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
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«:> 1997 by Prentice-Hall, Inc. A Pearson Education Company Upper Saddle River, NJ 07458
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Printed in the United States of America
JO 9 8 7 6 5 4 3 2
ISBN
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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:
<< c, Eq. (2.19) reduces to the linear (2.21)
p=mv
where the mass m depends on geometry and material properties and represents the classic newtonian mass. The energy stored in an ideal mass is
e=
p • pvdp= Lp -dp=-p om 2m Lo
2
1 =-mv2 2
(2.22)
and is always positive because it depends on the square of the momentum or velocity. For the ideal mass, an elemental equation describing the element in terms of power variables may be derived by differentiating Eq. (2.21 ):
dv F=mdt
(2.23)
This form of the elemental equation is a statement of Newton's law relating force to mass and acceleration. The schematic symbol for the primitive mass element, shown in Fig. 2.5, depicts a mass with velocity v referenced to a nonaccelerating coordinate system. The mass symbol has two terminals, one associated with the velocity v of the mass element and the second connected to the inertial reference frame to indicate that velocity is always measured with respect to the reference. Energy Dissipation Element-The Damper: A pure mechanical translational damper is defined as an element in which the force is a single-valued monotonic function of the velocity across the element. A relationship for a pure damper is shown in Fig. 2.6 and is of the form F =F(v)
(2.24)
where F() ·is a ·single-valued monotonic function. The symbol for a damper is shown in Fig. 2.6, where the velocity vis the difference in velocities across the damper terminals, that is, v v2 - Vt •
=
Sec. 2.2
27
Mechanical Translational System Elements
v2(t)
I
F(t)
0~
2
.I Reference Velocity v Figure 2.6:
Definition of the pure translational damper elemenL
For an ideal damper the constitutive relationship is linear and is written
F=Bv
(2.25)
where B is defined to be the damping constant (N-s/m). The power flow associated with an ideal damper is
'P= Fv
1 2 = Bv2 = -F B
(2.26)
and is always positive, so power always flows into the damper. Energy and power cannot be recovered from the damper; the mechanical work performed on the damper is converted to heat, becoming unrecoverable as mechanical energy and transferred from the system to its environment The constitutive equation for a damper, Eq. (2.25), is expressed directly in tenns of the power variables and is also the elemental equation. The damping constant B is a function of both geometry and material properties. Linear damping phenomena, as described by Eq. (2.25), are known as viscous damping effects. In mechanical systems frictional drag forces occur between two members in contact that are moving relative to each other. When the sliding surfaces are sufficiently lubricated, forming a hydrodynamic film, the drag force is proportional to the relative velocity between the surfaces and depends on the roughness of the two· surfaces, the material properties of the surfaces and lubricant, and the normal load force between the members. When there is very little lubrication, the friction forces tend to be relatively independent of speed, as shown in Fig. 2. 7, with an almost constant force. Such damping is nonlinear since the force is not proportional to velocity. Also shown in Fig. 2. 7 is the Coulomb friction characteristic which represents both stiction and sliding friction between two solids. Because this characteristic is not monotonic and a wide range of forces may occur at zero velocity, it is not strictly a pure damper. If the single-valued monotonic requirement is relaxed, the Coulomb characteristic can be considered a quasi-pure damper and can be used (albeit with care) in models of physical systems.
Energy and Power Flow in State-Determined Systems
28
Chap. 2
"- Aerodynamic drag
Slightly velocitydependent Velocity v
friction Figure 2.7:
Examples of frictional characteristics representing mechanical damping.
An additional form of damping is associated with the drag force on an object traveling through a fluid, such as the aerodynamic drag on an automobile or aircraft. This drag force is nonlinear and is often approximated as being proportional to the square of the velocity:
F = clvlv
(2.27)
where Ivi is the absolute value of velocity and c is a drag constant that depends on the geometry of the object and properties of the fluid. (The tenn IvI v is lmown as the absolute square, or absquare, and generates a force which changes sign as the velocity changes direction so that power is always dissipated as a result of the drag force.)
Source Elements Two source elements are defined in terms of the power variables for modeling mechanical translational systems: 1. The ideal force source is a source of energy in which the force exerted is an independently specified function of time Fs (t). The force produced by this element is independent of the velocity at the input port. The velocity produced by the source depends entirely on the system to which it is connected. 2. Conversely, the ideal velocity source is an element in which the velocity is an independently specified function of time ~r (t) and is maintained without regard to the force necessary to generate the velocity. The force produced by the source is detennined entirely by the reaction of the system to which it is connected. These sources are illustrated in Fig. 2.8. These two ideal sources may continuously supply or absorb. energy since in each one power variable is independently specified while the complementary· power variable is determined by the system to which the source is coupled. Ideal sources are capable of supplying infinite power and are idealizations of real sources, which have finite power and energy capability. In Chap. 4 models of real power-limited sources are developed by combin.ing ideal source and damper elements.
29
Mechanical Translational System Elements
Sec. 2.2
~(1)
f's(t)
(b) Velocity source
(a) Force source Figure 2.8:
Ideal mechanical translational sources.
Example2.1 To illustrate the power flow in mechanical systems, consider the distance required for an automobile to stop when the driver suddenly applies the brakes, locking the wheels so that a skid occurs. The system model is sketched in Fig. 2.9, where the system is defined as the automobile and the tire-pavement interaction and where the output variable of interest is the car velocity. A model for the car after the wheels are locked at time t = 0 may be formulated by considering 1. the car as a simple mass element m which at timet= 0 has velocity vo. 2. for timet > 0, that the only forces acting on the car are the tire-pavement resistance force Fr and the weight of the car which generates the normal force between the car and the pavement, and 3. a model for the tire-pavement interaction force after the wheels are locked to be represented by a damper with a constant resistance force equal to (i)
Fr = iJ.mg
where m is the mass of the car, g is the acceleration due to gravity, and iJ. is the coefficient of sliding friction for the tire-pavement interface.
v::: v0
v:::O
~
~
~/77//TM/T/717/T/T/.ITI//T~Jll: j I x =0
Stopping distance
x1
Car stopped
Brakes locked Figure 2.9:
11
Automobile stopping distance.
The stopping distance may be determined by equating the initiai energy of the car f:o to the energy dissipated Ed at the tire-pavement interface during the skid. When all the initial energy has been dissipated, the car velocity has decreased to zero. The initial energy of the mass is 1
2
Eo= -mv0 2
(ii)
30
Energy and Power flow in State-Detennined Systems
Chap. 2
The energy dissipated by the damper over the time period 0 to 1 is the integral of the power dissipated by the resistance force: (iii)
Since the resistance force is assumed to be constant. the energy dissipated is (iv) where xf is the distance traveled (displacement) in time 0 to 1. The stopping distance may be determined by equating the initial energy to the energy dissipated: (v)
and using Eq. (i), (vi) Equation (vi) indicates that. in terms of this model, the stopping distance during a skid is independent of the mass m. For a car traveling at 20 m/s with a pavement-tire friction coefficient of 0.8, the stopping distance is 202
Xj
= 2 X 0.8 X 9.8)
(m/s)l/(rn/s
2 )
(vii)
=25.5m The stopping distance is proportional to the square of the speed, and for an initial velocity of 10 mls the distance is reduced to 6.4 m.
2.3
MECHANICAL ROTATIONAL SYSTEMS 2.3.1 Definition of Power Flow Variables In rotational systems power is transmitted and energy is stored by rotary motion about a single axis. The power flow across a rotational system boundary is through angular motion of a shaft, as illustrated in Fig. 2.11. In a rotational system, the power flow at any instant may be expressed as the product of the torque acting about the fixed axis and the angular velocity about the axis: 'P(t) = T(t)O(t)
(2.28)
where T is the applied torque (N-m) and O(t) is the angular velocity (radians/secon
Sec. 2.3
Mechanical Rotational Systems
31
I
Force
Radi··I!~F r
Rm~·r ;~al
Center of rotation
velocity
Center of rotation
T=Fr
O=vlr
Rgure 2.10:
Definition of power variables in a rotational system.
Also shown in Fig. 2.11 is the convention for schematically depicting torques on shafts. An applied torque is shown as an arrow aligned along the shaft; with a length proportional to the magnitude of the torque and with a direction defined by the "right-hand grasping rule." If the shaft is grasped in the right hand with the fingers curling in the direction of the assumed positive angular velocity, the direction of the arrow representing a positive torque is in the direction pointed to by the thumb. Thumb points in direction of positive torque vector
Power into port Angular ~(t) = T(t) O(t) velocity Torque O(t) T(t)~ system
(a) An input port represented by a shaft
Rgure 2.11:
(b) Right-hand grasping rule
Power flow and torque sign convention in a rotational system.
The rotational work crossing the system boundary over a time increment dt is ~W
= T(t)Q(t) dt
(2.29)
In a manner analogous to mechanical translational systems, the energy may be expressed directly in terms of integrated power variables by defining angular displacement e as the integral of the angular velocity: 19(1)
=
fo' !J(t)dt
or
de= Q(r) dr
(2.30)
with units of radians, and angular momentum h as the integral of the torque: h(t)
with units of N -m-s.
= fo' T(t) dt
or
dh = T(t)dt
(2.31)
Energy and Power Flow in State-Detennined Systems
32 The increment in following three forms:
wor~
.~W
~w
~w
=
=
=
Chap.2
Eq. (2.29), across a boundary may then be expressed in the
=
T(Odt) n(Tdt) Tfldt
= =
Tde Odh OTdt
= = =
d£Potential df:kinetic
(2.32)
df:dissipated
Mechanical rotational power flow into a system. may result in a change in stored energy through changes in angular displacement de associated with a rotational spring, or may result in changes in angular momentum dh associated with a rotational mass or inertia, or may be dissipated by conversion of rotati~nal work to hea~ in a rotational damper.
2.3.2 Primitive Rotational Element Definitions Energy Storage Elements
Rotational Spring: A pure rotational spring, also known as a torsional spring, is defined as an element in which the angular displacement e is a single-valued monotonic function of the torque as illustrated in Fig. 2.12. A pure spring has a general constitutive relationship:
e = :F
(2.33)
The energy stored in a pure. rotational spring is
(e
E=
lo
(2.34)
Tde
and is illustrated by the shaded area in Fig. 2.12. An ideal spring is one in which the constitutive relationship Eq. (2.33) is linear and may be written (2.35)
where Kr is the rotational spring constant, or rotational stiffness, with units of N-m/rad.
T0 Torque T Figure 2.12: Definition of the pure rotational spring element
Sec. 2.3
33
Mechanical Rotational Systems
For an ideal spring the energy E stored with an applied torque T is (2.36)
The symbol for the rotational spring, shown in Fig. 2.12, is similar to that for the translational spring. The rotational displacement e is defined as the displacement from a rest displacement resulting from a finite torque and the resultant energy storage. As shown in Fig. 2.12, the relative displacement between the ends of the spring is the difference between displacements at each end measured with respect to a fixed reference, that is, 9 = 92 - 9 1· The rotational spring constant Kr in the constitutive equation is a function of material properties and geometry. In rotational systems, shafts and couplings often have rotational stiffness which can be determined analytically or have been measured experimentally and tabulated [4, 5]. An elemental equation for the ideal rotary spring may be derived by differentiating Eq. (2.35) to give an expression in terms of power variables: O= _1 dT Kr dt
(2.37)
Rotational Inertia Element: A pure rotational inertia element is defined as an element in which the angular momentum h is a single-valued monotonic function of angular velocicy: h
= .1'(0)
(2.38)
The energy stored in a pure inertia is illustrated in Fig. 2.13 and may be computed directly from Eq. (2.32). When the constitutive equation is linear, the ideal inertia may be defined as (2.39)
h=JO
where J is defined as the rotational moment of inertia (kg-m2 ). h
O=O:z-0 1 ---1
T(r)~~L_i! 2~~
Constant angular velocity reference
e
!a
€~
j ho c 0
I77'7'TT~~rr•.-rr..,-~.r7"71
nl
no Angular velocity Figure 2.13:
Definition of the pure rotational inertia elemenL
o
Energy and Power Flow in State-Determined Systems
34
Cbap.2
The value of the moment of inertia of a rotating body depends on the geomeny and mass distribution and has been tabulated for a variety of bodies in [4, 5]. The moment of inertia of a single lumped mass m rotating at a constant radius r about an axis is J = mr2 , and for a collection of n discrete mass elements the net moment of inertia is the sum of each elemental moment of inertia: n
J
= :Em;rf
(2.40)
i=l
For a general body with a spatially distributed mass rotating about a fixed axis, the inertia J may be determined by integrating over the body as shown in Fig. 2.14.
J=mr2 (a) Point mass at radius r
(b) Circular disk
Massm
£
\n mL2
J=w (c) Rod rotating about its centroid Figure 2.14:
Moments of inertia of typical bodies.
The energy stored in an ideal inertia is &=
ho Odh = -1 Lho hdh = J L 0
0
1 2J
1 2
-h~ = -JQ 2
(2.41)
The symbol for a rotary inertia is shown in Fig. 2.13, where the angular velocity is referenced to a fixed axis to indicate that the angular velocity of the inertia must be referenced to a fixed or nonaccelerating reference frame. The elemental equation for the ideal rotary inertia may be derived by differentiating Eq. (2.39) to obtain: T=JdQ . dt
(2.42)
Energy Dissipation Element A pure rotational damper is defined as an element in which the torque is a single-valued monotonic function of the angular velocity across the element: T =.1'(0)
(2.43)
Sec. 2.3
Mechanical Rotational Systems
35
The symbol for the rotary damper is shown in Fig. 2.15 where the angular velocity across the element is defined in terms of the difference in angular velocities at each end of the element, 0 = 02 - n •. The ideal rotary damper has a linear constitutive equation: T=B,Q
(2.44)
where B, is the rotary damper constant (N-mls). T n=~-n. ~ 1(t)
n
~ T(r)-~ -T(r) Reference
Angular velocity Figure 2.15:
n
Definition of the pure rotational damper element.
The power dissipated by an ideal damper is (2.45) and is always positive; that is, power is always dissipated. As indicated in Eq. (2.32), the work performed on a rotary damper element is converted to heat and transferred to the system environment Rotational damping occurs naturally from frictional effects in bearings an~ like translational friction, is often highly nonlinear. The ideal or linear form, given by Eq. (2.44), is often known as viscous rotary damping.
Source Elements Two energy source elements are defined for mechanical rotational systems:
1. The ideal torque source is an element in which the torque exerted on the system is an independently specified function of time Ts (t). The angular velocity produced by the source is determined by the reaction of the system to which it is connected.
2. The ideal angular velocity source is an element in which the angular velocity at a port is an independently specified function of time Sls(t). The torque produced by the source is determined by the system to which it is connected. These two ideal sources, illustrated in Fig. 2.16, are capable of supplying infinite power and thus are idealizations of real power-limited sources.
36
Energy and Power Flow in State-Determined Systems ~(t}
Chap. 2
c-> 2
n.(t)
~(t}-
n.(t} = nser>
(b) Angular velocity source
(a) Tmque source
Figure 2.16: Mechanical rotational sources.
Example2.2 A shaft-flywheel system, as shown in Fig. 2.17~ is an integral part of many mechanical systems such as pumps, turbines, and automobile transmissions. Long, lightweight shafts exhibit significant compliance and act as torsional springs. One of the issues in the design of these systems is the amount of ''windup" or twist that occurs in the drive shaft as the flywheel is accelerated or deaccelerated. A worst-case operating condition occurs when the input end of the drive shaft is stopped suddenly while the flywheel is turning at its normal operating speed. What is the maximum deflection across the shaft? With knowledge of the shaft deflection~ the stresses within the shaft and the possibility of failure can be determined. Flywheel inertial
Shaft deflection 8
=82- 8 1
Figure 2.17:
Shaft windup during ttansients.
A model can be formulated to determine the maximum shaft deflection as follows: 1. The flywheel is considered an ideal rotary inertia J with an initial angular speed of 0 0 • 2. The shaft is considered an ideal rotational spring with rotational stiffness Kr. 3. The bearings and all other dissipative features, for example aerodynamic drag on the flywheel, are neglected.
The shaft deflection is estimated by noting that the maximum shaft deflection E>m occurs when all the kinetic energy in the flywheel is transferred to potential energy in the shaft with no losses: (i)
which yields
·em= '/[(; /TOo
(ii)
Sec. 2.4
Electric System Elements
37
The maximum shaft deflection is therefore directly proportional to the initial flywheel angular velocity and is proportional to the square root of the inertia/stiffness ratio.
l.4
ELECTRIC SYSTEM ELEMENTS 2.4.1 Definition of Power Flow Variables In the electric domain the power flow through a port represented by a pair of wires is the product of cu"ent and voltage drop, as shown in Fig. 2.18: 'P(t)
= i (t)v(t)
(2.46)
where i (A) is the current flowing into the port, and v (V) is the voltage drop across the tenninals. The electric work flowing across a system boundary in an incremental time period dt is
fl. W = i (t)v(t) dt
(2.47)
The work may be expressed in terms of electric domain variables by defining the integrals of the two power variables:
q
= fo' j dt
or
dq
= i dt
(2.48)
where q is the electric charge in units of coulombs (C) (A-s) and
A=
fu' vdt
or
d'A = vdt
(2.49)
where J.. is defined to be the magnetic flux linkage (V-s or Wb). Current i(t) .....
-+
Voltage
-
Electric
Power into port
system
~(t)
v(t)
I
Reference voltage vrer
= v(t)i(t)
-
Figure 2.18: Electric power port.
The unit of charge, the coulomb, corresponds to the negative of the total charge associated with 6.22 x 1018 electrons and is equivalent to 1 A-s. The current i is the rate of flow of electric charge q and is positive in the direction of positive charge flow (or in the direction opposite electron flow). The voltage represents the potential difference between two points and is equivalent to the work performed in moving a unit charge between the two points with a potential difference of v. The distribution of charge within a medium leads to the establishment of an electric field as shown in Fig. 2.19a [7].
Energy and Power Flow in State-De~ined Systems
38
Chap.2
+ .. (a) Energy st~rage in an eleclric field
(b) Energy storage in a magnetic field
Figure 2.19: Energy storage mechanisms. (a) Electric field between two conducting plates, and (b) magnetic field associated with a conducting coil.
Flux linkage, the integral of voltage drop, is associated with the magnetic flux generated in a coil of wire carrying a current, as illustrated in Fig. 2.19b. The flux linkage is defined as the total magnetic flux passing through, or linking, the coil. The total flux linkage is a function of the magnetic flux density produced by the coil, the area of the coil, and the total number of turns in the coil through which the flux passes [7]. The electrical work passing through a system boundary in time dt may be written in terms of power variables and integrated power variables in the following three fonns: AW AW AW
= = =
i(vdt) v(idt) i v dt
= = =
idA. vdq vi dt
= =
dEma,;netic
=
dEmssipated
d£electrical
(2.50)
Electrical work crossing a system boundary may result in a change in the magnetic flux associated with the system through electromagnetic energy storage in inductors, a change in the total charge in the system associated with electrosfB:tic energy storage in capacitors, or dissipation in resistOrs through the generation of heat with no electric energy storage.
2.4.2 Primitive Electric Element Definitions Energy Storage Elements
Electric Inductor: The pure inductor is defined as an element in which the flux linkage is a single-valued monotonic function of the current: A.= :F(i)
(2.51)
The constitutive characteristic of an inductor is shown in Fig. 2.20. When the relationship between flux linkage and current is linear, the ideal constitutive equation is l=Li
(2.52)
where the constant of proportionality L is defined as the inductance in henrys (H) (V-s/A). Practical inductors used in electronic circuits are often specified in millihenries (mH) or Io-3 H, or microhenries (JLH) or to-6 H.
Sec. 2.4
39
Electric System Elements Shaded area is energy stored at flux
v2
i
L
f
v1
:; Xo
~ 2 v = v2 - v1 1
Xo )7.,..,.7~.~,,.;,
Jt io
Current Figure 2.20: Definition of the pure electric inductance elemenL
The energy stored in an ideal inductor is
£=
}.
1 /.}. 1 1 idJ...=J...dJ...=-J... 2 =-Li 2 o L o 2L 2
!.
(2.53)
The property ofinductance is associated with the magnetic fields generated by currents in a conductor, usually in the form of a coil of wire. The value of the inductance L is a function of the coil geometry and the material properties of the core on which it is wound. The circuit symbol for an inductor is shown in Fig. 2.20. The inductance for two coil geometries is shown in Fig. 2.21 where it is noted that the inductance is proportional to the permeability JL of the material through which the magnetic flux passes and to the square of the number of turns of wire. A simple coil of wire with an air core has a relatively low inductance because air has a low magnetic permeability. To achieve a higher inductance for a given coil geometry, ferromagnetic materials with high magnetic permeability are used as a core within the coil. Such materials are, however, subject to magnetic saturation and losses and are not suitable for all applications. N turns on length I
•I
14
CoreofareaA
p~~~~/Md_.wi~ ~ _p.N2A L- I
L=
(a) Helically wound coil
~
In(%)
(b) Two parallel conductors in hairpin configuration
Figure 2.21: Inductance of two conductor configurations.
An elemental equation for the inductor, expressed in terms of power variables, may be derived by differentiating the constitutive equation, Eq. (2.52): di
v=Ldt
(2.54)
Energy and Power flow in State-Determined Systems
40
Chap. 2
Electric Capacitor: A pure electric capacitor is defined as an element in which the electric charge q stored is a single-valued monotonic function of the voltage v across its tenninals: (2.55) q = F(v)
The constitutive relationship for a pure capacitor is illusttated in Fig. 2.22 where the stored energy is also indicated. When the constitutive relationship for a capacitor is linear, (2.56)
q=Cv
the capacitor is defined to be ideal, and C is the capacitance with units of farads (F) (CN). q
v0
v
Voltage · Figure 2.22: Definition of the pure electric capacitor element
A capacitor stores energy in an electrostatic field established in a nonconducting dielectric material between two conducting surfaces (plates). The value of capacitance C in the constitutive equation depends on the geometry of the plates and the material properties of the dielectric in the gap. For the parallel plate geometric configuration shown in Fig. 2.23, the capacitance is proportional to the plate area, inversely proportional to the separation of the plates, and directly proportional to the dielectric pennittivity €, which for air is 8.85 x to- 12 F/m. The energy stored in an ideal capacitor is
e= f.oqvdq =-CI f.qo qdq =
I 1 - q 2 = -Cv2
2C
2
Area A
Dielectric permittivity£ (a) Parallel plate capacitor Figure 2.23:
(b) Concentric cylinder capacitor
Capacitance of two plate configurations.
(2.57)
Sec. 2.4
Electric System Elements
41
The elemental equation for the ideal capacitor may be found by differentiating the constitutive equation, Eq. (2.56): i =Cdv dt
(2.58)
Electric capacitors are used in electronic and electric equipment They come in many forms with different plate geometries and dielectric materials. In practice the farad is too large a unit to be of practical use; in electronic circuits, capacitors are usually expressed in microfarads (J.LF) or 1o-6 F, nanofarads (nF) or Io-9 F, or picofarads (pF) or I o- 12 F. Energy Dissipation Element A pure electric resistor is defined as an element in which the current is a single-valued monotonic function of the voltage drop: i = F(v)
(2.59)
Many electric devices, including semiconductor diodes and resistors, exhibit pure resistance. A typical characteristic is shown in Fig. 2.24. For standard electric resistors, the relationship between voltage and current is linear and the ideal elemental equation becomes Ohm's law: .
1
1
= -v R
(2.60)
where R is the electric resistance in ohms (Q) (VIA). Electric resistance is a function of the bulk resistivity p of the conducting medium and the size and shape of the conductor. (Resistivity is easily visualized as the resistance between opposite faces of a unit cube of the material and has units of Q-m.) Materials with a low resistivity, such as copper which has a resistivity of 1.742 x 10-8 Q-m at room temperature, are known as conductors, while those with very high resistivity are collectively known as insulators. A conductor of length 1 and uniform cross section A has a resistance pi R=-
(2.61)
A
Electric resistors are important components in electronic circuits. They are typically constructed of bulk carbon or a composite material, of a carbon film, or as a coil of resistance wire.
v2
•
R
vi
c
1
u
~
2
v
=v2 -v1
~
::1
v Voltage Figure 2.24:
Definition of the pure electric resistance elemenl
Energy and Power Flow in State-Determined Systems
42
Chap.2
An ideal resistor dissipates power, with the electrical work being convened to heat and made unavailable to the electric system: 2
'P(t) = vi = i R
v2
= -R
(2.62)
Source Elements 1\vo ideal electric source elements may be defined, with symbols as shown in Fig. 2.25. An ideal voltage source is an element in which the voltage across its tenninals is an independently specified function of time Vs (t). It is capable of supplying, or absorbing, infinite current in order to maintain the specified voltage. An ideal current source is an element in which the current supplied to the system is an independently specified function of time ls(t). The terminal voltage of a current source is defined by the system to which it is connected. Both ideal sources are capable of supplying or absorbing infinite power and thus only approximate real power-limited sources.
2
2
Ys(t)
(a) Cunent source
(b) Voltage source
Figure 2.25: Ideal electric sources.
Example2.3 A wire-wound resistor with a nominal resistance of 10 0 is constructed as a coil as illustrated in Fig. 2.26. A feature of wire-wound resistors is that they have relatively good beat transfer characteristics and maintain a relatively constant value of resistance with temperature. However, because of their coil structure, wire-wound resistors also have a small inherent parasitic inductance associated with the magnetic field generated by the current in the coil. Both the resistive and inductive effects may be important in critical applications. To determine the importance of the inductance effect, the frequency of a sinusoidal current waveform at which the magnitude of the voltage drop associated with the inductance is 10% of that due to the resistance is estimated. rii'St, the inductance of the coil is computed assuming an air core. The inductance for the configuration illustrated is
J.LAN 2 L=--
1
(i)
where p, is the penneability of the core, A is the cross-sectional area. l is the length of the coil, and N is the number of turns distributed along the length.
Sec. 2.4
Electric System Elements
43 Area A ls(t) = 10 sin(wt)
Electric current source
Ntums Figure 2.26:
A wire-wound electric resistor.
If the coil has 100 turns. a diameter of 1 em. and a length of 0.02 em. the inductance (assuming a permeability of air of p, = 1.26 x to-6 Him) is _ (1.26 x I0-6)(H x o.oos2 )(IOe>2> _ L 0.02 - 49.5
X
10
_6 H
(ii)
Assume that the resistor is driven by a sinusoidal current source ls(t) = lo sin(wt)
(ill)
where lo is the magnitude and w is the angular frequency. If the resistor is an ideal resistance (without any inductance). the voltage drop VR is determined from the resistance elemental equation: VR
= R/0 sin wt
(iv)
and its magnitude is
lvRI =I Rio sin(wt)l =Rio which is independent of the frequency w. If the coil is an ideal inductor L with no resistance. the associated voltage drop determined from the elemental equation
VL
= L di dt
= IoLwcos(wt)
(v)
VL
is
(vi)
and the magnitude of the voltage drop due to the inductance is (vii)
which is directly proportional to the angular frequency w. Therefore. as the frequency of the applied current waveform is increased, the total voltage drop. which is the sum of the two effects. increases. The frequency CLJo at which the ratio of the magnitude VL to VR is 0.1 is (viii)
Energy and Power Flow in State-Determined Systems
44
Chap. 2
and for the values of L and R given above, Clio
=
0.1 X 10 _ x -6 = 202,020 rad/s 49 5 10
(ix)
As the applied frequency is increased, the inductive voltage drop becomes more significant and
the resistor behaves more like an inductor. At 202,020 radls the voltage drops due to resistance and inductance are equal, and at higher frequencies the inductive voltage drop exceeds the resistance voltage drop.
2.5
FLUID SYSTEM ELEMENTS 2.5.1 Definition of Power Flow Variables
The internal flow of fluids through pipes, vessels, and pumps and the external ftow around vehicles, aircraft, spacecraft, and ships are complex phenomena involving flow variables that are continuous functions of both space and time. As such they generally cannot be represented in tenns of pure lumped elements. With some simplifying assumptions, however, a number of significant characteristics of the dynamic behavior of fluid systems, particularly for one-dimensional pipe flows, can be adequately modeled with lumped-parameter elements. ·In this section we define a set of lumped-parameter elements that store and dissipate energy in networklike fluid systems, that is, systems that consist primarily of conduits (pipes) and vessels filled with incompressible fluid. These definitions are analogous to those for mechanical and electric system networks. The power flow thiough a port into a fluid system, shown in Fig. 2.27, is expressed as the product ·of two fluid variables: 'P(t)
= P(t) Q(t)
(2.63)
where Q(t) is the fluid volume flow rate (m3/s) and P(t) is the fluid pressure drop across the port. The unit of pressure is the pascal (N/m2 ), after Jacques Pascal who formulated the law that the pressure at any point in a fluid at rest has a single scalar value independent of direction [8]. Volume
flow rate
Pipe
Q(t)
Pressure/
Fluid . system
Power into pon ~(t)
=P(t)Q(t)
P(t)
Reference pressure Prr:f Figure 2.27:
Fluid power port.
The fluid pressure is the normal force F per unit area A of the port (pipe) cross section: P=dF dA
(2.64)
Sec. 2.5
Fluid System Elements
45
For a pipe with a uniform pressure profile the pressure is the total force acting across a cross section divided by the area. The volume flow rate Q is the total volume of fluid passing through the port per unit time. It is the integration of the fluid velocity v over the area of the port: Q
=
L
(2.65)
vdA
For a port with uniform pressure and fluid velocity profiles, the power flow may be written P= PQ
F = -(vA) = Fv A
(2.66)
which is directly related to mechanical power as defined in Sec. 2.2. The incremental fluid work crossing a system boundary at a port in time dt is t:,. W
= P(t)Q(t) dt
(2.67)
and may be expressed in terms of fluid system variables by defining a pair of integrated power variables. The totaljluid volume V (m 3 ) is the integration of volume flow rate; that is,
v = Ia' Q(t) dt
or
dV = Q(t)dt
(2.68)
and the integral of the pressure is defined to be the pressure momentum r (N-s/m 2 ):
r
=
fo' P(t)dt
or
dr = P(t)dt
(2.69)
The volume V represents the total volume of fluid passing through the port over a given time period. The pressure momentum r is the time integral of pressure, analogous to momentum in mechanical systems. The increment in work passing through a fluid port in time dt may be written in terms of the power variables P and Q and the integrated power variables r and v in the following three forms: t:,.W 6W t:,.W
=
= =
Q(Pdt) P(Qdt) PQdt
=
= =
Qdf PdV PQdt
=
dekinetic
=
dGpotential
=
dedissipated
(2.70)
Work done by a fluid crossing a system boundary may result in 1. a change in pressure momentum in the system which is usually associated with energy
storage in a fluid inertance, 2. a change in the energy stored in a fluid volume which is usually associated with fluid capacitance, or 3. simply a change in pressure and flow rate representing dissipation in a fluid resistance in which fluid work is converted to heat
Energy and Power Flow in State-Determined Systems
46
Chap. 2
2.5.2 Primitive Fluid Element Definitions Energy Storage Elements Fluid Inertance: A pure fluid inertance is defined as an element in which the pressure momentum is a single-valued monotonic function of flow rate:
r
= F(Q)
(2.71)
A representation of this characteristic is shown in Fig. 2.28 where the energy stored in the inertance is indicated as detennined from Eq. (2.70). The symbol for a fluid inertance shown in Fig. 2.28 is similar to that for an electric inductor. When the relationship betWeen pressure momentum and flow is linear, the constitutive equation (2.71) for an ideal inertance is (2.72) where I is defined as the fluid inertance (N-s 2/m5 ).
r
Qo
Q
Volume flow rate
Figure 2.28: Definition of the pure fluid inertance element
The energy stored in an ideal fluid inertance is
&=
J.r o
Qdr
=.!:.I
J.r r o
dr
2 2 = ..!..r = .!:.IQ 2I 2
(2.73)
The elemental equation for a fluid inertance may be detennined by differentiating Eq. (2.72) to obtain a first-order differential equation in terms of power variables:
dQ
(2.74)
P=ldt
The value of fluid inertance depends on the pipe geometry and ftuid properties [9]. For an incompressible fluid flowing iri a pipe of uniform area A. and length l with a unifonn velocity profile, the element of ftuid is accelerated by a force F equal to the pressure difference between the two ends of the pipe multiplied by pipe area, as shown in Fig. 2.29. The acceleration of this element of fluid is given by Newton's law:
dv
F=AP=mdt
(2.75)
Sec. 2.5
47
Fluid System Elements
The mass m of fluid in the pipe is the density of the fluid p multiplied by its volume V = Al, and the velocity is volume flow rate Q divided by the pipe area A, and so Eq. (2.75) may be expressed in terms of the fluid variables as 1 1 dQ pl dQ P =-(pAl)--=-A A dt A dt
(2.76)
which is the same form as Eq. (2.74). The fluid inertance of the pipe is therefore I= pl A
(2.77)
The fluid inertance increases with increasing fluid density and pipe length and decreases with increasing area.
I
Pipe
•
Length I ..
I Area
4;1s u:rA \
P2
P1
Figure 2.29: The fluid inertance of a pipe section.
Fluid Capacitance: A pure fluid capacitance is defined as an element in which the volume of fluid stored is a single-valued monotonic function of the pressure: (2.78)
V = :F(P)
The constitutive characteristic of a pure fluid capacitor is illustrated in Fig. 2.30 where the energy stored in the capacitor is shown as the shaded area. The symbol for a fluid capacitor always has one terminal connected to a fixed reference pressure because the pressure P and the energy E stored in a fluid capacitor are always measured with respect to a known pressure. For convenience the reference is usually selected as either atmospheric pressure or absolute zero pressure. Reference pressure
v
Po Pressure Figure 2.30:
Definition of the pure fluid capacitance elemenL
P
Energy and Power Flow in State-Determined Systems
48
Chap. 2
An ideal capacitor is one with a linear relationship between volume and pressure:
(2.79) where C1 is defined to be the fluid capacitance (m5 IN). The energy stored in an ideal fluid capacitor is
E=
VPdV = LV -1V dV = -1V o CJ 2CJ Lo
2
1 2 = -C 1P
_2
(2.80)
The elemental equation for an ideal fluid capacitor, expressed in tenns of power variables, is found by differentiating Eq. (2. 79):
dP Q=C,dt
(2.81)
The most common form of fluid capacitance is an open tank containing an incompressible fluid of density p in a gravity field g, as shown in Fig. 2.31. The pressure at the bottom of _the tank is related to the depth, and therefore the volume, of fluid in the tank. The pressure due· to the weight of a liquid column of height h is
P=pgh
(2.82)
where P is the difference between the pressure at_ the bottom of the tank and atmospheric pressure. For a tank with uniform cross section A, the volume of fluid is V = Ah, and so Eq. (2.82) may be written
P
= PK hA = PKv A
A
(2.83)
or
A V=-P pg
(2.84)
The fluid capacitance of the tank is therefore A
c,=pg
(2.85)
Compressible fluids (fluids whose density varies with pressure) in a rigid wall container, as shown in Fig. 2.31, also exhibit fluid capacitance. For slightly compressible fluids, the relationship between changes in density and pressure may be expressed as
dp dP -=p /3
(2.86)
where /3 is the fluid bulk modulus. The bulk modulus is a fluid property that expresses the degree of compressibility; for liquids such as oil and water it is large, on the order
Sec. 2.5
Fluid System Elements
49
Gravity Ambient pressure Paun
g
J
Rigid container Volume V Area A
i Compressible Fluid density p
p2 (a) Open tank in gravity field Figure 2.31:
'
p
fluid Reference pressure Prcr
(b) Fixed-volume tank with compressible fluid
Two fonns of fluid capacitance.
of 2.1 x 109 Pa (300,000 lb/in 2 or psi) and a significant change in pressure is required to change the fluid density, while for gases such as air the bulk modulus depends on the process employed to change the fluid state. For perfect gases the bulk modulus is ~
=kP
(2.87)
where k is a constant and P is the absolute pressure of the gas. The constant k is equal to 1 if the temperature remains constant as the gas changes state and is larger (1.4 for air) when the process is adiabatic, that is, with no heat transfer occurring between the gas and its environment [2]. The latter case is usually associated with rapid changes in the state of the gas that do not allow significant heat transfer to occur. The capacitance of the rigid chamber may be detennined from conservation of mass by equating the net mass flow rate into the chamber to the change of mass within the chamber: d dp dV (2.88) pQ = -(pV) = V - + p dt dt dt Since the volume of the chamber is constant, the derivative of the chamber volume with respect to time is zero and Eqs. (2.86) and (2.88) may be combined to yield
V dP dP Q=--=Ct~ dt dt with the result
v
Ct=-
/3
(2.89)
(2.90)
Energy Dissipation Element A pure fluid resistance is defined as an element in which the flow rate is a single-valued monotonic function of the pressure drop across the element as indicated in Fig. 2.32:
Q = :F(P)
(2.91)
50
Energy and Power Flow in State-Determined Systems .
Chap. 2
The symbol for a fluid resistor, shown in Fig. 2.32, has a pressure drop P across the tenninals and a flow Q through the element An ideal fluid resistor has a linear elemental relationship 1
(2.92)
Q=-P Rf
where R1 is defined to be the fluid resistance (N-s/m5 ). Q
p2 Q R/ PI ~
2
I P=P2 -P 1
p Pressure
Figure 2.32: Definition of the pure fluid resistance elemenL
The power dissipated by an ideal fluid resistor is (2.93) and is always positive, representing the conversion of fluid work to heaL Fluid resistance is associated with flow through pipes, orifices, and valve openings. The value of resistance depends on geometry and fluid properties as well as flow conditions. For a long, unifonn-area circular pipe with laminar flow, the fluid resistance is given by the Hagen-Poiseuille flow law [8, 9]: (2.94) where l is the pipe length, p, is the fluid viscosity, and d is the pipe diameter. This law assumes that the flow is laminar, which requires that a Reynolds number Ry defined as R = 4p Q y 1Cdp,
(2.95)
be less than 2000. As the .flow rate increases so that Ry >-2000, the flow in the pipe becomes
turbulent and the pipe resistance becomes nonlinear and a fu~ction of flow. For incompressible flows through orifices and valve opeirlngs, as illustrated in Fig. 2.33, the orifice equation relating p~ssure and flow .is of the form (2.96)
Sec. 2.5
Fluid System Elements
51
where the coefficient CR is a function of the fluid density p, the orifice area A, and the orifice discharge coefficient cd: p
CR
(2.97)
= 2C2A2 d
The discharge coefficients cd have been tabulated for orifices with different geometries [9].
~
Orifice area A
p1
Annular~~ Outlet
area A
Inlet (a) FlXed orifice Figure 2.33:
t
p2
P=P2 -P1
Q
(b) Annular metering valve
Fluid orifices and valve metering areas.
Fluid Source Elements
Two ideal fluid sources are defined and are illustrated in Fig. 2.34. An ideal pressure source is an element in which the pressure applied to a port is an independently specified function of time Ps (t). An ideal flow source is an element in which the flow through the port is an independently specified function of time Qs(t). Like the ideal sources in the other energy domains, ideal fluid sources are not power-limited.
2
2
Q,(t)
P1 (t)
(b) Pressure source
(a) Flow source Figure 2.34:
Ideal fluid sources.
Example2.4 In many hydraulic systems, including automotive power steering and transmission systems, water distribution systems, and medical assist systems, fluid is transmitted through pipes and passageways as shown in Fig. 2.35. When the flow is time-varying, the question of the relative importance of the resistive and inertance effects in the fluid passageway is important In this example, we first compute the fluid resistance and inertance for a typical fluid passageway, and then the pressure drop due to the two effects is compared for sinusoidally varying flows. The angular frequency at which the magnitude of the resistive pressure drop is equal to that due to the passageway inertance is to be determined.
52
Energy and Power Flow in State-Detennined Systems
I•
Lengtb~A
Chap.2
~I
/~'-
~
~
~
Fluid with ~ity p.
Rgure 2.35:
Long ftuid passageway.
A pipe of diameter d = 0.002 m and length I = 0.1 m is filled with a liquid that has fluid properties equivalent to those of water at room temperature. that is. a density of 996 kg/m3 and a viscosity of 7.98 x 10-4 N-s/m 2 • The resistance of the passageway is computed from Eq. (2.94) with the assumption that the flow is laminar and that the effects of flow nonuniformity at the entrance and exit of the passage may be neglected: R I
= 128JLl = 128 x 7.98 x J0-4 x 0.1 = 2 _03 x 10s N-s/ms 7rd4
3.14 X 16
(i)
X JO-ll
The fluid inertance may be computed directly from Eq. (2.77) as . _ pi _ 996 X 0.1 _ ~ 2 S I - A- _ x J0-6 -5.08 x hr N-s /m 3 14
(ii)
If the flow in the passageway is assumed to be sinusoidal with magnitude Q0 and angular frequency w. that is. Q
= Qo sin(c.c>t)
(iii)
then the pressure drop due to resistance is (iv)
and its magnitude is (v)
The pressure drop due to the inertance of the passageway is P1
dQ = I dt = I wQ0 cos(wt)
(vi)
and the magnitude is (vii)
The frequency c.c>o at which the magnitudes of the two pressure drops are equal is found from
.
~M~~
_ R1 _ 2.03 x 10 · ~ dl 400 ra s I - 5.08 x lOS
·-
.
8
c.c>o -
(viii)
Sec. 2.6
53
Thermal System Elements
For sinusoidal flow variations with frequencies above 400 rad/s, the magnitude of the pressure drop due to the inertance effect is greater than the magnitude of the pressure drop due to passageway resistance. For slowly varying flows, with frequencies much less than 400 radls, the primary pressure drop is due to the passageway resistance.
2.6
THERMAL SYSTEM ELEMENTS 2.6.1 Definition of Power Flow Variables
Thennal systems, in which heat is generated, stored, and transferred across boundaries, have historically been characterized in tenns of thermal energy and power flow and the relationships of these variables to temperature. In the lumped-parameter thennal system models considered in this text, the temperature T is selected as a fundamental thermal variable. The temperature of an object may be defined using several relative and absolute scales, including the Kelvin absolute temperature K associated with SI units and Rankine 0 ( R) absolute temperature associated with English units [1 0]. At absolute zero temperature, a body has no kinetic energy associated with its molecules. At standard pressure water freezes at a temperature of 273.2 K or 0°C, while at the boiling point of water the temperature is 373.2 K or l00°C. When bodies at two different temperatures are brought into contact, the basic laws of thennodynamics indicate that heat H flows from the body at a higher temperature to the body at a lower temperature. Heat continues to flow until the bodies are at equal temperatures. Heat has been quantitatively defined in tenns of pure substances. One calorie (cal) of heat is the amount of heat required to raise the temperature of 1 g of water one degree Celsius (pure water at atmospheric pressure raised from l5°C to l6°C). Heat is a measure of thermal energy, and the rate of change of energy with respect to time is heat flow rate q, or power P: dH
'P=q=dt
(2.98)
where H is the heat (J) and q is the heat flow rate (J/s or W). One calorie is equivalent to 4.187 J. The law of energy conservation, defined in Sec. 2.2, may be written specifically for thermal systems in a fonn that represents the first law of thermodynamics [2]: d£ q = dt - 'Pw
(2.99)
which states that the net flow of heat across a system boundary must equal the sum of changes in the internal system stored thermal energy and the ~eat converted to another fonn 'Pw. such as mechanical, electric, or fluid work. The thennal power variable is simply heat flow rate q, and thermal energy is represented by heat H. Unlike the mechanical, electric, and fluid systems, power flow in thermal systems is not commonly described as a product of two variables. While mathematically a complementary pair of variables could be defined, physical observation of simple thermal systems of engineering interest has not identified a pair of complementary energy storage mechanisms. Thermal systems are represented by a set of pure elements that includes only one energy storage element
Chap. 2
Energy and Power Flow in State-Determined Systems
54
2.6.2 Primitive Thermal Element Definitions Energy Storage Element Thermal Capacitance: Thermal energy storage may be defined in terms of a pure thermal capacitance in which the heat H is a single-valued monotonic function of the temperature T: (2.100) H = F(T)_
A symbolic representation of the pure thennal capacitance and the constitutive relationship is illustrated in Fig. 2.36. The constitutive relationship is in general a function of geometry and material properties. For an ideal thennal capacitance, the relationship is linear: (2.101)
H=C,T
where C, is the thennal capacitance (J/K). The energy stored in a thennal capacitance is simply the heat H. H
Reference temperature
T
Temperature Figure 2.36: Definition of the pure thennal capacitance.
The elemental equation for the ideal thennal capacitance is derived by differentiating Eq. (2.101): dT q=C,dt
(2.102)
The symbol for the thermal capacitance is shown in Fig. 2.36, where the variables associated with the element are identified as heat flow H and temperature difference T across the tenninals. Because the temperature associated with thermal energy must be referenced to a fixed temperature, one tenninal of the symbol is shown connected to a fixed reference. The reference temperature nominally is absolute zero (where the stored energy is zero); however, in many engineering applications the temperature and energy difference from a specific fixed ambient level is of greater interest In these cases a fixed nonzero temperature is often used as the reference. For example, the temperatures and energy level in many systems are referenced to the ambient environmental temperature. For an object made of a pure substance the-thermal capacitance may be expressed as (2.103) where CP is the specific heat of the substance and m is the mass of subs~ce present
Sec. 2.6
Thennal System Elements
55
Thermal Inductance: No significant physical phenomenon has been observed that corresponds to energy storage due to heat flow in a "thermal inductor.n Thus only one thermal energy storage element, the thermal capacitance, is defined. Pure Thermal Resistance Resistance to heat flow is characterized in terms of a pure thermal resistance in which the heat flow is a single-valued monotonic function of the temperature difference T between two points:
q = :F(T)
(2.104)
A number of thermal phenomena are characterized by thermal resistance, including heat transfer by conduction, convection, and radiation [10, 11 ]. In an ideal thermal resistance the relationship between heat flow and the temperature difference is linear: . (2.105) where CD is the heat conductance (KIW) or where
e-N/K) and the reciprocal R,
is the thermal resistance
T = R,q
(2.106)
1 R,=-
(2.107)
with the ideal thermal resistance simply
CD
The symbol for thermal resistance is shown in Fig. 2.37 with the temperature difference T across the terminals and heat flow q identified as the variables associated with the element. q
T2 q
R1
.T1
~ 2 1
T= T2 -T1
T Temperature
Figure 2.37: Definition of the pure thermal resistance.
A pure thermal resistance neither stores nor dissipates energy since the net heat flow into the element is zero, that is, the heat flow into the resistance equals the heat flow out of the element The thermal resistance simply acts to impede heat flow and is associated with a change in system thermodynamic entropy [2]. The values of thermal resistance depend on material properties and geometry and are determined for three heat transfer mechanisms · in the following paragraphs.
56
Energy and Power Flow in State-Determined Systems
Chap. 2
Conduction Heat 1hmsfer: The conduction of heat through a material such as a metal block was characterized by Jean Baptiste Joseph Fourier (1768-1830) [11] as (2.108) where T1 and T2 are the temperatures of the two sides of the block. For a uniform material, the conduction constant CD for the geometry shown in Fig. 2.38 is CD= PeA
I
(2.109)
where Pc is the thermal conductivity (W/K-m), A is the area of ihe body (m2 ), and l is the length of the body (m). The temperature difference (T2 - T1) across the body is the driving force for the heat flux. Materials with high thermal conductivities, such as metals, are classified as thermal conductors, while those with low conductivities such as wood and plastic foam are known as insulators. The thermal conductivity of various materials is tabulated in [ 11]. Conduction heat transfer occurs in solid, liquid, and gaseous states. It is usually the dominant form of heat transfer in stationary fluids and gases and in solids at moderate temperatures. In moving fluids heat transfer by convection may dominate, and at high temperatures radiation heat transfer may also become important Convection Heat Transfer: In moving fluids heat may be transferred between spatially separated regions by bulk transport of the fluid. This mechanism is known as convective heat transfer and is directly related to the fluid motion. Convection may be natural (when the fluid moves on its own as it is heated) or forced (when an external source is used to move the fluid). Heat transfer between a moving fluid and the walls containing the fluid has been studied extensively [11] and characterized by an overall convection heat transfer coefficient and the area of contact between the fluid and the walls: (2.110) where A is the contact area (m2 ) and Ch is the convection heat transfer coefficient (W/m2 - K). The equivalent thermal resistance associated with convection is 1
R-t - ACh
(2.111)
The convection heat transfer coefficient is strongly dependent on flow geometry and flow velocities. Radiation Heat Transfer: Heat may also be transferred from one body to a second body at a distance by thermal radiation. This type of heat transfer is described by the StefanBoltzmann law [11]: (2.112) where C, is the radiation heat transfer constant (W/K4 ) and T1. and T2 are the absolute temPeratures of bodies 1 and 2 (K). The constant C, depends on the geo~etry of the two
Sec. 2.6
Thermal System Elements
57
Plates at unifonn temperature
Heat flow --H-:----H-~
q
Material with thermal conductivity Pc (a) Heat transfer by conduction
~
T2
T1
T=T2 -T1
,~;
.' ~ \
Heat flow
~-~·~q
~~
~ , ""'-Fluid in motion transfers heat (b) Heat transfer by convection
T2,
~
J
~
I
! ~
~ ~
r,
T= T2 -T1
Heat flow ~q
~
~ ~
~
Heat transfer depends on the emissivity and absorption characteristics of surfaces
(c) Heat transfer by radiation
Figure 2.38:
Modes of heat transfer.
surfaces and material properties that influence the absorption and emission characteristics of the bodies as a function of wavelength. Radiation heat transfer is highly nonlinear as indicated in Eq. (2.112) and in fact cannot be defined in terms of pure thermal resistance since the heat transfer depends on the differences between absolute temperatures raised to the fourth power. Thus, care must be used in modeling thennal systems in which heat transfer by radiation is significant since radiation is not a direct function of the difference in temperatures.
58
Energy and Power Flow in State-Detennined Systems
Chap. 2
Source Elements
1\vo ideal thermal sources may be defined as shown in Fig. 2.39. An ideal temperature source is an element in which the temperature at a system port is an independently specified function of time Ts (t) without regard to the heat flow necessary to maintain the temperature. An ideal heat flow source is an element in which heat flow is an independently specified function of time Qs(t). The temperature at the port is determined by the system connected to the heat flow source. 2
2
QJ.t)
TJ.t)
(a) Heat flow source
(b) Temperature source
Figure 2.39:
Ideal thermal source elements.
Example 2.5 An electric resistor with a resistance of 10 f2 and a thennal capacitance of 0.1 J/K is subjected to short-duration (0.01-s) voltage pulses of magnitude 100 Vas shown in Fig. 2.40. The peak temperature rise of the element resulting from the conversion of electric energy to heat is to be detennined. The time required for the element temperature to decay to within 1 K of its original temperature is also required as an estimate of the minimum time period between pulses that will allow the temperature to moderate (so that the element will not burn out after a few pulses). i(r) V(t)
100 V(t)
0
0.01
t(s)
Figure 2.40:
Temperature variation in an electric resistor.
It is assumed that in the short period of the pulse duration no significant heat is transferred from the element to the environment The te~J~perature rise due to a single pulse may then be estimated ~y assuming that all the electric energy is converted to heat and stored in the element During the pulse, the power Pis constant: 1
'P= -v2 R
2
100 = --· = 10
1000W
(i)
Sec. 2.6
59
Thennal System Elements
The power flow occurs over a period of 0.01 s, generating total electric work of
W= 101 The heat transferred is therefore H respect to the initial temperature is
ll.T
(ii)
= 10 J. For the element the temperature rise ll.T with
H 10 = -c, = -0.1 = 100 K
(iii)
The resistor's temperature is therefore raised 100 K above the ambient environmental temperature. After the rapid temperature rise, heat transfer occurs between the element and the environment. As stored heat is transferred to the environmen4 the temperature of the element decreases. The rate of the temperature decay may be found from Eq. (2.1 02): (iv) where T is the temperature of the resistive element above the environmental temperature. Assume that the thennal resistance characterizing heat transfer to the surroundings is R, = 10 K/W: (v)
Combining Eqs. (iv) and (v), the equation for the temperature difference as a function of time is obtained as
C, dTr = _ Tr dt R,
(vi)
Equation (vi) may be rearranged and integrated to determine the time required to reach Tr = 1 K above the initial temperature:
dTr
dt
-=--Tr R,C,
1
1"
dTr= - -11dt Tr R,C, o T1 lf ln-=--To R,C,
Tf
To
(vii) (viii) (ix)
For To= 100 K and Tt = 1 K, the time is
t1
= -R,C, In
io = -10 x 0.1 x ln(0.01) = 4.6 s
(x)
Thus it takes 4.6 s for the tempera~ of the resistor to return to within 1 K of its original temperature.
Energy and Power Flow in State-Detennined Systems
60
Chap. 2
PROBLEMS 2.1. Consider a mechanical system in which a force F acts through an infinitesimal distance dx, where the force and the displacement are in the same direction. The infinitesimal amount of work Fdx. The power flow 'Pinto the system is 'P = dW fdt = Fdxfdt = Fv, where v done is dW is the velocity of the point at which the force is applied.
=
(a) Check the units of 'P = Fv to show that they are consistent {b) For an accelerating mass m moving at velocity v, show that the time rate of increase of stored energy is equal to Fv.
(c) For a spring with one end free, the force required to displace the spring an amount x from its rest length is F = Kx (Hooke's Law), where K is the spring constant Determine the amount of energy required to stretch the spring a distance x. 2.2. An external force applied to a point in a system, and the velocity of that point, are shown in Fig. 2.41. Assume that the velocity reference direction is the same as the force reference direction. Plot a graph of the power input to the system at this point, and a graph of the energy input to the system from timet= 0 to 3T. Force
2Fo
:-------------~
Fo..,__ _ _ ___.
Time 0~--------------------~----------,---~
-Fo/2 .__ ____________:[ _____________ ..~.. :r____.,.J3r
Velocity Vo
Tune 2T
3T
Figure 2.41: The force and velocity at a point in a system.
2.3. The velocity of a machine element over a 10 second period is:
v = 2.5t m/s
0
v = lOm/s
4
~
t
(a) If the element is an ideal damper with B = 10 N-s/m, determine the power absorbed as a function of time, and the total energy dissipated over the 10 s period. {b) If the element represents aerodynamic drag, approximated by the characteristic F = Cv2 , detennine the power absorbed as a function of time, and the total energy dissipated if C = 1.0 · N-s 2 /m2 •
Chap. 2
61
Problems
(c) In the time period t < 4 s, how does the power absorbed by the two dampers compare? Which damper dissipates the most energy in the 10 s period? (d) At a velocity of 20 rnls, which damper absorbs the most power? 2.4. Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained damage, while allowing the vehicle front end structure to deform and absorb impact energy at higher speeds. Consider a frontal impact test of a 1000 kg mass vehicle. (a) For a low speed test at 2.5 rnls, compute the energy in the vehicle just prior to impacl If the bumper is a pure elastic element, what is the effective design stiffness required to limit the bumper maximum deflection during impact to 4 em? (b) At a higher speed impact of 25 m/s, considerable deformation occurs. To absorb the energy the
front end of a vehicle is designed to deform while providing a nearly constant force. For this condition, what is the amount of energy that must be absorbed by the deformation [neglecting the energy stored in the elastic deformation in (a)]? If it is desired to limit the deformation to 10 em, what level of resistance force is required? What is the deacceleration of the vehicle in this condition? 2.5. Consider two mechanical springs: spring A is a simple linear spring with a characteristic F = K 1x, while spring B is a nonlinear hardening spring wbicb becomes stiffer as the deflection increases F8 = K2x 2 • (Hardening springs are often designed in this manner to prevent bottoming of the load.) (a) In a laboratory test a 100 N force was found to deflect both springs by 5 em. Fmd the values (and units) of K1 and K2. (b) Find the energy stored in each spring when a force of 100 N is applied to each.
(c) If the force is doubled to 200 N, find the deflection and energy stored in each spring. For what range of values of applied force is the energy stored in the nonlinear spring greater than that stored in the linear spring? 2.6. A computer hard disk stores data on a rotating cylindrical disk. Consider such a disk with a radius of 6 em and a mass of 0.02 kg. (a) What is the moment of inertia of the disk? (b) If the disk drive motor provides a torque of 0.1 N-m during spin-up, what is the rotational speed of the disk after 5 seconds? (c) A new disk design uses composite materials to make a disk of the same mass but with a radius 1.5 times that of the original disk. What is the inertia of the new disk? What motor torque would be required to spin the new disk to the same speed as the original in 5 seconds? 2.7. The torsional stiffness of a cylindrical shaft of diameter D and length lis 1r D4 K, = G32-l
where G is the shear modulus of the shaft material. Consider the problem in Example 2.2 for the case in which a steel shaft, 5 m long and 5 em in diameter, drives a cylindrical flyweel with a 30 em diameter and a thickness of 5 em at a rotational speed of 90 r/s. Steel has a density of p = 7.8 gm/cm 3 and a shear modulus of G = 83 GPa. (a) What are the values of the shaft stiffness and the flywheel moment of inertia? (b) What is the stored energy when the system is spinning?
(c) If, as in Example 2.2. the velocity input to the shaft is suddenly stopped, what will be (i) the maximum energy stored in the shaft and (ii) the maximum angular deflection of the shaft?
Energy and Power flow in State-Detennined Systems
62
Cbap.2
2.8. Consider the air-cored inductor shown in Figure 2.21. (a) For an inductor with a 1 em diameter and a l£mgth of 4 em, how many turns of wire are required to achieve an inductance of 0.1 mH? The penneability of air is JL 1.257 x 10-6 N/m.
=
(b) How much energy is stored in the coil when a current of 0.1 A is flowing?
(c) If an of the energy stored in the inductor is transferred to a capacitor, resulting in a voltage of 10 volts, what is the value of the capacitance? Assume the pennittivity of air is E 0.885 X 1o- 11
=
F/m. (d) If the capacitor consists of two parallel plates with an area of I cm2 , what distance between the
plates is required to achieve the capacitance computed in (c)? 2.9. Superconducting materials have essentially no resistance to current ftow below their critical temperature. An important application of superconductors is the generation of strong magnetic fields in applications such as magnetic resonance imaging (MRI) for medical diagnosis. For such an application a coil of 10 em diameter, and length 5 em is wound with 400 turns. (a) Compute the inductance of the coil. (b) How much energy is stored in the coil when a steady current of one ampere is flowing.
(c) When the machine is turned on, the current is ramped up linearly from zero to one ampere over
a period of two seconds and then held steady. Make a sketch of the tenninal voltage at the coil during and after the power-on phase. (d) What is the voltage across the coil when a current I
= 2 sin 4001 A is flowing?
(e) At room temperatute (above the critical temperature) the coil has a resistance of 1 ohm. What is
the voltage across the coil with the same current as in part (c)? 2.10. Electrostatic loudspeakers are sometimes used in high-quality audio systems. An idealized representation of an electrostatic speaker is given by a parallel plate capacitor with plates separated by a nominal distance h, aS shown in Fig. 2.42. One plate is rigidly mounted while the other, the diaphragm, can move. When a voltage is applied an electrostatic attractive force is generated between the plates, and the resulting movement x_ of the diaphragm generates acoustic waves. Rigid conductor Area=A
Movable diaphragm Area=A
l_ ___Ft._____ . . ~c:----~--0 h+x
+ + + + + ;
T /
Chargeq
Figure 2.42:
F
+ ++
V
J'\s__o i
An electrostatic loudspeaker.
Chap. 2
63
Problems
(a) Assuming that both plates are fixed and unable to move, compute the electrical energy stored and the force F on the plates as a function of the applied voltage V. (b) Find the capacitance as a function of the plate separation h + x. (c) Find the change in stored electrical energy if the plate separation is increased by a small amount dx while the applied voltage v is held constant
(d) Repeat (c) with the stored charge q held constant; that is, with i
= 0.
(e) For a small increase in the gap, from h to h + dx, assume that energy is conserved and equate
the mechanical work done to the change in stored electrical energy at constant charge; thereby compute F, the electrostatic force as a function of h, A, and v. 2.11. A pair of cylindrical vertical-walled tanks, open at the top, are used to store liquids. (a) The first tank has an area of I 0 m2 • H the tank contains water, compute th~ fluid capacitance and
the pressure at the bottom when the depth is 10m. What is the stored energy in the tank? (b) The second tank, with twice the area, contains a fluid with a density I.5 times that of water. What is the fluid capacitance of this tank? When the depth is I 0 m, what is the pressure at the bottom and the total energy stored in this tank?
2.12. The excavation for a reservoir used to store water is shown in Fig. 2.43. The reservoir bas base dimensions of 1000 m by 1000 m with vertical side walls and front and back walls that slope at 30° to the horizontal.
Figure 2.43: Excavation for a reservoir. (a) Derive the relationship between the pressure at the bottom of the reservoir and the volume of
fluid stored. (b) Determine the capacitance of the reservoir.
(c) If the height of the water in the reservoir is 20m, what is the energy stored? (d) If the water depth is doubled to 40 m, what is the energy stored?
2.13. We wish to estimate the capacitance of an automobile tire. (a) Estimate the volume of a typical automobile tire. (b) If the tire is pressurized to the car manufacturer's recommended value, determine the pressure
in SI units. (c) If we assume that air may be represented by a bulk modulus of 1.4 P where P is the absolute pressure, estimate the tire capacitance and the energy stored in the tire at its nominal pressure.
Energy and Power Flow in State-Determined Systems
64
Chap.2
2.14. Water hammer in pipes frequently occurs when the flow is interrupted by the closing of a faucet or valve. The energy stored in the fluid inertance is transferred to the pipe, generating a large pressure transient Consider a typical 1.0-cm internal diameter pipe, 50 m long. supplying water to a house. (a) Compute the ftuid inertance of the pipe. (b) If the fluid is flowing in the pipe with a velocity of 0.5 mls, what is the energy stored in the
moving fluid? (c) Assume that the flow velocity is decreased from its nominal value to zero linearly over a period of 0.05 s. Determine the maximum pressure generated. 2.15. Electric utility systems must be able to meet peak electricity demands at certain times of the day. One way in which these demands are met is to use pumped water storage. At night. when demand is low, water is pumped to storage reservoirs on a hill, and in the peak demand periods the water is used to generate power through hydroelectric turbines, as shown in Fig. 2.44. The tank is located on a hill with the tank bottom 300 m above the turbine. The tank area is 200 m2 •
300m
Control valve
Figure 2.44: Hydroelectric energy storage system.
(a) Compute the fluid capacitance of the tank. (b) The depth of the water in the tank is 100m. What is the pressure at the bottom of the tank'? What
is the pressure at the inlet of the turbine? (c) A valve at the turbine is automatically controlled to keep a constant flow rate to the turbine of 0.75 m3 /s. Detennine the pressure at the turbine inlet over a period of four hours as the tank powers the turbine. At the end of the four hour period, what is the depth of the water in the tank'? (d) If it is assumed that the energy conversion efficiency of the turbine is 75%, what is the electric power produced over the four hour period? What is the total electric energy produced over the four hour period? You may neglect any fluid losses in the pipe.
