)

To obtain the standard form we again divide through by -a: 1 dy

--+ y(t) = a dt

d du --a dt

+

ad - be u(t) a

(9.8)

278

FliSt- and Second-Order System Response

Chap.9

Compariso_n with Eq. (9.1) shows the time constant is again t' = -1/a, but in this case the forcing function is a combination of the input and its derivative d du a dt

f(t) = - - -

+

ad-be u(t) a

(9.9)

In both Eqs. (9.5) and (9.8) the left-hand side is a function of the time constant T only and is independent of the particular output variable chosen.

= -l/a

Example9.1 A sample of fluid, modeled as a thermal capacitance C, is contained within an insulating vacuum flask. Fmd a pair of differential equations that describe (1) the temperature of the fluid, and (2) the heat flow through the walls of the flask as a function of the external ambient temperature. Identify the system time constant. Solution The walls of the flask may be modeled as a single lumped thermal resistance R, and a linear graph for the system drawn as in Fig. 9.1. The environment is assumed to act as a temperature source Tamb(t). The state equation for the system, in terms of the temperature Tc

of the fluid, is 1 1 -dTc = ---Tc + --Tamb(t) dt R,C, R,C,

(i)

The output equation for the flow q R through the walls of the flask is 1

qR = -TR R, 1 1 = --Tc + -Tamb(t) R, R,

(ii)

The differential equation describing the dynamics of the fluid temperature Tc is found directly by rearranging Eq. (i): dTc (iii) R,C, ---;it+ Tc = Tmnb(t) from which the system time constant 1' may be seen to be 1' = R, C,. The differential equation relating the heat flow through the flask is found by writing the state equation, Eq. (i}, in operational form (iv)

( S + R,1C,) {Tc} = R,IC, (Tamb}

from which rearrangement and substitution into Eq. (ii) give qR(t)

-1/RlC,

I)

= ( S + 1jR,C, + R, (Tamb(t)} = -1/RlC, + ljR, [S + ljR,C,] {T. S + 1jR,C,

=

( )} amb t

(v)

(1/ R,)S (Tamb(t)} S+1JR,C,

The required differential equation is

dqR 1 1 dTamb -+--qR=--dt R,C, R, dt

(vi)

Sec. 9.2

279

FliSt-Order Linear System Transient Response

This equation may be written in the standard form by dividing both sides by 1/ R,C,:

(vii)

=

and by comparison with Eq. (9.8) it can be seen that the system time constant T R,C, and the forcing function is f(t) = C,dTamb/dt. Notice that the time constant is independent of

the output variable chosen.

R,

Tamb

Tc

Walls R, Heat

Fluid

flow~ qR

c,

Tret Figure 9.1: A first-order thermal model representing the heat exchange between a laboratory vacuum flask and the environmenL

9.2.1 The Homogeneous Response and the First-Order Time Constant The standard form of the homogeneous first-order equation, found by setting f(t) Eq. (9.1), is the same for all system variables: dy

T-+y=O dt

= 0 in

(9.10)

and generates the characteristic equation (Sec. 8.3.1):

-rJ..+ 1 = 0

(9.11)

which has a single root, "'- = -1/-r. From Sec. 8.3.1 the system response to an initial condition y(O) is Yh(t)

= y(O)eJ..t = y(O)e-t/r

(9.12)

280

First- and Second-Order System Response

Chap. 9

y(t)

4

3

I e

2

.!! UJ

:>..

ell

T=5 4

2

0

Tinfmite T= 10 8

6

10

Time(s) Figure 9.2: Response of a first-order homogeneous equation T y + y(t) = 0. The effect of the system time constant T is shown for stable systems ( t' > 0) and unstable systems (T < 0).

A physical interpretation of the time constant r may be found from the initial condition response of any output variable y(t). If r > 0, the response of any system variable is an exponential decay from the initial value y(O) toward zero, and the system is stable. If r < 0, the response grows exponentially for any finite value of yo, as shown in Fig. 9.2, and the system is unstable. Although energetic systems containing only sources and passive linear elementS are usually stable, it is possible to create instability when an active control system is connected to a system [5, 6]. Some sociological and economic models exhibit inherent instability [7]. The time constant r, which has units of time, is the system parameter that establishes the time scale of system responses in a first-order system. For example, a resistor-capacitor circuit in an electronic. amplifier might have a time constant of a few microseconds, while the cooling of a building after sunset may be described by a time constant of many hours. TABLE 9.1:

Exponential Components of First-order System Responses in Terms of Normalized Time t /T

Time

-t/r

0

0.0 1.0 2.0 3.0 4.0

f

2r 3r 4r

y(t)/y(O)

= e-r/r

1.0000 0.3679 0.1353 0.0498 0.0183

y,(t)

= 1- e-r/r 0.0000

0.6321 0.8647 0.9502 0.9817

281

First-Order Linear System Transient Response

Sec. 9.2

y(t)!y(O)

1

?A

= 0 Cl.

e '0

0.8

~

0.6

U)

.~

~

0.368

~_j__l_ I

I

H--

\

'

\

0.4

' """"

I

1'-

0.2

---

0.135 0.049 0.018 0

-

-,........_,

2

3

4

5 riT

Nonnalized time Figure 9.3:

Normalized unforced response of a stable first-order system.

It is common to use a normalized time scale, t lr:, to describe first-order system responses. The homogeneous response of a stable system is plotted in normalized form in Fig. 9.3, using both the normalized time and a nonnalized response magnitude y(t)ly(O): y(t) _

-(t/'r)

(9.13)

y(O)- e

The third column in Table 9.1 summarizes the homogeneous response after periods r:, 2-r:, ••.. After a period of one time constant (tlr: 1) the output has decayed to y( r:) e- 1y(O) or 36.8% of its initial value, and after two time constants the response is y(2r:) 0.135y(O). Several first-order mechanical and electric systems and their time constants are shown in Fig. 9.4. For the mechanical mass-damper system shown in Fig. 9.4, the velocity of the mass decays from any initial value in a time detennined by the time constant r: m I B, while the unforced deflection of the spring shown in Fig. 9.4 decays with a time constant r: = BI K. In a similar manner the voltage on the capacitor in Fig. 9.4 decays with a time constant r: = RC, and the current in the inductor in Fig. 9.4 decays with a time constant equal to the ratio of the inductance to the resistance r: = L I R. In all cases, if SI units are used for the element values, the units of the time constant will be seconds.

t

=

= =

=

=

Example9.2 A water tank with vertical sides and a cross-sectional area of 2m2 , shown in Fig. 9.5, is fed from a constant displacement pump which may be modeled as a ftow source Qin(t). A valve, represented by a linear ftuid resistance R1 , at the base of the tank is always open and allows water to flow out. In normal operation the tank is filled to a depth of 1.0 m. At time t = 0 the power to the pump is removed and the ftow into the tank is disrupted. If the ftow through the valve is 1o-6 m3 /s when the pressure across it is 1 N/m2 • determine the pressure at the bottom of the tank as it empties. Estimate how long it will take for the tank to empty.

282

First- and Second-Order System Response

~vm

~

F(t) -Yh=1 F(t) t m dvm

1

B dt + 'Vm =jj_ F(t)

V(t)

/vM:O

~1 W)V:=D K

B

B dFK Kdt + FK= BV(t)

T=i

T=~

~t)Q=D /(t)rnL l(t)~L

0

dv. RC dt'C + vc

8

Chap. 9

=V(t)

L diL • R dl + 'L :;;: /(t)

T=RC

Vref=O L

T=R

Flgure9.4: Tnne constants of some typical first-order systems.

Solution The tank is represented as a fluid capacitance c, with a value (Sec. 2.5.2) A c,=pg

(i)

where A is the area, g is the gravitational acceleration, and p is the density of water. In this case c1 = 2/(1000 x 9.81) 2.04 x 10-4 m5JN and R1 1/10-6 = tot» N-s/m5 •

=

=

The linear graph generates a state equation in terms of the pressure across the fluid capacitance Pc(t): (ii)

which may be written in the standard first-order form

(iii)

=

The time constant is T R1 c1 . When the pump fails, the input flow Qin is set to zero and the system is described by the homogeneous equation

(iv)

The homogeneous pressure response is [from Eq. (9.12)] (v)

Sec. 9.2

FJ.I'St-Order Linear System Transient Response

283

With the given parameters the time constant is r = R1 C1 = 204 sand the initial depth of the water h(O) is 1 m; the initial pressure is therefore Pc(O) = pgh(O) = 1000 x 9.81 x I N/m2 • With these values the pressure at the base of the tank as it empties is Pc(t)

= 9810e-'1204 N/m2

(vi)

which is the standard first-order form shown in Fig. 9.3.

Figure 9.5: Fluid rank example.

The time required for the tank to drain cannot be simply stated because the pressure asymptotically approaches zero. It is necessary to define a criterion for the complete decay of the response; commonly a period oft = 4r is used since y(t)fy(O) = e-4 < 0.02 as shown in Table 9.1. In this case after a period of 4r = 816 s the tank contains less than 2% of its original volume and may be approximated as empty.

9.2.2 The Characteristic Response of First-Order Systems In standard fonn the input-output differential equation for any variable in a linear first-order system is given by Eq. (9.1): r dy dt

+y =

f(t)

(9.14)

The only system parameter in this differential equation is the time constant r. The solution with the given f(t) and the initial condition y(O) = 0 is defined to be the characteristic first-order response. The first-order homogeneous solution is in the form of an exponential function Yh (t) = e-A.t, where A. = 1/r. The solution method described in Sec. 8.3.3 expresses the total response y(t) as the sum of two components y(t) = Yh (t)

+ Yp(t)

= Ce-tf't:

+ Yp(t)

(9.15)

where Cis a constant found from the initial condition y(O) = 0 and Yp(t) is a particular solution for the given forcing function f (t). In the following sections we examine the form of y(t) for the ramp, step, and impulse singularity forcing functions.

First- and Second-Order System Response

284

Chap. 9

The Characteristic Unit Step Response The unit step Us (t) is commonly used to characterize a system's response to sudden changes in its input It is discontinuous at time t = 0 and is defined in Sec. 8.2, f(t)

0 t< 0 t ;: O

= Us(t) = { l

The characteristic step response Ys(t) is found by determining a particular solution for the step input using the method of undetermined coefficients (Sec. 8.3.2). From Table 8.2, with a constant input fort > 0, the fonn of the particular solution is Yp (t) = K, and substitution into Eq. (9.14) gives K = 1. The complete solution Ys(t) is Ys(t)

= ce-t/T + 1

(9.16)

The characteristic response is defined when the system is initially at rest, requiring that at t = 0, Ys(O) = 0. Substitution into Eq. (9.15) gives 0 = C + 1, so the resulting constant C = -1. The unit step response of a system defined by Eq. (9.14) is Ys(t)

=1-

e-t/r:

(9.17)

Equation (9.17) shows that, like the homogeneous response, the ~e dependence of the step response depends only on t' and may be expressed in terms of a normalized time scale t /T. The unit step characteristic response is shown in Fig. 9.6, and the values at normalized time increments are summarized in the fourth column of Table 9.1. The response asymptotically approaches a steady-state value Yss

= ,_00 lim Ys(t) = 1

(9.18)

It is common to divide the step response into two regions:

1. A transient region in which the system is still responding dynamically 2. A steady-state region in which the system is assumed to have reached its final value Yss· There is no clear division between these regions, but the time t = 4T, when the response is within 2% of its final value, is often chosen as the boundary between the transient and steady-state responses. The initial slope of the response may be found by differentiating Eq. (9.17) to obtain

dyl dt

1

t=O

=;

(9.19)

The step response of a first-order system may be easily sketched with knowledge of (1) the system time constant t', (2) the steady-state value YsSt (3) the initial slope y(O), and (4) the fraction of the final response achieved at times equal to multiples of t'.

Sec. 9.2

285

First-Order Linear System Transient Response Ys(t)IYss 1.2 !lA

8. f1

I

I

&

0.8

--r--

1

2 3 Normalized time

0

4

Figure 9.6: The step response of a first-order system described by

5 tiT

t'J + y =

U 8 (t).

The Characteristic Impulse Response In Sec. 8.2 the impulse function 8(t) is defined as the limit of a pulse of duration T and amplitude 1IT as T approaches zero and is used to characterize the response of systems to brief transient inputs. The impulse may be considered to be the derivative of the unit step function. The derivative property of linear systems (Sec. 8.4.4) allows us to find the characteristic impulse response Y&(t) by simply differentiating the characteristic step response Ys(t). When the forcing function f(t) 8(t), the characteristic response is

=

Y&(t )

= -dys = -dtd {I dt 1 - tfr = -e t'

for

e

-t/T} (9.20)

t

>0 -

The characteristic impulse response is an exponential decay similar in form to the homogeneous response. It is discontinuous at timet = 0 and has an initial value y(O+) = 1/t', where the superscript o+ indicates a time incrementally greater than zero. The response is plotted in normalized form in Fig. 9. 7.

The Characteristic Ramp Response The unit ramp u,(t) = t fort ~ 0, described in Sec. 8.2.1, is the integral of the unit step function Us(t): u,(t) =

fo' u,(t) dt

(9.21)

The integration property of linear systems (Sec. 8.4.4) allows the characteristic response y,(t) to a ramp forcing function f(t) u,(t) to be found by integrating the step response Ys(t):

=

y,(t) =

fo' y,(t) dt = fo' (I - e-•1•) dt

FII'St- and Second-Order System Response

286

Chap. 9

T)'a(t)

1.2

i

1

le o.s ~

i .5

0.6

~

0.4

'ii

z§ 0.2

'

_\

\.

'·' ' "" ~

i

~

~

........_ 4

2 3 Normalized time

0

StiT

Figure 9.7: The impulse response of a first-order system described by TY + y = cS(t).

(9.22)

and is plotted in Fig. 9.8. As t becomes large, the exponential term decays to zero and the response becomes Yr(t)

~

t

-1"

fort>>

1"

(9.23)

9.2.3 System Input-Output Transient Response In the previous section we examined the system response to particular forms of singularity forcing functions f(t). We now return to the solution of the complete most general firstorder differential equation, Eq. (9.8):

1"

dy dt

+ y(t) =

du q1 dt

+ qou(t)

(9.24)

where 1" = -lfa, q1 = -d/a, and q2 =(ad- be) fa are constants defined by the system parameters. The forcing function in this case is a superposition of the system input u (t) and its derivative: du f(t) = q1 dt + qou(t) The superposition principle for linear systems allows us to compute the response separately for each term in the forcing function and to combine the component responses to form the overall response y(t). In addition, the differentiation property of linear systems allows the response to the derivative of an input to be found by differentiating the response to that input These two properties may be used to determine the overall input-output response in two steps:

Sec. 9.2

287

FirSt-Order Linear System Transient Response y,(t)

5

i I

I I I

I

!

I

I i i i I

I

I

!./l~ ~ I 0

./ '/"

1/

I./"I

/" ..............

1/

./"'

1 !./ T

./ l/' T

"""'"""'

./

./

·-

2 3 Normalized time

5 tiT

4

Figure 9.8: The ramp response of a first-order system described by

'r

y + y = u, (t).

1. Find the characteristic response Yu(t) of the system to the forcing function f(t) u(t), that is, solve the differential equation: -r

dyu

dt + Yu (t) =

u(t)

=

(9.25)

2. Fonn the output as a combination of the output and its derivative:

(9.26)

The characteristic responses Yu(t) are by definition zero for time t < 0. If there is a discontinuity in Yu(t) at t = 0, as in the case for the characteristic impulse response Y&(t) [Eq. (9.20)], the derivative dyufdt contains an impulse component, for example,

!!_ Y&(t) = ~ 8(t) -r

dt

_!_e-tlr -r 2

(9.27)

and if q 1 :f: 0, the response y(t) will contain an impulse function. The Input-Output Step Response The characteristic response for a unit step forcing function, f(t) = u 5 (t), is [Eq. (9.17)] Ys(t)

= 1-

e-t/T fort > 0

The system input-output step response is found directly from Eq. (9.26):

(9.28)

288

First- and Second-Order System Response

Chap. 9

If q1 :f;; 0, the output is discontinuous at t = 0, and y(O+) = q 1/T. The steady-state response Yss is (9.29) Yss lim y(t) qo

=

=

, ... 00

The output moves from the initial value to the final value with a time constant T.

The Input-Output Impulse Response The characteristic impulse response y 8 (t) found in Eq. (9.20) is

1

Y6(t) = -e-tf-c fort :::: 0 T

with a discontinuity at timet= 0. Substituting into Eq. (9.26), y(t )

dy8 = q1 dt + qoY& (t")

= q1 B(t) + (qo T

(9.30)

_ q1) e-t/-c

T

T2

where the impulse is generated by the discontinuity in y 8 (t) at t = 0 as shown in Eq. (9 .27).

The Input-Output Ramp Response The characteristic response to a unit ramp r{t) =tis Yr(t)

= t - T (1 -

e-tfr)

and using Eq. (9.22), the response is y(t)

= q1 dtd {[t- T (I - e-'1")] Us(t)} + qo [t- T (1- e-tf-r)] Us(t) = [qot + (ql - qoT) (1 - e-'1")] u (t)

(9.31)

8

9.2.4 Summary of Singularity Function Responses Table 9.2 summarizes the homogeneous and forced responses of the first-order linear system described by the classical differential equation T

dy dt

du

+ Y = q1 dt + qou

(9.32)

for the three commonly used singularity inputs. The response of a system with a nonzero initial condition y(O) to an input u(t) is the sum of the homogeneous component due to the initial condition and a forced component computed with zero initial condition, that is, Ytotai(t)

= y(O)e-tf-r + Yu(t)

(9.33)

where Yu (t) is the response of the system to the given input u(t) if the system was originally at rest

289

First-Order Linear System Transient Response

Sec. 9.2

The Response ofthe First-Order Linear System tjo +y = q,u +qou for the Singularity Inputs

TABLE 9.2:

Input

Characteristic Response

u(t) = 0

y(t) = y(O)e-t/T

u(t) = u,(t)

y,(t) = I - T {I

u(t) = u,.(t)

y,.(t) = y,.(t)

u(t)

= cS(t)

Y.s(t)

=

-

Input-Output Response fort;:: 0

e-I/T)

= 1- e-''" 1

-e-1/T T

The response to an input that is a combination of inputs for which the response is known may be found by adding the individual component responses using the principle of superposition. The following examples iUustrate the use of these solution methods. Example9.3 A mass m = 10 kg is at rest on a horizontal plane with viscous friction coefficient B = 20 N-s/m, as shown in Fig. 9.9. A short impulsive force of amplitude 200 Nand duration 0.01 s is applied. Determine how far the mass travels before coming to rest and how long it takes for the velocity to decay to less than 1% of its initial value.

F(tl

Rt)ts;m

200

~~r=O

0 0.01 Figure 9.9:

A mass element subjected to an impulsive force.

Solution The differential equ·ation relating the velocity of the mass to the applied force is m dvm

1

- + Vm = -Fin(t) B dt B

(i)

The system time constant is -r = mj B = ~ = 0.5 s. The duration of the force pulse is much less than the time constant, and so it is reasonable to approximate the input as an impulse of strength (area), 200 x 0.01 = 2 N-s. The system impulse response [Eq. (9.30)] is Vm (t)

so if u(t) =

2~(1)

1 = -e-Bt/m

m

(ii)

newton-seconds, the response is Vm(t)

= 0.2e-2t

(iii)

The distance x traveled may be computed by integrating the velocity:

x

=

1

00

0.2e- 21 dt

= 0.1 m

(iv)

290

First- and Second-Order System Response

Chap. 9

The time T required for the velocity to decay to less than 1% of its original value is found by solving Vm(T)/Vm(O) = 0.01 = e-2T, or T = 2.303 s. Example9.4 A disk flywheel J of mass 8 kg and radius 0.5 m is driven by an electric motor that produces a constant torque of Tin = 10 N-m. The shaft bearings may be modeled as viscous rotary dampers with a damping coefficient of BR = 0.1 N-m-s/rad. If the flywheel is at rest at t = 0 and power is suddenly applied to the motor, compute and plot the variation in speed of the flywheel and find the maximum angular velocity of the flywheel.

Solution The state equation for the system may be.found directly from the linear graph in

Fig. 9.10: dCJ

BR J

-=--OJ+ dt

1 J

-T~n(t)

(i)

which in the standard fonn is (ii)

T~ Figure 9.10: Rotary flywheel system and its linear graph.

For the flywheel J = mr2 /2 = 1 kg-m2 , and the time constant is J BR

~=-=lOs

(iii)

The characteristic response to a unit step in the forcing function is Ys(t) = 1 - e-r/10

(iv)

and by the principle of superposition. when the forcing function is scaled so that f(t) = (Tm/ BR)Us(t), the output is similarly scaled: (v)

291

FII'St-Order Linear System Transient Response

Sec. 9.2

!lj(t)

120

~ A

I

100

n~ I

I

I I

80

·s

60

--

'"'

,

/

40

l!ll)

~

__.

/

l;>

l a "S

./

I

......-

1

20

I 0

I

I

-

I

...

-.

--10

30

20

40

50 t

Time (s) Figure 9.11:

Response of the rotary flywheel system to a constant torque input, with initial condition 0 1 (0) = 0, in Example 9.4.

The steady-state angular velocity is (vi) and the angular velocity reaches 98% of this value in t shown in Fig. 9.11.

= 4't' = 40 s. The step response is

Example9.5 During nonnal operation the flywheel drive system described in Example 9.4 is driven by a programmed torque source that produces a torque profile as shown in Fig. 9.12. The torque is ramped up to a maximum of 20 N-m over a period of 100 s, held at a constant value for 25 s, and then reduced to zero. Find the resulting angular velocity of the shaft.

0

50

Figure 9.12:

100 Time (s)

150

Rotary flywheel system and the input torque function specified in Example 9.5.

Solution From Example 9.4 the differential equation describing the system is (i)

292

Fust- and Second-Order System Response

Chap. 9

and with the values given (J = 1 kg-m 2 and BR = 0.1 N·m·s/rad) dQJ

IOdr

+ 0.1

= 10T~n(t)

(ii)

The torque input shown in Fig. 9.12 may be written as a sum of unit ramp and step singularity functions: Tm(t)

= 0.2u,(t) -

0.2u,(t - 100)- 200u.r(t - 125)

(iii)

The response may be determined in three time intervals:

1. Initially, 0 ~ t < l 00 when the input is effectively 7ia (t) 2. For 100 ~ t < 125 s when the input is

3. Fort

~

125 when

T~n(t)

T~n(t)

= 0.2u, (I)

= 0.2u,(t) - 0.2u,(t - 100)

= 0.2u,(t)- 0.2u,(t -

100) - 20u.r(t - 125).

From Table 9.2 the response in the three intervals may be written 0 ~ t < 100 s: 0.1(t)

= 2 [t- 10 (1 - e-'1 10)] rad/s

100 ~ t < 125 s: QJ(t)

= 2 [t- 10 {1- e-'110)] -2 [Ct -100)- 10 {1- e-cr-I00)/ 10)] rad/s

fij{t) 250

~ g

200

~

ISO

~

100

i

-;

Cll)

~

so 0

so

100 Time(s)

ISO

200

I

Agure 9.13: Response of the rotary flywheel system to the torque input profile 7in(t) 0.2ur(t)- 0.2ur(t- 100)- 20u,(t) newton-meters, with initial condition Sl1 (0) = 0 radls.

=

Sec. 9.2

293

First-Order Linear System Transient Response t >

125 s: Q 1 (t)

= 2 [t- 10 (1- e- 110)] 1

-2 [Ct- 100)- 10 (1- e- 1), AI and A2 = -~wn - ../~ 2 - 1wn, this reduces to

= -~wn + ./~ 2 - lwn

(9.101)

where 'tt

= -1/A.t and 't2 = -1/A2·

314

First- and Second-Order System Response For the case of complex conjugate roots, 0
n

for 0 !:: ~ < 1. The phase angle

= tan- 1 ( r I Jt=" fi') for 0 ~ ~ < 1. For overdamped systems ( ~> 1) the time constants are

t:, =

1/ (~edn- ./r2 - led,) and 1:2 = 1/ (red,+ ./~ 2 - led,).

determined directly by superposition of characteristic responses. The complete differential equation (9.110)

in general involves a summation of derivatives of the input The principle of superposition allows us to detennine the system response to each component of the forcing function and to sum the individual responses. In addition, the derivative property tells us that if the response to a forcing function /(t) = u(t) is Yu(t), the other components are derivatives of Yu(t) and the total response is (9.111)

As in the case of first-order systems, the derivatives must take into account discontinuities at timet= 0.

Sec. 9.3

Second-Order System Transient Response

317

Example 9.11

Detennine the response of a physical system with the differential equation d2y dt2

to a step input u(t)

dy

du

+ 8 dt + 16y = 3 dt + 2u

= 2 fort ~ 0.

Solution The characteristic equation is (i)

which has roots >.. 1 = -2 and >..2 = -8. For this system Wn = 4 radls and ~ = 1.25; the system is overdamped. The characteristic response to a unit step is (from Table 9.3) (ii)

where T 1 =!and

T2

= l· or (iii)

The system response to a step of magnitude 2 is therefore

(iv)

For systems in which q2 =f: 0 a further simplification is possible. The system differential equation may be written in operational form: (9.112)

and rearranged as (t)

y

= q

2

{u}

+ (qt - 2~~wn) S + (qo- b2w;) {u} S2 + 2~wnS + w~

(9.113)

The response is then found from the characteristic response and the input: (9.114)

Fust- and Second-Order System Response

318

Chap. 9

Example 9.12

Find the response of a physical system with the differential equation d 2y dy dt2 +S dt

to a step input u(t)

+ 4y

d 2u

= dt2

du + 2 dt +u

= 2 fort ~ 0.

Solution The characteristic equation is 2 ).

+ 4). + 4 = 0

(i)

which bas a pair of coincident roots, A. 1 = A.2 = -2. The system is critically damped with = 2 radls. The characteristic impulse response is (from Table 9.3)

Wn

(ii)

Because q2 '::f: 0, we may write the system response as y(t)

= q28(t) + (qJ -

dy, 2b2Cwn) dt

+ (qo- b2wn2) Y&(t)

dy, = 8(t)- 4dt- 2y,

(iii)

= 8(t) - re- 21 - 2e-21 Example 9.13 An electric motor is used to drive a large-diameter fan through a coupling as shown in Fig. 9.26. The motor is not an ideal source but exhibits a torque-speed characteristic that allows it to be modeled as a Thevenin equivalent source with an ideal angular velocity source 0, (t) = 0 0 in series with a hypothetical rotary damper Bm as discussed in Chap. 5. The motor is coupled to the fan through aflexible coupling with torsional stiffness K,, and the fan impeller is modeled as an inertia J with the bearing and impeller aerodynamic loads modeled as an equivalent rotary damper B,.

Motor (nonideal source)

J9

Impeller

Figure 9.26: Electric motor fan drive system.

The response of the fan speed when the motor is energized is of particular interest since if the fan speed exceeds the design speed, the impeller can experience excessive stresses due to centrifugal forces. It is desired to select the system components so that the fan impeller reaches its operating speed with no overshoot. The torque in the coupling K r during the start-up transient is also of interest because excessive torque can lead to failure. The motor specifications indicate

Sec. 9.3

319

Second-Order System Transient Response

that Co = 100 radls and Bm = 1.0 N-m-s/rad. The inertia of the fan impelleris J = 1.0 kg-m 2 , and the net drag of the bearings and aerodynamic load is B, = 1.0 N-m-s/rad. The coupling stiffness needed to achieve an impeller response with no overshoot is to be determined, as well as the response of the system state variables. Solution The state equations for the system may be expressed in terms of the two state variables OJ, the fan impeller angular velocity, and TK, the torque in the flexible coupling:

= [-B,fJ [ ~J] Tx -K,

1/ J ] -K,JBm

[OJ] Tx

0]O + [ K, s

(i)

The system characteristic equation is B, +K, I A] =.i..2 + ( det [.i..-) J Bm

K, ( 1 +B,- ) A.+J Bm

=0

(ii)

and the undamped natural frequency and damping ratio are K, ( 1 + B, )

w, =

J

1 2w,

~=-

(iii)

Bm

(B,- +BmK,) -

(iv)

J

Notice that for this system the values of the two damping coefficients influence both the natural frequency and the damping ratio.

1. The differential equation describing the fan speed is d2yl

dyl

2

+ 2~Wn dt + W,YI

dtl

K,

= JOs

(v)

with a constant input Os(t) = 0 0 • For no overshoot in the step response on starting the motor, the system must be at least critically damped(~ ?! 1). Using Eqs. (ii) and (iii), the value of K, required for critical damping may be found by setting ~ = 1 in Eq. (iv), obtaining

Kr ( Br) 2 - 1+}

Bm

B, K, =-+J Bm

(vi)

and with the system parameter values this equation gives K, = 5.83 N-m!rad

With this value of K, the system parameters are~ Table 9.3 the unit characteristic step response is y,(t)

= ~ (1 -

(vii)

= 1 and w, = 3.41 rad/s. From

e-OJnt- w,te-a~n')

(viii)

n

and the impeller response to a step of 100 rad/s is 0 1 (t)

= 100q0 (t - e-a~nr- w,te-wn') w~

=50 {1 -

(ix) e-3 ·41 ' -

3.41te-3.4 1')

Fust- and Second-Order System Response

320

Chap.9

The response in fan speed is similar to the nondimensional fonn shown in Fig. 9.22 and is plotted in Fig 9.27. Note that the steady-state speed is 50 rad/s, which is one-half the motor no-load speed of 100 radls. 2. The differential equation relating the torque TK to the source velocity is

(x)

which contains both the input ns and its derivative. Then, Tg(t)

= 100 [ K,te-c&Jn' + ~i {I =50 {1 -

e-3•411

+

e-> ', where -t' is the system time constant.

.--.

10

I I

I I

I

I

-10

T Figure 9.38:

A first-order e]ectric circuit excited by a repetitive square wave.

9.18. In an automated machine, a force source is used to drive a mass supported on a guide as shown in Fig. 9.39. The force waveform is also shown in the figure. Derive an equation relating the mass velocity to the driving force and determine the system time constant for m = 10 kg, and B = S N-s/m. Derive an expression for the mass velocity for A = 2 N, and T = 2 s. Derive an expression for the mass displacement. Sketch the response of the mass displacement.

9.19. A common form of a second-order linear differential equation is:

Chap. 9

327

Problems

Sliding mass

0 Figure 9.39:

T/2

T

A machine drive and the force driving waveform.

where the as are constant coefficients. Determine the undamped natural frequency and the damping ratio in terms of the system parameters, and rewrite the equation in terms of the undamped natural frequency and damping ratio. If the output variable y(t) is a displacement, what are the appropriate units for the constant coefficients in (i) the SI system and (ii) the English system of units? 9.20. Determine the system undamped natural frequency and damping ratio for the mechanical and the electrical systems shown in Fig. 9.40(a) and (b). In each case, if it is desired to increase the undamped natural frequency, what parameters should be changed? How do these changes influence the system damping ratio? What parameters can be used to increase the system damping ratio without altering the undamped natural frequency?

Fs(t}

c

m

(b)

(a}

Figure 9.40:

A mechanical and an electrical second-order system.

9.21. Consider the system described in Example 9.8 and illustrated in Fig. 9.16. Assume that the system parameters yield an undamped natural frequency of I 00 rad/s and a damping ratio of 0.2. Derive an expression for the flywheel angular velocity as a function of time for the case that the motor angular velocity is changed stepwise from 0.0 to 50.0 rad/s at t = 0, and sketch the response. If the spring stiffness were increased by a factor of four what are the new values of undamped natural frequency and damping ratio? How much would the damping ratio have to be increased to obtain a response in flywheel angular velocity which has no overshoot, if the original spring stiffness were considered? 9.22. An apparatus to determine the stiffness and damping characteristics of foam-like materials consists of a mass which is released in a gravity field and allowed to compress the material. As shown in Fig. 9.41, the mass displacement is measured and analyzed to determine the equivalent stiffness and damping of the material. The mass is released at zero velocity from a point where it is in contact with the material but causes no deflection. (a) Assuming the material may be represented by equivalent linear stiffness and damping properties, formulate a system model and derive the state equations. What is the equivalent source in the model after the mass is released?

328

First- and Second-Order System Response

Chap. 9

)'(I)

r

Massm

0.4

released at r = 0 to defonn material

y(t)

E"

'E o.3 CJ

E CJ

Material specimen

~

0..

0"'

0.2 0.1

/

0

2

3

Time (s) Figure 9.41:

Apparatus for measuring material stiffness and damping properties and response to two samples.

(b) Derive a single differential equation for the mass displacement Determine the system undamped natural frequency and damping ratio in terms of system parameters. (c) Tests are conducted on two material specimens, labeled (a) and (b), using a 1.0 kg mass. The dynamic responses of the mass displacement are shown in Fig. 9.41. Use the response plots to estimate the stiffness and damping of each sample. 9.23. A simplified model of a vibration absorbing table for optical measurements is shown in Fig. 9.42. A massive table m is supponed on four legs with resilient mounts with parallel stiffness K and damping B. Measurements have shown that m = 320 kg, and for each leg K = 4000 N/m, and B = 300 N-s/m. The venical floor vibrations are modeled by velocity input Vs(t). The model combines the four legs into an equivalent stiffness K,q and damping B,q.

t

Vs(l)

Figure 9 .42:

Lumped model of a vibration isolating table.

(a) Derive a differential equation relating the venical table velocity to the floor velocity. Determine the system undamped natural frequency and damping ratio with the parameter values given. (b) Find and sketch the velocity response of the table to a sudden change in floor velocity of0.2 m/s. (c) What is the minimum value of the damping coefficient B of each leg that will prevent oscillatory responses to transient inputs.

Chap. 9

329

Problems

9.24. A model for a feedback control system employing both angular position and velocity feedback is shown in Fig. 9.43. The equation describing the system is:

Summing

amplifier

\.

t

Motor

J-

jJ

?:_z_b._~/._z_;;;_'//._z_;;;...~..'//._z_z_?/.__z_z_._ __,9

9 .__K._e_ _ _ _ _ _ Figure 9.43:

sensor

==n

A feedback control system.

where J is the rotary inertia, K 9 and Ko are the position and velocity feedback gains, and Kais the gain between the input voltage to the motor and the motor torque produced. (a) Derive expressions for the closed-loop system undamped natural frequency and damping ratio. (b) If the inertia is 10 kg-m2 , what values of K 9 and Ko are required with Ka = I to achieve a

response to a step input in voltage which reaches a peak value that is 1.25 times the steady-state response in 0.1 s? (c) How much would the velocity feedback gain K0 have to be increased to achieve a response with zero overshoot? (d) Sketch the responses corresponding to cases (b) and (c). 9.25. A pumping station consists of a pump, a long pipe, and a tank with an outflow valve as shown in Fig. 9.44. The goal is to design the system so that there wiU be no overshoot in the depth of fluid in the storage tank resulting from transient changes in the pump pressure Pp(t).

g~

Tank

c

Valve

Long pipe I

___

._

_. Fluid reservoir Figure 9.44:

A fluid pumping station.

(a) Assume the pipe resistance can be neglected and that the pump may be modeled as a pressure source. Formulate a system model and derive the state equations. (b) Develop a single differential equation relating the pressure at the bottom of the tank to the pump pressure, and determine the system undamped natural frequency and damping ratio.

330

FlrSt- and Second-Order System Response

Chap. 9

(c) A test is conducted on the system, in which the pump is suddenly energized to provide a pressure P,(t) = P0 • If the outflow valve is shut, that is R = oo, determine and sketch the tank pressure response. In the test it is observed that the peak values of pressure occur every 0.4 s. What is the system undamped natural frequency? (d) The goal is to fill the tank without a water spill caused by overshoot in the response. If the tank area A = 20 m2• what is the maximum value of valve resistance R that may be selected to prevent overshoot?

REFERENCES [1] Shearer, J. L.• Murphy, A. T., and Richardson, H. H., Introduction to System Dynamics, AddisonWesley, Reading, MA, 1967. [2] Ogata. K., System Dynamics, Prentice Hall, Englewood Cliffs, NJ, 1978. [3) Kamopp, D. C., Margolis, D. L., and Rosenberg, R. C., System Dynamics: A Unified Approach (2nd ed.), John Wiley, New York, 1990. [4] Shearer, J. L., and Kulakowski, B. T., Dynamic Modeling and Control of Engineering Systems, Macmillan, New York, 1990. · [5) Kuo, B. C., Automatic Control Systems (6th ed.), Prentice Hall, Englewood Cliffs, NJ, 1991. [6] Franklin, G. F., Powell, J.D., and Emami-Naeni, A., Feedback Control ofDyruunic Systems (2nd ed.), Addison-Wesley, Reading, MA. 1991. [7] Luenberger, D. G., Introduction to Dynamic Systems, Theory, Models, and Applications, John Wiley, New York. 1979.

10

General Solution of the Linear State Equations

10.1

INTRODUCTION In the previous chapters the response of Jinear systems was derived from models based on the classical input-output differential equation. In this chapter we examine the responses of linear time-invariant models expressed in the standard state equation form

x=Ax+Bu y=Cx+Du

(10.1)

(10.2)

The solution proceeds in two steps: First the state variable response x(t) is determined by solving the set of first-order state equations, Eq. ( 10.1 ), and then the state response is substituted into the algebraic output equations, Eq. ( 10.2) in order to compute y(t). As described in Chap. 8, the total system state response x(t) is considered in two parts: the homogeneous solution xh(t) that describes the response to an arbitrary set of initial conditions x(O) and a particular solution Xp(t) that satisfies the state equations for the given input u(t). The two components are then combined to form the total response. The solution methods encountered in this chapter rely heavily on matrix algebra. In order to keep the treatment simple, we attempt wherever possible to introduce concepts using a first-order system in which the A, B, C, and D matrices reduce to scalar values and then to generalize results by replacing the scalars with the appropriate matrices. 331

332 10.2

General Solution of the Linear State Equations

Chap.IO

STATE VARIABLE RESPONSE OF LINEAR SYSTEMS

10.2.1 The Homogeneous State Response The state variable response of a system described by Eq. (1 0.1) with zero input and an arbitrary set of initial conditions x(O) is the solution of the set of n homogeneous first-order differential equations (10.3) To derive the homogeneous response Xh(t), we begin by considering the response of a first-order (scalar) system with state equation

.i =ax +bu

(10.4)

with initial condition x(O).In Chap. 9 the homogeneous response Xh(t) is shown to have an exponential form defined by the system time constant r = -lfa, or (10.5) The exponential term lfl' in Eq. (10.5) may be expanded as a power series to give (10.6) where the series converges for all finite t. Let us now assume that the homogeneous response Xh (t) of the state vector of a higher-order linear time-invariant system, described by Eq. (1 0.3), can also be expressed as an infinite power series similar in fonn to Eq. (10.6) but in tenns of the square matrix A, that is, we assume (10.7) where x(O) is the initial state. Each tenn in this series is a matrix of size n x n, and the summation of all tenns yields another matrix of size n x n. To verify that the homogeneous state equation i Ax is satisfied by Eq. (I 0. 7), the series may be differentiated tenn by term. Matrix differentiation is defined on an element-by-element basis, and because each system matrix A lc contains only constant elements,

=

(10.8)

Equation (I 0.8) shows that the assumed series form of the solution satisfies the homogeneous state equations, demonstrating that Eq. (10.7) is in fact a solution of Eq. (10.3).

Sec. 10.2

State Variable Response of Linear Systems

333

The homogeneous response to an arbitrary set of initial conditions x(O) can therefore be expressed as an infinite sum of time-dependent matrix functions involving only the system matrix A. Because of the similarity of this series to the power series defining the scalar exponential, it is convenient to define the matrix exponential of a square matrix A as (10.9)

which is itself a square matrix the same size as its defining matrix A. The matrix form of the exponential is recognized by the presence of a matrix quantity in the exponent. The system homogeneous response xh (t) may therefore be written in terms of the matrix exponential (10.10)

which is similar in form to Eq. (1 0.5). The solution is often written as (10.11) where cl»(t) =eAt is defined to be the state transition matrix [1-5]. Equation (10.11) gives the response at any time t to an arbitrary set of initial conditions, thus computation of eAr at any t yields the values of all the state variables x(t) directly. Example 10.1 Determine the matrix exponential, hence the state transition matrix, and the homogeneous response to the initial conditions x 1 (0) = 2, x2(0) = 3 of the system with state equations

it= -2xl +u X2 = x,- X2 Solution The system matrix is

A=

[-2 o] 1

-1

From Eq. (10.9) the matrix exponential (and the state transition matrix) is

A2r2 A3r3 Alctk ) = ( I+At+--+--+···+-+··· 2! 3! k!

= [ 01

0] 1

+

[ -2 1

+ [ -8 7

0 ]

[ 4

-1 t + -3

l

0]

2

t

1 2!

(i)

3

o -+ t ...

-1 3! 4t 2 8t 1-2t+---+ ... 2! 3! - [ 3t 2 1t3 O+t--+-+··· 2! 3!

r2

0

r3

1-t+---+··· 2! 3!

]

334

General Solution of the Linear State Equations

Chap. 10

The elements ¢ 11 and ¢22 are simply the series representation for e-21 and e-1 , respectively. The series for ¢21 is not so easily recognized but is in fact the first four tenns of the expansion of 1 e- - e- 21 • The state transition matrix is therefore

0 ] e-t

• = [ e_,e-21 -e-21

(ii)

and the homogeneous response to initial conditions x 1(0) and x2 (0) is Xh(t) = fl»(t)X(O)

(iii)

or

= X1 (O)e-21 x2(t) = XJ (0) {e-

(iv)

Xt (t)

1

-

e-21 )

+ x2(0)e-r

(v)

With the given initial conditions the response is X1

(t)

x2(t)

=

(vi)

2e-2l

= 2 (e-

1

-

e-

21

)

+ 3e-' (vii)

In general the recognition of the exponential components from the series for each element is difficult and is not nonnally used for finding a closed form for the state transition matrix.

Although the sum expressed in Eq. (1 0.9) converges for all A, in many cases the series converges slowly and is rarely used for the direct computation of C»(t). There are many methods for computing the elements of cl»(t), including one presented in Sec. 10.4, that are much more convenient than the direct series definition [1, 5, 6].

10.2.2 The Forced State Response of Linear Systems we· now consider the complete response of a linear system to an input u(t). Consider first a first-order system with a state equation = ax + bu written in the form

x

x(t) - ax(t) = bu(t)

(10.12)

If both sides are multiplied by an integrating factor e-at, the left-hand side becomes a perfect differential

(10.13) which may be integrated by introducing a dummy variable of integration t: to give (10.14)

Sec. 10.2

State Variable Response of Linear Systems

335

and rearranged to give the state variable response explicitly: X

(t) = e"' X (0)

+

fo' e"(t-t)bu (T) dT

(10.15)

The development of the expression for the response of higher-order systems may be perfonned in a similar manner using the matrix exponential e-At as an integrating factor. Matrix differentiation and integration are defined to be element-by-element operations, so if the state equations i = Ax + Bu are rearranged and all terms premultiplied by the square matrix e-At, e-Atx. (t)- e-At Ax (t)

=

!

J

[e-Atx (t) = e-A'Bu(t)

(10.16)

As in Eq. (10.14), integration ofEq. (10.16) gives (10.17) and because e-Ao = I and [e-Ar]- 1 written in two similar fonns: x(t)

= eAt

the complete state vector response may be

= ,Arx(O) +,At fo' e-ArBu(T) dT

x(t) = eArx(O)

+

fo' eA

0 if x 1 (0) = 0 and x 2

= 0.

Solution This is the same system described in Example I 0.1. The state transition matrix was shown to be e-21

tl»(t) = [ _, -21 e - e

0 ] e

_,

.

336

General Solution of the Linear State Equations

Chap. 10

With zero initial conditions, the forced response is [Eq. (10.18)] X(t)

= eAl

1'

e-ATBU(T)dT

(i)

Matrix integration is defined on an element-by-element basis, and so (ii) (iii)

(iv)

10.3

THE SYSTEM OUTPUT RESPONSE

For either homogeneous or forced responses, the system output response y(t) may be found by substituting the state variable response into the algebraic system output equations (10.20) In the case of the homogeneous response, where u(t)

= 0, Eq. (10.20) becomes

= CeAtx(O)

Yh(t)

(10.21)

while for the forced response, substitution of Eq. (10.19) into t}te output equations gives y(t) = CeAix(O) + C

fo' eA..2t m,.. ][~''] .

m12 m22

l

(10.32)

m2n

mn2

..

..

mnn

eA.nt

and if a set of m;i and A; can be found that satisfy Eq. ( 10.32), the conjectured exponential form is a solution to the homogeneous state equations. It is convenient to write the n x n matrix M in terms of its columns, that is, to define a set of n column vectors mi for j = 1, 2, ... , n from the matrix M:

mljl

m2j

mi =

[

: mnj

340

Geneml Solution of the Linear State Equations

Chap. 10

so the matrix M may be written in a partitioned fonn: (10.33)

Equation ( 10.32) may then be written

(10.34)

and for Eq. (10.34) to hold the required condition is [AJMJ I A2M2

I ... I Anmn] =A [mt I m2 I ... I m,.]

(10.35)

=[Amt1Am21 ... 1Am,.]

The two matrices in Eq. ( 10.35) are square n x n. If they are equal, then the corresponding columns in each must be equal, that is, A;m; =Am;

i = 1, 2, ... , n

(10.36)

Equation (10.36) states that the assumed exponential fonn for the solution satisfies the homogeneous state equation provided a set of n scalar quantities A; and a set of n column vectors mi can be found that each satisfy Eq. (10.36). Equation (1 0.36) is a statement of the classical eigenvalue or eigenvector problem of linear algebra [7]. Given a square matrix A, the values of Asatisfying Eq. (10.36) are known as the eigenvalues, or characteristic values, of A. The corresponding column vectors m are defined to be the eigenvectors, or characteristic vectors, of A. The homogeneous response of a linear system is therefore determined by the eigenvalues and the eigenvectors of its system matrix A. Equation (10.36) may be written as a set of homogeneous algebraic equations [A;I- A]m;

=0

(10.37)

where I is the n x n identity matrix. The condition for a nontrivial solution of such a set of linear equations is that (10.38) 8(A;) = det [A; I- A]= 0 which is defined to be the characteristic equation of the n x n matrix A. Expansion of the determinant generates a polynomial of degree ninA, and so Eq. (10.38) may be written (10.39)

or, in factored fonn in tenns of its n roots A1 , ••• , An, (10.40)

Sec. 10.4

341

The State Transition Matrix

For a physical system the n roots are either real or occur in complex conjugate pairs. The eigenvalues of the matrix A are the roots of its characteristic equation, and these are commonly known as the system eigenvalues. Example 10.4 Determine the eigenvalues of a linear system with state equations

[Xl] = [ 0 -91 -40] [XI] + [OJ ~2

0

X3

-JO

0

1

x2

0

X3

]

Solution The characteristic equation is det [AI - A]

u(t)

= 0 or

~ ~~ ~1

] =0 9 l+4 l 3 +4l2 +9l+ 10 = 0 (l + 2) [l + (1 + j2)] [l + (1- j2)] = 0 det [

(i)

10

The three eigenvalues are therefore l 1 = -2, l 2 = -1

+ j2, and l2 =

(ii)

(iii)

-1 - j2.