Chap. 2
References
65
2.16. In many situations it is important to be able to make good engineering estimates of the parameters of a system. Consider a thermal system consisting of a freshly brewed cup of coffee in an insulated mug. Estimate the volume of a typical coffee mug, and assuming that it is full of water just below boiling temperature, compute the thermal energy and thermal capacitance of the water. 2.17. Steel parts are often heat treated to improve their properties. A heat treatment furnace, consisting of an electric resistance heater and an insulated furnace housing, is used to heat a steel bar with a cross sectional area of 0.01 m2 and length of 0.3 m. Steel has a density of 7. 75 glcm3 and a specific heat of 486 Jlkg-K. (a) What is the thermal capacitance of the steel bar? (b) The electric element of the furnace has a resistance of 10 0. If the voltage v(t) to the unit is prescribed as v(t) = 40t V
0
~
t < 50s
= 2000 v
50 ::: t < 200 s
=OV
200::;ts
what is the electric power drawn as a function of time'? What is the total electric energy supplied to the furnace'? (c) If it is assumed that after 200 seconds, 65% of the electric energy has been converted into thermal energy in the bar, what is the temperature rise of the bar above the ambient temperature?
REFERENCES [1] Paynter, H. M., Analysis and Design ofEngineering Systems, MIT Press, Cambridge, MA, 1961. [2] Cravalho, E. G., and Smith, J. L., Jr., Engineering Thermodynamics, Pitman, Marshfield, MA, 1981. [3] Crandall, S. H., Karnopp, D. C., Kurtz, E. F., Jr., and Pridemore-Brown, D. C., Dynamics of Mechanical and Electromechanical Systems, McGraw-Hill, New York, 1968. [4] Baumeister, T., et al., Handbook for Mechanical Engineers, McGraw-Hill, New York, 1967. [5] Roark, R. J., and Young, W. C., Formulas for Stress and Strain, McGraw-Hill. New York, J975. [6] Feynman, R. P., Leighton, R. B., and Sareds, M.• The Feynman Lectures on Physics, AddisonWesley, Reading, MA, 1977. [7] Halliday, D., and Resnick, R., Fundamentals of Physics, John Wiley, New York, 1988. [8] Sabersky, R. H., Acosta, A. J., and Hauptmann, E. G., Fluid Flow: A First Course in Fluid Mechanics, Macmillan, New York, 1989. [9] Streeter, V. L., Handbook of Fluid Dynamics, McGraw-Hill, New York, 1961. [10] Lienhard, J. H., A Heat Transfer Textbook, Prentice Hall, Englewood Cliffs, NJ, 1981. [11] Ronsenow, W. M., and Hartnett. J. P., Handbook of Heat Transfer, McGraw-Hill, New York, 1973.
3
Summary of One-Port Primitive Elements
3.1
INTRODUCTION In Chap. 2 we examined elementary physical phenomena in five separate energy domains
and used concepts of energy flow, storage, and dissipation to define a set of lumped elements. These primitive elements form a set of building blocks for system modeling and analysis and are known generically as lumped one-port elements because they represent the spatial locations (ports) in a system at which energy is transferred. For each of the domains with the exception of thermal systems, we defined three passive elements, two of which store energy and a third dissipative element In addition in each domain we defined two active source elements which are time-varying sources of energy. System dynamics provides a unified framework for characterizing the dynamic behavior of systems of interconnected one-port elements in the different energy domains, as well as in nonenergetic systems. In this chapter the one-port elements developed in Chap. 2 are integrated into a common description by recognizing similarities between the elemental behavior in the energy domains and by defining analogies between elements and variables in the various domains. The formulation of a unified framework for the description of elements in the energy domains provides a basis for development of unified methods of modeling systems spanning several energy domains. The development of a unified modeling methodology is based on analogies between the variables and elements in different energy domains. Several different types of analogs may be defined. In this text we have chosen to relate elements using the concepts of generalized through- and across-variables associated with a linear graph system representation introduced by F. A. F.rrestone [1] and H. M. Trent [2] and described in detail in several textS [3-5]. This set of analogs allows us to develop modeling methods that are similar to well-known techniques for electric circuit analysis. The set of analogies we have selected is not unique, for example, another widely used analogy is based on the concepts of "effort" and "flow" variables in bond graph modeling methods developed by H. M. Paynter [6] and 66
Sec. 3.2
Generalized Through- and Across-Variables
67
described in D. C. Karnopp et al. [7]. These two methods lead to different analogies, both of which are valid. For example, in this text we consider forces and electric currents to be analogous, while in the bond graph method forces and electric voltages are considered to be similar.
3.2
GENERAUZED THROUGH- AND ACROSS-VARIABLES
Figure 3.1 shows a schematic representation of a single one-port element, in this case a mechanical spring, as a generic element with two terminals through which power flows to be stored, supplied, or dissipated by the element. This two-terminal representation may be thought of as a mechanical analog of an electric element, in this case an inductor, with two connecting "wires." Hall system elements are represented in this form, the interconnection of elements may be expressed in a common "circuit" structure and a unified method of modeling and analysis may be derived for this fonn known as a linear graph. In Fig. 3.lc the linear graph representation of the spring element is shown as a branch connecting two nodes.
(a)
(c)
(b)
Figure 3.1: Schematic representation of a typical one-port element. (a) A translational spring, (b) as a two-terminal element, and (c) as a linear graph element.
With the two-tenninal representation, one of the two variables associated with the element is a physical quantity which may be considered to be measured "across" the terminals of the element, and the other variable represents a physical quantity which passes ''through" the element. For example, in the case of mechanical elements such as the spring in Fig. 3.1, the two defined variables are v, the velocity, and F, the force associated with the element. The velocity associated with a mechanical element is defined to be the differential (or relative) velocity as measured between the two terminals of the element, that is, v = v2 - VI in Fig. 3. I; notice that it must be measured across the element. Figure 3.2 shows a simple system with the same spring connected between a mass m and an applied force source F(t). In Fig. 3.2b the connection has been broken, and so the forces acting on the spring and the mass may be examined. Assume that the force transmitted to the mass is Fm(t). Because the spring element is assumed to be massless, Newton's laws of motion require that the sum of all external forces acting on it must sum to zero, or F(t) - Fm(t)
=0
Summary of One-Port Primitive Elements
68
Chap. 3
(a) Mass and spring elements driven by an external force.
F(t)~
(b) Force is measured by inserting an instnlment in series with the elements;
velocity is measured by connecting an instnlment across an elemenL
Figure 3.2: Definition of through- and across-variables in a simple mechanical system.
In other words, Fm(t) = F(t), and the external force applied to the spring element is transmitted through the spring to the mass element connected to the other side. Another way of looking at this is to say that in order to measure the force in a mechanical element,
the element must be broken and a sensing device, such as a spring balance, inserted in series with the element as in Fig. 3.2b. Such arguments lead us to define elemental velocity v to be an across-variable and force F to be a through-variable in mechanical systems. Figure 3.3 shows a simple electric circuit consisting of a battery and a resistor. The elemental variables in the electric domain are current i and voltage drop v. In order to measure the current flowing in the resistor the electric circuit must be broken and an ammeter inserted so that the current flows through it. To measure the voltage drop associated with the resistor a voltmeter is connected directly across its terminals. Current is defined as the through-variable for electric systems, and voltage drop is the across-variable. We may extend the concept of through- and across-variables to all the energy domains described in Chap. 2. Of the two variables defined for each domain, one is an across-variable because it is a relative quantity that must be measured as a difference between values at the two terminals of a network element. The other is designated as a through-variable that is continuous through any two-terminal element. Once the choice of this pair of variables has been made, generalized modeling and analysis techniques may be developed without regard to the particular energy domains associated with a system. The through- and across-variables for each energy domain discussed in this book are defined as follows. Mechanical Systems: In both translational and rotational mechanical systems the velocity drop of an element is the velocity difference across its terminals. In the case of a translational mass or rotary inertia one terminal is always assumed to be connected to a constant-velocity inertial reference frame. The force or torque associated with an element is assumed to pass through the element The elemental across-variable is therefore defined to be the relative velocity of the two terminals, and the elemental through-variable is defined to be the force or torque associated with the element.
Sec. 3.2
Generalized Through- and Across-Variables
69
An ammeter measures current "through" an element.
I
I v
A voltmeter measures voltage "across" an element.
I I
Rgure 3.3:
Definition of through- and across-variables in an electric system.
Electrical Systems: In an electric element, for example a capacitor, at any instant a potential (or voltage) difference exists between the terminals and a current flows through the element The across-variable is therefore defined to be the voltage drop across the element, and the through-variable is defined to be the current flowing through the element Fluid Systems: In the fluid domain the pressure difference across an element satisfies the definition of an across-variable, while the volume flow rate through the element is a natural choice for the through-variable.
Thermal Systems: While not strictly analogous to the other domains, thennal systems may be analyzed by defining heat flow rate as the through-variable, and the temperature difference across an element as the across-variable. The definitions of across- and through-variables for the energy domains introduced in Chap. 2 are summarized in Table 3.1. In describing generic systems, without regard to a specific energy domain, it is convenient to define a set of generalized variables. The generalized across- and through-variables are introduced as Generalized across-variable: v Generalized integrated across-variable: x
=
Generalized through-variable: f
£' £'
Generalized integrated through-variable: h =
v dt + x(O)
fdt
+ h(O)
With the exception of thermal elements, the power 'P passing into a lumped one-port element in terms of the generalized variables is
'P=fv
(3.1)
and the work performed by the system on the element over time period 0 ::: t ::: T may be expressed in terms of the generalized variables as
W=
loT Pdt= loT fvdt
(3.2)
TABLE 3.1:
Definition of Across- and Through-Variables in the Various Energy Domains
System
Across-Variable, v
Through-Variable, f
Integrated Across-Variable, x
Integrated Through-Variable, h
Translational Rotational Electric Fluid Thermal
Velocity difference, v Angular velocity difference, 0 Voltage drop, v Pressure difference, P Temperature difference, T
Force, F Torque, T Current, i Volume flow rate, Q Heat ftow rate, q
Linear displacement, x Angular displacement, e Flux linkage, A Pressure difference momentum, r Not defined
Momentum, p Angular momentum, h Charge, q Volume, V Heat, H
Sec. 3.3
Generalization of One-Port Elements
71
For thennal elements, while an across-variable, temperature T, and a throughvariable, heat flow rate Q, may also be defined, the product is not power since Q is a power variable itself.
3.3
GENERAUZATION OF ONE-PORT ELEMENTS In each of the energy domains described in Chap. 2, several primitive elements are defined: one or two ideal energy storage elements, a dissipative element, and a pair of source elements. For one of the energy storage elements, the energy is a function of its acrossvariable (for example, an ideal mass element stores energy as a function of its velocity; £ = !m v2 ), while in the other energy storage element the stored energy is a function of the through-variable; in a translational spring the stored energy is £ = 2 ~ F 2 • The dissipative elements, which store no energy, and the source elements, which may supply energy or power continuously, complete the set of one-port elements. In this section these elements are classified into generic groups.
3.3.1 A-Type Energy Storage Elements Energy storage elements in which the stored energy is a function of the across-variable are defined to be A-type elements and are collectively designated generalized capacitances. All A-type energy storage elements have constitutive equations of the form
h = :F(v)
(3.3)
where h is the generalized integrated through-variable, vis the generalized across-variable, and :F() designates a single-valued monotonic function. In general Eq. (3.3) may represent a nonlinear relationship, but a linear (or ideal) A-type element has a linear form of Eq. (3.3):
h=Cv
(3.4)
where the constant of proportionality C is defined to be the ideal generalized capacitance of the element Differentiation of Eq. (3.4) gives the generalized A-type elemental equation
f=Cdv dt
(3.5)
The definition of the lumped elements in Chap. 2 shows that the capacitive elements are the translational mass, rotational inertia, electric capacitance, fluid capacitance, and thermal capacitance. The collection of A-type elements is shown in Fig. 3.4, and their elemental relationships are summarized in Table 3.2.
Summary of One-Port Primitive Elements
72
(a) TranslationaJ mass
(b) RotationaJ inertia
Pnt~ p
a-fl (d) fluid capacitance
Chap. 3
(c) Electric capacitor
q~ Trd' (e) Thennal capacitance
Figure 3.4: The A-type elements in the five energy domains descn'bed in this book.
TABLE3.2:
Summary of Elemental Relationships for Ideal A-Type Elements
Element
Constitutive Equation
Elemental Equation
Energy
Generalized A-type
h=Cv
Translational mass
p=mv
f=Cdv dt dv F=mdt
1 £= -Cvl 2 1 £= -mv2 2
Rotational inertia
h=JO
T=JdQ dt
£=
~JQ2
Electric capacitance
q=Cv
£=
-cvl
fluid capacitance
V=C1P
Thermal capacitance
H=C,T
i=Cdv dt dP Q=C,dt dT q=C,dt
£=
2
1
2
1 2 -c 1P
2
£=C,T
The two-terminal representation of A-type elements in systems often requires a connection to a known reference value of the across-variable. Figure 3.5 shows two A-type elements, a translational mass and a fluid capacitance. In a newtonian mechanical system the momentum of a mass element m is measured with respect to a nonaccelerating inertial reference frame h = m (Vm - Vref) where Vm is the velocity of the element and Vref is the velocity of the reference frame. Then differentiation gives dh d dvm F=- =m-(Vm -vrer) =m-dt dt dt
since Vref is constant. There is an implied connection to the reference velocity that defined the momentum, and one terminal must be connected to this reference value (usually assumed
Sec. 3.3
Generalization of One-Port Elements
73
to be zero velocity). Similarly, the angular velocity of an rotary inertia must be measured with respect to a nonaccelerating rotating reference frame.
~· ..~...,[
:
h
P= pgh relative toPref
Inertial reference frame Vrer (a)
(b)
Figure 3.5: Implicit connection of typical A-type elements to a reference node. (a) A translational mass, and (b) a fluid capacitance.
1 In the case of the fluid capacitance defined by a vertical walled tank., the constitutive relationship relating volume to pressure is V
= CJ (P -
Prer)
where Pref is the constant external pressure at the fluid surface and P is the pressure at the base of the tank. The elemental equation may be written Q
dV
d
dP
= -dt = CJ(P- Pret) = CJdt dt
if Pret is constant. The two-terminal representation requires an implicit connection to the reference pressure. Similarly, the temperature associated with a thermal capacitance is measured with respect to a fixed reference temperature. The electric capacitor, however, does not require connection to a fixed reference voltage and may have its two terminals connected to points of arbitrary voltage. With the exception of the thermal capacitance, the energy stored in a pure or ideal A-type element is given by
L h
£=
vdh
(3.6)
For an ideal element with a constitutive equation given by Eq. (3.4), the stored energy can be expressed as
(3.7) resulting in a form in which the energy is a direct function of the across-variable. For the ideal thermal capacitance the energy is simply £ = H = C, T and is a function of the across-variable.
74
Summary of One-Port Primitive Elements
Chap. 3
Example3.1
Show that an A-type element is capable of both absorbing and supplying power. Solution For an A-type element the instantaneous power flow is
dv 'P=fv=C-v dt
(i)
Our sign convention is that if 'P > 0, power is flowing into the element, while if 'P < 0, power is flowing from the element Thus the direction of power flow is defined by Eq. (i); if v and dvjdt have the same sign, the element is absorbing power and storing energy, while if the signs are opposite, the element is returning stored energy to the system. Consider a mechanical mass element Equation (i) states that the element accumulates energy whenever it is accelerated in the direction of its travel and returns energy as it is decelerated.
Equation (3.7) shows that any change in the stored energy in an A-type element results from a change in the across-variable. In order to change the energy in a stepwise fashion the across-variable must change instantaneously. Equation (3.5) shows that the through-variable is proportional to the derivative of the across-variable, and therefore an instantaneous change in the stored energy requires an impulse in the through-variable. (An impulse is a mathematical function of infinitesimal duration and infinite amplitude and is introduced formally in Chap. 7.) The stored energy in any A-type element cannot change instantaneously unless infinite power is available in the fonn of an impulse in force, torque, current, or volume flow. Physical energy sources are generally power-limited and are therefore incapable of providing an instantaneous change in the across-variable or stored energy of an A-type element. Figure 3.6 shows the relationships between across- and through-variables for an A-type element
Lf-GEJ-v~ 1=, f-liEr- ]L, v
(a) Generalized capacitance
(b) Response of capacitance to discontinuous changes
in the through-variable Figure 3.6: Across- and through-variable relationships in an ideal A-type elemenL
Example3.2
A satellite circling the earth every 90 minutes (min) is subjected to cyclic heating by the sun as it passes in and out of the earth's shadow. Measurements have shown that it is reasonable to model the net solar heat flow rate Q(t) into the satellite as a cosinusoidal function with the orbital period, assuming that at time t = 0 the satellite is at the position of peak sunlight Find the time in the orbit at which the internal temperature within the satellite is a maximum.
Sec. 3.3
Generalization of One-Port Elements
75
Solution Let the time t be measured in seconds so that the heat flow rate is Q(t)
= Qmax COS ( 90': 60 t)
(i)
where Qmax is the peak heat flow rate (J/s). The satellite is modeled as a lumped thermal capacitance C, and stores energy as an A-type thermal element. For a general A-type element the elemental equation is (ii)
and for the thermal capacitance the relationship is dT
Q=C,-
(iii)
dt
In this case we require the value of T(t), given Q(t), and so Eq. (iii) must be written in integral form: T(t)
l('
= C, Jo
Q dt
+ T(O)
2n ) = C,1 Jo(' Qmax cos ( 5400 t
dt
(iv)
+ T(O)
. ( 27r ) = 5400Qmax 27rC, sm 54001 + T(O) The system input Q(t) and the response T(t) are shown in Fig. 3.7. The temperature and the input heat flow are not in synchrony; the response lags the input by one-quarter of a cycle. Since sin8 is a maximum when 8 = n/2, T(t) is a maximum when 2ntj5400 = Ir/2 or when t = 1350 s. The satellite therefore reaches its maximum temperature 22.5 min after passing the point of maximum brightness, and the maximum temperature is T. max
at
e a
~
e
,g
~
= 5400Qmax 27rC,
+
T(O)
(v)
o.s
l:l!!
]]
0
:a :a ~ e -o.s 0 0 zz
Tune (s)
-1
Figure 3.7: The input heat flow rate Q(t) and the temperature response T(t) of the satellite.
Summary of One-Port Primitive Elements
76
Chap. 3
3.3.2 T-Type Energy Storage Elements
Energy storage elements in which the stored energy may be expressed as a function of the through-variable are designated as T-type elements and are collectively known as generalized inductances. The T-type energy storage elements are defined by generalized constitutive equations of the form (3.8) X= F(f) where x is the generalized integrated across-variable, f is the generalized through-variable, and F() designates a single-valued monotonic function. For a linear, or ideal, T-type element the constitutive relationship Eq. (3.8) reduces to a simple linear equation X=
Lf
(3.9)
where the constant of proportionality L is defined to be the ideal generalized inductance. Differentiation of the constitutive equation gives the generalized elemental equation df
V=L-
(3.10)
dt
Figure 3.8 shows the four T-type elements defined in Chap. 2; there is no known thermal energy storage phenomenon that defines a T-type element for thermal systems. The generalized inductance is equivalent to the reciprocal of the mechanical translational and rotational spring constants and is equivalent to the electric inductance and the fluid inertance. Table 3.3 summarizes the elemental relationships forT-type elements. v2
v1
a.~
~K~
F~~....,_F 2 1 (a) Translational spring
~~-T T~·
£
(b) Torsional spring
P 1
; ~
v2 L VJ 1 o----rnT'-----0 ~ 2 1 (c) Electric inductance
~.-,-Q ~~ p,
Q
2 (d) Fluid inertance
Figura 3.8: The T-type elements in the energy domains described in this book. There is no T-type element for the thermal domain.
The energy stored in a T-type pure or ideal element is given by fi= fox fdx
(3.11)
For an ideal element with a constitutive equation ofEq. (3.11), the energy is a direct function of the through-variable f: (3.12)
Sec. 3.3
77
Generalization of One-Port Elements TABLE 3.3:
Summary of Elemental Relationships for Ideal T-type Elements
Element
Constitutive Equation
Generalized T-type
X=
Elemental Equation v = L df/dt
Lf
Translational spring
1 x = -F K
Torsional spring
9=-T K,
Electric inductance
').=Li
Fluid inertance
r= I1 Q
1
Energy E=
1
-Lf 2
v=--
1 dF K dt
E= _l_Fz 2K
O= _!_dT
E= _l_Tz 2K,
K, dt di v=Ldt dQ P=/1 dt
£= .!_Li 2 2 1
£= -I,Q 2
2
As in the case of an A-type element, it is not possible to change the stored energy or the through-variable in aT-type element instantaneously without an infinite source of power. Example3.3 It is commonly observed in electric circuits containing inductances that when a switch is opened a brief electric arc may develop across the air gap, causing the switch contacts to become pitted. In severe cases arcing may occur between the turns of the coil itself, causing breakdown of the electric insulation and perhaps destruction of the inductor. Explain why this arcing occurs. Solution Consider the circuit shown in Fig. 3.9. An inductor is aT-type element and has an elemental equation ·
di v=Ldt
(i)
If a current i is flowing just before the switch is opened, the energy stored in the magnetic field of the inductor is = Li 2 • When the current is interrupted, the magnetic field "collapses" and the stored energy must be either returned to the system or dissipated. The rapid change in the magnetic flux as the field decays generates a large inductive voltage in the coil. This induced voltage is sufficient to cause an arc, a short current pulse across the gap, that is potentially damaging to the switch and the coil itself. The inductive back electromotive force (emf) attempts to maintain the current through the coil so as to dissipate the stored energy.
e !
-r1 I
I
v
L
I ..........._
Figure 3.9: An electric circuit containing an inductance.
Summary of One-Port Primitive Elements
78
Chap. 3
This phenomenon may be described in terms of the elemental equation Eq. (i). An attempt to decrease the current instantaneously creates a large negative value of the derivative di1dt, generating a correspondingly large value of the across-variable v. The arcing allows the cwrent to continue briefly after the switch is opened and therefore to decay in a finite time. In practice engineers often connect semiconductor diodes or capacitors across inductors to provide an alternate current path and reduce inductive voltage spikes and arcing.
3.3.3 0-Type Dissipative Elements The elements that dissipate energy are collectively known as D-type elements. They are defined by an algebraic relationship between the across- and through-variables of thefonn
v
= F(f)
or
f
= :F 1(v)
(3.13)
where f and v are the generalized through- and across-variables, respectively. For linear (ideal) dissipative elements the relationship is commonly expressed in one of two forms: 1 (3.14) v Rf . or f=-V R
=
where R is defined to be the generalized ideal resistance. It is also common to define the conductance G = 1/R as the reciprocal of the resistance and to write Eq. (3.14) as
f=Gv
or
1
V=-f G
(3.15)
The generalized resistances are equivalent to the reciprocals of the mechanical and rotational damping constants and are equivalent to the electric, fluid, and thennal resistances. For all D-type elements except the thennal resistance element, power supplied to the element is converted to heat and dissipated. For the ideal elements the power may be expressed as (3.16) The power 'P is always a positive quantity and flows into a D-type element. In the thermal D-type element power is not dissipated In this case, because the through-variable is power, the element simply acts to impede heat flow. Table 3.4 summarizes the algebraic D-type relationships for resistances. The dissipative elements store no energy, and instantaneous changes in the power dissipated by these elements are associated with instantaneous changes in the through- and across-variables as indicated by the ideal elemental equation in which the through- and across-variables are directly related by the constant R.
Sec. 3.3
Generalization of One-Port Elements TABLE 3.4:
79
Summary of Elemental Relationships for Ideal 0-Type Elements
Element
Elemental Equations
Power Dissipated
Generalized D-type
1 f= -v R
P=
Translational damper
F=Bv
Rotational damper
T = B,rl
Electric resistance FJuid resistance Thermal resistance
1 R 1 Q=-P Rf 1 q=-T R, i = -v
V=
Rf
1 v=-F B 1 0=-T B,
1
-Vl = Rf 2
R 1 P= Bv2 = -F2 B
1 P= 8,02 = -T2 B,
P= .!.v2 = Ri 2 R
v= Ri P=RJQ
P=
1
-p2
Rf
= Q2RJ
T=R1q
Example3.4 An electric resistance of value R is connected to a voltage source supplying a sinusoidal voltage of the fonn V(t) = Vm sin(wt), as shown in Fig. 3.10. Find the average power dissipated in the resistor over one period of the voltage input.
Solution The sinusoidal applied voltage V(t) repeats itself with a period T = 27r/w seconds. The instantaneous power dissipated in the resistance is (i)
The average power dissipated over one period T is found by integrating the power over one period and dividing by the period:
(ii)
This expression shows that the average power dissipated in R over one period of the sinusoidal voltage is the same as would be dissipated by a constant applied voltage of value v = Vm 1.Ji = 0.101Vm volts.
i
=;
sinwt
V.,sinwUR Figure 3.10: An electric resistance.
Summary of One-Port Primitive Elements
80
Chap. 3
3.3.4 Ideal Sources In each energy domain two general types of idealized sources may be defined:
• The ideal across-variable source in which the generalized across-variable is a specified function of time f(t), Vs(t)
= /(t)
and is independent of the through-variable • The ideal through-variable source in which the generalized through-variable is a specified function of time Fs(t)
= /(t)
and is independent of the across-variable. An example of a through-variable source is an idealized positive displacement pump in a fluid system, in which the flow rate is a prescribed function of time and is independent of the pressure required to maintain the flow, while an example of an across-variable source is a regulated laboratory electric power supply in which the output voltage is independent of the current drawn by the circuit to which it is connected. These ideal sources are not poweror energy-limited and theoretically may supply infinite power and energy. The symbols for the ideal sources are shown in Fig. 3.11, where in the through-variable source the arrow designates the assumed positive direction of through-variable flow and in the across-variable source the arrow designates the assumed direction of the across-variable decrease or drop. For each source type one variable is an independently specified function of time.
(a) Through-variable source Figure 3.11:
(b) Across-variable source
Idealized source elements.
The value of the complementary variable of each source is determined by the system to which the source is connected. A source may provide power and energy to a system or may absorb power and energy, depending upon the sign of the complementary source variable. Table 3.5 defines the source types in each of the energy domains.
81
Generalization of One-Port Elements
Sec. 3.3
TABLE 3.5:
Definition of Ideal Sources
Energy Domain
Across-Variable Source
Through-Variable Source
Generalized Mechanical translational Mechanical rotational Electric Fluid Thermal
Across-variable, V,(t) Velocity source, V6 (t) Angular velocity source, 0,~(1) Voltage source, V,.(t) Pressure source, P.r(t) Temperature source, T, (t)
Through-variable, F, (t) Force source, F,(t) Torque source, Ts(t) Current source, /,r(l) Flow source, Q,(t) Heat flow source, Q 6 (t)
Example3.5 A force source is used to accelerate and decelerate a mass in a cyclic motion as shown in Fig. 3.12. The force source provides a square wave in force cycling between values of +Fo and - F0 with a total cycle time of To. In this example the velocity of the mass as a function of time and the power flow into the mass as a function of time are to be detennined. As shown in Fig. 3.12. the mass velocity is defined as positive when the force is positive. The velocity of the mass m is detennined from the elemental equation
dv Fs=mdt
(i)
The problem solution may be found by solving Eq. {i) in each fraction of the total time. Over the time period 0 ~ t < To/4. the elemental equation may be expressed as
Fodt = mdv
(ii)
and integrated to yield
v
1 = -Fot m
(iii)
Over the period To/4 ~ t < 3T0 j4, the equation may also be integrated, noting that at time t = To/4 the mass has velocity v0 , yielding
(-Fo)dt
= mdv
(iv)
and integrated to yield v(t)
To) = vo - Fo m (t - 4
Over the period 3T0 /4 ~ t < To. the equation may be integrated, noting that at timet the velocity is -vo:
Fodt
= mdv
(v)
= 3To/4 . (vi)
and integrated to yield
Fo ( t ·- 4 3To) v = -vo + -;;
(vii)
82
Summary of One-Port Primitive Elements
FR
Fo -Fo
Force
.._ _ _
Chap. 3
,.-:
__.li'o •
t
~~ .. ~ot
-vo
Rgure 3.12: A mass element driven by a force source.
Using the results from integration of the elemental equation, the velocity of the mass is plotted over one period of time To in Fig. 3.12. For subsequent periods of time the velocity may ·be determined in a similar fashion. The velocity curve is a sawtooth function, that is, an alternating series of linear curves with positive and negative slopes with values of ±Fo/m, the mass acceleration. The maximum and minimum velocities are
0.25FoTo vo = ± m
(viii)
The power delivered to the mass is (ix)
and may be determined by multiplying the force and velocity curves together as shown in Fig. 3.12. During the period 0 to To/4, the source provides positive power to the mass, accelerating it in the positive direction. In the period To/4 to To/2. the force source opposes the motion of the mass, absorbing power and decreasing its velocity, and then in the period To/2 to 3T0 f4, the negative force results in a velocity that has increasingly negative values and again supplies power to the mass. During the period 3T0 /4 to To, the force is in the positive direction and the mass velocity is negative, and so the source absorbs power from the mass. Over a full cycle of period T0 , the integral of the power supplied by the source is zero; for half of the cycle the source supplies energy, while for the other half it absorbs the kinetic energy stored in the mass.
3.4
CAUSALITY Each of the primitive elements is defined by an elemental equation that relates its throughand across-variables. This equation represents a constraint between the across-variable and the through-variable that must be .satisfied at every instant. An immediate consequence is that the across-variable and the through-variable cannot both be independently specified at the same time. One variable must be considered to be defined by the system or an external input, and the other variabJe is defined by the elemental equation. This is known as causality.
Sec. 3.5
83
Linearization of Nonlinear Elements
In the energy storage elements the constraint is expressed as a differential or integral relationship that defines the element as having integral or derivative causality. For example, a mass element m has an elemental relationship that is normally written in the form dv F=mdt
If a mass element is driven by a defined velocity v(t), the required force F is detennined by the above elemental equation; solution for the through-variable F (t) requires differentiation of the velocity v, and the element is said to be in derivative causality. On the other hand, if the element is driven by a specified force F(t), its resulting velocity is determined by rewriting the elemental equation: dv dt
= !_F m
or
v(t)
= -m1 f.to F dt + v(O)
which is known as the integral causality fonn. In Example 3.3.2 the thermal capacitance of the satellite is in integral causality because the heat flow rate is specified by the solar flux. Dissipative elements always operate in algebraic causality because the through- and across-variables are related by algebraic equations. The concept of causality becomes important in developing models of systems of interconnected elements. When an element is part of an interconnected system, its causality is determined by the system structure. It will be shown later that all independent energy storage elements in a system can be expressed in integral causality.
i.5
LINEARIZATION OF NONLINEAR ELEMENTS In many physical systems the constitutive relations used to define model elements are
inherently nonlinear. The analysis of systems containing such elements is a much more difficult task than that for a system containing only linear (ideal) elements, and for many such systems of interconnected nonlinear elements there may be no exact analysis technique. In engineering studies it is often convenient to approximate the behavior of nonlinear pure elements by equivalent linear elements that are valid over a limited range of operation. In many practical situations an element operates at a nominal, nonzero value of its through- or across-variable and is subjected to small deviations about this equilibrium value. For example, the springs in the suspension of an automobile may be inherently nonlinear over the full range of operation, but in nonnal use they are subjected to a nominal load force of the weight of the car, with "small" perturbation forces superimposed by the normal road conditions. We may, with care, use a linearized model of the spring that is valid over a limited range of operation. While any dynamic analyses based upon such models are at best an approximation to the behavior of the real system, for preliminary analyses such mode]s frequently capture the dominant features of the overall system response.
Summary of One-Port Primitive Elements
84
Chap. 3
Assume that a pure element is operating with an equilibrium value vo of its acrossvariable or fo of its through-variable. For small deviations about these values a pair of incremental variables v• and f* may be defined: (3.17) (3.18)
v• =v-vo
f*
= f- fo
Similarly~ if under equilibrium conditions one or both of the integrated through- or acrossvariables is constant with a value ho and xo, respectively, incremental values may be defined as perturbations from the nominal values:
= h- ho
(3.19)
x*=x-Xo
(3.20)
h*
The linearized elemental behavior is defined in tenns of these incremental variables.
A-Type Elements The A-type element defined in Eq. (3.3) has a single-valued monotonic relationship between the integrated through-variable and the across-variable, that is, h = F(v)
(3.21)
Under e.quilibrium conditions both h and v are constant with values ho and vo. When v is perturbed from equilibrium, the nonlinear function .r(v) may be expressed as a Taylor series about vo: h=
dF(v) .r (v)lv=Vo + dv
I .
I
(v- vo) V=Vo 2
2
I
d F(v) 2 (v - vo) + .. · 2. dv2 V=Vo
+ 9I
I .
(3.22)
dF(v) 1 d F(v) 2 =ho + - v+v+··· dv V=Vo 2! dv2 V=Vo
For small changes in v, v* is small and higher-order terms in the series may be neglected. If second and higher terms may be neglec~ only the first two tenns of the series are retained and an approximate linear relationship results: h - ho
I
~
dF(v) v• dv V;;No
(3.23)
= c•v•
(3.24)
I
(3.25)
or h*
where
c• =
dF(v) dv V=Vo
Sec. 3.5
Linearization of Nonlinear Elements
85
(a) A-type elements
(b) T-type elements
Agure 3.13: Linearization of constitutive relationships for A-type and T-type elements.
Equation (3.25) is a constitutive relationship for an ideal A-type element with capacitance C* and represents the elemental behavior of the nonlinear element in the region of the equilibrium point. The linearized generalized capacitance C* is the slope of the constitutive characteristic at the operating point, as shown in Fig. 3.13a. This linear approximation is used to define the elemental equation of an equivalent linear A-type element in the region of the equilibrium point by differentiation:
f* = dh* ~ c•dv* dt
(3.26)
dt
The linearized elemental equation may be used as an approximation to the behavior of the nonlinear element. Example3.6 A conica1 tank with angle 60° at the base drains through an orifice into the atmosphere as shown in Fig. 3.14. In normal operation the tank contains a fluid volume Vo. Find an expression for a linearized fluid capacitance that may be used to represent the tank for sma11 deviations about its nominal operating point.
Figure 3.14: A nonlinear fluid system and its linear graph.
Summary of One-Port Primitive Elements
86
Chap. 3
Solution Consider an elemental disk of fluid of width dh at a height h above the base. The radius of the disk is r = h tan(n'/6) = hf.J3. Its volume dV is (i)
If the tank is fi11ed to height h, the total volume of fluid V stored is
(ii)
and the pressure at the outlet is P acceleration due to gravity. Then
= pgh, where p is the density of the fluid and g is the 9)1/3
p = ( 1f
pgVI/3
(iii)
or
v
= _1C_p3 9 (pg)3
(iv)
which is the constitutive relationship of a pure but nonideal A-type fluid element. At the operating point V = Vo, and the corresponding pressure at the base of the tank is Po, which may be found directly from Eq. (iii). The equivalent linear fluid capacitance c• is found by differentiating Eq. (iv):
c• =
I
dV dp P=Po
= 3-1C-PJ 9 (pg)3
(1C
(v)
- 3 )1/3 2/3 --Vo pg
9
The equivalent linear elemental equation is
(vi)
T-Type Elements Nonlinear pure T-type elements may be linearized in a similar manner. Equation (3.8) defines a T-type element as a single-valued monotonic relationship between the integrated across-variable and the through-variable: X=
F(f)
(3.27)
Sec. 3.5
Linearization of Nonlinear Elements
87
If there is a nominal operating point defined by >
(3.28)
The first two tenns may be used to define an approximate linear relationship:
d:F(f) I f* df f:::fo
•
X =X-Xo~ - -
(3.29)
An elemental relationship may be found by differentiating both sides: v*
~ d:F(f)
I
df
df* = L• dt* dt
f=fo dt
(3.30)
where L* = d:F(f) I df f=fo
(3.31)
is a linearized generalized inductance representing the elemental behavior of the pure element at the equilibrium point Figure 3.13b shows the linearizing approximation of the constitutive relationship at the operating point
Example3.7 The measured force-extension characteristic of a spring has been found to closely approximate F = 0.125 x l06 x 3 • In its normal operating mode the spring is subjected to a static load F0 with a small sinusoidal force superimposed. Find the equivalent linearized stiffness of the spring. Solution The stiffness of a spring is the reciprocal of the generalized inductance. The constitutive relation may be rewritten X=
2
X
Then I
K•
=
dx dF
- 2
-3
X
10-2 F 113
(i)
I F=Fo
(ii)
I o-2 F.-213 0
or (iii)
88
Summary of One-Port Primitive Elements
Chap. 3
0-Type Elements D-type elements are characterized by an algebraic relationship between the the across- and through-variables:
v = F(f)
(3.32)
The nonlinear function may be expanded as a Taylor series, and the linear terms retained to fonn an approximation to the elemental behavior:
v• = v - vo
~
dF (f) dx
I
f*
(3.33)
f=fo
Then
v•
~
R*f*
(3.34)
where (3.35)
is a linearized resistance. An expression for a linearized conductance G* may be developed similarly. The linearization of lumped elements is summarized in Table 3.6. TABLE3.6:
Summary of Linearized Lumped-Parameter Elements
Element
Constitutive
Linearized Elemental Equation
A-type
h = J='(v)
r =C* dv*
T-type
X= J='(f)
v• = L* df"
D-type
v =.?='(f)
v* = R*f"
dt
dt
Elemental Value
C* = dF(v) dv L* = dF(f) df
IVc.Vo If=fo I
R* = dF(f) df f=fo
Example3.8 A set of measurements made on a test vehicle traveling along a straight road showed that the aerodynamic drag force is approximately described by a quadratic relationship (i)
where c0 is an overall drag coefficient and v is the velocity. In its normal operation the vehicle is known to travel at a nominal speed v0 but is subjected to small variations in this speed. Find a linearized D-type element that approximates the behavior of the drag force for vehicle speeds that are close to v0 •
Chap. 3
89
Problems
Solution The aerodynamic drag is a pure dissipative element which may be expressed as an equivalent nonlinear damper: (ii) The linearized representation of this damper is
F; = Bv*
(iii)
where v* = v - vo and FJ = Fd - Fo represent excursions from the nominal operating point. The value of the equivalent linear damper coefficient B* is
B*- dFd
I
dv ll=vo
-
(iv)
= 2covo The value of the linearized damper coefficient B* is directly proportional to the equilibrium velocity and at high velocities is relatively large, while at low velocities it is relatively small. The value of the drag force computed by Eq. (iii) is the excursion from the nominal operating value, and the total drag force acting on the vehicle is given by Fd:::::::
Fo + B*v*
::::::: cov~
+ B* (v -
(v)
vo)
PROBLEMS 3.1. A mass of 4 kg and a spring with stiffness 50 N/m are connected as shown in Fig. 3.15. The mass is displaced so as to compress the spring by 5 em and then released Displacement ..,.._
_. .. _x(t)
Figure 3.15: A mass-spring system.
(a) Compute the energy stored in the spring at the time of release. (b) What is the maximum velocity the mass will achieve? (c) Assume that as the spring expands it breaks when the tension reaches 2 N. What is the velocity of the mass at the moment after the break? What is the energy of the mass just prior to and just after the break?
Summary of One-Port Primitive Elements
90
Chap. 3
3.2. Failure of a component in an electrical system may generate damaging transient voltages and currents that may destroy other components. In analyzing failure modes of circuits it is often of interest to determine the state of the remaining components. Assuming that a single circuit element fails by open or short circuit, generating a transient current or voltage, and that the nonfailed elements do not change their energy storage at the moment of failure, what can you state about: (a) The currents and voltages associated with inductors in the circuit? (b) The currents and voltages associated with capacitors in the circuit? (c) The currents and voltages associated with resistors in the circuit? 3.3. Generalize the answers to Problem 3.2 to failure conditions in a generalized system. What can you state about the instantaneous values of through- and across-variables immediately after a system transient for: (a) An A-type element? (b) A T-type element? (c) A D-type element? 3.4. In many cases, the parameters that describe an ideal element are determined from experimental measurements. Consider placing a source that provides a sinusoidal input across an element. as shown in Fig. 3.16. Through variable
Source
Ideal
element
Across variable
~~ Figure 3.16:
A source connected to an ideal element
(a) The source provides a sinusoidal across-variable v(t) = v0 sin(wt). Determine and sketch the resulting through-variable f(t) if the element is (i) an A-Type, (ii) aT-type, and (iii) aD-type element. (b) The source provides a sinusoidal through-variable f(t) = fo sin(wt). Determine and sketch the resulting across-variable v(t) if the element is (i) an A-Type, (ii) aT-type, and (iii) aD-type element. 3.5. In many applications, engineers must decide how to store energy in a system. If we wish to store 100 joules of energy, find the value (in SI units) of the element in each of the following systems: (a) A pneumatic system in which compressed air is stored in a tank at a pressure of 10,000 N/m2 • Assume the bulk modulus of the compressed air is 1.41 P 1. (b) A rotational flywheel spinning at 10,000 rad/s. If the flywheel is made of steel plate with a thickness of 2 c~ what diameter is required? (c) An air dielectric parallel plate capacitor, charged to a voltage of 1000 v with plates spaced 0.1 em apart. What plate area is required? How do the energy densities-that is, the energy stored per unit volume-compare for each of these systems. 3.6. Electric cars have been proposed as a method of easing air pollution in urban areas. One proposal to make cars more efficient is to use regenerative braking, in which the motor is turned into a generator and used to charge the batteries during braking. The kinetic energy of the car is thus converted to electrical energy and stored for later use. Consider a car with a mass of 1000 kg.
Chap. 3
91
References
(a) If the car is traveling at 30m/sand regeneratively braked to a standstill, how much energy would be returned to the battery'? Assume 100% efficiency. (b) If the battery voltage is regulated at 24 volts, how much charge expressed in ampere-hours is imparted to the battery'? 3.7. Consider an electric home heating system. A resistive element R is connected to the 110-volt, 60-Hz electric supply. (a) Determine the current flow through the element as a function of time. (b) Determine the average thermal power generated per cycle. (c) If the heater is rated at 1000 watts, what is the value of the heater resistance R? 3.8. An electrical resistor, with the characteristic that the voltage drop is proportional to the square of the current, is used in a communications circuit. (a) Under normal operating conditions the voltage across the resistor is 10 v, with a current of 2 rnA. Determine the constitutive equation for the resistor. What is the power dissipated at this operating point? (b) Determine the linearized value of the resistance of the element at this nominal operating point. (c) If the applied voltage is increased to 12 v, what current would be predicted by the linearized model? What is the actual current in the nonlinear resistor? (d) Compare the predicted (from the linear model) and the actual power dissipation with 12 v applied. 3.9. In many fluid systems flow is controlled by valves or orifices, and is described by a quadratic orifice equation
where C 1 is a constant that depends upon the valve or orifice geometry. We wish to derive the equivalent linearized resistance of a quadratic orifice. (a) Derive the equivalent linear resistance at flow Q. (b) In a fluid system a pressure source of 100 N/m 2 drives a pipe with resistance of R = 5 N-s/m 5 and discharges through an orifice. Without the valve the flow through the pipe is 20 m3 /s, while with the orifice installed the flow reduces to 10 m3 /s. What are the equilibrium flow and pressure across the orifice? (c) What is the linearized resistance of the orifice?
REFERENCES [1] Firestone, F. A., "A New Analogy Between Mechanical and Electrical Systems," Journal of the Acoustic Society ofAmerica, 3, 249-267, 1933. [2] Trent, H. M.• "Isomorphisms Between Oriented Linear Graphs and Lumped Physical Systems," Journal of the Acoustic Society ofAmerica, 27, 500-527, 1955. [3] Shearer, J. L., Murphy, A. T., and Richardson, H. H., Introduction to System Dynamics, AddisonWesley, Reading, MA, 1967. [4] Koenig, H. E., Tokad, Y., Kesavan, H. K, and Hedges, H. C., Analysis of Discrete Physical Systems, McGraw-Hill, New York, 1967. [5] Blackwell, W. A., Mathematical Modeling of Physical Networks, Macmillan, New York. 1968. [6] Paynter, H. M., Analysis and Design of Engineering Systems, MIT Press, Cambridge, MA, 1961. [7] Kamopp, D. C., Margolis, D. L, and Rosenberg, R. C., System Dynamics: A Unified Approach (2nd Ed.), John Wiley, New York, 1990.
4
Formulation of System Models
4.1
INTRODUCnON TO LINEAR GRAPH MODELS
Graphical techniques are widely used to aid in the formulation and representation of models of dynamic systems. Several techniques, including linear graphs [1-5], bond graphs [6-7], operational block diagrams [8], and signal flow graphs [9-10], have been developed and are used extensively by engineers. In this text we use the linear graph to represent the structure of lumped-parameter systems and to provide basis for the generation of a state equation description for lumped-parameter system models. Linear graphs represent the topological relationships of lumped-element interconnections within a system. The tenn linear in this context denotes a graphical lineal (or line) segment representation as shown in Fig. 4.1 and is not related to the concept of mathematical linearity. Linear graphs are used to represent system structure in many energy domains and are a unified method of representing systems that involve more than one energy medium. They are similar in form to electric circuit diagrams. A graph is constructed from the following:
a
1. A set of branches each of which represents an energy port associated with a passive or source system element Each branch is drawn as an oriented line segment. 2. A set of nodes (designated by dots) that represent the points of interconnection of the lumped elements. All graph branches terminate at nodes. The nodes define points in the system where distinct across-variable values may be measured (with respect to a reference node), for example, points with distinct velocities in a mechanical system, or points in an electric system that have distinct voltages. 92
Sec. 4.2
Linear Graph Representation of One-Port Elements
93
r]Node
vmc ).Wble Across-
Through-
Branch
Figure 4.1: Linear graph representation of a single passive element as a directed line segment
A typical complete linear graph, representing a simple mechanical system with a single source and three one-port elements, is shown in Fig. 4.2. In this case there are two nodes representing points in the system at which distinct velocities may be measured. In practice it is common, but not necessary, to designate one of the nodes as a reference node and to draw this node as a horizontal line (sometimes cross-hatched) as shown. In mechanical systems the reference node is usually selected to be the velocity of the inertial reference frame, while in electric systems it commonly represents the system "ground" or zero-voltage point In fluid systems the reference node designates the reference pressure (often atmospheric pressure) from which all system pressures are measured. Apart from this special interpretation the reference node behaves identically to all other ~odes in the graph. v
Figura 4.2: Linear graph representation of a simple mechanical system.
In a linear graph one-port elements are represented in the two-terminal form introduced in Chap. 3. Each element generates a branch in the graph and is drawn as a line segment between the two appropriate nodes. Associated with each branch is an elemental throughvariable, assumed to pass through the line segment, and an elemental across-variable which is the difference between the across-variable values at the two nodes. Each linear graph branch thus represents the functional relationship between its across- and through-variables as defined by the elemental equation. Linear graph segments may be used to represent pure or ideal elements.
4.2
LINEAR GRAPH REPRESENTATION OF ONE-PORT ELEMENTS
Graph branches that represent one-port elements are drawn as oriented line segments with an arrow designating a sign convention adopted for the through- and across-variables. Figure 4.3 shows branches for the generalized passive energy storage and dissipation elements. Each branch is labeled with the generalized element type, and the across- and through-variables in the branch are related by the elemental equation. For the three generalized ideal (linear) elements the relationships.are
Formulation of System Models
94
Chap. 4
• For a generalized ideal A-type element (capacitance) C, dv dt
= .!_1
(4.1)
c
• For a generalized ideal T-type element (inductance) L, df dt
1
(4.2)
=LV
• For a generalized ideal D-type element (resistance) R, v=Rf
or
f
= .!.v R
(4.3)
where for energy storage elements the equations are expressed with the derivative on the left-hand side.
Y1
(a) A-type elements
(b) T-type element
(c) D-type element
Figure 4.3: Linear graph representation of generalized one-port passive elements.
As described in Chap. 3, A-type elements (with the exception of electric capacitors) must have their across-variable defined with respect to a constant reference value. For example, the velocity difference for a mass element is defined with respect to a constantvelocity inertial reference frame. The branches representing these A-type elements therefore must have one end connected to the reference node. Some authors use a dotted line to indicate this implicit connection to ground, as shown in Fig. 4.3. Apart from this notational difference, A-type branches are treated identically to all other branches. Each branch contains an arrow designating the sign convention associated with the across- and through-variables. The arrow on the graph element is drawn in the direction in which • v, the across-variable associated with the branch is defined to be decreasing, that is, in the direction of the assumed across-variable "drop," and • the through-variable f is defined as having a positive value. With this convention, when the elemental across- and through-variables have the same direction (or sign) power, 'P = fv, is positive and flows into the element The choice of arrow direction for passive branches simply establishes a convention to define positive and negative values of the through- and across-variables and is arbitrary. The arrow direction does not affect the equation formulation procedures described in Chap. 5 or any subsequent system analyses; the effect of reversing an arrow direction is simply to reverse the sign of the defined across- and through-variable on the element The choice of sign convention is discussed more fully in Sec. 4.4.
Sec. 4.3
Element Interconnection Laws
95
Ideal source elements are represented by linear graph segments containing a circle as shown in Fig. 4.4. In all source elements one variable, either the across- or through-variable, is a prescribed independent function of time. For source elements the arrow associated with the branch designates the sign associated with the source variable: 1. For a through-variable source the arrow designates the direction defined for positive through-variable flow. 2. For an across-variable source the arrow designates the direction defined for the across-
variable drop. The arrow on an across-variable source branch is commonly drawn toward the reference node since that is usually the direction of the assumed drop in an across-variable value.
il
Direction of
V, \
::-variable
(a) Across-variable source Figure 4.4:
4.3
F.r
{ t
Direction of through-variable
flow
(b) Through-variable source
Linear graph representation of ideal source elements.
ELEMENT INTERCONNECTION LAWS
Linear graphs represent the structure of a system model and specify the manner in which elements are connected. The general interconnection laws for linear graph elements are derived in this section, with one set of laws relating across-variables and a second set relating through-variables, following the developments of several authors [1-3].
4.3.1 Compatibility The compatibility law represents a set of constraints on across-variables on a graph that may be related to physical laws governing the interconnection of lumped elements. It may be stated: The sum of the across-variable drops on the branches around any closed loop on a linear graph is identically zero, or (4.4)
for any N elements forming a closed loop on the graph.
96
Fonnulation of System Models
Cbap.4
A compatibility equation may be written for any closed loop on a graph, including inner loops or outer loops, as shown in Fig. 4.5. Because the arrows on the branches indicate the direction of.the across-variable drop, they are used to assign the sign to terms in the summation; if the loop traverses a branch in the direction of an arrow, the term in the summation is positive, while if a branch is traversed against an arrow, the term in the sum is assigned a negative value.
A
(b)
(a)
Figure 4.5: Compatibility equations defined from loops on a linear graph. (a) Some possible loops on a graph, and (b) a loop containing four nodes and four branches.
Figure 4.5b shows a single loop with four branches and four nodes. With the arrow directions as shown, the compatibility equation for this loop is 4
LV;= VJ- v2 +v3 -v4 =0
(4.5)
i=l
We can demonstrate the compatibility law using the loop in Fig. 4.5b. The across-variable drop on an element is the difference between the value of the across-variable at the two nodes to which it is connected, for example, VJ = v A - v s is the drop associated with element 1. If all the nodal values are substituted into Eq. (4.5), then 4
LV;= (vA- vs)- (vc- Vs)
+ (vc- VD)- (vA- VD) = 0
(4.6)
i=l
The physical interpretation of the compatibility law in the various energy domains is Mechanical systems: The velocity drops across all elements sum to zero around
any closed path in a linear graph. Compatibility in mechanical systems is a geometric constraint which ensures that all elements remain in contact as they move. Electric systems: The compatibility law is identical to Kirchoff's voltage law which states that the summation of all voltage drops around any closed loop in an electric circuit is identically zero. Fluid systems: Pressure is a scalar potential which must sum to zero around any closed path in a fluid system. Thermal systems: Temperature is a scalar potential which must sum to zero around any closed path in a thermal system.
Sec. 4.3
Element Interconnection Laws
97
4.3.2 Continuity The continuity law specifies constraints on the through-variables in a linear graph that may be related to physical laws governing the interconnection of elements. It may be stated: The sum of through-variables flowing into any closed contour drawn on a linear graph is zero, that is,
(4.7) for any N branches that intersect a closed contour on the graph. Continuity is applied by drawing a closed contour on the linear graph and summing the through-variables of branches that intersect the contour, as shown in Fig. 4.6. The arrow direction on each branch is used to designate the sign of each tenn in the summation.
-
(a)
(b)
Figure 4.6: The definition of continuity conditions at (a) a single node on a linear graph. and (b) the extended principle of continuity applied to any closed contour on a graph.
For the special case in which a contour is drawn around a single node, the continuity law states that the sum of through-variables flowing into any node on a linear graph is identically zero. The law of continuity at a single node is illustrated in Fig. 4.6a. In this case f 1 - f2 - f3 = 0. The extended principle of continuity for a general contour may be demonstrated by considering the example containing three nodes shown in Fig. 4.6b. The continuity conditions at the three nodes are
+ fs = 0
at node A
f2- fs- f6 = 0
at node B
(4.8) (4.9)
+ f4 + f6 = 0
at node C
(4.10)
ft- f4 -f3
For the contour enclosing all three nodes, the sum of through-variables into the contour is (4.11) The principle of continuity applied to any node states that there can be no accumulation of the through-variable at that node. If this principle did not hold, it would imply that the integrated through-variable is nonzero at the node, and the node would either store or dissipate energy, thus acting as one of the primitive elements described in Chap. 3. In each of the energy domains, the principle of continuity corresponds to the following physical constraints:
Formulation of System Models
98
Chap.4
Mechanical systems: In a translational (or rotational) mechanical system continuity at a node arises as a direct expression of Newton's laws of motion, which require that the sum of forces (or torques) acting at any massless point must be identically zero. Electric systems: The principle of continuity at an electric node is Kirchoff's current law, which states that the sum of currents flowing into any node Gunction) in a circuit must be identically zero. Fluid systems: A node represents a junction of elements in a fluid system. The continuity principle requires that the sum of volume flow rates into the junction must be zero; if this were not true, then fluid would accumulate at the junction. Thermal systems: In a thennal system the continuity of heat flow rate ensures that there is no accumulation of heat at any junction between elements. 4.3.3 Series and Parallel Connection of Elements
Figure 4.7 shows two possible connections of elements within a linear graph. In Fig. 4.7a several elements are connected in parallel, that is, they are connected between a common pair of nodes. Compatibility equations may be written for the loop containing any pair of branches to show Vt = v2 = V3 = V4 = -vs. Similarly, the continuity condition applied to node B shows that ft + f2 + f3 + f4 - fs = 0. In general elements connected in parallel share a common across-variable, and the through-variable divides among the elements at the two nodes.
(a) Parallel connection Figure 4.7:
(b) Series connection
System elements connected in parallel and series.
Figure 4.7b shows four elements connected in series. In this configuration, with the arrows as indicated, the continuity condition may be applied to each of the internal nodes to show that f1 = f2 = -f3 = f4. If this series chain of elements is part of a loop, the compatibility condition requires that the across-variable drop across the chain is the sum of the individual drops of the branches, that is, VAB = v 1 + v 2 - v 3 + v 4 • Elements that are connected in series share a common through-variable.
4.4
SIGN CONVENTIONS ON ONE-PORT SYSTEM ELEMENTS
A simple electric system consisting of a battery and a resistor is shown in Fig. 4.8. The battery is modeled as an across-variable (voltage) source, and the resistor is modeled as an ideal D-type element. The positive (+) and negative (-)battery terminals are indicated. This
Sec. 4.4
Sign Conventions on One-Port System Elements A
99 A
(b)
(a)
Agure 4.8: Dlustration of passive element sign conventions using a simple electrical model.
simple system has only two nodes: the voltage reference node, arbitrarily chosen as the battery's negative terminal, and a node corresponding to the battery's positive terminal, which is the only other distinct voltage in the system. Branches corresponding to the source and resistive elements are connected in parallel between these nodes. The sign convention for the source requires that the arrow point in the direction of the assumed voltage drop. We have assumed that positive voltage corresponds to a positive across-variable value, and therefore the arrow must point downward, that is, from node A toward the reference node as shown. The sign convention for the resistor may be arbitrarily assigned, and in the figure the two possibilities are shown. In Fig. 4.8b the arrow is aligned in the direction of the assumed voltage drop, that is, toward the reference node. In this case the compatibility equation from the graph is -V.s +vR =0
(4.12)
which together with the D-type elemental equation for the resistor v R = RiR gives an expression for the current in the resistor: .
lR
IV.
= R
.S'
(4.13)
In Fig. 4.8c the same system graph is redrawn with the arrow on the resistor branch reversed. The compatibility equation then becomes
V.s +vR =0
(4.14)
and the current through the resistor is therefore (4.15)
which is opposite in sign to the first case. The direction defined as positive current flow is opposite in the two systems. A positive value of a computed through-variable implies that the flow is in the direction of the arrow, and a positive across-variable means that the drop is in the direction of the arrow. In this example the negative result implies that the direction of the current ftow is opposite that of the arrow. The results of both models are physically equivalent. The power flow into the resistor is positive regardless of the arrow direction.
100
Fonnulation of System Models
Chap.4
Figure 4.9 shows a simple mechanical system consisting of a mass resting on a frictionless plane and moving under the influence of an external prescribed force source. Four possible assumed positive force and velocity conditions are shown together with the corresponding linear graphs. In each case the upper node represents the velocity of the mass in the defined direction. An increase in the value of the across-variable indicates an increase in velocity in that direction. The sign convention assigned to the force source defines whether a positive force increases or decreases the velocity of the mass. In Fig. 4.9a and d the force and velocity directions are aligned and a positive force accelerates the mass in the direction of the applied force. In practice it is often convenient to adopt a convention directing all arrows on passive elements away from sources and toward the reference node and then to assign a source convention that is compatible with the convention defined in the physical system. Define positive velocity
(a)
Define positive velocity
(b)
Define positive velocity
Define positive velocity
Fm=F(t)
Fm=-F(t)
m
m
(c)
Figure 4.9:
(d)
Possible force and velocity orientations for a simple translational mass.
Sec. 4.5 4.5
Linear Graph Models of Systems of One-Port Elements
101
LINEAR GRAPH MODELS OF SYSTEMS OF ONE-PORT ELEMENTS
The representation of a physical system as a set of interconnected one-port linear graph elements is a system graph. The construction of a system graph usually requires a number of modeling decisions and engineering judgments. The general procedure may be summarized by the following steps: 1. Define the system boundary and analyze the physical system to determine the essentiaJ features that must be included in the model, including the system inputs, the outputs of interest, the energy domains involved, and the required elements. 2. Form a schematic, or pictorial, model of the physical system and establish a sign convention for the variables in the physical system.
3. Determine the necessary lumped-parameter elements representing the system sources, energy storage, and dissipation elements. 4. Identify the across-variables that define the linear graph nodes and draw a set of nodes. 5. Determine the appropriate nodes for each lumped element and add each element to the graph. 6. Select a set of sign conventions for the passive elements and draw the arrows on the graph.
7. Select the sign conventions for the system source elements to be consistent with the physical model and add them to the graph. The formulation of the model in steps 1-3 is perhaps the most difficult part of the modeling process because it requires a detailed knowledge of the system configuration and the physics of the energy domain. Usually engineering approximations and assumptions are required in the model formulation. Care must be taken to include all the essential elements so as to capture the required dynamic behavior of the physical system while not making the model overly complex. Whenever practicable, model responses should be verified against measurements made on the physical system, and the model modified if necessary to ensure fidelity of the response. In the remainder of this section we develop modeling procedures for deriving linear graphs in the five energy domains. 4.5.1 Mechanical Translational System Models
Translational system models utilize mass (A-type), spring (T-type), and damper (D-type) one-port passive elements, together with velocity (across-variable) and force (throughvariable) ideal source elements. The graph nodes represent points of distinct velocity with respect to an inertial reference frame. All A-type (mass) elements in a mechanical system must be connected to the inertial reference node.
102
Formulation of System Models
Chap.4
Example4.1 A mass m supported on a cantilever beam and subjected to a prescribed force F, (t) is shown in Fig. 4.1 Oa. In the figure positive velocity is defined as downward and is aligned in the direction of the positive force. It is assumed that the displacement of the mass is small, and so the system may be represented as a translational system in which all velocities are in the vertical direction. The schematic model, shown in Fig. 4.1 Ob, includes the following elements:
1. A force source F,(t) to represent the system input. 2. A mass element m to represent the mass.
3. A spring element K that models the effective force-displacement characteristic of the end point to represent the beam, which is assumed to be massless. There are only two nodes required in this example (the reference node and a node representing the velocity of the mass). The elements are added to the graph noting the following. 1. The velocity of the mass must be referenced to the fixed reference node.
2. The force source Fs (t) acts on the mass and acts with respect to the same fixed reference node.
3. One end of the spring moves with the velocity of the mass, and the other end is connected to the zero-velocity reference node. The sign orientation of the force source Fs (t) is chosen so that a positive force yields a positive mass velocity, as shown in the pictorial representation. The completed linear graph appears in Fig. 4.10c. F8 (t)
F,(t)
v
[
v
m
vrer= 0 (a) Physical system
(b) Schematic representation
(c) Linear graph
Figure 4.10: A mechanical system consisting of a mass element on a cantilevered beam.
The system graph indicates that all three branches are connected in parallel, with a compatibility condition indicating that Vm =Vx
(i)
And at the top node a continuity equation may be written to show F,- Fx- Fm =0
(ii)
As in any parallel connection, the across-variable (velocity) of the mass and spring are identical and the applied through-variable [force F,(r)] divides between the mass and the spring.
Sec. 4.5
103
Linear Graph Models of Systems of One-Port Elements
4.5.2 Mechanical Rotational Systems The construction of a linear graph model for a mechanical rotational system is similar to that for translational systems. Nodes on the graph represent points of distinct angular velocity, with respect to an inertial reference angular velocity, and the passive elements are rotary inertias (A-type), torsional springs (T-type), and rotary dampers (D-type). The across-variable source is an angular velocity source, and the through-variable source is a torque source. As in the case of translational systems, all A-type (inertia) elements are referenced to an inertial reference frame. Example4.2 A power transmission driving a large flywheel is shown in Fig. 4.11 a. The flywheel is supported on bearings and is driven through a frictional drag cup transmission by a motor acting as an angular velocity source rls(t). Clockwise angular rotations are defined as positive. The following elements are used to represent the system.
1. The system input from the motor is modeled as an angular velocity source rls(t). 2. The flywheel is modeled as a rotary inertia J.
3. The shaft bearings are modeled as a rotary damper B1 to account for energy dissipation due to friction as the shaft rotates. 4. The drag cup transmission is modeled as a rotary damper B2 connecting the motor to the flywheel. It is assumed that the shafts are rigid and massless. and so they do not deflect and do not add significant rotational inertia to the system.
J
(a) Physical system Figure 4.11:
(b) Linear graph
A rotational system consisting of flywheel driven through a drag cup.
Figure 4.11 shows that there are two distinct angular velocities with respect to the reference, labeled points A and B in Fig. 4.11 a, and therefore three nodes are necessary on the linear graph. The reference node is defined to be stationary; that is Orer = 0. The elements may be added to the graph by noting the following:
1. The angular velocity 0 1 of the flywheel must be defined relative to the fixed reference node. 2. The inner bearing race rotates at the same angular velocity as the flywheel, and the housing is fixed; thus, the damper B 1 is inserted in parallel with the flywheel.
104
Fonnulation of System Models
Chap. 4
3. For the transmission drag cup element B2, one end rotates at the angular velocity of the input shaft, OA, and the other end rotates at the angular velocity of the flywheel, 08 0 1 • It is therefore inserted between the nodes A and B.
=
4. The source angular velocity O.r(t) is defined with respect to the reference node. The sign of the angular velocity source is selected to provide a positive angular velocity to the damper, requiring the arrow to point toward the reference node. The completed linear graph is shown in Fig. 4.11 b.
4.5.3 Linear Graph Models of Electric Systems Electric system models consist of capacitors (A-type), inductors (T-type), and resistors (D-type) as passive elements, and voltage (across-variable) and current (through-variable) ideal sources. Electric circuits are usually easily translated into linear graphs because the topology of the linear graph is similar to the circuit diagram. The wires and connections between components in the circuit diagram are implicitly the nodes on the graph because they represent points of defined voltage. The following example illustrates the conversion of an electric circuit to a linear graph form. Example4.3 Figure 4.12 shows an electric filter designed to minimize the transmission of high-frequency electrical noise from an alternator to sensitive electronic equipment. The linear graph is generated by the fo11owing steps:
1. The alternator is represented by an ideal voltage source, and the electrical noise is modeled as variations of the voltage about its nominal value. 2. The electronic instrument is modeled as a resistive load RL. The value of the resistance is detennined from the manufacturer's specification of the nominal operating voltage and current for the instrument. Alternator
Filter
Instrument
l::t:tl
~~·~~~
(a) Physical system
V,(t)cf3:2 f A
L1
B
~
C
fR£
G (b) Electrical model
Figure 4.12:
An electric filter shown as (a) the physical system, and (b) an elecnical equivalent circuit modeling the source and the load.
Linear Graph Models of Systems of One-Port Elements
Sec.4.5
105
3. The circuit diagram in Fig. 4.12b is used to generate the system model. 4. The passive electric elements, the two coils and two capacitors, are each represented by single lumped elements.
S. The circuit diagram has four nodes, the reference ground node G and three others labeled A, B, and C in Fig. 4.12b. Each node represents a point in the circuit where a distinct voltage can be measured. 6. The elements are inserted between the nodes as shown in Fig. 4.13. 7. Sign conventions for the passive elements are established by directing the arrows away from the source and toward the reference node. 8. The sign convention for the voltage source is established as shown in Fig. 4.13 to correspond with that shown for the source in Fig. 4.12b.
Figure 4.13: Linear graph representation of the electrical filter.
4.5.4 Fluid System Models
Linear graph models for fluid systems are based on pressure drop Pas the across-variable and volume flow rate Q as the through-variable. Nodes on the graph represent distinct ·points of fluid pressure with respect to a constant reference pressure, and the passive elements are fluid capacitances (A-type), fluid inertances (T-type), and fluid resistances (Dtype). The across-variable source is a pressure source, and the through-variable source is a flow source. Fluid A-type elements are referenced to a fixed-pressure node. Example4.4 A water storage system consisting of a large reservoir, two control valves, and a tank is illustrated in Fig. 4.14. The system is fed by rainfall and may be represented with the following elements: 1. Rainfall-a flow source Qs(t) 2. The reservoir-a fluid capacitor C1
3. The two valves-in a partially open state modeled as linear ftuid resistances R1 and R2 4. The storage tank-a fluid capacitor c2 It is assumed that the connecting pipes are sufficiently short so that pressure drops associated with piping resistances and fluid inertances may be neglected The figure shows that there are two independent pressures in the system, at the base of the reservoir, point A, and at the base of the tank point, B. The graph therefore requires three nodes: the reference node representing atmospheric pressure and the two capacitance pressures.
106
Formulation of System Models
Chap.4
(\Rmn ~Q,(t)
'''
Reservoir
~
c.
A Valve (a) Physical system Figure 4.14:
(b) Linear graph
A fluid system with two storage tanks.
The two fluid capacitances (A-type elements) are placed between the appropriate nodes and the reference node Patm. The outlet valve R2 discharges between the storage tank pressure PA and the reference pressure Paun and so is connected in parallel with C2 • The pressure drop across valve R1 is PA- P8 , and so it is inserted between the two nodes A and B. Finally the flow source Q,. is inserted between the capacitance C1 and the reference node. The sign convention for the flow source Q,. is selected to give an increase in reservoir pressure when the source flow is positive. Figure 4.14b shows the completed linear graph.
In the next example we examine a simple lumped equivalent model of the distributed inertance and resistance effects in a long pipe. Example4.5 In the system shown in Fig. 4.15a fluid is pumped into a tank through a long pipe. The tank discharges to atmospheric pressure through a partially open valve. The model is formed to study the dynamic response of the flow through the outlet valve in response to changes in the pressure generated by the pump. The pump is represented as a pressure source P,.(t). The open tank is represented as a fluid capacitance C. The discharge valve is modeled as an ideal fluid resistance R1 • In the previous example it was assumed that pressure drops associated with the connecting pipes could be ignored; in this example the pipe is of sufficient length that internal pressure drops need to be included in the model. The pipe is assumed to
1. dissipate energy through frictional losses at the walls. and 2. store energy associated with the motion of the fluid within the pipe.
Tank
B
Cj
Long pipe Rp, lp A
Palm
Fluid reservoir (a) Physical system
(b) Linear graph
Figure 4.15: A fluid system that includes pipe effects in its model.
Sec. 4.5
Linear Graph Models of Systems of One-Port Elements
107
While these two effects are distributed throughout the length of the pipe, they may be approximated by a combination of a single lumped resistance Rp and a fluid inertance lp. The two elements have a common flow Q and are described by the elemental equations PRp P1p
= RpQ =
dQ
/p
dt
for the resistance for the inertance
(i)
(ii)
It is reasonable to assume that the total pressure drop across the pipe is the sum of the two effects and that the pipe should be modeled as a series connection of the elements. A nonphysical node is created in the linear graph to represent the point of connection of the two Jumped elements used to model the effects of distributed resistance and inertance in the pipe. With the addition of the pseudonode the linear graph requires a total of four nodes, representing the reference pressure, the pressure at the base of the tank A, the pressure at the end of the long pipe B, and the junction of the pipe resistance and inertance elements at the node C. The fluid capacitance is inserted between node A and the reference node. The pipe elements are inserted in series between the tank A and the pump B in arbitrary order. The discharge resistance R1 is connected to the reference node, indicating that the flow is to atmospheric pressure. The standard sign convention for passive elements is adopted, and the flow source direction is established to ensure that a positive flow from the pump establishes a positive pressure in the tank. Figure 4.15b shows the completed model.
4.5.5 Thermal System Models
Thermal systems are inherently different from the other energy domains because ( 1) the product of the across- and through-variables (temperature and heat ftow rate) is not power, (2) there is no defined T-type energy storage element, and (3) the D-type element does not dissipate energy. The two passive elements are a thermal capacitance (A-type) and a thermal resistance (D-type). The sources are a temperature source (across-variable source) and a heat ftow source (through-variable source). Example4.6 A laboratory furnace used to heat cylindrical metal specimens is illustrated in Fig. 4.16. The system model elements include the following: 1. The metal specimen, modeled as a thermal capacitance element C
2. The space between the specimen and the furnace coil element, modeled as a thermal resistance R1 3. The heating element, modeled as a heat (through-variable) source Q, 4. The outer insulation around the element, modeled as a thermal resistance element R2. In the selection of these elements, the representation of the metal specimen as a single thermal capacitance assumes that temperature gradients between its surface and center may be neglected, that is, the cylinder may be represented as a single lump at a uniform temperature. In addition the resistance R1 between the heater coil and specimen represents the combined effects of the air gap and any insulation around the inner surface of the coil, while the resistance R2 from the heater coil to the furnace exterior wall represents the heat losses to the environment and includes the total effective resistance due to the coil insulation interface, the insulation itself, and the insulation-atmosphere interface.
108
Formulation of System Models
Cbap.4
Heating element Q,r(t)
Air gap Rl
Outer insulation
source (a) Physical system
Figure 4.16:
(b) Linear graph
A laboratory furnace system.
The model contains two distinct temperatures with respect to the ambient environmental temperature Trcr: the temperature associated with the furnace heat source and the temperature of the specimen itself. The graph therefore contains three nodes, including the reference node. The thennal capacitance CT is referenced to the ambient temperature and is connected to the source node through the resistive element R1• Resistance R2 represents direct heat loss to the environment through the outer insulation and is connected directly across the sowce node. The sign convention adopted for the heat source ensures an increase in the temperature of the capacitance for a positive heat flow. Figure 4.16b shows the linear _graph.
4.6
PHYSICAL SOURCE MODELING
The ideal source elements introduced in Chap. 2 are capable of supplying infinite power to a system. Physical energy sources, on the other hand, have a limit on the power that they can supply. For example, the terminal voltage of an electric battery decreases as the current demand from the system increases. A battery is limited in the power it can supply even if the terminals are short-circuited. For small current loads it may be satisfactory to model a battery as a voltage source, but in more demanding situations with large and varying current requirements, the model of the battery must represent the variation of the terminal voltage. In general, physical energy sources are represented by a nonlinear relationship between acrossand through-variables such as shown in Fig. 4.17, and only over a limited range of operation may a real source be represented by an ideal across- or through-variable source. The power-limited characteristic of a real source can often be approximated by coupling an ideal source element with a D-type resistive element A typical power-limited source characteristic is represented in Fig. 4.18. It has a maximum value of its output across-variable Vs when the supplied through-variable is zero, corresponding to an opencircuit condition of an electric source, and a maximum value of the supplied through-variable Fs when the across-variable is zero, corresponding to a short-circuited electric source. If the characteristic is a straight line, with a slope -R, the relationship between the across- and through-variable at the source terminals at any point on the characteristic may be expressed as a linear algebraic equation in either of the following two forms: v=V,-Rf 1 f= Fs- -v R
(4.16)
(4.17)
Sec. 4.6
Physical Source Modeling
109 Region of approximate ideal across-variable
v
_/___=---
/
General cbanlcteristic
I
I
:
Region of approximate
'V source ideal through-variable 1 I
f
Through-variable (a) Typical real physical source characteristic
n
v
Current (b) A battery Figure 4.17:
p
Torque (c) A motor
T
Flow rate
(d) A pump
General characteristics of real sources.
F.r
Output through-variable Figure 4.18:
Characteristic of a simple linear source.
f
Q
110
Formulation of System Models
Chap.4
where v is the source across-variable when it is supplying through-variable f to the system. The first form states that iff= 0, the across-variable is equal to Vs, and as f increases, the output across-variable v decreases linearly. The second form states that if v = 0, the output through-variable f is equal io Fs, and as v increases, the through-variable decreases linearly. The two forms generate two possible models for a power-limited source with a linear characteristic: 1. Equation (4. 16) may be implemented by an ideal across-variable source of value Vs in series with a resistance element with a value R as shown in Fig. 4.19a. This series equivalent source model is known as a Thivenin equivalent source. 2. Equation (4.17) may be implemented by an ideal through-variable source of value Fs in parallel with a resistance of value R as shown in Fig. 4.19b. This configuration is known as a Norton equivalent source model.
These two models of real sources are equivalent and have identical characteristics as measured at their terminals. Either may be used in the modeling of systems involving physical
sources that may be approximated by a linear characteristic. The load power 'P delivered by an equivalent source model depends on the acrossand through-variables at the terminals. For the Thevenin source the power is (4.18) and for the Norton source it is 1
'P = vf = vFs - -v
2
R
(4.19)
The maximum power an equivalent source can provide is found by differentiating Eq. (4.18) with respect to for Eq. (4.19) with respect to v and equating the derivative to zero. In either case the maximum power is supplied when f = Vs/2R and v = RFs/2. The maximum power supplied is Pmax = VsFs/4.
Throughvariable source
F1(t)
(a) Th6venin equivalent source Figure 4.19:
(b) Nonon equivalent source
Th6venin and Nonon models of power-limited physical sources.
Sec. 4.6
Physical Source Modeling
111
Torque (a) The motor drive system
(b) The motor source characteristic
Figure 4.20: A rotary flywheel drive system, and the source characteristic of the electric motor.
Example4.7 It has been found that the pedonnance of the rotational flywheel drive model derived in Example 4.2, with the linear graph shown in Fig. 4.11, does not adequately reflect the dynamic response of the physical system over the full operating range of interest. Measurements on the system show that it is not valid to represent the motor as an ideal angular velocity source over the full speed range. With a fixed supply voltage and no load, the motor spins at an angular velocity of Omu, but as the torque load is increased, the motor speed decreases linearly until the shaft is stationary and generates a torque Tmax. Extend the linear graph model in Example 4.2 to include (a) a Th6venin and (b) a Norton source equivalent for the motor. Solution The measurements on the motor indicate that the source characteristic is as shown in Fig. 4.20. The equivalent source resistance is found from the slope of the characteristic, that is, R = Omax/ Tmax· Figure 4.21 shows the two modified system linear graphs using a Th6venin (Fig. 4.21a) and a Norton source (Fig. 4.21b) equivalent model for the motor where B = 1/ R. Both are equivalent with respect to the dynamic behavior of the flywheel.
(a) Thevenin source-based model
(b) Nonon source-based model
Figure 4.21: The rotational system in Fig. 4.11 redrawn with (a) a Thevenin equivalent source and (b) a Norton equivalent source.
Formulation of System Models
112
Chap.4
Thevenin and Norton source models may also be used to approximate the behavior of nonlinear physical sources when the range of variation of the source variables is small. Consider a source with a nonlinear characteristic (4.20)
v = F(f)
and assume that the source normally operates with small excursions about a nominal operating point v = vo and f = fo. IfEq. (4.20) can be expanded as a Taylor series and the first two terms are retained, v~vo+
If we define a D-type element
dF(f) df
I
f=fo
(4.21)
(f-fo)
R* = _ dF(f)l df
(4.22) f=fo
we can write an approximate source characteristic v = v0 - R* (f - fo)
(4.23)
= Vs- R*f
where Vs = vo + R*fo. Equation (4.23) defines a Thevenin equivalent source with an ideal source Vs and a series resistance R*. A linearized Norton source can also be expressed as a through-variable source Fs = fo + (1/R*)vo in parallel with aD-type element R*. PROBLEMS 4.1. For each of the systems shown in Fig. 4.22, establish a sign convention and construct the linear graph.
(b)
(c)
Figure 4.22: Four mechanical systems.
Chap. 4
Problems
113
4.2. A test apparatus for measuring the frictional characteristics of materials is shown in Fig. 4.23. A mass is released on an inclined plane and the terminal velocity is measured and used to estimate the friction between the material sample and the surface of the plane.
Figure 4.23:
An inclined plane for measuring friction.
(a) Draw a sketch identifying (i) major system elements, (ii) the forces acting, and (iii) a velocity reference direction. How is the effect of gravity represented in your sketch? (b) Construct the system linear graph.
(c) How is the system influenced by changes in the angle of inclination, 8? How is the system influenced by changes in the frictional properties of the sample? (d) If the frictional effects are modeled as linear, derive an expression for the terminal velocity of
the mass element. 4.3. A locomotive puiJing a single car on a straight, flat track is sketched in Fig. 4.24. The locomotive may be considered to generate a prescribed force F(t). The coupling between the locomotive and the car has both stiffness and damping properties. The car is subject to both rolling friction at the wheels and aerodynamic drag. Car
Figure 4.24:
A railroad locomotive and car.
(a) Make a sketch of the system, identifying the major system elements and a velocity reference direction. Make a suitable model of the coupler. Do you think that the coupler elements should be in series or in parallel? (b) Construct the system linear graph.
4.4. A four-story parking building, shown in Fig. 4.25, is located in an earthquake zone. Model each floor as a lumped mass element connected by steel girders to the floors above and below. Assume that during an earthquake the ground moves horizontally with a specified velocity, and that the girder structure has a finite translational stiffness, so that a sideways displacement on a floor results in a restoring force proportional to the displacement. Generate a linear graph model of the parking building.
Fonnulation of System Models
114
~---
Chap. 4
Slender suppon columns
'/j -
Ground motion V(r)
Figure 4.25:
~
A parking building in an earthquake zone.
4.5. A pans assembly station on a production line exhibits a severe vibration problem. A simplified schematic representation is shown in Fig. 4.26. Two large tables of mass m 1 and m 2 are each mounted to a sliding metal plate on resilient rubber mounts with shear stiffness K 1 and K2 , as shown. The tables are each subjected to a vibrational excitation force, F 1 (t) and F2 (t ). The plates are able to slide viscously on a second pair of defonnable rubber mounts, with shear stiffnesses K3 and K4. The viscous sliding coefficients are 8 1 and 8 2 • The two plates are coupled by a shaft with longitudinal stiffness Ks. Draw a lin ear graph for the system using the two forces F 1 and F2 as inputs.
K I !:-:=::--~7!1::1rtti
8I
lf"?ij Jf,M;;fl1 Figure 4.26:
A parts assembly station.
4.6. For each of the rotational systems shown in Fig. 4.27, establish a sign convention and draw a linear graph.
n,
0
r,(3)
n,
81
0~"' (d)
(b)
Fi gure 4.27:
Four rotational systems.
Chap. 4
Problems
115
4.7. Consider a wind driven electric generator, as discussed in Problem 1.1 and shown in Fig. I.7. Assume that the wind exerts a torque proportional to the wind speed on the rotating blades. The turbine is mounted in bearings and connected to an electric generator. A simple model of the electric generator is that it produces a load torque that is proportional to the shaft rotational velocity. (a) Sketch the system and define a reference angular velocity. (b) Identify a set of lumped elements that may be used to model the wind effects on the turbine shaft, the turbine rotating parts, the bearings, and the electric generator. (c) Construct a linear graph for the system. 4.8. A compact disc (CD) player uses a de electric motor to rotate the disk at a constant speed. The system is shown in Fig. 4.28, together with the motor torque-speed characteristic. T
Disc
Drive motor
To
~
E' ~
+ 0 Angular velocity Figure 4.28: A compact disc drive system and its motor characteristics.
(a) What elements are necessary to describe the motor, disc, and bearings? (b) CoD'struct a linear graph for this system.
4.9. R-C electric filters are often used in measurement and control systems to reduce high-frequency noise. Construct linear graphs for the first and second-order R-C filters shown in Fig. 4.29. R
(b)
(a)
Figure 4.29:
Two electrical filters.
4.10. Figure 4.30 shows a resistance welding system that uses the electrical discharge of energy stored in an inductive circuit to create a transient high-current flow between two pieces of metal. The localized beat generated forms a weld. A constant voltage source Vs is connected through a switch to a high-inductance coil of wire and the two pieces to be welded. When a high current is established in the coil the switch is opened. Because the current in the coil cannot change instantaneously, a high current is forced through the workpieces generating the heat necessary for welding.
116
Formulation of System Models
Chap. 4
Energy storage Switch
coil
Metallic
workpieces
Figure 4.30: An electric welding system.
(a) How would you represent the energy storage coil using lumped elements? What element would
be used to represent the metal workpieces? (b) Construct a linear graph representing the system when the switch is closed. What will determine the current flowing in the coil and the resistive workpieces after any transients have decayed to zero?
(c) Construct a linear graph representing the system after the switch is opened. At the instant after the switch is opened. how much cummt flows through the workpieces? {d) What is the direction of current flow (i) just before the switch is opened and (ii) just after the switch is opened? 4.11. For each of the fluid systems shown in Fig. 4.31, establish a sign convention and construct a linear graph.
J
c
c
~
J /,R 1
R2 (a)
(b)
(c)
(d)
Figure 4.31:
Four fluid systems.
4.12. In a biomechanics laboratory an experimental apparatus, shown in Fig. 4.32, is used to simulate the basic property of blood flow in arteries. The experiment uses a water-based fluid and a pump that may be approximated as a pressure source. A long, straight small-diameter plastic tube is used to simulate the arteries.
Chap. 4
Problems
117
/
Soft rubber section [inserted for part (c)]
Figure 4.32: Experimental apparatus to simulate blood flow.
(a) In the first experiments the pump pressure is set to a fixed, constant value. What lumped elements are required to represent the experimental system under steady flow conditions? Construct a linear graph that represents the system. If the tube internal surface is coated to simulate the effects of arterial plaque, what parameter in the model is changed? (b) In a second set of experiments the pump pressure is varied sinusoidally. It is noticed that the ratio of the amplitude of the flow to the amplitude of the pump pressure varies with the frequency of the input pressure. What additional elements are needed to model nonsteady flow conditions? Revise your linear graph to include any additional elements.
(c) In a third experiment, it is desired to simulate the effect of an aneurism; that is, a section of the artery where the wall is weak. A short piece of soft rubber tubing is inserted at a point half way down the tubing. Tests on the tube have shown that the change in volume of fluid stored in the section is proportional to the change in pressure. How might the rubber tubing be represented in a lumped-parameter model? Construct a linear graph for the system with the aneurism. _4.13. The air flow system in many fossil-fuel power plants consist of (i) a forced-draft fan, (ii) a furnace/boiler volume, (iii) an induced-draft fan, and (iv) an exhaust stack, connected by long ducts as shown in Fig. 4.33. In initial plant qualification tests the system is tested without firing the boiler, so that air at nonnal temperature is the system fluid.
Vertical
Furnace volume
exhaust stack
Forced-draft fan Duct
--
t
Duct
Figure 4.33: The air and exhaust-gas system in a power plant.
(a) We wish to form a simplified model of the fluid system. Consider each fan as a prescribed pressure rise, and the ducts and stack as having fluid inertance and resistance. What elements might be used to model the fans, ducts and stack, and the unfired furnace? (b) Construct a linear graph of the system.
Formulation of System Models
118
Chap. 4
4.14. Fonn a simple model of a household heating system. Consider a small house with a furnace that provides a prescribed heat-ftow rate within the house. (a) What elements are required to represent (i) the house and its contents and {ii) the beat ftow to and from the house from the ambient temperature outside? How can changes in the ambient outside temperature be represented? What is an appropriate element to represent the furnace? (b) Construct a linear graph for the system, and write continuity equations for all nodes on the graph. 4.15. In electronic camera ftash units a capacitor is charged from an electronic power supply, and its stored energy is discharged through the flash tube to create the flash. The electronic power supply is not an ideal source, that is, it cannot be modeled as a voltage source or a current source. The charging system is shown in Fig. 4.34. Construct linear graphs for the system using (i) a Thevenin source model and {ii) a Norton source model for the electronic power supply. Assume the switch is open.
Figure 4.34: The charging system for a camera flash-tube.
4.16. The source characteristics of (i) an electric power source and {ii) an electric motor have been measured and are plotted in Fig. 4.35. Determine equivalent Thevenin and Norton source models for each.
v
T
2S
45
e
i' ]
-r
~ u
::I
e-
~
~
0
3
1
Current (amps) Figure 4.35:
0
1000
n
Angular velocity (rad/s) Electrical and motor source characteristics.
4.17. AD-type element is driven by a source that is modeled as a Thevenin or Norton equivalent. Show that the power dissipated in the element is a maximum when its resistance R is equal to the source resistance R0 • This is known as the maximum power transfer theorem. 4.18. Long telecommunication cables have capacitance, inductance, and resistance properties distributed along their length. Simple lumped parameter models of such cables do not adequately predict the ability to faithfully transmit information over long distances. Consider a cable of length l with measured resistance R, capacitance C, and inductance L. A simple single-section model might lump all of the capacitance at the mid-point, leading to a model shown in Fig. 4.36a, where half of the inductance and resistance has been assigned to each end A less approximate model may be found by dividing the cable into several sections, using a lumped approximation for each section. as shown in Fig. 4.36b. As the number of sections is increased, the dynamic response of the model approaches
-
Chap. 4
119
References
that of the real distributed system. Use the sectional model of Fig. 4.36a to generate (i) a two-section and (ii) a three-section model of a telecommunication cable. Express your model in schematic form and as a linear graph. Include a source model and a load resistance. Identify the parameters in terms of R, L, and C. Rl2
L/2
L/2
Rl2
(a)
Section 1
(b)
Figure 4.36:
Lumped model of a long communications cable.
4.19. Garden hoses have flexible walls that expand under pressure. The system parameters are distributed along its length. Use the approach taken in the previous problem to construct a two-section lumped linear graph representation of a long garden bose connected to a lawn sprinkler. Take into account distributed fluid inertance, capacitance, and resistance. Include elements to represent the sprinkler and the source.
REFERENCES [1] Koenig, H. E., Tokad, Y., Kesavan, H. K., and Hedges, H. G., Analysis of Discrete Physical Systems, McGraw-Hill, New York, 1967. [2] Blackwell, W. A., Mathematical Modeling of Physical Networks, Macmillan, New York, 1967. [3] Shearer, J. L., Murphy, A. T., and Richardson, H. H.,lntroduction to System Dynamics, AddisonWesley, Reading, MA, 1967. [4] Kuo, B. C., Linear Networks and Systems, McGraw-Hill, New York, 1967. [5) Chan, S. P., and Chan, S. G.,Analysis ofLinear Networks and Systems, Addison-Wesley, Reading, MA,l972. [6] Paynter, H. M., Analysis and Design ofEngineering Systems, MIT Press, Cambridge, MA, 1961. [7] Kamopp, D. C., Margolis, D. L., and Rosenberg, R. C., System Dynamics: A Unified Approach (2nd ed.), John Wiley, New York, 1990. [8) Shearer, J. L., and Kulakowski, B. T., Dynamic Modeling and Control of Engineering Systems, Macmillan, New York, 1990. [9] Mason, S. J., "Feedback Theory: Further Properties of Signal Flow Graphs," Proceedings of the IRE, 44, 7, 1956. [10] Chow, Y., and Casignol, E., Linear Signal Flow Graphs and Applications, John Wlley, New York, 1962.
5
State Equation Formulation
5.1
STATE VARIABLE SYSTEM REPRESENTATION Linear graph system models, described in Chap. 4, provide a graphical representation of a system model and the interconnection of its elements. A set of differential and algebraic equations that completely define the system may be derived directly from the linear graph model. In this chapter we develop a procedure for deriving a specific set of differential equations, known as state equations, from the system linear graph. These equations are expressed in terms of a set of state variables and provide a basis for determining the system response to external inputs. The state equations are derived from the set of elemental equations representing the dynamics of each system element, together with a set of compatability and continuity equations defined from the linear graph model structure.
5.1.1 Definition of System State The concept of the state of a dynamic system refers to a minimum set of variables, known as state variables, that fully describe the system and its response to any given set of inputs [1-3]. In particular, a state-determined system model has the characteristic that A mathematical description of the system in terms of a minimum set of variables x1 (t), i = 1, ... , n, together with knowledge of those variables at an initial time to and the system inputs for time t ~ to, are sufficient to predict the future system state and outputs for all time t > to.
This definition asserts that the dynamic behavior of a state-determined system is completely characterized by the response of the set of n variables x; (t), where the number n is defined to be the order of the system. The system shown in Fig. 5.1 has two inputs u1 (t) and u2(t) and four output variables Yl (t), ... , Y4 (t). If the system is state-determined, knowledge of its state variables ·[xi (to), x2(to), ... , Xn(to)] at some initial time to and the inputs Ut (t) and u2(t) fort ~to is
120
Sec. 5.1
State Variable System Representation
121 Output vector y
___ .,._ u,(t)
System described by state variables (xl,x2, ••. Xn}
~ ~--~
t
t Figura 5.1:
System inputs and outputs.
sufficient to determine all future behavior of the system. The state variables are an internal description of the system that completely characterizes the system state at any time t and from which any output variables y;(t) may be computed. Large classes of engineering, biological, social, and economic systems may be represented by state-determined system models. System models constructed with the pure and ideal (linear) one-port elements defined in the preceding chapters are state-determined system models. For such systems the number of state variables n is equal to the number of independent energy storage elements in the system. The values of the state variables at any time t specify the energy of each energy storage element within the system and therefore the total system energy, and the time derivatives of the state variables determine the rate of change of the system energy. Furthermore, the values of the system state variables at any time t provide sufficient information to determine the values of all other variables in the system at that time. There is no unique set of state variables that describe any given system; many different sets of variables may be selected to yield a complete system description. However, for a given system the order n is unique and is independent of the particular set of state variables chosen. State variable descriptions of systems may be formulated in terms of physical and measurable variables or in terms of variables that are not directly measurable. It is possible to mathematically transform one set of state variables to another; the important point is that any set of state variables must provide a complete description of the system. In this text we concentrate on a particular set of state variables based on physical variables derived directly from linear graph models.
5.1.2 The State Equations A standard form for the state equations is used throughout system dynamics. In the standard form the mathematical description of the system is expressed as a set of n coupled first-order ordinary differential equations, known as the state equations, in which the time derivative of each state variable is expressed in terms of the state variables Xt (t), ... , xn(t) and the system inputs u 1 (t), ... , ur(t). In the general case the form of then state equations is it = /1 (x, u, t) i2 = /2 (x, u, t)
. .
·-· .-.
Xn
= fn (X, U, t)
(5.1)
Chap. 5
State Equation Fonnulation
122
where i; = dx; 1dt and each of the functions /; (x, u, t) (i = 1, ... , n) may be a general, nonlinear, time-varying function of the state variables, the system inputs, and time. 1 It is common to express the state equations in a vector fonn in which the set of n state variables is written as a state vector x(t) [Xt (t), x2(t), ... , Xn (t)]T and the set of r inputs is written as an input vector u(t) = [u 1(t), u2(t), ... , u,(t)]T. Each state variable is a time-varying component of the column vector x(t). This form of the state equations explicitly represents the basic elements contained in the definition of a state-determined system. Given a set of initial conditions (the values of . the x; at some time to) and the inputs for t ~ to, the state equations explicitly specify the derivatives of all state variables. The value of each state variable at some time ll.t later may then be found by direct integration. The system state at any instant may be interpreted as a point in an n-dimensional state space, and the dynamic state response x(t) can be interpreted as a path or trajectory traced out in the state space. In vector notation the set of n equations in Eqs. (5.1) may be written
=
i = f (x, u, t)
(5.2)
where f (x, u, t) is a vector function with n components/; (x, u, t). In this text we restrict our attention primarily to a description of systems that are linear and time-invariant (LTI), that is, systems described by linear differential equations with constant coefficients. For a LTI system of order n and with r inputs, Eqs. (5.1) become a set of n coupled first-order linear differential equations with constant coefficients:
X) X2
= =
a))X} a2JXJ
+ at2X2 + ... + atnXn + b]JUJ + ... + btrUr + a22X2 + ... + a2nXn + b2tUt + ... + b2rUr
(5.3)
where the coefficients a;i and bii are constants that describe the system. This set of n equations defines the derivatives of the state variables to be a weighted sum of the state variables and the system i~puts. Equations (5.3) may be written compactly in matrix form:
!!_ [xtl ~2 = [au a~t dt
:
:
Xn
anl
Q}2
a22
an2
Xt. l + [bu.
a1n a2n ] [ x2
~~
ann
bnt
.. .
..
Xn
..
b2r btr
b:,
l
LJ UJ
(5.4)
which may be summarized as
i=Ax+Bu
(5.5)
In this text we use boldface type to denote vector quantities. Uppercase letters are used to denote general matrices, while lowercase letters denote column vectors. See App. A for an introduction to matrix notation and operations.
Sec. 5.1
State Variable System Representation
123
where the state vector x is a column vector of length n, the input vector u is a column vector of length r, A is an n x n square matrix of the constant coefficients aii, and B is an n x r matrix of the coefficients biJ that weight the inputs.
5.1.3 Output Equations
A system output is defined to be any system variable of interest A description of a physical system in terms of a set of state variables does not necessarily include all the variables of direct engineering interest An important property of the linear state equation description is that all system variables may be represented by a 1inear combination of the state variables x; and the system inputs u;. An arbitrary output variable in a system of order n with r inputs may be written
(5.6) where the c; and d; are constants. H a total of m system variables are defined as outputs, the m such equations may be written as YI Y2
= =
Ym
=
l [en
cuXt C2JX1
CmtXl
+ CJ2X2 + ··· + CJnXn + duul + ·· · + dtrUr + C22X2 + ·· •+ C2nXn + d21U1 + · •· + duUr + Cm2X2 + · · · + CmnXn + dmtUt + · · · + dmrUr
or in matrix form as
[ Y2 Yt
.. .
Ym
Ct2
.. .
C22
Cml
Cm2
C2t
(5.7)
Ctn C2n ] [ Xt X2
..
..
Cmn
Xn
.
.
l[ +
dn d21
..
.
dml
f][]
(5.8)
The output equations, Eqs. (5.8), are commonly written in the compact form:
y=Cx+Du
(5.9)
where y is a column vector of the output variables y; (t), C is an m x n matrix of the constant coefficients c;i that weight the state variables, and D is an m x r matrix of the constant coefficients dij that weight the system inputs. For many physical systems the matrix D is the null matrix, and the output equation reduces to a simple weighted combination of the state variables: y=Cx
(5.10)
124
State Equation Formulation
Chap.5
5.1.4 State Equation-Based Modeling Procedure The complete system model for a linear time-invariant system consists of (1) a set of n state equations, defined in terms of the matrices A and B, and (2) a set of output equations that relate any output variables of interest to the state variables and inputs and are expressed in terms of the C and D matrices. The task of modeling the system is to derive the elements of the matrices and to write the system model in the form
i=Ax+Bu y=Cx+Du
(5.11)
(5.12)
The matrices A and B are properties of the system and are determined by the system structure and elements. The output equation matrices C and D are determined by the particular choice of output variables. The overall modeling procedure developed in this chapter is based on the following steps: 1. Determination of the system order n and selection of a set of state variables from the linear graph system representation. 2. Generation of a set of state equations and the system A and B matrices using a welldefined methodology. This step is also based on the linear graph system description. 3. Determination of a suitable set of output equations and derivation of the appropriate C and D matrices.
5.2
LINEAR GRAPHS AND SYSTEM STRUCTURAL PROPERTIES
5.2.1 Linear Graph Properties The derivation of the state equations in this chapter is based on the use of the system linear graph model. A linear graph with B branches represents B system elements, each with a known elemental equation or source function. The graph also represents the structure of the element interconnections in terms of the continuity and compatibility constraint equations described in Chap. 4. In the following sections we use the properties of linear graphs to (1) derive the system structural constraints, (2) define the set of state variables, and (3) provide a systematic technique for deriving the system state equations [4-8]. The following definitions are introduced: System graph: The oriented linear graph model of a system.
Connected graph: A system graph in which a path exists between all pairs of nodes. A path is said to exist if the node pair is joined by at least one branch. Figure 5.2 shows a connected graph along with a system graph that is not connected. System graphs for systems consisting of one-port elements are usually connected graphs, while systems that include the two-port elements introduced in Chap. 6 may not generate connected graphs. In this chapter we assume that all system graphs are connected graphs.
Sec. 5.1
121
State Variable System Representation Output vector y
___ ..,
System described by state variables (Xt, X2, ••• Xn}
,_
__
~
Y1(t)
~
t
t Figure 5.1:
System inputs and outputs.
sufficient to determine all future behavior of the system. The state variables are an internal description of the system that completely characterizes the system state at any time t and from which any output variables y; (t) may be computed. Large classes of engineering, biological, social, and economic systems may be represented by state-detennined system models. System models constructed with the pure and ideal (linear) one-port elements defined in the preceding chapters are state-determined system models. For such systems the number of state variables n is equal to the number of independent energy storage elements in the system. The values of the state variables at any time t specify the energy of each energy storage element within the system and therefore the total system energy, and the time derivatives of the state variables determine the rate of change of the system energy. Furthermore, the values of the system state variables at any time t provide sufficient information to determine the values of all other variables in the system at that time. There is no unique set of state variables that describe any given system; many different sets of variables may be selected to yield a complete system description. However, for a given system the order n is unique and is independent of the particular set of state variables chosen. State variable descriptions of systems may be formulated in terms of physical and measurable variables or in terms of variables that are not directly measurable. It is possible to mathematically transform one set of state variables to another; the important point is that any set of state variables must provide a complete description of the system. In this text we concentrate on a particular set of state variables based on physical variables derived directly from linear graph models.
5.1.2 The State Equations A standard form for the state equations is used throughout system dynamics. In the standard form the mathematical description of the system is expressed as a set of n coupled first-order ordinary differential equations, known as the state equations, in which the time derivative of each state variable is expressed in terms of the state variables x1 (t), ... , Xn (t) and the system inputs Ut (t), ... , ur(t). In the general case the form of then state equations is
XI = !l {X, U, t) i2
= /2 {x, u, t)
. . ·-· .-. Xn = fn (X, U, t)
(5.1)
122
Chap.5
State Equation Fonnulation
=
where .i; = dx;/dt and each of the functions/; (x, u, t) (i I, ... , n) may be a general, nonlinear, time-varying function of the state variables, the system inputs, and time. 1 It is common to express the state equations in a vector fonn in which the set of n state variables is written as a state vector x(t) [x 1(t), x 2(t), ••• , Xn (t)]T and the set of r inputs is written as an input vector u(t) = [u 1(t), u 2(t), •.. , u,(t)]T. Each state variable is a time-varying component of the column vector x(t). This form of the state equations explicitly represents the basic elements contained in the definition of a state-determined system. Given a set of initial conditions (the values of the x; at some time to) and the inputs fort ~to, the state equations explicitly specify the derivatives of all state variables. The value of each state variable at some time llt later may then be found by direct integration. The system state at any instant may be interpreted as a point in an n-dimensional state space, and the dynamic state response x(t) can be interpreted as a path or trajectory traced out in the state space. In vector notation the set of n equations in Eqs. (5.1) may be written
=
(5.2)
i = f (x, u, t)
where f (x, u, t) is a vector function with n components /; (x, u, t). In this text we restrict our attention primarily to a description of systems that are linear and time-invariant (LTI), that is, systems described by linear differential equations with constant coefficients. For a LTI system of order n and with r inputs, Eqs. (5.1) become a set of n coupled first-order linear differential equations with constant coefficients:
XJ X2
= =
a11XJ a21X1
+ a12X2 + ·· · + atnXn + bJJUJ +
··· +
btrUr
+ a22X2 + · · · + a2nXn + hltUt + · · · + b2rur
(5.3)
where the coefficients a;i and b;i are constants that describe the system. This set of n equations defines the derivatives of the state variables to be a weighted sum of the state variables and the system inputs. Equations (5.3) may be written compactly in matrix form:
!!_ [Xtl ~2 = [au a~1 dt
:
:
Xn
ani
a12 a22
Otn][Xtl [bu . . + . D2n
..
an2
ann
X2
b21
..
..
Xn
bnt
b~, b2r btr
l
[JJ UJ
(5.4)
which may be summarized as
i=Ax+Bu
(5.5)
In this text we use boldface type to denote vector quantities. Uppercase letters are used to denote general matrices, while lowercase letters denote column vectors. See App. A for an introduction to matrix notation and operations.
Sec. 5.1
123
State Variable System Representation
where the state vector xis a column vector of length n, the input vector u is a column vector of length r, A is an n x n square matrix of the constant coefficients a;i, and B is an n x r matrix of the coefficients bii that weight the inputs.
5.1.3 Output Equations A system output is defined to be any system variable of interest. A description of a physical
system in terms of a set of state variables does not necessarily include all the variables of direct engineering interest. An important property of the linear state equation description is that all system variables may be represented by a linear combination of the state variables x; and the system inputs u;. An arbitrary output variable in a system of order n with r inputs may be written
(5.6) where the c; and d; are constants. If a total of m system variables are defined as outputs, the m such equations may be written as
Y2
= =
Ym
=
Yt
C11X1 C2JX1
CmtXt
+ CJ2X2 + ··· + CtnXn + d11U1 + ·· · + dtrUr + C22X2 + · •· + C2nXn + d21U1 + •••+ d2rUr
(5.7)
+ Cm2X2 + · · · + CmnXn + dmiUJ + ·· · + dmrUr
or in matrix form as
[Yll [ .. .. Y2
C21 cu
.
.
Ym
Cmt
Ct2
C22
C(n C2n
..
.
Cm2
Cmn
][XI. l X2
. +
.
Xn
[.
du d21
..
dml
!][J
(5.8)
The output equations, Eqs. (5.8), are commonly written in the compact form: y=Cx+Du
(5.9)
where y is a column vector of the output variables y; (t), C is an m x n matrix of the constant coefficients c;i that weight the state variables, and D is an m x r matrix of the constant coefficients dij that weight the system inputs. For many physical systems the matrix D is the null matrix, and the output equation reduces to a simple weighted combination of the state variables: y=Cx
(5.10)
124
State Equation Fonnu1ation
Chap. S
5.1.4 State Equation-Based Modeling Procedure The complete system model for a linear time-invariant system consists of ( 1) a set of n state equations, defined in terms of the matrices A and B, and (2) a set of output equations that relate any output variables of interest to the state variables and inputs and are expressed in terms of the C and D matrices. The task of modeling the system is to derive the elements of the matrices and to write the system model in the form
x=Ax+Bu y=Cx+Du
(5.11) (5.12)
The matrices A and B are properties of the system and are determined by the system structure and elements. The output equation matrices C and D are determined by the particular choice of output variables. The overall modeling procedure developed in this chapter is based on the following steps:
1. Determination of the system order n and selection of a set of state variables from the linear graph system representation. 2. Generation of a set of state equations and the system A and B matrices using a welldefined methodology. This step is also based on the linear graph system description. 3. Determination of a suitable set of output equations and derivation of the appropriate C and D matrices.
5.2
LINEAR GRAPHS AND SYSTEM STRUCTURAL PROPERTIES
5.2.1 Linear Graph Properties The derivation of the state equations in this chapter is based on the use of the system linear graph model. A linear graph with B branches represents B system elements, each with a known elemental equation or source function. The graph also represents the structure of the element interconnections in terms of the continuity and compatibility constraint equations described in Chap. 4. In the following sections we use the properties of linear graphs to (1) derive the system structural constraints, (2) define the set of state variables, and (3) provide a systematic technique for deriving the system state equations [4--8]. The following definitions are introduced: System graph: The oriented linear graph model of a system.
Connected graph: A system graph in which a path exists between all pairs of nodes. A path is said to exist if the node pair is joined by at least one branch. Figure 5.2 shows a connected graph along with a system graph that is not connected. System graphs for systems consisting of one-port elements are usually connected graphs, while systems that include the two-port elements introduced in Chap. 6 may not generate connected graphs. In this chapter we assume that all system graphs are connected graphs.
Sec. 5.1
121
State Variable System Representation Output vector y
---~
System described by state variables (.xl• xl• ... X'n}
t ~--~
t
t Rgure 5.1:
System inputs and outputs.
sufficient to determine all future behavior of the system. The state variables are an internal description of the system that completely characterizes the system state at any time t and from which any output variables y;(t) may be computed. Large classes of engineering, biological, social, and economic systems may be represented by state-determined system models. System models constructed with the pure and ideal (linear) one-port elements defined in the preceding chapters are state-determined system models. For such systems the number of state variables n is equal to the number of independent energy storage elements in the system. The values of the state variables at any time t specify the energy of each energy storage element within the system and therefore the total system energy, and the time derivatives of the state variables determine the rate of change of the system energy. Furthermore, the values of the system state variables at any time t provide sufficient information to determine the values of all other variables in the system at that time. There is no unique set of state variables that describe any given system; many different sets of variables may be selected to yield a complete system description. However, for a given system the order n is unique and is independent of the particular set of state variables chosen. State variable descriptions of systems may be formulated in terms of physical and measurable variables or in terms of variables that are not directly measurable. It is possible to mathematically transform one set of state variables to another; the important point is that any set of state variables must provide a complete description of the system. In this text we concentrate on a particular set of state variables based on physical variables derived directly from linear graph models.
5.1.2 The State Equations A standard form for the state equations is used throughout system dynamics. In the standard form the mathematical description of the system is expressed as a set of n coupled first-order ordinary differential equations, known as the state equations, in which the time derivative of each state variable is expressed in terms of the state variables Xt (t), ... , Xn(t) and the system inputs u 1 (t), ... , ur (t). In the general case the form of the n state equations is it
=It (X, U, t)
i2 = l2 (x, u, t)
. . ·-· .-. Xn
= In (X, U, t)
(5.1)
122
State Equation Formulation
=
Chap.5
=
where .i; dx;/dt and each of the functions/; (x, u, t) (i I, ... , n) may be a general, nonlinear, time-varying function of the state variables, the system inputs, and time. 1 It is common to express the state equations in a vector fonn in which the set of n state variables is written as a state vector x(t) [XI (t), x2(t), ... , Xn (t)]T and the set of r inputs is written as an input vector u(t) = [u 1(t), u 2 (t), ... , u,(t)]T. Each state variable is a time-varying component of the column vector x(t). This fonn of the state equations explicitly represents the basic elements contained in the definition of a state-determined system. Given a set of initial conditions (the values of . the x; at some time to) and the inputs fort :::: to, the state equations explicitly specify the derivatives of an state variables. The value of each state variable at some time at later may then be found by direct integration. The system state at any instant may be interpreted as a point in an n-dimensional state space, and the dynamic state response x(t) can be interpreted as a path or trajectory traced out in the state space. In vector notation the set of n equations in Eqs. (5.1) may be written
=
i = f(x, u, t)
(5.2)
where f (x, u, t) is a vector function with n components fi (x, u, t). In this text we restrict our attention primarily to a description of systems that are linear and time-invariant (LTI), that is, systems described by linear differential equations with constant coefficients. For a LTI system of order n and with r inputs, Eqs. (5.1) become a set of n coupled first-order linear differential equations with constant coefficients:
XJ i2
=
=
alJXl
a21X1
+ al2X2 + ... + atnXn + bJJU) + ... + btrUr + a22x2 + · ·· + a2nxn + b21u1 + · · · + b2rur
(5.3)
where the coefficients aii and bii are constants that describe the system. This set of n equations defines the derivatives of the state variables to be a weighted sum of the state variables and the system inputs. Equations (5.3) may be written compactly in matrix form:
!!_ [x1] ~2 = [au a~1 dt
:
:
Xn
anl
at2 a22
an2
a1nl [XI].. + [bu.. .. a2n
x2
b21
.
.
.
ann
Xn
bnl
!l []
(5.4)
which may be summarized as
x=Ax+Bu
(5.5)
In this teXt we use boldface type to denote vector quantities. Uppercase letters are used to denote general matrices, while lowercase letters denote column vectors. See App. A for an introduction to matrix notation and operations.
Sec. 5.1
123
State Variable System Representation
where the state vector xis a column vector of length n, the input vector u is a column vector of length r, A is an n x n square matrix of the constant coefficients aii, and B is an n x r matrix of the coefficients bii that weight the inputs.
5.1.3 Output Equations
A system output is defined to be any system variable of interest A description of a physical system in terms of a set of state variables does not necessarily include all the variables of direct engineering interest An important property of the linear state equation description is that all system variables may be represented by a linear combination of the state variables x; and the system inputs u;. An arbitrary output variable in a system of order n with r inputs may be written
(5.6) where the c; and d; are constants. If a total of m system variables are defined as outputs, the m such equations may be written as
Y2
= =
Ym
=
Yt
Cl1Xl
C2)X)
CmtXt
+ CJ2X2 + ··· + CtnXn + d11U1 + ·· · + dtrUr + C22X2 + •••+ C2nXn + d21U1 + ••• + d2rUr + Cm2X2 + ··· + CmnXn + dmtUt + · · · + dmrUr
or in matrix form as Ct2
[ Y2Y1] [ cu. .
en
Cm)
Cm2
..
Ym
C2t
..
(5.7)
C2n CJn
...
Cmn
l l [ !][J [XI..
X2
.
+
Xn
d21 du
...
(5.8)
dml
The output equations, Eqs. (5.8), are commonly written in the compact fonn:
y=Cx+Du
(5.9)
where y is a column vector of the output variables y; (t), C is an m x n matrix of the constant coefficients c;i that weight the state variables, and D is an m x r matrix of the constant coefficients dij that weight the system inputs. For many physical systems the matrix D is the nulJ matrix, and the output equation reduces to a simple weighted combination of the state variables:
y=Cx
(5.10)
124
State Equation Formulation
Chap.5
5.1.4 State Equation-Based Modeling Procedure The complete system model for a linear time-invariant system consists of (1) a set of n state equations, defined in tenns of the matrices A and B, and (2) a set of output equations that relate any output variables of interest to the state variables and inputs and are expressed in terms of the C and D matrices. The task of modeling the system is to derive the elements of the matrices and to write the system model in the form
i=Ax+Bo y=Cx+Do
(5.11) (5.12)
The matrices A and B are properties of the system and are determined by the system structure and elements. The output equation matrices C and D are determined by the particular choice of output variables. The overall modeling procedure developed in this chapter is based on the following steps: 1. Determination of the system order n and selection of a set of state variables from the linear graph system representation. 2. Generation of a set of state equations and the system A and B matrices using a welldefined methodology. This step is also based on the linear graph system description. 3. Determination of a suitable set of output equations and derivation of the appropriate C and D matrices.
5.2
LINEAR GRAPHS AND SYSTEM STRUCTURAL PROPERTIES
5.2.1 Linear Graph Properties The derivation of the state equations in this chapter is based on the use of the system linear graph model. A linear graph with B branches represents B system elements, each with a known elemental equation or source function. The graph also represents the structure of the element interconnections in terms of the continuity and compatibility constraint equations described in Chap. 4. In the following sections we use the properties of linear graphs to (1) derive the system structural constraints, (2) define the set of state variables, and (3) provide a systematic technique for deriving the system state equations [4-8]. The following definitions are introduced: System graph: The oriented linear graph model of a system. Connected graph: A system graph in which a path exists between all pairs of nodes. A path is said to exist if the node pair is joined by at least one branch. Figure 5.2 shows a connected graph along with a system graph that is not connected. System graphs for systems consisting of one-port elements are usually connected graphs, while systems that include the two-port elements introduced in Chap. 6 may not generate connected graphs. In this chapter we assume that all system graphs are connected graphs.
Sec. 5.1
State Variable System Representation
121 Output vector y
---~
u,(t)
System described by state variables {x 1,x2, ••• Xn}
~ ~--~
t
t Figure 5.1:
System inputs and outputs.
sufficient to determine all future behavior of the system. The state variables are an internal description of the system that completely characterizes the system state at any time t and from which any output variables y;(t) may be computed. Large classes of engineering, biological, social, and economic systems may be represented by state-determined system models. System models constructed with the pure and ideal (linear) one-port elements defined in the preceding chapters are state-determined system models. For such systems the number of state variables n is equal to the number of independent energy storage elements in the system. The values of the state variables at any time t specify the energy of each energy storage element within the system and therefore the total system energy, and the time derivatives of the state variables determine the rate of change of the system energy. Furthermore, the values of the system state variables at any time t provide sufficient information to determine the values of all other variables in the system at that time. There is no unique set of state variables that describe any given system; many different sets of variables may be selected to yield a complete system description. However, for a given system the order n is unique and is independent of the particular set of state variables chosen. State variable descriptions of systems may be formulated in terms of physical and measurable variables or in terms of variables that are not directly measurable. It is possible to mathematically transform one set of state variables to another; the important point is that any set of state variables must provide a complete description of the system. In this text we concentrate on a particular set of state variables based on physical variables derived directly from linear graph models.
5.1.2 The State Equations A standard form for the state equations is used throughout system dynamics. In the standard form the mathematical description of the system is expressed as a set of n coupled first-order ordinary differential equations, known as the state equations, in which the time derivative of each state variable is expressed in terms of the state variables x 1(t), ... , xn(t) and the system inputs u 1 (t), ... , u, (t). In the general case the form of the n state equations is it
=It (X, u, t)
i2 = /2 (x, u, t)
. .
·-· .-.
Xn = fn (X, U, t)
(5.1)
State Equation Formulation
122
=
Chap.5
=
dx; /dt and each of the functions /; (x, u, t) (i 1, ... , n) may be a general, where i; nonlinear, time-varying function of the state variables, the system inputs, and time. 1 It is common to express the state equations in a vector fonn in which the set of n state variables is written as a state vector x(t) = [xt (t), x2(t), ... , Xn (t)]T and the set of r inputs is written as an input vector u(t) = [u 1(t), u 2(t), ..• , u,(t)]T. Each state variable is a time-varying component of the column vector x(t). This fonn of the state equations explicitly represents the basic elements contained in the definition of a state-determined system. Given a set of initial conditions (the values of the x; at some time to) and the inputs fort ~to, the state equations explicitly specify the derivatives of all state variables. The value of each state variable at some time ~t later may then be found by direct integration. The system state at any instant may be interpreted as a point in ann-dimensional state space, and the dynamic state response x(t) can be interpreted as a path or trajectory traced out in the state space. In vector notation the set of n equations in Eqs. (5.1) may be written
(5.2)
i = f (x, u, t)
where f (x, u, t) is a vector function with n components /; (x, u, t). In this text we restrict our attention primarily to a description of systems that are linear and time-invariant (LTI), that is, systems described by linear differential equations with constant coefficients. For a LTI system of order n and with r inputs, Eqs. (5.1) become a set of n coupled first-order linear differential equations with constant coefficients:
X] = =
i2
Q)]X] a21X1
+ a]2X2 + ... + atnXn + h]JUJ + ... + btrUr + a22x2 + · · · + a2nxn + h2tUt + · · · + b2rur
(5.3)
where the coefficients a;i and b;1 are constants that describe the system. This set of n equations defines the derivatives of the state variables to be a weighted sum of the state variables and the system i~puts. Equations (5.3) may be written compactly in matrix form:
[x'] = [a"
!!_ ~2 dt
a~1
:
:
Xn
ant
a12 a22
a,.. ][x,. a2n
..
an2
ann
x2
..
Xn
l [ !l[] +
bu b21
..
.
b,,
(5.4)
bnl
which may be summarized as
i=Ax+Bu
(5.5)
In this text we use boldface type to denote vector quantities. Uppercase letters are used to denote general matrices, while lowercase letters denote column vectors. See App. A for an introduction to matrix notation and operations.
Sec. 5.1
State Variable System Representation
123
where the state vector x is a column vector oflength n, the input vector u is a column vector of length r, A is an n x n square matrix of the constant coefficients a;i, and B is an n x r matrix of the coefficients bu that weight the inputs.
5.1.3 Output Equations
A system output is defined to be any system variable of interest A description of a physical system in tenns of a set of state variables does not necessarily include all the variables of direct engineering interest An important property of the linear state equation description is that all system variables may be represented by a linear combination of the state variables x; and the system inputs u;. An arbitrary output variable in a system of order n with r inputs may be written
(5.6) where the c; and d1 are constants. If a total of m system variables are defined as outputs, the m such equations may be written as
Y2
= =
Ym
=
Yt
ClJXt C2JXJ
CmtXt
+ CJ2X2 + ··· + CJnXn + dnUt + · · · + dlrUr + C22X2 + •·· + C2nXn + d21U1 + •••+ d2rUr + Cm2X2 + ··· + CmnXn + dmtUl + · · · + dmrUr
or in matrix form as
[
Y2 Yt
.. .
Ym
l[
Ct2
C2t cu
en
Cml
Cm2
.. .
(5.7)
Ctn ] [ Xt C2n X2
..
.
Cmn
l [ !][]
.. + .
Xn
du d21
...
(5.8)
dml
The output equations, Eqs. (5.8), are commonly written in the compact form: y=Cx+Du
(5.9)
where y is a column vector of the output variables y; (t), C is an m x n matrix of the constant coefficients c;i that weight the state variables, and D is an m x r matrix of the constant coefficients d;i that weight the system inputs. For many physical systems the matrix D is the null matrix, and the output equation reduces to a simple weighted combination of the state variables: (5.10)
State Equation Formulation
124
Chap. 5
5.1.4 State Equation-Based Modeling Procedure
The complete system model for a linear time-invariant system consists of (1) a set of n state equations, defined in terms of the matrices A and B, and (2) a set of output equations that relate any output variables of interest to the state variables and inputs and are expressed in terms of the C and D matrices. The task of modeling the system is to derive the elements of the matrices and to write the system model in the form x=Ax+Bu y=Cx+Du
(5.11) (5.12)
The matrices A and B are properties of the system and are determined by the system sttucture and elements. The output equation matrices C and D are determined by the particular choice of output variables. The overall modeling procedure developed in this chapter is based on the following steps: 1. Determination of the system order n and selection of a set of state variables from the linear graph system representation. 2. Generation of a set of state equations and the system A and B matrices using a welldefined methodology. This step is also based on the linear graph system description. 3. Determination of a suitable set of output equations and derivation of the appropriate C and D matrices. 5.2
LINEAR GRAPHS AND SYSTEM STRUCTURAL PROPERTIES
5.2.1 Linear Graph Properties
The derivation of the state equations in this chapter is based on the use of the system linear graph model. A linear graph with B branches represents B system elements, each with a known elemental equation or source function. The graph also represents the sttucture of the element interconnections in terms of the continuity and compatibility constraint equations described in Chap. 4. In the following sections we use the properties of linear graphs to (I) derive the system structural constraints, (2) define the set of state variables, and (3) provide a systematic technique for deriving the system state equations [4-8]. The following definitions are introduced: System graph: The oriented linear graph model of a system. Connected graph: A system graph in which a path exists between all pairs of nodes. A path is said to exist if the node pair is joined by at least one branch.
Figure 5.2 shows a connected graph along with a system graph that is not connected. System graphs for systems consisting of one-port elements are usually connected graphs, while systems that include the two-port elements introduced in Chap. 6 may not generate connected graphs. In this chapter we assume that all system graphs are connected graphs.
Sec. 5.2
125
Linear Graphs and System Structural Properties
(a)
(b)
Figure 5.2: Examples of (a) a connected system graph, and (b) an unconnected system graph.
Consider a system represented by a connected linear graph with B branches of which S branches are active source elements and the remaining B - S branches represent passive one-port elements. Each branch has an across-variable and a through-variable, giving a total of 2B variables within the system. Of these, S are prescribed source variables, and so there are 2B - S unknowns in the system; we therefore require a total of 2-B - S independent equations in these unknowns in order to determine all system variables. There are B - S elemental equations relating the across- and through-variables for the passive branches. In addition the system structure, defined by the compatibility and continuity conditions, may be used to generate an additional B linearly independent constraint equations. The total set of 2B - S elemental and structural equations may be algebraically manipulated to produce the state and output equations. The following example illustrates these relationships for an electric system model. Example 5.1 The electric system shown in Fig. 5.3 consists of a capacitor C, an inductor L, and a resistor R connected as shown and driven by a voltage source Vs(t). Its linear graph contains four branches each of which has an across-variable and a through-variable, giving a total of eight system variables of which seven (ic, vc. h. VL, iR, VR, and is) are unknown. Each of the three passive elements is described by an elemental equation: (i) (ii) (iii)
giving three equations in six unknowns. The system structure defined by the linear graph imposes additional constraints; the three variables on the right-hand side may be eliminated by using (1) a continuity equation (iv) to define ic, and (2) two compatibility equations VL
= VC
VR
= Vs- vc
(v) (vi)
126
State Equation Formulation
Chap.5
to define VL and VR· Substitution of Eqs. (ivHvi) into Eqs. (iHili) and some algebraic manipulation produce a pair of coupled differential equations:
dve 1 1. 1 -=--ve+-rL+-Vs dt RC C RC
dit
1
dt =
(vii) (viii)
Lve
which are a pair of coupled first-order differential equations in the form of Eqs. (5.8). These are a set of two state equations for this system in tenns of state variables ve and i L. Equations (vii) and (viii) allow the system A and B matrices to be written
A= [-1/RC -1/C] 1/L
0
,
B = [ 1/RC] 0
(ix)
Equations CiHvi) may be used to generate output eqWuions in the fonn of Eqs. (5.7). For example, if the variables iR. VR, VL. and ic are of interest,
. lR
1 1 = -live + li Vs
(X)
= -ve + Vs VL = Ve
(xi)
VR
(xii)
. 1 . 1 v. ze =-livelL + R s
(xiii)
l
(xiv)
and so if the output vector is defined to be y = [i R, vR, vL, ie ]T, the C and D matrices are
-1/R -1 C=
[
1
-1/R
00 o ' -1
D=[T] l/R
R
R
L
Figure 5.3:
An electric system and its linear graph.
Sec. 5.2
Linear Graphs and System Structural Properties
127
5.2.2 Graph Trees
A systematic procedure for generating the system equations is based on relationships defined by a graph tree. A tree is defined to be a subgraph of the system graph containing 1. all of the graph nodes, and 2. the maximum number of branches of the system graph that can be included without creating any closed loops. Branches of the system graph that are not included in a tree are known as links. In general several different graph trees may be formed from any linear graph, as shown in Fig. 5.4. The number of branches in a tree Br and the number of tree links BL depend on the number of nodes N and the number of branches B in the graph. For a connected graph, the first branch entered into a tree connects two nodes, while all of the subsequent branches connect one additional node until the maximum number (N- 1) of branches is entered without forming any closed loops. The number of tree Jinks is simply the total number of branches Jess those in the graph tree: Br
= N -1
Bt = B - Br = B - N
(5.13) (5.14)
+1
Each tree link is a system graph branch that forms a closed loop when added to a graph tree.
\l)
~
v T
\l l) Figure 5.4: A system graph with four nodes and five branches. and severaJ trees derived from the graph.
128
State Equation Formulation
Chap. 5
5.2.3 System Graph Structural Constraints A graph tree may be used to generate a set of B linearly independent compatibility and continuity equations that are particularly useful in system equation development. We define the following classes of variables based on a given tree: Primary variables: The system primary variables are the across-variables on tree branches and the through-variables on tree links. Secondary variables: Conversely, the secondary variables are the through-variables on tree branches and the across-variables on tree links. The compatibility constraint equations relate across-variables around any closed loop in the system graph. Consider the tree of a linear graph with N = 5 and B = 6 shown in 4 branches and B - N + 1 2 links in any tree of this Fig. 5.5. There are N - 1 graph. Each time a link is placed in the tree, a closed loop is formed, and a compatibility equation may be written for the loop. For the tree shown in Fig. 5.5 the two compatibility equations generated by replacing the links are
=
=
In this tree structure v2 and v 4 are secondary variables; each of the above compatibility equations specifies one of these secondary variables in terms of the primary across-variables. In general, each time a loop is formed by placing a link in a tree, the across-variable on that link may be written in terms of the across-variables on the tree branches; in other words, the resulting compatibility equation contains only one secondary variable.
Figure 5.5: A system graph tree, with the links shown as dotted lines, and a set of contours used to generate continuity equations.