For each eigenvalue of a system there is an eigenvector, defined from Eq. (10.37). If the eigenvalues are distinct, the eigenvectors are linearly independent and the matrix M is nonsingular. In the development that follows it is assumed that M has an inverse and therefore the development applies only to systems without repeated eigenvalues. An eigenvector m; associated with a given eigenvalue A.; is found by substituting into the equation (10.41) [.A.;I- A]m; = 0 No unique solution exists, however, because the definition of the eigenvalue problem, Eq. (1 0.36), shows that if m is an eigenvector, then so is am for any nonzero scalar value cr. The matrix M, which is simply a collection of eigenvectors, defined in Eq. ( 10.33) therefore is not unique and some other information must be used to fully specify the homogeneous system response. Example 10.5 Detennine the eigenvalues and corresponding eigenvectors associated with a system having an A matrix

A= [-2 I] 2 -3 Solution The characteristic equation is det [ll - A] = 0 or

det[l~22

l~3] =0 l 2 +5l+4=0

(l + 4)(l + 1) = 0

(i)

. . 342

General Solution of the Linear State Equations

=

Chap. 10

The two eigenvalues are therefore A. 1 -1 and A. 2 = -4. To find the eigenvectors these values are substituted into the equation (A.1I - A] m1 0. For the case A. 1 = -1 this gives

=

[ 1 -1] [mu] m21 =[OJ -2

2

Both of these equations give the same result: m 11 defined, and although one particular solution is

m•

(ii)

0

=

= m 21 • The eigenvector cannot be further

[!]

the general solution must be defined in terms of an unknown scalar multiplier a 1 • (iii)

provided a 1.:;6 0.. Similarly, for the second eigenvalue,

A.2

= -4, the equations are

-1][m12]=[0] m22 0 both of which state that -2m 12 = m22 .The general solution is -2 [ -2 -1

(iv)

(v)

for cr2 :;6 0. The following are all eigenvectors corresponding to

A.2

= -4:

Assume that the system matrix A has no repeated eigenvalues and that the n distinct eigenvalues are A1, A2, .•• , An. Define the modal matrix M by an arbitrary set of corresponding eigenvectors m;:

M

= [m1 I m2 I ...

I mn ]

(10.42)

From Eq. (10.29) the homogeneous system response may be written

(10.43)

Sec. 10.4

343

The State Transition Matrix

for any nonzero va1ues of the constants a;. The rules of matrix manipulation allow this expression to be rewritten

(10.44)

where a is a column vector of length n containing the unknown constants a; and eAt is an x n diagona1 matrix containing the modal responses eJ.;t on the leading diagonal:

n

J,]

(10.45)

At time t = 0 all the diagonal elements in eA0 are unity, and so eA0 = I is the identity matrix and Eq. (10.44) becomes x(O) =Mia

(10.46)

For the case of distinct eigenvalues the modal matrix M is nonsingular and the vector a may be found by premultiplying each side of the equation by M- 1: (10.47)

specifying the values of the unknown a; in terms of the system initial conditions x(O). The complete homogeneous response of the state vector is (10.48) Comparison with Eq. (10.11) shows that the state transition matrix may be written

( 10.49)

leading to the following important result: Given a linear system of order n described by the homogeneous equation i = Ax, where the matrix A has n distinct eigenvalues, the homogeneous response of any state variable in the system from an arbitrary set of initial conditions x(O) is a linear combination of n modal components ehi', where the A.; are the eigenvalues of the matrix A.

General Solution of the Linear State Equations

344

10.4.3 A Method for Determining the State Transition

Cbap.JO

~atrix

Equation (10.49) provides the basis for determination of the state transition matrix for systems with distinct eigenvalues: 1. Substitute numerical values into the system A matrix and compute the system eigenvalues, a modal matrix M, and its inverse M- 1• A computer-based linear algebra package provides a convenient method for doing this.

2. Form the diagonal matrix diagonal.

eAt

by placing the modal components

3. The state transition matrix f!J(t) is f!J(t)

e'A;r

on the leading

= MeAtM- 1 •

Example 10.6

Determine the state transition matrix for the system discussed in Example 10.2 and find its homogeneous response to the initial conditions x 1(0) = 1 and x2(0) = 2. Solution The system is described by the matrix

A=

[-22 -31]

and in Example 10.2 the eigenvalues are shown to be >..1 = -1 and >..2 = -4 and a pair of corresponding eigenvectors are

A modal matrix M is therefore

M

and its inverse M -• is

=

[1 1] 1 -2

[2 1]

M_ 1 = ~ 3 1 -1

The matrix eAt is found by placing the modal responses e-1 and e-41 on the diagonal: (i)

The state transition matrix tl»(t) is 1] -2

[e-r0

0 ] [2 1 ] 1 -1

e-41

(ii)

Sec.10.4

345

The State Transition Matrix

The homogeneous response for the given pair of initial conditions is

or (iii)

or (iv) (v)

10.4.4 Systems with Complex Eigenvalues

The modal components of the response are defined by the eigenvalues of the matrix A as found from the roots of the characteristic equation det [AI -A] = 0 which is a polynomial of degree n in A with constant and real coefficients. Any polynomial equation with real coefficients must have roots that are real or which appear in complex therefore either real or occur as pairs of the form conjugate pairs. The eigenvalues of A Ai,i+l = u ± jw, where j = J=T. In such cases there are modal response components in the matrix eAl of the fonn e 0, the integral may be rewritten

x(r) = .,Atx(O)

+ fo' e"sKdT

= eAt x(O) + eAt

(L' e-M dT)

(10.84) BK

where the order of the matrices must be preseJVed. If A-l exists, the element-by-element integration of the matrix exponential may be found using the integration property described

Sec. 10.5

The Response of Linear Systems to the Singularity Input Functions

357

in Table 10.1: x(t)

= eAtx(O) +eAtA-l (I- e-At) BK (10.85)

= eAtx(O) +A-I (eAt- I) BK since AeAI response is

= eAt A, which may be shown directly from the series definition. The output y(t) = Cx+Du

(10.86)

= CeAtx(O) +CA-l (eAt- I) BK + DKus(t)

If A is nonsingular, that is if A does not have an eigenvalue l; response reaches a steady-state constant value Xss as t ---+> oo and

Xss

=lim x(t) =lim t-oo

t-oo

= 0, then the step

[~'x(O) +A- 1 (eAt -I)BK]

= -A- 1BK

(10.87)

because lim,_oo [eAt] = 0. The steady-state output response is

(10.88)

The steady-state response may be confinned directly from the state equations. When the steady state is reached, all derivatives of the state variables are, by definition, identically zero: O=Ax.u +BK

giving the same result, Xss =-A - 1BK.

(10.89)

358

General Solution of the Linear State Equations

Chap. 10

Example 10.11

=

The hydraulic system shown in Fig. 10.5 is driven by a constant-pressure pump. At time t 0 the tank is empty and the pump is turned on at a constant pressure of 10 N/m2 • Fmd an expression for the flow in the inlet pipe for t > 0.

c

Inertance I

Resistance Rl

I

Rl

Fluid accumulator

R2ij~u

patm

(a)

Agure 10.5: (a) A hydraulic system, and (b) its linear graph.

Solution The system is modeled as shown. Lumped fluid inertance I and resistance R 1 elements are used to account for pressure drops in the pipe. The tank is modeled as a fluid capacitance C, and the outlet valve is modeled as a linear fluid resistance. The following values are assumed: I 0.25 N-s 2 /m5 , R1 1 N-slm5 , R2 = 1 N-slm5 , and C = 5 1 m IN. The system state equations and the output equation for the inlet flow are

=

=

(i)

(ii)

With the values given,

B=

[~]

c = [0

1]

The step response with zero initial conditions is y(t)

= CA- 1 (~-I) BK (iii)

= ( CA - 1M) eA' (M- 1BK)- CA- 1BK The system eigenvalues are A- 1 weighting matrix K [10], and

=

M

=

-2.5

+ 1.323j

= [ -0.375 ~ 0.33tj

M-t = [ 1.512j -1.512j

j]

0.5 + 0.567 0.5- 0.567j

and A-2

-0.375

= -2.5 -

1.323j. The input

7

0.33lj]

A_ 1 = [-0.5 0.5

-0.125] -0.125

Sec.l0.5

359

The Response of Linear Systems to the Singularity Input Functions

The following matrices are computed:

CA -tM = [-0.3125- 0.1654j -0.3125 + 0.1654j 1 M-tBK = [-20+22.678j] -20- 22.678j

CA- BK = [-5.0] 1

and used in the solution: QI(I)

= [ -0.3125- 0.1654j -0.3125 + O.t654j] 223 1 e(-2.S+I.3 1> 0 ] [ -20 + 22.678j] [ 0 e(-2.5-t.322JJ)I -20- 22.678j + 5·0

= 5.0 + e-2-'

1

[

(iv)

-5 cos (l.323t) + 20.8 sin (1.323t)]

which is plotted in Fig. 10.6. QJ(t)

8 ~------~--------~---------r--------~

...,s

6

£

4

~

e ~

~ 2

0

~------~~------~--------_.

0.0

0.5

1.0

_________

1.5

2.0 t

Time (s)

Figure 10.6: Response of the hydraulic system to a 1O-N/m 2 step in pump pressure.

10.5.3 The Ramp Response If the input vector is a weighted series of unit ramp functions distributed among the r inputs,

that is (10.90)

The ramp response may be found without solving the full response equation Eq. (10.19) by using the integration property of linear systems described in Chap. 9, namely, that if the

360

General Solution of the Unear State Equations

Chap. 10

response to an input u(t) is y(t), the forced response to an input J~ u(t) dt is J~ y(t) dt. The ramp function t is the integral of the unit step, therefore, the forced component of the ramp response is simply the integral of the forced component of the step response:

x(t) =

~'x(O) + fo' A -I (~•- 1) BKdT 1

=eAtx(O)+A- [A-

1

(eAt -I) -It]BK

(10.91)

The output ramp response is found by substituting into the output equations: y(t) =CeAtx(O)+CA- 1 [A- 1

(eAt -I) -It]BK+DKt

(10.92)

Example 10.12 Find the ramp response of a first-order system written in the standard form

dx

T-

dt

where

T

+x = u(t)

is the system time constant.

Solution In state space form the first-order system is

.

l

l

T

T

x=--x+-u

(i)

The first-order ramp response [Eq. (10.91)] reduces to x(t) =

e"' x(O) + a-•

where in this case a= -1/T and b

[a-• (e01 - I)- t] b

= 1/T. Substitution of these values gives

= e- /r x(O) + t - T (1 - e- /r) which, if the initial condition x(O) = 0, is identical to the result given in Sec. 9.2. x(t)

(ii)

1

1

(iii)

PROBLEMS 10.1. The rate of convergence of the series defining the matrix exponential eAt depends on the elements of the A matrix and on the time t. Evaluate the series expression for the matrix exponential for the first-order system

x =ax +bu including the first four terms of the series expansion. Compare the result of using a four-term expansion with the exact value of the exponential~~ fort -0.1/a, -1/a. and -10/a s. What conclusion do you reach concerning the conditions under which a four-term expansion provides a reasonable approximation to the exact value of the exponential? 10.2. A second-order system with a damping ratio t 0.5 has the following A matrix:

=

=

Chap. 10

361

Problems

where a>n is the system undamped natural frequency. Assume Wn = 1 radls and evaluate the first four terms of the matrix exponential at times t = 0.1, 1, and 10.0 s. Compare these values with those computed using an available computer program. What conclusion do you reach regarding the accuracy of the four-term expansion in comparison with the computed results from a standard computer program?

10.3. Suppose that a linear system has a state transition matrix ~(t). (a) Show that if the state transition matrix is evaluated at time· t response at time t = nT. n = 1, 2, 3, 4, ... is X(nT)

(b) If .x, (0)

=

(~(T))n

= T, the homogeneous system

X(O)

= 2, x2 = -1, and ~(0.1) = [ 0.3

0.0

0.5] 0.4

find

i. X(O.l) iL x(0.3) (c) Show that, in general,

10.4. A system has a state transition matrix, evaluated at time t

~(1) = [0.25 0.0

=1s

0.5] 0.4

At timet = 2 s the homogeneous response is observed to be x 1 (2) = 4, x 2 (2) = -2. Find the initial conditions that existed at time t = 0.

10.5. Consider the second-order system described in Example 10.1, with the state transition matrix

and with initial conditions x 1(0)

= x2(0) = 1.0.

(a) Evaluate the state transition matrix at time t = 0.2 s. (b) Determine the system solution for the period from t = 0 to I s by successively computing the values of the state variables at increments in time of 0.2 s, and using the result of each step as the initial condition for the next step.

10.6. Consider the general solution x(t) to the linear state equations for a system

i=Ax+Bu which has a single input u(t), and has initial values of all state variables equal to zero. Derive an expression for the response to each of the inputs shown in Fig. 10.7.

362

General Solution of the Linear State Equations

Chap. tO

u(t)

1.0 ,...._

0

__,.

T

0

T

2T

Figure 10.7: Two input functions.

10.7. Derive the characteristic equation and determine the eigenvalues for the matrix

A-[ - 0 1] -(.()~

-2~a>,

10.8. Determine the system eigenvalues, the system undamped natural frequency ratio ~ for second-order systems represented by the following A matrices:

(a)

(b)

[-7 1] -12 0

[I~

(c) [ -1

4

(d)

Wn

[

~3] -1] -1

~9 ~]

10.9. Determine the stability of the systems represented by the following A matrices: (a)

(b)

[~2 ~3] [~2 !]

(c)

[~4 ~]

(d)

[~ ~2]

and damping

Chap. 10

363

Problems

10.10. Detennine the eigenvalues and a set of eigenvectors of the following matrices:

~6 ~7] (b) [ ~3 !] (c) [ ~4 ~o] (d) [~5 ~2] (a)

(e)

[

[0 I [-10 0 0 -6 -11 1

(f)

0 -6

-5

-4

IJ

:J

10.11. Figure 10.8 shows an overhead suspension system that is subjected to vibration from the support structure. The A matrix for the system is:

l

K2

V,(r)

l~ Kl

ml

l··

Rgure 10.8: An overhead mechanicaJ suspension system. (a) Use a computer to determine the system eigenvalues and eigenvectors for parameter values

B = 0, m 1 = 0.1 kg, m2 = 1 kg, and K1 = K2 = 9 N/m. Comment on the modal response components and the frequencies of vibration associated with the system.

(b) Determine the eigenvalues and eigenvectors when the damping parameter B is increased to

B = 20 N-s/m, and comment on the influence of damping on the frequencies and modes of vibration.

364

General Solution of the Linear State Equations

Chap. 10

10.12. For the following A matrices, determine the eigenvalues, a set of eigenvectors, the modal matrix M, and the state transition matrix ~(t).

(b)

A= [ ~2 A= [ ~3

(c)

A=[-~ ~o]

(a)

~3]

~4]

(d) A= [ 0

1 ]

-5 -2

10.13. Show that when a system with state transition matrix ~(t) and state vector x undergoes a p-l x where P is a square nonsingular matrix, the state transformation to a new state vector r transition matrix in the new coordinates is

=

~'(t)

= p- 14t(t)P

10.14. Determine the state transition matrix~· and find the response of the system

when x 1 (0)

= 5, x2 (0) = 15, and x3 (0) = 2.

10.15. Transform the system

to diagonal form, and find the homogeneous response to initial conditions xa (0) = 2, x2(0) = -1.

10.16. Show that for any square matrix A, det(A) is the product of the n eigenvalues of A. (Hint: Assume that the matrix A has been transformed to its Jordan diagonal form A by the transform A = M- 1AM where M is the modal matrix. What is det(A)? You might also use the identities det(A -t) = 1/ det(A) and det(AB) = det(A) · det(B).)

10.17. The matrix

A=[~ ~ ~] . -6 -11 -6 has a Jordan diagonal form

Find the transformation matrix M such that A= M- 1AM.

Chap. 10

365

References

10.18. Use the state transition formulation (Sec. 10.5) to compute the impulse, step, and ramp responses of the first-order system

x = -3x+2u y = 2x +u 10.19. Determine the state variable response of the system described by the following matrices

B=

[0 2] 2

-1

with initial conditions and inputs U(t)

= [~] =

10.20. For the rotational system described in Example 9.8, with parameter values J 1 kg-m 2 , K = 9 N-llllrad, and B 0.6 N-s/rad (a) Compute the state transition matrix for the system, based on a time increment of 0.2 s. (b) Compute and plot the solution for (i) the flywheel angular velocity S'l1 and (ii) the spring torque TK as a function of time when the motor speed increases suddenly from 0 to 50 radls and then remains constant (c) Compute and plot the response of

=

i. the flywheel speed ii. the spring torque as a function of time when the motor speed increases from 0 to 50 radls in a linear ramp over a period of 10 s and then remains constant (d) Compare the two solutions with respect to i. the maximum torque in the spring ii. the maximum flywheel angular velocity iii. the steady-state angular velocity and torque

REFERENCES [1] Schultz, D. G., and Melsa, J. L., State Functions and Unear Control Systems, McGraw-Hill, New

York, 1967. [2] Luenberger, D. G., Introduction to Dynamic Systems: Theory, Models, and Applications, John Wiley, New York, 1979. [3) Skelton, R. E., Dynamic Systems Control-Linear Systems Analysis and Synthesis, John Wiley, NewYork, 1988. [4] Chen, C.-T., Unear System Theory and Design, Holt, Rinehart and Winston, New York, 1984. [5] Reid, J. G., Unear System Fundamentals, McGraw-Hill, New York, 1983. [6] Moler, C., and Van Loan, C., "Nmeteen Dubious Ways to Compute the Exponential of a Matrix," SIAM Review, 20(4), Oct. 1978, 801-836. [7] Strang, W. G., Unear Algebra and Its Applications (2nd ed.), Academic Press, New York, 1980.

11

Solution of System Response by Numerical Simulation

11.1

INTRODUCTION Numerical simulation of the response of state and output variables using a digital computer is an important engineering tool for the analysis of both linear and nonlinear systems [1-6]. Many numerical algorithms have been developed that allow a set of first-order ordinary differential equations to be numerically integrated at a set of discrete times to define the system response. These techniques have been implemented in commercially available computer codes that run on personal computers, engineering workstations, and mainframe computers [7-11 ]. Numerical simulation methods compute the numerical values of the output variables at a set of discrete times and produce tabular or plotted output. They are useful for highorder and nonlinear systems when closed-form solutions are intractable. They also allow the system response to be determined for complex input waveforms and can use experimental data that have been digitized and stored in a computer as the system input Real-time dynamic simulations of physical systems in which the computer is programmed to drive an external object, such as an aircraft training simulator, also use similar numerical algorithms to endow the object with lifelike dynamics. Numerical methods usually consist of an algorithm, or procedure, that is applied repetitively to compute the system response at a series of discrete times, often spaced apart by a fixed interval ll.t. In such melhods the results of the previous time step are used as a set of initial conditions for the current computation. The problem therefore reduces to sequentially detennining the system response at a time t + lit, given the system state at time t and the inputs over the interval. In general, numerical methods generate approximate solutions to the system equations and inherently generate errors from both the algorithm itself and from the finite precision of the arithmetic operations in the digital computer. The choice of algorithm for a particular application is determined by the level of accuracy required and the computational costs associated with the method. There is frequently a 366

Sec. 11.2

Solution of State Equations by Numerical Integration

367

trade-off between the complexity of an algorithm and the size of the time step required to achieve a given accuracy. In this chapter two basic numerical integration techniques are introduced to illustrate the basis of numerical simulation and to demonstrate the influence of the integration time step and the order of the numerical integration technique on the simulation errors. The first method is based on direct numerical integration of the state equations, and the second is based on the state transition matrix discussed in Chap. 10. The pwpose of this chapter is to introduce the concepts of numerical simulation; it is not intended to be a complete description of available simulation methods. Commercially available simulation packages may use numerical algorithms that are more sophisticated than those described here [7-11]. 11.2

SOLUTION OF STATE EQUATIONS BY NUMERICAL INTEGRATION

11.2.1 Numerical Integration Techniques

Numerical solution of the state variable response may be considered an initial value problem, that is, given the state vector x(to) at some initial time to together with the input u(t), we seek the value of the response x(to + Llt) at a time At later. Initially the problem is started at timet = 0, using the given initial conditions x(O) to estimate the response at time t1 = ll.t. Repeated application of the numerical algorithm produces response values at a series of discrete times, x(t1), x(t2), x(t3), ... , which may be plotted or tabulated as the system response history. Various algorithms differ in the choice of time step; some automatically choose and vary the time step at each interval, while others produce responses at fixed intervals, that is, t; = t;_ 1 + ll.t, where Llt is a constant. The system state equations, summarized in nonlinear form as i

= f (x, u, t)

(11.1)

or in standard linear time-invariant form as (11.2)

i=Ax+Bu

express the time derivative of each state variable explicitly in terms of the state variables and inputs. Thus, given the current values of ~e state variables and inputs at a time t, the values of the state variables at a timet+ Llt may be determined by direct integration ofEq. (11.1) or (11.2), for example, x (t

+ .6.t) =

x(t)

l + l

+

t+lu

idt

1

= X(t)

. (11.3)

t+At

f (X, U, t) dt

1

The simulation methods described in this section use numerical algorithms to approximate the integral in Eq. (11.3). There are many numerical integration methods; all are approximate and vary in the accuracy of solution. In general the choice of a small time step leads to greater accuracy in the response at the expense of requiring more computational steps. For a given value of the time step Llt, different algorithms generate different integration errors and there

Solution of System Response by Numerical Simulation

368

Chap. II

is frequently a ttade-off in the choice of algorithm between computational complexity and the time step required to achieve a given accuracy of response [2, 3]. 11.2.2 Euler Integration of a First-Order State Equation

We first consider a first-order system with a single input of known form u(t) and an initial condition x(O) = xo. The system may be nonlinear and have a state equation

x=

(11.4)

f(x, u, t)

If it is assumed that the response x(to) and its derivatives are continuous at timet = to, the response x(to + dt) at a time dt later may be written as a Taylor series expansion:

I

I

I

2 2 dx d x dt dnx dtn x (to+ dt) = x (to) + dt + + ··· + dt to dt2 to 2! dtn to n!

+ ···

(11.5)

The infinite series provides the exact value of the function at time to + 8t in terms of the value of the function and its derivatives evaluated at t 0 • The series may be written in terms of a finite number of terms, for example, (11.6)

where 0 (dtk) is the sum of all terms of order k and above. If the series is truncated after a finite number of terms n, the Taylor series provides an approximation to the value of the function; the error is O(dtn+ 1). For example, if the series is terminated after four terms, x (to

I 8t + + 8t) ~ x (to)+ -dx dt to

2 -d x2 dt

I -dt2!

2

10

+ -ddtx3 I -8t 3! 3

3

(11.7)

10

The truncated series solution of Eq. (11. 7) is exact if the fourth and all higher-order derivatives of x(t) are identically zero; otherwise, it is an approximation to the true value. The lowest-order approximation to the Taylor series includes terms through first order in 8t in the expansion, that is, x (to+

~t) ~

dxl 8t x (to)+ -d t to

or

(11.8) x (to+ 8t) ~ x (to)+

f

(x, u, t)lto 8t

where the system state equation x = f (x, u, t) has been used to define the time derivative of x at to. Equation (11.8) is used to define the numerical algorithm known as the Euler integration method by treating the approximation as an estimate .i of the exact solution: .i (to + ~t) = x (to) + f (x, u, t) Ito

~t

(11.9)

Sec. 11.2

Solution of State Equations by Numerical Integration

369

x

The method estimates the response (to+ ~t) at a time increment ~t using a first-order approximation to the response over the interval, that is, by assuming that the derivative x is constant over the interval dx dt =

f

to~

(x, u, t)lto

t

t'. For larger time steps, for example, l::t.t > 2'l'.