The continuity equations represent a set of constraints on through-variables passing through any closed contour on the graph. If a closed contour is drawn on a tree such that it cuts only one tree branch, then the through-variable on that branch (a secondary variable) may be expressed in terms of the through-variables on the tree links, which are primary variables. In Fig. 5.5 a set of four contours has been drawn so that each cuts a single tree branch. For the contours shown the following continuity equations may be written: fl = -f2
f3 = f2- f4 fs
= f4
f6 = f4
Sec. 5.2
Linear Graphs and System Structural Properties
129
The secondary through-variables on the tree branches f1, f3, fs, and f6 are expressed directly in tenns of the through-variables on the links (f2 and f4). This particular set of contours has generated a set of continuity equations that express the secondary through-variables in tenns of the primary through-variables. In summary, each of the B - N + 1 tree links generates a compatibility equation involving a single secondary across-variable, and each of the N- I tree branches generates a continuity equation expressing its secondary through-variable in terms of the link primary through-variables. The tree therefore generates a set of B constraint equations that can be used in the equation formulation method to eliminate secondary variables from elemental equations. Figure 5.6 shows the division of the system variables into primary and secondary variables using a graph tree.
.
System linear graph: Bbranches Nnodes I
_i
N-ltree branches
B-N-ltree links
I
'
N-l primary across-variables I
I
t I
I
t
l
N- 1 secondary through-variables
•• I Figure 5.6:
l
B- N + 1 primary through-variables
~t
Totals: B prim~ variables B secon ary variables
I
B - N + 1 secondary across-variables I
I
System graph variables.
5.2.4 The System Normal Tree
We may fonn a special graph tree, known as a normal tree, and use it to define ( 1) the system primary and secondary variables, (2) the system order n, {3) the state variables, and (4) a set of independent compatibility and continuity equations. A nonnal tree for a connected system graph is formed using the following steps: Step 1: Draw the system graph nodes. Step 2: Include all across-variable sources as tree branches. (If all across-variable sources cannot be included in the nonnal tree, then the across-variable sources must fonn a loop and compatibility is violated.) Step 3: Include as many as possible of the A-type energy storage elements as tree branches. (Any A-type element that cannot be included in the nonnal tree is a dependent energy storage element as described below.)
130
State Equation Formulation
Chap. S
Step 4: Attempt to complete the tree, which must contain N - 1 branches, by including as many as possible D-type dissipative elements. It may not be possible to include all D-type elements. Step 5: If the tree is not complete after the addition of D-type elements, add the minimum number ofT-type energy storage elements required to complete it. (Any T-type element included in the tree at this point is a dependent energy storage element) Step 6: Examine the tree to determine if any through-variable sources are required to complete it. If any through-variable source can be inserted into the normal tree, then that source cannot be independently specified and continuity is violated. The normal tree effectively defines a set of independent energy storage elements, that is, the set of elements whose stored energy may be independently set and controlled. If in step 3 an A-type element cannot be included in a normal tree, it implies that a loop has been formed from a combination of across-variable sources and A-type elements. The compatibility equation for that loop specifies that the across-variable on the A-type element link is dependent on the across-variables on other A-type elements and sources. This is illustrated in Fig. 5.7a where an electric system and a normal tree are shown. It is not possible to include both capacitors in the normal tree; if capacitor C2 is placed in the tree, the resulting compatibility equation will be
The energy storage variables (voltage) on the two capacitors are therefore linearly dependent In this case either C 1 or C2 , but not both, can be included in the normal tree. Simi1arly, in step 5 any T-type element that is a branch of a normal tree is a dependent energy storage element When the tree is complete, the resulting continuity equation generated by a contour cutting this branch expresses its through-variable in terms of link primary variables. The continuity equation for this branch therefore specifies a linear dependence between its through-variable and other T-type or source through-variables. This is illustrated for a mechanical system in Fig. 5. 7b where a spring K is used to transmit energy between a force source Fs(t) and a mass m. The spring can be inserted into the normal tree. Continuity applied to the spring element requires that
or in other words, the energy stored in the spring is defined entirely by the force source Fs (t). The spring K is a dependent energy storage element Dependent energy storage elements do not generate independent state equations. From the normal tree we make the following definitions: 1. The system primary variables are the across-variables on the normal tree branches and the through-variables on its links. 2. The system secondary variables are the through-variables on the normal tree branches and the across-variables on its links.
Sec. 5.2
Linear Graphs and System Structural Properties
V1 (t)
"-----+---' R
131
r
Vm(t)
F,(t)-:Q~
Linear graph
Linear graph
Normal tree
NormaJ tree
K
K
(b)
(a)
Figure 5.7: Two systems with dependent energy storage elements and their normaJ trees. (a) An electric system with dependent capacitors, and (b) a mechanical system
with a dependent spring.
3. The system order is the number of independent energy storage elements in the system, that is, the sum of the number of A-type elements in the normal tree branches and the number ofT-type elements among the links. 4. The n state variables are selected as the energy storage variables among the primary variables, that is, the across-variables on A-type elements in the normal tree branches and the through-variables on T-type elements in the links.
Figure 5.8 shows two systems along with their system graphs and normal trees. The mechanical system in Fig. 5.8a has four energy storage elements, the normal tree indicates that they are all independent, and the system order is 4. The normal tree defines the state variables to be the velocities (vm 1 and Vm 2 ) of the two mass elements and the forces (FK 1 and FK2 ) on the two springs. The electric system in Fig. 5.8b contains five energy storage elements, but the normal tree indicates that capacitors c2 and c3 are dependent. The order of this system is 4, and the state variables for the normal tree shown are vc 1 , vc2 , i L 1 , and h 2 • ·
State Equation Fonnu1ation
132
Chap. S
Linear graph
Linear graph
K2
Rl
Normal tree
Figure 5.8:
Two systems with their linear graphs and normal trees.
Special Cases in which the Procedure Results in "Excess" State Variables Two special cases in which the procedure outlined above 2 results in a formulation in which the number of state variables exceeds the number of independent energy storage elements are the following:
1. Systems containing two or more A-type elements in direct series connection 2. Systems containing two or more T-type elements in direct parallel connection. 2 In these special cases, the procedure results in a state variable formulation which is uncontrollable, that is, the across-variables on each of the direct series A-type elements and the through-variable on each of the direct parallel T-type elements may not be independently specified and controlled [2, 3].
Sec. 5.2
Linear Graphs and System Structural Properties
133
m V,(t)
Linear graph
Linear graph
Normal tree
Normal tree
I I I ~
Figure 5.9: Examples of two systems with excess state variables.
Illustration of these two special cases is provided in Fig. 5.9. In the two cases the number of state variables identified by forming a normal tree exceeds the number of independent energy storage elements. The difficulty represented by the two special cases described may be avoided by combining all A-type energy storage elements directly in series into a single equivalent A-type element, and all T-type elements directly in parallel into a single equivalent element. The general expressions for the equivalent elements are indicated in Fig. 5.1 0. When the elements are combined and the equivalent system linear graph is used to generate the normal tree, the number of state variables in the combined system is equal to the number of independent energy storage elements in die combined system. Since controllable representations of systems are desired, all A-type elements directly in series and all T-type elements directly in parallel should be combined before determining the system normal tree.
/
1 Ceq=-n--
I
i=l
Figure 5.10:
5.3
Chap. 5
State Equation Fonnu1ation
134
l.cq=
1 -n--
1/C;
I
i;;;J
liL;
Combination of elements in series and in parallel to eliminate excess state variables.
STATE EQUAnON FORMULATION The system normal tree may be used to generate a set of state equations in terms of the energy storage variables on the n independent energy storage elements. In a system graph with B branches, of which S represent ideal source elements, there are 2(B - S) system variables associated with the passive branches: one across- and one through-variable on each branch. On each branch one of these variables is a primary variable, while the other is a secondary variable. The desired n state variables are a subset of the B - S primary variables. There are B - S elemental equations relating these primary and secondary variables for the passive branches; the nonnal tree is used to generate B - S continuity and compatibility equations that can be used to eliminate the secondary variables associated with the passive elements. The state equations are formulated in two steps:
I. Derivation of a set of B - S differential and algebraic equations in terms of primary variables only by starting with the passive elemental equations and using B - S compatibility and continuity equations to eliminate all secondary variables. 2. Algebraic manipulation of this set of B - S equations to produce n differential equations in the n state variables and the S specified source variables. Since sources have one variable independently specified, only B - S elemental equations for the passive elements need to be written. It is convenient to divide the number of sources S into across-variable sources SA and through-variable sources ST, and so S = SA +ST. The secondary variables may be eliminated from these equations by using a total of B - S independent compatibility and continuity equations formed from (1) N -1 -SA continuity equations, ~d (2) B - N + 1 - ST compatibility equations. (The secondary variables associated with sources do not enter directly into the state equation formulation; therefore, SA c~ntinuity and Sr compatibility equations do not need to be considered.)
Sec. 5.3
State Equation Formulation
135
A systematic procedure for deriving the n state equations is as follows:
Step 1: Generate a normal tree from the connected system graph. Step 2: From the branches and links of the normal tree identify the primary and secondary variables. Define the system order n as the number of independent energy storage elements. Select the state variables as across-variables on A-type energy storage elements in the normal tree branches and through-variables on T-type energy storage elements in the links. Step 3: Write the B- S elemental equations for the passive (nonsource) elements explicitly in terms of their primary variables, that is, with the primary variable on the left-hand side of the equations. Note that the derivatives of the n state variables should appear on the left-hand side of the elemental equations for all independent energy storage elements. Step 4: Write N- 1 -SA independent continuity equations involving only one secondary through-variable (a tree branch through-variable) by applying the continuity condition to a set of N - 1 -SA contours that each "cut" only one passive branch of the normal tree. Express each equation explicitly in terms of the secondary through-variable. StepS: Write B - N + 1 - ST independent compatibility _equations involving only one secondary across-variable (a tree link across-variable) by placing each of the passive links back in the tree to form a loop and writing the resulting compatibility equation. Express each equation explicitly in terms of the secondary across-variable. Step 6: Use theN- 1 -SA continuity and the B- N + 1 - ST compatibility equations (a total of B- S equations) to eliminate all secondary variables from the B - S elemental equations by direct substitution. Step 7: Reduce the resulting B- S equations in the primary variables ton state equations in the n state variables and the S source variables. Step 8: Write the resulting state equations in the standard form. The application of this procedure is illustrated in the examples that follow, which show that after the normal tree is formed the subsequent steps in the procedure may be completed in a straightforward manner. Example5.2 A set of state equations describing the mechanical system in Example 4.1 may be derived directly from the system graph, shown in Fig. 5.11. The system graph has three branches (B = 3), two nodes (N = 2), and a single force source (S = 1). The normal tree therefore contains N - 1 I branch. The rules for constructing the tree specify that the branch should contain the mass element m. The spring element K and the source element F therefore form the tree links. From the normal tree in Fig. 5.llc, where the link elements are indicated by dotted lines,
=
Primary variables: F.J(t), Vm, FK Secondary variables: Vs' Fm. v K System order. 2 State variables: Vm, FK
State Equation Formulation
136
Cbap.S
Contour used to continuity
. ; - genera1e
(~~:' equation
v
m
F,(t)
I
\
\
I I I /
cb \
fK
m
~ (a) Physical system
(b) Linear graph
(c) Normal tree
Figure 5.11: Mechanical system model and its normal tree.
1. The B - S
= 2 elemental equations may be written in tenns of the primary variables as
Primary variables
dvm dt { dFg dt
= =
!Fm}
Secondary variables
(i)
Kvg
2. The single continuity equation (N- 1 - S~o = 2- 1 - 0 = 1) required to eliminate Fm may be written using the closed contour shown in Fig. S.llc: Fm
= F,- Fg
(ii)
3. The single compatibility equation (B - N + 1 - S7 = 3 - 2 + 1 - 1 = 1) required to eliminate the secondary variable v K may be found by placing the spring element in the nonnal tree (Fig. S.llc), and writing the loop equation: (iii)
4. The secondary variables Fm and vg may be eliminated by substituting Eqs. (ii) and (iii) into the elemental equations to yield dvm 1 -=-(F,-Fg) dt
m
dFg
(iv)
- - =Kvm dt
Equations (iv) express the derivatives of the state variables directly in terms of the state variables and the input They are the desired pair of state equations. 5. The pair of state equations may be written in the standard matrix notation: (v)
If the output variable of interest is the velocity of the mass, then an output equation may be written in matrix fonn: (vi)
Sec. 5.3
137
State Equation Formulation
Example5.3 A mechanical system and its oriented linear graph are shown in Fig. 5.12. In the graph two T-type energy storage elements are connected direct1y in parallel; therefore, to eliminate the generation of an excess state variable, the two spring elements are combined into one equivalent spring with an effective spring constant K = Kt + K2 • The equivalent linear graph is shown in Fig. 5.12b, with the system normal tree in Fig. 5. l 2c with N - 1 = 1 branch containing the mass m. From the normal tree, Primary variables: F.,(t), Secondary variables:
v.~.
Vm,
FK, Fs
Fm.
VK,
vg
System order: 2 State variables:
Vm,
FK
Total viscous friction B (b) Linear graph
(a) Physical system Figura 5.12:
1. The B - S
(c) Normal tree
A mechanical system with two parallel springs, its linear graph and a nonnal tree after combining the two springs.
= 3 elemental equations written in terms of the primary variables are
: : ~: I ! 1
dvm
Primary variables
dt d FK dt
Fs
Secondary variables
=
(i)
Bvs
2. The single (N - 1 - SA = I) continuity equation required to eliminate the secondary through-variable F,, is found by drawing a closed contour that cuts the ma'is tree branch as shown in Fig. 5.12: (ii)
+ 1 - ST = 2 compatibility equations required to eliminate the secondary variables VK and v 8 are written by considering the loops formed by adding the two links back into the tree shown in Fig. 5.12c:
3. The B - N
(iii) Vg
=
Vm
(iv)
138
State Equation Formulation
Chap.5
4. All secondary variables may be eliminated from the elemental equations by direct substitution of the three continuity and compatibility equations to yield
dvm
1
dFx dt Fs
= Kvm = Bvm
dt =;;; (F,- Fs- Fx)
(v)
(vi) (vii)
=
5. These B - S 3 equations in terms of primary variables may be combined to give the two desired state equations:
dvm 1 = - (-Bvm- Fx dt m
-
dFx dt
+ F,)
(viii)
= Kv
(ix)
m
6. Finally the pair of state equations may be written in the standard matrix fonn: (x)
Example5.4 The mechanical rotational system described in Example 4.2 has the linear graph shown in Fig. 3). The normal 5.13b. The linear graph contains four elements (B 4) and three nodes (N tree, shown in Fig. 5.13c, contains the angular velocity source and the rotary inertia. From the normal tree,
=
=
Primary variables: O,(t), 0;. T8 " Ts,. Secondary variables: T,, TJ, Os1 , Os2 System order: I State variable: OJ
(a) Physical system Figure 5.13:
(b) Linear graph
(c) Normal tree
A rotational system, its linear graph and normal tree.
Sec. 5.3
139
State Equation Formulation
1. The 8 - S = 4 - 1 = 3 elementaJ equations written in terms of the primary variables
are
I
r81
=
2.J 1, 8 1Q 81
TTJz
=
82!2s2
dn, dt
Primary variables
=
I
Secondary variables
2. The secondary variable T1 may be eliminated by the single (N- 1 -
s~.
(i)
= 1) continuity
equation found from the closed contour shown in Fig. 5.13c: (ii)
3. The two secondary variables !2 81 and 0 82 maybeeliminatedfromthe 8-N +1-Sr = 2 compatibility equations formed by placing the two Jinks in the normal tree:
(iii)
4. The secondary variables may be eliminated from the elemental equations (i) using Eqs. (ii) and (iii): dO, dt Ts 1
= .!_J (TB2 -
Ts) I
= BtQJ
(iv)
Ts 2 = 82 (Q.f- !2,) 5. By combining the elemental equations in Eq. (iv) to eliminate Ts 1 and Ts2 • the single state equation may be written
(v)
6. The state equation may be expressed in matrix form as
(vi)
7. If the flywheel angular velocity n, is selected as the output variable, the output equation in matrix form is (vii)
140
State Equation Formulation
Chap. 5
Example5.5 An electric circuit is shown in Fig. 5.14a together with its oriented linear graph model in Fig. 4 and 8 4. The normal 5.14b. For the model the numbers of nodes and branches are N tree, shown in Fig. 5.J4c, has N- 1 3 branches which include the voltage source Vs. the capacitor C, and the resistor R. The inductor branch L remains in the tree links. From the normal tree,
=
=
Primary variables: Vs(t), vc, VR,
=
h
Secondary variables: is. ic. iR, VL System order: 2 State variables: vc.
R
h
c
c
R
Vs(t)
(a) Physical system Agure 5.14:
(b) Linear graph
(c) Normal tree
An electric circuit, its linear graph, and a normal tree.
The state equations may be derived in the following steps:
I
1. The 8 - S = 3 elemental equations are written in terms of the primary variables:
Primary variables
[
d;; dh
= ~ic = .!..vL
dt VR
Secondary variables
(i)
L
=
RiR
2. TheN- 1 -SA = 2 continuity equations required to eliminate the two secondary through-variables ic and i R are found by drawing a pair of closed contours that cut only one normal tree branch:
ic=h iR =it
(ii) (iii)
3. The B - N + 1 - Sr = 1 compatibility equation to eliminate the secondary variable VL is written by adding the inductor back into the tree and writing the equation (iv)
Sec. 5.3
State Equation Formulation
141
4. The secondary variables may be eliminated from the right-hand side of the elemental equations by direct substitution from the continuity and compatibility equations to generate three equations in the primary variables: dvc
1.
diL
1
Tr = c'L
(v)
dt = L (Vs VR
(vi)
VR - Vc)
=Rh
(vii)
5. The B - S = 3 equations in the primary variables may be combined to yield two first-order differential equations in terms of the state and source variables:
1.
dvc
Tr
=
c'L
(viii)
diL 1 , = - (-vc - R1L
dt
L
+ V.)s
(ix)
6. The pair of state equations may be written in the standard matrix form: 0 [ vc h ] = [ -1/L
1/ C ] [ vc ] -R/L h
+[
0 ] 1/L Vs(t)
(x)
VL,
ic. and v,. are defined to be output variables for this example, a set of algebraic output equations may be formed in the state and input variables by using the elemental, compatibility, and continuity equations. In particular, If the system variables
(xi)
= Vs- RiL- vc ic = iL
(xii)
vc = vc
(xiii)
The output equations y = Cx + Du become
~
·[VL]
=
~ -R] ~. [ ;~] + [1] ~ V,(t)
[-]
(xiv)
Example5.6 The fluid system described in Example 4.5 is shown in Fig. 5.15 together with its linear graph. The linear graph contains five elements (B 5) with two energy storage elements and one source (SA= 1). The graph contains four nodes (N = 4); thus, N- 1 = 3 elements must be placed in the normal tree. The order of placement of elements into the tree is ( 1) the pressure source (an across-variable source), (2) the fluid capacitance C (an A-type element), and (3) the resistive element Rp. At this point the tree is complete and is as shown in Fig. 5.15c. The independent energy storage elements are then the fluid capacitance C and the fluid inertance of the pipe /p:
=
142
State Equation Formulation
Chap. 5
Valve Long pipe
Rf
~j
I-Fl-uidreserv-oir _ _ P_
(a) Physical system
(b) Linear graph
Figure 5.15:
(c) Normal tree
A fluid system, its system graph, and a normal tree.
Primary variables: P,(t), Pc. PR,• QJ,, QR 1 Secondary variables: Qs. Qc. QR,. P,P, PR 1 System order: 2 State variables: Pc, Qlp
1. The B - S = 4 elemental equations are
Primary variables
dPc dt dQ1 dt
=
PRp QRI
1 -Qc
=
c
I -QI lp RpQRp
=
Secondary variables
(i)
1 Rl PR.
2. The two secondary flow variables Qc and QRp may be expressed in terms of primary variables using the N - 1 - S~a 2 continuity equations. Using the contours shown in Fig. 5.15c,
=
Qc
= Q,- QR1
QRp = Ql
(ii)
Sec. 5.3
State Equation Formulation
143
3. The two secondary pressure variables P1 and PR, may be expressed in terms of primary variables from the B - N + 1 - Sr = 2 compatibility equations formed by placing the links back into the tree:
P1 = Ps - PR, - Pc PR1
(iii)
= Pc
4. The secondary variables in the elemental equations (i) may be eliminated by direct substitution from the continuity and compatibility equations: (iv) (v)
(vi) (vii)
5. Equations (vi) and (vii) may be used to eliminate PR, and QR, from Eqs. (iv) and (v) to yield two state equations: (viii) (ix) 6. The pair of state equations may be written in the standard matrix fonn:
Pc] [-l/R1C 1/C ] [Pc] [ Q., = -1/1 -Rp/ I Q1
+ [ 1/0 I ]
Ps(t)
(x)
Example5.7 The thermal system model for Example 4.6 has the linear graph shown in Fig. 5.16 consisting of four elements (B = 4) with one energy storage element (a thermal capacitance), a throughvariable source ( Sr = I ). and three nodes N = 3. The order of entry of elements into the nonnaJ tree is ( 1) the capacitance C, and (2) one of the two thermaJ resistances R 1 or R2 • It is arbitrary which one is chosen; in this example we select R2 as a normal tree branch. Primary variables: Q.,(t), Tc. QR,. TR 2 Secondary variables: Tu Qc. TR,. qR2 System order: I State variable: Tc
144
State Equation Fonnulation
Chap. 5
Heating element Q,(t)
source
(a) Physical system
(b) Linear graph
(c) Nonnal tree
Figure 5.16: A thennal system, iiS linear graph. and a nonnal tree.
Primary variables
I
d:,c
=
TR2
=
QRJ
=
Secondary variables
(i)
2. The secondary heat flow variables qc and q R2 may be expressed in tenns of primary variables using the N- 1 -SA = 2 continuity equations derived from the contours shown in Fig. 5.16c: qc =qR• QR2
(ii)
= Qs -qR 1
3. The secondary temperature variable TR 1 may be expressed in terms of primary variables from the B - N + 1 - Sr 1 compatibility equations fanned by placing the link back into the tree: (iii)
=
4. Using the continuity (ii) and compatibility (iii) equations, the secondary variables may be eliminated from the elemental equations (i):
dTc
1
Tt = cqR. TR2
(iv)
= R2 (Q,- qR,) 1
qR1 = Rt (TR 2
-
Tc)
(v)
(vi)
Sec. 5.4
145
Systems with Nonstandard State Equations
5. Equations (v) and (vi) may be used to eliminate all but state and source variables from Eq. (iv). Combining Eqs. (v) and (vi).
1
qR 1
= R1 + R2 (R2Q8- Tc)
(vii)
and substituting into Eq. (iv). dTc
1
dt = C (R1 + R2)
(R Q 2
8
T. -
c)
(viii)
6. If temperature is selected as the output variable. the matrix forms of the state and output equations are
[tc] = [ C
(R~: R2)] Tc + [ C (R~~ R2)] Qs(t)
Tc = [1] Tc
5.4
+ [0] Q.t(t)
(ix) (x)
SYSTEMS WITH NONSTANDARD STATE EQUATIONS
5.4.1 Input Derivative Form Occasionally system models generate a set of state and output equations that contain terms in the derivative of the system inputs. In such cases. which generally result from models containing dependent energy storage elements, the standard form of the state equations must be extended to include these derivative terms:
i = Ax+ Bu + Eti y=Cx+Du+Fti
(5.15) (5.16)
The n x r matrix E and the m x r matrix F are introduced to include the input derivative terms ti(t). Assume, for example, that a system contains a dependent T-type element L. The element is a branch in the normal tree. and its primary variable is the across-variable v. The elemental equation is written with the derivative of the through-variable f on the right-hand side: df (5.17) V=Ldt
If the continuity equation used to eliminate the secondary variable f contains any throughvariable source terms F3 (t), substitution into Eq. (5.17) generates a term involving the derivative of the input:
v
d
= L(.. · + Fs + · · ·) dt
(5.18)
This input derivative term remains throughout the subsequent algebraic manipulations that generate the state and output equations.
146
State Equation Formulation
Chap.5
The most common situations that generate the extended form of the state equations are the following: 1. When a compatibility equation includes the across-variable on a dependent A-type element and an across-variable source term 2. When a continuity equation includes the through-variable on a dependent T-type element and a through-variable source term. Example5.8 A lead-lag electric circuit, used in control systems, is shown in Fig. 5.17 together with its system graph and normal tree. Find a state equation describing the system and an output equation for the current in the capacitor C1•
V.r(t)
(b) Linear graph
(a) Circuit
(c) Normal tree
Figure 5.17: Simple first-order electric lead-lag network with a dependent energy storage element.
Solution The system is first-order and contains a dependent capacitor. 'IWo normal trees are possible; with the choice of the nonnaJ tree as shown: Primary variables: Vs(t), vc2 • iR,. iR2' ic, Secondary variables: System order: 1
i.r,
ic2 , vR., VR2 , vc,
State variable: vc2 The elemental equations are dvc2 dt
=
c2'c2
iR1
=
Rt VR,
jR2
=
1 R2 VR2
ic,
=
Primary variables
1 •
I
Secondary variables
{i)
C dvc 1
ldr
The normal tree generates one continuity equation: {ii)
Sec. 5.4
Systems with Nonstandard State Equations
147
and three compatibility equations: VR2
=
VR 1
= V.s(t) - vc2
(iv)
vc1 = Vs(t) - vc2
(v)
Vc2
(iii)
When Eq. (v) is substituted into Eqs. (i) to eliminate the secondary variable vc1 , the elemental equation becomes ic = Ct [dVs(t) _ dvc2 ] I dt dt
(vi)
The derivative propagates through the subsequent algebraic steps, and the state equation is (vii) The output equation for ic 1 may be found from Eqs. (i) and (v): (viii)
and substituting Eq. (vii), (ix)
Both the state and output equations include the derivative of the input Vs(t).
5.4.2 Transformation to the Standard Form If the modeling procedure generates state equations in the form of Eq. (5.15), an algebraic transfonnation may be used to generate a new set of state variables and state equations in the standard fonn of Eq. (5.5). For a system with state equations,
i=Ax+Bu+Eo
(5.19)
a new set of state variables x' may be defined:
x' =x-Eu
(5.20)
Differentiation of Eq. (5.20) gives
i' = i - Eo= (Ax+Bu+Eti)- Eo
= A (x' + Eu) + Bu =Ax' +B'u
(5.21)
148
State Equation Formulation
Chap.5
x
where B' = AE + B. Although the elements of the new state vector are no longer the physical variables that were used to generate the original equations, the complete vector r satisfies all the requirements of a state vector. Any physical variable in the system may be found from a modified set of output equations written in terms of the transformed state vector:
y=Cx+Du+FU = C(x'+Eu) +Du+Fti
(5.22)
= Cx' +D'u+Fii whereD'
= CE+D.
Example5.9 Figme 5.18 illustrates a mechanical translational system. The system graph and normal tree demonstrate that although there are three energy storage elements, the two springs are dependent and the system order is 2. Find a set of state equations that does not involve the derivative of the input vector and an accompanying output equation for the velocity of the dash pot B 1•
m
lSS,
;;;;;;;,;;;;;;;;;;;;;;;;;;;. (a} Mechanical system
(b} Linear graph
Figure 5.18:
(c) Normal tree
Second-order mechanical system with dependent springs.
Solution With the choice of nonnal tree as shown~ the state variables are Vm and FK 1 and the elemental equations are
dvm _ _!_F. dt - m m dFKt ---;it = KlVKt
(ii)
= B2vs2
(iii)
Fs2
1
vs 1 = B Fs 1
(i)
(iv)
1
I dFK
VK 2
=K2- -dt-2
(v)
Sec. 5.4
Systems with Nonstandard State Equations
149
There are two compatibility equations: VK 1
=
VB2
= Vm
VK2 -
(vi)
VB 1
(vii)
and three continuity equations: FK2
= Fs(t)- FK1
Fs 1
= FK
Fm
(viii) (ix)
1
= F.s(t) -
Fs2
(x)
Equation (viii) relates the through-variable FK2 on a dependent T-type element K 2 to the input Fs(t). When the secondary variables are eliminated and the two state equations are written in matrix form, the result is
[
-B2/m
Vm ] _ [
FK 1
0
-
0
-K, K2/ [Bt (Kr
+ K2)]
][
Vm ]
+
FK 1
[
1/m] F. (1) 0
s
(xi)
+ [KI/(K~ + K2)] Fs(t)
showing the dependence on the derivative of the input F,, (t). The output equation for v81 is found from Eqs. (iv) and (ix): 1 (xii) VBl = -FKI Br and in matrix form, (xiii) The state variables may be transformed using Eq. (5.20); that is,
i'
= Ax' + (AE +B) u
or
(xiv)
and the corresponding output equation in terms of the transformed state vector is found directly from Eq. (5.22): vs 1
= Cx' + (CE +D)u +Fil
= [0
1/ B1 ]
[~i] + [[o
1/81] [ Kt/
+ [0] Fs.(t) 1 ,
Kr
= Bt x 2 + 81 (Kt + K2) Fs(t)
(K~ + K
2
)]
+ [0]] Fs(t) (xv)
150 5.5
State Equation Formulation
Chap. 5
STATE EQUATION GENERATION USING LINEAR ALGEBRA For linear systems, the state equations may also be found directly from the equations in the primary variables and inputs by matrix methods. After the elemental equations have been written with causality defined by the normal tree and the substitutions for the secondary variables made through the continuity and compatibility equations, the primary variables may be divided into three groups and expressed in three vectors: • x-the vector of state variables (primary variables associated with independent energy storage elements) • d-the vector of primary variables associated with dependent energy storage elements • p-the vector of primary variables associated with nonenergy storage elements [Dtype elements and two-port branches (Chap. 6)]. The set of equations in the primary variables and inputs can be written as three matrix equations in x, d, and p:
i = Px+Qp +Rd+So d=Mi+Nti
(5.23)
p = Hx+Jp+Kd +Lu
(5.25)
(5.24)
The state equations are found by solving explicitly for p and d and substitution into Eq. (5.23). If there are no dependent energy storage elements in the system, that is, d = 0 and N = K = R = 0, the state equations are found by solving Eq. (5.25) for p and M substituting into Eq. (5.23). In the simplest case, the matrix J = 0 and the state equations may be found by substituting Eq. (5.25) directly into Eq. (5.23):
=
X = [P + QH] X + [S + QL] u
(5.26)
Otherwise p must be found by rearranging and solving Eq. (5.25):
[1-J]p=Hx+Lu
(5.27)
where I is the identity matrix, or
p
= [I - J]-1 Hx + [I = H'x+L'u
where B' = [I- J]- 1 Hand L' (5.23) generates the state equations
i
J]- 1 Lu
(5.28) (5.29)
= [I- J]- 1 L. Substitution of Eq. (5.29) into Eq.
= {P + QB') X + {S + QL') u
(5.30)
Sec. 5.5
State Equation Generation Using Linear Algebra
151
Example 5.10
The electric circuit shown in Fig. 5.19, together with its linear graph and a normal tree, generates the following equations in the primary variables: 1. 11 -c'R• +c ,
.
(i)
vc = . 1 h = LVR2 VR2
. 'R1
= R2iRl
=
(ii} (iii}
- R2iL
1 1 R, vc- R
1
(iv)
VR 2
Use the matrix method to find the state equations in the variables vc and h. Soludon Divide the primary variables into two vectors, X = [vc iLlT and p = [ VR2 Then the primary equations [Eqs. (iHiv)] can be written
[ ~c]=[O0 lL
O][~c]+[
0
lL
0
1I L
-I/C][~R2]+[1/C] 1s(t) 0
0
l R,
iRl
r. (v)
(vi)
Equation (vi) may be rearranged:
(vii)
(viii)
R2] [ 0 1
1/Rt
-R2] [vc] h 0
(ix) (x)
When Eq. (ix) is substituted into Eq. (v), the state equations are found to be
152
State Equation Formulation
(b) Linear graph
(a) Electric circuit
Figure 5.19:
Chap.5
(c) Nonnal tree
An electric system, its linear graph. and a normal tree.
In the most general case, when there are dependent energy storage elements in the system so that d :F 0, the state equations are found by substituting Eqs. (5.29) and (5.24) into Eq. (5.23), giving
[I- (QK' +R)M]i = (P + QH')x + (S + QL') u + (R + QK') Nti which may be solved by premultiplying both sides by the state equations. 5.6
(5.31)
[I- (QK' + R) M]- 1 to generate ·
NONLINEAR SYSTEMS 5.6.1 General Considerations
In the general case of nonlinear state-determined system models, relationships may exist in which a single system variable is a nonlinear function of several other variables. For example, the torque output of an automobile engine may be represented as a nonlinear function of the fueVair ratio, spark advance, and engine speed. Systematic equation formulation procedures are not generally available for such systems. However, for nonlinear systems that can be represented by linear graph models containing combinations of pure (nonlinear) and ideal Oinear) one- and two-port elements, the equation formulation procedures described in Sees. 5.1-5.4 provide an effective starting point for the formulation of a set of nonlinear state equations. For such models, the system constraints represented by a set of compatibility and continuity equations are valid. Thus, the procedure of forming a normal tree and identifying the independent variables and state variables is unchanged from the method described for linear system models, as are the procedures for generating the compatibility and continuity equations. With models containing nonlinear elements the algebraic steps in the subsequent elimination of secondary variables are not always straightforward and may not always be accomplished with simple analytical techniques. However, for many nonlinear systems containing pure elements, the methods developed for linear system models may be used with some modification to derive a set of nonlinear state equations. The formulation of nonlinear state equations is illustrated in the following section, where a set of nonlinear state equations is derived in the general form of Eqs. (5.2), that is, i
= f(x, u, t)
Sec. 5.6
Nonlinear Systems
153
where the derivative of each state is related by a nonlinear function to the system state variables and inputs. Some nonlinear systems models may not be written explicitly as a set of nonlinear state equations in the standard form of Eqs. (5.1) because it may not be possible to solve directly for the derivative of each state variable independently.
5.6.2 Examples of Nonlinear System Model Formulation To illus trate the equation derivation procedures for systems containing nonlinear elements, state equations are derived for three representative systems in the following examples. Each example contains one or more nonlinear dissipative or energy storage elements. Example 5.11 An automobile, with mass m, traveling on a flat, straight section of road is shown in Fig. 5.20. Derive a model for the forward speed Vm of the car in response to a time-varying propulsive force F,(t) generated by the engine. The model is restricted to the moderate- to high-speed range where the two primary resistance forces (in the absence of braking) are road resistance F, and aerodynamic drag Fa , approximated as (i)
(ii) Solution The road resistance is represented as a constant resistance force Fo. and the aero· dynamic drag Fa is represented as a resistive force which increases with the square of vehicle velocity with the constant ca dependent on the automobile drag coefficient.
F,(r)'fJ!Fd~ F, m\
\
-1----F,
' (b) Linear graph
(a) Nonlinear mechanical system Figure 5 .20:
(c) Normal tree
Linear graph model of automobile with nonlinear drag forces.
A system graph model for the car is shown in Fig. 5.20b, in which the car mass m is represented as an ideal element, the propulsive force is represented as an ideal force source F,(r), the road resistance is represented as an ideal force source Fo, and the aerodynamic drag is represented as a nonlinear damper element. The system contains one independent energy storage element as shown by the normal tree. From the normal tree (Fig. 5.20c), Primary variables: Vm, F,, F, , Fa Secondary variables: Fm . v, , v,, va System order: I State variable: Vm The elemental equations for the two passive elements are
dvm I = - Fm
(iii)
Fa= cav~
(iv)
dr
m
State Equation Formulation
154
Chap. 5
The nonnal tree generates one continuity equation: (v)
and one compatibility equation: (vi) Substitution to eliminate the secondary variables from the elemental equations yields (vii) (viii)
Combining these two equations generates the system nonlinear state equation: 1 ( -dvm dt = m F.P -
2)
(ix)
Fo- cdvm
Example 5.12 Part of a sensitive torque-measuring instrument, shown in Fig. 5.21a, consists of a pendulum of length l with a proof mass m mounted in support bearings in a gravity field g .
-
, ' /'
n I
I
T1 (t)
T,(t)
~~~ I
I
....,, r
~
t
...
\
\
' tB
Jl K J \ \ I I ', \ I I ,\I /
~
(a) Nonlinear pendulum Figure 5.21:
(b) Linear graph
(c) Normal tree
A mechanicaJ torque-measuring instrument based on a pendulum.
The input to the instrument is a time-varying torque T,(t) which acts on the shaft to displace the pendulum by an angle 8. Under static input conditions where Ts(t) = T0 , the gravitational restoring torque is a function of the angular displacement: (i)
To= mglsin8
This is a algebraic relationship between a torque (through-variable) and a displacement (integrated across-variable) and represents the constitutive relationship of a nonlinear torsional spring. The elemental equation for the equivalent spring K is found by differentiating Eq. (i): dTK
-
dt
df)
= (mglcos8)dt
(ii)
Sec. 5.6
155
Nonlinear Systems
The support bearings for the shaft have a resistive torque T8 proportional to the shaft angular velocity Sl B: Ts = B!ls
(iii)
In addition to the gravitational restoring torque, the pendulum also has a rotational inertia with a point mass m concentrated at a radius I from the center of rotation, and so the moment of inertia is (iv) By considering the inertia, the equivalent spring, the damper, and the input source, a system graph model may be formulated, as shown in Fig. 5.21b, that has the normal tree shown in Fig. 5.21c. From the linear graph and normal tree,
Primary variables: Sl;, Ts, TK, Ts Secondary variables: T;, ns, nK, ns System order: 2 State variables:
nJ, TK
There are three elemental equations: dSl; 1 --=-T; dt J dTK d9 dr (mgl COS 9) dt
=
(v)
= (mgl COS 9) QK
(vi)
(vii)
Ts = B!ls
The normal tree generates a single continuity equation:
TK = - T; - TB
+ 1's
(viii)
and two compatibility equations: (ix) (x)
Elimination of the secondary variables yields
-dQJ = -J1 (Ts dt dTK
B!l; - TK)
dr = (mgl cos 9) nJ
(xi) (xii)
State Equation Formulation
156
Chap. 5
The equations may be expressed directly in terms of the state variables alone by substituting directly for 8 from the constitutive equation [Eq. (i)}:
9
= sin-• (~) mgl
(xiii)
and Eq. (xii) may be written
dJ,K = mgl cos [sin- 1 (;;l)] Q 1
(xiv)
Equations (xi) and (xiv) represent a pair of state equations for the nonlinear system, while
Eq. (xiii) allows the computation of 8. the system output. The nonlinearity in this system arises from the trigonometric relationship of the pendulum restoring torque to its displacement. While the basic derivation method used is similar to that for linear systems, additional manipulation is required to eliminate the secondary variables and to form the state equations. Example 5.13 The fluid distribution system shown in Fig. 5.22 consists of a ftow source Qs (t) which feeds a storage tank with nonvertical walls. The output from the tank is distributed into a fluid network consisting of a short pipe discharging through an orifice and a long pipe discharging through another orifice. The fluid flow through an orifice obeys a quadratic relationship:
Q = CoJI~PI sgn (~P)
(i)
where Q is the flow through the orifice, ~ P is the pressure drop across the orifice, and Co is an orifice coefficient that is dependent on the geometry of the orifice. The signum function, sgn (),is used to indicate that the flow changes sign when the sign of ~p changes. The tank is shaped like the frustum of a cone and therefore has a volume that is a nonlinear function of the height of the fluid in the tank. From the tank geometry shown in Fig. 5.22, the volume is
V=
1h
1rr
2
dh
=
1h
1r (r1
+ Kch) 2 dh
= foh 1r (rr + 2r1Kch + K~h 2 ) dh =
1r
(ii)
(r~h + r 1Kch 2 + K~h 3 /3)
In an open tank in a gravity field, the pressure at the base is directly related to the height of the fluid:
Pc
= pgh
(iii)
so that the following constitutive relationship may be written
V
1rr~ p,
=Pi c +
1r K~ pl
1rr1Kc pl (pg)2 c
+ 3(pg)3
Kn
Kn
2
= KroPc + TPc
·
3
+ 3Pc
c
(iv) (v)
=
where Kro 1rrfjpg, Kr1 2Hr 1Kcf(pg) 2, and Kn HK'f:j(pg) 3 • The tank has the constitutive relationship of a pure nonlinear fluid capacitance.
=
=
Sec. 5.6
Nonlinear Systems
157
Long pipe/
Orifice
(a) Nonlinear fluid system I
(b) Linear graph Figure 5.22:
(c) Normal tree
Fluid distribution system with nonlinear elements.
The system graph model, including the nonlinear capacitance, the two nonlinear resistances of the orifices, and a fiuid inertance I of the long fiuid line (assuming that the pipe resistance is small compared to the orifice resistance), is shown in Fig. 5.22b. From this graph and the normal tree shown in Fig. 5.22c, the system model includes the following: Primary variables: Pc, PR2 • Q.r. QJ, QR 1 Secondary variables: Qc. QR2 , Ps, P1, PR 1 System order: 2 State variables: Pc. Q1 The system has four elemental equations. For the fiuid inertance dQ1 1 --=-PI
dt
I
(vi)
For the nonlinear fiuid capacitance the elemental equation must be found by differentiating the constitutive equation (v): (vii) or (viii)
State Equation Formulation
158
Chap.S
For orifice R 1 the elemental equation is (ix)
while for orifice R2 , which has the pressure drop PR2 as its primary variable, the equation may be written (x)
where QR2 IQR2 1 is the absolute square (absquare), which ensures that the pressure drop changes sign when the flow changes direction through an orifice. The two continuity equations are Qc = Q,- QRs QR2 = Ql
(xi) (xii)
and the two compatibility equations are
PR 1 = Pc P1 = Pc- PR2
(xiii)
(xiv)
Elimination of the secondary variables yields (xv)
(xvi) (xvii) (xviii)
and substituting for QRs and PR2 results in two nonlinear state equations: (xix)
(xx)
5.7
LINEARIZATION OF STATE EQUATIONS
A set of nonlinear state equations for a system may be linearized directly to obtain a set of linear state equations that approximate the system response over a limited range of operation. In the process of linearization a new set of variables is defined that describe the response about a system equilibrium point. The method is useful for systems containing
Sec. 5.7
Linearization of State Equations
159
nonlinear elements where the excursion of the system variables about the equilibrium point is small. Because the response of a linearized model only approximates that of the nonlinear system, care must be taken in interpreting any analyses, and judgment must be used in defining the applicable range of operation. To illustrate the linearization process, it is convenient to initially consider a first-order nonlinear system with a single state equation
x=
(5.32)
f(x, u)
Assume that with a known constant input uo, the system ultimately reaches a steady response xo, after which time the derivative x is identically zero. Any such equilibrium value for the state variable must therefore satisfy the algebraic equation f(xo, uo)
=0
(5.33)
In general more than one value or no values of x may satisfy a nonlinear equilibrium equation, and the solution of Eq. (5.33) may prove to be a difficult task. However, for many
nonlinear systems containing only ideal and pure elements an equilibrium condition can be found. The linearization of I (x, u) may be performed by expanding the function as a Taylor series in two variables x and u about the equilibrium condition:
aj(x, u) a
l(x, u) = l(xo, Uo) + .
X
I X=Xo
(X - Xo)
U=Uo
+ al(x, aU u) IX=Xo (U -
uo)
+ ·· ·
(5.34)
U=Uo
By definition, l
=
aI (x' u) I ax
X*
X=Xo
u=uo
u) + af (x' au
I
U*
(5.35)
X=Xo
u=uo
Equation (5.35) is a linear first-order state equation with constant coefficients which may be written as a standard state equation
x* =ax* +bu* where a= a1(x, u)
ax
I X=Xo
U=Uo
and
b = a1(x, u) au
(5.36)
I
(5.37)
X=Xo
O=Uo
The procedure for linearizing a higher-order system about an equilibrium point is analogous to that described for a first-order system. The set of n state equations with r inputs (5.38) :X= f(x, u)
160
State Equation Fonnulation
Chap. 5
is first solved for an equilibrium state xo with the given input uo by setting the derivative of the state vector to zero and solving the resulting set of nonlinear algebraic equations
f(xo, no)= 0
(5.39)
Each of the n state equations (5.40)
may then be linearized by generating a Taylor series expansion in the n + r variables about the operating point and ignoring all but the linear terms. A set of incremental variables x; = x; - (xo); and ui = u1 - (uo); is defined. Because there are n + r such variables, n + r partial derivatives are required in each equation. The result is
(5.41)
which is a linear state equation of the form (5.42)
The collection of all such equations can be written in matrix form:
x• = Ax• +Bn•
(5.43)
where the elements of the matrices A and B are a;j
= a/;(x. u) I axj
X=Xo U=Uo
.
and
.. _ aj;(x, u) bI J auj
I
(5.44)
X=Xo U=Uo
Matrices containing partial derivatives in the form of A and B are known as Jacobian matrices. Example 5.14 In Example 5. I J a nonlinear state equation was derived for a vehicle of ma'is m traveling at moderate speed Vm with propulsive force Fp(t). The model. including the effects of aerodynamic drag (Fd = cdv!) and constant road resistance (Fr)• generated a single nonlinear state equation
Derive a linearized state equation to describe the dynamics of the vehicle for small deviations in speed from a nominal operating speed Vm = vo.
Solution First the equilibrium condition is found by solving the equation (i)
Chap. 5
Problems
161
and at equilibrium (ii)
The linearized equations may be derived in terms of incremental variables
.• Vm
v:, F;, and F,*:
aF I . aF 1 . aF 1 . = avm 0 vm + BFp 0 Fp + aF, 0 F,
and when each tenn is evaluated and the substitution ( Vm )o
• = [ -2cdvo] v. m -m- vm• + [ -m1 ] F*p
(iii)
= v0 is made,
+ [ -m1 ]
F*
(iv)
r
When it is noted that F,* is zero because F, is constant, the linearized equation may be written
B • IF'* v.•m = --v m m +m P
(v)
where B = 2cdvo is the equivalent damper coefficient. In interpreting and using this equation, is the deviation from the equilibrium speed and that the input it is noted that the solution F;(t) is the deviation from the equilibrium propuJsive force.
v:
PROBLEMS
5.1. A metal block of mass m sits on a tabJe. Vibrations in the floor cause the table to move horizontally with a velocity V (t). A thin film of lubricant allows the block to slide on the table with an effective viscous frictional coefficient B. as shown in Fig. 5.23.
L I
Input V(t)
vm
Metal block /
m
r
Lubricant B
sssss;ssss/ Table
Figure 5.23: A mass element sliding on a table.
(a) Draw the system linear graph and normal tree. (b) Derive a state equation for the system. (c) Derive an output equation for the force accelerating the mass.
I
State Equation Formulation
162
Chap.5
5.2. For each system shown in Fig. 5.24: (a) Draw the linear graph. (b) Determine the system order and identify a set of state variables.
(c) Derive the state equations and express them in matrix form. (d) Write the output equation for the velocity of the mass Vm.
(a)
(b)
Figure 5.24:
Two mechanical systems.
5.3. For the friction measurement system in Problem 4.2. with the assumption that all elements are linear, derive a state equation and an output equation for the frictional force between the sliding element and the inclined plane. 5.4. A one-quarter car model is useful for studying the effects of the road surface on the vertical motion of a car body. The model includes the tire stiffness as well as the suspension damping and stiffness, and supports a mass of one quarter of that of the car, as shown in Fig. 5.25. The roadway is modeled as providing a vertical velocity input to the tire as the car travels along the road. {Because the gravitational force is constant, it may be omitted when modeling incremental motions about an equi1ibrium point.)
Body velocity
Axle velocity
l l
114 car body mass Suspension stiffness/damping Wheel/axle mass Tire stiffness
....__ _.,..,_ forward velocity Figure 5.25:
(a) (b) {c) {d)
A quaner-car model of an automobile suspension.
Construct the system linear graph. Identify the system state variables. Derive the state equations and express them in matrix form. Derive output equations for the total force acting on the car body mass from the suspension spring and damper.
Chap. 5
163
Problems
5.5. Seismometers are used to measure the motion of the earth's surface. A schematic drawing of a simple seismometer is shown in Fig. 5.26. A proof mass is suspended in springs and slides horizontally on a viscous friction material. The relative displacement of the proof mass with respect to the instrument case is used as a measure of the severity of an earthquake. Input velocity
I
B
Figure 5.26: A seismometer. (a) Construct a linear graph model of the system. (b) How many independent energy storage elements are there? What are the system state variables?
(c) Derive the system state equations and express them in matrix form. (d) Derive an output equation for the instrument reading, that is, the relative displacement of the proof mass with respect to the instrument case. 5.6. A tugboat tows a heavy barge at the end of a long elastic cable in smooth water. The tug's propellers generate a controlled propulsive force; hydrodynamic drags may be represented by linear viscous drag effects. Generate a third-order linear graph model, a set of state equations in matrix form, and an output equation that describes the dynamics of the velocity of the barge. 5.7. Generate one or more state equations to describe the dynamics of the wind-driven electric generator described in Problem 4.7. 5.8. For each of the rotational systems shown in Fig. 5.27:
(b)
(a)
Agure 5.27: Two rotational systems.
(a) Draw the linear graph. (b) Derive the state equations.
(c) Write the output equation for the angular velocity of the rotary inertia J.
164
State Equation Fonnulation
Chap.S
5.9. A simple tachometer is shown in Fig. 5.28. The rotation of the input shaft is coupled to the indicator, with inertia J, through a rotary drag cup B1. The indicator is supported in bearings with viscous friction 8 2 , and is connected to the instrument housing through a torsional spring K.
Input shaft
Bt Drag cup Figure 5.28:
A tachometer mechanism.
(a) Draw a linear graph and normal tree for this system.
(b) Derive a set of state equations for this system. (c) Derive an output equation for the indicated rotational speed, that is, the angular displacement of the inertia J. 5.10. In a two-color printing press, two pairs of large printing drums are rotated from a single drive shaft as shown in Fig. 5.29. Each drum pair has total rotary inertia J. and is supported in bearings with a linear rotational drag coefficient B. The drive-shaft sections each have a torsional stiffness K. The system is driven by a motor that may be considered as an angular velocity source. Derive a set of state equations for this system.
Motor
n.
J,B
Figura 5.29:
J,B Longsbaft K
Long shaft K
A rotary drive system.
5.11. Derive a set of state equations for each of the electrical networks in Problem 4.9. Write an output equation for the voltage across capacitor C2 • 5 ••2. L-C circuits are commonly used as tuning circuits in radio receivers. In Fig. 5.30 the remote radio transmitter, the propagation path, and the antenna are modeled as a Thevenin source with source resistance R,. The circuits consist of an inductor L (with a small but finite resistance RL) and a variable capacitor C connected in series or parallel as shown. The output voltage v0 is amplified and demodulated into an audio signal. For the series L-C circuit in Fig. 5.30a. and the parallel L-C circuit in Fig. 5.30b: (a) Draw the linear graph and normal tree for the tuning circuit. (b) Derive a set of state equations for the circuit (c) Develop an output equation for the voltage V0 (t).
Chap. 5
Problems
165
+
+ V,(t)
Vs(t)
(a)
c
(b)
Figure 5.30:
Series and parallel L-C tuning circuits.
5.13. An electric "bridged-T., filter is shown in Fig. 5.31. The output voltage is the voltage across R2 • Draw a system linear graph and derive a set of state equations. Derive an output equation for the output voltage.
Figure 5.31:
An electrical filter network.
5.14. An electric circuit containing three inductive devices is shown in Fig. 5.32.
Figure 5.32:
An inductive network.
(a) Construct the system linear graph and normal tree. (b) Identify the system primary variables and state variables. What is the order of this system? Are there any dependent energy storage elements in the system? (c) Derive a set of state equations. (Note that you may have to solve a pair of simultaneous equations to generate state equations in the standard form.) (d) Derive an output equation for the voltage across resistor R2 •
166
State Equation Formulation
Chap. 5
5.15. A current source /1 (I) drives a parallel pair of electromechanical machines, each of which may be modeled as a series connection of resistive and inductive elements, as shown in Fig. 5.33.
Figure 5.33: Two inductive electrical machines connected in parallel.
(a) Construct the system linear graph. (b} Identify the system order.
(c) Derive a set of state equations. (d) Explain why this system's state equations are not directly in the form X. =
Ax+ Bu.
5.16. For each of the two fluid systems shown in Fig. 5.34:
(a)
(b)
Figure 5.34:
Two fluid systems.
(a) Construct the system linear graph. (b) Identify the state variables. (c) Derive the state equations. (d) Write the output equations. 5.17. Consider the fluid apparatus used to study the effect of an arterial aneurism in Fig. 4.32. (a) Derive a set of state equations for the system as shown. (b) If the expandable rubber section is removed from the midpoint of the tube and placed at the pump outlet, how is the linear graph altered'? How many state variables are required to represent the system? Derive a new set of state equations.
Chap. 5
167
Problems
5.18. A chemical plant uses a pair of open tanks to store chemicals prior to use, as shown in Fig. 5.35. Each tank is fed by a positive displacement pump modeled as a flow source. The chemicals in the tanks are combined in a mixing valve with a check valve at each of its inlets to prevent back flow. When the plant is operating normally these check valves are open and may be modeled as a linear ftuid resistance that is large compared to the pipe resistances. The pipe fluid inertances cannot be neglected. The flow of the mixed chemicals is controlled by a valve at the outlet, which may be modeled as a linear resistance.
Check valves
r"
--~~== Outlet valve
. . hamber / Mrunge
Figure 5.35:
i
Qout
An industrial chemical mixing system.
(a) Construct a linear graph model of the system.
(
(b) Derive a set of state equations in matrix form.
(c) Derive an output equation for the output flow rate. 5.19. A fluid system consists of a tank. fed from a flow source Qs(t), that discharges through a valve R at the end of a vertical pipe of height h as shown in Fig. 5.36. Draw a linear graph for the system, and explain how you have accounted for the pressure increase pgh at the base of the pipe. Derive a state equation for the system.
Tank
r h
L Figure 5.36:
Valve
A fluid system with a vertical pipe section.
168
State Equation Formulation
Chap. 5
5.20. Consider the thermal model of the home heating system described in Problem 4.14. Consider the system to have two inputs: the outside ambient temperature and the heat generated by the furnace. (a) Construct the system linear graph and identify the state variables. (b) Derive the state equations. ' 5.2L It is common to mount machines and rotating equipment on shock isolation pads to reduce the transmission of time varying forces to the ground. In the system shown in Fig. 5.37 a machine, represented as a mass m, has a vibrational force acting at its center of mass in addition to the gravitational force. The pad that supports the machine is made of a damping material with nonlinear stiffness given by the constitutive equation F cx 2 , where c = 25 N/m 2 • The material also has damping properties that are approximately linear, that is, the damping force is proportional to the velocity across the pad.
=
Machine m
&
+
F(t)
Nonlinear
_ _Ef~i~·ii!;['1:1J "'·.~·t:~~~"iij§:ii::~~~·-dam=ping pad Figure 5.37: A vibrating machine mounted on a resilient pad. (a) Construct a linear graph for the system and identify the state variables. (b) Derive a set of nonlinear state equations for the system. (Hint: it is useful to differentiate the constitutive equation to obtain an elemental equation in terms of the power variables.) (c) If the mass of the machine tool is 1000 kg, and the vibrational force has a zero average value, what is the nominal equilibrium condition for the system? (d) Derive a set of linearized state equations for small excursions from the nominal equilibrium condition.
REFERENCES [1] Kalman, R. E., "On the General Theory of Control Systems," Proceedings of the First IFAC Congress, 481-493, Butterworth, London, 1960. [2] Schultz, D. G., and Mel sa, J. L., State Functions and Unear Control Systems, McGraw-Hill, New York, 1967. [3] Timothy, L. K., and Bona, B. E., State Space Analysis, McGraw-Hill, New York. 1969. [4] Koenig, H. E., Tokad, Y., Kesavan, H. K., and Hedges, H. G., Analysis of Discrete Physical Systems, McGraw-Hill, New York, 1967. [5] Blackwell, W. A., Mathematical Modeling of Physical Networks, Macmillan, New York, 1967. [6] Chan, S. P., Chan, S. Y., and Chan, S. G., Analysis of Linear Networks and Systems, AddisonWesley, Reading, MA. 1972. [7] Shearer, J. L., Murphy, A. T.• and Richardson, H. H., Introduction to System Dynamics, AddisonWesley, Reading, MA, 1967. [8] Durfee, W. K., Wall, M. B., Rowell, D., and Abott, F. K., "Interactive Software for Dynamic System Modeling Using Linear Graphs," IEEE Control Systems Magazine, 11(4), 60-66, June 1991.
6
Energy-Transducing System Elements
6.1
INTRODUCTION
One-port model elements are used to represent energy storage, dissipation, and sources within a. single energy domain. In many engineering systems energy is transferred from one energy medium to another. For example, in an electric motor electric energy is converted to mechanical rotational energy, while in a pump mechanical energy is converted to fluid energy. The process of energy conversion between domains is known as transduction, and elements that convert the energy are defined to be transducers. Within a single energy domain power may be transmitted from one part of a system to another, for example, a speed reduction gear in a rotational system. In this chapter we define two types of ideal energy transduction elements which can be used to represent the process of energy transmission. We also develop methods to derive a set of state equations for these types of systems. A wide range of devices and mechanisms have been developed to transform energy from one medium to another. Figure 6.1 illustrates the transduction between mechanical translation, mechanical rotational, electric, and fluid systems. Energy transduction devices include the following: • Rack and pinions, ball screws, and linkages for transduction between mechanical translation and mechanical rotational systems • Motors and generators for transformation between electric and mechanical rotational systems • Electromagnetic, magnetostrictive, and piezoelectric devices for transduction between electric and mechanical translational systems • Magnetohydrodynamic and electrohydrodynamic energy transfer for transductions between electric and fluid systems • Pumps, compressors, and turbines for transduction between fluid and mechanical rotational systems 169
170
Energy-Transducing System Elements
Chap. 6
Figure 6.1: Transduction between different energy domains.
• Ram and piston-cylinder systems for transduction between fluid and mechanical translational systems. Several of these transducers are illustrated in Fig. 6.2. Energy may also be transmitted within a single energy domain through transducers such as the following: • Levers and linkages for transmission between one part of a mechanical rotational system and another
• Gear trains for transmission between one part of a mechanical rotational system and another • Electric transformers for transmission between one part of an electric system and another • Fluid transformers for transmission between one part of a fluid system and another. Several of these transducers are illustrated in Fig. 6.3. The basic energy transduction processes occurring in these types of devices can be represented by a two-port element, as shown in Fig. 6.4, in which energy is transferred from one port to another. Each port has a through- and an across-variable defined in its own energy domain. Power may flow into either port. The ideal two-port transducer is a lossless element; for many physical systems it is necessary to formulate a model that consists of an ideal two-port transducer coupled with one or more one-port elements to account for any energy storage and dissipation that occurs in the real transducer. In the following sections general ideaJ energy transduction processes are defined, yielding two ideal two-port transducers foJiowing the development given by H. M. Paynter [1]. Other two-port model elements that represent dependent sources and energy sources and dissipation elements have also been defined [2, 6].
Sec. 6.1
171
Introduction
V=rfi F=-j_T r
(a) Rack and pinion
(b) Slider-crJnk
E= Kv i= -
iF
(c) Rotary positive displacement pump
(d) Moving-coil loudspeaker
(e) A uid piston-cylinder
(f) Electric motor-generator
Figure 6.2:
Examples of systems and devices using two-port transducers between different energy domains.
Energy-Transducing System Elements
172
/
Ft =
~
F2
v1 =-2v2
(b) Gear train
(a) Block and tackle
Vt
=-l
v2
Ft =LF2
(d) Belt drive
(c) Mechanical lever
For ac inputs
Vt
1 = Nv2
i 1 = -Ni2 (e) Electric transformer
Figure 6.3:
(f) Fluid transformer
Examples of two-port transducers within a single energy domain.
Chap. 6
Sec. 6.2
Ideal Energy Transduction
173
Two-port transducer
Figure 6.4:
6.2
1\vo-port element representation of an energy transducer.
IDEAL ENERGY TRANSDUCTION
We define an ideal energy transduction process as one in which the transduction is • Lossless-power is transmitted without loss through the transducer with no energy storage or dissipation associated with the transduction process. • Static-the relationships between power variables are algebraic and independent of time. There are no dynamics associated with the transduction process. • Linear-the relationships between power variables are represented by constant coefficients and are linear. A two-port transducing element as illustrated in Fig. 6.4 identifies a power flow P 1 into port 1, defined in terms of a generalized through-variable f1 and an across-variable v 1: (6.1)
and a power flow P2 into port 2, also defined in terms of a pair of generalized variables f2 and v2: (6.2)
The condition that the two-port transduction process is lossless requires that the net instantaneous power sums to zero for all time t: (6.3)
where it is noted that power flow is defined to be positive into both ports. The most general linear relationship between the two pairs of across- and throughvariables for the two-port transducer may be written in the following matrix form: (6.4)
where Ctt, c12, c21, and c22 are constants that depend on the particular transducer. With the specification of the constants in this form, a unique relationship is established between the power variables. When the additional condition is imposed that the transduction be lossless as well as linear and static, Eqs. (6.3) and (6.4) may be combined to yield a single equation in terms of the variables f2 and v2 (or, equivalently, in terms of f1 and Vt) and the four transducer parameters: (6.5)
174
Chap.6
Energy-Transducing System Elements
[If the condition were derived in tenns of f1 and VJ, it would be of the same fonn as Eq. (6.5).] Only two nontrivial solutions exist in which Eq. (6.5) is satisfied for arbitrary values of f 2 and v 2 and which correspond to power transfer between ports 1 and 2. By appropriate selection of the constants c 11 , c 12 , c21, and c22 these solutions yield two possible ideal two-port transducers:
Transforming Transducer: CJ2
= C2J = 0
and
c22 = -1/cu
(6.6)
and
C2J
= -1/Ct2
(6.7)
Gyrating Transducer: C]J
= C22 = 0
These two general solutions are the only nontrivial solutions with power transfer and give two distinct forms of ideal two-port transduction. The ideal two-port transformer relationship, defined by the conditions in Eq. (6.6), may be written by substituting c 11 = TF: (6.8) where TF is defined to be the transformer ratio. Equation (6.8) states that in a transformer the across-variable v 1 is related by a constant TF to the across-variable v2 on the other side and through-variable f 1 is related by the negative reciprocal of the same constant ( -1 /TF) to through-variable f2:
v 1 = TFv2 ft
(6.9)
=-(~)t2
(6.10)
This transduction process is called a transfonner because it relates across-variables to acrossvariables and through-variables to through-variables at the two ports. The symbol for the ideal transfonner is shown in Fig. 6.5.
..
fl
:~·
lt l-
--
J
(a) Transformer Figure 6.5:
.
f2
fl
~:
f2
+
+
(b) Gyrator
Symbolic representation of transforming and gyrating transducers.
Sec. 6.2
Ideal Energy Transduction
175
Similarly, the ideal two-port gyrating transducer, described by the conditions in Eq. (6.7}, may be written by substituting c12 = GY:
[ 0 [ VJ] f1 = -1/GY
GY] 0
[v2] f2
(6.11)
where GY is defined to be the gyrator modulus. Equation (6.11) states that the acrossvariable VJ at port 1 is related by a constant GY to the through-variable f 2at port 2 and the through-variable f1 is related by the negative reciprocal constant -1/GY to the acrossvariable v2: Vt = GYf2
(6.12)
f1 =- (a~)v2
(6.13)
The transduction process is termed gyration because it relates across-variables to throughvariables, and vice versa. The two-port symbol for the gyrator is shown in Fig. 6.5. 6.2.1 Transformer Models
Many engineering transduction devices and mechanisms are transformers. In this section we examine ideal models of a gear train, a rack-and-pinion drive (commonly employed in automotive steering systems), and a permanent-magnet electric motor-generator (commonly used in control systems). The Gear Train A pair of mechanical gears used to change the torque and speed relationship between two rotational power shafts is illustrated in Fig. 6.6.
Figure 6.6: A gear train as a transforming transducer.
If gear 1 has n 1 teeth at an effective radius r1 and meshes with gear 2 which has n2 teeth at a radius r2 , the gear ratio N for the two gears is defined as the ratio of the number of teeth on the two gears or, equivalently, the ratio of the two radii: (6.14)
176
Energy-Transducing System Elements
Chap. 6
The torque and angular velocity sign convention for each gear is defined in Fig. 6.6, with the assumption that power flow is defined as positive into each power port. For rotation of the two gears without slippage, the linear velocity v, tangent to the pitch radius at the meshing teeth must be identical for each gear. (6.15) Therefore, the angular velocity of the two shafts must be related: (6.16) where the sign conventions are chosen according to the definitions of positive angular velocity for each gear. Furthermore, at equilibrium the linear force F1 tangent to the pitch radius on the meshing teeth of gear 1 must be equal and opposite to the linear force F2 tangent to the pitch radius on gear 2: (6.17)
consistent with the definition of torques identified in Fig. 6.6, or (6.18)
Equations (6.16) and (6.18) define a transformer relationship which may be written in the matrix form ofEq. (6.8) as (6.19) where it is noted that to establish consistency with the torque and angular velocity sign conventions in Fig. 6.6, the transformer modulus TF is -1 1N.
A Rack-and-Pinion Drive Rack-and-pinion drive systems are used to convert rotary motion to translational motion, and vice versa, as shown in Fig. 6.7. The rotary input at the pinion shaft is expressed in terms of its angular velocity n and the applied torque T, while the translational power associated with the rack is expressed in terms of its linear velocity v and force F. The analysis of the system as an ideal transducer assumes that both elements are massless and that no frictional losses occur at the contact between the teeth. With the further assumption that no slippage occurs between the rack and pinion, the linear velocity of the rack is related to the angular velocity of the pinion directly by the pinion radius r, that is, v=rrl
(6.20)
Sec. 6.2
177
Ideal Energy Transduction
For equilibrium, the force F on the rack must be related to the torque on the pinion shaft T at the meshing teeth by the relationship (6.21) where a negative force is required to balance the positive torque as shown in Fig. 6.7. Equations (6.20) and (6.21) define the rack and pinion as a transformer because of the direct relationship between the across-variables v and n and between the through-variables F and T at the two ports. In this case the transformer ratio TF is simply the pinion radius r and the transduction equations are (6.22) The rack and pinion can be used either to transform rotary motion to translational motion if the shaft is driven from an external source or to transform translational motion to rotary motion if the rack is connected to a translational source.
Electromagnetic Transducers Electromagnetic transducing elements include electric motors, electric meters, solenoids, microphones, and phonograph pickups. Their operation depends on two complementary laws: ( 1) the Lorentz law, which states that an electric charge moving in a magnetic field experiences a force, and (2) Faraday's law, which states that an electric voltage is induced in a coil of wire that moves in a magnetic field [4]. Figure 6.8 shows a particle with electric charge q moving with vector velocity v along a wire in a magnetic field of flux density B. The particle experiences a force ~F given by the vector cross-product ~F
= q (v x
B)
(6.23)
which states that the direction of the force is at right angles to the direction of the motion v and the magnetic flux B and has a magnitude proportional to the component of the vector v that is perpendicular to B, that is, to Ivi sin 8 in Fig. 6.8. If the wire has length t. and there are N such discrete charged particles within the wire, the total transverse force is F
= Nq (v x B) =(I
X
B) t.
(6.24) (6.25)
178
Energy-Transducing System Elements
Chap. 6
Electric conductor
Figure 6.8: Electromagnetic forces on a moving charged panicle.
where the the vector current I = N qvI i. is the total charge passing through any cross section of the wire in unit time. If a coil is constructed by winding the wire into a circular or rectangular configuration, the total electromagnetic force on the coil is equal to the sum of the forces acting on each elemental section of the coil. Equation (6.23) also implies that if a mobile charge within a conductor is transported physically in a magnetic field it experiences a force that tends to displace it along the conductor. The resulting migration of the charged particles induces a voltage difference proportional to the velocity between the two ends of the wire. The induced voltage V; in a wire of length i. traveling with velocity v peq>endicular to a magnetic field with flux B is equal to
V; = i.Bv
(6.26)
If the conductor is wound into a coil configuration, the total voltage induced by motion is found by integrating the elemental potential differences along all sections of the coil. Motion-induced voltages sum to zero in a coil subjected to a translational motion in a uniform magnetic field but sum to a finite value if the coil is rotated in the same field. Equations (6.25) and (6.26) indicate that there is a direct relationship between force (or torque) and current and a similar relationship between velocity (or angular velocity) and voltage. Electromagnetic transducers as a class therefore may be represented as transformers. The particular transformer ratio for any device must be derived from the system structure and the magnetic flux geometry.
The Permanent-Magnet De Electric Motor A permanent-magnet direct current (de) motor-generator transforms energy between the electric and mechanical rotational energy domains. Its basic structure is illustrated in Fig. 6.9. In applications where the transduction is primarily electric to mechanical rotational energy, the transducer is designated a motor, while in applications where the transduction is from mechanical to electric energy, the transducer is usually designated a generator. In both cases the electromechanical coupling determines the motor-generator characteristics. The motor-generator consists of a rotor containing an electric coil that rotates in the magnetic field produced by a pair of permanent magnets. The electromechanical coupling constant may be derived by assuming that the magnets generate a radial magnetic flux with constant flux density B which is assumed to be uniform around the circumference of the rotor. If the radius of the rotor is r, its length is i and it is wound with an electric coi1 with N turns carrying a current i, a force F Bii perpendicular to the local magnetic field is produced on each length of wire. (No force is exerted on the ends of the coil.) Each of
=
Sec. 6.2
Ideal Energy Transduction
179
+
Commutator/ ,.....__,_............ brushes
Figure 6.9:
The de permanent-magnet motor-generator.
the N turns in the coil has two sides interacting with the field, so the total torque exerted on the rotor is its radius multiplied by the number of wire lengths 2N multiplied by the magnetic force on each wire, or T
= -2rNBli
(6.27)
Additionally, as the rotor spins with angular velocity Q, each wire section passes through the magnetic field with a tangential velocity v = rs-2, generating an induced voltage between the two ends of each wire. The total voltage v induced in all wire sections is equal to the total number of wire lengths (2N) multiplied by the voltage induced in each wire traveling at the rotational velocity: v = 2NBlrfJ.
(6.28)
In summary the relationship for the transducer may be written (6.29) where the transformer constant is
TF= 2NBlr
(6.30)
The sign convention for the motor-generator is consistent with the idea that power is defined as positive into the transducer at both ports, and the relationship in Eq. (6.30) represents the energy transduction for both motor and generator applications. In practice, many motor manufacturers adopt a sign convention with electric power into the transducer as positive and mechanical power out as positive and express the transduction relationships in terms of two constants as
T* = K,i V=Kvfl
(6.31) (6.32)
180
Energy-Transducing System Elements
Chap.6
where K1 is the motor torque constant and Kv is the motor back-emf voltage constant. If the manufacturer's sign convention is adopted with T* = - T, these two constants may be expressed as (6.33) and by comparing Eqs. (6.33) and (6.29), the standard motor constants are identified as
Kv
= K, = 2NBlr
(6.34)
When a consistent set of units such as the SI system is utilized in specifying Kv and K,. the two constants are numerically equal; however, common practice for manufacturers
is to utilize English units, which are not a consistent set, and thus in many motor and generator specifications the values for Kv and K, are not numerically equal and conversion to a consistent set of units is required in model formulation. 6.2.2 Gyrator Models The Hydraulic Ram A common engineering example of a gyrating transducer is the hydraulic ram, consisting of a cylinder and piston as shown in Fig. 6.1 0, in which conversion occurs between mechanical translational power and fluid power. For the ram illustrated in Fig. 6.1 0, the mechanical force F on the piston is related directly to the fluid pressure P by the piston area A: F=AP
(6.35)
and with the sign convention adopted in the figure, the velocity v of the shaft is related to the fluid volume flow rate Q by the negative reciprocal of the area: 1 A
v=--Q
(6.36)
Equations (6.35) and (6.36) show the linear relationship between an across-variable in one energy domain and a through-variable in the other. The ram therefore acts as a gyrator, with · the gyration constant 1 A
OY=--
(6.37)
In matrix form the hydraulic ram equations are
(6.38)
Multipart Element Models
Sec. 6.3
181
F~r====l
Figure 6.10: The hydraulic ram as a gyrating transducer.
p
The ram is one example of a class of positive displacement fluid-mechanical devices represented by gyrators. Other similar transducers inc1ude gear pumps and hydraulic motors, vane pumps and hydraulic motors, and piston pumps and hydraulic motors.
6.3
MULTIPORT ELEMENT MODELS The linear graph symbols used to represent the ideal transformer and gyrator two-port elements are illustrated in Fig. 6.11. The sign convention adopted for each of the two-port elements is that power flowing into the element at either port is defined to be positive. The graph element implies a direct coupling between the across- and through-variables aSsociated with each branch of the graph as defined by the two-port elemental equations (6.8) and (6.1 1). The two-port elements are inherently four-terminal elements ·and are in general connected to four distinct nodes in a system graph. Thus, a system model containing an electromechanical transducer contains two reference nodes, one mechanical and one electric, as illustrated in Example 6.1. Similarly, in Example 6.2, which illustrates the use of a gyrating transducer to couple the fluid and mechanical domains, the model contains two reference nodes. Systems with two-port transducing elements often generate system graphs containing separate connected graphs because of the multiple energy domains represented. The variables associated with the two separate graphs are related by the transducer equations. Many physical transducing elements cannot be modeled directly with a simple energyconserving two-port element. They have implicit energy dissipation and storage phenomena associated with the transduction, and these must be accounted for by the inclusion of additional lumped-parameter elements in the model. Consider, for example, the transduction between the electric and rotational energy domains in a permanent-magnet de motor-generator. The ideal energy conversion relationships are described by Eq. (6.29). The modeling of a real motor, however, may need to account for the following additional phenomena:
v] 1~\. I
f1
I \
\ I
~2
f2
\
I
\
I
(a) Tran!;fonner
v; 1~\. I
\
I I
I I
\
I
\ I
I
f1
~2
f2
I I
I \ \
I
\
I (b) Gyrator
Figure 6.11: Linear graph representation of two-pon elements.
182
Energy-Transducing System Elements
Chap. 6
• The windings in the motor armature will have finite electric resistance R. The voltage drop v R = i R across this resistance may be significant in a given modeling situation. • The motor windings consist of many turns of wire, usually on a high permeability ferrous core, and will therefore exhibit properties of inductance associated with energy storage in the magnetic field. It may be important to account for the voltage drop VL = Ldi/dt in a given model. • The rotating armature will have a finite moment of inertia J; it may be important to include the kinetic energy storage in a model. • The internal bearings in the motor may have significant frictional losses that need to be described by a viscous damping coefficient B. The two electrical phenomena share the common motor current i, and the voltage drop may be represented by series lumped R and L elements in the electric circuit The energy storage and dissipation in the rotational phenomena may be represented by lumped J and B elements in parallel with the mechanical side of the two-port element The complete model for the de motor is shown in Fig. 6.12a. L
R
(a) Electric motor with armature inductance. resistance, inertia. and friction ~~"iJii?l'~·r::.~Q p
(c) Rack and pinion with inenia and friction effects Figure 6.12: Examples of two-pon energy conversion devices with associated lumped elements to account for internal energy storage and dissipation.
Sec. 6.3
183
Multipart Element Models
The decision whether or not to include these additional elements in a system model must be based on an analysis of the complete system and its expected operating mode. For example, a decision to include the armature inductance L in a model should be based on the estimated significance of the voltage drop vL during the normal operation of the motor. If the motor is expected to act in a mode where the torques and drive current change slowly, it may be acceptable to ignore the inductive effects, but if the same motor is to be used in a "high-performance" dynamic system where the voltage VL is significant, the inductance L may need to be included in the model. Two additional examples of elements that might be included in power conversion devices are shown in Fig. 6.12. In Fig. 6.12b a positive displacement fluid pump is shown with additional elements to account for the inertia J of the shaft and rotor, viscous damping B to account for frictional effects in the bearings and rotating fluid seals, and a fluid leakage resistance R1 to account for the fact that some flow does leak past the seals. Figure 6.12c shows a rack-and-pinion drive with additional elements to account for the mass m and sliding friction B of the translational rack element, and inertia J and rotational damping coefficient B, to account for the rotating pinion and its bearings. Example6.1 A de electric motor is used to drive a turntable in a high-quality audio reproduction system. Because very small variations in the turntable speed can have an audible effect on the sound quality, the variation in turntable speed in response to changes in the input voltage is of interesL Form a system graph model of the electromechanical system shown in Fig. 6.13.
R
:Motor R 1-------------...1
L
(b) Linear graph
(a) System
Figure 6.13: Linear graph model of a electric motor drive with an inertial load.
Solution The construction of the linear graph model begins with the selection of elements. The input to the motor is assumed to be a prescribed voltage and is represented as a voltage source Vs (t). The motor armature (rotor) coil has both inductive energy storage and energy dissipation and is modeled by a series inductance L and resistance R. The electromechanical conversion of energy is represented by an idea] two-port transformer with the electromechanical coupling relating torque T to current im and angular velocity Sl to the motor internal voltage (back emO Vm generated. The two-port transformer relationships are T =-Kim
(i)
1 0=-Vm K
(ii)
184
Energy-Transducing System Elements
Chap.6
and are represented explicitly by the two-port linear graph elemenL Figure 6.13 shows the turntable to be supported by external bearings. We assume that there are significant frictional effects in the internal motor bearings and in the external bearing and that these may be combined to form an effective rotational viscous damping coefficient B. Similarly, we assume that the moments of inertia of the armature and the turntable may be combined to fonn an equivalent inertia J. The construction of the system graph may start with the mechanical system. The inertia and damper share a common angular velocity n and therefore are inserted in parallel between a node representing the turntable angular velocity and the rotational reference node. The rotational side of the two-port transformer element (motor) also has the same angular velocity as the turntable and is inserted in parallel with the inertia and damper. The electrical side of the two-port transfonner is placed with one node at the electric reference node and with the second node representing the internal motor voltage. The motor coil, represented by the inductor and resistor elements, has a common current with the electric branch of the transformer and thus is represented as a series connection. The voltage source acts with respect to ground and is assigned a sign convention to ensure that positive input voltage generates a positive current ftow to the motor. The completed linear graph is drawn in Fig. 6.13b. The linear graph demonstrates that in normal operation the voltage v 1 applied to the ideal transfonner is not the motor terminal voltage Vs(t). The dynamic behavior of the drive system will be affected by the ·resistance and inductance elements.
Example6.2 Fonn a dynamic model of a system consisting of a positive displacement pump which drives a hydraulic ram to move a mass sliding on a surface as represented in Fig. 6.14. The pump is driven from a constant angular velocity source D.r(t). The dynamic response of the mass to variations in the angular velocity D.r(t) is of interesL
(a) System Figure 6.14: Linear graph model of a hydraulic linear actuator system.
Solution The hydraulic part of the system is represented by an angular velocity source driving a transfonner-based pump model with a leakage resistance R1 included to account for internal fluid flow around the seals in the pump. A fluid resistance R1 is included to account for pressure drops in the pipe connecting the pump and hydraulic ram. The hydraulic-mechanical interface is represented by a gyrator with F=AP 1 V=--Q A
(i) (ii}
Sec. 6.3
185
Multipart Element Models
where A is the piston surface area. By convention power is defined as positive into both branches, and thus a positive fluid volume flow Q generates a negative mechanical velocity. The system graph may be constructed by first considering the mechanical system in which the mass is referenced to the mechanical reference node. The mechanical subsystem consists of three elements: a mass m, a spring K, and a damper B, all sharing a common velocity (across-variable) and therefore connected in parallel across the gyrator port. The hydraulic branch of the gyrator (the piston-cylinder) has flow supplied by the pump through the resistance R1 ; therefore, the three hydraulic branches are connected in series. The signs associated with the hydraulic system are selected so that positive pressure with respect to the-hydraulic reference node generates a positive volume flow to the ram. The complete linear graph is shown in Fig. 6.14. The two-port elements divide the system into three sections, each with its own reference node; the system graph is not a single connected graph, but contains three connected graphs, each with its own reference node in its energy domain.
Two-port transducers representing energy transfer within a single energy domain may share a common reference across-variable on each side and effectively be reduced from a fourterminal element to a three-terminal element. In such cases two of the nodes are implicitly joined to form a common reference as shown in Fig. 6.15. A common reference, or equivalent three-terminal representation, is required for two port transducers that represent 1. mechanical translational to mechanical translational energy transfer, such as occurs in a mechanical lever, since both port velocities must be referenced to a common inertial reference frame; 2. mechanical rotational to mechanical rotational energy transfer, such as occurs in a gear train, since both port angular velocities must be referenced to a common frame; and 3. fluid to fluid energy transfer, such as occurs in a fluid transformer, since both port pressures must be referenced to a common constant pressure.
a
b
c
(a) Transfonner
a
b
c
(b) Gyrator
Figure 6.15: Linear graph representation of two-port elements with a common node (VJ = V0 - Vc and V2 = Vb- Vc)•
When a common reference node is shared between the two sides of a two-port element, the two sides fonn a connected region of the overall graph. The formation of a linear graph model incorporating a common node is illustrated in Example 6.3.
Energy-Transducing System Elements
186
Chap. 6
Example6.3 In many engineering applications, de motors are connected to loads through a gear train, as shown in Fig. 6.16, in which a de motor is coupled to an inertial load though a speed-reducing gear train with ratio N . It is desired to form a system model relating the motor voltage to the angular speed of the flywheel. Motor R
L
+
Gear ratio n
(a) System Figure 6.16:
(b) Linear graph
Electric motor-speed reducer drive syste m.
Solution The de motor is represented as described in Example 6.1, using an armature resistance R and an inductance L to account for voltage drops in the coil and a two-port electric to mechanical rotational transformer with coupling constant K to model the electromechanical conversion. The gear train is represented as a rotational transformer with a gear ratio N. The fl ywheel is represented as a rotational inertia J. All the shafts in the system are assumed to be rigid, and two sets of bearings represented by rotational dampers B 1 and B2 are shown in Fig. 6.1 6. The system has an electric reference node and an angular velocity reference node. On the electric side, the voltage source defines the voltage between the reference node and the series connection of the resistance, inductance, and the electric side of the two-port transduction element. The rotational part of the graph contains the inertia J and bearing damper element B2 in parallel with branch 2 of the gear train. Branch 1 of the gear train is in paral lel with bearing damper B1 and branch 2 of the electric motor element. The linear graph model is shown in Fig. 6.16, where the two-port transducer for the electric motor is connected to four distinct nodes and is effectively a four-terminal device, while the two-port element representi ng the gear train is connected effectively as a three-terminal element since two terminals are j oined at a common reference node. The system graph contains two connected graphs because the electromechanical two-port transducer couples two separate energy domains.
Example6.4 A single massless, fri ctionless pulley with radius r is attached to a mechanical system as shown in Fig. 6.17a and is driven through a flexible line of fi xed length L . The two ends of the line are connected to two independent velocity sources V0 (t ) and Vb(r). Derive a linear graph-based model that will describe the system dynamics.
Sec. 6.4
187
State Equation Formulation
Solution The kinematics of the pu11ey drive system are shown in Fig. 6.17a. If the distances of the two ends of the line from a reference point are Xa(t) and xb(t), then
which when differentiated and rearranged gives the velocity of point c as (i)
Because the pu11ey is massless and there are no frictional forces in the bearings, the force in the line FL is continuous. The net force acting on the mechanical system at point cis therefore (ii)
The system linear graph must embody Eqs. (i) and (ii). The linear graph is constructed by realizing that the velocity (across-variable) of point cis the sum of two components, Va/2 and Vb/2. and that the force (through-variable) is twice that ac;sociated with either source, that is, 2F. These facts indicate that a pair of transformer relationships exist between the forces and the velocity components associated with the sources and that experienced by the mechanical system. Figure 6.17b shows how these relationships may be expressed by the use of two transformers, each driven by a velocity source, and with the two branches 2 and 4 connected in series. Note that for drawing convenience the reference node v = 0 has been drawn as three separate nodes; they should an be considered one node.
a
Va(l)~o---~..... 1
b
:
I I I I
I I I
i
Vb(t)~O-----t-_ ____.;::..,........
1 I
:
xb
Xa
~·•.---~------------------------------------~
l· (a) System
Figure 6.17:
6.4
I I I I
Vb(t)l
I I I
._ (b) Linear graph
A mechanical system driven through a floating pulley.
STATE EQUATION FORMULATION The generation of state equations for systems containing two-port transduction elements is similar to the method described in Chap. 5 for systems with one-port elements. The major difference lies in the stepwise procedure for constructing the normal tree because of the four-terminal nature of the two-port elements.
188
Energy-Transducing System Elementc;
Chap. 6
6.4.1 Graph Trees for Systems of Two-Port Transduction Elements
1\vo-port elements are essentially four-terminal elements and in general are connected to four distinct nodes on the system graph. When there is no common reference node between the two sides of a two-port element, the system graph is effectively partitioned into two distinct connected graphs, each with a separate reference node. When a system graph with a total of N nodes and B branches is divided into Nd separate connected linear graphs by two-port elements, the overall system tree consists of Nd section trees. When the ith such section contains N; nodes, as defined in Chap. 5, the tree for that section contains N; - I branches. The total number of branches Br in any tree of the system graph is then NJ
Br
= L (N; -
1)
=N -
Nd
(6.39)
+ Nd
(6.40)
1=1
and the number of links B L in any system tree is BL = B - Br = B - N
For a system graph containing SA across-variable sources and Sr through-variable sources, and containing Nd distinct connected graphs, there is a total of B - SA - Sr passive branches, each with an elemental equation. The B - SA - Sr constraint equations required to solve the system may be found from B L - Sr = B- N + Nd - Sr compatibility equations formed by replacing the passive links in the tree and Br - SA = N - Nd - SA continuity equations formed by creating contours that cut a single tree branch. 6.4.2 Specification of Causality for Two-Port Elements
The elemental equations for transformers and gyrators impose causality constraints across the two-port element and generate a pair of rules that specify how the branches of a twoport element may be entered into a tree. In Chap. 5 we defined a primary and a secondary variable associated with each elemental equation in the system graph. Furthennore, the primary variables are defined to be the across-variables on tree branches and the throughvariables on tree links. The elemental equations for the two-port elements relate variables across the element, so when the condition is imposed that only one of the two variables on each branch may be considered primary, only two possible causal conditions may be defined for a two-port element.
The Transformer: For the transformer shown in Fig. 6.18, branch I has acrossand through-variables Vt and f 1, while branch 2 has variables v 2 and f 2 • The transformer equations Vt
= TFv2
fl =-
(~ )t2
(6.41) (6.42)
Sec. 6.4
189
State Equation Formulation
specify that if Vt (or v2) is considered to be a primary variable, then v2 (or v 1) must be a secondary variable. Similarly, if ft (or f2) is chosen as a primary variable, f2 (or f 1 ) is by definition a secondary variable. The transformer equations allow only one across-variable and one through-variable to be used as primary variables.
Primary variables:
v1, f 2
Secondary variables: f 1, v2
Primary variables:
f 1, v2
Secondary variables: v1, f 2
Figure 6.18: The two allowable tree configurations for a transfonner. The links are shown as dotted lines.
Since only one across-variable may be a primary variable for a transformer, only one branch of a transformer may appear as a tree branch. In other words, either 1. branch 1 appears in the tree and branch 2 is a link, in which case v 1 and f2 are the two primary variables and v2 and f 1 are secondary variables, or 2. branch 2 appears in the tree and branch 1 is a link, and so v2 and ft are the two primary variables and v 1 and f2 are secondary variables.
These two allowable causalities are shown in Fig. 6.18. The Gyrator: The generation of a set of independent compatibility and continuity equations from a tree structure containing a gyrator requires a different set of causal conditions. For an ideal gyrator, such as shown in Fig. 6.19, with elemental equations V]
f,
= GYf2
(6.43)
=-(a~ )v2
(6.44)
it can be seen that 1. ifv 1 is taken as a primary variable, then v2 must also be considered a primary variable since f 2 is a secondary variable, or 2. iff 1 is considered a primary variable, then because v2 is then by definition a secondary variable, f2 is also a primary variable. To satisfy the first case, with two primary across-variables, both gyrator branches must be placed in a tree. To satisfy the second possibility, where both through-variables are primary, the two gyrator branches must both be tree links. The two allowable tree structures for gyrators are illustrated in Fig. 6.19.
190
Energy-Transducing System Elements
,
~
\ \ \
I I I
I
\ \
Chap. 6
lqxp2 I I
•
\
• Figure 6.19:
Secondary variables: f 1, f 2
Secondary variables: v1, v2
The two allowable tree con-
figurations for a gyrator. The links are shown as dotted lines.
6.4.3 Derivation of the Normal Tree The derivation of a normal tree for a system containing two-port elements is an extension of the procedure described in Sec. 5.2 for systems of one-port elements. The system normal tree for a system graph model containing two-port transducers should be formed in the following steps:
Step 1: Draw the system graph nodes. Step 2: Include all across-variable sources as tree branches. (If all across-variable sources cannot be included in the normal tree, then the sources must form a loop and compatibility is violated.)
Step 3: Include as many as possible of the A-type elements as tree branches such that the completion of the tree does not require the placement of both branches of a transformer or one branch of a gyrator in the tree. (Any A-type element that cannot be included in the normal tree is a dependent energy S!orage element) Step 4: Include one branch of each transformer and both or neither branch of each gyrator in the tree so that the maximum number of T-type energy storage elements remain out of the tree. If this step cannot be completed, the system model is invalid. Step 5:
Attempt to complete the tree by including as many as possible D-type dissipative elements in the tree. It may not be possible to include all D-type elements.
Step 6: If the tree is not complete after the addition of D-type elements, add the minimum number ofT-type energy storage elements required to complete it (Any T-type element included in the tree at this point is a dependent energy storage element.) Step 7: Examine the tree to determine if any through-variable sources are required to complete it. If any through-variable source can be inserted into the normal tree, then that source cannot be independently specified and continuity is violated. System graphs and their normal trees for some simple systems containing one-port energy storage elements and an ideal two-port transducer are illustrated in Fig. 6.21. In some cases a choice of two-port causality exists in formulating the normal tree. For example, in cases 2, 3, and 4 in Fig. 6.21 either causality results in the identification of only one state variable. The two energy storage elements are dependent, and the energy storage
191
State Equation Formulation
Sec. 6.4
variable on either may be used as the state variable. In cases 1, 5, and 6, however, the choice of two-port causality leads to either two state variables or none. The procedure outlined above usually chooses the config uration that maximizes the number of selected state variables. In more complex systems, with many sources and energy storage elements, system structural constraints may require use of the representations 1 B , 5 A, and 6 B in order to identify the maximum number of independent energy storage elements in the complete system. Example 6.5 Derive the normal tree for the system shown in Fig. 6.20a in which an electric motor is coupled to an inertial load through a speed-reducing gear train.
Motor
(a) System
R
R
L
L
'~\ \I J
\
'~/,'
2
{
\ \t /
/
/
J
-, nrer =0 (b) Linear graph
(c) Normal tree
Figure 6.20: Generation of the normal tree for a electromechanical system consisting of an electric motor driving an inertial load through a gear train.
Solution The system linear graph in Fig. 6.20b contains two two-port elements to represent the motor and the gear train. The motor is represented as a four-terminal transducer, while the gear train has each branch referenced to a common reference angular velocity and is represented as a three-terminal element. The system graph consists of two connected graphs and has N 7 nodes and two distinct sections ( Nd = 2). The number of branches in the two sections of the system graph normal tree is therefore Br = N - Nd = 5. The normal tree is constructed by inserting first the voltage source V,, followed by the inertia element J. In this case there is no choice of causality assignment because branch 2 of the gear train transformer must be in the links, requiring that branch I be a normal tree branch. With this assignment, branch 2 of the motor must also be in the links. The only D-type element that may be included in the tree branches is the electric resistor R.
=
192
Energy-Transducing System Element~
Case 1:
T-type
1~4
A-type
1{ 2 34 8:(~' A:'?,) I 2\ 3t4 I
L\
\
C
1~4
A-type
A-type
1~4
T-type
T-type
T-type
1~4
A-type
1~4
A-type
A-type
c\
\
\
c
\
1~4
T-type
T-type
1C
C
I
1C
I
I
State variables: v1
1{ 2 34 ~ I(~' A:'?,) 2\ 3t4 I
L\
\
\
L
L
\
I
I
I
1L
State variables: f 4
A: t1~\ 8:·(~)4 I
\
1
2
L\
3 t4
\
I
/c
l
1
\
\
I
C
State variables: v4
A: t1~\ 8:·(~)4 I
\
1
2
3 t4
\
C\
I
\
I
I C
State variables: none Case6:
I
I
State variables: f 1 CaseS:
1
1{ 2 34 8:(~' A:'?,) I 2\ 3t4
State variables: f 1 Case4:
I
State variables: none
State variables: v4 Case 3:
L
\
\
State variables: f 1• v4 Case2:
Chap. 6
C
1
\
I
\
C
State variables: v1, v4
A: t1~\ 8:1(~)4 I
\
1
2
3 t 4
L\
}L
\
I
State variables: f 1• f 4
L
I
1
\
\
State variables: none
Figure 6.21: Examples of two-port system normal trees when energy storage elements are connected across both branches. The links are shown as dotted lines.
L
Sec. 6.4
193
State Equation Formulation
6.4.4 State Equation Generation
With the nonnal tree, we can begin the derivation of the state equations for systems with twoport elements by choosing the primary variables as the across-variables on all branches of the normal tree and the through-variables on all nonnal tree Jinks, including those associated with the two-port elements, As defined in Chap. 5 the system state variables are those primary variables associated with the n independent energy storage elements defined by the normal tree, that is, 1. the across-variables of the A-type energy storage elements in the normal tree, and 2. the through-variables of the T-type energy storage elements in the normal tree links. The procedure for deriving n state equations in tenns of the n state variables and S source variables is similar to that described in Chap. 5 for systems of one-port elements, with the addition that the two elemental equations for each two-port element must be included in the derivation. The procedure is illustrated in the following examples. Example6.6 A sketch of a fixed-field de motor drive system, with its system graph model containing B = 7 branches and N = 6 nodes, is shown in Fig. 6.22. The motor is represented as a four-terminal element, generating two distinct connected graph sections in the system graph (Nd = 2}, so there are Br = N- Nd = 4 branches in the normal tree and two branches in the links. The system normal tree is shown in Fig. 6.22c. From the normal tree in Fig. 6.22c: Primary variables: V1 (t), QJt VR, h. Ts. Vt. T2 Secondary variables: ls(t), TJt iR. VL, Os. i., !12 System order: 2 State variables: QJt hThe B - S = 6 elemental equations written in terms of primary variables are dOJ
dt
= .!.rJ J
dh
1
dr = LVL VR
= RiR
Ts = Brls Vt T
l2
1
(i)
(ii) (iii)
(iv)
= -Q2 Ka
(v)
1 •
(vi)
=
--ll
Ka
TheN- Nd- SA = 3 continuity equations are
= -T2- Ts iR =it
T1
it
=it
(vii) (viii)
(ix)
194
Energy-Transducing System Elements
Chap.6
:Motor R 1-------------...1 (a) System
R
L
R
L
I~ 2
, \
jsl
,,/
'7IJ77 fi=O (b) Linear graph
(c) Normal tree
Figure 6.22: An electric motor drive, a system graph, and a normal ttee.
The 8- N
+ Nd- ST = 3 compatibility equations are VL
= Vs -
VR -
(x)
V1
0a=0 1
(xi)
02=0}
(xii)
The secondary variables may be directly eliminated from the elemental equations: d01
1
dh
1
dt = J(-T2-Ts)
(xiii)
Tt =I (V.r- VR- VJ) VR
(xiv)
= RiL
(XV)
(xvi)
Ts = B01 I
v, =-OJ Ka
'I"'
'2
=
(xvii)
1 .
(xviii)
--IL
Ka
By direct substitution the six elemental equations may be reduced to two state equations and placed in the standard form:
[ -B/J [OJ] h =
-l/KaL
1/KaJ] -RJL
·[OJ] [ 0]V,~(t) h + 1/L
(xix)
Sec. 6.4
State Equation Formulation
195
If the dynamic study required computation of the torque to accelerate the inertia T1 , the motor current iL. and the viscous bearing torque T8 as output variables, the elemental and constraint equations may be used to write a set of three output equations in terms of the state and input variables: (XX) (xxi)
(xxii) In output matrix form these equations are written
~
T1 ]
[
~
[ - B =
1I~Ka ]
0]
~] + ~ [
[
V,(t)
(xxiii)
Example6.7 A hydraulically actuated mechanical system and its linear graph model are shown in Fig. 6.23. The hydraulic actuator is a gyrator and divides the system into the fluid and mechanical domains. On each side a separate reference node is established. The system graph has B = 7 branches and N = 5 nodes. The gyrator, representing the hydraulic ram, divides the overall graph into two distinct connected graphs Nd = 2. The normal tree for the system has BT = 5 - 2 = 3 branches; the pressure source, the mechanical mass and the fluid resistor are shown in Fig. 6.23c. The two gyrator branches, together with the spring and mechanical damper form the links of the tree. From the normal tree shown in Fig. 6.23c: Primary variables: Ps(t), PR, Vm, FK, Fs, Q1, F2 Secondary variables: Qs(t), QR, Fm. VK, Vs, P., ~ System order: 2 State variables: Vm, FK The B - S
= 6 elemental equations written in terms of primary variables are dvm _ 2_F. dt - m m dF1c -=KvK dt PR = RQR Fs = Bvs Q1 = -Av2 F2 =APt
(i) (ii) (iii)
(iv) (v) (vi)
From the normal tree the N - Nd - SA = 2 continuity equations are Fm = - F2 - FK QR = Q1
-
FB
(vii) (viii)
Energy-Transducing System Elements
196
Chap. 6
(a) System
P=Parm v=O (b) Linear graph Figure 6.23:
and the B- N
Hydraulic actuator system. its linear graph. and a normaJ tree.
+ Nd- ST = 4 compatibility equations are Vm
(ix)
VB= Vm
(x)
. VK
V2
Pt
=
= Vm
(xi)
= P,- PR
(xii)
The secondary variables may be eliminated from the elemental equations to yield
dvm
dt =
1
;; (F2 - FK
dF~:
dt =KFK PR = RQl Fs = Bvm Ql = -Avm F2 = A (P, - PR)
-
Fs)
(xiii) (xiv)
(xv) (xvi)
(xvii)
(xvili)
By direct substitutio~ the six elemental equations may be reduced to two state equations: (xix)
(x.x)
with the matrix form
Chap. 6
197
Problems
PROBLEMS
6.1. The dynamic performance of actuator systems is affected by the equivalent inertia driven by the motor. Consider the two systems illustrated in Figure 6.24. In (a) a servo motor, which may be considered as a torque source, drives a load inertia J through a 10:1 frictionless gearbox. In (b) a similar servo motor drives a translating mass m through an ideal rack and pinion of radius rP.
Mass m
(b)
(a)
Figure 6.24: 1\vo inertial loads on a servo motor.
(a) Draw the linear graph for each system. (b) Determine the equivalent rotational inertia Jeq reflected through the transduction element to the drive motor. 6.2. A cam follower may be modeled as a lever with ann lengths l1 and 12 as shown in Figure 6.25. A spring, with stiffness K. is used to keep the follower in contact with the cam. Detennine the effective stiffness of the follower as seen at the contact between the follower and the cam. The mass of the follower may be ignored.
Figure 6.25: A cam follower.
198
Energy-Transducing System Elements
Chap.6
6.3. A ball-screw linear actuator is shown in Fig. 6.26. The actuator uses a de electric motor, with an electromotive coupling constant Ka, armature resistance R, and inductance L, and driven by a voltage source V.r(t}. The motor drives a threaded shaft with N threads/em on which the ball-nut moves linearly. The mechanical load is a mass m which slides on a surface with viscous damping coefficient B. Motor +
m
DampingB
/ Figure 6.26: A ball-screw drive mechanism.
(a) Construct a model for the ball-screw that relates the shaft torque and angular velocity to the nut linear velocity and force. Write the equations for the transduction process and show that they are energy conserving. Is this a transforming or gyrating relationship? (b) Generate a set of state equations that describe the dynamics of the system.
6.4. The drive train for a front wheel drive car is shown in Fig. 6.27. The wheels interact directly with the groun~ and if no slip occurs provide a force to accelerate or decelerate the car mass. Engine
"""/n,(t)
/
Wheel
/
~(r)
/ ,.L..
I
l
~·
Axle
""" ._
r
I """"'
17-'
'
~"
Transmission
(N:l)
Figure 6.27: An automotive front-wheel drive.
(a} Consider the pair of drive wheels as an ideal rotary-to-linear transducer. Construct the transducer linear graph and determine the relationships among the translational force and velocity and the wheel angular velocity and torque. Show that the wheels are a power-conserving transducer in the ideal case. (b) Construct a linear graph relating the torque input through the wheel axle to the linear velocity of the car, including the effects of the car mass, and assuming that the car aerodynamic drag and rolling resistance may be represented as a single equivalent linear damper.
Chap. 6
199
Problems
(c) The transmission in a car is used to match the engine output to the car speed. If the transmission is a lossless gear train with a step-down ratio N, what element would you use to represent the
transmission? If it is desired to operate the engine at a constant speed of approximately 6001r r/s {approximately 1800 rpm) what are the gear ratios required for {i) traveling in town at 15 m/s and {ii) highway driving at 30 m/s? Assume that the tire diameter is 0.75 m.
= 0 r/s has a static torque of 1000 N-m and at a top speed of 2001r r/s develops no torque. Construct a model for the engine. (e) Construct the linear graph model for the complete vehicle drive system.
(d) Consider the engine as a power limited source, which at S"2
(t) Derive the system state equations.
6.5. A steamroller, with total mass m, has a pair of large rear drive wheels with radius r 1 and a combined inertia J 1• The roller at the front has radius r2 and inertia J 2 • Assume that all wheels roll without friction or slip. Find the effective moment of inertia reflected to the rear
Hammer
Rotational bearing Spring between shaft and bearing
Drive shaft Sliding bearing B
Figure 6.28: A mechanical hammer mechanism.
{a) Draw a linear graph for the system. (b) Derive a set of state equations for the system and express them in matrix form. 6.7. A permanent magnet de motor may be modeled by the lumped system shown in Fig. 6.29, where R1 and L1 are the electrical resistance and inductance of the field winding, eb is the back-emf, and J and B are the inertia and viscous damping of the rotor and its bearings. The motor performance has been specified by the manufacturer in terms of the relationships T = Kmi and vb = KuO.. (a) Draw a linear graph and normal tree for the system, assuming that the motor is driven from a voltage source. (b) Derive a set of state equations.
Energy-Transducing System Elements
200
Chap. 6
Rotor
+
J
Electrical
Rotational
Figure 6.29: Equivalent circuit of a de motor.
(c) Manufacturers typically specify Km in units of oz-inlamp, and Kv in units of voltsllapm (1 krpm I000 revs/minute). These are not consistent units. Consider the motor as an ideal powerconserving two port. If a sample motor has a specified Km 6 oz in/amp, find its value of K v
=
=
in voltsllapm. (d) Measurements on a motor with the values of Km and Kv as found in part (c) showed that
R1 = 3Sl, and L 1 = O.OOSH. When connected to a constant current source of 1 amp, this motor reached a constant speed of I000 ~pm. i. What is the effective viscous damping B in the bearings? ii. What is the voltage measured at the motor terminals?
iiL What fraction of the total power being supplied by the source is dissipated on the electrical side, and on the mechanical side of the motor. Identify the elements that are dissipating the power? 6.8. Explain why a de motor becomes hot, and may bum out. if the shaft jams so that it may not rotate. Assume that the motor described in Problem 6.7 is driven by a 15 volt source. What power is dissipated if the shaft is jammed? Where is the power dissipated? 6.9. Fig. 6.30 shows a loudspeaker voice coil. The magnet structure setc; up a uniform radial magnetic field of strength B weber/meterl in the air gap. The coil has N turns, length l, and radius r. Voltage v1 appears across the coil and current i flows through il (a) Assume that the coil is massless and has neither resistance nor inductance. Also assume that the
coil is mounted so that it is always radially centered in the air gap but can move freely in the axial direction. Derive relationships between the electrical variables i and v 1 , and the mechanical variables v and F. (b) Is this ideal element a transformer or a gyrator?
(c) Draw the linear graph and normal ttee for a real loudspeaker. Add the following nonideal effects
to the model: assume that the coil has resistance R and inductance L and the cone has mass m and is supported in a structure that resists linear motion with a damping coefficient B, and linear stiffness K. Assume that the input is an ideal voltage source V, (t). (d) Derive the state equations for your model and derive an output equation for the velocity of the
coil v.
Chap. 6
201
Problems SupportK
+
Magnetic
flux
(b)
(a)
Figure 6.30: A loudspeaker voice coil.
6.10. A voice coil, similar to that described in Problem 6.9, is modified to act as an adjustable dashpot. A push rod is connected to the coil, and a variable resistor R is connected across the coil as shown in Fig. 6.31. Using the same electromagnetic parameters as in Problem 6.9, find the equivalent viscous damping coefficient for the dashpot. The inductance of the coil and the mechanical parameters may be ignored.
Variable resistance R
(a)
Figure 6.31:
An electro-mechanical dashpot
6.11. A gyrator may be constructed from electronic components. Fig. 6.32 shows an electronic gyrator in which i 1 = Gv2 , connected in a circuit with two capacitors C1 and C2 , and a resistor R. Derive a set of state equations for the system and an output equation for the voltage across capacacitor C 1 • R
Electronic gyrator G
V(t)
Figure 6.32:
An electronic gyrator circuit
202
Energy-Transducing System Elements
Chap. 6
6.12. Electrical transfonners consist of two coils of wire, designated the primary and secondary windings, wound on a common high-penneability magnetic core. For alternating current (ac) inputs the system behaves approximately as an ideal transfonner with a voltage relationship v,.jv 1 = N2/ N 1• where N 1 and N2 are the number of turns on the primary and secondary windings. In an ideal electrical transfonner the penneability of the core would be infinite, and all of the magnetic flux would link both coils. Practical transformers have cores with finite premeabiJity, and inevitably not all of the flux associated with current flowing in one winding links the other coil. A lumped equivalent model of a transformer is shown in Fig. 6.33 in which L 1 and L 2 represent the leakage inductances associated with the flux components that do not link both coils, Lm represents the magnetic field necesary to maintain the flux in the finite penneability core, and R1 and R2 are the resistances of the primary and secondary windings.
\
Ideal
transformer
Secondary
High-penneability magnetic core (b)
(a)
Figure 6.33: (a) An electrical transformer; (b) its lumped electrical equivalent model.
(a) Derive a set of state equations for an electrical transfonner connected to a voltage source and driving a resistive load R. · (b) Does the model given in Fig. 6.33 predict that an electrical transformer will not work for de inputs, that is, when v (I) is constant. Explain your answer. 6.13. An automotive shock absorber consists of a fluid-filled closed cylinder of cross sectional area A, with a moving piston containing a small orifice as shown in Fig. 6.34. As the piston moves, fluid is forced through the orifice. Laboratory measurements have shown that the fluid flow rate through the orifice is approximately proportional to the pressure difference across it, and that it may be modeled as a linear fluid resistance R1 . Find the translational viscous damping coefficient for the shock absorber. (The cross sectional area of the piston rod may be neglected.) Orifice
Area A
Figure 6.34: A fluid filled mechanical shock absorber.
6.14. A fluid accumulator, shown in Fig. 6.35, consists of a cylinder of cross sectional area A containing a movable piston that is supported against the housing by a spring of stiffness K. Find the fluid capacitance of the system.
Chap. 6
203
Problems Area A
Q_....
Figure 6.35:
A spring loaded fluid accumulator.
6.15. A high-performance hydraulic actuator is shown in Fig. 6.36. The de motor, with the torque/current relationship T = - Kai, is controlled from a voltage source Vs, and has winding resistance R and inductance L. The positive displacement pump displaces D m 3 of fluid per radian of shaft rotation. The mass m is driven through a ram of area A. A bypass valve, with fluid resistance R 1 , returns the fluid to the reservoir tank.
Figure 6.36:
A high-performance hydraulic drive system.
(a) Construct the system linear graph and identify the state variables. (b) Derive the state equations.
(c) Write an output equation for the displacement of the mass. (d) Determine the relati onship between the ram force on the mass to the motor torque for the case when the mass is clamped so that it cannot move. 6.16. A rotary turntable with radius r 1 and moment of inertia 1 1 , is driven by a permanent magnet de motor through a rubber belt, as shown in Fig. 6.37. The drive pulley has radius r 2 , and the motor, which has an armature with a moment of inertia h. is driven by a current source I (t) . Each side of the rubber drive belt may be approximated as a linear spring of stiffness K . The turntable is mounted in bearings with a small but significant viscous drag coefficient B . (a) Draw a linear graph representation of the system. (b) Determine the system order.
(c) Generate a set of state equations fo r this system.
204
Energy-Transducing System Elements
Chap. 6
Motor J2
Source /(t)
figure 6.37: A flexible belt rotary drive system.
6.17. A permanent magnet de generator is driven by a line around a pulley of radius r as shown in Figure 6.38. One end of the line is connected to a spring of stiffness K, and the other is attached to a source of known force F (t). The line is taut at all times. The generator is connected to a large capacitor C. Derive a set of state equations and an output equation for the voltage across the capacitor. You may ignore the inductance of the generator winding but should include the rotor inertia J and the winding resistance R.
c Figure 6.38: A line driven electric generator.
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-Hill, New York, 1967. [3] Blackwell, W. A .• Mathematical Modeling of Physical Networks, Macmillan, New York, 1967. [4] Shearer, I. L., Mwphy, A. T., and Richardson, H. H., Introduction to System Dynamics, AddisonWesley, Reading, MA, 1967. [5] Kuo, B. C., Linear Networks and Systems, McGraw-Hill, New York, 1967. [6] Chan, S. P., Chan, S. Y.• and Chan, S. G., Analysis of Linear Networks and Systems, AddisonWesley. Reading, MA. 1972.
7
Operational Methods for Linear Systems
7.1 .INTRODUCTION There are many ways to express the relationship among variables in a system. Engineers often use block diagrams to indicate both the structure of a system and its interactions with the environment. Mathematical operational and transform methods provide tools for the manipulation of system equations and allow engineers to develop different system representations. In this chapter we develop graphical and mathematical methods for depicting and manipulating the structure of a system through time domain operational methods. In particular we introduce a notation that allows the differential equations representing a system to be manipulated as if they were algebraic equations. It is often useful to consider the functional or mathematical relationship between a single system input and a single output variable. Figure 7.1 shows a simple block diagram, containing one block, for a system. The block implies that there is a causal relationship between the input u(t) and the system output y(t), that is the output is defined by an operation on the input. In general the relationship is defined by the system itself through its differential equations. A single block description, as in Fig. 7.1, may be expanded to include multiple interconnected blocks representing the dynamics of single lumped elements, blocks representing a combination of system elements, or blocks interconnecting systems [1-3]. The operational block diagram also provides a basis for developing computational methods for solving system responses using both analog and digital computers and is a particularly useful system representation when coupled with measurement and control systems [4-6]. 205
Operational Methods for Linear Systems
206
System input u(t)
Mathematical "operation" relating output to input
Chap. 7
System output y(t) = H{u(t)}
H{u(t)}
Figure 7.1: Block diagram of a system operator.
For linear time-invariant systems, the input-output operational descriptions relate a single output variable y(t) directly to a single system input u(t). The system dynamics are often expressed in the fonn of a single nth-order differential equation that relates the output variable to the input variable:
(7.1)
where the constant coefficients a; and b; are defined by the system parameters. Equation (7. 1) is known as a classical form of system description. It is the basis for the classical feedback control theory for single-input single-output (SISO) systems that was developed in the 1930s and 1940s [4-6]. On the other hand, the state space modeling methods, developed in Chaps. 4-6, represent a system by a set of n coupled first-order differential state equations and m associated algebraic output equations: i =Ax+Bu
y=Cx+Du
(7.2) (7.3)
where the state vector x(t) is an internal representation of the system. This system representation, developed in the 1950s and 1960s, is the modem form of representation and is the basis for many multiple-input multiple-output (MIMO) control system design techniques [7, 8]. Both state variable and classical system descriptions are used commonly in dynamic system analysis and design. The state variable descriptions are in a fonn that may be solved directly by numerical integration in a digital computer (as described in Chap. 11), while the classical descriptions are in a fonn for which analytical solutions are readily available, especially for low-order systems (as described in Chap. 9). For linear systems the two forms of description are directly related, and both provide complete representations. In this chapter we develop the necessary operational tools and block diagram representations for transfonning system models between the system representations. We consider linear timeinvariant systems in which all system parameters contained in the A, B, C, and D matrices are constants and develop operational methods for transfonning between the state variable and classical input-output system descriptions.
Sec. 7.2
7.2
Introduction to Linear Time Domain Operators
207
INTRODUCTION TO LINEAR TIME DOMAIN OPERATORS
7.2.1 System Operators
A linear time-invariant system with one input and one output may be described in terms of afunctional operator relating the output y(t) to the input u(t) through a well-defined mathematical relationship. The system behavior may be written operationally as y(t) = H {u(t)}
(7.4)
with the interpretation that the output function y(t) is the result of a mathematical operation H (} on the input function u(t), as illustrated in Fig. 7.1. The operator H {} is defined to be the dynamic transfer operator and forms an alternative description of the dynamics of the system. Equation (7.4) is a causal relationship since it implies that the system response y(t) is defined by a direct operation on the input. The dynamic transfer operator represents the system dynamics contained in a system model, for example, a differential equation such as Eq. (7.1). Dynamic transfer operators are illustrated in Fig. 7.2 for three simple systems, each under the influence of a single external input. For each the system response y(t) is related to the input u(t) by the system dynamic model, expressed as a differential or algebraic equation together with any necessary initial conditions. For example, in the case of the fluid capacitance in Fig. 7.2c, the mathematical operation relating the output variable Pc(t) to the input flow Q(t) is Pc(t) = -I
c
f.' 0
Q(t) dt
+ Pc(O)
(7.5)
which involv~s the elementary operations of scaling (multiplication by 1I C) and integration. If the tank is empty at time t = 0, the time history of the flow rate Q(t) into the tank defines the pressure Pc(t) for all timet > 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 + <.C2 + .C3) .c1 + .c2.c3] Ix
(7.28)
generates an equivalent second-order operator in .C 1{). Cascaded linear operators may be combined into an equivalent single operator using the standard rules of algebra. For example, the operational expression y(t) = [S + 2] [S + 1] {x(t)} = [S2
+ 3S + 2] {x(t))
(7.29)
implies that (7.30)
Similarly, it is often possible to factor a polynomial operator into a cascaded set of lowerorder terms. The classical single nth-order differential equation [Eq. (7.1)] relating a system output variable y(t) to a single input u(t),
d"y dt"
=
+ an-1
dmu bm -
dtm
d"- 1y drn-1
+ ... +at dm- 1u
+ bm-1 1 + drm-
dy dt
+ aoy
dy .. · + bt -
dt
(7.31)
+ bou
may be written in operational form using polynomial operators as
[S" + an-tsn-t + · .. + a1S + ao] {y) = [bmSm + bm-lsm- 1 + · .. + btS + bo] {u}
(7.32)
7.5.3 The Inverse Operator
Every linear operator C, {}that is not the null operator N {}implicitly has an inverse, written
.c- 1 (}, which has the property
.c-• {.C {}}a I{}
(7.33)
so that for any x(t), y(t) =
e,- 1 {£ {x(t)}} = x(t)
(7.34)
The inverse .c- 1 {} therefore "undoes" any action of£ {} and reproduces the original input x(t).
224
Operational Methods for Linear Systems
Chap. 7
The concept of an inverse operator is important in the description of dynamic models. For example, assume that a differential equation describing a system may be operationally described as in Eq. (7 .32) and summarized by a pair of linear operators £1 and £2 acting on the input and output variables: £2 {y(t)}
= £1 {u(t)}
(7.35)
Then an explicit operational expression for the response y(t) may be found by operating on both sides of the equation using the inverse operator r; 1 (7.36) or I {y(t)}
= y(t) = £2 1£1 {u(t)}
(7.37)
For historical reasons it is common to write an expression with an inverse scalar operator y(t) = £2 1£ 1 {x(t)} as a quotient: £1 2
£2-1 £1 {x(t)}
y(t) = £ {x(t)} =
(7.38)
Both notations may be used, however, when an operator appears in a denominator, it does not imply algebraic division; it denotes application of the inverse of the operator. The inverse of the primitive scaling operator a {} is simply another scaling operator with a value 1/a, or 1 a-t{}=-{}
(7.39)
a
The derivative operator S and the integral operator s-t are by definition natural inverses. It is for this reason that the integrator is commonly written as an inverse operator. Example7.6 Show that the mathematical operation implied by [S + y(t)
Solution Assume that u(O)
= e-at
1'
ar
1
{u(t}} is
e01 u(t) dt
=0, so du [S +a] {u(t)} = dt
+ au(t)
and substitute into the assumed fonn of the solution: [S + ar' [S + a]{u(t)} =e-ar
1'
e"' [ ~;
+ au(t)] dt
Sec. 7.6
The System Transfer Operator
225
Evaluating the first term in the integral by parts gives [S +
ar• [S +a] {u(t)} =e-at [ea'u(t)- L' e(J'au(t)dt + L' e(J'au(t)dt] = u(t)
Thus, the sequential application of the operator and its assumed inverse yields the original function.
7.6
THE SYSTEM TRANSFER OPERATOR
The operational form of the classical nth-order differential equation [Eq. (7 .32)], is
[S" + an-Isn-l + ·· · + a1S + ao] {y} = [bmSm + bm-1sm-t + ·· · + b1S + bo] {u}
(7.40)
1
If the inverse operator [S" + an-1S"- 1 + · · · + a1S + ao]- exists, it may be applied to each side to produce an explicit operational expression for the output variable: y(t) =
[S" + an-tsn-t + ·· · + a1S + ao]- 1 [bmsm + bm-l sm-l + ... + bl s + bo] {u(t)}
(7.41)
or using the quotient notation for the inverse, (7.42)
The operational description of the system dynamics has been reduced to the form of a single linear operator, defined in Eq. (7.4) to be the dynamic transfer operator H {}, y(t) = H {u(t)}
(7.43)
H {} = bmSm +.bm-lsm- 1 + · · · + btS + bo {} S" + an-tS"- 1 + · · · + a1S + ao
(7.44)
where
Equation (7 .44) is a system description that is equivalent to the differential equation from which it was derived. Figure 7.17 shows this input-output relationship in block diagram fonn.
Operational Methods for Linear Systems
u(t)
bmsm + bm-1sm-l + •·· + b 1S + b0
--.~
Chap. 7
.....,_~y(t)
sn + a,_ 1S""'1 +···+ a aS+ Do
Figure 7.17: The input-output dynamic transfer operator of a system represented by a single differential equation.
Example7.7
Derive the dynamic transfer operator relating the mass position y(t) to the input force f's(t) for the mechanical system described in Example 7.S. Solution From Example 7.S the second-order differential equation for the system is d 2y
B dy
1
K
-dt2 + - + -y= -F,(t) m dt m m
(i)
This differential equation may be written in operational form,
[
B K]
S2 + -S + - {y} m m
= -m1 {F,}
(ii) 1
and applying the inverse operator [S2 + (Bjm)S + K/m]- to solve for y yields
K]-• 1
B
y
= [ S 2 + ;;;S + m
;;; {F,}
1/m
= S2 + (B/m)S + K/m {F,}
(iii)
and so the transfer operator is
1/m
H {}
7.7
= S2 + (Bjm)S + Kjm
(iv)
TRANSFORMATION FROM STATE SPACE EQUATIONS TO CLASSICAL FORM
The transfer operator may be used to derive the classical input-output differential equation for any system variable directly from a state space representation. The following example illustrates the general method for a first-order system. Example7.8
Find the transfer operator and a single first-order differential equation relating the output y(t) to the input u(t) for a system described by the first-order linear state and output equations dx -d
. t
= ax(t) + bu(t)
(i)
+ du(t)
(ii)
y(t) = cx(t)
Sec. 7.7
Transfonnation from State Space Equationsto Classical Fonn
227
Solution The state equation in operational form is S {x(t)} =a {x(t)} + b {u(t)}
(iii)
which may be rewritten with the state variable x(t) on the left-hand side: [S-a] {x(t}} = b {u(t)}
Then using the inverse operator [S-
(iv)
ar'. solve for the state variable:
x(t) = [S-
ar' b {u(t)}
(v)
and substitute into the output equation y = e {x} + d {u} y(t) = [e (S- a)- 1 b + d] {u(t}}
(vi)
The transfer operator may be found by extracting the inverse y(t)
= (S- a)- 1 [eb + d (S-a)] {u(t)} = dS +;be- ad) {u(t)}
(vii)
= dS +(be- ad){}
(viii)
-a
and so
H {)
S-a
The differential equation is found by operating on both sides with [S-a] (S-a) {y(t)}
= [dS +(be- ad)] {u(t}}
(ix)
and rewriting as a differential equation: dy - - ay dt
= ddu - +(be- ad) u(t) dt .
(x)
Classical representations of higher-order systems may be derived in an analogous set of steps by using matrix operational algebra. A set of linear state and output equations written in standard form
x=Ax+Bu y=Cx+Du
(7.45) (7.46)
Operational Methods for Linear Systems
228
Chap. 7
may be rewritten in operational form by interpreting the matrices A, B, C, and D as matrix operators acting on the state and input vectors and writing them in operational form as A {}, B {}, C {}, and D {}. The system equations are then
S {x(t)} =A {x(t)} + B {u(t)}
(7.47)
y(t) = C {x(t)} + D {u(t)}
and the state equations may be rewritten (7.48)
S {x(t)}- A {x(t)} = [SI- A] {x(t)} = B {u(t)}
where the term SI creates an n x n matrix operator with S on the leading diagonal and the null operator elsewhere. (This step is necessary because matrix addition and subtraction are defined only for matrices of the same dimension.) The matrix operator [SI- A] {) appears frequently throughout linear system theory; it is a square n x n operator with elements directly related to the A matrix: (S- au) 21
-~
[SI- A]{}=
-atn -a2n
.
l
{}
(7.49)
[ -ani
(S -·ann)
The state equations, written in the form of Eq. (7.48), are a set of n simultaneous operational expressions. The common methods of solving linear algebraic equations, for example, gaussian elimination, Cramer's rule, the matrix inverse, elimination, and substitution, may be directly applied to linear operational equations such as Eq. (7.48). For low-order single-input single-output systems the transformation to a classical formulation may be performed in the following steps:
1. Rewrite the state equations in operational form. 2. Reorganize each operational state equation so that all terms in the state variables are on the left-hand side. 3. Treat the operational state equations as a set of simultaneous algebraic equations and solve for the state variables required to generate the output variable. 4. Substitute for the state variables in the output equation. 5. Write the output equation in operational form and identify the transfer operator. 6. Use the transfer operator to write a single differential equation between the output variable and the system input. This method is illustrated in the following two examples.
Sec. 7.7
229
Transformation from State Space Equationsto Classical Form
Example7.9 Use the operator method to derive a single differential equation for the capacitor voltage vc in the series RLC electric circuit discussed in Example 5.5. Solution The two state equations are {from Example 5.5) (i)
and the required output equation is (ii)
Step 1: In operational form the state equations are S {Vc} = 0 {Vc}
=
S{iL}
1
~
1 . +C {1 L} + 0 {Y.r}
±
(iii)
C {zd = O{Vs}
(iv)
{vc}-
~ {h} +
{V.r}
Step 2: Reorganize the state equations: S {vc}-
1 .
I{vel+ [s+ ~] =I {h}
(v)
{V.r}
Step 3: In this case we have two simultaneous operational equations in the state variables vc and iL· The output equation requires only vc. If Eq. (iv) is operated upon by [S + R/ L]. Eq. (v) is operated upon by 1/C. and the equations added. iL is eliminated: (vi) Step 4: The output equation is y = vc. Operate on both sides of Eq. (vi) using [S2 + (R/ L)S + 1/Lcr 1 and write in quotient form v _
1/LC
c- S2 + (R/L)S + lfLC
{V.} .r
\
~[_ J~
(vii)
Step 5: The transfer operator is H
_ 1/LC {} - S2 + (R/ L)S + 1/ LC
Step 6: The differential equation relating vc to
V,~
{viii)
is (ix)
Operational Methods for Linear Systems
230
Chap. 7
Cramer's rule for the solution of a set of linear algebraic equations, described in App. A, is a useful method to apply to the solution of operational equations. In solving for the variable x1 in a set of n linear algebraic equations such as Ax = b, the rule states
x;
=
det [A] det[A]
(7.50)
where A Ci) is another n x n matrix formed by replacing the i th column of A with the vector b. Cramer's rule for solution of the operational equations describing a single-input linear system is rewritten in terms of the inverse operator. H
[SI- A] {x}
= B {u}
(7.51)
then the relationship between the i th state variable and the input is x;
= (det [SI- A])- 1 det [lSI- A]<1>] {u(t)}
(7.52)
or, in quotient form, x;
=
det [lSI - A]J det [SI- A] {u(t)}
(7.53)
where (SI- A) isdefinedtobethematrixfonnedbyreplacingtheithcolumnof(SI- A) with the column vector B. The differential equation is det [SI- A] {x;} = det [ (SI- A)
(7.54)
Example 7.10 Use Cramer•s rule to solve for vL(t) in the electric system in Examples 5.5 and 7.9.
Solution From Example 5.5 the state equations are
(i)
and the output equation is (ii)
Sec. 7. 7
231
Transformation from State Space Equationsto Classical Form
In operational form the state equations are S [ 1/L
-1/ C ] [
S+R/L
vc] = [ 1/L 0 ] "in(t)
(iii)
h
The voltage vc(t) is given by
Vc(t)
=
>]
det [(sl- A)< 1
det[(SI-A)]
_
det [ (V;,.(t)}
=
det
[
0
-1/C ] S+ R L 1 I /C ] S+R/L
1/L
s
_
l/L
[V~n(t)J
(iv)
1/LC
- S2 + (R/L)S + (1/LC)
{Vin(t)}
The current h(t) is
[(SI-A)(2)] iL(I)=
det
det((SI-A)j
l"in(t)}
detL~L -1/C i~L] ] (V~a(t)).