Solution of System Response by Numerical Simulation

372

Chap. 11

the method results in a solution in which the computed· velocity increases at each time step. At this point the solution method is unstable. This example illustrates that the integration time step must be selected to be a fraction of the system characteristic time in order to achieve reasonable accuracy in the numerical computation.

12 10 ~-----+-------+

-

Ana1ytica1 lit= 1 sec - - - - - - t

-

.....e- lit= 2 sec

I

~

]

8

--- lit= 4 sec

------t

IS

20

6

Q)

>

4

2

s

00

10 Time(s}

Figure 11.3: Plotted result~ of the Euler integration of the automobile coast-down model with state equation = -I Ov and initia1 condition v(O} 10 mls. The results are plotted for time steps of ll.t 1, 2, 4 s.

v

=

=

Numerical integration methods are usually implemented as digital computer software. The Euler method is easy to implement, as is shown in the next example. Example 11.2

Write a short demonstration computer program that computes 100 response values of the state and output responses for the system

x=

-2x+3u y =4x +2u

(i) (ii)

when subjected to an input u = cos 3t. Solution The following program is written in the Fortran language but should be understandable to readers familiar with any high-level programming language. It is written for tutorial clarity rather than computational efficiency.

Sec. 11.2

Solution of State Equations by Numerical Integration

373

(-----------------------------------------------(-----------------------------------------------( Fetch the time-step and initial condition

( Simple Program to Demonstrate Euler Integration

READ{*,*) DELTAT READ(*,*) X T c: 0.0 C Write the initial output record U = COS(3.0*T) Y = 4.0*X + 2.0*U WRITE(*,*) T, X, Y C Iterate 100 times, printing at each step: DO 10 I = 1,100 C Determine the system input: U = COS(3.0*T) C The following is the Euler integration step: X = X + (-2.0*X + 3.0*U)*DELTAT C Evaluate the output equation: Y = 4.0*X + 2.0*U C Display the results: T = T + DELTAT WRITE{*,*) T, X, Y 10 CONTINUE END Figure 11.4 shows the output data from the above program plotted for x(O) = 1 and a time step l:l.t = 0.05 s. This program was compiled and executed so that the output data were stored in a file for subsequent plotting. 8 6

4 ~

c

2

0

Q.

rn

~

0 -2 ~

2

3

5 t

4

Time (s) Figure 11.4: Ploued results from the Fortran program for Euler integration of the first-order system described in Example 11.2. The simulation parameters are x(O) 1 and fir 0.05 s.

=

=

Solution of System Response by Numerical Simulation

374

Chap.11

11.2.3 Euler Integration Methods for a System of Order n

The Euler method of simulation. may be directly extended to the general case of n state equations represented in vector form: i

= f{x, u,t)

(11.12)

Each component state equation is an explicit expression for the derivative of a state variable, which may be substituted into a truncated Taylor series to generate an Euler integration rule: it (to + At)

= i1 (to) + /1 (i, u, t) 1,

0

lit

i2(to +At) = i2(to) + /2(i, o, t) 1,0 ll.t

. .

(11.13)

:=:

where each scalar function /; (i, u, t) is evaluated at time to. Example 11.3

Write a simple computer program to demonstrate E~ler numerical solution of the second-order system

. [-3 1] [2]

X=

0.5

-1 X+ 0 U

with arbitrary initial conditions and an exponential input u(t) = e-1 • Solution The two state equations are

i1

= -3xl +x2 +2u

(i)

which generate the Euler integration equations: i1 (to + ~t) = i1 (to) + [ -3it (to) + i2(to) + 2u(to)] ~t x2(to + ~t)

= x2Cto) + [O.Sit(to)- x2(to)] ~t

(ii)

The program below is similar to the one in Example 11.2. In this case, however, care must be taken to use the previous values of the state variables to evaluate the state derivatives and not to update the state vector until all derivatives have been computed.

Sec. 11.2

Solution of State Equations by Numerical Integration

375

c------------------------------------------------------------

c Program to Demonstrate Euler Integration C for a Second-Order System

c-----------------------------------------------------------c Fetch the time-step and initial condition READ(~,~)

READ(~,~)

T

DELTAT XlOLD,X20LD

= 0.0

T, XlOLD, X20LD C Iterate 100 times, printing at each step: DO 10 I ::: 1,100 C Compute the value of the input: WRITE(~,~)

U

= EXP{-T)

C The following is the Euler integration: X1NEW ::: XlOLD + (-3.0~X10LD + 1.0~X20LD + 2.o~u)~OELTAT X2NEW::: X20LD + ( O.S~X10LD - 1.0~X20LD)~OELTAT C Type the results: T = T + DELTAT WRITE(~,~) T, X1NEW, X2NEW X10LD ::: XlNEW X20LD = X2NEW 10 CONTINUE END This Fortran program was compiled and executed with initial conditions x 1(0) = x2(0) = 0 and a time step !::&t = 0.1 s. The tabulated output from the program is plotted in Fig. 11.5.

~

0.5

..-----------r-----r------r------,

0.3

tf--·-t---t------1

·~-------1---------r-------,

·-------t------1

I

&

~ 0.2

! ---·

.. - - · · - - · · - · · - · ·

I

0.1 ~-#----~~-----+-------r'-----~r-----~

Time (s) Figure 11.5: Plotted results from the Fortran program for Euler integration of the second-order system described in Example 1J.3 with initial conditions x 1(0) = x2 (0) 0 and lit 0. l s.

=

=

11.2.4 Higher-Order Integration Techniques

The Euler integration technique is not widely used in engineering practice because it usually requires small time steps relative to the system characteristic times to solve the state equations with acceptable accuracy. A number of higher-order integration techniques have

376

Solution of System Response by Numerical Simulation

Chap. 11

been developed to provide integration methods with increased levels of accuracy [1, 2, 6]. Among these, the family ofRunge-Kutta methods are based on the inclusion of higher-order terms in the Taylor series expansion, Eq. (11.5). The higher-order derivatives are computed by evaluating the state equations at intermediate values of time within the integration interval. The nth-order Runge-Kutta integration techniques provide an approximation that is exact up to the term in ll.tn in the Taylor series expansion of Eq. (11.5). The first-order Runge-Kutta integration formula includes terms of order ll.t and is identical to the method of Euler integration. A second-order Runge-Kutta integration algorithm includes terms in the Taylor series expansion up to ll.t 2 , and the fourth-order technique, which is described below, corresponds to a Taylor series with terms up to ll.t4 • In the fourth-order technique the value of i(to + ll.t) is given by [2]

(11.14)

where each of the K's represents a sequential evaluation off (x, u, t). The first coefficient is evaluated at time to:

Kt

=f

(x (to), u (to), to)

(11.15)

while the second and third coefficients are evaluated at to + ll.t /2:

ll.t ( to+2 ll.t) ,to+2 ll.t] K2=! x(to)+Kt2'" A

[

ll.t

(

ll.t) ,to+2 ll.t]

K 3 =f x(to)+K22'" to+2 A

[

(11.16) (11.17)

and the fourth coefficient is evaluated at to + ll.t:

K4

=f

(~ + K3 ll.t, u (to + ll.t) , to + ll.t)

(11.18)

The fourth-order technique provides a good compromise between computational complexity and significantly increased levels of accuracy in comparison to the first-order technique. It is generally accepted that there is little to be gained from the use of higher-order methods beyond fourth order. The fourth-order Runge-Kutta integration method is a commonly used numerical solution technique and is representative of higher-order algorithms for the solution of ordinary differential equatio~s.

Sec. 11.2

Solution of State Equations by Numerical Integration

377

Example 11.4 Write a demonstration computer program illustrating the use of the fourth-order Runge-Kutta method to compute the response of the system described in Example 11.2. Solution The system in Example 11.2 is described by the state and output equations i = -2x+3u

(i)

y=4x+2u

(ii)

and is subjected to an input u = cos 3t. The following Fortran program uses three function subprograms, DERIV (T, X) to compute the derivative of the state variable, AINPUT (T) to compute the input, and OUTPUT (T, X) to compute the output at time t. In this way the problem studied may be modified without altering the main program.

(-----------------------------------------------------( Simple program to demonstrate Runge-Kutta integration C of a single first-order state equation.

(-----------------------------------------------------REAL DELTA, T, X, Y, AK1, AK2, AK3, AK4 REAL DERIV, OUTPUT INTEGER I C Fetch the time-step, number of steps, and initial condition READ(*,*) DELTA READ(*,*) NSTEPS READ(*,*) X T = 0.0 C Write the initial output record

Y = OUTPUT(T,X) WRITE(*,*) T, X, Y C Iterate 100 times, printing at each step: DO 10 I = 1,NSTEPS C Determine the system input: AK1 = DERIV(T,X) AK2 = DERIV(T + DELTA/2., X+ DELTA*AK1/2.) AK3 = DERIV(T + DELTA/2., X+ DELTA*AK2/2.) AK4 = DERIV(T + DELTA, X + DELTA*AK3) X= X+ DELTA*( AKl + 2.*AK2 + 2.*AK3 + AK4)/6. Y = OUTPUT(T,X) T "" T + DELTA WRITE(*,*) T, X, Y 10

CONTINUE END

Solution of System Response by Numerical Simulation

378

Chap.ll

(------------------------------------------------------

( User-supplied function to compute the state derivative C This version is specific to the system dx/dt = $-$2x + 3u

c

C Inputs: C

T X

(REAL) (REAL)

Time State variable x(t)

(-----------------------------------------------------REAL FUNffiON DERIV(T, X) REAL T, X, INPUT DERIV = -2.0*X + 3.0*INPUT(T) RETURN END

(-----------------------------------------------------( User-supplied function to evaluate the system output C This version computes the output y

4x + 2u

a

c

C Inputs: C

T X

(REAL) Time (REAL) . Current state value

(-----------------------------------------------------REAL FUNffiON OUTPUT(T ,X) REAL T,X, INPUT OUTPUT = 4.0*X + 3.0*INPUT(T) RETURN END

(-----------------------------------------------------( User-supplied function to specify the system input C This version computes the input u cos(3t) g

c·

C Inputs:

T

(REAL)

Time

(-----------------------------------------------------REAL FUNCTION INPUT(T) REAL T INPUT = COS(3.0*T) RETURN END Example 11.5 Compare the results obtained using Euler integration of the first-order system in Example 11.1 with those computed using the founh-order Runge-K.utta integration rule. Solution Example 11.1 studied the coast-down behavior of an automobile. The state equation was

-B

f(x,u,t) = - x m

1 = --x -r

(i)

Sec. 11.2

379

Solution of State Equations by Numerical Integration

with t' = m/ B = 10 s. Assume~~= 2 sand an initial condition v(O) to determine the value of x (2) using the fourth-order technique is

= v0 ; the computation (ii)

Equations ( 11.15) through ( 11.18) allow computation of the four terms:

K1 = -0.1vo K2 = -0.1 [vo + (-0.1) vo (1)] = -0.09v0 K3 = -0.1 [vo + (-0.09) vo (1)] = -0.091vo K4 = -0.1 [vo + (-0.091) vo (2)] = -0.0818vo and

v(2) =

vo + ~ [-0.1vo + 2 (-0.09vo) + 2 (-0.09lvo) + (-0.0818vo)]

= 0.8187333vo

=

The exact solution is v(2) v0 e- 0•2 = 0.8187307v0 • and the first-order Euler method in Example 11.1 gave a solution of v(2) = 0.8vo. The fourth-order solution is in agreement to the fifth decimal place and shows much greater accuracy than the Euler method solution. The process may be repeated to generate subsequent time steps. Table 11.2 compares the exact solution with the first- and founh-order solutions out to 20 s. The use of the fourthorder technique with ~~ = 2 s provides a more accurate solution than the first-order technique throughout the response. TABLE 11.2:

Comparison of Fourth-Order RungeKutta and First-Order Integration Methods•

Time

Exact Solution

Computed Runge-Kutta

Computed Euler

1.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0

10.0000 8.18730 6.70320 5.48811 4.49329 3.67879 3.01194 2.46597 2.01896 1.65298 1.35335

10.0000 8.18733 6.70324 5.48816 4.49334 3.67885 3.01199 2.46602 2.01901 1.65303 1.35339

10.0000 8.00000 6.40000 5.12000 4.09600 3.27680 2.62144 2.09715 1.67772 1.34217 1.07374

• Results of numerical integration of the state equation iJ = - 1Ov with initial condition v(O) = 10 mls and a time step lit = 2 s.

Solution of System Response by Numerical Simulation

380

Chap. 11

Like the Euler method, the Runge-Kutta method may be extended to the solution of a system of n state equations. H the n state equations are represented in vector form

i

= f(x, u,t)

(11.19)

each state equation may be solved by the Runge-Kutta method:

" (to+ At)= x; ,.. (to)+ At (Ku x;

6

+ 2K2i + 2K3; + K4;)

(11.20)

where each of the K;'s represents a sequential evaluation of the derivative/; (x, u, t). Ku = f;

[i (to), u (to), to]

(11.21)

At u ( to+ 2 At) , to+ 2 At] = f; [ x... (to)+ K1T'

(11.22)

... At ,u ( to+2 At) ,to+2 At] K 3; =/; [ x(to)+K2

(11.23)

K2i

2

K4;

= /;

(i + K3 At, u (to+ At), to+ At]

(11.24)

In the following example some practical considerations concerning the numerical simulation of a nonlinear system are presented. Example 11.6 Example 5.11 describes a torque-measuring instrument that uses a pendulum of length l and mass m suspended in bearings with a viscous damping coefficient B in a gravity field g. The input torque 1's is indicated by the shaft angle (J, which under equilibrium conditions is (J

• -1 =sm

(

- T, )

(i)

mgl

In the example, the selected state variables are the shaft angular velocity Q 1 and the shaftrestoring torque TK, giving a pair of nonlinear state equations dQJ

1

dt = J (1',- Brl1- Tg)

(ii)

!;

(iii)

dJrx

= mgl cos [ sm-1 (

)] g 1 1

where J = ml2 is the moment of inertia about the shaft. The torque sensor dynamic characteristics are important and are dependent on the choice of the damping ratio B. If B is too small, the indicator needle oscillates when the input changes suddenly, while if it is too large, the instrument respons.e is sluggish. In the absence of an analytical solution for the nonlinear model, the system is studied through numerical integration methods. The task in this example is to study the dynamic response using fourth-order Runge-Kuna numerical simulation techniques and to consider the following: 1. The selection of an appropriate time step for use with the fourth-order Runge-Kuna method 2. The difference between the response of the nonlinear model and a simplified linear pendulum model 3. The selection of an appropriate damping ratio to minimize overshoot in the response.

Sec. 11.2

Solution of State Equations by Numerical Integration

381

Solution It is convenient to rewrite the state equations directly in terms of the output variable 8 by substituting into Eq. (iii): 1

dOJ

dt = J (T, -

BOJ - mgl sin 8)

d(}

(iv) (v)

-=OJ

dt

The values l =1m, m = 1 kg, g = 9.81 mls 2 may be assumed.

1. The first step is to decide on an appropriate integration time step for the numerical simulation; this must be done on the basis of the expected characteristic times associated with the pendulum. A linearized model of the pendulum, based on a small-angle approximation sin 8 ~ 8 substituted into Eq. (iv), generates a pair of linear state equations dOJ

-

dt

1 = - (T, - BOJ - mgl8) J

d(}

(vi) (vii)

-=OJ

dt

Using methods developed in Chap. 9, it is possible to show that if the input torque Ts = 0 and the damping coefficient B = 0, the linearized pendulum oscillates sinusoidally when released from an initial displacement 80 : 21l't

8(t) = 80 cosT

(viii)

where the period T = 21r .JfTi is independent of the initial displacement. While this response is an approximation to that of the nonlinear model, it may be used to define a "characteristic time" for the system that allows an initial guess at an appropriate integration time step tit. With the values given,

T = 21r

J l ~ 2.0 s '/9.81

The integration time step should be a fraction of this period. A common method of selecting a time step is to make an initial guess based on a simplified analysis of the system, such as above, and to then run several numerical simulations, each time halving the step size until a pair of successive simulated responses converge within an acceptable tolerance. As a starting point a choice of tit = 0.4 s is made. For the purpose of this study a simple Fortran program was written and the simulation was initially run with B = 0 and T, = 0 from an initial displacement 80 = 1r /2 with OJ (0) = 0. The time step was started at 0.4 s, and the simulation run to a maximum of 10 s (25 iterations). The process was then repeated with time steps of 0.2, 0.1, 0.05, and 0.025 s. The computed responses at times t = 4.8 sand 9.6 s are listed in the accompanying table.

Solution of System Response by Numerical Simulation

382

Chap. 11

9(t) and number of steps to reach t

Tune Step At

t =Os

0.400 0.200 0.100 0.050 0.025

1.570 1.570 1.570 1.570 1.570

t = 9.6 s

t =4.8 s

1.230370 (12) 1.537368 (24) 1.548812 (48) 1.549155 (96) 1.549164 (192)

0.225661 (24) 1.444810 (48) 1.485482 (96) 1.486648 (192) 1.486672 (384)

The data show that as the time step is decreased from 0.4 s to 0.025 s the solutions converge. Depending on the application, any choice of At at any value less than 0.1 s would probably be acceptable. Other factors must be considered, such as the time resolution required, computational expense, and amount of data produced. In this case a decision was made to use a time step of 0.025 s in order to generate enough points to produce an acceptable resolution for plotting the results. · 2. To illustrate the difference between the nonlinear pendulum model in Eqs. (iv) and (v) and the simplified model in Eqs. (vi) and (vii), the undamped responses (B 0) to initial values 8o 7r/4,7r/2, and 37r/4, with 0 1 (0) = 0, are plotted in normalized form in Fig. 11.6. In addition the response of a Runge-Kutta model of the linearized pendulum is included in the figure. The nonnalized data (plotted as 8 /80 ) show that while the simple pendulum model bas a fixed period T that is independent of the amplitude

=

=

O(t)/6(0) i

•••• 8(0) = 'D'/4 ••••••. 8(0) 'D'/2

=

1.0

-

0

8(0) = 31r/4

•

3

2

4

Tune (s)

Figure 11.6: Undamped initial condition response of a nonlinear pendulum model for various angles of initial displacement compared to the response of a linearized pendulum model.

5 t

383

Solution of State Equations by Numerical Integration

Sec. 11.2

of the motion, the period of the nonlinear model is highly dependent on the ampliwde of the motion. From the nonlinear data the measured values of the period are 3.07 s, 2.37 s, and 2.08 s for initial displacements of 311' /4, 1rf2, and 1r/4 rad, respectively. For comparison, the period of the linear model is 2.00 s. 3. The choice of a value of B to ensure a satisfactory transient response to sudden changes in the input torque T, is studied by examining the response of the shaft angle 8(t) to a step input ~ = T0 u,(t), where u,(t) is the unit step function, with the pendulum initially at rest (80 = 0, 0 1 = 0). Typical response curves are plotted in Fig. 11.7 for an input step of To = 5 N-m, with values of B = I. 2, 4. 8 N-m-s. The response data show that while all the solutions converge to an equilibrium value of 8 = 0.53 rad, the responses for B = 1 and 2 N-m-s both exhibit significant overshoot. that is. they have maximum values that exceed the final value and sustain several oscillations. The solution for B = 4 N-m-s has a single peak that overshoots the final value. while the solution for B = 8 N-m-s monotonically increases toward the final value. From this simulation data a set of bearings with a value of B = 8 N-m-s would be appropriate for this instrument. It should be noted, however, that because this system is nonlinear and its response is amplitude-dependent. a value of B that produces no overshoot at one value of input amplitude may produce undesirable responses at other input levels. Thus, a number of simulations with different input levels should be made to verify that the final selection of B provides satisfactory performance over the full operating range of the instrument. 8{1)

1.0

0.8

!

0.6

~

a

e :g= c

0.4

d!

I

---r----n-·-l·--·· ' I .

0.2

0.0

0

I

2

I

I

3

4

5 t

T'une (s)

Figure 11.7: The effect of damping coefficient B on the step response of the nonlinear pendulum model for an input step in torque of amplitude 5 N-m.

384

11.3

Solution of System Response by Numerical Simulation

Chap. II

NUMERICAL SIMULATION METHODS BASED ON THE STATE TRANSmON MATRIX The forced response of a linear system may be expressed in terms of the state transition matrix, as described in Sec. 10.2. For linear systems the convolution integral, Eq. (10.19) may be used to derive algebraic equations for the numerical simulation without direct integration of the state equations. Assume that the response is to be computed at fixed intervals nT, that is, we require x(nT) and y(nT) for n 0, 1, 2, ... , N, and that the input u(nT) is known or can be computed at these discrete times. Equation (1 0. 73) may be rewritten as

=

X [(n

+ 1) T] = ~T x(nT) +

(n+l)T

L

eA[(n+J)T-T1Bu('r) dr:

(11.25)

nT

Equation (11.25) expresses the new value of the state vector x [(n + 1) T] at the end of the interval using the initial condition x (nT), provided an assumption is made about the behavior of the input function during the interval T. In this section two sets of simulation equations are derived. The first method assumes that the input vector is piecewise constant during the time step nT ~ t < (n + 1)T and that the continuous input u(t) is approximated as a series of step functions that form a "staircase" approximati()n to the continuous function. The second method assumes that the input vector is a series of straight line ramp functions joining the discrete input points u(nT) and u[(n + 1)T]. Figure 11.8 illustrates these methods of approximating the continuous input H the time step T is chosen to be small enough, both methods are capable of p~viding high-accuracy simulations. /(t)

Step

f(t)

approximation ~~lllllllil.-

Time Figure 11.8:

Time

Step and ramp approximations to a continuous system input function.

11.3.1 Step-Invariant Simulation H the input u(t) is assumed to be constant over each interval, that is, u(t) = u(nT) for nT ~ t < (n + l)T, the system step response developed in Sec. 10.5.2 may be used to compute x [(n + 1) T] using x (nT) as the initial conditions:

x [(n + 1) T] = ~T x(nT)

+ A- 1 (eAT- 1) Bu(nT)

(11.26)

Numerical Simulation Methods Based on the State Transition Matrix

Sec. 11.3

385

This process may be applied repetitively, at each time step using the results of the previous step as a new set of initial conditions to compute a new state vector. The complete output simulation requires substitution of the state and input vectors into the output equations: Xn+l

= Qxn +Pun

(11.27)

Yn+l

= Cxn+l + Don+I

(I 1.28)

p = A-I [eAT - I] B

(11.29)

where

Q=

~(T)

(11.30)

are both constant matrices that need to be computed only once and the subscript notation designates the simulation time step. Equation (11.27) is a recursive difference equation relationship that allows the response to be computed in a stepwise fashion from a given set of initial conditions x(O) and known inputs. It is recursive because each computation requires the result of the previous time step. The algorithmic structure is as follows: Given matrices A, B, C, D and time step T: Compute matrices P and Q. Given initial conditions x(O), input o(nT), and the number of steps N: Repeat for n = 0 to N - 1: Compute x[(n + l)T] and y[(n + I)T]: Xn+l = Qxn +Pun Yn+I = Cxn+l + Dun+l Plot or tabulate results. The response computed using Eqs. ( 11.27) and ( 11.28) is exact if the input is a sequence of step functions and is therefore known as a step-invariant simulation. Simulation errors occur for nonstepwise inputs but can be kept small by controlling the time step interval T. Example 11.7

Develop a set of step-invariant simulation equations with a time step T system described in Example 10.11.