= det [ i/L S
S+R/L
(v)
s/L
= S2 + (R/L)S + (ljLC) {Vin(t)} The output equation may be written direct1y from Eq. (ii): vL(t) = -vc - RiL
+ Vs(t)
-1/LC -(R/L)S ] [ S2 + (R/L)S + (1/LC) + S2 + (R/L)S + (1/LC) + 1 {Vs(t)} = -1/LC-(R/L)S+[S2 +(R/L)S+(l/LC)] {V.( )} S2 + (R/L)S + (lfLC) I t
=
(vi)
s2 = S2 + (R/L)S + (ljLC) {Vs(t)} giving the differential equation 2
2
1 () d V, d vL - +-RL -dvL + -LC VL t = - dt2 dt dt2
(vii)
For a single-input single-output system the transfer operator may be found directly by evaluating the inverse matrix x = (SI- A)- 1 B {u}
(7.55)
232
Operational Methods for Linear Systems
Chap. 7
Using the definition of the matrix inverse operator (App. C),
[SI- A]- I
{}
= adj [SI -A] {} det[SI -A]
x(t)
=
adj [SI- A]B det [SI- A] {u(t)}
(7.56)
(7.57)
and substituting into the output equations gives
+ D {u(t)}
y(t) = C [SI- A]- 1 B {u(t)}
= (C[SI -A]- 1 B +D] {u(t)}
(7.58)
Expanding the inverse in terms of the detenninant and the adjoint matrix yields
() =
y t
Cadj(SI-A)B+det[SI-A]D { ( )} det[SI -A] u t
(7.59)
= H {u(t)} and so the required differential equation is
det [SI- A] {y(t)}
= (C adj (SI- A) B + det [SI- AJD] {u(t)}
(7.60)
Example 7.11
Use the inverse matrix operator to find a differential equation relating VL(t) to Vs(t) in the system described in Example 7.1 0. Solution The state vector, expressed operationa11y as (i)
from the previous example is
[
S
;: ] = [ 1/L
-1/C
S+R/L
]-l [ 0 ]
1/L Vm(t)
(ii)
The determinant of [Sl - A] is det[SI -A]= (S2 + (R/L)S+ (1/LC)]
(iii)
Sec. 7.8
Transformation from Classical Form to State Space Representation
233
and the adjoint of [Sl - A] is
•
·[ S adJ 11L
-liC ]=[S+RIL S + RIL -11L
liC] S
(iv)
From Example 5.5 and the previous example, the output equation VL (t) = -vc - Rh + V, (t) specifies that C = [ -1 - R] and D = [1]. The transfer operator, Eq. (7.59) is
H {}
=
C adj (Sl - A) B + det [Sl - A] D det[SI
-Ar
I
(v)
{}
Since
C adj (Sl - A) B = [ -1
R
- R 1[ S + R I L -tiL
1I C] [ 0 ] S
liL
I
(vi)
=-Ls- LC the transfer operator is H {}
= -(RIL)S- 1/LC + [S + (RIL)S + (IILC)] [I] 2
S2 + (RI L)S + (II LC) S2
(vii)
+ (RI L)S + 01 LC)
which is the same result found by using Cramer's rule in Example 7.9.
7.8
TRANSFORMATION FROM CLASSICAL FORM TO STATE SPACE REPRESENTATION The block diagram provides a convenient method for deriving a set of state equations for a system specified in terms of a single input-output differential equation. A set of n state variables can be identified as the outputs of integrators in the diagram, and state equations can be written from the conditions at the inputs to the integrator blocks (the derivative of the state variables). There are many methods for doing this; we present here one convenient state equation formulation that is widely used in control system theory. Let the differential equation representing the system be of order n and without loss of generality assume that the order of the polynomial operators on both sides is the same:
[anS"
+ Gn-lsn-l + · · · + ao] {y(t)} = [bnSn + bn-lsn-l + · · · + bo] {u(t)}
(7.61)
234
Operational Methods for Linear Systems
Chap. 7
We may operate on both sides of the equation using s-n to ensure that all differential operators have been eliminated:
[an+ an-1s-t + ... + ats-
(7.62)
from which the output may be specified in terms of a transfer operator. If we define a dummy variable z(t) and split Eq. (7.62) into two parts,
Z (t)
= [an + an-tS-l +···+at s-
(7.63)
y (t)
= [bn + bn-tS- + · · · + bts- + bos-n] (z (t)}
(7.64)
1
Equation (7.63) may be solved for u(t): U
(t)
=[an+ an-ls-I+···+ ats-
(7.65)
and rearranged to generate a feedback structure that can be used as the basis for a block diagram:
Z (t)
= _!_ {u (t)}- [an-I s-l + · · · + al s-
an
an
an
(7.66)
The dummy variable z(t) is specified in terms of the system input u(t) and a weighted sum of successive integrations of itself. Figure 7.18 shows the overall structure of this directform block diagram. A string of n cascaded integrator (S- 1) blocks, with z(t) defined at the input to the first block, is used to generate the feedback terms, (z (t)}, i = 1, ... n, in Eq. (7.66). Equation (7.64) serves to combine the outputs from the integrators into the output y (t).
s-i
A set of state equations may be found from the block diagram by assigning the state variables x; (t) to the outputs of the n integrators. Because of the direct cascade connection of the integrators, the state equations take a very simple form. By inspection,
(7.67) Xn-1
.
Xn
=
Xn
ao at an-I 1 = - -an X t - -X2 • • • - - - X n + -u(t) an an an
Sec. 7.8
Transformation from Oassical Fonn to State Space Representation
235
y(t)
s-1
u(t)
Figure 7.18:
BJock diagram of a system represented by a classical differentia] equation.
In matrix form these equations are
it i2 Xn-2 Xn-1 Xn
0 0
=
1
0
0 0 -aofan
Xn
0 0
I
0 1 -an-tfan
0 -an-2/an
-a Ifan
(7.68)
0 0
XJ X2 Xn-2 Xn-1
0 0
0 0
+
0 0 lfan
u (t)
The A matrix has a very distinctive form. Each row except the bottom one is filled with zeroes except for a l in the position just above the leading diagonal. Equation (7 .68) is a common form of the state equations used in control system theory and known as the phase variable or companion form. This form leads to a set of state variables that may not be physical variables within the system. The corresponding output relationship is specified by Eq. (7.64) by noting that X; = s-(n+l-i) {z(t)}. (7.69)
236
Operational Methods for Linear Systems
Chap. 7
But z (t) = dxnfdt, which is found from the nth state equation in Eq. (7.67). When substituted into Eq. (7.69), the output equation is
XI] bn . + -u(t)
(7.70)
X2
[:
an
Xn
Example 7.12 Draw a direct-form realization of a block diagram and write the state equations in phase variable fonn for a system with the differential equation
. d3 y d 2y -d +7-d t3 t2
dy
+ 19-dt +
du 13y = 13-d t +26u .
(i)
Solution The system order is 3, and using the structure in Fig. 7.18. the block diagram is as shown in Fig. 7.19. The state and output equations are found directJy from Eqs. (7.68) and (7.70):
[it] = [. 0 0
0
1
x2
0
X3
-13
-19
-7
X3
1
y(t)
u(t)
1 0] [Xt] + [0]
~2
= [~
13 0]
[::J
(ii)
u(t)
+ [O]u (t)
+
~
Figure 7.19: Block diagram of the transfer operator of a third-order system found by a direct realization.
(iii)
y(l)
Sec. 7.9
7.9
The Matrix Transfer Operator
237
THE MATRIX TRANSFER OPERATOR
For a multiple-input multiple-output system Eq. (7 .55) is written in terms of the r component input vector u(t): x(t)
= [SI- A]- 1 B {u(t)}
(7.71)
generating a set of n simultaneous linear operator equations where B is n x r. Themcomponent system output vector y(t) may be found by substituting this solution for x(t) into the output equation as in Eq. (7 .58): y(t)
= C [SI- A]- 1 B (u(t)} + D {u(t)}
= [C[SI- A]- 1 B +D] {u(t)}
(7.72)
and expanding the inverse in terms of the determinant and the adjoint matrix, y(t)
= (det [SI- A])- 1 (C adj (SI- A) B + det [SI- A] D) {u(t)}
= H{u(t)}
(7.73)
where H {} is defined to be the matrix transfer operator: B (}
= C adj (SI -
A) B + det [SI - A] D det[SI- A]
(7.74)
For a system with r inputs UJ (t), ... , u,(t) and m outputs Yl (t), ... , Ym(t), H {}is am x r linear matrix operator whose elements are individual scalar transfer operators relating a given component of the output y(t) to a component of the input u(t). Expansion ofEq. (7.74) generates a set of equations: Yt(t) Y2(t) [
.. .
Ym(t)
l[
Hu H21
Hlr Htr
.. .
Hml
...
Hm2
Hmr
l["I l Ul(t) (t)
.. .
(7.75)
u,(t)
where the ith component of the output vector y(t) is y;(t) = H;1 {u1 (t)}
+ Hn [u2(t)} + ·· · + H;, (u,(t)}
(7.76)
The elemental operator Hij () is the scalar transfer operator between the i th output component and the jth input component All the elemental scalar operators in B {} have the same operator factor (det [SI- A])- 1 {} associated with them. The result is that all input-output differential equations for a system have the same coefficients on the left-hand side. If the system has a single input and a single output, H (} is a scalar operator H (} and the procedure generates the input-output transfer operator directly.
Operational Methods for Linear Systems
238
Chap. 7
PROBLEMS 7.1. Consider the mechanical system consisting of a mass suspended on a cantilevered beam as shown in Fig. 7.20. Use the linear graph to draw an operational block diagram for this system, and derive the system state equations directly from the block diagram.
v
m
m
Figure 7.20: A mechanical system and its linear graph.
7.2. A mechanical cam drive system is shown in Fig. 7.21. The cam may be represented as a velocity input V (t) to the cam follower, which is connected to a mass supported on a horizontal surface. The rod connecting the follower and the mass has finite stiffness K, and the mass slides on the horizontal surface with effective viscous friction B. Assume that the follower is always in contact with the cam. Use an operational block diagram to derive a set of state equations for this system.
K
m
B
Figure 7.21:
Cam drive system.
7.3. Construct a linear graph model for the electrical circuit shown in Fig. 7.22. Draw the operational block diagram for the circuit, and derive the system state equations from the block diagram.
Figure 7.22: Electrical circuit
7.4. Consider the nonlinear fluid system described in Example 5.13 and shown in Fig.·5.22. From the linear graph model given in the example, construct the system operational block diagram using linear and nonlinear elements. Derive the system state equations from the operational block diagram and compare them with the results given in the example.
Chap. 7
239
Problems
7 .S. Operational block diagrams are often used to model physical systems that are coupled to instrumentation and control systems. Consider the motor drive system described in Example 5.4 and illustrated in Fig. 5.13. (a) Use the linear graph to draw an operational block diagram for the system. (b) A tachometer is attached to the shaft and produces a voltage v, proportional to the angular velocity of the flywheel. that is. v, = K1 0. Modify the system block diagram to include the tachometer and derive the differential equation relating the sensor output voltage to the motor rotational speed.
(c) An angular position sensor is installed on the flywheel shaft. The output voltage v4 of this sensor is proponional to the shaft displacement 8. that is, v4 = K 28. Modify the block diagram for the system to include the position sensor and derive a set of system state equations which reflect the addition of the position sensor. Has the addition of this sensor changed the order of the system? 7.6. Feedback control, illustrated in Fig. 7.23, is used to modify the dynamic behavior of a system by monitoring its response y(t), comparing the response with a desired response r(t), and using the error e(t) = r(t)- y(t) as the input to drive the system. Example 6.1 describes an electric motor drive system for a turntable and derives the linear graph for the system. In this problem we examine feedback control of the motor speed.
e(t)
System
Error
Figure 7.23:
l----4-~
y(t)
Response
Feedback control applied to a system.
(a) Draw the operational block diagram for the motor drive.
=
(b) A tachometer, which provides an output voltage, v0 K 10, proponional to the turntable angular velocity 0, is installed on the shaft. The output voltage from the tachometer is compared to a reference voltage vd in an electronic amplifier-summer and the difference is amplified to generate the input voltage to the motor terminals V.s = K2 (vd- v 0 ), as shown in Fig. 7.24. Modify the system block diagram to include the tachometer and the amplifier-summer.
(c) Use the block diagram to derive the system state equations for the complete system with the feedback control. What is the influence of the angular velocity feedback on the system? 7.7. Draw an operational block diagram for the system described by the state and output equations:
u: J ~ ~ n[~: J ~ n[::] = [
[~] [~ ~ ~] [~;] =
+[
240
Operational Methods for Linear Systems
Chap. 7
Turntable/
Amplifier
+
K2
Motor Figura 7.24:
R
Feedback speed conttol for a rotary drive system.
7.8. A system is described in terms of three linear operators
= 2S + 1
.C.1
.c.2 = .C.3
4
= 3S2 + 2S + 1
as
.C.]1.c.2.c.~ {y} = u (a) Determine the system transfer operator relating the output y to the input u. (b) Write the differential equation relating y to u. 7.9. In Example 7.6 it is shown that if f(t) = 0 fort:: 0, the inverse operator (S- a)- 1 is satisfied by the integral (S- a)- 1{/(t)} e
[oo ez
f(r:)dr:
A first-order system has a state equation
x=-3x+2u Find the transfer operator between the input u(t) and the state variable response x(t). Use the result of Example 7.6 to find the response of the system to a step input, that is u(t) = 0
fort:: 0
=1
fort> 0.
7.10. A linear system is described by the second-order differential equation
d2y dt2
dy
+ 7 dt + 12y = u
241
Problems
Chap. 7
(a) Find the transfer operator between the input u(t) and the response y(t). (b) Express the system as a cascade connection of two first-order systems. Draw a block diagram of this configuration. (c) Express the system as a parallel connection of two first-order systems. Draw a block diagram of this configuration. (d) Use the result of Example 7.6 to find the response of the second-order system to a step input (see Problem 7.9) using both the cascade and the parallel system representations. Show that the results are the same. 7 .11. Find a single differential equation for each of the systems represented in the block diagrams in Fig. 7.25. SS+ 1
u(t)
y(t)
S+S
3S+2
u(t)
S+2
y(t)
Sl+2S+l
3 S+3 (a)
Figura 7.25:
(b)
Systems represented by cascade and parallel operators.
7.12. Consider the feedback control of the motor drive system described in Problem 7.6. The motor/turntable system is a second-order system with a measured transfer operator relating angular velocity 0 to input voltage Vs
100 - S2 +21S+20
H (} _
The tachometer gain K 1 = 0.05 v/rad, and the amplifier gain K2 feedback control system is shown in Fig. 7.26.
= 100.
A block diagram of the
r-----------------------------
1 I
I I I I+
Amplifier
100
-
Vs !
Motor 50 S2 + 21S + 20
~
Tachometer 0.05
---------------------~o~~~E=~J Figure 7.26:
Block diagram of the motor/turntable speed control system.
(a) Find the differential equation representing the motor. (b) Find the transfer operator relating the actual speed of the motor 0 to the input voltage vd for the closed-loop controlled motor.
(c) Find the differential equation that describes the cJosed-loop controlled motor.
242
Operational Methods for Linear Systems
Chap. 7
7.13. A system that has been modeled is described by the following system matrices:
C=[l
0],
D=[O]
(a) Draw a block diagram representation of the system. (b) Produce a single input-output differential equation for the system using
L Direct manipulation of the differential equations (that is by substitution and elimination). ii. Cramer's Rule. iiL Expansion of the matrix relationship y(t)
= [C(SI- A)- 1B + D] U(t).
7.14. Consider the mechanical system described in Example 5.2 and represented by the state equations:
(a) Using operator methods, derive a single differential equation relating the force input to the mass velocity. (b) Derive a differential equation relating the mass displacement from its at-rest position to the force input. 7.15. For the fluid system described in Example 5.6, derive the system transfer operator relating the tank pressure to the pump output pressure, and write a single differential equation relating the tank pressure to the pump output pressure. 7.16. For the mechanical system described in Example 5.9, write a single differential equation relating the velocity of the mass element to the input force, and derive the transfer operator relating the velocity of the mass to the input force. 7.17. Consider the system described in Example 6.5. Derive the system transfer operator relating the flywheel speed to the input voltage, and write a single differential equation relating the flywheel speed to the input voltage. 7.18•. An electro-mechanical translational drive system is used to position a mass on a horizontal surface as shown in Fig. 7.27. The drive system provides a translational force output that is a function of the drive system input voltage. The mass is driven through a rigid rod and rests on a surface that has a viscous friction retarding force. A differential equation for the drive system has been constructed from experimental measurements on the system relating the driving force Fs to the voltage input Y.r in tenns of several constant coefficients:
Figure 7.27: Electro-mechanical drive.
243
References
Chap. 7
(a) Draw an operational block diagram representing the drive system differential equation. (J)) Draw the-operational block diagram for the mass driven by the force and couple the two block
diagrams to generate the system operational block diagram. (c) Derive a set of state equations for the complete system from the system block diagram. 7.19. Write each of the following differential equations in state-space form. (a)
~; + 3y = d 2y
(b) dt 2
dy
du
+ 4 dt + 2y = u +? dt
d 3y
(c) 3 dtl
2u
d2y
dy
d 2u
du
+ 2 dt2 + 2 dt + Y = 7 dt 2 + 5 dt + u REFERENCES
[1] Shearer, J. L., Murphy, A. T., and Richardson, H. H., Introduction to System Dynamics, Addison-
Wesley, Reading, MA, 1967. [2] Shearer, J. L., and Kulakowski, B. T., Dynamic Modeling and Control of Engineering Systems, Macmillan, New York, 1990. [3] Ogata, K., System Dynamics, Prentice Hall, Englewood Cliffs, NJ, 1978. [4] Ogata, K., Modem Control Engineering (2nd ed.), Prentice Hall, Englewood Cliffs, NJ, 1990. [5] Dorf, R. C., Modem Control Systems (5th ed.), Addison-Wesley, Reading, MA, 1989. [6] Franklin, G. F., Powell, J.D., and Emami-Naeni, A., Feedback Control of Dynamic Systems, Addison-Wesley, Reading, MA, 1986. [7] Maciejowski, J. M., Multivariable Feedback Design, Addison-Wesley, Reading, MA, 1989. [8] Schul~ D. G., and Melsa, J. L., State Functions and linear Control Systems, McGraw-Hill, New York, 1967. [9] Kaplan, W., Operational Methods for linear Systems, Addison-Wesley, Reading, MA. 1962. [10] Yoshida, K., Operational Calculus, Springer-Verlag, New York, 1984. [11] Liverman, T. P. G., Generalized Functions and Direct Operational Methods, Prentice Hall, Englewood Cliffs, NJ, 1964.
8
System Properties and Solution Techniques
8.1
INTRODUCTION Knowledge of the way dynamic systems respond to external inputs is central to their design and evaluation. In order to detennine the response of a state-determined system we require the following:
1. The mathematical description of the system, that is, the set of system state equations 2. The specification of the n state variables at an initial time to, that is, conditions
th~
initial
3. The specification of the system inputs for all time t ::! to. These are necessary and sufficient conditions to determine the system behavior for all time t > 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
(8.2)
fort> 0.
The impulse function (also known as the Dirac delta function) 8(t) is defined as the limiting form of the unit pulse tST(t) as the duration T approaches zero. As the duration of tSr(t) decreases, the amplitude of the pulse increases to maintain the requirement of unit area under the function, and (8.3)
The impulse bas the properties that tS(t)
= 0 for all t =F 0 and bas unit~ and so
L:
&(t)dt =I
(8.4)
Sec. 8.2
System Input Function Characterization
247
The impulse is therefore defined to exist only at time t = 0, and although its value is strictly undefined at that time, it must tend toward infinity to maintain the property of unit area in the limit. The strength of a scaled impulse K 8(t) is defined by its area K. The impulse functions are designated graphically by an arrow, with the length indicating the strength, as shown in Fig. 8.2. &(t)
liT 1 -
liT2-11/T3 liT4
I I
I
Time
0 Time
(a) Unit pulses of different widths
(b) The impulse function
Figure 8.2: The unit impulse defined as the limit of a pulse with unit area.
Although true impulse functions are not found in nature, they are approximated by short-duration, high-amplitude phenomena such as a hammer impact on a structure or a lightning strike on a radio antenna. The unit impulse may be considered informally to be the derivative of the unit step Us (t). Although the step function is not formally differentiable because it is discontinuous at time t = 0, its derivative is zero for all t =F 0, and if the unit impulse is integrated from t = -oo to t, the result is a unit step:
u,(t) = [
&(r)ddor t > 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<••+.rl)' = e'•'e'2' fll" = (e"
t
d - e'' =se'' dt
1 T
e'' dt
0
=.!. (e''- 1} s
lime"= 1
t-+0
e
= ea' [cos (wt) + j sin (wt)]
Alternatively, the sinusoidal waveforms may be represented as the real and imaginary parts of the complex exponential, that is, sin (wt) = tJ { ei"'1 } cos (wt) = m{eiOJt}
(8.13)
(8.14)
Sec. 8.3
251
Classical Solution of Linear Differential Equations
where the operator m{} extracts the real part of a complex function and ~ (} extracts the imaginary part. In its most general form, the exponents is complex, that is, s = u + j(J). Then u(t) =est = e
=
8.3
eat
(cos (J)t
(8.15)
+ j sin wt)
CLASSICAL SOLUnON OF LINEAR DIFFERENTIAL EQUAnONS In this section we briefly review the classical method for solving a Jinear nth-order ordinary differential equation with constant coefficients [5-6]. We showed in Chap. 7 that any output variable y(t) in a single-input single-output system represented by state equations
= Ax(t) + Bu(t) y(t) = Cx(t) + Du(t)
(8.16)
i(t)
(8.17)
may be expressed as a single nth-order differential equation relating the output y(t) to the input u(t): (8.18)
where in general m ::: n. Because the right-hand side involves only the input u(t) and its derivatives, it is a known function of time and it is convenient to rewrite the differential equation: (8.19)
where f (t) is defined to be the forcing function: dmu
f(t)
dm-lu
dy
= bm -dtm + bm-1 - + · · · + bt -dt + bou drm-l
(8.20)
The task is to find a unique function y(t) that satisfies the differential equation (8.18), given the forcing function f (t) and a set of initial conditions. The existence and uniqueness theorem [5, 6] guarantees that such a solution can be found.
252
System Properties and Solution Techniques
Chap. 8
Theorem: If the forcing function I (t) is continuous on the interval to !:: t !:: T, a unique solution to the nth-order linear differential equation exists: dny dtn
+ an-1
dn-ly drn-l
+ an-2
dn-2y dtn-2
+ ... + al
dy dt
+ aoy = l(t)
on the interval satisfying a set of n initial conditions dyl y(to) =co dt to
= c1 •••
dn-lyl dtn- 1 to
= Cn-1
(8.21)
where to is a fixed time and the constants c; have known values.
This theorem provides assurance that (1) a solution can be found if the value of the response y(t) and its first n- 1 derivatives are known at some time, and (2) there is only one such solution. In system dynamics we commonly assume that to = 0, defining an initial condition problem. The theorem also implies that the forcing function need be known only fort ~ to, a requirement for state-determined systems, and alJows us to find the response to discontinuous forcing functions. If l(t) is discontinuous at time t 1, the current values of y(tt) and its derivatives may be used as a new set of n initial conditions for the solution in the interval t > tr. The general solution to Eq. (8.19) may be derived as the sum of two solution components [5-6]: y(t)
= Yh(t) + Yp(t)
(8.22)
where Yh (t) is the solution to the homogeneous equation, that is, Eq. (8.19) with I (t) = 0, and where Yp(t) is a particular solution that satisfies Eq. (8.19) for the specific input I (t). As is shown in the following sections, the homogeneous solution contains n arbitrary constants that must be determined in order to make the total solution y(t) satisfy the system's initial conditions. 8.3.1 Solution of the Homogeneous Differential Equation
A homogeneous differential equation is one in which the forcing function I (t), and therefore the input u(t), is identically zero: dny dn-ly +an-I - dtn drn-l
-
dy
+ · .. + a1 -dt + ilOY =
0
(8.23)
The homogeneous response Yh (t) represents the system response to a finite set of initial conditions in the absence of any system input.
Sec. 8.3
Classical Solution of Linear Differential Equations
253
The standard method of solving differential equations assumes there exists a solution of the fonn Yh (t) = Ceu, where Aand C are both constants. The validity of this assumption may be verified by substitution into the homogeneous equation: dn -d ce'·'
rn
dn-l
+ an-1 -rndI Ce}J +
d ... + al -d Ce).t + aoCeAI t
=0
(8.24)
or (8.25) For any nontrivial solution, C is nonzero, and because eu is never zero, we require that An+ an-JAn- I+···+ ao
=0
(8.26)
In other words, the assumed fonn Yh (t) = Cel-1 , for any nonzero value of the constant C, is a solution to the differential equation if A satisfies Eq. (8.26). This equation is defined to be the characteristic equation of the system, and the left-hand side is defined to be the characteristic polynomial. The characteristic equation is found by substituting An for the nth derivative in the homogeneous equation. For an nth-order system there are n roots of the characteristic polynomial and n possible solution tenns C;eAi 1 (i = 1, ... , n), each with its associated arbitrary constant Any linear combination of these n solutions is also a solution of the homogeneous differential equation. If all the roots of the characteristic polynomial are distinct, the most general form of the homogeneous solution is a sum of all such terms:
n
= LC;eA.,t
(8.27)
i=l
We can therefore make the following important statement: The homogeneous, or unforced. response of any linear system to a set of nonzero initial conditions consists of a superposition of n components of the form e1'', where the A.; are the roots of the characteristic equation.
The n coefficients C; are arbitrary constants and must be found from then initial conditions. Example8.1
Fmd the homogeneous response of the differential equation d2y dt2
dy dt
-+5-+6y=0 given that at timet= 0, y(O)
= 1 and dyfdt = 0.
Solution The system characteristic equation is
>..2 +5>..+6=0
(i)
System Properties and Solution Techniques
254
Chap. 8
which has roots l = -3, -2. The two solution components of the homogeneous equation are therefore C1e-3' and C2e-21 , and the most general form is (ii)
At time t
= 0, substitution of the initial conditions gives = 1 = c. + c2 -dy' = 0 = -3Cl - 2C2 dt 0 y(O)
which is a pair of simultaneous equations in
c.
and
(iii) (iv)
c2 and which gives cl = -2 and
c2 = 3. The complete homogeneous response is therefore Yh(t)
= 3e-21 -
2e-3'
(v)
If the characteristic polynomial has repeated roots, that is, A.; = Aj when i =f: j, there are not n linearly independent terms in the general homogeneous response [Eq. (8.27)]. It is then necessary to supplement the general solution with additional terms. For example, if a double root occurs, A; = Aj = A, and then the two linearly independent components in the general solution are
In general if a root A of multiplicity m occurs, the m components in the general solution
are
Example8.2 Find the general solution to the homogeneous differential equation
Solution The characteristic equation is )..4
+5)..3 +9)..2 +7)..4 +2 = 0
(i)
().. + 2)().. + 1) = 0
(ii)
3
=
and so there is a single root at l -2 and a root of multiplicity 3 at ).. = -1. The general homogeneous solution is therefore (iii)
which contains four arbitrary constants to be evaluated from the initial conditions.
Sec. 8.3
Classical Solution of Linear Differential Equations
255
In a differential equation such as Eq. (8.18), representing the dynamics of a physical system, the coefficients a; are always real. The coefficients of the characteristic polynomial are therefore also real, which requires that the roots l; be either real or appear in complex conjugate pairs. When complex conjugate roots are found, for example, A.;
= u + jw and l;+t = u- jw
the corresponding response terms e
= eu' [cos(wt) + j
sin(wt)] and e
always combine to generate a purely real response component Example8.3 Find the homogeneous response of a system described by the differential equation
with initial conditions y(t) = 1 and dyfdt = 0 at timet= 0 and f(t) = 0. Solution The characteristic equation is (i)
with roots A= +2j and A= -2j. The homogeneous response is therefore
(ii) At time t = 0 the initial conditions require Yh co> =
I
dyh dt '""0
c1+ c2 =
= j2C1 -
(iii)
1
j2C2
=0
(iv)
.. C1 = 1 • andC2 = 21 or gtvmg 2 (v)
from the Euler relationship cos 8 = (ei 6 + e-i 6 ) /2. This system is a pure harmonic oscillator.
8.3.2 Solution of the Nonhomogeneous Differential Equation
A particular solution Yp (t) of a differential equation satisfies the equation for the given input u(t). There are several methods that may be used to determine a particular solution [5-6]; for a differential equation with constant coefficients, the method of undetermined coefficients is often used. This method assumes a solution of a given form, dependent on the
256
System Properties and Solution Techniques
Chap. 8
nature of the forcing function f(t), and then by substitution into the differential equation matches a set of coefficients in the assumed solution to achieve equality of the left- and right-hand sides of the equation. Table 8.2 defines particular solutions for forcing functions that are the singularity, exponential, and sinusoidal functions described in Sec. 8.2.1. TABLE 8.2:
Definition of Particular Integrals Yp(t) Using the Method of Undetermined Coefficients
Term in u(t) k kt11
(n
= 1, 2, 3, ...)
K,.rn
k~
keiOII kcos(wt) k sin(wt)
Assumed Form for yp(t)
Test Value
+ K,._ 1tn-t + · ·· + K1t + Ko
K,
0 0
x.~
.A
x.elt»t K1"cos{wt) + K2 sin(Cdt) K 1 cos{wt) + K 2 sin(wt)
jw jw jw
Example8.4 Find a particular solution for the nonhomogeneous equation d 2y -
dt 2
dy
+- +2y = lOcost dt
Solution In this case the forcing function f(t) is sinusoidal. From the second column in Table 8.2 assume a solution of the form (i)
which contains two unknown coefficients K 1 and K 2 • Then dyp . dt = -K1 smt + K 2cost dlyp
dt 2
. = -K 1 cost- K 2 smt
(ii) (iii)
Substitution of these values into the differential equation gives (iv) and equating coefficients gives K 1 - K 2 particular solution is therefore
= 10, K2 + K 1 = 0, so K 1 = 5 and K2 = -5. The
Yp(t) = 5cost- 5sint
(v)
Because the method of undetermined coefficients is basically a table "look-up" method, care must be taken to ensure that the particular solution is not simply a component of the homogeneous response. The third column in Table 8.2 contains a test value
Sec. 8.3
257
Classical Solution of Linear Differential Equations
that should be checked before the solution in the second column is used. If the value in the third column is equal to any root of the characteristic polynomial, then the assumed solution in the second column must be multiplied by a factor tm, where m is the multiplicity of that root Example8.S Find a particular solution for the differential equation dy
- +3y =e- 3' dt
Solution The characteristic equation is A+ 3 = 0, which has a single root A = -3, and the homogeneous response contains a tenn ce-3'. The proposed particular solution for an exponential forcing function, from the second column in Table 8.2, is yp(t) = K 1e- 3' and is a component of the homogeneous response. We need to use a modified fonn for the assumed solution and choose (i)
Differentiation and substitution give (ii)
or K 1 = 1. The particular solution is therefore (iii)
8.3.3 The Complete Solution
The complete solution consists of the sum of the homogeneous response in the form of Eq. (8.27) with n arbitrary constants C; (i = 1, ...• n) and the particular solution Yp(t): n
y(t)
=
L C;ei..
1 '
+ Yp(t)
(8.28)
i=l
The value of the constants are found from the n initial conditions
dyl = CJ ••• dn-lyl d n-l =
y(to) =co-d
t -
t
Example8.6 A physical system is described by a differential equation dy
- +Sy dt
= u(t)
~
Cn-1
258
System Properties and Solution Techniques Find the system response to a ramp input u(t) when y(O) 0.
=
Chap. 8
= 2t when the system is initially at rest, that is,
Solution The characteristic equation is A.+ 5
= 0, which has a single root A. = -5. The
homogeneous solution is therefore (i)
For the ramp forcing function the particular solution is selected from Table 8.2: (ii)
Substitution into the differential equation gives Kt + 5 (K 1t + Ko) cients K 1 ~ and K2 ~. The particular solution is therefore:
=
=
Yp(t)
= 2t and equating coeffi-
= -52 t -252-
(iii)
The complete solution is the sum of the homogeneous solution and the particular solution: y(t)
and when t
2 2 = ce-St + ·5'25
(iv)
= 0, the initial condition y(O) = 0 requires 2 O=C-25
(v)
or C = ~. The complete solution is then y(t)
2 2 = -t + (1 5 25
e-5')
(vi)
Example8.7 A second-order system with a linear differential equation d 2y dt2
is subjected to a step input u(t) initially at resL
dy
+ 7 dt + 12y = u(t)
= 4 at time t = 0.
Find the response y(t) if the system is
Soludon The system characteristic equation is A. 2 +7A.+ 12
=0
(i)
which has roots A. 1 = -3 and A. 2 = -4. The homogeneous solution is therefore (ii)
Sec. 8.4
259
System Properties
=
From Thble 8.2, for a step (or constant) input the particular solution is Yp(t) K, and substitution into the differential equation gives 12K = 4 or K = i· The complete solution is therefore (iii)
and when t
=0, the initial conditions y(O) = 0 and dyfdt = 0 at t =0 are substituted:
or C1 = - ~ and C2
I 3
(iv)
0=-3Ct-4C2
(v)
= 1 The complete solution is therefore y(t)
8.4
0= Ct +C2+-
4 3 1 ' + e-4' + = --e3 3
(vi)
SYSTEM PROPERTIES A number of general system properties, which complement the solution techniques described above, are useful in defining system performance. For linear systems, described by differential equations with constant coefficients, these properties may be derived directly from the system response characteristics, but the concepts can be extended to more general formulations. In this section the properties of system stability and time invariance are defined for general state-determined system models, and the principle of superposition is defined for linear models.
8.4.1 System Stability The concept of system stability is related to the response of the state vector x(t) to perturbations from an equilibrium condition. A system is defined to be in equilibrium if all derivatives of the state vector are identically zero, that is, if none of the state variables are changing and the syst~m remains at a fixed point in the n-dimensional state space. If a system is in equilibrium in the absence of an input, it is defined to be at rest. Consider a linear system defined by the state equations
x=Ax+Bu
(8.29)
y=Cx+Du where the matrices A, B, C, and Dare constant In the absence of an input, that is, when u(t) = 0, the system has an equilibrium state xo 0 at the origin of the state-space because Axo 0 implies that xo 0. The property of asymptotic stability is defined in terms of the system response to a disturbance from an equilibrium condition. If the state vector is
=
=
=
System Properties and Solution Techniques
260
Chap. 8
perturbed from equilibrium by an external influence, the response x(t) 1. asymptotically returns to the equilibrium point, in which case the system is defined to be asymptotically stable, 2. diverges from the equilibrium point, in which case the system is unstable, or 3. remains at its perturbed state or undergoes constant amplitude oscillations about the original equilibrium point In this case the system is defined to be neutrally stable or marginally stable because the system response neither converges to the original equilibrium condition nor diverges away from it Figure 8.8 illustrates the three stability conditions using a rolling ball on an undulating surface as a graphic example. Assume that the horizontal position and velocity of the ball are a pair of state variables describing this system. In a concave region of the surface, as shown in Fig. 8.8a, the base of the hollow is an equilibrium point H the ball is displaced a small distance from this position and released, it oscillates but ultimately returns to its rest position at the base as it loses energy as a result of friction; this is therefore a stable equilibrium point. (Without energy dissipation the ball would roll back and forth forever and exhibit neutral stability.) On a convex portion of the surface, as shown in Fig. 8.8b, the ball is in equilibrium if placed exactly at the top of the surface, but if it is displaced an infinitesimal distance to either side, the net gravitational force acting on it will cause it to roll down the surface and never return to the equilibrium point This equilibrium point is therefore unstable. H the ball is displaced along the flat portion of the surface, as shown in Fig. 8.1 c, it neither moves away nor returns; the flat portion· represents a neutrally stable equilibrium region.
Gravitational force acts to restore a displaced ball to its equilibrium position.
~· (a) A stable system
Gravitational fora: acts to move a displaced ball away from its equilibrium position.
____ill_____---.... ~///////~
No forces act to move a displaced .ball.
-
////.1///g;////ff//, (b) An unstable system
(c) A neutrally stable system
Figure 8.8: Demonstration of system stability using the response of a ball moving under gravity on a surface. In (a) the system is stable and gravitational forces and frictional losses eventually cause the ball to return to its equilibrium position. in (b) the system is unstable and the ball never returns. while in (c) the ball neither converges or diverges from its original position.
For linear systems, stability is a global system property; a system is stable, neutrally stable, or unstable. In the absence of an input a stable linear system relaxes to the origin of the state space from any arbitrary initial state. Nonlinear systems may be globally stable or may exhibit stable characteristics in one region of the state space and unstable behavior in another region. For nonlinear systems a system may be asymptotically stable for small perturbations from a given equilibrium point but unstable for larger displacements [1].
261
System Properties
Sec. 8.4
The concept of the state space is useful in visualizing the dynamic response of a system. For a system of order n the response at any time t is defined by the instantaneous value of the n components of the state vector x. As the system response evolves from its initial state x(O) it may be considered to trace out a trajectory through an n-dimensional space' known as the state space. At any instant the values of the state variables define the position of the system in its state space.· For example, the responses of typical secondorder systems are shown in Fig. 8.9 where the trajectories of the two state variables x1 (t) and x2(t) from a pair of arbitrary initial conditions XI (0) and x2(0) are shown. In this case the state space is two-dimensional and the system response evolves along paths that lie in a plane. The response of the stable system converges toward the origin where it reaches equilibrium, while the unstable second-order system response diverges away from the origin. The neutrally stable system either remains at the initial point in the state space or oscillates continuously around a closed path, neither converging nor diverging from an equilibrium point. 2 Stable N ~
lo >
----
I
5
I I I
~
-2
0 State variable 1 Figure 8.9:
~
.g
----
'i 0 >
'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-
Example9.6 The first-order electric circuit shown in Fig. 9.14 is known as a lead network and is commonly used in electronic control systems. Find the response of the system to an input pulse of amplitude 1 V and duration 10 ms if R1 = R2 = 10,000 Q and C = 1.0 p.F. Assume that at timet = 0 the output voltage is zero. Solution From the linear graph the state variable is the voltage on the capacitor vc(t), and the output is the voltage across R2• The state equation for the system is (i)
and the output equation is (ii)
The input-output differential equation is Rt R2C dvo R1R2C -----+vo= R1
+ R2
dt
R1
+ R2
d~n +
dt
R1
Rt
"
+ R2 Yin
(iii)
with the system time constant t" = Rt R2C/(Rt + R2) = s X w-3 s. The input pulse duration (1 0 ms) is comparable to the system time constant, and therefore it is not valid to approximate the input as an impulse. The pulse input can, however, be written as the sum of two unit step functions: Vm(t)
(iv)
= Us(t)- Us(t- 0.01)
and the response determined in two separate intervals: 0 :S t < 0.01 s, where the input is Us(t), and t ~ 0.01 s, where both components contribute.
c
c
Figure 9.14: Electric lead network and its linear graph.
294
First- and Second-Order System Response
Chap. 9
The input-output unit step response is given by Eq. (9.28):
(v)
At time t = o+ the initial response is Vo(O+) = 1 v and the steady-state response (vo)ss = 0.5 V. The settling time is approximately 4r, or about 20 ms. The response to the tO-ms-duration pulse may be found from Eqs. (iv) and (v) by using
the principle of superposition:
=
vpu~se(r)
(vi)
vo(t) - vo(t - 0.01)
In the interval 0 !S t < 0.01, the initial condition is zero and the response is (vii) = (0.5 + o.5e-''0·oos) In the second interval t ~ 0.01, when the input is V1n = u,(t)- u,(t- 0.01), the response is
vpu~se(t)
the sum of two step responses:
vpu1sc(t) = (0.5 + 0.5e-''0·oos)- (0.5 + o.se-
= 0.5 (etfO.OOS _ e-(t-D.Ol)/O.OOS) = 0.5e'l0 ·005 (1- e2 ) = -3.195e-tfO.oos V
(viii)
The step response [Eq. (v)] and the pulse response described by Eqs. (vii) and (viii) are plotted in Fig. 9.15. vou,(t)
1.2 1.0
0.8
~
!
0.6
0
~
0.4
~
0.2
:;
g
' ""'
~
Step response
~---- ~----t---------
-·--
-·-,..----
0.0
-0.2
'L_
-0.4 -0.6 0.000
0.005
0.010
+-.l--·-
-.......--;;rise
0.015 Time (s)
I
respons~
0.020
I
0.025
0.030
Figure 9.15: Response of the elecbic lead network to a unit step in input voltage and to a unit amplitude pulse of duration 10 ms.
t
Sec. 9.3
9.3
295
Second-Order System Transient Response
SECOND-ORDER SYSTEM TRANSIENT RESPONSE
Second-order state-determined systems are described in tenns of two state variables. Physical second-order system models contain two independent energy storage elements which exchange stored energy and may contain additional dissipative elements; such models are often used to represent the exchange of energy between mass and stiffness elements in mechanical systems, between capacitors and inductors in electric systems, and between fluid inertance and capacitance elements in hydraulic systems. In addition second-order system models are frequently used to represent the exchange of energy between two independent energy storage elements in different energy domains coupled through a two-port element; for example, energy may be exchanged between a mechanical mass and a fluid capacitance (tank) through a piston, or between an electric inductance and mechanical. inertia as might occur in an electric motor. Engineers often use second-order system models in the preliminary stages of design in order to establish the parameters of the energy storage and dissipation elements required to achieve a satisfactory response. Second-order systems have responses that depend on the dissipative elements in the system. Some systems are oscillatory and are characterized by decaying, growing, or continuous oscillations. Other second-order systems do not exhibit oscillations in their responses. In this section we define a pair of parameters commonly used to characterize second-order systems and use them to define the conditions that generate nonosciUatory, decaying or continuous oscillatory, and growing (or unstable) responses. In the following sections we transform the two state equations into a single differential equation in the output variable of interest and then express this equation in a standard form. Transformation of State Equations to a Single Differential Equation The state equations x = Ax + Bu for a linear second-order system with a single input are a pair of coupled first-order differential equations in the two state variables:
[!~] = [:~: =~~][~~] + [:~] u
(9.34)
or dx1
dt = GJ)Xl +a12X2 + b1u dx2
(9.35)
dt = G2)Xl + a22X1 + b2U The state space system representation may be transformed into a single differential equation in either of the two state variables by using the linear operator methods introduced in Chap. 7. In particular, if the state equations, Eqs. (9.35), are interpreted as operational expressions, they may be written [S- au] {xJ} ( -a2tl {XJ}
+ [-a12l {x2} =
+ (S -
[bi] {u}
a22l {X2} = (b2] {u}
(9.36) (9.37)
where S is the differential operator and the matrix elements aij and b; are constant scaling operators. All the operators are linear, and the state equations may be written in matrix
296
FJrSt- and Second-Order System Response
Chap.9
[b1]
(9.38)
operational fonn:
[
S- a11
-a12 ] { Xt }
S - an
-a21
x2
=
{u}
b2
The solution of Eq. (9.38) using the operational form of Cramer's rule, as described in Sec. 7.7 and App. A, is a convenient method for transforming the state space formulation to a classical second-order equation. Each of the two state variables can be expressed in terms of two operational determinants:
-a12 S-an
det [ bt
~
XI=
det [ S - au -a21
J {u}
-a12 ]
S-an
and det [ S X2
=
au
-a21 det [ S - a11 -a21
bt ] b2
( }
J S-an
U
-a12
(9.39) Each of these expressions is in the form of X;
= £;; 1£; {u(t)}
(9.40)
=
where .Cc det [SI -A] is the characteristic operator and £; is the forcing function operator associated with the i th state variable. The differential equation in the state variable is found by operating on both sides using £c
.Ccx;
= £; {u(t)}
(9.41)
The expression for the first state variable is det [. S- au -a21
-al2 ] {xt} S - an
[S2 - (au +an)S+ (aua22 -at2a21)] {y} and in differential equation form is
= det [bt
b2
-at2 ] {u} S - an
= [ b1S+ (a12b2 -anbt)] {u}
(9.42) (9.43)
Sec. 9.3
Second-Order System Transient Response
297
Similarly, the differential equation for the second state variable x2(t) is d 2x2 dx2 dt 2 - (au + a22) dt + (aua22 - a12a21) x2
du
= b2 dt
+ (a21b1 - au~)u
(9.45)
The characteristic operator £c is fundamental to the response of every variable in a secondorder system and is commonly defined in tenns of a pair of parameters used to characterize the response: £c{y)
= det [SI- A] (y) dly
= dt2 + 2~ Wn
dy dt
(9.46) 2
+ WnY
(9.47)
where Wn is the undamped natural frequency (radls) and ~ is the system dimensionless damping ratio. This definition may be compared to Eq. (9.45) to give the following relationships: Wn
= Jaua22- a12a21
(9.48)
1 ~=--(au +a22) 2wn - (a11 + a22)
= =2...;--;:::a=u=a=22=-=a=12=a2=1
(9.49)
The undamped natural frequency and damping ratio play important roles in defining secondorder system responses, similar to the role of the time constant in first-order systems, since they completely define the system homogeneous equation. Example9.7
Use Cramer's rule to detennine the differential equations in the state variables x1 (t) and x2(t) for the system (i)
Fmd the undamped natural frequency ClJn and damping ratio ' for this system. Solution For this system, [Sl - A]
= [ S-2 +1
2 ] S+3
(ii)
and therefore for state variable x1(t), (iii)
or d 2x1
dx1
du
- - + 4 - + 7x1 = - + 3u dtl dt dt
(iv)
298
First- and Second-Order System Response
Chap.9
Similarly, for x 2 (t), det [
S~ 1
S! 3 ] {x2}
= det [ S-=; 1 ~] u
(v)
or d 2x2 dt2
dx2
+4dt +1x2 = 2u
(vi)
By inspection of Eq. (iv) or (vi), w~ = 7 and 2('(.r)11 = 4, giving (' = 2/.../7 = 0.755.
(.r)11
= ./7 radls and
Generation of a Differential Equation in an Output Variable The output equation y
= Cx + Du for any system variable is a single algebraic equation: y(t)
=I Ct C2] [;~]+[d) u(t)
(9.50)
H the operational expressions for Xt and x2 are substituted and both sides are operated on by the characteristic operator .Cc, (9.51) then operating on both sides using .Cc gives
det
[S-
~]
au -a21
(9.52) The detenninants may be expanded and the resulting equation written as a differential equation:
d2y dy dtl -(au+ a22) dt + (ana22- a12a21) y
d 2u
= q2 dt 2 + q1
du dt + qou
(9.53)
or in terms of the standard system parameters,
d 2y dy dt2 + 2~wn dt
2
+ WnY =
d 2u du q2 dtl + ql dt + qou
(9.54)
where the coefficients qo, q 1, and q2 are
qo = q1
c1
(-a22b1 + a12~)
+ c2 (-au~+ a21b1) + d (aua22- a12a21)
= c1b1 + c2b2- d (au+ a22)
q2=d
(9.55)
Sec. 9.3
Second-Order System Transient Response
299
Notice that the left-hand side of the differential equation, .Cc {y}, is the same for all system variables and that the only difference between any of the differential equations describing any system variable is in the constant coefficients q2, q1, and qo on the right-hand side. Example9.8 A rotational system consists of an inertial load J mounted in viscous bearings B and driven by an angular velocity source Oin(t) through a long light shaft with significant torsional stiffness K, as shown in Fig. 9.16. Derive a pair of second-order differential equations for the variables 0; and OK. K
Flywheel J
Motor (velocity source)
nr;J Rgure 9.16: Rotational system for Example 9.8.
Solution The state variables are 0 1 and TK, and the state equations are (i)
The output variable 0 1 is a state variable; Cramer's rule may be used directly: 0 det [ K
0;=
det
-1/J] S
[S+B/J -1/J](Om} K
S
(ii)
·
and reorganizing and expanding the determinants gives (iii)
The required differential equation is therefore (iv)
The differential velocity of the spring QK is defined by the output equation (v)
300
First- and Second-Order System Response
Chap. 9
The. operational expression for this variable is
[S + B/J
Slx =I {Slm(t)}-
-l/J] {Om(t)}
K
det
S
det [S+B/J -1/J]-d [0 -1/J] K S etK S =
[S+B/J
K
det
-l/J]
(vi)
{Slm}
S
S2 + (B/J)S
= S2 + (B/J)S+ K/J {Sltn} and the resulting differential equation is (vii)
The undamped natural frequency and damping ratio is found from either differential equation. For example, from Eq. (vii), ~ K 1J and ~ Cl), B1J. From these relationships,
=
Cl),
=~
and
=
BfJ
B
~ = 2.JKT] = 2~
(viii)
9.3.1 Solution of the Homogeneous Second-Order Equation
For any system variable y(t) in a second-order system, the homogeneous equation is found by setting the input u(t) = 0, so Eq. (9.54) becomes d2y dt2
+ 2~ Wn
dy dt
2
+ WnY = 0
(9.56)
The sol~tion, Yh (t), to the homogeneous equation is found by assuming the general exponential form described in Sec. 8.3.1, that is, (9.57)
where Ct and C2 are constants defined by the initial conditions and At and l2 are the roots of the characteristic equation det [SI- A]= l 2 + 2~w,l +
w; = 0
(9.58)
found from the quadratic formula: (9.59)
Sec. 9.3
Second-Order System Transient Response
301
If~ = 1, the two roots are equal (At = A2 =A) and, as described in Sec. 8.3.1, a modified form for the homogeneous solution is necessary:
(9.60)
In either case the homogeneous solution consists of two independent exponential com-
ponents with two arbitrary constants, cl and c2, whose values are selected to make the solution satisfy a given pair of initial conditions. In general the values of the output y(O) and its derivative y(O) at timet= 0 are used to provide the necessary information. The initial conditions for the output variable may be specified directly as part of the problem statement, or they may have to be determined from knowledge of the state variables Xt (0) and x2(0) at timet = 0. The homogeneous output equation may be used to compute y(O) directly from elements of the A and C matrices: y(O) = CtXt (0)
+ C2X2(0)
(9.61)
and the value of the derivative y(0) may be determined by differentiating the output equation and substituting for the derivatives of the state variables from the state equations:
+ C2X2(0) = CJ (OUXJ (0) + OJ2X2(0)] + C2 (a21X1 (0) + 022X2(0))
y(O) = CtXt (0)
(9.62)
To illustrate the influence of damping ratio and natural frequency on the system response, we consider the response of an unforced system output variable with initial output conditions of y(O) = Yo and y(O) = 0. If the roots of the characteristic equation are distinct, imposing these initial conditions on the general solution of Eq. {9.57) gives y(O) =Yo
dy dt
I
= C1 + C2
= 0 = AJCI
+ A2C2
(9.63)
t==O
With the result that and
(9.64)
For this set of initial conditions the homogeneous solution is therefore (9.65) (9.66)
302
FJISt- and Second-Order System Response
H the roots of the characteristic equation are identical, A. 1 based on Eq. (9.60) and is
Chap. 9
= "-2 = A., the solution is (9.67)
The system response depends directly on the values of the damping ratio undamped natural frequency (J)n. Four separate cases are described below.
Overdamped System > 1): When the damping ratio two roots of the characteristic equation are real and negative:
r and the
r is greater than 1, the (9.68)
From Eq. (9.66), the homogeneous response is
(9.69)
which is the sum of two decaying real exponentials, each with a different decay rate that defines a time constant 1 1 (9.70) 't'l = - - T 2 = - AJ A.2 The response exhibits no overshoot or oscillation and is known as an overdamped response. Figure 9.17 shows this response as a function of r using a normalized time scale Of Wnl. y(t)ly(O)
1.2 u ll.l
1.0
= 0.8 8. ll.l
e ]
:a
0.6
e 0.4
z0
0.2 0.0
5
10
15
20 w11 t
Nonnalized time Agure 9.17: Homogeneous response of an overdamped and critically damped second-order system for the initial condition y(O) = 1 and j(O) = 0.
Sec. 9.3
Second-Order System Transient Response
303
Critically Damped System(~= 1): When the damping ratio~ = 1, the roots of the characteristic equation are real and identical, (9.71) The solution to the initial condition response is found from Eq. (9.67): (9.72) which is shown in Fig. 9.17. This response fonri is known as a critically damped response because it marks the transition between the nonoscillatory overdamped response and the oscillatory response described in the next paragraph. Underdamped System (0:::; ~ < I): When the damping ratio is greater than or equal to 0 but less than 1, the two roots of the characteristic equation are complex conjugates with negative real parts: (9.73) where j
= .J=T and where Wd is defined to be the damped natural frequency: (9.74)
The response may be detennined by substituting the values of the roots in Eq. (9.73) into Eq. (9.66):
(9.75) When the Euler identities cos a= (e+ia + e-ia) /2 and sin a= (e+ia- e-ia) j2j (defined in App. B) are substituted, the solution is
(9.76)
where the phase angle '1/f is
""= tan-• J1-~2 ~
(9.77)
The initial condition response for an underdamped system is a damped cosine function oscillating at the damped natural frequency Wd with a phase shift '1/f and with the rate of
304
Fmt- and Second-Order System Response
Chap. 9
decay determined by the exponential term e-~w,r. The responses for underdamped secondorder systems are plotted against normalized time CIJnt for several values of damping ratio in Fig. 9.18. y(r)/y(O)
105~~---+~~--~------~----~ e
1 :a
0.0
8-05 ~---~~'-+----t----T------1
z
-1.0 '------'----~---'----___. 0 5 10 15 20 Nonnalized time
Wnt
Figure 9.18: Nonnalized initial condition response of an underdamped second-order system as a function of the damping ratio r.
For damping ratios near unity, the response decays rapidly with few oscillationsy but as the damping is decreased and approaches zero, the response becomes increasingly oscillatory. When the damping is zero, the response becomes a pure oscillation (9.78)
Yh(t) =Yo cos (CIJnt)
and persists for all time. (The term undamped natura/frequency for CIJn is derived from this situation because a system with ~ = 0 oscillates at a frequency of CIJn .) As the damping ratio increases from zero, the frequency of oscillation CIJd decreases, as shown by Eq. (9.74), until at a damping ratio of unity, the value of CIJd 0 and the response consists of a sum of real decaying exponentials. · The decay rate of the amplitude of oscillation is determined by the exponential term e-~w,r. It is sometimes important to determine the ratio of the oscillation amplitude from one cycle to the next. The cosine function is periodic and repeats with a period Tp = 2n-I CIJd, so if the response at an arbitrary timet is compared with the response at timet+ Tp, an amplitude decay ratio {DR) may be defined as
=
DR= y(t + Tp) y(t) e-t(J)n(t+21r/Q)d)
= - e-~w,r ----
provided y(t)
:f= 0 (9.79)
=e-21rC/~ The. decay ratio is unity if the damping ratio is zero and decreases as the damping ratio increases, reaching a value of zero as the damping ratio approaches unity.
Sec. 9.3
Second-Order System Transient Response
305
Unstable System (~ < 0): If the damping ratio is negative, the roots of the characteristic equation have positive real parts, and the real exponential tenn in the solution, Eq. (9.57), grows in an unstable fashion. When -1 < t < 0, the response is oscillatory with an overall exponential growth in amplitude, as shown in Fig. 9.19, while the solution for ~ < -1 grows as a real exponential. y(t)/y(O)
1.0
l
0.5
~
~~~-~~~---+--~~-+------------~
]
0.0
1z ~-5~~~~~----~-----------+--------~ -1.0
1.-----li..-------L-----'------'
0
5
10
15
20
Wnt
Normalized time Figure 9.19:
A typical unstable oscillatory response of a second-order system when the damping ratio Cis negative.
Example9.9 Many simple mechanical systems may be represented by a mass couj,led through spring and damping elements to a fixed position as shown in Fig. 9.20. Assume that the mass has been displaced from its equilibrium position and is allowed to return with no external forces acting on it. We wish (I) to find the response of the system model from an initial displacement so as to determine whether the mass returns to its equilibrium position with no overshoot, (2) and to determine the maximum velocity that it reaches. In addition we wish (3) to determine which system parameter we should change in order to guarantee no overshoot in the response. The values of the system parameters are m = 2 kg. K = 8 N/m, and B = 1.0 N-s/m, and the initial displacement Yo= 0.1 m.
F(t)
Figure 9.20:
Second-order mechanica1 system.
Soludon From the linear graph model in Fig. 9.20 the two state variables are the velocity of mass x1 = Vm and the force in the spring x2 = FK. The state equations for the system, with
306
Fll'St- and Second-Order System Response
Chap. 9
an input force Fm(t) acting on the mass are (i)
The output variable y is the position of the mass, which can be found from the constitutive K y and therefore the output equation is relation for the force in the spring Ftc
=
y (t) = (O)vm +
(~) FK + (0) Fm(t)
(ii)
The characteristic equation is (iii)
or
B m
K m
A.2 +-.A.+-=0
(iv)
The undamped natural frequency and damping ratio are therefore and
B
B
~=-=-2mCdn
(v)
2.J'Km
With the given system parameters, the undamped natural frequency and damping ratio are
Cdn
= If = 2 rad/s
~
1 = -=0.125 4x2
Because the damping ratio is positive but less than unity, the system is stable but underdamped; the response Yh (t) is oscillatory and therefore exhibits overshoot The solution is given directly by Eq. (9.76): (vi) and when the computed values of Cdd and ClJd
1/1 are substituted,
= ClJn~ = 2/1- (0.125) 2 = 1.98 rad/s
and
l/1 = tan- 1
0 125 • ,/1- (0.125) 2
= 0.125 rad
the response is Yh(t)
= O.l0le-o.2St cos(1.98t -
0.125) m
(vii)
The response is plotted in Fig. 9.21a where it can be seen that the mass displacement response y(t) overshoots the equilibrium position by almost 0.1 m and continues to oscillate for several cycles before settling to the equilibrium position.
Sec. 9.3
y(t)
,.
0.15
0.10
g cu eu u
'
0.05
I
I
~\
\I
-0.05
_.
__
t=1
(\ ~
"S.. 10
.....
-·----·--
'1. \ 1\ I \ II \.../ 1-
0.00
ca
iS
307
Second-Order System Transient Response
1
-
-I
.I
I
·-r-
Vt =0.125
!
-0.10
0
5
10 Time (s)
15
t
20
(a) Vm(t)
0.15 0.10 0.05
I
0.00
~>
-0.05
~
-0.10 -0.15 -0.20 0
5
10 Time (s)
15
20
t
(b)
Rgure 9.21:
The displacement (a) and velocity (b) responses of the mechanical second-order system.
The velocity of the mass Vm(t) is related to the displacement y(t) by differentiation of Eq. (vi): (viii)
308
First- and Second-Order System Response
Chap. 9
The velocity response is plotted in Figure 9.21b, where the maximum value of the velocity is found to be -0.17 mls at a time of0.75 s. In order to achieve a displacement response with no overshoot, an increase in the system damping is required to make ~ ~ 1. Since the damping ratio ~ is directly proportional to B, the value of the viscous damping parameter B would have to be increased by a factor of 8, that is, to B = 8 N-s/m, to achieve critical damping. With this value the response is given by Eq. (9.72): (ix)
The critically damped displacement response is also plotted in Fig. 9.2la, showing that there is no overshooL As before, the velocity of the mass may be found by differentiating the position response: v(t)
= 0.1 ( -2e-2t + 2e-21 + 4te-21 ) = 0.4te-21
(x)
The velocity response is plotted in Fig. 9.2lb where it can be seen that it reaches a maximum value of 0.075 mls at a time of 0.5 s. The maximum velocity in the critically damped case is less than 45% of the maximum velocity when the damping ratio~= 0.125.
9.3.2 Characteristic Second-Order System Transient Response The Standard Second-Order Form The input-output differential equation in any variable y(t) in a linear second-order system is given by Eq. (9.54): d 2y .·dt2
+ 2~{L)n
dy dt
d 2u dt2
2 + {L)nY = q2
+ ql
du dt
+ qou
where the coefficients qo, q1, and q2 are defined in Eqs. (9.55). Because the input u(t) is a known function of time, a forcing function
d 2u f(t) = q2 dt 2
+ q1
du dt
+ qou
(9.80)
may be defined. The forced response of a second-order system described by Eq. (9.54) may be simplified by considering in detail the behavior of the system in response to various forms of the forcing function f(t). We therefore begin by examining the response of the system
d2y dt2
+ 2~{L)n
dy dt
2
+ {L)nY = f(t)
(9.81)
The response of this standard system fonn defines a characteristic response for any variable in the system. The derivative, scaling, and superposition properties of linear systems allow the response of any system variable y;(t) to be derived directly from the response y(t):
d 2y
y;(t) = q2 dt2
+ ql
dy dt
+ qoy(t)
(9.82)
In the sections that follow, the response of the standard fonn to the unit step, ramp, and impulse singularity functions are derived with the assumption that the system is at rest at
Sec. 9.3
Second-Order System Transient Response
309
time t = 0, that is, y(O) = 0 and y(O) = 0. The generalization of the results to responses of systems with derivatives on the right-hand side is straightforward. The Step Response of a Second-Order System We start by deriving the response Ys (t) of the standard system, Eq. (9.81), to a step of unit amplitude. The forced differential equation is
(9.83) where Us(t) is the unit step function. The solution to Eq. (9.83) is the sum of the homogeneous response and a particular solution. For the case of distinct roots of the characteristic equation, A. 1 and A.2, the total solution is Ys(t)
= Yh(t) + Yp(t)
= C1ei. t + C2ei. t + Yp(t) 1
(9.84)
2
The particular solution may be found using the method of undetermined coefficients, and from Table 8.2 we take Yp(t) = K and substitute into the differential equation, obtaining (9.85) or (9.86) The ~onstants C1 and C2 are chosen to satisfy the two initial conditions: (9.87) (9.88) which may be solved to give (9.89)
The solution for the unit step response when the roots are distinct is therefore (9.90) (9.91)
310
Fust- and Second-Order System Response
Chap. 9
It can be seen that the second and third terms in Eq. (9.91) are identical to those in the homogeneous response, Eq. (9.66), so the solution may be written for the overdamped case as (9.92) where TJ = -IfJ.. 1 and T2 = -l/J..2 are time constants as previously defined. For the underdamped case, when AJ = -~Ct)n + j{t)n.Jl - ~ 2 and J..2 = ~Ct)n- jCt)nJI- ~ 2 , from Eq. (9.76) the solution is Ys(t)
= __1
?
wn
[
1-
1 e-Cta~n ] ~ COS(Ct)dt- 1/f)
v 1- ~ 2
forO<~<
1
(9.93)
-I ( J
~1 1 - ~2). where, as before, the phase angle Y, = tan When the roots of the characteristic equation are identical ( ~ = I) and At = J..2 = -wn, the homogeneous solution has a modified form (Sec. 8.3.1) and the total solution is (9.94) The solution that satisfies the initial conditions is
(9.95)
In all three cases the response settles to a steady equilibrium value as time increases. We define the steady-state response as (9.96) The second-order system step response is a function of both the system damping ratio ~ and the undamped natural frequency wn. The step responses of stable second-order systems are plotted in Fig. 9.22 in terms of nondimensional time Wnt and normalized amplitude y(t)/Yss· For damping ratios less than I, the solutions are oscillatory and overshoot the steadystate response. In the limiting case of zero damping the solution oscillates continuously about the steady-state solution Yss with a maximum value of Ymax = 2yss and a minimum value of Ymin = 0 at a frequency equal to the undamped natural frequency Ct)n· As the damping is increased, the amplitude of the overshoot in the response decreases until at critical damping, ~ =·I, the response reaches steady state with no overshoot For damping ratios greater than unity, the response exhibits no overshoo4 and as the damping ratio is further increased, the response approaches the steady-state value more slowly.