= 0.1 s for the hydraulic

Solution With the system parameters given in the example the state equations in the tank pressure and inlet flow rate are

[ :~ ] = [

=!

ql = [ 0

~4 ] [ :~ ] + [ ~] 1] [

(i)

Pin (t)

:~]

(ii)

The following matrices were computed using a linear algebra computer program: cll(O l) ·

=

O.IA

e

= [ 0.8885

0.0777] -0.3106 0.6555

A-• = [-0.5 -0.125] 0.5

-0.125

Solution of System Response by Numerical Simulation

386

Chap. 11

from which

P=A-1 [AT -I]B= [0.0169] e

(iii)

0.3275

Then the simulation equations are

= [ 0.8885

Pc,.+1 ] [ qln+l

0.0777] [ Pc,. ] -0.3106 0.6555 ql,.

qln+l

= [0

+ [0.0169] p,. 0.3275

(iv)

Uln

(v)

1] [PCn+l] qln+l

11.3.2 Ramp-Invariant Simulation If the input waveform is known at discrete times n T and the assumption is made that the input is a linear function of time over each time step, then

u(t)

= u(nT) + u [ (n + 1) T] T

u (n T)

(t - nT)

nT =::: t < (n

+ 1)T

(11.31)

Equation (11.31) shows that the input can be considered as having two components: a constant term and a term that is linearly related to time t. The system response may be computed by combining the system step and ramp responses. Over a given interval n T =::: t < (n + l)T the response of the state vector is found from Eqs. (10.83) and (10.89): x [(n

+ 1) T] = ~T x(nT) +A - 1 (eAT- I) Bu(nT) A-1

+T

[A-

1

(~-I) -It]B{u[(n + l)T] -u(nT)} A-1

=eATx(nT) + T + A;l

[A-

1

(~T- I)- IT] Bu[(n + 1) T]

(11.32)

[reAT -A- 1 (~T -I)]Bu(nT)

If the following constant matrices are defined from Eq. ( 11.32), Q=~T

R= A;l [A- 1 (eAT -1) -IT]B S = A;t

[r~T -A-1 (~T- I)]B

the complete simulation, including the output response, may be written Xn+l

Yn+l

= Qx,. + Run+l +Sun = CXn+l + Dun+l

(11.33) (11.34)

This is also a recursive simulation equation, but in this case two values of the input function are required for each time step. The computational complexity is only slightly greater than

Sec. 11.3

Numerical Simulation Methods Based on the State Transition Matrix

387

the step-invariant method, requiring one extra matrix multiplication and addition at each time step. Given matrices A, B, C, D and time step T: Compute matrices Q and R and S. Given initial conditions x(O), input u(nT), and the number of steps N: Repeat for n = 0 to N - 1: Compute x[(n + l)T] and y[(n + l)T] : Xn+l = Qxn + Run+l + Sun Yn+I = Cxn+l +Dun+ I Plot or tabulate results. In this case the computed response is exact for all inputs that are ramps between the time

steps and is known as a ramp-invariant simulation. It is also exact for piecewise step input functions. In general this simulation method provides more accurate results than the stepinvariant method. Example 11.8

Repeat Example 10.9 using a ramp-invariant simulation with a simulation time step of T 0.05 s.

=

Solution As in Example 10.9, the system is described by state equations

[ :~ ] = [ qI

=!

= [0

~4 ] [ :~] + [ ~] Pin (t) l] [

:~]

(i) (ii)

With the smaller time step,

Q = o.osA = [ 0.9467

0.0441 ] -0.1764 0.8144

e

R= A;' [A- (.,AT -1) -IT]B =[~::!:] 1

S=A-I [rAT -A-1 (~T -l)]B = [0.0030] T e 0.0874 from which the simulation equations are Pen+ I [ qln+l

]

_

-

[

0.9467 0.0441 ] [ Pen ] -0.1764 0.8144 qln 0.0016]

+ [ 0.0936 qr,+l

= [0

Pinn+l

1] [PCn+l] qrn+l

+

(iii)

[0.0030] 0.0874 Pian (iv)

388

Solution of System Response by Numerical Simulation

Chap. 11

PROBLEMS 11.1. Derive a Taylor series expansion for the step response of the first-order state-equation i =ax +bu

with the initial condition x(O)

= 0. Show that this series is equivalent to the analytical solution x(t)

= _!!_a (1 - e"')

11.2. Consider the initial condition response of the first-order linear system with a state equation i

= -2x+u

with u(t) = 0 and x(O) = 2. Use Euler integration to compute the response at times t = 0 ... 0.5 s in steps of 0.1 s. 11.3. A signal generator is connected to the simple electrical circuit shown in Fig. 11.9. The circuit parameters are R = 10000, C = 0.2JJ.F. R

V,(t)

Agure 11.9: FII'St-order electrical circuit.

(a) Assume that the capacitor is initially uncharged and that the input is a voltage step of 10 V. Use the Euler integration procedure to compute the voltage across the capacitor at discrete time increments of i. 50 IJ.S and H. 100 p,s to a time of 400 IJ.S. (b) Derive the analytical solution for the step input and compare the results with the two numerical computations at a time of 400 IJ.S. 11.4. Write the simulation equations for an Euler integration-based solution of the following secondorder system

y = [0

1]

[~:]

with a time step of 0.1 s. 11.5. A technique known as extrapolation to the limit may be used to increase the accuracy of many numerical procedures with little computational overhead. In this problem we examine its application to the Euler method for solving differential equations. We start with the assertion that the error in the .Euler method is approximately proportional to the step size. Then if the error in the solution XT (n T) using a step-size T is s, that is, x(nT)

= xr(nT) + s

where x(nT) is the true solution, we can assume that if we computed the response at the same time xr!2(nT) using half the step size T /2, the error will be halved x(nT) ~ XTf2(nT)

+ £/2

Chap. 11

389

Problems

(a) Show that these two expressions can be combined to give a better estimate of i 0 (nT)

io(nT)

= 2Xr12(nT) -

ir(nT)

(b) Euler integration was used to compute the state-variable step response of the system

x = -2x+3u

=

=

at time 1 1 s, with step sizes ofT= 0.2 and 0.1 s, giving io.20) 1.38336 and i 0.1 (1) = 1.33894. Use extrapolation to the limit to compute a better value and compare the three numerical results to the analytical solution for the system step response at 1 = 1 s. 11.6. Extrapolation to the limit may be used within a recursive solution procedure. Consider the first-order system = f(x, 1). The Euler method, with time step T, gives the recursive formula

x

ir [(n + 1) T] = xr(nT)

+ Tf [xr(nT), nT]

Derive a recursive procedure for solving the first-order system using the Euler method with extrapolation to the limit in the following steps: (a) Write a single recursive expression for computing ir12CnT) in two steps. That is, write an expression for ir12(nT + T /2) and use that value to find Xrf2(nT) in a second step. (b) Combine your recursive procedures (as in the previous problem) to show

i 0 [(n + 1) T]

= i 0 (nT) + Tf [x 112 (nT + T /2), nT + T /2]

where x1 12 (n T + T /2) is estimated in a separate Euler step. Notice that this procedure uses the derivative estimated at the midpoint of the interval. (c) Write a fragment of computer code (in any language) that uses the extrapolated Euler method to solve a general first-order differential. 11.7. Other ways of numerically solving initial-value problems are based upon different fonnulas for numerical differentiation. The midpoint-method is based on the symmetrical formula

'()

xt =

x (I -liT) -x (t +lit) 2 lit

+ O(A.at2 )

Derive a recursive computational procedure for solving a first-order differential equation based on the midpoint formula. Do you see any problems in starting the procedure?

11.8. The following formula, the trapezoidal method for the system x =

x [(n + 1) T]- x (nT)

1

= 2r [/ [x(nT), nT] + f

f (x, t),

. {x [(n + 1) T], (n

+ 1) T}]

is an implicit solution method because x [(n + 1) T] (which is to be computed) appears on the righthand side. The Euler formula may be used to predict the value, that is, x [(n + 1) T] = x(nT) + Tf [x(n T), T] and substituted into the corrector formula above. The whole procedure is an elementary predictor-corrector method. Derive a recursive computational procedure for solving a first-order system using the trapezoidal ru1e with Euler prediction.

11.9. Considerthesystemx = -5x+2u withx(O) to compute x(0.2) with a step size T 0.1 s.

=

= 1 andu(t) = 1. UsetheRunge-Kuttamethod =

11.10. Consider the rotational system described in Example 9.8 with the parameter values J 1.0 kg-m 2 , K 9.0 N-m/rad, B 2.0 N-m-slrad. Use a Runge-Kutta based computer simulation package to explore the response of the system. If you do not have access to such a package, you might consider writing your own using the code in Example 11.4 as a prototype.

=

=

390

Solution of System Response by Numerical Simulation

Chap. 11

(a) Write the state equation for the system in a form appropriate for solution by numerical integration. (b) Detennine the system undamped natural frequency and select a time step equal to 0.1 of the system period T = 2rrfwn. (c) Use a computer-based numerical simulation package to detennine the response for a step input of 50 rad/s.

(d) Vary the value of B to find a response that has no overshoot in the angular velocity of the flywheel. Compare the maximum torque found in this solution with that using the nominal value of B . 11.11. Mechanical systems are often influenced strongly by friction forces. Consider the simple mecban.ical system shown in Fig. 11.1 0, where m 10 kg, and K 40 N/m. The friction force generates a force that opposes the motion of the mass. We wish to use numerical simulation to examine the effect of linear and nonlinear friction characteristics on the system response when the mass is initially displaced a distance x 0 and released with initial velocity v0 0. Using any convenient numerical integration package, or one that you have written yourself, determine the response as described in the following.

=

=

=

K m

/

Frictional contact

Figure 11.10: Mass on a friction al surface. (a) For the case of viscous friction (F1 = Bv ), with B = 4.0 N-s/m, write a set of state equations

for the system. (b) Determine the system undamped natural frequency and select an appropriate integration time step, and use a numerical simulation software package to determine the mass position response x (t) from an initial displacement of 0.2 m. (c) Modify your model and state equations to include nonlinear friction F1 = F0 sgn (u), where F0 = I N, and sgn() is the signum function,

sgn(v) =

I~

-)

v>O v =O vm)- (91 + · · · + 9n)

(12.29) (12.30)

Chap. 12

The Transfer Function

406

S(s)

S(s)

s-plane

9t(s)

9t(s)

IH(s)l = LH(s)

K-#i-

2

=ct» 1 -0 1 -02

(b)

(a)

Agure 12.7: (a) Definition of s-plane geometric relationships in polar form . . (b) Geometric evaluation of the transfer function from the pole-zero plot.

The transfer function at any value of s may therefore be determined geometrically from the pole-zero plot, except for the overall "gain" factor K. The magnitude of the transfer function is proportional to the product of the geometric distances on the s-plane from each zero to the point s divided by the product of the distances from each pole to the point The angle of the transfer function is the sum of the angles of the vectors associated with the zeros minus the sum of the angles of the vectors associated with the poles. Example 12.5

=

A second-order system has a pair of complex conjugate poles s -2 ± j3 and a single zero at the origin of the s-plane. Fmd the ttansfer function and use the pole-zero plot to evaluate the transfer function at s = 0 + j 5. Solution From the problem description,

s

H(s) = K [s- (-2 + j3)][s- (-2- j3)]

=K

(i)

s s2 +4s + 13

The pole-zero plot is shown in Fig. 12.8. From the figure the transfer function evaluated at s = 0+ j5 is IH(s)l

=K J[O- (-2)]2

5>2 Jco2

+ (5- 3)

J[O- (-2)] 2 + [5- (-3)]2

(ii)

5 =K--

4J34

and LH(s) =tan

-1(5)O -1(2)2 -1(8)2 -tan

-tan

(iii)

Sec. 12.6

Transfer Functions of Interconnected Systems

407

S(s)

Figure 12.8: The pole-zero plot for a second-order system with a zero at the origin.

12.6

TRANSFER FUNCTIONS OF INTERCONNECTED SYSTEMS

Complete systems are often described by combinations of transfer functions [3-7] of subsystems. 1\vo specific types of interconnections are described here: the cascade and parallel connections. Using these relationships, complex systems may be represented by combinations of simple transfer functions. Figure 12.9 shows two systems with transfer functions H1 (s) and H2(s) connected in cascade, that is, with the output of the first system acting as the input to the second. It is assumed that this interconnection does not "load" the first system, that is, the output variable YI (t) is not affected by the connection to the second system. Then if the first system has an input u(t) = U(s)t!', its particular solution component is Ypl (t) = Ht (s)U(s)es'

We assume u 2 (t) ponent

(12.31)

= Ypi (t) and determine the combined system particular response com(12.32)

so the overall transfer function relating Yp2(t) to u(t) is (12.33) In general, the transfer function of cascaded connections of "nonloading" systems is the product of the individual transfer functions. An assumption of the lack of any loading means

that the output variable of the first system must act as a pure source to the second system.

408

The Transfer Function

y(t)

u(t)

u(t)

Equivalent system H 1(s)H2(s)

Chap. 12

y(t)

Figure 12.9: Cascade connection of two systemS and the equivalent transfer function.

Similarly, if two systems are connected so that they are driven with a common input and the outputs summed together in a parallel connection as shown in Fig. 12.1 0, with the particular responses of the two systems to an exponential input given by Ypt (t) Ht (s)U(s)es' and Yp2(t) = H2(s)U(s)~', the output is

=

Yp(t) = Ypl (t)

+ Yp2(t) = [H1 (s) + H2(s)] U(s)es'

(12.34)

The overall input-output transfer function for the parallel combination of the two systems is therefore (12.35) and the overall system transfer function is the sum of the two component transfer functions.

u(t)

y(t)

u(t)

Equivalent system H 1(s) + H2(s)

y(t)

Figure 12.10: Parallel connection of two systems and the equivalent transfer function.

12.7

STATE SPACE-FORMULATED SYSTEMS The transfer function may be derived directly from the state space system representation for single-input single-output systems. The concept may also be extended to include a matrix representation from state space descriptions of multiple-input multiple-output systems.

12.7.1 Single-Input Single-Output Systems State Variable Response Consider an nth-order linear SISO system described by a set of n state equations and a single-output equation: i(t) = Ax(t) y(t)

+ Bu(t)

(12.36)

= Cx(t) + Du(t) =

U (s)e 51 is found using The particular solution Xp(t) for an exponential input u(t) the method of undetermined coefficients by assuming that each of the state variables x; (t)

Sec. 12.7

409

State Space-Formulated Systems

also has an exponential response similar to the input, that is, x;(t)

= X;(s)est

(12.37)

where the value of X;(s) is the complex amplitude of the response of x 1(t). A set of n transfer functions may be defined between the input and each of the state variables: R·(s) '

= X;(s)

fori= 1, ... , n

U(s)

(12.38)

The forced response may be written in vector form as (12.39)

where the vector X(s) can be found by substituting the assumed solution into the state equations: sX(s)es'

= AX(s)est + BU(s)es'

(12.40)

The vector X(s) must therefore satisfy sX(s) = AX(s)

+ BU(s)

or (sl- A)X(s)

= BU(s)

(12.41)

This vector equation is a set of n linear algebraic equations in the unknowns X; (s). Any of the classical methods of linear algebra may be used to find the response magnitudes, including substitution, elimination, Cramer's rule, and Gauss elimination. They may also be solved explicitly by premultiplying both sides by (sl- A)- 1: X(s) = (sl- A)- 1BU(s)

(12.42)

Example 12.6 A second-order linear system has a set of state equations

(t)] = [ -30 [ ~1X2(t)

1][X1 (t)] + [0]

-2

3

X2(t)

Vin(t)

Find the transfer functions between state variables x 1{t) and x 2 (t) and the input u(t).

Solution For this system,

[sl- A]= [s

-I ]

{i)

3 s+2

so the set of equations to be solved [Eq. {12.41)] is s [3

-1 ][

s +2

X 1(s) ] X2(s)

=

[

0] 3

(ii)

410

TheTnmsferFunction

Chap. 12

It is convenient to use Cramer's rule (App. A) to derive the transfer function

X;(s) U(s)

=

det [ (sl- A)] det [sl- A]

(iii)

where (sl- A)

(}(HUw>

0 Figure 14.2:

Definition of the magnitude and phase angle of the complex frequency response.

Since H(jw) is complex, we can also define its complex conjugate H(jw) as H(jw) = ffi {H(jw)}- j::J {H(jw)} = IH(jw)l e-t/J(jw)

(14.17) (14.18)

Examination of N(jw) and D(jw) in Eq. (14.12) shows that the real part of each polynomial comes from the even powers of jw, while the imaginary part comes from the odd powers. Furthermore, when N(- jw) and D(- jw) are evaluated, the reaJ parts are the same as those of N(jw) and D(jw), while the imaginary parts are negated, with the result that H(- jw)

N(- jw) - N(jw) - D(jw)

= D(- jw) = H(jw)

= IH(jw)l e-t/J(jw)

(14.19)

Sec. I 4.4

14.4

457

The Sinusoidal Frequency Response

THE SINUSOIDAL FREQUENCY RESPONSE

The steady-state response of a linear single-input single-output system to a real sinusoidal input of the form of Eq. (14. 1), that is, u(t) = A sin (wt + Y,), where A is the amplitude of the input and t/1 is an arbitrary phase angle, is found directly from the system complex frequency response function H(jw). The Euler fonnulas [Eqs. (14.6) and (14.7)] may be rearranged to allow real sinusoidal and cosinusoidal waveforms to be expressed as the sum of two complex exponentials: sin(wt) =

1 . j (e1 wr 2 1 .

cos(wt) = -(e1w' 2

-

.

e-Jw')

.

+ e-1w1 )

(14.20)

(14.21)

and so the system input may be written u(t) =A sin (wt

+ t/1)

= ~ (eiJ _ 2

=A

IH(jw)l sin(wt + t/1

e-j(wt+T/f+cfJ(jw)))

(14.24)

+ t/J(jw))

The steady-state sinusoidal response is a sinusoidal function of the same angular frequency was the input but modified in its amplitude by the factor IH U w) I and shifted in phase

458

Sinusoidal Frequency Response of Linear Systems

Chap. 14

by the quantity t/J (j w). Thus, in general, the steady-state response of a linear SISO system to a sinusoidal input u(t) =A sin wt can be characterized in terms of the magnitude of the frequency response function IH(jw)l and the phase shift t/J(jw) = LH(jw). With knowledge of IH(jw)l and t/J(jw), the response may be determined directly from Eq. (14.24). Figure 14.3 shows the typical steady-state sinusoidal input and output of a linear system, demonstrating the modification to the amplitude and phase. The magnitude of the frequency response represents the ratio of the output amplitude to the input amplitude as a function of frequency and is known as the gain of the system. A system that responds to low-frequency inputs but attenuates high-frequency inputs is known as a low-pass system, while a system that does not respond to low frequencies but responds to high frequencies is known as a high-pass system.

u(t) y(t)

Amplitude ratio:

Figure 14.3:

IHUw)l = ly(t)l

lu(t)l

Steady-state response of a linear system with a sinusoidal forcing function.

The phase angle ·q,(jw) represents the temporal shift of the response relative to the input, measured in either degrees or radians. If t/J(jw) < 0, the system is said to exhibit a phase lag at that frequency because the output waveform effectively lags behind the input. On the other hand, if t/J(jw) > 0, the system is said to exhibit a phase lead.

Example 14.1 A tidal pond, with area A = lOS m 2 , is connected to the ocean by a culvert under a causeway as shown in Fig. 14.4. The normal tidal excursions in the ocean are sinusoidal, with a period of 11.5 hand a peak-to-peak height variation of 1.5 m. Measurements at the culvert show that the flow rate in and out of the pond is Q = 0.1 m 3 /s for each centimeter of height difference !ih between the pond and the ocean. Detennine the peak-to-peak tidal excursion in the pond and the time differential between the time of high tide in the ocean and in the pond.

Sec. 14.4

459

The Sinusoidal Frequency Response

Solution The pond is modeled as a fluid capacitance C1, and the culvert as a linear fluid resistance R1 . The ocean tidal variation represents a pressure source P(t) = pgh(t) = pgh 0 sin wt + P"'r' where lt 0 is the tidal amplitude (0.75 m) and P,.r is the average pressure at the entrance to the culvert. The linear graph model in Fig. 14.4 has a transfer function H(s) between the tidal depth hp(t) in the pond to the ocean depth lt 0 (t) of (i)

The system frequency response is therefore

H(j w)

= H (s)ls=Jw = I+ jRfCfw

(ii)

and the magnitude and phase functions are I

IH (j w) I = --;::::.==;:===:;o==::;;:2

(iii)

= tan- 1 (-RfC!w)

(iv)

Jl + (RfCfw)

LH(jw)

Ocean tidal excursion

Pond area A

1.5m

t

cI . excursion Figure 14.4:

Tidal pond and linear graph.

The 11.5 h tidal period corresponds to an angular frequency of w = 27r IT = 27r /( 11.5 x 3600) = 0.000152 rad/s. From Sec. 2.5 the fluid capacitance of the pond is C1 = A/pg, and from the given measurements on the flow through the culvert R1 = pg 6.hfQ. Then pg 6.h A 2rr

R!Cfw

= -Q- pg T = 2

I OS ( 11.5

X 7f X

IQ-1 X

X

X

2rr A 6./z

QT

10-2 3600)

(v)

= 1.517 From Eqs. (iii) and (iv) the magnitude and phase of the response are IH(jw)l = 0.55 and LH(jw) = -0.987 rad. The tidal fluctuation in the pond hp(t ) is therefore h p(t)

= 0.75 x

0.55 sin (O.OOO J52t- 0.987)

(vi)

and so the peak-to-peak tida.l excursion is 2 x 0. 75 x 0.55 = 0.825 m and the time lag between high tide in the pond and in the ocean is 0.987 x 11 .5/27r = 1.8 h.

460 14.5

Sinusoidal Frequency Response of Linear Systems

Chap. 14

THE FREQUENCY RESPONSE OF FIRST- AND SECOND..QRDER SYSTEMS

14.5.1 First-Order Systems The first-order system with time constant t' and an input-output differential equation t'

dy dt

+ y = Kou(t)

(14.25)

where Ko is a constant, has a transfer function

Ko rs + 1

(14.26)

H(s)=--

The frequency response function is found by direct substitution of s

= j{J):

H(j{J)) = . Ko j{J)T: + I

(14.27)

The magnitude and phase angle functions of the frequency response are

.

IH (JW) I = rjJ(jw)

so if an input u(t)

Ko

-;=::==:::;::=

J({J)t')2

+1

= tan- 1( -wr)

(14.28) (14.29)

= sin({J)t) is applied to the input, the steady-state response is Yss(t)

=

Ko J(j{J))]

(14.30)

As the input frequency becomes small and approaches zero, the magnitude of H (j {J)) approaches Ko. As the angular frequency w increases, the value of IH(j{J))l approaches

zero, indicating that the system attenuates high-frequency sinusoidal inputs. A first-order system of this type is therefore a low-pass system. The phase response tf> (j {J)) has a lag characteristic because tf> (j w) < 0 for all frequencies. At low frequencies (w 15

0

-15

'DO

!

~

-30

~

-45

c..

-60

= &. u ~ .c

~\

!

I I i I

I

'' I

0

Figure 14.5:

I

I

II I

I I

!

-

I

I ,_.._,

I

I

I

i

!

I

--

!

6 Nonnalized frequency 4

2

I

r-

I I

-·

i

I!

I

I

~

-75

-90

I Ir

I

I

I

8

JO

WT

The magnitude and phase of the frequency response of a first-order system.

shown in Fig. 14.6, inserted between the transducer preamplifier and the oscilloscope would reduce the effects of the 60-Hz interference. The circuit values for the RC circuit must be selected to attenuate the 60-Hz noise without significantly affecting the pressure signals in the 0-3 Hz range. The specifications require that any sinusoidal signal with a frequency in the 0-3 Hz range be attenuated by no more than a factor of0.9. Solution The transfer function for the filter as shown is

H(s)

= Vout(s) = Vm(s)

1

RCs + 1

(i)

462

Chap. 14

Sinusoidal Frequency Response of Linear Systems

Oscilloscope

Mixing aank

0

0

0

Pressure sensor Amplifier

Filter

•______________ .! Figure 14.6: An electric filter inserted in a measurement system to attenuate high-frequency interference.

and the magnitude of the frequency response is (Eq. 14.28) (ii)

where r = RC is the system time constant. The filter design requires choosing an appropriate time constant r and then selecting values for R and C to meet the time constant. The response of a first-order system is a monotonically decreasing function of {l). The time constant can be selected directly from Eq. (ii) by setting the magnitude to 0.9 at the specified frequency of 3 Hz, or {l) = 61r rad/s:

which gives a value of r = 0.026 s. With this value the response to all sinusoids below 3 Hz is in the range 0.9 ~ IH (j w) I ~ 1.0. The magnitude response function is shown in Fig. 14.7. With this value for r the response of the filter to a 60-Hz noise signal, ~n sin(120Ht), is

IVout I I'Vinl

-- =

y'(0.0257

X

1201r)2 + 1

=0.093

and so the noise amplitude at the filter output is reduced to Jess than 10% of its input value. The values of R and C must then be chosen so that RC = 0.026 s. Many combinations may be selected, for example, C = 0.2 JJ,f and R = 130 kQ. Practical considerations such as the input impedance of the oscilloscope and the output impedance of the transducer dictate the final choice of component values.

Sec. 14.5

463

The Frequency Response of First- and Second-Order Systems

IHUw)l 1.0

.g 0.8

·~CIO ~

0.6

~

c:

0

i....

0.4

re

0.2

!!

0.0

20

40

60

80

100 w/2'fr

Input frequency (Hz) Rgure 14.7: The frequency response magnitude of the electric filter with a time constant of 0.026 s.

14.5.2 Second-Order Systems

Consider a second-order system with an input-output differential equation d2y

+ 2~wn

dt 2

dy

dt

2

+ WnY =

Kou(t)

(14.31)

where wn is the undamped natura) frequency,~ is the damping ratio as defined in Chap. 9, and Ko is a constant. The system transfer function is Ko

H(s) =

82

+ 2~wns + W~

(14.32)

and by substitution of s = j w the frequency response is Ko

H(jw)

= (w~- wl) + j2~WnW Ko/w; = [1- (w/wn) 2]

(14.33)

+ j (2~w/wn)

The magnitude and phase functions of the frequency response H (j w) are

Ko/w; IHUw)l = rp(jw)

=

J[!- (w/w.)2)2 + [2{(wfwnll2 tan

-1

-2~(w/wn)

1 - (wfwn)

2

(14.34)

(14.35)

464

Sinusoidal Frequency Response of Linear Systems

Chap. 14

It is convenient to plot the magnitude response in a normalized form by dividing by the factor Ko/w~ and to define a normalized frequency scale wfwn. Figure 14.8 shows the normalized magnitude and phase plots for this second-order system with the damping ratio ~ as a parameter. At low frequencies the normalized magnitude response is approximately unity, and for values of~ > 1 (that is, when the system is overdamped) the gain function monotonically decreases with frequency. For lightly damped systems(~ n

T

+ l)w~

(iv)

= ,.fl[Jm, ~ = 0.5 BI .../i(i;i, and t' = B I K. For the system parameters given, Wn

and

(iii)

=

[2SX104

y --wr- = 50 radls

= 0.01 s as determined previously.

_

~-

0.5

X

.../25 X

25

102 X 102

X

}()4

_ Q

- .25

Sec. 14.7

Frequency Response and the Pole-Zero Plot

483

20 log 101Fs(ioo)/Fu(ioo)l

ai'

s

.e~

10

0 ~--------~=

fe -to ~~-20 ~

~

g-30 CT

~~ ~_._.~~~--._~~~~_._.~~LU 1

2

s

10 20 50 100 200 Angular frequency (radls)

500 1000 (I)

(c)

(b)

figure 14.19: A model of the vertical motion of the pump system. (a) Lumped-parameter model, (b) linear graph, and (c) plot of the system frequency response magnitude of the forces transmitted to the floor using experimentally detennined parameters.

The magnitude of the frequency response of the force transmitted to the floor is sketched in Fig. 14.19c as a composite of straight lines for the first-order numerator term and the secondorder denominator term. The resultant low-frequency response asymptotically approaches a gain of unity. At a frequency of w = 50 radls there is a resonant peak value of approximately 7 dB (or a factor of 2.24) in the response, and for high frequencies the response decreases. The transfer function zero has an effect at frequencies above 100 rad/s. At the normal operating speed of w = 2H(900j60) = 94.2 radls the frequency response magnitude is IH U94.2) I ~ 0.5, indicating that the shock mounts reduce the force transmitted to the floor by one-half under normal operating conditions. However, when the pump is started from rest after maintenance, it must be brought up to its operating speed slowly through its resonant speed of w = 50 rad/s (477 rpm). At this speed the vibratory force transmitted to the floor is 2.24 times that which would exist if the motor were mounted rigidly to the floor. The shock mounts have a beneficial effect in reducing the transmitted forces in normal operation but could potentially cause structural problems during start-up or slow-down.

14.7

FREQUENCY RESPONSE AND THE POLE-ZERO PLOT

The frequency response may be written in terms of the system poles and zeros by substituting directly into the factored form of the transfer function given in Eq. (12.11): HUw)

=K

Uw- Zt)Uw- Z2) · · · (jw- Zm-tHiw- Zm) Uw- pt)Uw- P2) · · · (jw- Pn-tHiw- Pn)

(14.76)

Because the frequency response is the transfer function evaluated on the imaginary axis of the s-plane, that is, when s = j w, the graphical method for evaluating the transfer function described in Sec. 12.6 may be applied directly to the frequency response. Each of the vectors

484

Sinusoidal Frequency Response of Linear Systems

from the n system poles to a test point s

liw- p;l

Chap. 14

= j w has a magnitude and an angle:

= Jul + (w- w;)2

(14.77)

((I)--u;(l)•)

L(s- p;) =tan- 1 - - '

(14.78)

as shown in Fig. 14.20a, with similar expressions for the vectors from them zeros. The magnitude and phase angle of the complete frequency response may then be written in tenns of the magnitudes and angles of these component vectors:

IHUw)l

= K nr:::tiUw- Z;)l ni=t

LHUw)

(14.79)

IUw- p;)l

m

n

1=1

i=l

= L LUw- Z;)- L LUw- p;)

(14.80)

As defined in Sec. 12.6, if the vector from the pole p; to the points = jw has length q; and an angle 8; from the horizontal, and the vector from the zero z; to the point j w has a length r; and an angle t/J;, as shown in Fig. I4.20b, the value of the frequency response at the point j w is

IHUw)l = K' 1 •• ·Tm

(14.81)

q1 • · ·qn

LHUw) = (t/Jt

+ · ·· + t/Jm)- (8t + ·· · + 8n)

(14.82)

jw

s-plane

s-plane

a

(a)

(b)

Figure 14.20: Definition of the vector quantities used in defining the frequency response function from the pole-zero plot. In (a) the vector from a pole (or zero) is defined. and in (b) the vectors from all poles and zeros in a typical system are shown.

The graphical method can be very useful in deriving a qualitative picture of a system frequency response. For example, consider the sinusoidal response of a first-order system with a pole on the real axis at s -1/T as shown in Fig. 14.21a and its Bode plots in Fig. 14.21 b. Even though the gain constant K cannot be determined from the pole-zero plot,

=

Sec. 14.7

Frequency Response and the Pole-Zero Plot

485

the following observations may be made directly by noting the behavior of the magnitude and angle of the vector from the pole to the imaginary axis as the input frequency is varied: 1. At low frequencies the gain approaches a finite value and the phase angle has a small but finite lag. 2. As the input frequency is increased, the gain decreases (because the length of the vector increases) and the phase lag also increases (the angle of the vector becomes larger). 3. At very-high-input frequencies the gain approaches zero and the phase angle approaches 1t /2. jw

IH(iw)l

K Klql Klq2

LH(iw) -----~~--._~~---.a

0~~----------------------------------------~-w

-90 (a)

(b)

Figure 14.21: (a) The pole-zero plot of a first-order system, and (b) its frequency response functions.

As a second example consider a second-order system with the damping ratio chosen so that the pair of complex conjugate poles are located close to the imaginary axis as shown in Fig. 14.22a. In this case there are a pair of vectors connecting the two poles to the imaginary axis, and the following conclusions may be drawn by noting how the lengths and angles of the vectors change as the test frequency moves up the imaginary axis: 1. At low frequencies there is a finite (but undetermined) gain and a small but finite phase lag associated with the system. 2. As the input frequency is increased and the test point on the imaginary axis approaches the pole, one of the vectors (associated with the pole in the second quadrant) decreases in length and at some point reaches a minimum. There is an increase in the value of the magnitude function over a range of frequencies close to the pole. 3. At very high frequencies, the lengths of both vectors tend to infinity and the magnitude of the frequency response tends to zero, while the phase approaches an angle of 1r radians because the angle of each vector approaches 1r /2.

486

Sinusoidal Frequency Response of Linear Systems

Chap. 14

IHUw)l

jw

OL-J,......---J_ __:::::::::~:...---....

LHUw) 0

t-.lli::::::::::=---1----------~w

-180 (a)

(b)

Figure 14.22: (a) The pole-zero plot for a second-order system. and (b) its frequency response functions.

The following generalizations may be made about the sinusoidal frequency response of a linear system, based upon the geometric interpretation of the pole-zero plot 1. If a system has an excess of poles over the number of zeros, the magnitude of the frequency response tends to zero as the frequency becomes large. Similarly, if a system has an excess of zeros, the gain increases without bound as the frequency of the input increases. This cannot happen in physical energetic systems because it implies an infinite power gain through the system. 2. If a system has a pair of complex conjugate poles close to the imaginary axis, the magnitude of the frequency response has a "peak" or resonance at frequencies in the proximity of the pole. If the pole pair lies directly upon the imaginary axis, the system exhibits an infinite gain at that frequency. 3. If a system has a pair of complex conjugate zeros close to the imaginary axis, the frequency response has a "dip" or "notch" in its magnitude function at frequencies in the vicinity of the zero. Should the pair of zeros lie directly upon the imaginary axis, the response is identically zero at the frequency of the zero and the system does not respond at all to sinusoidal excitation at that frequency. 4. A pole at the origin of the s-plane (corresponding to a pure integration term in the transfer function) implies an infinite gain at zero frequency. 5. Similarly, a zero at the origin of the s-plane (corresponding to a pure differentiation) implies a zero gain for the system at zero frequency.

14.7.1 A Simple Method for Constructing the Magnitude Bode Plot Directly from the Pole-Zero Plot The pole-zero plot of a system contains sufficient infonnation to define the frequency response except for an arbitrary gain constant It is often sufficient to know the shape of the

487

Frequency Response and the Pole-Zero Plot

Sec. 14.7

magnitude Bode plot without knowing the absolute gain. The method described here allows the magnitude plot to be sketched by inspection without drawing the individual component curves. The method is based on the fact that the overall magnitude curve undergoes a change in slope at each break frequency. The first step is to identify the break frequencies, either by factoring the transfer function or directly from the pole-zero plot Consider a typical pole-zero plot of a linear system as shown in Fig. 14.23a. The break frequencies for the four first- and second-order blocks are all at a frequency equal. t9 the radial distance of the poles or zeros from the origin of the s-plane, that is, Wb = J u 2 + w 2 • Therefore, all break frequencies may be found by taking a compass and drawing an arc from each pole or zero to the positive imaginary axis. These break frequencies may be transferred directly to the logarithmic frequency axis of the Bode plot. I

20 log 10IHUw)l

jw ::

30~~---r--~--~~~-r--~~

, .. ,

...,.----

5 rad/s

20

;'

,, / I

-to

s-plane

~

1.414 rad/s ';;;' 0 ~-+---+----+----f---+~--+----+----i

I

·a

I

f

0_10

I I I I

-20

-5

CT

(a)

-30 0.01 0.03 0.1

(J)

0.3

3

10

30

100

Angular frequency (rad/s) (b)

Figure 14.23: Construction of the magnitude Bode plot from the pole-zero diagram. (a) A typical third-order system and the definition of the break frequencies, and (b) the Bode plot based on changes in slope at the break frequencies.

Because all low-frequency asymptotes are horizontallines with a gain of 0 dB, a pole or zero does not contribute to the magnitude Bode plot below its break frequency. Each pole or zero contributes a change in the slope of the asymptotic plot of ±20 dB/decade above its break frequency. A complex conjugate pole or zero pair defines two coincident breaks of ±20 dB/decade (one from each member of the pair), giving a total change in the slope of ±40 dB/decade. Therefore, at any frequency w, the slope of the asymptotic magnitude function depends only on the number of break points at frequencies less than w, or to the left on the Bode plot. If there are Z break points due to zeros to the left and. P break points due to poles, the slope of the curve at that frequency is 20(Z - P) decibels/decade. Any poles or zeros at the origin cannot be plotted on the Bode plot because they are effectively to the left of all finite break frequencies. However, they define the initial slope. If an arbitrary starting frequency and an assumed gain (for example, 0 dB) at that frequency are chosen, the shape of the magnitude plot may be easily constructed by noting the initial slope and constructing the curve from straight-line segments that change in slope by units of ±20 dB/decade at the break points. The arbitrary choice of the reference gain results in a vertical displacement of the curve.

1/'

488

Sinusoidal Frequency Response of Linear Systems

Chap. 14

Figure 14.23b is the straight-line magnitude plot for the system shown in Fig. 14.23a constructed using this method. A frequency range of0.01 to 100 rad/s was selected, and a gain of 0 dB at 0.01 radls was assigned as the reference level. The break frequencies at 0, 0.1, 1.414, and 5 radls were transferred to the frequency axis from the pole-zero plot The value of N at any frequency is Z - P, where Z is the number of zeros to the left and P is the number of poles to the left. The curve was drawn simply by assigning the value of the slope in each of the frequency intervals and drawing connected lines. PROBLEMS 14.1. The phase and amplitude relationships between two sinusoidal variables in a dynamic system are important in fields ranging from sound and video signal processing to elements of automated machinery. Many simple dynamic systems with a single energy storage element have a transfer function: H(s)

= Y(s) = ___..!__ U(s}

TS

+1

=

=

Consider a system with K = 2 and T 0.1 sand with an input u(t) 5 sin(wr). (a) Derive the frequency response function HUw>. and express it in terms of the magnitude and phase functions. (b) Determine the steady-state sinusoidal response for input frequencies i. w = 5 radls il. (I) = 10 rad/s iii. (I) 40 rad/s For each case sketch both u(t) and y(t) on the same graph. (c) Comment on the influence of increasing frequency on the amplitude and phase of y(t) with respect to u(t). 14.2. In the processing of signals, it is often desired to amplify or attenuate signals in a given frequency range. Two electrical circuits illustrated in Fig. 14.24 are identified as either low-pass (transmitting low-frequency signals while attenuating high frequencies), or high-pass (transmitting high frequency signals and attenuating low frequencies).

=

c

R

"/1M

0 u(t)

CI I

0

(a)

0

y(t)

0

I

u(r)

R

y(l)

0 (a)

Figure 14.24: Low- and high-pass circuits.

(a) Derive the transfer functions between the input and output voltages for each circuit. (b) Sketch the amplitude and phase of the frequency response functions for the circuits when R 10,000 0, and C 1 JLF. (c) Identify which circuit is low-pass and which is high-pass. 14.3. Viscoelastic materials are often characterized by measuring their frequency responses. Models for different classes of materials have been developed. 1\vo commonly considered models are depicted in Fig. 14.25.

=

=

Chap. 14

489

Problems

8

(a) Kelvin model Figure 14.25:

(b) Maxwell model

Common viscoelastic material models.

(a) Derive the transfer function between the force and the velocity for each model.

(b) Determine the frequency responses for the two models for the case K N-s/m.

= I00 N/m, and B = 20

(c) Comment on the characteristics of the frequency responses in terms of how materials represented by the models behave at high and low frequencies. 14.4. The solar pond shown in Fig. 14.26 is used to store energy in the form of hot water. Assume that the system input is Q,(r) , the solar heat input, and that the output is the pond temperature T . Measurements and analysis have shown that it is reasonable to model the pond as a first-order system with the transfer function: H(s)

T(s) Q,(s)

= -- =

__1~ /..:...(m_c.r..P..:...)_ s (hA / (mcp))

+

where m is the mass of the water, cp is the specific heat, A is the surface area of the pond, and h is the heat transfer coefficient

Water: mass m , specific heat cP Figure 14.26:

Solar pond.

(a) Use the transfer function to write a differential equation relating the pond temperature to the solar input (b) If the solar heat flow is a constant, that is, Q,(r) = Q0 , find the steady-state temperature that the pond will reach in terms of system parameters.

490

Sinusoidal Frequency Response of Linear Systems

Chap. 14

(c) Determine the magnitude and phase of the system frequency response. (d) Now assume that the solar flux undergoes annual seasonal variations about a mean value that is Q 6 (1) Q 0 sin((l)l- rr/2) + Qaug• where (I) 2rr/365 radlday and the phase shift of -rr /2 ensures maximum solar flux in the summer. What is the annual fluctuation, maximum to minim~ of the pond temperature? (e) What day of the year does the pond reach its maximum temperature? 14.5. A simple seismometer is shown in Fig. 14.27. As the base moves vertically, the relative movement between the mass m and the drum is recorded by a moving pen. Assume that earthquakes may be modeled as a sinusoidal vertical motion u(t) A sin((l)t). The resulting displacement of the mass in the inertial reference frame is x(t), and the pen indicates the instantaneous difference

=

=

=

y(t) = x(t) - u(t).

y(t)

x(t)

Inenial reference Figure 14.27: A simple seismometer.

(a) Find the transfer function relating y(t) to u (t). Note that displacement is the integral of velocity. (b) Determine expressions for the magnitude and phase of the frequency response of the system. (c) Show that for frequencies (I)>> (1),, where (1), = .jl{"/m, the device may be used for measuring the amplitude of the base displacement u(t), while at frequencies (I) ot

+ tPn)

then fn(t

+ t') =.A, sin [nwa(t + t') + tf>n) =.A, sin (nwot + tf>n + nwor)

The additional phase shift nwor, caused by the time shift t', is directly proportional to the frequency of the component nwo. S. Interpretation of the zero-frequency term: The coefficients Fo in the complex series and ao in the real series are somewhat different from all the other tenns because they correspond to a harmonic component with zero frequency. The complex analysis equation shows that 1

Fo =T

lr,+T f(t)dt IJ

and the real analysis equation gives 1 =] -ao 2 T

l''+T

f(t) dt

11

which are both simply the average values of the function over one complete period. If a function f(t) is modified by adding a constant value to it, the only change in its series representation is in the coefficient of the zero-frequency tenn, either Fo

orao.

. Frequency Domain Methods

512

Chap. 15

Example 15.5

Find a Fourier series representation for the periodic "sawtooth" waveform with period T, /(1)

2 = -1 T

Ill
oT

=

-T/2

+lT/2 -jnWo1-e-illflJOI dt -T/2

_j_ (e-Jmr + eimr) + 0 2mr

(ii)

j = -cos(mr) mr j(-1)"

=-mr since cos(mr)

n;l:O

= (-1 )n. The zero-frequency coefficient must be evaluated separately: 11T/2 Fo = -

T -T/2

(2 ) dt =0 -1

(iii)

T

The Fourier series is /(I)=

f: n=l

j(-l)n (einwor _ e-Jnwor) n1r

oo 2(-l)n+l

=L

n=l

sin(ner>ot) n1r

= 1r~ [sin(er>ot) - !2 sin(2wot) + !3 sin(3Wot) - !4 sin(4wot) + ·· ·]

(iv)

Fourier Analysis of Periodic Wavefonns

Sec.l5.2

513

Example 15.6

Find a Fourier series representation of the periodic function 1 J

2

5T T - - )l =

1

(ii)

J(a>RC) 2 + 1

LHUa>) = tan- 1(-a>RC)

(iii)

From Example 15.5, the input function u(t) may be represented by the Fourier series u(t)

=

00

2(-1)"+1

n=I

n1r

L

(iv)

sin(nc.oot)

At the output the series representation is y(t)

oo

2(-l)n+l

n=l oo

n1r

= LIHUn(I)Q)l

sin [nc.oot

(v)

2(-l)n+l

=L

n=l mrJ(nc.ooRC)2 + 1

+ LH(jnc.oo)] 1

sin[nc.oot+tan- (-nc.ooRC)]

As an example consider the response if the period of the input is chosen to be T that a>o = 2/ RC, then y(t)

= Loo 11... 1

2{-l)n+l

n7rJ(2n) 2 + 1

[ 2n sin RC'

+ tan-•(-2n)

= 1r RC so

]

Figure 15.8 shows the computed response found by summing the first 100 terms in the Fourier

series. Yout(t)

-

0.4

:3

0

~

0.2

:;

3::s 0 0.0

t/RC

-o.2

Figure 15.8:

Response o~ a first-order elecbic system to a sawtooth input

Sec. 15.3

517

The Response of Linear Systems to Periodic Inputs

Equations (15.24) and (15.23) show that the output component Fourier coefficients are products of the input component coefficient and the frequency response evaluated at the frequency of the hannonic component. No new frequency components are introduced into the output, but the form of the output y(t) is modified by the redistribution of the input component amplitudes and phase angles by the frequency response H (j nwo). If the system frequency response exhibits a low-pass characteristic with a cutoff frequency within the spectrum of the input u(t), the high-frequency components are attenuated in the output, with a resultant general "rounding" of any discontinuities in the input. Similarly, a system with a high-pass characteristic emphasizes any high-frequency component in the input. A system having lightly damped complex conjugate pole pairs exhibits resonance in its response at frequencies close to the undamped natural frequency of the pole pair. It is entirely possible for a periodic function to excite this resonance through one of its harmonics even though the fundamental frequency is well removed from the resonant frequency, as is shown in Example 15.8. Example 15.8 A cart, shown in Fig. 15.9a, with mass m = 1.0 kg is supported on low-friction bearings that exhibit a viscous drag B = 0.2 N-s/m and is coupled through a spring with stiffness K = 25 N/m to a velocity source with a magnitude of 10 m/s but which switches direction every 1r seconds as shown in Fig. 15.9b.

10 m/s

Vm(t)

= { -10 m/s

0 ~ t < 1r ~ t < 21r

1r

The task is to find the resulting velocity of the mass Vm(t).

K

(b)

(a)

Rgure 15.9:

(a) A second-order system with its linear graph, and (b) its input waveform "in(t).

Solution The system bas a transfer function H(s)

= -

K/m

s2

+ (B/m)s + K/m

s2

25 +0.2s +25

.

and an undamped natural frequency Wn = 5 rad/s and a damping ratio lightly damped and has a strong resonance in the vicinity of 5 rad/s.