Second-Order System Transient Response
Sec. 9.3
311
y(t)/yss
2 ~----r-----~----~-----r----~----~------r---~
1.5 ~
= 0
1.25
~
]
1z
0.75 0.5 0.25
2.5
0
5
10
7.5
12.5
15
17.5
20
(J)nl
Normalized time Figure 9.22: Step response of stable second-order systems with the differential equation y + ~WnY + w!Y;:: u(t).
Example 9.10
The electric circuit in Fig. 9.23 contains a current source driving a series inductive and resistive load with a shunt capacitor across the load. The circuit is representative of motor drive systems and induction heating systems used in manufacturing processes. Excessive peak currents during transients in the input could damage the inductor. We therefore wish to compute response of the current through the inductor to a step in the input current to ensure that the manufacturers stated maximum current is not exceeded during start-up. The circuit parameters are L = 10-4 H. C = 1o-s F, and R = 500. Assume that the maximum step in the input current is to be 1.0 A. Solution From the linear graph in Fig. 9.23 the state variables are the voltage across the capacitor vc(t) and the current in the inductor h (t). The state equations for the system are
~-------
r-----t---+-....,
L R I
lLoad • ______ J Figure 9.23: A second-order electrical system.
312
Fust- and Second-Order System Response
0 [ ~c] = [ 1/L lL
-1/C] [~c] + [1/C]
-R/L
0
lL
I,
Chap. 9
(i)
The differential equation relating the current h to the source current I, is found by Cramer's rule:
1/C ] .
S det [ -1/L
[
1/C]
S
S+R/L {IL}=det -1/L
0
{I,}
(ii)
or (iii)
And the undamped natural frequency
CIJn
and damping ratio
~
are
1 6 = -vLC r;-;:; = 10 rad/s
CIJn
~=
R/L
21-JLC
=
(iv)
!!_ {£ = 0.25 2
VI
(v)
The system is underdamped ( ~ < 1), and oscillations are expected in the response. The differential equation is similar to the standard form and therefore has a unit step response in the form ofEq. (9.93):
iL(t)
=
1 [
2
CIJn ••?
Wft
e-Ceunt 1- ~ v 1- ~ 2
COS(CIJJt
-l/f)
= 1- 1.033e-o.2Sxlo6t cos {0.968 x
]
I06 t - 0.2527}
(vi) (vii)
which is plotted in Fig. 9.24. The step response shows that the peak current is 1.5 A. which is approximately 50% above the steady-state current.
Impulse Response of a Second-Order System The derivative property of linear systems, described in Sec. 8.4.4, allows the impulse response Y&(t) of any linear system to be found by differentiating the step response y,(t): Y&(t)
dy, dt
=-
because o(t)
= dtd u,(t)
(9.97)
where u,(t) is the unit step function. For the standard system defined in Eq. (9.81) with f(t) = c5(t), the differential equation is (9.98)
Sec. 9.3
313
Second-Order System Transient Response
i£(1)
1.6 r-------~-----~-------,-------
I
1.4
~----~-~--~--------~~--------~---------1
1.2
$
= ~ ...u
1.0 0.8
g u
-6
.s
0.6 0.4 0.2
0.0
5
10
15
20
I
Time (fLS) Figure 9.24: Response of the inductor current ; L (t) to a 1-A step in the input current I,.
When the roots of the characteristic equation, AJ and A2, are distinct, the impulse response is found by differentiating Eq. (9.91):
(9.99)
(9.100)
since w; = AtA2· For the case of real and distinct roots(~ > 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 <
w
~
Chap.9
< 1, Eq. {9.101) reduces to
e-~cu..t
Y«J(t) = ~ sin (cudt) 1- ~2
{9.102)
For a criticaJiy damped system ( ~ = I), the impulse response may be found by differentiating Eq. (9.95), obtaining (9.103) Y&(t) te--~»n'
=
Figure 9.25 shows typicaJ impulse responses for overdamped, critically damped, and underdamped systems. y(t) I
t=O.l 0.8 0.6 0.4 0.2 ~
c
&.
Ill
0
~
-0.2
-0.8 ~--------~----~------_.-------~------~----~---~~----~ 0 2.5 5 7.5 10 12.5 15 17.5 20 Normalized time Figure 9.25:
wnt
Typical impulse responses for overdamped. critically damped. and
underdamped second-order systems.
The Ramp Response of a Second-Order System: The integral property of linear systems, defined in Sec. 8.4.4, allows the ramp response Yr(t) to a forcing function f(t) t to be found by integrating the step response Ys(t):
=
y,(t)
=fa' y,(t) dt
because u,(t)
=fa' u,(t)dt
(9.104)
where u8 (t) is the unit step function. For the standard system defined in Eq. {9.81) with (t) = t, the forced differentia] equation is
f
d 2 yr
dt 2
dyr
2
+ 2~Wn dt + WnYr = t
(9.105)
Sec. 9.3
Second-Order System Transient Response
315
When the roots of the characteristic equation are distinct, the ramp response is found by integrating Eq. (9.91), that is,
(9.106)
./s
2 - Iwn and For an overdamped system with real distinct roots, lt = -swn + 2 - lwn, the ramp response may be found from Eq. (9.106) directly A-2 -swn or by making the partial substitutions for and Wn:
=
./s
s
(9.107)
which consists of a term that is itself a ramp, a pair of decaying exponential terms, and a constant-offset term. When the system is underdamped with complex conjugates roots, Eq. · (9 .1 06) may be written
(9.108)
which consists of a ramp function, a damped oscillatory term, and a constant offset When the roots are real and equal ( = 1), the response is found by integrating Eq. (9.95):
s
(9.109)
Summary of Singularity Function Responses The characteristic responses of a linear system to the ramp, step, and impulse functions are summarized in Table 9.3. 9.3.3 Second-Order System Transient Response The characteristic response defined in the previous section is the response to a forcing function f(t) as defined in Eq. (9.80). The response of a system to an input u(t) may be
Fust- and Second-Order System Response
316 TABLE 9.3:
Summary of the Characteristic Transient Responses of the System y + ~w,y + f(t) to the Unit Ramp u,.(r), the Unit Step u,{t), and the Impulse &(r)•.
~Y
Damping ratio
0!::
r<
Chap.9
1
=
Characteristic Response y(r)
Input /(t)
/(t)
=u,(t)
f(t)
=u,(t)
f(t) =&(I) ~
= 1
/(t)
= u,(t)
f(t) = u,(t)
r>
1
f(t)
= &(t)
/(t)
= u,(t)
f(t)
=u,(t)
/(t)
== &(t)
• The damped na~ral frequency wd
t/t
= Jt="fi't.~>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-<»n'- Wnte-,.')]
8.22te-3•411 )
(xi)
N-m
which is plotted in Fig. 9.27. Notice that in this case, although the system is critically damped ((' = 1), the response overshoots the steady-state value. This behavior is common for output variables that involve the derivative of the input in their differential equation.
TIC
e
~
80
u
=
§ 'g Cll
60
~ g -~ ~
40
~
a
= < bO
c
20
0
0.5
1.5
2
2.5
Time(s) Figure 9.27:
Step response of shaft coupling torque TIC and fan angular velocity 0 1 •
3
t
Chap. 9
321
Problems
PROBLEMS 9.1. The standard fonn of a first-order linear differential equation, expressed in tenns of the characteristic time constant t', is: dy t'-+y=f(t) dt (a) For a first-order equation of the fonn: a1
dy dt
+ aoy =
g(t)
determine the time constant and write the equation in standard form. (b) For a first-order system, expressed in state equation fonn:
x =ax+bu y =ex +du determine the time constant and derive a single differential equation in the output variable y(t) in the standard fonn. 9.2. The time constant t' of a first-order system has units of seconds when system parameters are expressed in a consistent set of units. The characteristic time constants of the four first-order physical systems shown in Fig. 9.4 are respectively: t'
= RC;
L R
t'=-
(a) Verify that when the physical system parameters are expressed in SI units, the time constants are in units of seconds for all four systems. (b) Verify that when mechanical system parameters m, B, and K are expressed in appropriate English units, the time constant of the two mechanical systems in Fig. 9.4 are expressed i~ seconds. 9.3. Consider the four first-order systems illustrated in Fig. 9 .4. For each case derive the system state equation and verify the given system time constant 9.4. For the electrical system depicted in Fig. 9.28,
Figure 9.28:
A first-order electrical system.
(a) (b) (c) (d)
Derive the system state equation and determine the system time constant Determine the value of the time constant if R 1 = R2 = 10 0 and C 1 p,F. What is the limiting value of the time constant as the resistance R2 is increased to infinity? As the value of the resistance R 1 approaches zero, what value does the time constant approach? What occurs in the system as R 1 approaches a short circuit, that is, R 1 ~ 0? 9.5. For the system shown in Fig. 9.29, derive the system state equation and determine the system 5 kg. What is the asymptotic value of the time time-constant when Bt = B2 = 10 N-s/m and m constant as the damping coefficient 82 approaches zero? Explain what happens to the time constant as the damping coefficient B1 approaches infinity.
=
=
First- and Second-Order System Response
322
Rgure 9.29:
Chap.9
A first-order mechanical system.
9.6. Each of the three physical systems depicted in Fig. 9.30 is initially at rest, with no stored energy. A unit step input is applied to each at t = 0. For each system derive a first-order differential equation relating the output to the input, and detennine the system time constant. Find the final value of the system response and the time it will take for the system to respond to within 2% of the final value.
Input Output:
V,(t) i(t)
(a)
Input: Output:
V,(t) vm(t)
{b)
Input Output
(c)
Figure 9.30: Three first-order systems.
9.7. Consider the thawing of frozen food in a freezer after a electric power failure. Form a system model representing the food as a lumped thermal capacitance, and the walls as having an effective thermal resistance between the frozen food mass temperature and the external environment (assumed to be at a constant temperature). Derive the system state equation and detennine the system time constant. If the thermal capacitance of the food is 100 JIK., and the thermal resistance is 1000 KIW, what is the value of the time constant? If the initial temperature of the food is -l0°C (2631{) and the room temperature outside the freezer is 30°C (303K), how long will it take for the food to reach the thawing temperature of ooc (273K)? 9.8. An experimental apparatus to determine the friction characteristics of various materials is shown in Fig. 9.31. A mass element is released with a known initial velocity and slides on the test sample. The velocity decay, due to the frictional resistance, is monitored and used to estimate the effective viscous drag coefficient. Derive the system state equation and detennine the system time constant. Derive an
Chap. 9
323
Problems
expression for the velocity of the mass assuming that it is released with an initial velocity vo at time = 0. If the mass is 10 kg, and after a time of 5.0 seconds the mass is at 50% of its initial velocity, determine the value of the damping parameter.
t
r
Vm(t)
Unknown coefficient
~ Figure 9.31: Friction coefficient test apparatus.
9.9. Practical electrical capacitors are nonideal energy storage elements because of charge leakage through the insulating dielectric material between the plates. In a test to measure the leakage resistance, a switch is closed to charge a capacitor to a voltage V0 , and then opened, as shown in Fig. 9.32. The open circuit voltage decay v(t) is monitored. Fonn a model of a real capacitor accounting for charge leakage and detennine the system time constant Derive an expression for the decay in capacitor voltage from the initial voltage Vo. In a test a capacitor C = 5JLF is charged to an initial voltage of 10 V. After 100 s the voltage is 3.68 V. What is the value of the dielectric leakage resistance of this capacitor?
Figure 9.32: Capacitor leakage test circuit
9.10. Electric powered vehicles must have sufficient acceleration capability to maneuver well in traffic. Consider a simplified model of an electric powered car of mass m in which the electric motor generates a thrust F(t), and assume that the effective total roadway and aerodynamic resistance is represented as a viscous drag with coefficient B. (a) Fonn a model for the car and derive the state equation relating the car velocity to the thrust What is the system time constant? (b) A car with a mass of 3200 kg, is allowed to coast down from an initial velocity on a horizontal surface. The data shows that in 20 s the car speed decreases to 62.5% of its initial value. Determine the system time constant and the value of the viscous resistance coefficient B. (c) If it is assumed with the car initially at rest, the maximum thrust from the electric motor is applied to the car, derive an expression for the car velocity for t > 0. What value of thrust is required for the car in (b) to reach a velocity of 100 kmlhr in 10 s? 9.11. A boat with a mass of 1000 kg is driven by an engine with a maximum thrust of 1500 N. The water resistance may be represented as viscous drag with a coefficient of B = 250 N-s/m. For the system shown in Fig. 9.33:
~--··-
..
v(t)
FT(t)-....j-~~-1!! !!2~Figure 9.33:
Boat propulsion system.
324
First- and Second-Order System Response
Chap. 9
(a) Form a system model and derive differential equations for the following output variables: (i) the velocity of the boat, (ii) the force acting to accelerate the boat, and (iii) the force across the dissipative element in the model.
(b) Assume that the boat is initially at rest. and that at timet = 0 the engine is turned on to its maximum thrust. Derive the solution for the boat velocity as a function of time. Sketch the
velocity as a function of time and determine the maximum velocity reached. Also sketch the response of the force accelerating the boat mass and the force across the dissipative element.
(c) After 40 s the engine is throttled back to a thrust of 500 N. Determine and plot the response of the boat velocity.
(d) Determine and plot the engine power required by the boat as a function of time.
9.12. A sky diver, with mass m, is experimenting with a new double parachute system shown in Fig. 9.34. When the sky diver first jumps a small parachute opens immediately, and total air resistance may be approximated as a linear drag force F8 that is proportional to the vertical velocity v. The sky diver may subsequently activate a second larger parachute which also has a linear drag characteristic, and essentially acts to increase the parachute drag coefficient to a factor of ten times that of the small parachute.
Initial small parachute
Large parachute deployed at time t
r
Massm
v
Figure 9.34: Prototype dual parachute system.
Chap. 9
325
Problems
(a) Form a pair of models for the system (under the two conditions), and derive the state equations.
=
(b) Assume that the sky diver jumps at t 0 and deploys the small parachute. A speed sensor sends a signal that automatically deploys the larger parachute at a time when the descent speed is equal to 98% of the maximum descent speed with the small parachute. Using the system parameters determine the following: (i) the time at which the large parachute is deployed, (ii) the velocity at which the large parachute deploys, (iii) the time at which the sky diver reaches a terminal velocity after releasing the large parachute, and (iv) the value of the terminal velocity?
(c) Sketch the velocity of descent. and the vertical restraining force on the sky diver from the parachute system as a function of time. 9.13. In a manufacturing process a motor drives a surface grinding wheel with diameter d and mass m. The friction from the workpiece produces a torque load which is proportional to the wheel speed. The motor is driven by a current source, and produces a torque proportional to the input current / 8 {1). Derive the differential equation relating the grinding wheel speed to the input current Determine the system time constant and determine the grinding wheel angular velocity a~ a function of time for the current input shown in Fig. 9.35. Make a sketch of the angular velocity response.
0
T
Figure 9.35: Surface grinder and excitation current waveform.
9.14. Derive the differential equation relating the output voltage to the input current in the electric circuit shown in Fig. 9.36. The circuit values are R1 0.5 0, R2 1 0, and L 3 H. Determine the system time constant and the value of the output voltage at times t = 0, 1, 2, 3, 4 s. Sketch the output voltage response.
=
=
=
Is (f)
l,(t)r
10
Rl
Lf
!R,
[•(I)
~
c ~
u= 0
2
3
4
5
6
Time (s) Figure 9.36: An electric circuit and excitation current waveform.
9.15. A water reservoir with an area A = 1000 sq. m contains· water with an initial depth of 20 m. Water is removed from the reservoir by constant flow pumps to service a town. There is no inflow of water into the reservoir in the time period of interest, and the outflow is 20 m 3 /s. While the pumps are running, and the water level is at a depth of l 0 m, a gate valve at the bottom of the reservoir breaks and provides a leakage path to the environment. The gate valve leakage flow is proportional to the
326
Fust- and Second-Order System Response
Chap. 9
pressure drop across the valve and has a value of resistance, R = 0.5 N-s/m5 • Form a model of the system that is valid after the leakage develops from the faulty gate valve. What is the system time constant? Determine and sketch the height of the water in the tank as a function of time. Compare· the time it would take to empty the reservoir in the presence and absence of the leakage.
9.16. A test is conducted to spin-up the rotor in the flywheel energy storage system illustrated in Fig. 9.37. Construct a model for the system using linear elements,and derive the differential equation relating the flywheel angular velocity to the input angular velocity. Find the system time constant if J 10kg-m2 , B1 = IOON-m-s/rad,and B2 ION-m-s/rad. Att =Otheinputangularvelocityis changed instantaneously from 0.0 to I 00 radlsec. Detennine the time history of the flywheel angular velocity. Sketch the flywheel angular velocity as a function of time.
=
=
Drag-cup transmission
n.r
Figure 9.37:
A ftywhee1 energy storage system.
9.17. The electric circuit shown in Fig. 9.38 is excited by a repetitive square wave with a period T s. The circuit parameters are R = 10 kn, and C = 100 J.LF. When the squarewave is initially started, the circuit has no stored energy. Determine the solution for the output for two periods of the squarewave (i) when T = 8.0 s, and (ii) when T = 2.0 s. Comment on the nature of the solution for the cases: (i) T << 1'. (ii) T = 2r, and (iii) T >> ', 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
(10.18) (10.19)
· The full state response, described by Eq. (10.18) or (10.19), consists of two components. The first is a term similar to the system homogeneous response xh(t) = eAtx(O) that is dependent on only the system initial conditions x(O). The second term is a convolution integral that is the particular solution for the input u(t) with zero initial conditions. Evaluation of the integral in Eq. (10.19) involves matrix integration. For a system of order nand with r inputs, the matrix eAt is n x n, B is n x r, and u(t) is an r x 1 column vector. The product eABu(r) is therefore ann x 1 column vector, and the solution for each of the state equations involves a single scalar integration. Example 10.2 Find the response of the two state variables of the system
.Xt
= -2xt + u
X2 =Xt -X2
to a constant input u(t)
= 5 fort >
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<•-•lBu("r) d~ + Du(t).
(10.22)
Example 10.3 Fmd the response of the output variable
in the system described by state equations it= -2xt
+u
X2 =Xt -X2
to a constant input u(t) = S fort > 0 if x 1 (0) = 0 and x2 = 0. Solution This is the Same system described in Example 10.1 with the same input and initial conditions used in Example 10.2. The state variable response is (Example 10.2) (i)
337
The State Transition Matrix
Sec. 10.4
The output response is y(r)
= 2x1 (r) + x2(t) = 15 - ~e-2r -se-t 2
10.4
(ii)
2
THE STATE TRANSITION MATRIX
10.4.1 Properties of the State Transition Matrix Table 10.1 shows some of the properties that can be derived directly from the series definition of the matrix exponential eA'. For comparison the similar properties of the scalar exponential ea' are also given. 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 ~(t). There are many methods for computing the elements of cl»(t), including one presented in the next section, that are much more convenient than the series definition. The matrix exponential representation of the state transition matrix allows some of its properties to be simply stated:
1. 4»(0)
=
I, which simply states that the state response at time t = 0 is identical to the initial conditions.
2. 4»(-t) = 4»- 1(t). The response of an unforced system before timet = 0 may be calculated from the initial conditions x(O}, x( -t) = 4»(-t)x(O) = 4»- 1(t)x(O)
(10.23)
and the inverse always exists.
=
cl»(t1 + t2). With this property the state response at time t may be defined from the system state specified at some time other than t = 0, for example, at t t0 • Using property 2, the response at timet= 0 is
3. cl»(t1)cl»(t2)
=
x(O) = «<»( -to)x(to)
(10.24)
and using the properties in Table 10.1, x(t) = cl»(t)x(O) = cl»(t)cl»( -to)x(to)
(10.25)
xh (t) = «<»(t - to)x(to)
(10.26)
or
4. If A is a diagonal matrix, then eAt is also diagonal and each element on the diagonal is an exponential in the corresponding diagonal element of the A matrix, that is, eaur. This property is easily shown by considering the terms An in the series definition and noting that any diagonal matrix raised to an integer power is itself diagonal.
338
General Solution of the Linear State Equations TABLE 10.1:
Cbap.lO
Comparison of Properties of the Scalar and Matrix Exponentials
Scalar Exponential
Matrix Exponential
A2r2 Alr3 ~=I+At+-+-+··· 2! 3! ~=I e-At= (~)-1
eA(tJ+f2) =~I~ er = eA1'eA2' only if A,A2 = A2A1
!_~=~=~A dt
1' ~dt=A-• (~ -1) = (~ -I)A-
1
if A- 1 exists; otherwise defined by the series
10.4.2 System Eigenvalues and Eigenvectors
In Example I 0.1 each element of c!»(t) = eAt is a sum of scalar exponential terms, and the homogeneous response of each of the two state variables is a sum of exponential terms. In Chap. 8 it was shown that the homogeneous response of first- and second-order systems is exponential in form, containing components of the form eu, where l is a root of the characteristic equation. The first-order homogeneous output response is of the form y (t) = C eAt, where C is determined from the initial condition C = y(O), and the second-order response consists of two exponential components y(t) = C 1e'-•' + C2eA21 , where the constants Ct and C2 are determined by a pair of initial conditions, usually the output and its derivative. It is therefore reasonable to conjecture that for an nth-order system the homogeneous response of each of the n state variables x; (t) consists of a weigh~d sum of n exponential components: n
x;(t)
=L
mijeA.i'
(10.27)
j=l
where the mij are constant coefficients that are dependent on the system structure and the initial conditions x(O). The proposed solution for the complete state vector may be written in matrix form:
xt(t)l [mu ... [Xn(t)... mnl x2(t)
m21
(10.28)
Sec. 10.4
339
The State Transition Matrix
or (10.29)
where M is an n x n matrix of the coefficients mij. To detennine the conditions under which Eq. (10.29) is a solution of the homogeneous state equations, the suggested response is differentiated and substituted into the state equation. From Eq. (10.27) the derivative of each conjectured state response is
(10.30)
or, in the matrix form,
[
~1]
x2
.. .
-
[
...
A2m12 A2m22 .:
... ...
·. .
Anmln Anm2n .:
Atmnl
A2mn2
...
Anmnn
Atmu Atm2t
Xn
l[
eA.•'] e>..2t
.:
(10.31)
e>..nt
Equations (1 0.28) and (1 0.31) may be substituted into the homogeneous state equation, Eq. (10.3),
[ Atm2t A1m11 AJmnl
=A
[mu m21 .
mnl
A2m12 A2m22
... ... ...
A2mn2
111
Anm2n e>..2t A.m'"] .. [ e..
.
.
Anmnn
eAnl
e>..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
are
sin (l)t
. = 2j1 (e'wr
COS (l)f
= -1 ( 2
.
eJWI
. ) e-Jwt
. ) + e-)WI
(10.50) (10.51)
may be used to express the elements of the state transition matrix in a purely real form sin ((l)t) or eat cos (wt).
eat
Example 10.7
Determine the state transition matrix for an undamped mechanical oscillator with a mass m = 1 kg suspended on a spring with stiffness K = 100 N/m as shown with its linear graph in Fig. 10.1.
Solution The state equations for this system are (i)
346
General Solution of the Linear State Equations
Chap. 10
With the vaJues given the state equations become
(ii)
A] = 0 is
The characteristic equation det [AI -
(iii)
so the eigenvaJues are purely imaginary: At = j I 0 and A2 = - j I 0. A pair of corresponding eigenvectors is found by substituting these values into the equation [A; I - A] m; = 0 and choosing an arbitrary scaling constant:
The modaJ matrix M
M=[
and its inverse are
1 0.1j
M-t
1 ] -O.Ij
= _1_ [-O.lj -0.2j
-1]
-O.lj
1
and the state transition matrix is
1
•
[ 1 O.lj
1
1 ] [ei 0t -O.lj 0
eliOt+ e-i10t
.
[
_ [
-
eJtOr _ e-JlOt]
2
-
-
2.
10
eitOt _ e-J10t
-0.1
eilOt
2j
cos(lOt)
0 ] [-O.lj e-itOt -O.lj
) + e-ilOt
(iv)
2
10 sin (lOt)]
-0.1 sin(lOt)
cos(lOt)
The homogeneous response to arbitrary initial values of FK and Vm is therefore
FK(t)] [ cos(lOt) [ Vm(t) = -0.1 sin(tOt)
lOsin(lOt)] [FK(O)] cos(tOt) Vm(O)
(v)
and the individuaJ state responses are FK(t)
= FK(O) cos( lOt)+ tOvm(O) sin( lOt)
Vm(t) = -O.lFK(O) sin(lOt) + Vm(O) cos(tOt)
which are purely real responses despite the imaginary system eigenvaJues.
(vi) (vii)
Sec. 10.4
The State Transition Matrix
347
F(t)
(a)
Figure 10.1: A simple spring-mass mechanical oscillator, with (a) schematic representation, and (b) its linear graph.
10.4.5 Systems with Repeated Eigenvalues
The method for deriving the state transition matrix presented in the previous section is dependent on the existence of the inverse of the modal matrix M; that is, it must be nonsingular. In general if there are two or more eigenvalues with the same value, the eigenvectors are not linearly independent and M- 1 does not exist It is possible to extend the method to handle repeated eigenvalues, as described in references on linear system theory. When there is a pair of repeated eigenvalues, instead of a linear combination of simple exponential terms, the state response may have a pair of components eA11 and teA11 corresponding to the two identical eigenvalues. This does not imply that the state transition matrix fll(t) does not exist for systems with repeated eigenvalues; it may be computed by other methods. It may, however, lose the simple exponential form assumed throughout this chapter. We do not discuss this case further in this introductory book, and readers are referred to more advanced texts [1-5]. 10.4.6 Stability of Linear Systems
The concept of the stability of first- and second-order dynamic systems was introduced in Chaps. 8 and 9. A system is said to be asymptotically stable if the homogeneous response of the state vector x(t) returns to the origin of the state space from any arbitrary set of initial conditions x(O) as timet -+ oo. This definition of stability is equivalent to stating that the homogeneous response of all state variables must decay to zero in the absence of any input to the system, or (10.52) lim X;(t) = 0 r-.oo
for all i
= 1, ... , n. This condition may be rewritten in terms of the state transition matrix: lim fll(t)x(O) = 0
t-.oo
(10.53)
for any x(O). All the elements of the state transition matrix are linear combinations of the modal components e}.11 , therefore, the stability of a system depends on all such components decaying to zero with time. For real eigenvalues this requires that A.; < 0, for all i, since any positive eigenvalue generates a modal response that increases exponentially with time. If eigenvalues appear in complex conjugate pairs Ai.i+l =a± j{J), the state homogeneous response contains components of the form err' sin({J)t) or errt cos({J)t). If a > 0, these
348
General Solution of the Linear State Equations
Chap. 10
components grow exponentially with time and the system is by definition unstable. The requirements for system stability may therefore be summarized: A linear system described by state equations i = AX + Bu is asymptotically stable if and only if all eigenvalues of the matrix A have negative real parts. Three other separate conditions should be considered:
1. If one or more eigenvalues, or pair of conjugate eigenvalues, has a positive real part, there is at least one corresponding modal component that increases exponentially without bound from any finite initial condition, violating the definition of stability. 2. Any pair of conjugate eigenvalues that are purely imaginary, Ai.i+l = ±jw with a real part a= 0, generate an.undamped oscillatory component in the state response. The magnitude of the homogeneous system response neither decays or grows but continues to oscillate for all time at a frequency (.1). Such a system is defined to be marginally stable.
3. An eigenvalue with a value A = 0 generates a modal component e01 that is a constant The ~ystem response neither decays nor grows, and again the system is defined to be marginally stable. Example 10.8 Discuss the stability of an invened pendulum consisting of a mass m on the end of a long, light rod of length I mounted in a rotational bearing that exhibits a viscous rotational drag, as shown in Fig. 10.2.
K (-mgl)
(a)
(b)
Figure 10.2: An inverted pendulum. (a) the physical system. and (b) a linear graph with a spring having negative stiffness.
Solution The system may be treated as a rotational system. The moment of inertia J of the m/ 2 , and when the rod is displaced from the vertical, gravity exerts a torque mass is J mgl sin 9 about the bearing as shown. This system is inherently nonlinear because of the angular dependence of the torque, but it may be linearized by using the small-angle approximation sin 6 ~ 9 for small 8. Then the restoring torque after a displacement 9 is -mgl9. This is the constitutive relationship of an ideal T-type element and allows the gravitational effect to be -mgl. Let the frictional modeled as a torsional spring with a negative spring constant K drag about the bearing be represented by the damping coe~cient B. The linear graph for the system is shown in Fig. 10.2. Define the states to be the torque in the spring TK and the angular velocity of the shaft n. The linearized state equations for this homogeneous system are
=
=
(i)
349
The State Transition Matrix
Sec. 10.4
The characteristic equation is (ii)
or B J
K J
l 2 +-l+-=0 (iii)
2 B g l +-l--=0
R
J
The quadratic equation may be used to determine the two eigenvalues
1 (B) 4g·R +-
B l12=--±. 2J 2
2
(iv)
J
The following may be noted directly from Eq. (iv):
1. The quantity under the radical is positive, therefore, both eigenvalues are real. 2. (v)
and so there is always a positive eigenvalue. We conclude, therefore, that the inverted pendulum is an unstable system and in its linearized form will exhibit an exponential growth in angular displacement from the vertical for any finite initial offset or angular velocity.
10.4.7 Transformation of State Variables The choice of a set of state variables used to represent a system is not unique. In Chap. 7 operational methods for transferring between representations were discussed. It is possible to define an infinite number of different representations by transforming the state vector using linear operations. If a system is described by a state vector x, a new set of state variables x' may be generated from any linearly independent combination of the original state variables x; (t), for example, x;(t)
= PilXt(t) + Pi2X2(t) + · ·· + PinXn(t)
(10.54)
where the Pii are constants. This linear transformation may be written in matrix form:
x=Px'
(10.55)
where P is a nonsingular n x n square matrix. (With this definition the elements Pii in Eq. (10.54) are elements of p- 1.) The state equations in the new state variables become
:X
= P:X' = APx' + Bu
(10.56)
350
General Solution of the Linear State Equations
Chap. 10
and after premultiplying each side by the inverse of P, the new set of state equations gives (10.57) The output equation must be similarly transformed: y
= (CP)x' +Do
(10.58)
The system is now represented by modified A, B, and C matrices. The state variable representation is an internal system representation that is made observable to the system environment through the output equations. The input-output system dynamic behavior should be independent of the internal system representation. For the transformed state equations the characteristic equation is defined in terms of the transformed A matrix:
=o
(10.59)
det (Ap-Ip- p- 1AP)
=0
(10.60)
det (p-t [AI- A] P]
=0
(10.61)
det [u- (P- 1AP}] If the substitution I = p-I IP is made,
Since the determinant of a product is the product of the determinants, the characteristic equation is (10.62) det [r- 1] det [AI- A] det [P] = 0 and because Pis not singular, det [P] #: 0 and det [P- 1] # 0, the transformed characteristic equation is det[AI -A]= 0 (10.63) which is the same as that of the original state equations, leading to the following important conclusion: The characteristic equation, and hence the eigenvalues J..; and the modal response components e'-11 , are invariant under a linear transformation of the system state variables. These quantities are properties of the system itself and are not affected by the choice of state variables used to describe the system dynamics.
If a transformation of the state variable representation is made, it is necessary to similarly transfonn any initial conditions to the new representation using x'(O)
= p- 1x(0)
(10.64)
Transformation to Diagonal Form A transformation that results in a diagonal form of the system matrix A can provide insight into the internal structure of a system. Consider a system with distinct eigenvalues A1, ••• , An and a modal matrix M formed by adjoining columns of eigenvectors as described in Sec. 10.4.2. Let x' be the transformed state vector, defined by x = Mx', so the new set of state and output equations are
i' = (M- 1AM)x' + (M- 1B)u y = (CM)x' +Dn
(10.65) (10.66)
Sec. 10.4
351
The State Transition Matrix
The new system matrix is (M- 1AM). As in Eq. (10.35), the product AM may be written in terms of the eigenvalues and eigenvectors: (10.67)
because Am; = A.;m; is the relationship that defined the eigenvalue A.;. Equation (10.67) can be rearranged and written 0 A.2
r
[At
AM=[mtlm21 ... lm.l
0
...
=MA
il
(10.68)
where A is the diagonal n x n square matrix containing the system eigenvalues on the leading diagonal:
A=
[
A.t 0
A.2
0
•••
.. .
0 0
.:
.:
•. .
.:
0
0
... An
l
(10.69)
If both sides ofEq. (10.68) are premultiplied by M- 1,
M- 1AM
= M- 1MA =A
(10.70)
the transformed state equations are
x =Ax' +B'u
(10.71)
where B' = (M- 1B). Equation (10.71) represents a set of n uncoupled first-order differential equations, each of which may be written r
x; = A.;x; + Lb;jui
(10.72)
j=l
and does not involve any cross-coupling from other states. The homogeneous state equations
x = AX are simply
x; = A.;x;
(10.73)
The state transition matrix for the diagonal system is «
= eA1, as given by Eq.
(10.45),
0
0
l.l
(10.74)
352
General Solution of the Linear State Equations
Chap. 10
= .Z,(t)x(O) has the simple uncoupled form
and the homogeneous response XII (t)
X;(t) = X;(O)e"''
(10.75)
for each state variable. Systems with repeated eigenvalues may not be reducible to a diagonal form but may be represented in a closely related form known as the Jordan form [1-5]. Example 10.9
Transform the system
to diagonal form and find the homogeneous response to an arbitrary set of initial conditions.
Solution The A matrix in this example is the same as that examined in Examples 10.2 and 10.3. that is.
A= [-2 1] 2 -3 In the previous examples it was shown that for this system the eigenvalues are A. 1 A.2 = -4 and that a suitable modal matrix and its inverse are
M=
[1
M-•
J ]
1 -2
= -1 and
=.!. [2 1] 3
1 -1
The transformed state variables are
x' =M- 1x= ~ [~ ~ 1 ]x
(i)
or 2
x:(t)
,
X2 (t)
1
.
= 3 x 1(t) + 3X2(t)
(ii)
1 1 = 3X1 (t)- 3x2(t)
(iii)
The transformed state equations are (iv) or
[ i2i~]=.!.[2 3 1 .=
[-10
1 ] [ -2 -1 2. 0 ] -4
1 ] [ 1 1 ] [ xl ] -3 1 -2 xi
[x!] + [_:-} xi
+ 31 [ 21
1 ] [ 0] -1 1 u(t)
(v)
! ] u(t)
which is clearly diagonal in fonn with the eigenvalues on the leading diagonal of the transfonned
A matrix. The state transition matrix is found by plaCing the modal components on the diagonal: (vi)
Sec. 10.5
The Response of Linear Systems to the Singularity Input Functions
353
and the homogeneous state response in the transformed system representation is .(t)X'(O):
Xia (t)
=
X~ (t)
= X~ (O)e-t
(vii)
x~(t)
= x2(0)e--4t
(viii)
where the transformed initial states are found from Eqs. (ii) and (iii): (ix) (x)
10.5
THE RESPONSE OF LINEAR SYSTEMS TO THE SINGULARITY INPUT FUNCTIONS In Chap. 8 the singularity input functions (the impulse, step, and ramp functions) were introduced as a set of functions commonly used to characterize the transient response characteristics of linear time-invariant systems. The forced system response, given by Eqs. (10.18) and (10.19), is x(t) = eAtx(O)
=
+~
~x(O) +
£
(10.76)
e-ATBu(r)dt
L' .,A<~-<>Bu(t)dt
(10.77)
and may be used to derive simple expressions for the response of linear time-invariant systems to the individual singularity functions.
10.5.1 The Impulse Response Assume that the input vector u(t) is a weighted series of impulse functions the r system inputs:
u(t)
= K8(t) =
[j:
]8(1)
~(t)
applied to
(10.78)
The vector K is used to distribute impulses among the r inputs. For example, if the response to a single impulse on the jth input is required, then k; = 0 for all k :1= j and ki = 1. The state vector impulse response is found by substituting u(t) = K~(t) into Eq. (10.19): (10.79)
Chap. 10
General Solution of the Linear State Equations
354
The sifting, or sampling, property of the delta function states that for any finite value of fl.,
1
0+~
f(t) ~(t) dt
= /(0)
0-~
which allows Eq. (10.79) to be simplified: x(t)
= ~x(O) + eAtBK =
eAt
(10.80)
[x(O) + BK]
The effect of impulsive inputs on the state response is similar to a set of initial conditions; the response for t > 0 is defined entirely by the state transition matrix. For an asymptotically stable linear system the impulse response decays to zero as time t -+ oo. The system output equations may be used to find the impulse response of any output variable y(t)
= CeAt [x(O) + BK] + DK8(t)
(10.81)
which shows a direct feedthrough of the impulses for any system in which D :F 0. Example 10.10 The electric filter shown in Fig. 10.3 is installed in the power line of a computer in a factory environment to reduce the possibility of damage caused by impulsive voltage spikes created by other equipmenL Measurements have shown that the transients are approximately 100 V in amplitude and have a typical duration of 10 p,s. The values of the filter components are L 1 = 10 mH, L 2 = 10 mH, and C = 1 p,F. The load resistance is 50 0. Find the effect of the filter on the voltage spikes. V1(t) 100
0 010 (JLS) (b)
(a) Figure 10.3:
(c)
(a) A third-order electric filter, (b) its linear graph, and (c) a typical voltage pulse at its input
Solution The system output is the voltage vc. The state and output equations are
VR.
The system has state variables
h,.
iL 2 ,
and
(i)
VR
= [0
R
(ii)
Sec. 10.5
The Response of Linear Systems to the Singularity Input Functions
w-s
355
= 10-3 V-s. that is, Vs(t) and the input is modeled as an impulse of strength 100 X 3 10- 6(1) as detennined from the area under the typical voltage spike shown in Fig. 10.3c. If the values of the system parameters are substituted into the matrices, they become
0
A=
c = [0
-100]
0 -5000 -1000000
0 [ 1000000
50 0]
B=[TJ
100 0
and
K
=
= [0.001]
From Eq. (10.81) the output impulse response is y(t) =C~'BK
(iii) (iv)
= (CM) eAI (M- 1BK)
using Eq. (10.49). A linear algebra software package is used to find the system eigenvalues:
= -1210 + 13867j
lt
l2 = -1210 -13867j
l3
= -2581
and the modal matrix and its inverse:
M
0.0006 + 0.0072j 0.0006- 0.0072j 0.0388] 0.0067 j 0.0018 + 0.0067i 0.0413 1 1 1 -6.2301 - 36.673j -6.6450 + 35.4j 0.5161 - 0.042j] -6.2301 + 36.673j -6.6450- 35.4j 0.5161 + 0.042j [ 12.4603 13.2900 -0.0322
= [ 0.0018 -
M-1 =
The matrices M- 1BK and CM are -0.6230 - 3.667 j]
M- 1BK = -0.6230 + 3.667 j [
0.1246
CM = [0.0917- 0.3355j 0.0917 + 0.3355j 2.0666] so the solution is
VR(t)
= CM [
eC-I2IO+I3867J)r
0
0
0
e<-t210+13867j)r
0
0
0
e-2SB1t
]
M-IBK
(v)
= 2.575e-2581 ' + (-1.2875- 0.1273j) e<- 1210+13867Jlt (vi) =
+ (-1.2875 + 0.1273j) e<-J2IO-IJ867J>r 2.575e-25811 + e- 1210' ( -2.575 cos(13867t) + 0.2546 sin(13867t)]
(vii)
The impulse response is plotted in Fig. 10.4. The maximum voltage reached is approximately 3 V. a significant reduction from the input amplitude of 100 V.
356
General Solution of the Linear State Equations
Chap. 10
0
0.001
0.002
0.003
0.004 t
Tune(s)
Figure 10.4: Response of the filter to an impulsive input of 100 V amplitude
and 10-5 s duration.
10.5.2 The Step Response
Assume that the input vector is a.weighted series of unit step functions us(t) applied to the r inputs, that is,
(10.82)
The vector K is used to distribute the step among the r inputs, and if the response to a unit step on a single input channel is desired, for example, the jth input, then k1 = 0 for all k :/; j and ki = 1. The state step response is found by substituting u(t) = Kus(t) into Eq. (10.19): x(t) = .,Aix(O) +
and because us(t)
= 1 for all t
fo' e"<•-•>sKu,(T) dT
(10.83)
> 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
(11.10)
and extrapolating the response with this slope across the interval, as illustrated in Fig. 11.1. Using the method of separation of variables, x(to+.6t)
dx
1x(to)
or
=
1to+.6t to
=f x(to + ~t)- x(to) = f
f
(x, u, t)lto dt to+.6t
(x, u, t)l 10
(x, u, t)l 10
1 to
dt
~t
The error resulting from the first-order approximation depends on the time step ~t, and this time interval must be selected to be sufficiently small to yield acceptable results. x(t)
--- ---·
x(to) +x!M x(t0 + /M)
Integration error
x(to)
t0
t0 +At
Tune Figure 11.1: Illustration of first-order integration using the Euler method. The computed response x (to) + x (to) At in general contains an error that depends on the size of the time step At ..
The system response at a series of discrete time intervals T = k ~t, where k = 1, 2, 3, ... , may be obtained by repeatedly applying the Euler integration rule, first over the time period 0- ~t using the given initial condition and then successively for the k steps, at each interval using the value of x computed from the previous interval as the initial condition for the new interval. Then, if tn = to+ n ~t, Xn = x(to+ n ~t) and Un = u (to + n ~t), the Euler integration algorithm may be written for n
= 1, 2, 3, ... , N
(11.11)
370
Solution of System Response by Numerical Simulation
Chap. 11
The application of Euler integration in a sequence of steps to compute a total system response is illustrated in the following example. Example 11.1 A simple model of the coast-down test of a vehicle of mass m with an initial velocity Vo is illustrated in Fig. 11 .2. It is assumed that the retarding forces on the vehicle are proportional to the vehicle velocity, Fb = Bv , and that no other propulsion or braking forces are present. Use Euler integration to determine the velocity of the vehicle as it slows down.
Automobile
Equivalent viscous rolling resistance B
Figure 11.2: Coast-down test of a vehicle. Solution The state equation for the vehicle in terms of the vehicle velocity v is
.
B
v= - - v m
(i)
with the initial condition that at t = 0, v(O) = v 0 • We wish to compute the vehicle velocity at discrete time intervals by integrating the state equation. The Euler method states that at time t + D. t the velocity is (ii) (iii)
For this system, it is convenient to define a new quantity m B
r=-
(iv)
which has units of time and is defined in Chap. 9 as the system time constant. In terms of the time constant,
v• (t + D.t)
= (I
D.t) v• (t) - 7
(v)
Sec. 11.2
Solution of State Equations by Numerical Integration
371
Using Eq. (v), the vehicle velocity may be computed as a function of time from the initial velocity vo. Starting at t = 0, successive values of velocity are
v(0) = vo v.. (l::t.t) = ( 1 - l::t.t) vo = ( 1 - l::t.t) vo
v(2/.l.t) =
7 7 2 ( l.l.t) . !::t.t) ( 1 - 7 0 (t) = 1 - 7 vo
(vi)
3
llt)..v (2t) = ( I - 7 llt) vo v.. (3/lt) = ( l - 7
The value of v at time k llt resulting from successive integrations may be written
v(k llt) = ( 1 - 7l::t.t)k Vo
(vii)
The results of the Euler integration are tabu1ated in Table 11.1 and plotted in Fig. 11.3, for 10 s and an initial velocity of vo = 10 a vehicle with a time constant of t' = mI B mls using integration time intervals of llt = 0.1, 1, 2, and 4 s. In addition the analytical solution to the first-order state equation, which is derived using methods described in Chap. 9 as v (t) = v0 e-tlr, is included for comparison.
=
TABLE 11.1:
Results of Euler Integration of the Coast-down Model of an Automobile, with State Equation v = -10v and Initial Condition v(O) 10 m/s•
=
Tune
Exact Solution
Computed for !:l.t 0.1
=
Computed for !:l.t = 1
Computed for !:l.t=2
Computed for !:1.1 =4
0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0
10.00 8.187 6.703 5.488 4.493 3.679 3.012 2.466 2.019 1.653 1.353
10.00 8.179 6.690 5.472 4.475 3.660 2.994 2.449 2.003 1.638 1.340
10.00 8.100 6.561 5.314 4.305 3.487 2.824 2.288 1.853 1.501 1.216
10.00 8.000 6.400 5.120 4.096 3.277 2.621 2.097 1.678 1.342 1.074
10.00
• The results are tabulated for time steps !:1.1
6.000 3.600 2.160 1.296 0.778
= 0.1, l, 2, 4 s.
The data in Fig. 11.3 demonstrate that as the integration time interval At is increased and approaches the system characteristic time, the accuracy of the numerical solution decreases. For example, Eq. (vi) shows that for llt = t', the Euler method indicates that the velocity is zero after one time step and remains at zero for all 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 v
Use simulation to detennine the mass position response. As the response velocity approaches zero, does the nonlinear representation of the fri ction cause any numerical difficulties? (d) Compare the responses in (b) and (c) and comment on the waveform shapes. 11.12. A central issue in designing the main landing gear for an aircraft is its ability to absorb the shock at wheel contact during landing. A simple lumped model of a landing gear is shown in Fig. 11.11. This simple model assumes that the aircraft is descending and the tire has just come in contact with the pavement so that the initial vertical velocities of the wheel mass and body mass are equal, and have the values v1 (0) = 112(0) = vo. (a) Derive a state space model for the system that is valid after wheel contact. (b) For a small aircraft m 1 = 1500 kg, m2 = 20 kg, K1 = 2,000 N/m, and K2 = 20,000 N/m. (c) We wish to select a value of damping B which yields a good response that minimizes the shock loads to the aircraft Thus, we wish to select a value of B that keeps the deacceleration level experienced to a reasonable value while allowing the shock absorbing wheel strut to deflect and absorb the aircraft weight.
Chap. 11
391
Problems
Aircraft mass
Wheel strut
Wheel mass
Tire stiffness
Figure 11.11:
A simple aircraft landing gear model.
i. One proposal bas been to select a value of damping B that would yield a critical damping ratio for the second-order aircraft mass and shock strut system (m 1, B. K 1) assuming the
wheel mass is fixed, that is, v2 = 0. Determine the value of B required to achieve this critical damping ratio. Using numerical simulation, determine and plot the aircraft body acceleration response with this value of B. Assume that at the moment of touchdown the aircraft has a vertical descent velocity of 0.5 rnls. ii. A second proposal is to use a value of B which is 20% of the value used in (i). Compute and plot the acceleration response using this reduced value of B. iii. Compare the acceleration levels of the aircraft body during landing with the two different values of B. Which yields the lowest level of peak acceleration? 11.13. Derive recursive solution equations for the first-order system
x =ax +bu y =ex +du
using (a) a step invariant simulation model (b) a ramp invariant model with a time-step T. Show that the step invariant model accurately predicts the system step response at time T, and that the ramp invariant model accurately computes the system ramp response at time
T. 11.14. In Example 10.1 the matrix exponential for the matrix
A= is shown to be
Derive (a) step (b) ramp invariant
[-2 o] 1
-1
392
Solution of System Response by Numerical Simulation
Chap. 11
simulations of the second-order system
Y=X2
and write out the recursive equations. 11.15. Derive an exponential invariant simulation procedure for the first-order system i = ax + bu; that is, a simulation that is accurate if the input is of exponential form u(t) = K e'AI. 11.16. An electrical circuit is shown with an input waveform in Fig. 11.12. Use the numerical state transition method to compute the system state response to the given waveform. The parameter values are R = 2 0, L = 0.25 mH, and C = 100 J.LF.
V.r(t)
R
3 2
c
1
V.r(t)
L
0 -1
-2 -3 Figure 11.12: An electrical circuit and its input waveform.
(a) Compute the appropriate system P and Q matrices for the numerical state transition matrix method for a time step of 0.00025 s. (b) Represent the waveform as a series of steps at increments of 0.00025 s.
(c) Develop a set of step invariant simulation equations. (d) Compute the response to the input waveform shown in Fig. 11.12. (e) Would you expect that a ramp invariant simulation would give a more accurate response for the given input waveform? 11.17. For the fluid system showninFig.11.13, with parameter values C = 8 m5 /N, I= 2 N-s 2 /m5 , R1 = R2 = 2 N-s/m 5 , use the ramp invariant method to determine the response of the pressure at the base of the tank to the flow input shown.
(a) Determine the system A, B, C matrices. (b) Determine a suitable time step from the input waveform and derive the ramp invariant
Q, R,
and S matrices. (c) Use a computer (or calculator) to compute the response of the system to the given input flow. Assume that the system is initially at rest.
Chap. 11
393
Problems
Tank C
Q(t)
~Longpipe
Valve
R 1.1
Q(t) -
30
1
20
~
10
~
li::
Time (s) 20 Figure 11.13:
40
60
80
100
A fluid storage system and its input flow.
11.18. A part of a high performance drive in an automated manufacturing plant is modeled as a mass
m = 10 kg, coupled with a nonlinear spring with characteristic F = cx 3 , where c = 107 N/m, and a linear damper B = 200 N-s/m, as shown in Fig. 11.14. A step input in force Fs (t) acts on the system, which is initially at rest
Figure 11.14:
A mechanical positioning system with a nonlinear spring.
(a) Derive a set of state equations using mass position and velocity as state variables. (b) Use a computer simulation to determine the response for step inputs of amplitude
i. ION iLlOON (c) Comment on the differences in the response of the system to the two input amplitudes. 11.19. The stability of nonlinear systems may be a function of the initial conditions. A common form of nonlinear damping for second-order systems is described by a nonlinear differential equation of the form d 2y
dt2
=
+ a(l -
dy
byl) dt
+ w~y =
0
=
Assume a= -1, b 1, and Wn 10 rad/s. (a) Write the nonlinear equation in state variable form. (b) Use a numerical simulation package to find the response when y(O) 0.5 and dyfdt timet= 0. (c) Repeat the simulation with initial conditions y(O) = 5.0 and dyfdt = 0 at timet= 0. (d) Comment on the stability in the two cases.
=
= 0 at
394
Solution of System Response by Numerical Simulation
Chap. 11
REFERENCES [1] Press, W. H., Flannery, B. P., Teukolsky, S. A., and Vetterling, W. T., Numerical Recipes: The Art
of Scientific Computing, Cambridge University Press, Cambridge, 1986. [2] Dahlquist. G., and Bjorck, A., Numerical Methods, Prentice Hall, Englewood Cliffs, NJ, 1974. (3] Bathe, K.-J., Finite Element Procedures in Engineering Analysis, Prentice Hall, Englewood Cliffs, NJ, 1982. [4] D'Sousa, F., and Garg, V. K.,Advanced Dyntunics, Prentice Hall, Englewood Cliffs, NJ, 1984. [5] Forsythe, M. A., et al., Computer Methods ofMathematical Computations, Prentice Hall, Englewood Cliffs, NJ, 1977. [6] Edwards, C. H., Jr., and Penney, D. E., Elementary Differential Equations with Boundary Value Problems (2nd ed.), Prentice Hall, Englewood Cliffs, NJ, 1989.
The following are typical of commercially available software packages that are either designed for analysis and simulation of dynamic systems or are general-purpose linear algebra packages with system simulation capabilities: [7] Matlab, The MathWorks, Sherborn, MA. [8] Simulab, The Mathworks, Sherborn, MA. [9] Matrix-X, Integrated Systems, Inc., Palo Alto, CA. [1 0] CTRL-C, Systems Control Technology, Palo Alto, CA. [11] CC, Systems Technology, Inc., Hawthorne, CA.
12
The Transfer Function
12.1
INTRODUCTION
The transferfunction is a widely used representation of the input-output dynamics of linear single-input single-output (SISO) system dynamics [1-2] and system theory [3-7]. The transfer function provides an algebraic representation of the single nth-order differential equation and is closely related to the time domain transferoperatordiscussed in Chap. 7. The two representations are similar in form but differ in their interpretation and use. The transfer function may be developed from a number of starting points, including the Laplace transform (described in Chap. 15) [3-7], the differential operator (described in Chap. 7), and the generalized exponential input method [1-2] described in this chapter. Each of the methods leads to a slightly different interpretation of the transfer function, but all approaches provide a rationale for algebraic manipulations of the system representation. The concept of the transfer function is developed here in terms of the particular solution component of the total system response when the system is excited by a given exponential input waveform of the form u(t) = U(s)est
(12.1)
where s = u + jw is a complex variable with real part u and imaginary part jw and the amplitude U (s) is in general complex. Then the input u(t)
= U(s)e
(12.2)
is itself complex and is shown in Chap. 8 to represent a broad class of input functions of engineering interest including growing and decaying real exponential waveforms as well as growing and decaying sinusoidal waveforms. This derivation of the transfer function provides a basis for determining the steady-state response characteristics of periodic waveforms. In Chap. 14 the transfer function is used as the basis for describing the steady-state 395
The Transfer Function
396
Chap. 12
response to sinusoidal inputs, and in Chap. 15 the restriction to exponential classes of waveforms is removed through use of the Fourier and Laplace transforms.
12.2
SINGLE-INPUT SINGLE-OUTPUT SYSTEMS The transfer function may be developed by considering the dynamics of a system described by a single input-output differential equation, such as
dny dn- 1y dy dmu dm-lu du an -+an-1--+···+at-+aoy = bm -+bm-t--+···+bi-+bou dtn drn- 1 dt dtm drm-l dt
(12.3) where m ~ n and all coefficients are real constants. If this system is subjected to an exponential input of the fonn
= U(s)es'
u(t)
(12.4)
where s is a known constant, the method of undetennined coefficients (Sec. 8.4.2) assumes that the particular solution yp(t) is also exponential in fonn, that is,
Yp(t)
= Y(s)es'
(12.5)
where Y (s) is a complex amplitude to be determined. The relationship between Y (s) and U(s) is found by substituting the exponential input, Eq. (12.4), and the assumed fonn of the response, Eq. (12.5), into Eq. (12.3), obtaining
dn-1 d dn Y (s)e st +an-I - d1 Y(s)e...sf + · · · + a1 -d Y(s)e st + aoY(s)est an -d p pt m dm-l d d bm-d U (s)eSf +bm-1-d1 U(s)e Sl +···+bt-d U(s)e Sl +boU(s)eSf tm rmt
(12.6)
=
or
+ an-ISn-l + ... + atS + ao) Y(s)esf = (bmsm + bm-JSm- 1 + · · · + bts + bo) U(s)esf
(ansn
(12.7)
The transfer function H (s) is defined to be the ratio of the response amplitude Y (s) to the input amplitude U (s) and is
H(s)
= Y(s) = bmsm + bm-1Sm- 1 + · · · + bts + bo U(s)
ansn
+ an-lsn- 1 + ... + atS + ao
(12.8)
The transfer function, Eq. ( 12.8), is an algebraic. rational function of the variable
s. The numeratQr and denominator p.olynomials are formed by replacing the derivatives in the differential equation with corresponding powers of s, with the left-hand side forming the denominator and the right-hand side the numerator. There is, therefore, a direct one-toone relationship between the differential equation and the transfer function. The transfer
Single-Input Single-Output Systems
Sec. 12.2
3fJ7
function may be considered an alternate input-output system representation containing all the information of the system dynamics of a classically formulated system. Figure 12.1 shows the input-output relationship in block diagram form. For a given value of the exponent s the relationship between the two amplitudes, Y (s) and U (s), is completely defined by the differential equation coefficients. For a physical system all the coefficients are real, so if s is real, the value of H (s) is real, but if s is complex, H (s) is in general itself complex. Unear SISO system u(t); U(s)e" ___...
u(); .O:IS
Y(s) U(s)
1---...,._ Yp(t) ;
Y(s)e" ; H(s)U(s)e"
Figure 12.1: Block diagram fonn of ttansfer function representation for a single-input single-output system.
Example 12.1 The mechanical mass-spring-dashpot system shown in Fig. 12.2 has a differential equation
m
d2 dt
d
- 2 Vm
+ B -dt Vm + K Vm =
d
-d F(t) t
Find the transfer function relating the velocity of the mass Vm(t) to the applied input force F(t). Solution The transfer function may be found by assuming an exponential input F(t) F(s)e'' and a particular solution ttm(t) = Vm(s)e":
(ms2 + Bs + K} Vm(s)e'' = F(s)se''
= (i)
so
H (s)
Vm(S)
S
= -F-(s-) = -m-s2_+_B_s_+_K_
(ii)
F(t)
F(t)
z (a)
(b)
Figure 12.2: A second-order mass-spring-dashpot system. (a) The system structure, and (b) its linear graph.
The Transfer Function
398
Chap.l2
The transfer function, as defined here, is derived from the particular solution corresponding to an exponential input It represents the differential equation, however, to completely describe the system response the initial conditions must be included 12.3
RELATIONSHIP TO THE TRANSFER OPERATOR
The transfer function for a SISO system is identical in appearance to the transfer operator, defined in Chap. 7, written in ratio form. While they have the same appearance, the two quantities are interpreted differently. The transfer operator is a time domain-based description of a mathematical (operational) relationship between the input and output of a system expressed in terms of the differential operator S {} and is independent of the form of the input It implies a causal relationship between the system input and output On the other hand, the transferfunction is an algebraic quantity, expressed as a rational function of an algebraic variable s, that describes the particular solution component of the response for an exponential form of the system input It may be manipulated using the standard rules of linear algebra. The use of S {} as the differential operator in Chap. 7 and s as the exponent in the exponential input function to define the transfer function creates the similarity in appearance. When consideration is restricted to linear time-invariant systems, these two system representations are frequently used interchangeably. The similarity results directly from the derivative operator relationship for an exponential waveform (12.9) The transfer function transforms a differential equation based system model into a rational algebraic function, and the normal algebraic rules for function manipulation apply. The transfer function has dimensions established by the units of the physical input and output variables. 12.4
SYSTEM POLES AND ZEROS
The transfer function provides a basis for determining important system response characteristics without solving the complete differential equation. As defined in Eq. (12.8), the transfer function is a rational function in the variable s, that is, H(s) = bmsm
ans"
+ bm-tSm-t + · ·· + btS + bo + an-Isn-t + ·· · + a,s + ao
(12.10)
It is often convenient to factor the polynomials in the numerator and denominator and to write the transfer function in terms of those factors: H(s)
= N(s) = K D(s)
.
(s- Zt)(s- Z2) ... (s- Zm-I)(s- Zm)
(s- PtHs.- P2) ... (s- Pn-tHs- Pn)
(12.11)
where the numerator and denominator polynomials, N(s) and D(s), have real coefficients defined by the system's differential equation and K = bmfan. As written in Eq. (12.11),
Sec. 12.4
System Poles and Zeros
399
the z; 's are the roots of the equation N(s) = 0
(12.12)
and are defined to be the system zeros, and the p; 's are the roots of the equation D(s) =0
(12.13)
and are defined to be the system poles. In Eq. ( 12.11) the factors in the numerator and denominator are written so that when s = z;, the numerator N (s) = 0 and the transfer function vanishes, that is,
and similarly when s = p;, the denominator polynomial D(s) transfer function becomes unbounded:
= 0 and the value of the
lim H(s) = oo
s-p;
All the coefficients of polynomials N (s) and D(s) are real, therefore, the poles and zeros must be either purely real or appear in complex conjugate pairs. In general for the poles, either p; = a1 or p;, Pi+l = a;± jw;. The existence of a single complex pole without a corresponding conjugate pole would generate complex coefficients in the polynomial D(s). Similarly, the system zeros are either real or appear in complex conjugate pairs. Example 12.2 A linear system is described by the differential equation d2y dy du - 2 +5- +6y=2-+1 dt dt dt
Find the system poles and zeros. Solution From the differential equation the transfer function is
2s+ 1 ) - s 2 +Ss +6
H(s -
(i)
which may be written in factored form as 1 s+!2 - 2(s+3)(s+2)
H(s)-
s-(-1) =-21 [s- ( -3)][s2 (-2)]
(ii)
The system therefore has a single real zero at s = - ~ and a pair of real poles at s = -3 and s =-2.
The Transfer Function
400
Chap. 12
The poles and zeros are properties of the transfer function and therefore of the differential equation describing the input-output system dynamics. Together with the gain constant K they completely characterize the differential equation and provide a complete description of the system. Example 12.3
A system bas a pair of complex conjugate poles PI, P2 = -1 ± j2, a single real zero Z1 and a gain factor K = 3. Find the differential equation representing the system.
= -4,
Solution The transfer function is H(s) = K
s
-z
(s- P•>
(i)
= 3 rs- (-1 + j2)][s- (-1- j2)] = 3 s+4 s 2 +2s +S
and the differential equation is d 2y dy du - 2 +2- +Sy = 3 - + 12u dt dt dt
(ii)
12.4.1 The Pole-Zero Plot A system is characterized by its poles and ~rosin the sense that they allow reconstruction of the input-output differential equation. In general, the poles and zeros of a transfer function may be complex, and the system dynamics may be represented graphically by plotting their locations on the complex s-plane whose axes represent the real and imaginary parts of the complex variable s. Such plots are known as pole-zero plots. It is usual to mark a zero location by a circle ( o) and a pole location by a cross ( x ). The locations of the poles and zeros provide qualitative insights into the response characteristics of a system. Many computer programs are available to determine the poles and zeros of a system from either the transfer function or the system state equations [8]. Figure 12.3 is an example of a pole-zero plot for a third-order system with a single real zero, a real pole, and a complex conjugate pole pair, that is,
3s+6
H(s)=
s3
+ 3s2 + 1s +
5
= 3
s-(-2) [s- (-l)][s- (-1- 2j)][s- (-1
+ 2j)]
12.4.2 System Poles and the Homogeneous Response Because the transfer function completely represents a system differential equation, its poles and zeros effectively define the system response. In particular the system poles directly
Sec. 12.4
System Poles and Zeros
401 9(s)
X- pole 0 - zero
s-plane
~----
j2
1
I I
I I
I I I I I I
-2
0
-1
9t(s)
I I I I I I I
~----
-j2
Figure 12.3: The pole-zero plot for a typical third-order system with one real pole, a complex conjugate pole pair, and a single real zero.
define the components in the homogeneous response. It was shown in Sec. 8.3.1 that the unforced response of a linear SISO system to a set of initial conditions is n
Yh(t) =
L C;e
111
(12.14)
i=l
where the constants C1 are determined from the given set of initial conditions and the exponents Ai are the roots of the characteristic equation or the system eigenvalues. Comparison of Eqs. (8.24) and (12.11) shows that in terms of the transfer function the characteristic equation is D(s) = s" + an-)Sn-l + ... + ao = 0 (12.15) and its roots are the system poles, that is, "-i = Pi, leading to the following important relationship: The transfer function poles are the roots of the characteristic equation and also the eigenvalues of the system A matrix (as discussed in Chap. 10).