(i)

s = 0.02. It is therefore

518

Frequency Domain Methods

Chap. 15

=

The input 0 (t) has a period ofT = 2Jr seconds and a fundamental frequency of CcJo 2n 1T = 1 rad/s. The Fourier series for the input may be written directly from Example 15.4 and contains only odd harmonics: 20

u(t)

= -;; L co

1 2n _

sin [(2n - l)CcJot]

(ii)

= 20 [sin(OJot) + ! sin(3Wot) + ! sin(5CI10t) + ···] 1r 3 s

(iii)

n=l

1

From Eq. (i) the frequency response of the system is

HU(t)>

= (25- or.~+ jO.'lCIJ

(iv)

which when evaluated at the harmonic frequencies of the input nCII() = n radians/second is

HUnOJo)

= (25 -n~+ j0.2n

(v)

The accompanying table summarizes the first five odd spectral components at the system input and output. Figure 15.10a shows the computed frequency response magnitude for the system and the relative gains and phase shifts (rad) associated with the first five tenns in the series.

n~

""

IHUn~)l

LHUnOJo)

Yn

1 3

6.366sin(t) 2.122sin(3t) 1.273 sin(5t) 0.909 sin(7t) 0.707 sin(9t)

1.041 1.561 25.00 1.039 0.446

-0.008 -o.038 -1.571 -3.083 -3.109

6.631 sin(t- 0.008) 3.313 sin(t - 0.038) 31.83sin(t- 1.571) 0.945 sin(t - 3.083) 0.315sin(t- 3.109)

5

7 9

IH(iw)l G) 30 '0 ::I

·~ 25

gp

8G) 20 :g

8.

15

r-

~

I I

I I

- I I

I I 8 - I I [ 5 -1

0

2

40

I --, n ~~t.~ 0

'0

J "\\

~

3

4

5

!

::I

=

6

7

Angular frequency (radls) (a)

--40 9

10

(I)

I

J

u

...;. I

8

J ~~~

= HUw>UUw) 1 1/T =-...;_1/T + jw a + jw

(iii)

In order to compute the time domain response through the inverse transform, it is convenient to expand YU w) in terms of its partial fractions:

YUw>

= _1_ (

1

a-r - 1 1/-r + jw

- _1- ) a + jw

(iv)

.. .

534

Frequency Domain Methods

Chap. 15

provided a¥- 1/T, and using the linearity property of the inverse transfonn, y(t)

= :F- 1{YUCd)}

= aT~ 1 [:F-I { 1/T ~ jw} - :F-I {a : jw}]

(v)

Using the results of Example 15.12 once more, 1 e-aT.L._ .,._...,..-a+jw

(vi)

the desired solution is (vii)

These input-output relationships are summarized in Fig. 15.20.

1

aT-1

0

2

3

4

5

tiT

Nonnalized time Ffgure 15.20: First-order system response to an exponential input

Before the 1960s frequency domain analysis methods were of theoretical interest, but the difficulty of numerically computing Fourier transforms limited their applicability to experimental studies. The problem lay in the fact that numerical computation of the transform of n samples of data requires n 2 complex multiplications, which took an inordinate amount of time on the existing digital computers. In the 1960s a set of computational algorithms, known as the fast Fourier transform (FFf) methods, that require only n log2 n multiplications for computing the Fourier transform of experimental data were developed [6, 7]. The computational savings are great; for example, in order to compute the transform of 1024 data points, the FFr algorithm is faster by a factor of more than 500. These computational procedures revolutionized spectral analysis and frequency domain analysis of system behavior and opened up many new analysis methods that had previously been impractical. FFT-based system analysis is now routinely done in both software and in dedicated digital signal-processing (DSP) electronic hardware. These techniques are based on

· Sec. 15.5

Fourier Transform-Based Properties of Linear Systems

535

a "discrete time" version of the continuous Fourier transforms described above and have some minor differences in definition and interpretation. 15.5.2 The Frequency Response Defined Directly from the Fourier Transform

The system frequency response function H (j w), defined in Chap. 14 for sinusoidal inputs, may be defined directly using the transform property of derivatives. Consider a linear system described by the single input single output differential equation

(15.54)

and assume that the Fourier transfonns of both the input u(t) and the output y(t) exist Then the Fourier transform of both sides of the differential equation may be found by using the derivative property (property 7 in Sec. 15.4.2):

F { ~:~} = (jw)" F(jw) to give {an(jw)n + an-1 (jw)n-l + · · · + a1 (jw) +

ao} Y(jw)

= {bm(jw)m +bm-J(jw)m-l +~ .. +bt{jw) +bo} U(jw)

(15.55)

which has reduced the original differential equation to an algebraic equation in jw. This equation may be rewritten explicitly in terms of Y (j w) and in terms of the frequency response H (j ~), Y(j ) = bm(jw)m +bm-tUw)m-l +· .. +bt(iw)+boU(" ) W an(jw)n + an-l (jw)n-l + · · · + a1 (jw) + Q() JW = H(jw)U(jw)

(15.56) (15.57)

showing again the generalized multiplicative frequency domain relationship between input and output. 15.5.3 Relationship between the Frequency Response and the Impulse Response In Example 15.10 it is shown that the Dirac delta function c5(t) has a unique property; its Fourier transform is unity for all frequencies: J='{c5(t)} = 1

536

Frequency Domain Methods

Chap. 15

The·impulse response of a system h(t) is defined to be the response to an input u(t) the output spectrum is then Y&UCJJ) = F{h(t)}, Y(j(l))

= .1"{8(t)} HU(I))

= 8(t); (15.58)

= H(j(l))

·or h(t)

= .1"~

1

(15.59)

{HUCJJ)}

In other words, the system impulse response h (t) and its frequency response H (j (I)) are a Fourier transfonn pair: h(t)

~ H(jCJJ)

(15.60)

In the same sense that H(jCJJ) completely characterizes a linear system in the frequency response, the impulse response provides a complete system characterization in the time domain. Example 15.15

An unknown electric circuit is driven by a pulse generator, and its output is connected to a recorder for subsequent analysis, as shown in Fig. 15.21. Oscilloscope

Electronic pulse generator

L. ~·

®

® ®

Unknown circuit

Figure 15.21:

Setup for estimating the frequency response of an electric circuiL

The pulse generator produces pulses of 1-ms duration and an amplitude of 10 V. When the circuit is excited by a single pulse. the output is found to be very closely approximated by a damped sinusoidal oscillation of the fonn y(t)

= 0.02e-' sin(lOt)

Estimate the frequency response of the system. Solution The input u(t) is a short, rectangular pulse, much shorter in duration than the observed duration of the system response. The impulse function 8(t) is the limiting case of a unit area rectangular pulse as its duration approaches zero. For this example we assume that the

Sec. 15.5

Fourier Transform-Based Properties of Linear Systems

537

duration of the pulse is short enough to approximate a delta functio~ and because this pulse has an area of 10 x 0.001 = 0.01 V-s, we assume u(t) = O.OlcS(t)

(i)

and therefore that the observed response is a scaled version of the system impulse response, y(t)

= O.Olh(t)

(ii)

or h(t) = 100y(t)

= 2e-'' sin(12t)

The frequency response is HUw)

= F{h(t)}

(iii)

= V: {e-st sin(12t)}

In Example 15.13 it is shown that F {e-at sin CtJQt} =

CtJo

(u

and substituting CtJo = 12, u

+ jw)2 + Cd~

= 5, gives 24

HU)

"' = U"')2 + j20w + 169

(iv)

We therefore make the substitution s = j w and conclude that our unknown electric network is a second-order system with a transfer function 24

H(s)

= s2 + 20s + 169

which has an undamped natural frequency Wn input-output differential equation is

dt; + J, +

d2

20

d

(v)

= 13 radls and a damping ratio ~ = 10113. The

169y

= 24u(t)

(vi)

15.5.4 The Convolution Property The time domain input-output convolution property for linear systems was described in Chap. 8. A system with an impulse response h(t), driven by an input u(t), responds with an output y(t) given by the convolution integral: y(t)

= h(t) * u(t) =

L:

(15.61) U(T)h(t- T)dT

538

Frequency Domain Methods

Chap. 15

or alternatively, by changing the variable of integration, y(t) =

L:

u(t- T)h(T)dT

(15.62)

In the frequency domain the input-output relationship for a linear system is multiplicative, that is, YU(I)) UU(I))H(j(l)). Because by definition

=

y(t)

k

YU(I))

h'(t) * u(t)

k

HU(I)>UU(I)>

we are lead to the conclusion that (15.63)

The computationally intensive operation of computing the convolution integral has been replaced by the operation of multiplication. This result, known as the convolution property of the Fourier transfonn, can be shown to be troe for the product of any two spectra, for example, F U (I)) and G U w):

FUw)G(jw) =

= and with the substitution t

L: L: L: L: L: [/_: L: [/ f(v)e-i"'• dv

g(T)e-i"" dT

j(v)g(T)e-J"'(•+Jtloo o

CIJO (u + jCIJ)2 + CIJ~ Cl1o s2 + CIJ~

for all u > 0.

These and other common Laplace transform pairs are summarized in Table 15.2.

(iii)

Sec. 15.6

545

The Laplace Transform TABLE 15.2:

Table of Laplace Transforms F(s) of Some Common One-Sided Functions of Time f(t)

f(t)

fort~

0

F(s)

8(t) u,(t)

s 1

$2 k! si+l

1

e-ar

s+a

k! (s +a)k+l a s(s +a)

1 + _b_e-a' - _a-e-bt

a-b

a-b

ab s(s + a)(s +b)

s s2 +oil (I)

coswt sinwt e-ar (wcoswt- a sinwt)

s2

+ w2 WS

(s

+a) 2 + w2

15.6.2 Properties of the Laplace Transform

1. Existence of the Laplace transform: The Laplace transform exists for a much broader range of functions than the Fourier transform. Provided the function /(t) has a finite number of discontinuities in the interval 0 < t < oo and that all such discontinuities are finite in magnitude, the transform converges for a > a provided there can be found a pair of numbers M and a such that

1/(t)l =::Meat for all t > 0. Like the Dirichelet conditions for the Fourier transform, this is a sufficient condition to guarantee the existence of the integral, but it is not strictly necessary. While there are functions that do not satisfy this condition, for example, e12 > Mea' for any M and a at sufficiently large values of t, the Laplace transfonn does exist for most functions of interest in the field of system dynamics.

2. Linearity of the Laplace transform: Like the Fourier transform, the Laplace transform is a linear operation. Htwo functions of time g(t) and h(t) have Laplace trans-

546

Frequency Domain Methods

Chap. IS

fonns G(s) and H(s), that is, g(t)

£ G(s)

h(t)

4

H(s)

then

£ {ag(t) + bh(t)} =a£ {g(t)} + b£ {h(t)}

(15.82)

which is easily shown by substitution into the transfonn integral. 3. Time-shifting: If F(s) = £f(t), then £{f(t + T)} = tfT F(s)

(15.83)

This property follows directly from the definition of the transform, C. (f(t + T)) =

fo"" f(t + T)e- 0

LAPLACE TRANSFORM APPLICATIONS IN LINEAR SYSTEMS

15.7.1 Solution of Linear Differential Equations

The use of the derivative property of the Laplace transform generates a direct algebraic solution method for determining the response of a system described by a linear input-output differential equation. Consider an nth-order linear system completely relaxed at time t = 0 and described by

(15.91)

Sec. 15.7

Laplace Transfonn Applications in Linear Systems

551

=

In addition assume that the input function u(t) and all its derivatives are zero at timet 0 and that any discontinuities occur at time t 0+. Under these conditions the Laplace transfonns of the derivatives of both the input and output simplify to

=

c { ~n = s"F(s) and so if the Laplace transfonn of both sides is taken, {ansn +an-JSn-l

= {bmsm

+ · · · +ats +ao} Y(s)

+ bm-tSm-l + · · · + bts + bo} U(s)

(15.92)

which has had the effect of reducing the original differential equation to an algebraic equation in the complex variable s. This equation may be rewritten to define the Laplace transform of the output: Y(s) = bmsm + bm-tSm-l + · · · + bts + bo U(s) ansn + an-tSn-l + ·· •+ atS + Cl() H(s)U(s)

=

(15.93) (15.94)

where H (s) is the same rational function of s that was defined as the system transfer function in Chap. 12. Whereas in Chap. 12 the transfer function was defined in tenns of the particular response component to an exponential input u(t) =If', the Laplace transform generalizes the definition of the transfer function to a complete input-output description of the system for any input u(t) that has a Laplace transform. The system response y(t) = r,-J {Y(s)} maybefoundbydecomposingtheexpression for Y(s) = U(s)H(s) into a sum of recognizable components using the method of partial fractions as described above and using tables of Laplace transform pairs, such as Table 15.2, to find the component time domain responses. To summarize, the Laplace transform method for determining the response of a system to an input u (t) consists of the following steps:

1. If the transfer function is not available, it may be computed by taking the Laplace transform of the differential equation and solving the resulting algebraic equation for Y(s).

2. Take the Laplace transform of the input.

3. Form the product Y(s)

= H(s)U(s).

4. Find y(t) by using the method of partial fractions to compute the inverse Laplace transform of Y (s). Example 15.23

Find the step response of the first-order linear system with a differential equation dy

T dt

+ y(t) = u(t)

552

Frequency Domain Methods

Chap. 15

Solution It is assumed that the system is at rest at time t = 0. The Laplace transform of the unit step input is (Table 15.2)

£ {U.r(t)}

= -s1

(i)

Taking the Laplace transform of both sides of the differential equation generates Y(s)

1 =- U(s) Ts+ 1

(ii)

1/T =--s(s + 1/T)

(iii)

Using the method of partial fractions (App. C), the response may be written 1 s

1

(iv)

Y(s)=---S

+ 1/T

= £ {u.r(t)} + £ {e-r/'r}

(v)

from Table 15.2, and we conclude that y(t)

=l -

e-'1"

(vi)

Example 15.24

Fmd the response of a second-order system with a transfer function 2

H(s)

to a one-sided ramp input u(t)

= s 2 + 3s+ 2

= 3t fort >

0.

Solution From Table 15.2, the Laplace transform of the input is U(s) = 3£ {t}

= s23

(i)

Taking the Laplace transform of both sides, Y(s)

= H(s)U(s) =

6

s2 (s2

+ 3s + 2) 6

- s2 (s

+ 2) (s + 1)

(ii)

The method of partial fractions is used to break the expression for Y (s) into low-order components, noting that in this case we have a repeated root in the denominator: Y(s)

1 ) - ~ (1) = _!2 (!) +3(.!.) +6(s s2 s +1 2 s +2

(iii)

9 3 = -2£ {1} + 3£ {t} + 6£ {e-'}- r. {e-21 } 2

(iv)

Sec. 15.7

553

Laplace Transfonn Applications in Linear Systems

from the Laplace transfonns in Table 15.2. The time domain response is therefore

9 3 y (t ) = -- + t

2

+ 6e_, -

3 -21 -e 2

(v)

H the system initial conditions are not zero, the full definition of the Laplace transform of the derivative of a function defined in Eq. (15.89) must be used:

{, {

~~} = snY(s)- tsn-i (~~~~;I 1=1

t=O

)

For example, consider a second-order differential equation describing a system with nonzero initial conditions y(O) and y(O), d 2y dy du a2 d 2 +at-d +aoy = bt- +bou t t dt

(15.95)

The complete Laplace transform of each term on both sides gives a2 {s 2Y(s)- sy(O)- y(O)} +a1 {sY(s)- y(O)} +aoY(s)

where as before it is assumed that at time t

= btsU(s) + boU(s)

(15.96)

= 0 all derivatives of the input u (t) are zero. Then (15.97)

where Ct = a2 [y(O) + y(O)] and co= aty(O). The Laplace transform of the output is the superposition of two tenns: one a forced response due to u(t), and the second a function of the initial conditions: Y(s) =

b1s + bo a2s + a1s + 2

ao

U(s) +

c2s + c0 2 a2s + a1s + ao

(15.98)

The time domain response also has two components: (15.99) each of which may be found using the method of partial fractions. Example 15.25

A mass m = 18 kg is suspended on a spring of stiffness K = 162 N/m. At time t = 0 the mass is released from a height y(O) = 0.1 m above its rest position. Fmd the resulting unforced motion of the mass. Solution The system has a homogeneous differential equation d2y K -+-y=O dt 2 m

(i)

554

Frequency Domain Methods and initial conditions y(O) equation is

= 0.1 and y(O) = 0. {s2 Y(s)- O.ls}

The Laplace transform of the differential

+ 9Y(s) = 0

{s2 + 9} f(s)

Chap. 15

(ii)

= O.ls

(iii)

s2: 9}

(iv)

and so y(t)

= O.l.c-J {

= 0.1 cos3t

(v)

from Table 15.2.

15.7.2 Solution of State Equations

The Laplace transform solution method may be applied directly to a set of dynamic equations expressed in state space form. Consider a linear system described by its state and output equations

= Ax(t) + Bu(t) y(t) = Cx(t) + Du(t)

i(t)

and assume initially that the system is at rest at time t the Laplace transform of both sides gives

(15.100)

= 0, so that x(O) = 0. Then taking

sX(s)

= AX(s) + BU(s)

Y(s)

= CX(s) + DU(s)

(15.101) (15.102)

The state equations may be rearranged to solve explicitly for X (s), [sl- A] X(s) = BU(s) X(s)

= [sl- A]-

(15.103) 1

BU(s)

(15.104)

and substituted into the output equation: Y(.r) = C [sl- A]- 1 BU(s)

+ DU(s)

1

(15.105)

= (C [sl- A]- B +D) U(s)

The response of the system is the inverse Laplace transform of Y (s): y(t) = c,-I { (C [sl- A]- 1 B +D) U(s)}

(15.106)

For a single-input single-output system, the Laplace domain system response can be written Y(s)

= H(s)U(s)

Sec. 15.7

Laplace Transform Applications in Linear Systems

555

H(s) = C[sl -A]- 1 B +D

(15.107)

where

is the system transfer function as described in Chap. 12. Then y(t)

= r.- 1 {Y(s)}

as before. If the initial conditions on the state variables are not zero, so the initial condition vector x(O) = XQ, the Laplace transform of the state equations must be modified to include the initial term in the Laplace transform of the derivative:

= AX(s) + BU(s) [sl- A]X(s) = BU(s) + Xo sX(s) - Xo

X(s)

(15.108)

= [sl- A]- 1 BU(s) + [sl- A]- 1 Xo

The output equation then becomes Y(s)

= {C [sl- A]- 1 B + D} U(s) + C [sl- A]- 1 Xo

(15.109)

which involves two terms, a forced component and an initial condition component. Then the time domain response is the sum of the two inverse Laplace transforms: (15.110) 15.7.3 The Convolution Property

The Laplace domain system representation has the same multiplicative input-output relationship as the Fourier transform domain. If a system input function u (t) has both a Fourier transform and a Laplace transform u(t)

k

UUw)

£.

u(t) U (s)

then we observed in Sees. 15.5.2 and 15.7.1 that a multiplicative input-output relationship between system input and output exists in both the Fourier and Laplace domains:

= UUw)HUw) Y(s) = U(s)H(s)

Y(jw)

Since in the time domain the system response is defined by the convolution of the input and the system impulse response h(t), y(t)

= h(t) * u(t)

556

Frequency Domain Methods

Chap. 15

the duality between the operations of convolution and multiplication therefore hold for the Laplace domain: £ (15.111) h(t) * u(t) H(s)U(s) As in the Fourier transfonn domain, this property holds for any pair of functions: £{/(t) * g(t)} = F(s)G(s)

(15.112)

15.7.4 The Relationship between the Transfer Function and the Impulse Response

The impulse response h(t) is defined as the system response to a Dirac delta function 8(t). Because the impulse has the property that its Laplace transform is unity, in the Laplace domain the transform of the impulse response is h(t)

= £- 1 (H(s)U(s)} = .c-1 (H(s)}

In other words, the system impulse response and the transfer function form a Laplace transform pair h(t)

~ H(s)

(15.113)

which is analogous to the Fourier transform relationship between the impulse response and the frequency response as shown in Sec. 15.5.3. Example 15.26 Fmd the impulse response of a system with a transfer function 2

H(s)

= (s + l){s + 2)

Solution The impulse response is the inverse Laplace transform of the transfer function H (s): h(t)

= r,-l {H(s)}

(i)

= r,-l { (s + l:(s + 2) }

(ii)

= r,-l { s

!

1 } - £-l { s

!

2}

= 2e-' - 2e-21

(iii)

(iv)

15.7.5 The Steady-State Response of a Linear System

The final value theorem, introduced in Sec. 15.6.2, states that if a time function has a steadystate value, then that value can be found from the limiting behavior of its Laplace transform as s tends to zero, lim f(t) =lim ,_00 s-o sF(s) .

Chap. 15

557

Problems.

This property can be applied directly to the response y(t) of a system:

lim y(t) = lim sY(s)

r-oo

s-o

(15.114)

=lim sH(s)U(s)

s-o

if y(t) does come to a steady value as time t becomes large. In particular, if the input is a unit step function Us(t), then U(s) 1/s and the steady-state response is

=

lim y(t) = lim sH(s)~ ,_00 s-o s

(15.115)

=lim H(s)

s-o

Example 15.27 Fmd the steady-state response of a system with a transfer function

H(s)

=

s +3 (s + 2)(s 2 + 3s + 5)

to a unit step input. Solution Using the final value theorem lim y(t) =lim [sH(s)U(s)]

s-o

r-oo

=lim [sH(s)!] s-o s =lim H(s)

s-o 3

= 10

(i)

(ii) (iii)

(iv)

PROBLEMS 15.1. Express each of the following functions as a complex exponential waveform. (a) /(t) = 3 sin(4wot) {b) /(t) 2e-3' COS(Cd()t -rr/4) 15.2. Amplitude modulation (AM) radio transmitters impress the audio information on a radio frequency carrier waveform sin(wct) by varying its amplitude in synchrony with the audio signal. If the audio signal to be transmitted is f(t), the transmitted radio signal r(t) is

=

r(t) = (1

+ Kf(t)) sin(wct)

where K is a constant. Assume that a radio station is transmitting the sound of a flute, which may be approximated as a sinusoid /(t) = sin(wat + t/J) (with a>a l = e-GIQ)I Determine F(jw) and f(t) when (a) f(t) is known to be an odd function of time. (b) f(t) is known to be an even function of time. 15.17. For the first-order system with the transfer function H (s) = 4I (2s+ 1), determine the response to an impulse input u(t) = 5cS(t) using the Fourier transform. Also determine the response to an exponential wavefonn u(t) = toe-' where u(t) = 0 fort < 0. 15.18. The impulse responses of two second-order systems have been determined to be (a) h(t) = tore-2S' and (b) h(t) = 4e-2t sin(20t). Determine the transfer function of each system and comment on their pole locations. 15.19. Consider the two systems H 1(s) = 2/(3s + 1) and H2 (s) = 1/(2s + 1). Refemng to Fig. 15.22, determine the frequency response function of (a) the cascade (b) the parallel connection 15.20. Derive the Laplace transform of the waveforms f(t) (a) /(t) = -5 (b) /(t) = 2 COS(W()t) (c) f(t) = -2e-41 where in all cases f(t) = 0 fort ~ 0. 15.2L Determine the Laplace transform of the function r(t) 4/(t) + 2g(t- T) where /(t) is the unit step function and g(t) is the unit ramp function. 15.22. Determine the inverse Laplace transforms of 5 (a) F(s) = s 2 + 2s

=

(b) F(s)

4s + 1

= s2 + 5s+ 6

562

Frequency Domain Methods

Chap.1S

15.23. Detennine the Laplace transfonn of the general second-order differential equation

subject to arbitrary initial conditions and input functions y(O) = co, y(O) t

= c1, and u(t) = 0 for

~0.

15.24. Determine the unit impulse response h(t) for the two systems 36

(a) H(s)

= s2 + 36

(b) H(s)

=s+1

s

15.25. Consider the system

dy 3-+y=2u dt (a) Detennine the response when the input is a unit ramp u(t)

(b) Determine the response when the input u(t)

= t and y(O) = 0.

= 0 and the initial condition is y(O) = 2.

(c) Derive the unit step response from the ramp response found in part (a). 15.26. Detennine the response of the system H(s) = 4/(2s + 1) with zero initial conditions to the input shown in Fig. 15.29. Show that the steady-state solution is correct by comparing it to the value

computed from the final value theorem. /(t)

2

0

2

4

6

8 t(s)

Figure 15.29: Fmite rise-time system input

15.27. For the system described by

use the Laplace transform to detennine the solution to the initial conditions y(O) 15.28. Determine the unit step response of the system

[~:] = [ ~2 ~3] [::] + [~]u y = [ 1 0] with zero initial conditions.

[~:]

= 0 and y(O) = 1.

Chap. 15

563

References

REFERENCES [1] Grauon-Guiness, I., Joseph Fourier 1768-1830, MIT Press, Cambridge, MA, 1972. [2] Bracewell, R.N., The Fourier Transform and Its Applications (2nd ed.), McGraw-Hill, New York, 1978. [3] Papoulis, A., The Fourier Integral and Its Applications, McGraw-Hill, New York, 1962. [4] Dy~ H., and McKean, H. P., Fourier Series and Integrals, Academic Press, New York. 1972. [5] Oppenheim, A. V., Willsky, A. S., and Young, I. T., Signals and Systems. Prentice Hall, Englewood Cliffs, NJ, 1983. (6] Brigham, E. 0., The Fast Fourier Transform, Prentice Hall, Englewood Cliffs, NJ, 1974. [7] Oppenheim, A. V., and Schafer, R. W., Digital Signal Processing, Prentice Hall, Englewood Cliffs, NJ,l975. [8] Rainville, E. D•• The laplace Transform: An Introduction, Macmillan, New York. 1963. [9] Holbrook. J. G.• Laplace Transforms for Engineers, Pergamon Press, Oxford, 1966. [ 10] McCollum, P. A., and Brown, B. F., Laplace Transform Theorems and Tables, Holt Rinehart and Winston, New York, 1965.

A

Introduction to Matrix Algebra

Modem system dynamics is based upon a matrix representation of the dynamic equations governing the system behavior. A basic understanding of elementary matrix algebra is essential. for the analysis of state space fonnulated systems. A full discussion of linear algebra is beyond the scope of this book, and only a brief summary is presented here.

A.1

DEFINmON A matrix is a two-dimensional array of numbers or expressions arranged in a set of rows and columns. An m x n matrix A has m rows and n columns and is written

[au A= a21

a22

... ...

aml

am2

...

a12

.

at.] a2n

(A.l)

amn

where the element a;1 , located in the ith row and jth column, is a scalar quantity: a numerical constant or a single-valued expression. If m = n, that is, there are the same number of rows as columns, then the matrix is square; othelWise, it is a rectangular matrix. A matrix having either a single row (m 1) or a single column (n 1) is defined as a vector because it is often used to define the coordinates of a point in a multidimensional space. (In this book the convention has been adopted of representing a vector by a lowercase boldface letter such as x, and a general matrix by a boldface uppercase letter such as A.) A vector having a single row, for example,

=

X

564

= (Xtt

XJ2

=

•••

Xtn ]

(A.2)

Sec. A.2

Elementary Matrix Arithmetic

565

is defined as a row vector, while a vector having a single column is defined as a column vector:

Y=

[~: Yml

l

(A.3)

Two special matrices are the square identity matrix I, which has zeros in all its elements except those on the leading diagonal (where i = j) which have the value of unity:

1 0 ... 0] I= [. . . . 0 1 ... 0 . . . . •

41

•

0 0 . ..

(A.4)

•

1

and the null matrix N, which has the value of zero in all its elements:

(A.5)

It is common to use the symbol 0 to represent a null matrix or vector.

~.2

ELEMENTARY MATRIX ARITHMETIC Matrix Addition The operation of addition of two matrices is defined only when both matrices have the same dimensions. H A and B are both m x n, then the sum C=A+B

(A.6)

is also m x n and is defined to have each element the sum of the corresponding elements of A and B; thus, (A.7)

Matrix addition is both associative, that is,

A+ {B +C)= (A+B) +C

(A. B)

A+B=B+A

(A.9)

and commutative,

Introduction to Matrix Algebra

566

App.A

The subtraction of two matrices is similarly defined; if A and B have the same dimensions, then the difference

(A.lO)

C=A-B implies that the elements of C are

(A.ll)

Multiplication of a Matrix by a Scalar Quantity If A is a matrix and k is a scalar quantity, the product B = kA is defined to be the matrix of the same dimensions as A whose elements are simply all scaled by the constant k, (A.12)

Matrix Multiplication Two matrices may be multiplied together only if they meet conditions on their dimensions that allow them to confonn. Let A have dimensions m x n and B be n x p (that is, A has the same number of columns as the number of rows in B), then the product (A.13)

C=AB is defined to be an m x p matrix with elements n

Cij

= L aucbkj

(A.14)

k=l

The element in position ij is the sum of the products of elements in the ith row of the first matrix (A) and the corresponding elements in the jth column of the second matrix (B). Notice that the product AB is not defined unless the above condition is satisfied; that is, the number of columns in the first matrix must equal the number of rows in the second. Matrix multiplication is associative, that is, A (BC)

= (AB) C

(A.15)

but is not commutative in general,

(A.l6)

AB':/=BA

In fact, unless the two matrices are square, reversing the order in the product causes the matrices to be nonconfomtal. The order of the terms in the product is therefore important. In the product C = AB, A is said to premultiply B, while B is said to postmultiply A. It is interesting to note that unlike the scalar case, the fact that AB 0 does not imply that either A= 0 orB= 0.

=

Sec. A.3

'·3

Representing Systems of Equations in Matrix Form

567

REPRESENTING SYSTEMS OF EQUATIONS IN MATRIX FORM

Linear Algebraic Equations The rules given above for matrix arithmetic allow a set of linear algebraic equations to be written compactly in matrix form. Consider a set of n independent linear equations in the variables x; fori = 1, ... , n, a11X1

a21X1

+ +

+ +

a12X2 a22X2

+

+

alnXn

=

a2nXn

=

b1 b2

(A.17)

and write the coefficients aii in a square matrix A of dimension n,

A= [

au

a12

.. •

a21

a22

•••

...

anl

...

an2

..

.

•••

the unknowns x; in a column vector x of length N,

and the constants on the right-hand side in a column vector,

b=[il then equations may be written as the product, all

a12

a21

a22

ani

an2

(A.l8)

[

which may be written compactly as Ax=b

(A.19)

Introduction to Matrix Algebra

568

App.A

State Equatio.ns The modeling procedures described in this book generate a set of first-order linear differential equations:

bnmUm

(A.20)

H the derivative of a matrix is defined as a matrix consisting of the derivatives of the elements of the original matrix, the above equations may be written

XJ] .!!_ ~2 dt

[:

Xn

[au

= . a~1

aln a2n

...

: ani

an2

l[XI] x2

[bu ~~

... + ...

ann

Xn

bnl

:f][] (A.21)

or in the standard matrix form

i=Ax+Bu

(A.22)

where .

d

X=-X

dt

A.4

(A.23)

FUNCTIONS OF A MATRIX A.4.1 The Transpose of a Matrix The transpose of an m x n matrix A, written AT, is then x m matrix fonned by interchanging the rows and columns of A. For example, if

[ 1 40 32]

B = -1

then the transpose is

BT =

[1-1]j ~

Notice that the transpose of a row vector produces a column vector, and similarly the transpose of a column vector produces a row vector. The transpose of the product of two matrices is the reversed product of the transpose of the two individual matrices: (A.24)

Sec. A.4

569

Functions of a Matrix

The rules of matrix multiplication show that the product of a vector and its transpose is the sum of the squares of all the elements: n

= Els:::-J

4+3j I = (3s- 1)(s- j) s=-J = - -102

s +s-2

and A3 = F(s)(s - i>ls=J 2

=

I

s +s -2 (3s - l)(s + j) s::::J

4-3j

= - -10-

Notice that A2 and A3 are complex conjugates. The complete expansion is 7 1 1 4 + 3j 1 4 - 3j - s-+-s --2- - = ---+--- +-(3s- 1)(s + j)(s - j) 5 3s - 1 10 s + j 10 s - j 2

Example C.6

Find the partial fraction expansion of the rational function F(s)

=

s+2

s3 + 5s2 + 7s + 3

s+2 -

(s

In this case there is a repeated factor (s A1

F(s)

+ 1)2 (s + 3)

+ 1)2 , and so the expansion is A21

A22

= s + 3 + s + 1 + (s + 1)2

Then,

1

4

The complete expansion is I

F(s)

1

1

1

1

1

= -4 s + 3 + 4 s + 1 + 2 (s + 1)2

App.C

Index

Absolute: acceleration, 26, 72 pressure. 47 temperature, 53 Accelerometer, 490 Accumulator, 202 Across-variable, 61-71

sketching methods, 4 79-83, 486-88 Bond graphs, 66, 92 Branches of linear graph, 92 Break frequency, 473, 476 Bbl, 13

Bulk modulus, 48 ButterWorth filter, 492

source. 80

Admittance. 422-26 driving point. 422

Ampliblde decay ratio, 304 Analogy, 66 Angular:

acceleration. 34 displacement. 32 fmluency, 248 momentum. 31 Angular velocity, 30 source. 30 Aperiodic function. 519

Asymptotic stability, 259 Athans, M., 18 A-type element energy storage, 71-75 source, 80 Automobile suspension, 162, 447, 496 Back-emf', 178-80 Bathe, K. J., 394 Baumeister, T., 65 Blackwell, W. A.. 17 Block diagram, 119, 205, 209-18 Bode, H. W., 468 Bode plots, 468-83 first-order system, 471 second-order system, 471

Capacitance:

electrical, 40-41 fluid, 47-48

general, 71 thermal, 54 Cascade operators, 221 Causality, 82, 188, 205, 207, 213 differential, 212 integral, 212

Celsius, 14 Characteristic equation, 253, 279, 300, 340, 571 multiple roots, 253 Characteristic operator, 296 Characteristic polynomial, 253, 340 Characteristic response, 283 Charge, 37 Cofactor, 569 Commutativity, 222 Companion form, 235 Compatibility, 95 equations, 95 Complex numbers, 574-79 arithmetic, 575-77 conjugates, 576 polar representation, 577 Compressibility, 48 Conduction: electrical, 41 thermal, 56 587

Index

588 Connected graph, 124 Conservation of energy, 19-20 Constitutive equation, 32 Continuity, 97 equations, 97 Controllability, 132 Convection, 56 Conversion factors, 14 Convolution, 264 properties, 268 relationship to Fourier transform, 537 relationship to Laplace transform, 555 Comer frequency, 473, 476 Coulomb friction, 27 Cramer's rule, 230, 572 Crandall, S. H., 17 Critical damping, 303 Current, 37 source, 42

Damped natural frequency, 303 Damper: square law, 28 rotational, 34-35 translational, 26 viscous, 27 Damping ratio, 297 critical, 303 overdamped. 302 underdamped. 303 Decade, 470 Decibels, 469 De Moivre's theorem, 579 Dependent energy storage element, 129 Derivative causality, 212 Determinants, 569 Diagonal form of matrix, 350 Differential equations: analytical solution, 252-59 classical form, 251-52 homogeneous, 252 linear, 123 matrix form, 121-23 numerical solution, 366-87 solution of matrix representation, 331-37 Differentiation, 210 property, 263 Dirac delta function, 246 Direct current motor, 178 Discharge coefficient, 51. Displacement, 22 Disttibuted parameter model, 21 Dorf, R. C., 243 D-type elements, 78-79 Dynamic transfer operator, 207-8

Eigenvalue, 338-43, 571 complex, 345 repeated. 347 Eigenvector, 338-43, 571

Electric: generator, 169, 178 motor, 169, 178-80 Electrical: charge, 37 elements, 38-44 energy, 38 sources, 42 transformer, 172 Electromechanical transducer, 177-80 Elemental equation, 25 Energy: conservation, 19-20 electrical, 38 fluid. 45 general dissipation element, 23, 78-79 general source elements, 80-81 general storage elements, 71-72, 76-77 mechanical, 22-23, 32-33 thermal, 53 Energy storage element: dependent, 130, 145 independent, 130 Environment, 5 Euler's fonnula, 249, 579 Euler integration, 368-75 first-order system, 368 high-order system, 374 Existence theorem, 252 Exponential functions: inputs,· 248 linear system response. 395-98 properties, 250 Extrapolation to the limit, 388 Fahrenhei~

14

Farad. 40 Faraday's law, 177 Feynman, R. P., 65 Filter: Butterworth, 492 high-pass, 488 low-pass, 488 Fmal value theorem, 548 FJ.restone, F. A., 66, 91 First law of thermodynamics, 20 First-order system: equations, 277-79 standard form, 277 First-order system response: forced, 283-86 impulse, 285 ramp, 285 step, 284 frequency, 460-63 homogeneous, 279-83 Flow, 45 source, 51 Fluid: compressibility, 48 elements, 44-53

589

Index Fluid: (continued) energy, 45 power, 44 transfonner, 172 viscosity, SO Flux linkage, 37 Force, 22 source, 28 Forrester, J. W., 17 Fourier: analysis: periodic functions, 502-14 transient functions, S 19-28 coefficients, 503-9 Fourier series: analysis fonnulas, 507 properties, 509 synthesis formulas, 507 system response methods, S 14-18 Fourier's law, 56 Fourier transform, 522 convolution property, 537-38 properties, 528-31 system response methods, 531-35 Franklin, G. R, 243 Frequency: ~

473-74, 479

response, 453-59 Bode plots. 467-83 from pole-zero plots, 483-88 Friction: coulomb, 27 square-Jaw, 28 viscous, 27 Function: exponential, 248 periodic, 250 singularity, 246-48 Gear, 175-76 Generalized: capacitance, 71-75 inductance, 76-78 resistance, 78-79 source, 80

Generalized variables, 67 Graph: linear, 92-94 tree, 127-33, 188-90 Gravitational constant, 13-14 Gyrator, 180-81

Hagen-Poiseuille law, 50 Harmonics, 517-19 Hartnett. J. P., 65 Heat, 53 conduction, 56 convection. 56 flow, 20, 53 flow source, 58

radiation, 56 storage, 54 Heaviside, 0., 208 Helmholtz resonator, 496 Henry, 38

Homogeneous equation, 252, 279, 300 Horsepower, 13 Hydraulic: accumulator, 202 pump, 182 ram, 180

Ideal element, 24 Imaginary numbers, 574-78 Imaginary part of complex number, 25 I Impedance: driving-point, 422-26 element, 424-26 fonnation of transfer function, 434-35 parallel and series combinations, 426-31 two-ports, 431-34

Impulse function, 24 sifting property, 270 Impulse response: first-order system, 285, 288 general linear system, 354 second-order system, 312 Independent energy storage element, 129-30 Inductance: · general, 76 ideal, 39 nonlinear, 38 Inertance, 46 Inertia, moment of, 33 Initial conditions, 6 Input, S vector, 121-22 Input derivative systems, 145 Input-output models, 219 Integral causality, 212 Integrated across-variable, 69 Integrated through-variable, 69 Integration property, 263 International system of units (ISO), 12 Inverse Fourier transform, 523 Inverse Laplace transform, 548 Inverse operator, 223 Joule, J. P., 13, 14 Kalman, R. E., 6, 18, 168

Karnopp, D. C., 17 Kelvin model. 489 Kilogram. 12, 14 Kinetic energy, 21, 23 Kirchoff's laws: current, 98 voltage, 96 Koenig, H. E.. 17 Kulakowski, B. T., 17 Kuo, B. C., 119

Index

590 Laminar ftow, SO Laplace transform, 539-50 applied to state equations. 554 convolution property, 555 inverse. 548-50 one-sided. 541 properties, 545-48 region of convergence, 541 solution to linear differential equations, 550-56 table of transfonn pairs, 545 two-sided. 540 Linear graph, 92-112 compatibility and continuity constraints, 128-33 connected, 124 tree. 127 Linearization, 83-89, 158-61 Linear operator, 207-11 inverse. 223 properties, 211 Linear time-invariant system. (LTI), 122 properties, 263 superposition, 262 Linear transfonnation, 349 Link of graph tree, 127 Logarithmic frequency response. 467-83

Loop: method, 95-96 variables, 94 Lorentz's law, In Luenberger, D. G., 8 Lumped parameter model, 20-21 Magnetic field, 38 flux linkage. 38 Margolis, D. L., 17 Mass: ideal, 26 pure, 25 Matrix, 564 adjoint, 570 A-matrix, 122, 567 characteristic equation, 340 determinant, 569 eigenvalues, 57 I inverse, 570-7I modal, 342 operations, 565-66 properties, 564-66 state transition, 333-53 transfer operator, 237 ttansfonnation to diagonal fonn. 350 ttanspose,568 Matrix exponential, 333 Maximum power transfer theorem, 449 Maxwell model, 489 Mechanical: energy, 23, 32 power flow, 22 rotational one-port elements, 30-37

translational one-port elements, 2I-30 two-port elements, I 72, 175, 177 Melsa, J. L., 18 Millman's theorem. 449 Minor, 569 Moment of inertia, 33 Momentum, 23, 25 Motor, D. C., 178-80 Multipons. 169-72 Murphy, A. T., 17 Natural frequency: damped, 303 undamped, 297 Newton's law, 13 Node. 92 equations, 97-98 reference, 93-94 Nonhomogeneous equation, 255 Nonlinear: equations, 122, 380, 383 models. 153-58, 216 solution techniques, 380-83 Nonminimum phase system, 490 Nonnal tree, 129, 190 Norton equivalent, 110. 443 Numerical integration, 367-83 Euler, 368-75 Runge-Kutta. 376-83 state transition matrix, 384-87 ramp-invariant simulation, 386 step-invariant simulation, 384 Octave, 470 Ogata, K., 17 Ohm's law, 41 One-port element, 66 Operator: differential, 210 identity, 2 I0 integral, 210 inverse, 223 linear, 211 matrix, 228 scaling, 209 Oppenheim, A V., 563 Order of system. 131 Ordinary differential equations, 251 Oriented graph, 67-70 Orifice equation, 50 Output, 5 vector, I23 Overdamped response, 302, 316 Overshoot, 310, 316

Parallel connection. 98 Parallel operators, 222 Parseval's theorem, 530 Partial fraction expansion, 580-86 Particular integral, 256

591

Index Pascal, J., 13, 44 Paynter, H. M., 17 Period. 248 Periodic extension, 519 Periodic function, 250, 453, 502-3 Permittivity, 40 Phase, 248, 453 angle, 457-59 plane, 261

shift. 458 • Phase variable form, 235 Poles. 398-400 Pole-zero plot, 400-4 Polynomial operators, 222 Port, 19 Potential energy, 21, 23 Powell, J. D., 243 Power: electrical, 37 fluid. 44 mechanical rotational, 30 mechanical ttanslational, 22 thermal. 53 Power Oow, 19-20 Predator-prey model, 8 Pressure, 44 source, 51 Pressure-momentum. 45 Primary variables, 128 Principle of superposition, 211, 262 Pulse, 246 Pump, 171 Q of system, 495 Rack-pinion. 176-77 Radian, 248 Radiation heat transfer, 56 Ramp function, 247 Ramp-invariant simulation method, 386 Ramp response: first-order system, 285, 288 general linear system, 359 second-order system, 314 Rankine, 14 Real part of complex number, 251 Relativistic mass, 26 Repeated roots, 347 Resistance: electrical, 41 fluid, 49 generalized, 78

thermal.

ss

Resonance, 464 Resonant frequency, 464 Reynolds num~ 50 Richardson, H. H., 17 Roark. R. J., 65 Robsenow, W. M., 6S Rosenberg, R. C., 17

Runge-Kutta integration. 37~3 for first-order systems, 376 fourth-order method, 376 for higher-order systems, 380 Schultz, D. G., 18 Secondary variables, 128 Second-order system, equations, 295-300 · standard form. 308 state equation transformation, 295 Second-order system response: forced: impulse, 312 ramp, 314 step, 309 frequency, 463-66 homogeneous, 300-8 Shearer, J. L., 17 Sign conventions, 98, 181 S.l units, 12-14 Single-input single-output, (SISO), 395 Singularity functions. 246-48 Sinusoidal: functions, 248 response, 453-59 Bode plots, 467 first-order system, 460 from the pole-zero plot, 483-86 second-order system, 463 Slotine, J. J., 275 Sources: ideal, 28, 35, 42, 51, 58 models, 108-11 Norton equivalent, 110, 443 Tbevenin equivalent, 110, 441 Spatial lumping, 20 Specific heat, 54 Spectra. 505-7 continuous, 522 line, 508 Spring: ideal, 24 nonlinear, 25 rotational, 32 translational, 24 Square law damping, 28 Stability, 259, 347, 404 State: determined system, 6, 120 equation, 6, 121 fonnation of, 135 linearization of, 150 nonstandard form, 145 standard form. 124

space. 261 State transition matrix, 333-53 computation of, 344 diagonal fonn, 351 exponential form, 333 properties, 337-38

Index

592 State variable response, forced, 334 homogeneous, 332 State variables, 6, 120, 135 companion form. 235 diagonal form, 350 phase variable form, 235 transfonnations, 349 State vector, 1~22 Steady-state frequency response, 453-54 Steady-state response, 284 Stefan-Boltzman law, 56 Step function, 246 Step-invariant simulation method, 384 Step response: first-order system, 284 general linear system, 356 second-order system, 309 Strang, W. G., 365 Streeter, V. L.. 65 Summer, 209 Superposition principle, 211, 262 System: definition, 5 dynamics, J-4 graph, ] 24-33 linear, 122 linear time-invariant, 122 nonlinear, 122 System transfer operator, 207, 225

Temperature, 53 source, 58 Thermal: elements, 53-59 energy, 53 power, 53 Th~venin equivalent model, 110, 441 Through-variable, 67-71 source, 80 Tame constant, 277. 279-83 Tame invariant, 261 Tame shift, 248, 267 Torque, 30 source, 35 Transducer, 169-72 Transfer function, 395-98 defined from exponential response, 396 formation from impedance, 434-45 geometric interpretation, 405 for interconnected systems, 407-8 defined from Laplace transform, 551 multiple input-output, 412-13 single input-output, 396-98, 408-12 Transfer operator, 225

Transformers, 174-80 Transient response: first-order system, 284 general linear system, 356 second-order system. 309 Translational elements, 23-28 Tree of graph, ·127-33, 188-90 branch. 127 link, 177 T-type element energy storage. 76-78 source, 80 Tuned mass damper, 3, 498 Turbulent flow, 50 1\vo-port:

elements, 169-73 models, 181-87 UncontroUable system. 132 Undamped natural frequency, 297 Underdamped systems, 303-5 Undetermined coefficients, method of, 256 Uniqueness theorem. 251-52 Units: English, 13-14 S.I., 12 Unit singularity functions: impulse, 246 ramp, 247 step, 246 Unstable systems, 280, 305

Vector: input, 120-22 output, 123 state, 120-22 Velocity, 22 absolute, 22-23 angular, 30 linear, 22 relative, 24, 67 source, 28 Vibration absorber, 480 Vibration damper, 498 Voltage, 37 source, 42 Volume, 45 Volume flow, 45 source, 51

Watt, J.. 13 Work. 22 Zero's, 398-400