The homogeneous response may therefore be written n
Yh (t)
=L
CieP11
(12.16)
i::::l
The locations of the poles in the s -plane therefore define the n components in the homogeneous response as follows. 1. A real pole Pi = -u in the left half of the s-plane defines an exponentially decaying component ce-(7' in the homogeneous response. The rate of the decay is determined by the pole location; poles far from the origin in the left half-plane correspond to components that decay rapidly, while poles near the origin correspond to slowly decaying components.
The Transfer Function
402
Chap. 12
2. A pole at the origin p; = 0 defines a component that is constant in amplitude and defined by the initial conditions. 3. A real pole in the right half-plane corresponds to an exponentially increasing component C eur in the homogeneous response, thus defining the system to be unstable. 4. A complex conjugate pole pair rr ±jw in the left half of the s-plane combine to generate a response component that is a decaying sinusoid of the form Ae-ur sin (wt + ¢), where A and¢ are determined by the initial conditions. The rate of decay is specified by rr; the frequency of oscillation is determined by w. 5. An imaginary pole pair, that is, a pole pair lying on the imaginary axis ±jw generates an oscillatory component with a constant amplitude determined by the initial conditions. 6. A complex pole pair in the right half-plane generates an exponentially increasing oscillatory component. These results are summarized in Fig. 12.4. ~(s)
~----- ---- --
- - --
Stable region - -- - - - -
- -- - - - Unstable region - -- - -
Figure 12.4: The specification of the form of components of the homogeneous response from the system pole locations on the pole-zero plot.
Example 12.4 Comment on the expected form of the response of the system with a pole-zero plot shown in Fig. 12.5 to an arbitrary set of initial conditions. Solution The system has four poles and no zeros. The two real poles correspond to decaying exponential terms C te - 31 and C2 e-0 · 11 , the complex conjugate pole pair introduce an oscillatory component Ae- ' sin (21 +¢),and so the total homogeneous response is (i)
Sec. 12.4
System Poles and Zeros
403 9(s)
¥-------
j2
I
s-plane
I
I I
I I I I
I I
I I
-3.0
-1.0
-0.1 0
9t(s)
I I I I I I
I I
I I
I I
~-------
Figure 12.5:
-j2
Pole-zero plot of a fourth-order system with two real and two complex conjugate poles.
Although the relative strengths of these components in any given situation is determined by the set of initial conditions, the following general observations may be made: 1. The term e-Jt, with a time constant T of 0.33 s, decays rapidly and is significant for only approximately 4T or 1.33 s.
+ rfJ) defined by the complex conjugate pair and exhibits some overshoot. The oscillation will decay in approximately 4 s because of the damping term.
2. The response bas an oscillatory component Ae-t sin(2t
e_,
3. The term e-O.lt, with a time constant T = 10 s, persists for approximately 40 s. It is therefore the dominant long-term response component in the overall homogeneous response. The pole locations of the classical second-order homogeneous system d2y
dt2
+ 2~ Wn
dy
dt
2
+ WnY = 0
(12.17)
described in Sec. 9.3 are given by PI• P2
= -~Wn ±wn~
(12.18)
If ~ ~ 1, corresponding to an overdamped system, the two poles are real and lie in the left half-plane. For an underdamped system, 0 ~ ~ < 1, the poles fonn a complex conjugate pair (12.19) and are located in the left half-plane, as shown in Fig. 12.6. From this figure it can be seen that the poles lie at a distance Wn from the origin and at an angle ± cos- 1 (~) from the
The Transfer Function .
404
Chap. 12
negative real axis. The poles for an underdamped second-order system therefore lie on a semicircle with a radius defined by wn and at an angle defined by the value of the damping ratio~.
5(s)
t decreasing
s-plane
+jw,~
t decreasing
t --+ 0
Figure 12.6: Definition of the parameters w11 and r for an underdamped second-order system from the complex conjugate pole locations.
12.4.3 System Stability The stability of a linear system may be determined directly from its transfer function. Ann thorder linear system is asymptotically stable only if all the components in the homogeneous response from a finite set of initial conditions decay to zero as time increases, or (12.20)
where the p; are the system poles. In a stable system all components of the homogeneous response must decay to zero as time increases. If any pole has a positive real part, there is a component in the output that increases without bound, causing the system to be unstable. In order for a linear system to be stable, all its poles must have negative real parts, that is, they must all lie within the left half of the s-plane. An "unstable" pole, lying in the right half of the s -plane, generates a component in the system homogeneous response that increases without bound from any finite initial conditions. A system having one or more poles lying on the imaginary axis of the s-plane has nondecaying oscillatory components iJ_l its homogeneous response and is ~efined to be marginally stable.
Sec. 12.5
12.5
Geometric Evaluation of the Transfer Function
405
GEOMETRIC EVALUATION OF THE TRANSFER FUNCTION
The transfer function may be evaluated for any value of s = a + jw, and in general, when s is complex, the function H (s) itself is complex. It is common to express the complex value of the transfer function in polar form as a magnitude and an angle: H(s)
= IH(s)l ei~
(12.21)
with a magnitude IH (s) I and an angle tjJ (s) given by IH(s)l =
Jm
{H(s)} 2
+!) {H(s)} 2
t/J( ) = tan-• ( ~ {H(s)}) fJt {H(s)}
s
(12.22) (12.23)
where fJt {} is the real operator and ~ {} is the imaginary operator. If the numerator and denominator polynomials are factored into terms s- p; and s- z;, as in Eq. (12.11), H(s)
=K
(s- Zl)(s- Z2) • • · (s- Zm-tHs- Zm) (s- PI)(s- P2) · · · (s- Pn-l)(s- Pn)
(12.24)
each of the factors in the numerator and denominator is a complex quantity and may be interpreted as a vector in the s-plane originating from the point Z; or p; and directed to the point s at which the function is to be evaluated. Each of these vectors may be written in polar form in terms of a magnitude and an angle; for example, for a pole p; = a; + w;, the magnitude and angle of the vector to the point s = a + w are Is-
pd =/(a- a;) 2 + (w- w;) 2
((J)W;) a -a;
L(s- p;) =tan- 1 - -
(12.25) (12.26)
as shown in Fig. 12.7a. Because the magnitude of the product of two complex quantities is the product of the individual magnitudes and the angle of the product is the sum of the component angles (App. B), the magnitude and angle of the complete transfer function may then be written IH(s)l = K
nr=l l(s- Z;)l ni=l l(s- p;)l
m
LH(s)
=L
(12.27)
n
L(s- z;)-
i=I
L L(s- p;)
(12.28)
i=l
The magnitude of each of the component vectors in the numerator and denominator is the distance of the points from the pole or zero on the s-plane. Therefore, if the vector from the pole p; to the points on a pole-zero plot has a length q1 and an angle 8; from the horizontal, and the vector from the zero z; to the point s has a length r; and an angle t/>;, as shown in Fig. 12.7b, the value of the transfer function at the point s is IH(s)l
= Kr 1 ••• Tm q1 · · ·qn
LH(s)
= (c/JI + · · · + t/>m)- (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)
X1(s)
U(s)
=d
[s
-1 ]
et 3 s +2
(iv)
or 3
H 1 (s)
= s2 +2s +-
and det [;
X2(s)-
~]
-1 ] U(s) - det [; s +2
or
(v)
3s
H2 (s)
= s2 + 2s +
Output Response For a single-input single-output system, substitution of the state variable response Xp(t) into the output equations y(t) Cx(t) + Du(t) gives
=
Yp(t)
= Cxp(t) + Du(t)
(12.43)
= (C (sl- A)= H(s)U(s)es'
(12.44)
1
B + D] U(s)~'
(12.45)
and so H(s)
y (s) = _P_ =
U(s)
[C (sl- A)- 1 B +D)
(12.46)
is the input-output transfer function for the system in terms of the matrices A, B, C, and D. The matrix inverse (sl - A)- 1 may be expressed in terms of the detenninant and the adjoint matrix as described in App. A: (sl _A)-I
= adj (sl- A) det (sl- A)
and so H(s)
= Cadj (sl- A) B + Ddet (sl- A) det(sl- A)
(12.47)
(12.48)
State Space-Fonnulated Systems
Sec. 12.7
411
For an nth-order single-input single-output system, the C matrix has one row and n columns, the B matrix has n rows and one column, and the D matrix is scalar, with the result that expansion of the numerator generates a polynomia1 in s. Since the det(sl - A) is also a polynomial in s, the transfer function H (s) is a rational function in s of the same form as
Eq. (12.8): H(s) = Ksm + bm-tSm-l + · · · + b1s + bo sn + an-JSn-l + · · · + Q}S + ao
The characteristic equation is therefore det (sl- A) function are the eigenvalues of the matrix A. ·
(12.49)
= 0, and the poles of the transfer
Example 12.7
The electric network shown in Fig. 12.11 together with its linear graph has a set of state equations
[ 0 [vc(t)] hCt) = -1/L Vou 1(t)=[-1
0]
1/C ] -R/L
[vc(t)] [0] h(t) + 1/L Vin(t)
vc(t)] [iL(t) +[l]Vm(t)
Determine the transfer function relating V001 (t) to the input \1 0 (t).
c
c L
R
(a)
Figure 12.11:
(b)
A second-order electric system. (a) The circuit diagram, and (b) the
linear graph.
Solution For this system.
[si-A]
s
= [ 1/L
-1/C ] s+R/Ls
(i)
and (ii) (iii)
Then H (s) = _C_ad_~..;_(s_I-_A.:....,}B,..---+_D....,.de_t....:..(s_I_-_A~)
det(sl -A)
(iv)
412
The Transfer Function _ -
s2
s(s + R/L) + (R/L)s + 1/LC
Chap. 12 (v)
12.7.2 Multiple-Input Multiple-Output Systems
The scalar transfer function is an input-output system description relating the dynamics of a single output to a given input and is therefore restricted to a single-input single-output system. The state equation description for a system in which the input is a vector u(t) having r components and the output is a vector y(t) having m components is i(t)
= Ax(t) + Bu(t)
(12.50)
y(t) = Cx(t) + Du(t)
In this case the B matrix has n rows and r columns, the C matrix has m rows and n columns, and the D matrix has m rows and r columns. As in the single-input single-output case, we consider the particular response to exponential inp1:1ts, but in this case it is assumed that all the r components of the input are similar, that is, u; (t) = U; (s )~ for i = I, ... , n. Then 1
u(t)
= U(s)e.r
1
where U(s) ·is a column vector of complex input amplitudes. The method of undetermined coefficients assumes the state response Xp(t) to be Xp(t) = X(s)e.r 1
giving the solution X(s)
= (sl- A)- 1BU(s)
(12.51)
The particular response of the output vector YP (t) is found by substituting into the output equation: Yp(t) = Cxp(t)
+ Du(t)
= (C (sl- A)- 1 B +D) Ues
(12.52) 1
= H(s)Ue.r'
(12.53) (12.54)
In this case the quantity H(s) is a matrix of scalar rational functions of s with m rows and r columns known as the transfer matrix. The matrix inverse may be expressed in terms of the adjoint matrix and the detenninant, and the transfer matrix written as Cadj (sl- A) B + Ddet (sl- A) H(s)=--~----------------------det (sl- A)
(12.55)
The denominator det (sl - A) is the characteristic polynomial in s, as in the single-input single-output system, but the numerator has dimensions m x r, each element of which is a
Chap. 12
413
Problems
polynomial in s. The transfer matrix is effectively a matrix of transfer functions:
(12.56)
where the ijthelementis the transfer function between the ith output and the jth input Each of these transfer functions has the same denominator polynomial D(s) = det (sl- A), which is the characteristic polynomial for the overall system, but a separate numerator polynomial N;j (s), and so H(s) = [
~;!)) ·.. ~;!))] !lml.!!l D(.r)
•••
(12.57)
!!mrJ!l D(s)
The individual transfer functions therefore have the same poles or eigenvalues, and therefore the same modal components in the homogeneous response, and differ only in their zeros. PROBLEMS 12.1. Detennine the system transfer function and the system poles and zeros for each of the systems described by the differential equations that follow. Also write each transfer function in factored form, with its poles, zeros, and gain constant. (a) dy +3y
= du +2u
dt dt d2y dy du (b) dt2 + 7 dt + 12y = 2 dt + u (c)
d3y dt3
+ 5d2y + 1dy = du dt 2
dt
dt
12.2. 1\vo first-order systems have transfer functions:
2 s+l
HJ(S)=--
s s+2
H2(s)= - -
(a) Detennine the poles and zeros of both H1 (s) and H2 (s).
=
(b) Consider the parallel combination of the two transfer functions Hp (s) H1 (s) + H2 (s). Determine Hp(s) as a rational function of s, and find the poles and zeros of Hp(s).
(c) Consider the cascade combination of the two transfer functions He (s) = H1 (s) H2 (s). Detennine Hc(s} as a rational function of s, and the poles and zeros of Hc(s). (d) Consider the parallel and cascade combination of two general transfer functions H1(s) = Nl(s)
D1(s}'
Derive the overall transfer function of each combination, and comment on the relationship of · the poles and zeros of H 1(s} and H2 (s) to those of the overall system.
414
The Transfer Function
Chap. 12
12.3. Figure 12.12(a) shows two first-order electrical circuits connected together by an electronic buffer amplifier with the characteristics that it draws no input current from the circuit to which it is connected and the output is a voltage that is equal to the input voltage. In Fig. 12.12(b) the same two circuits are connected together directly. Derive the transfer function relating output voltage to input voltage for each configuration and compare them. Explain any differences. Is it generally true that physically connecting two systems H 1(s) and H2 (s) in cascade will result in a transfer function He (s) = H1 (s) H2 (s)?
(a)
Figure 12.12: Two second-order systems configured from first-order circuits.
12.4. Consider a system represented by the block diagram shown in Fig. 12.13. u(t)
y(t)
Figure 12.13:
System block diagram.
(a) Determine the system transfer function relating y(t) to u(t). (b) Assume the transfer functions are: HJ(S)
2
=s+l
1
H3(s) = - -
s+3
Determine the transfer function, and the poles and zeros, and differential equation for the system. 12.5. Control systems often employ dynamic elements in feedback loops, such as in the system depicted in Fig. 12.14.
u(t)
Figure 12.14:
System with a feedback loop.
(a) Determine the system closed-loop transfer function between y(t) and u(t) in tenns of H 1 (s) and H2 (s).
Chap. 12
415
Problems
(b) If the functions H1(s) and H2 (s) are given by the expressions in Prob. 12.2. detennine the closed-
loop transfer function. How do the poles and zeros of the original transfer functions H 1(s) and H2 (s) appear in the new closed-loop transfer function? (c) The product H1 (s) H2 (s) is called the system open-loop transfer function. How do the open-loop poles and zeros relate to the closed-loop poles and zeros? Does feedback modify the poles and zeros? 12.6. Express the second-order system 3 H(s)---2
- s +3s+2
as a parallel combination of two first-order systems. Derive the unit step response of the second-order system and the two first-order systems. Combine the two first-order step responses and compare the result with the second-order response. 12.7. Make pole-zero plots of the systems with the differential equations given in Prob. 12.1.
X
-5
-2
9t(s)
jl2
-10
9t(s)
X (b)
(a)
X
-jl2
2(s)
2(s)
j2
j7 jS
-1
9t{s)
9t(s)
-jS
X
-}7
-j2
(c)
(d)
Figure 12.15:
Four pole-zero plots.
416
The Transfer Function
Chap. 12
U.S. Fmd the differential equations for systems represented by the pole-zero plots shown in Fig. 12.15. Assume that all systems have a steady-state response y,,(t) = 2 to a unit step input u,(t). 12.9. Consider the unit step response of a stable linear system with transfer function H(s). (a) Express the unit step function u,(t) as a limiting case of an exponential wavefonn. (b) Use the result of (a) to write the system step response Ystep(t) in terms of the transfer function H(s).
(c) Show that the steady-state value of the step response y,{t) is
Yss(t) = lim Ystep(t) =lim H(s). t-+oc
s-+0
12.10. Consider two first-order systems with transfer functions that differ in the location of a zero. H1(s)
s+2 = 2 - -, s+ 3
H2(S)
s-2 = - 2s+3 --
A system with a zero in the right-half s-plane is known as a nonminimum phase system. (a) Make pole-Zero plots for each system. (b) Use the result of Prob. 12.9 to compare the steady-state response of each system to a unit step
input. (c) Derive, and sketch, the step response of the two systems. (d) Does the zero location affect the stability of either system? 12.11. A linear system has a transfer function
s2 +5s+6
Hs _ ( ) -
s3
+ 12s2 + 4ls + 30
(a) Express the transfer function in terms of its poles and zeros.
(b) Fmd the particular solution to this system's differential equation when the input is
L u{t) = ce-3' ii. u(t) = ce-2t where C is an arbitrary constant. (c) Generalize this result to the particular solution of any system with a zero at s = a to an exponential input u(t) Ctfl'.
=
12.12. Two linear systems have transfer functions H s 1
( ) -
s2 + 16 s 3 + 12s2 + 41s + 30
H s s2 + 4s + 29 2( ) - s3 + 12s2 + 41s + 30 (a) Fmd the zeros of the first system and show that the particular solution to the system's differential equation is zero when u(t) = sin4t. (b) Fmd the zeros of the second system and show that the particular solution to the system's differential equation is zero when u(t) = e-21 sin St. (c) Generalize these results to the particular response of any system with a pair of complex conjugate zeros at Z1, Z2 = -a ± j(J) to an input u(t) = e<-a±iw>t.
Chap. 12
Problems
417
12.13. Derive the unit step response of the system H(s)
Assume a < 0, and b
~
= s +b s+a
0. Sketch the response for the conditions
(a) b = 0 (b) a< b (c) a> b
(d) a= b
12.14. At time t
= 0 the two systems s+4
(a) H (s)
= s2 + 6s + 5
(b) H(s)
= s 2 + 4s- 5
s+4
have initial conditions y(t) = 1 and dy I dt = 0. Make a pole-zero plot for each system. Determine the initial condition response of each and comment on its stability. 12.15. Consider the systems with the transfer functions given below. For each system, construct a pole-zero plot, comment on the system stability, and describe the nature of the system response to a set of finite initial conditions. s (a) H(s) = -
s+4
(b) H(s) = (c) H(s)
2
s
2
+ 5s+ 4
2 s -4
= - 2-
s-2 (d) H(s)= s2+2s+4
12.16. A first-order system, at rest at t = 0, with transfer function H(s)
is subjected to an exponential input u(t)
a
= -s+a
= e-ht.
(a) Find the response y(t). (b) What happens to H(s) as the value of the exponent in the waveform approaches the value of the pole? What is H(s)ls=-a? What happens to the particular solution as b-+ -a? (c) Does the overall system response become unbounded when b =-a? Consider limb-.-a y(t) in the solution to (a) to find the response of the system to the input u(t) = e-at. (Hint: L'Hospital's rule may be useful.) · 12.17. An n-th order linear system has a causal impulse response n
h(t)
=L
C;el.;t
+ K 8(t)
t ::: 0
i=l
where J...; are then distinct roots of the characteristic equation; Ci and K are constants. (In most physical systems K = 0.) Assume the system to be at rest at time t = 0.
418
The Transfer Function
Chap. 12
(a) Use the convolution integral (Chap. 8) to show that the response to an exponential input u (t) = may be written
es'
= H(s)e' + Yh(t)
1
00
(Hint:
f(x)dx =
1a
f(x)dx
+
1
00
f(x)dx.)
(b) Thus, show the important relationship between the system impulse response and its transfer function
(This fundamental property oflinear systems is a Laplace transfonn relationship, and is explored formally in Chap. 15.) (c) Evaluate the integral of (b) to show that the transfer function may be found in a parallel form directly from the impulse response
C
n
H(s)
=I:-'-+ K s-A.; i=l
Does the integral converge for all s? 12.18. Use the results. of Prob. 12. I 7 to determine the transfer function and differential equations of systems with the following impulse responses:
= 3e-21 h(t) = 2e-5' + 8(t)
(a) h(t) (b) (c) h(t)
(d) h(t) (e) h(t)
= Se-41 + 3e-r
= sin(wt) = Joe-' sin(50t)
U.l9. In an experiment two systems, both initially at rest, are excited by an exponential wavefonn u(t) e-21 for t ~ 0. The responses y(t) are digitally recorded and analyzed using curve fitting software. The results of the curve fitting are
=
System 1:
System2:
s
y(t) = -e- 4'
3
+ -10 e ' 3
Se
- 21
Use the results of Prob. 12.17 to detennine the transfer function and differential equation for each system.
U.lO. In Fourier and Laplace analysis (Chap. I 5) a general waveform u (t) is expressed as a superposition of exponential functions II
u(t)
= L C;e';' i=l
where the C1 are constants.
Chap. 12
Problems
419
(a) Use the results of Prob. 12.17 to find the response of a linear system with transfer function H (s) to this form of input. (b) Use the Euler relationships to express the input u(t) = 3 cos(lOt) as a sum of exponential waveforms. Find the response of the system
5 s+5
H(s)=-
to this input Express your answer as a real function of time. 12.21. Consider the system described by the state equations:
(a) Derive the transfer function relating x 1 (t) to u(t), and determine the system poles and zeros. (b) Derive the transfer function relating x 2 (t) to u (t), and determine the system poles and zeros. 12.22. Determine the system poles for the systems with the A matrices given below. Identify and express any real poles in terms of equivalent time constants. Identify and express any complex conjugate roots in terms of equivalent undamped natural frequencies and damping ratios. Write the components (as real waveforms) that would be found in the system's homogeneous response. Comment on the stability of each system. (a) A
= [~I
(b) A= [
~2 J
~4 ~1]
!J
(c)
A= [ ~4
(d)
A=[~2 -2~ ~]1
(e)
A= [
~ ~ ~]
-2 -2 -1 12.23. For each of the following systems System 1:
System 2:
A
= [ ~~ ~ 1 ] ,
B = [ ~] , C = [ 1 0]
(a) Determine the system eigenvalues and characteristic equation. (b) Determine the system transfer function between the output y and the input u, and identify the
poles and zeros. (c) What is the relationship between the eigenvalues and the transfer function poles?
420
The Transfer Function
Chap. 12
12.24. A feedback control system in the configuration shown in Fig. 12.14 consists of a system H 1 (s) described by the following state equations
[ xX2 ] = [ 1
0 -61][.x•] + [-1] 0 u
-8
y = [0 and a system H2(s)
Xl
1]
[:~]
= K.
(a) Determine the transfer function H 1(s) and its poles and zeros. (b) Detennine the complete closed-loop transfer function relating y(t) to u(t).
(c) Plot on the s-plane the locus of the poles of the closed-loop system as the feedback gain K is increased from 0 to oo. (d) What value of K is required to achieve a critically damped system(~ = I)? 12.25. For each of the two systems described below, determine the transfer function between the output y(t) and the input u(t): (a)
(b)
12.26. Determine the system transfer function matrix between the output vector y(t) and the input u(t) for the system described below:
1] [-60 -1][.x1] -5 + [1] 2 u
[ xX2 =
[ ~] = [ ~
X2
~] [::]
REFERENCES [I] Shearer, J. L., Murphy, A. T., and Richardson, H. H., Introduction to System Dynamics, AddisonWesley, Reading, MA, 1967. (2] Reid, J. G., Linear System Fundamentals, McGraw-Hill, New York, 1983. [3] Ogata, K., Modem Control Engineering (2nd ed.), Prentice Hall, Englewood Cliffs, NJ, 1990. [4] Dorf, R. C., Modem Control Systems (5th ed.), Addison-Wesley, Reading, MA, 1989. [5] Karnopp, D. C., Margolis, D. L., and Rosenberg, R. C., System Dynamics: A Unified Approach (2nd ed.), John Wiley, New York, 1990. [6] Kuo, B. C., Automatic Control Systems (6th ed.), Prentice Hall, Englewood Cliffs, NJ,
1991.
Chap. 12
References
421
[7] Franklin, G~ F., Powell, J. D., and Emami-Naeni, A., Feedback Control of Dynamic Systems (2nd ed.), Addison-Wesley, Reading, MA, 1991. [8] Matlab, The MathWorks, Sherborn, MA.
13
Impedance-Based Modeling Methods
13.1
INTRODUCTION The terms impedance and admittance are commonly used in electrical engineering to describe algebraically the dynamic relationship between the current and voltage in electric elements. In this chapter we e?C-tend the definition to relationships between generalized across- and through-vari~bles within an element, or a connection of elements, in any of the energy modalities described in this book. Algebraic impedance-based modeling methods may be developed in terms of the transfer functions described in Chap. 12, the linear operators defined in Chap. 7, or the Laplace transform introduced in Chap. 15. All three methods rationalize the algebraic manipulation of differential relationships between system variables. In this chapter we have adopted the transfer function as the definition of impedance-based relationships between system variables. The impedance-based relationships between system variables associated with a single element may be combined to generate algebraic relationships between variables in different parts of a system [1-3]. In this chapter we develop methods for using impedance-based descriptions to derive input-output transfer functions, and hence system differential equations directly, in the classical input-output form.
13.2
DRIVING POINT IMPEDANCES AND ADMITIANCES Figure 13.1 shows a linear system driven by a single ideal source, either an across-variable source or a through-variable source. At the input port the dynamic relationship between the across- and through-variables depends on both the nature of the source and the system to which it is connected. If the across-variable "'in is defined by the source, the resulting source through-variable Fm depends on the structure of the system; conversely, if the through-variable Fin is prescribed by the source, the across-variable "'in at the port is defined by the system. In either case a differentia] equation may be written to describe the dynamics
422
Sec. 13.2
Driving Point Impedances and Admittances
423
Source Figure 13.1: Definition of the drivingpoint impedance of a pon in a system.
of the resulting variable, and it is possible to define a transfer function that expresses the dynamic relationship between the dependent and independent input variables. For a system driven by an across-variable source "in (s )e1 the resulting particular solution Fin (s )est for the through-variable is defined by the transfer function Y (s): Fin(s) = Y(s) 'i-'in(s)
(13.1)
where Y (s) is defined to be the generalized driving-point admittance, the input admittance, or simply the admittance of the system. Similarly, for a system driven by a through-variable source, a transfer function, written Z(s), defines the resulting across-variable particular solution: "io(s) = Z(s)fin(s)
(13.2)
The transfer function Z(s) is defined to be the generalized driving-point impedance, the input impedance, or more usually, the impedance of the system. Both Z (s) and Y (s) are properties of the system, and can be used to define a differential equation relationship between "in and fin· From Eqs. (13.1) and (13.2) it can be seen that
Z(s)
= ltin(S)
Y(s)
and
Fm(s)
= Fin(s) "in (s)
(13.3)
and while they have been defined in tenns of different causalities, the impedance and admittance are simply reciprocals:
1 Y(s) = Z(s)
(13.4)
As defined in Chap. 12, any transfer function is a rational function in the complex variable s; therefore, if Z(s) = P(s) Q(s)
where P(s) and Q(s) are polynomials ins, then the admittance is Y( ) = Q(s) s
P(s)
Impedance-Based Modeling Methods
424
Chap. 13
Example 13.1 Find the input impedance and admittance of the first-order electric circuit shown in Fig. 13.2, consisting of a series-connected inductor L and resistor R. Solution Assume that the system is driven by a voltage source Vm (t) as indicated in the linear graph shown in Fig. 13.2b. The continuity condition applied to the input node requires that l~n(t) iR(t), and the resulting differential equation is
=
L dl~n 1 - - + / i n = -Vm(t)
R dt
(i)
R
The admittance transfer function is found by substituting s for the derivative and rearranging (Chap. 12): Y(s)
= lm(s) = Vm(s)
1/R
(ii)
(L/ R)s + 1
Similarly, if the system is assumed to be driven by a current source / 1 (t), compatibility around the loop shown in Fig. 13.2c shows that the tenninal voltage Vm(t) is Vm = VR
+ VL = L
dl·
(iii)
d; +Rim
and the impedance transfer function is Z(s)
~n(S) =- = Ls+ R lm(s)
(iv)
It is easy to show that Z(s) and Y(s) are reciprocals. R
R
I,.(r) R
f
L
, Vrer= 0
• Vrer= 0 (a)
(b)
(c)
Figure 13.2: (a) A series-connected inductor and resistor. (b) driven by a voltage source. and (c) driven by a current source.
13.2.1 The Impedance of Ideal Elements Consider a system that consists of a single ideal element connected to an ideal source as shown in Fig. 13.3. The elemental relationship between the across- and through-variables, as defined in Chap. 3, may be used to define the impedance or admittance of the element
Sec. 13.2
Driving Point Impedances and Admittances
425
V(s)
Source
V(s)
(arbitrary
Z(s)
= F(s)
causality)
Figure 13.3: Definition of the impedance of a single generalized ideal element.
, Vrer=O
The Generalized Capacitance: The elemental relationship for an A-type element, or generalized capacitance C, in any energy domain is (13.5)
where Vc is the across-variable on the capacitance and fc is the through-variable. Because \tin = Vc, and fin = fc, the definition of the impedance as a transfer function results in Z(s)
= Vin(s) = _1 Fin(s)
sC
( 13.6)
and the admittance of the capacitance is Y(s)
The Generalized Inductance:
= sC
(13.7)
For a T-type element, or generalized inductance
L, the elemental equation is
dfL L-=VL dt .
(13.8)
which gives the impedance and admittance transfer functions: Z(s) = sL
(13.9)
1 Y(s) = -
(13.10)
sL
The Generalized Resistance: For any dissipative D-type element R the elemental equation relating the through- and across-variables is an algebraic relationship, v = Rf, and so the impedance and admittance transfer functions are also static or algebraic functions:
= R 1 Y(s) =-
Z(s)
R
(13.11) (13.12)
Impedance-Based Modeling Methods
426
Chap. 13
Table 13.1 summarizes the elemental impedances of theA-type, T-type, andD-typeelements within each of the energy domains. TABLE 13.1:
0
13.3
The Impedance of Ideal Elements•
Energy Modality
Capacitance
Inductance
Resistance
Translational Rotational Electric Fluid Thermal
ljsm ljsJ ljsC lfsC1
s/K s/Kr sL
l/B l/Br
sl
Rf
1/sC,
R R,
The admittance is the reciprocal of the value given.
THE IMPEDANCE OF INTERCONNECTED ELEMENTS
For a system of interconnected lumped-parameter elements, the system input impedance (or admittance) may be found by using a set of simple rules for combining impedances (or admittances) directly from the system linear graph. 13.3.1 Series Connection of Elements
Elements sharing a common through-variable are said to be connected in series. For example, the linear graph shown in Fig. 13.4 consists of a through-variable source fin connected to three branches in series. The driving-point impedance Z (s) for the complete system is specified by the across-variable at the input port "in and the corresponding through-variable Fin. The continuity condition applied to any node in the graph requires that all elements, including the source, share a common through-variable, or /1 = !2 = /3 = Fin. The compatibility condition applied to the single loop in the graph requires that (13.13) The across-variable v; on each branch may be written in tenns of the elemental impedance Z; (s) and the common through-variable: Vin(s) =
/1 (s)ZJ (s) + /2(s)Z2(s) + /3(s)Z3(s)
= Fm(S) [ZI (s) + z2 (s) + z3 (s)]
(13.14)
= Fm(s)Z(s) where (13.15)
is the system driving-point impedance.
Sec. 13.3
The Impedance of Interconnected Elements
427
Figure 13.4: Linear graph of a system with three series·connected elements.
In general, if N branches in a linear graph are connected in series, the equivalent impedance of the group of branches is the sum of the individual branch impedances: N
Z(s) =
L Z;(s)
(13.16)
i=l
The equivalent admittance of the series combination can also be found directly from Eq. (13.13) since ltin(s) = /1 (s)
+
f1 (s)
= fin(s)
/2(s)
+ !J(s)
f2(s)
Y3(s)
[Yt~S) + Y2~s) + Y3~s)]
(13.17)
Fin(s)
= Y(s) so the equivalent admittance of the series elements is
1
1
I
1
-=--+--+-Y(s) f1 (s) Y2(s) f3(s)
(13.18)
and in general for N elements connected in series,
1
N
I
-=L:Y(s) Y;(s)
(13.19)
i=l
When two elements are connected in series, Eq. (13.19) reduces to the convenient form
(13.20) Example 13.2
The second-order mechanical system shown in Fig. 13.5 is driven by a velocity source Yin (t). Use impedance methods to derive a differential equation relating the force at the input to the input velocity.
Impedance-Based Modeling Methods
428
sl~
lllJ
(b)
(a)
Figure 13.5:
Chap. 13
Series-connected mechanical system with (a) system model and (b) linear graph using electrical impedances.
Solution The linear graph in Fig. 13.5 shows that the three elements are connected in series and share a common through-variable. The relationship between the source force and velocity is given by
where Y(s) is the system driving-point admittance. For the mass, spring, and dashpot elements the elemental impedances are (from Table 13.1) 1 s 1 Zm(s) = Zx(s) = Zs(s) = K B ms so the total impedance at the input port is Z(s) = Zm(s)
+ ZK(s) + Zs(s) s
1
1
=-+-+ms K B s2 + (K/B)s + K/m
=
(i)
Ks
The overall admittance could be computed directly from Eq. (13.18), but we note that
1 Y(s)
From the definition of the admittance, 2
(
s
Ks
= Z(s) = s2 + (K/B)s
+ K/m
fin (s) = Y (s) \lin (s)
(ii)
and so
K) fin(s) = KsVin(s) +K B s +;;;
which generates the differential equation 2
d Fm dt 2
+K
dFin B dt
+ K Fi = K dVin m
m
dt
(iii}
13.3.2 The Impedance of Parallel-Connected Elements
The linear graph in Fig. 13.6 shows an across-variable source \10 (1) connected to a parallel combination of three elements. The compatibility equation for any loop in the graph requires
Sec. 13.3
The Impedance of Interconnected Elements
429
that all elements have a common across-variable, that is, continuity condition at the top node requires
VI
= v2 = v3 = Vin' while the
Figure 13.6: A system containing three parallel-connected elements.
fin
= /I + /2 + /3
(13.21)
Using the impedance relationship F(s) = V(s)/Z(s) [Eq. (13.2)] for each of the passive branches,
(13.22)
and so the equivalent driving-point impedance of the system Z (s) is
1
1
1
1
= -Z1-(s)+ - +Z3(s) -Z(s) Z2(s)
(13.23)
Equation 13.23 may be generalized to a parallel connection of N branches in a linear graph:
1
N
1
-=2:Z(s) Z;(s)
(13.24)
i=l
As in the case of series admittances, a convenient form of Eq. (13.24) can be written for two parallel impedances: (13.25) The admittance of a set of N parallel branches in a linear graph may be found by substituting into Eq. (13.21), with the result N
Y(s)
= LY;(s) i=l
(13.26)
430
Impedance-Based Modeling Methods
Chap. 13
Example 13.3 Find the impedance and the admittance of the electric system shown in Fig. 13.7. Solution For the capacitor, the inductor, and the resistor elements the elemental impedances are (from Table 13.1) 1
Zc(s) = sC
ZL(s)
= sL
so the overall impedance is
1
1
1
1
= -Zc-(s-) + -ZL-(s-) + -ZR-(s-)
-Z(-s)
I
1
=sC+-+sL R s 2 + (1/ RC)s + 1/LC = s/C
(i)
or
s/C
Z(s)
= s2 + (1/ RC)s + 1/LC
(ii)
The admittance of the parallel combination is simply the sum of the individual admittances:
Y(s) = Yc(s) + YL(s) 1
+ YR(s)
1
=sC+-+sL R s 2 + (1/ RC)s + 1/ LC = s/C
(iii)
which is the reciprocal of the impedance found in Eq. (ii).
R
c
L
(a)
(b)
Figure 13.7: An electrical system with parallel-connected elements shown as (a) circuit diagram and (b) system impedance graph.
13.3.3 General Interconnected Impedances Impedances that represent combinations of lumped elements may be combined and reduced to a single equivalent impedance using the above ruJes for combining series and parallel combinations.
Sec. 13.3
The Impedance of Interconnected Elements
431
For example, the linear graph in Fig. 13.8 contains four branches and a single source. Each branch in this graph is described by an impedance and may represent a single element or a combination of elements. The two parallel impedances may be combined using Eq. (13.23), and the two resulting series branches may then be combined using Eq. (13.16) to give an equivalent system consisting of two series impedances Z 5 (s) and Z6 (s) as shown in Fig. 13.8b where
These two impedances may then be combined as in Fig. 13.8c, giving the equivalent system input impedance Z (s): Z(s)
= Zs(s) + Z6(s) =
[Zt (s)
(13.27)
+ Zz (s)] [Z3 (s) + Z4 (s)) + Z3 (s) Z4 (s)
Z=Zs+~
/ Vrer = 0 (a)
(b)
(c)
Figure 13.8: System reduction by combining series and parallel impedances showing (a) elemental impedance, (b) first reduction, and (c) reduction to a single impedance.
13.3.4 Impedance Relationships for Two-Port Elements Energy-conserving two-port elements, defined in Chap. 6, are used as transducers between different energy domains. The impedance of a subsystem on one side of a transforming or gyrating two-port element may be "reflected" to the other side using the two-port constitutive relationships. Figure 13.9 shows a transformer and a gyrator, each with a single impedance element Z3 connected to one side. The equivalent impedance for the two cases, as seen from the input port, may be derived as follows.
Transformer: Let the transformer shown in Fig. 13.9a have a ratio TF, defined by the constitutive relationship
[~] = [~ -1~] [~]
(13.28)
Impedance-Based Modeling Methods
432
Chap. 13
/
I
I
I
Vin(t)
cD (b)
(a)
Figure 13.9: The input impedance of two-port elements with a single load elemenL (a) A transformer, and (b) a gyrator.
From the linear graph in Fig. 13.9, the compatibility and continuity conditions give l-1n = VJ, Fin= ft, v3 = v2, and /3 =- /2· Substitution of these conditions into Eq. (13.28) gives
[
Vt (s)]
It (s)
and since by definition Z1 (s)
= [TF 0
0 ][Z3(s)/3(s)] -
-1/TF
(13.29)
j3(s)
=VI (s)//J (s),
21
(s)
= (TF)Z3(s)/3(s) /3(s)/TF
= (TF)
2
(13.30)
Z3(s)
The input impedance of a two-port element is therefore a factor of (TF) 2 times the impedance of the system on the other side.
Gyrator: The input impedance of a gyrator connected to a system with a known impedance is derived in a similar manner, with the difference that for a gyrator there is a proportionality between the across-variable on one side and the through-variable on the other side. Let the gyrator shown in Fig. 13.9b have a ratio GY, defined by the constitutive relationship (13.31)
The compatibility and continuity conditions yield Vin = !3 = - !2, with the result
Vt(S)] [ 0 [ /J(s) = -1/GY
VJ,
Fin = /It
GY] [Z3(s)/3(s)] 0 - /3(S)
V3
=
v2,
and
(13.32)
Sec. 13.3
The Impedance of Interconnected Elements
433
and since Z1(s) = V((S)/fi(S), 21
(s) =
-GY j3(s) -Z3(s)j3(s)JGY
= (GY) 2 __!_
( 13.33)
Z3
= (GY) 2 Y3 The input impedance at one side of a gyrator is therefore a factor of (GY) 2 times the admittance of the load connected to the other side. The gyrator effectively changes the nature of the apparent load. For example, a capacitive element C with an impedance Z = 1/sC connected to one side of a gyrator appears as an equivalent inductive element with impedance Z = s(GY) 2 C when reflected to the other side. Example 13.4
A permanent magnet de motor, modeled as shown in Fig. 13.1 0, drives a load that is a pure inertia J. If the motor produces torque Tm = - Kmim and generates a back emf Vnr = Km Om, find the equivalent electric impedance at the motor terminals. R
~n
+~---~ '-
llsJ
(b)
(a)
· Figure 13.10:
A second-order system consisting of a de motor with an inertial load represented as (a) model and (b) an impedance graph.
Solution The motor is a transforming transducer between the electric and rotational domains, and with the linear graph shown in Fig. 13.10 has a ratio TF = 1/ Km:
= [Km [ ~m] lm 0
0 ] [Om] Tm
-l/Km
(i)
With the substitution TJ = -Tm and Slm = TJZJ, where Z1 = ljsJ, the impedance of the two-port transducer referred to the electric side is
_ Zm-
K;,
Vm _
im
-
sJ
(ii)
The overall equivalent electrical impedance of the motor, as seen from the terminals, is therefore K2
Z=R+sL+.....!!!.
sJ _ LJs 2 + RJs + K,~
-
Js
(iii)
434
Impedance-Based Modeling Methods
Chap. 13
Example 13.5 Find the mechanical impedance as reflected to the piston rod in the translational-hydraulic system shown in Fig. 13.11. Assume that the piston has area A and that the pipe in the hydraulic system has inertance It and frictional losses modeled as fluid resistance R,.
Piston area A
(a)
(b)
Figure 13.11: A fluid system containing a gyrator represented as a (a) fluid model and (b) impedance graph.
=
Solution For the piston vp -qp/ A and F = Ap, which define a gyrator relationship with ratio -1 I A. The hydraulic system impedance is 1
z1 = R1 +sft + -c s I _ 11 c1 s 2 + R1 c1 s + 1 -
(i)
sc,
From Eq. (13.34) the mechanical impedance is
(ii)
13.4
TRANSFER FUNCTION GENERATION USING IMPEDANCES Impedance-based modeling methods provide a convenient way of generating the transfer function of a linear system directly from its linear graph. Consider a single-input singleoutput system with a linear graph containing N branches described by impedances Z;(s), i = I, ... , N, and a single source, either an across-variable source Vs or a through-variable source Fs. A total of 2N system variables are associated with the passive branches: the N across-variables v; and the N through-variables /;. In each branch the impedance, or admittance, provides an algebraic relationship between the through- and across-variables: v;(s)
= Z;(s)/;(s)
or
/;(s)
= Y;(s)v;(s)
(13.34)
In order to solve the system N additional independent equations are needed. If a tree, as defined in Chap. 5, containing K branches is constructed from the linear graph, the remaining N - K branches form the set of links. Compatibility and continuity equations. based on the tree define the additional required equations. It is not necessary that the tree be the system's normal tree, as defined in Chap. 5. The only restriction is on the location of the input source: If the system contains an across-variable source, it must
Transfer Function Generation Using Impedances
Sec. 13.4
435
be represented as a branch tree and a through-variable source must be contained in the links. The N independent linear equations may be fonned by the following steps: 1. On each branch in the graph define either the across-variable or the through-variable as a primary variable. The set of equations is expressed in terms of these N variables, and the remaining N secondary variables are eliminated in the next steps. 2. Generate N - K compatibility equations by placing the links in the tree one at a time and write the resulting loop equation in terms of the across-variable drops around the loop. If any of the across-variables in a compatibility equation are not primary variables, the impedance relationships ofEq. 13.34 are used to eliminate the secondary across-variable. 3. Generate K continuity equations by applying the principle of extended continuity to each open node in the tree and fonn an equation in tenns of the through-variables entering the closed volume around the node. If any through-variable is not a primary variable, it is eliminated by substitution using the admittance relationship in Eq. (13.34).
Each of the N equations generated in the above steps is a linear algebraic equation in the N primary variables. While there is considerable freedom in choosing whether the across- or the through-variable should be the primary variable, it is often convenient (but not essential) to select the system output variable as a primary variable. The equations may be rearranged and written in the form (13.35)
Zx=U
where Z is a square n x n matrix of impedance-based coefficients, xis a column vector of the n primary variables, and U is a column vector with elements related to the system input. Any of the standard algebraic methods for solving a system of linear equations may be used to find the transfer function between one of the primary variables and the input. Example 13.6
Find the transfer function relating the capacitor voltage Vc to the input voltage circuit shown in Fig. 13.12.
Figure 13.12:
Vio
in the electric
(a) A second-order electric circuit. (b) its Jinear graph, and (c) a ttee.
Solution The three impedances shown in the linear graph in Fig. 13.12b are
Z1 =R
Z3
= IfsC
436
Impedance-Based Modeling Methods
Chap. 13
The choice of the three primary variables is somewhat arbitrary. In this case select the voltage v3 (because it is the output variable) and the currents i 1, i 2 • From the tree given in Fig. 13.12c the two compatibility equations are
Vm
=0
(i)
t12 + v3 - Vm
= 0
(ii)
Vt
+ VJ -
and the single continuity equation is (iii)
Eqs. (i)-(iii) may be written in tenns of the primary variables using the impedance relationships
Z1i1 + V3 = \1o Z2i2 + V3 = \1n it+i2-Y3113=0
(iv) (v) (vi)
·where the s dependence has been omitted for convenience. The three equations may be written in matrix form:
1] [i
-~3
a]
:
[
V~n]
= ~m
(vii)
and Cramer's rule (App. A) may be used to solve for v3(s):
(viii)
(ix)
The required transfer function is H(s)
= Vc(s) = v3(s) Vm(s)
(x)
Vm(s)
Za+Z2 ZaZ2Y3 + z, + Z2 R+sL = --,.----RLCs2 + R+sL R+sL 1 ) = ( RLC s 2 + (1/RC)s + 1/LC
=------
(xi)
(xii) (xiii)
and the system differential equation is d 2vc dt 2
1 dvc
1
1 d~n
1
+ RC dt + LC Vc = RC dt + LC Vm
(xiv)
Sec. 13.4
Transfer Function Generation Using Impedances
437
A system containing N passive elements requires the solution of N linear equations if each element is represented as a discrete branch in the graph. There is, however, no restriction on the form of the impedances that may be represented in the branches of a linear graph used to generate the compatibility and continuity equations. The rules for combining series and parallel elements, described in Sec. 13.3, may be used to reduce the number of branches in a linear graph and therefore the number of equations to be solved. Care should be taken, however, not to mask the output variable through any graph reduction. If the output variable is a through-variable, the branch specifying that variable should not be eliminated; if the output is an across-variable, the nodes related to that variable should be retained. Example 13.7 The hydraulic system shown in Fig. 13.13 has a pump, characterized as a Thevenin source with a pressure source Pin and a series resistance Rs, connected to a long pipe with lumped inertance lp and resistance Rp and a vertical walled tank c,. A discharge valve is partially open and is modeled as a linear resistance Ro. Find the transfer function relating the pressure at the bottom of the tank Pc to the source pressure Pin.
Tank
c,
Pipe Thevenin source
Rp
/P
Outlet R0
(b)
(a)
Figure 13.13: (a) A fluid system, and (b) its linear graph.
Solution The system as shown has five passive elements and woulc;l require the solution of a set of five linear equations to generate the transfer function directly. If however, the three series elements Rs. Rp. and lp are combined into a single impedance
Zt = Rs
+ Rp +sip
and the two parallel elements C1 and R0 are combined into an equivalent impedance
Z2=
Ro RnCs + 1
the system may be represented by a reduced linear graph containing just two passive elements, as shown in Fig. 13.14. The reduced system may be represented by just two linear equations. If the primary variables are selected as q1 and p 2 , the compatibility and continuity equations are Q1Z1
+ P2 =Pin
Q.- p2y2 =0
(i) (ii)
which may be written in matrix form: (iii)
.
Impedance-Based Modeling Methods
438
Chap. 13
(b)
(a)
Rgure 13.14: (a) A reduced linear graph for the fluid system. and (b) its tree.
The required output variable Pc
= Pl• and Cramer's rule gives the solution det
P2(s)
= det
[~I
[Z
I
1
(iv)
Pin(s)
= Z1Y2 +I When the values of the impedance Z 1 and the admittance Y2 are substituted,
- (Rs + Rp +sip) (RoCts +I} /Ro +I _ R" 2 - lpCtRos + [lp + (Rs + Rp) RoC!] s + (Rs + Rp + Ru)
(v)
The two-port transducing elements may be incorporated into the procedure by using the constitutive relationships described in Chap. 6. For the transformer there is a direct a1gebraic relationship between the across-variables associated with the two branches, that is,
[;:] =
[~ -l~TF] [~]
(13.36)
where TF is the transformer ratio. Thus, while there are four variables associated with the two branches of the transformer, on1y two of them are independent One across-variable and one through-variable should be chosen as the two primary variables and the elemental relationships ofEq. ( 13.36) used to eliminate the secondary variables from the compatibility and continuity equations. The same rules as specified in Chap. 5 are applied in constructing the tree, namely, that one and only one of the branches of a transformer should be included in the tree.
Sec. 13.4
439
Transfer Function Generation Using Impedances
Similarly, the constitutive equations for a gyrator are
[
VJ ]
It
=
[
0
-1/GY
GY] [
0
V2 ]
(13.37)
/2
where GY is the gyrator ratio. In this case there is a direct algebraic relationship between the across-variable on one branch and the through-variable on the other branch. The two primary variables should therefore be selected as the two across-variables or the two through-variables. The relationships in Eq. (13.37) may be used to eliminate the secondary variables. The tree should contain either both branches of a gyrator or neither of the two branches. Example 13.8 Figure 13.15 shows a model of a moving-coil de voltmeter. The principle of operation is similar to that of a permanent-magnet de motor. A coi1 is wound on a rotating armature and mounted in a magnetic field so that it generates a torque proportional to the current. As the coi1 moves, a back emf is generated that is proportional to the angular velocity of the annature. The annature is modeled as series lumped inductance and resistance elements in a manner similar to the motor. The mechanical side of the meter is modeled as an inertia J. representing the armature, and a rotational spring K. When a constant current is applied, the steady-state deflection of the spring is directly proportional to the current. A series resistor is used to limit the current through the meter.
(b)
(a)
Figure 13.15: (a) A de voltmeter. and (b) its linear graph. Assume that the meter is connected to a voltage source Vin· Find the transfer function relating the angular velocity of the armature OJ to the input voltage ~n· Solution The meter is modeled as a transformer between the electric and rotational domains because there is a direct relationship between the through- and across-variables on the electric and mechanical sides. The constitutive relationships are therefore Om ]· = [ 1I Km [ Tm 0
0
-Km
] [
~m ] lm
where Km is the meter torque constant and the transformer ratio TF = 1I Km. The three series electric elements R.,. R0 , and La may be combined into a single impedance element (i)
440
Impedance-Based Modeling Methods
Cbap.l3
and the parallel mechanical elements J and K may also be reduced to an equivalent impedance
(ii)
generating the reduced Jinear graph and the tree shown in Fig. 13. J6.
(a)
(b)
Figure 13.16: (a) Reduced linear graph, and (b) tree for the moving-coil meter.
Choose as primary variables iJ, i2. 03, and r4 and note that the output variable is OJ 04. The tree shown in Fig. 13.16 generates the following compatibility and continuity equations:
=
VJ
+ Vz =
(iii)
Vin
(iv)
-03 +04 =0 ia- i2 = 0 -T3-T4=0
(v)
(vi)
with the following constraint equations to eliminate the four secondary variables:
v• = z.;.
(vii)
Vz
= Km03
(viii)
T3 T4
= -Kmi2
(ix)
= Y4~
(X)
If these constraints are substituted into Eqs. (ill)-( vi), the equations become
0
_t_
l[~:l [~ l i1
..
Vm
(xi)
Sec. 13.5
441
Source Equivalent Models
Cramer's rule generates the solution
[z, ~
del
[z,
04= del
=
~
0
Km
-1
0 0 -1 Km 0 0 Km -1 0 0 -1 Km 0
-Km Vm
!l Ll
-z.r4- K~ sKm\.'in 2 0 ) (Js + K) +sK~
= (R.f + Ra +sL
(xii)
(xiii) (xiv)
and so (xv) H(s)
13.5
= J Las 3 + J (Rs + Ra) s2 + ( K La + Km2) s + (Rs + Ra) K
(xvi)
SOURCE EQUIVALENT MODELS
In Chap. 4 the conceptofThevenin and Norton equivalent sources was introduced to account for the power limitation of physical sources. In that chapter the observed "droop" in the characteristic of a physical source was modeled by creating an equivalent source containing an ideal source and a dissipative D-type element. The concept ofThevenin and Norton source models may be extended and used as an aid in modeling systems that have a defined load impedance, that is, an impedance that defines the output variable. 13.5.1 Thevenin Equivalent System Model
Thevenin's theorem may be stated as follows: Any linear system of arbitrary complexity excited by a single active source and driving an external load ZL may be modeled as a single across-variable source Vs connected in series with a single impedance element Z0 u1(s).
Figure 13.17 shows the structure of the Thevenin model. Regardless of the internal complexity of the system, the theorem allows the overall system to be reduced to just three elements; the source Ys and two passive impedances Zout and the load Z L. The values of the equivalent source and the series impedance are found as follows: 1. The across-variable source Vs is the "no-load" across-variab1e at the output port obtained when the load impedance ZL is removed from the system. It may be found from the transfer function relating the output across-variable to the input when ZL is disconnected from the system.
442
Impedance-Based Modeling Methods Source impedance
Designated "load" element Source (through- or across-variable)
I
I
v,.
System vL
t
+
Chap. 13
Zout
~or impedallce ZL
Across-variable source impedance (b)
(a)
Figure 13.17: Thevenin equivalent of a system containing a single source and a load element ZL represented by a (a) block diagram model and (b) graph model.
2. The series impedance Zout is the system output impedance, found by setting any source (either through- or across-variable) to zero and determining the driving-point impedance of the system at the output port. · An across-variable source is set to zero by replacing it with a "short circuit"; that is, the nodes to which the source is connected are joined together. Conversely, a through-variable source is set to zero by removing the branch from the linear graph, leaving the nodes intact but creating an "open circuit." The following example serves to demonstrate how Thevenin 's theorem may be used to derive a system model. Example 13.9 A rotational power transmission system consists of a velocity source Qin coupled to a shaft through a flexible coupling with torsional stiffness K. The shaft is supported in a bearing with viscous frictional coefficient B and is connected to a machine tool. modeled as an unknown linear impedance Z L (s). F'md the Thevenin equivalent source model for the shaft transmission system and derive the transfer function for the complete model.. s/K
Flexible Bearing Load
Motor
(a~crolss-variable solu~rce) kling B ZL .Om
0-7///////J)~
J n (r) (b)
(a)
Figure 13.18:
L
Rotational power transmission system: (a) The physical system, and (b) the linear graph.
Solution Figure 13.18 shows the power transmission system. its linear graph. and the Thevenin equivalent system. The Thevenin across-variable source element Os is found by removing the load ZL from the output port and determining the no-load across-variable using the modified · linear graph shown in Fig. 13.19a. The compatibility and continuity conditions on this linear graph yield Os(S)
Zs = Z s+ Z Qin(S) K
(i)
Sec. 13.5
443
Source Equivalent Models
The output impedance Zo is found from Fig. 13.19b by setting the source Oin to zero and computing the impedance seen at the output port with ZL removed. The output impedance is the impedance of the elements Z K and Z 8 in parallel, that is,
Zout
ZsZK = Zs+ZK
(ii)
The Thevenin equivalent system is shown in Fig. 13.19c. When the load impedance ZL is connected, the output angular velocity OL computed from the Thevenin model is (iii)
(iv) (v)
which is the result we seek. Although the values Z B = 1I B and Z K = s I K may be substituted, the impedance of the load Z L must be known in order to find the complete transfer function. Thevenin velocity source s/K
(a}
s/K
(b)
(c)
Figure 13.19: Reduced linear graphs for determination of (a) the Tbevenin equivalent source OJ'. (b) the system output impedance Z0111 , and (c) the Thevenin source equivalent system.
13.5.2 Norton Equivalent System Model Norton's theorem, which is analogous to the Thevenin theorem, states: Any linear system connected to a single external load Z L may be represented by an equivalent through-variable source Fs connected in parallel with an impedance Zout across the output port.
Figure 13.20 shows the structure of a Norton source equivalent model. The difference between the Norton and Thevenin source models lies only in the nature of the assumed source and the series or parallel connection of the impedance element In all respects the systems are equivalent; no measurement at the output can distinguish between them.
lmpedance·Based Modeling Methods
444
Source (through or across-variable)
Chap. 13
Designated "load" element ~or impedance
I
V.s
System
~
l
ZL
r
(b)
(a)
Figure 13.20: Norton equivalent of a system containing a single source and a load element ZL shown as (a) a system and (b) a Nonon source equivalent system.
The values of the source and impedance elements are found as follows: 1. The value of the through·variable source F$ is the value of the through-variable at the output port when the load impedance ZL is reduced to zero. This may be considered the short-circuit output through-variable. 2. The value of the parallel impedance element Zout is identical to that of the Thevenin equivalent source; it is the system output impedance, found by setting all internal sources to zero and measuring the system impedance at the output port. In Norton 1I Zout (s) is often used. source equivalent systems the output admittance Yout (s)
=
The Norton and Thevenin models are equivalent descriptions of the system dynamic be· havior as measured at the output port. However, neither model is a representation of the internal structure of the system. Example 13.10 Find the Norton equivalent source model of the rotational power transmission system in Ex· ample 13.8 and show that it produces the same transfer function as the Thevenin model.
Solution The Norton through-variable source is found by setting ZL = 0, which effectively short-circuits Z 8 • Then the Norton source torque T.s is equal to the through-variable TL in the load branch: (i)
It was shown in Example 13.8 that the system output impedance is
z _ 001
-
ZsZx Zs + Zx
(ii)
The Norton equivalent system is shown in Fig. 13.21. At the single node the continuity equation is (iii)
or (iv)
Chap. 13
Problems
445
A compatibility condition shows that OL
= Oou1, and substituting for Zout gives the result (v)
which is the same as that derived using the Thevenin method in Example 13.8. sK
Norton torque nout(t) Load impedance
:\source
"Short circuit"
Z=O
\\ Ts sK
ZL
(b)
(a)
Figure 13.21:
(a) Reduced linear graphs, and (b) Norton equivalent for the system in
Example 13.8.
PROBLEMS 13.1. Determine the equivalent impedance and admittance of (a) Two identical capacitors of value C J.LF connected in parallel. (b) Two springs of stiffness K 1 and K 2 N/m connected in series.
(c) Two identical mechanical dampers of value B N-s/m connected in parallel. (d) Two identical electrical resistors of value R 0 connected in parallel. 13.2. Determine the equivalent impedance and admittance of (a) Two capacitors, C 1 = 0.1 J.LF and C 2 = 0.5 p.F, connected in series. (b) Two long water filled pipes of length 10 m, diameter 1 em, filled with water connected in parallel. (Ignore fluid resistance.)
(c) Two rotary dampers connected in series; the first produces a drag torque of 1 N-m when its angular velocity is 50 radls; the second produces a torque of 2 N-m at 25 radls. 13.3. For each of the systems shown in Fig. 13.22, construct a linear graph model and determine the system input impedance and the input admittance.
(a)
{b)
Figure 13.22: Two systems.
446
Impedance-Based Modeling Methods
Chap. 13
13.4. Use the results ofProb. 13.3 to derive the following differential equations: (a) Assume the electrical circuit in Fig. 13.22 is driven by a voltage source V (t). Derive a differential equation relating the input current i (t) to the source voltage.
(b) Assume the rotational system is driven by a torque source T(t). Derive a differential equation relating the input shaft angular velocity 0 to the input torque.
(c) Now assume that the rotational system is driven by an angular velocity source O(t). Derive an equation relating the required torque to the input angular velocity. Is the result a differential equation?
13.5. Determine the input impedance of each of the systems shown in Fig. 13.23.
v
F~~K m
~
8
P,
)j)))///77?/7/)/?
(a)
~=J=- C/- - !1= !': :'R='==~92 (b)
R
(a)
(b)
Figure 13.23: Four systems.
13.6. Does the system input impedance uniquely define the system? Answer the question by comparing the input impedance of the two electrical circuits in Fig. 13.24. R
L
L
2R
L
R
(a)
(b)
Figure 13.24: 1\vo electrical R-L systems.
13.7. A mechanical system with input impedance Z(s) is driven by an exponential force source F(t) = e-at, with a > 0. Write expressions for (a) The particular solution for the input velocity. (b) The complete input velocity response.
Chap. 13
447
Problems
=
13.8. A series R - L circuit has an input impedance Z (s) R + s L. Assume R L 100 mH, and that the system is at rest at timet= 0. Determine
=
= 1000 Q
and
(a) The input current response when the input is a voltage Vm(t) = lOe-51 • (b) The input voltage response when the circuit is driven by a current lm(t) = o.se-5'. 13.9. For each of the four systems depicted in Fig. 13.23, use impedance methods to derive a transfer function and differential equation, assuming the following input and output variables: (a) In the translational system (a), the input is a velocity source Vs(t), and the output is the mass velocity Vm(t). (b) In the fluid system (b) the input is a flow source Qs(t), and the output is the flow rate through the outlet valve R2 • (c) In the electrical system (c), the input is a voltage source V.s(t), and the output is the voltage drop across the resistor R. (d) In the rotational system (d), the input is a torque source 1's (t), and the output is the angular velocity of the flywheel J 2 • 13.10. Figure 13.25 shows two simplified models used to study the vertical motion of vehicle suspensions. Each model allows for two inputs: a velocity input representing Vs(t) the effects of the road profile as the vehicle moves, and a force input Fs (t) representing external forces acting directly on the body of the vehicle. For each model
(b)
(a)
Figure 13.25:
1\vo simple vehicle suspension models.
(a) Set V1 (t) = 0 and use impedance methods to find the transfer function relating the mass velocity to the force input F1 (t). Compare the poles and zeros associated with each model. (b) Set Fs (t) = 0 and use impedance methods to find the transfer function relating the mass velocity to the velocity input V1 (t). Compare the poles and zeros associated with each model. 13.11. A motor drive system is shown in Fig. 13.26. A transmission clutch is used to couple the motor to a shaft, with torsional stiffness K, that drives a flywheel J supported in a set of bearings B. The motor may be considered to be a velocity source rls.
01-=KT}.f Motor
Clutch
Figure 13.26:
A motor drive system.
448
Impedance-Based Modeling Methods
Chap. 13
(a) Consider the clutch to be disengaged, so that the motor is decoupled from the shaft. Determine the input impedance of the shaft-flywheel system as viewed from the clutch output. (b) Assume that the clutch is fu lly engaged, so that the shaft is rigidly coupled to the motor. Use impedance methods to derive the transfer function between the fl ywheel angular velocity Q 1 and the motor angular velocity n.,. Detennine the system undamped natural frequency w11 and the damping ratio
s.
(c) If the clutch is partially e ngaged it transmits a torque proportional to the angular velocity difference between the motor and the shaft Use impedance methods to find the transfer function between the flywh eel angular velocity Q 1 and the motor ang ular velocity n,. Determine the system undamped natural frequency w 11 and the damping ratio
s.
13.12. A ripple-filter circuit in an electronic power supply is shown in Fig. 13.27. The network is connected to a resistive load at its o utput terminals. R
Figure 13.27:
L
Power supply filter circuit.
(a) Generate a The venin equi valent source model for the filter; specify the equivalent voltage source and source impedance. (b) Generate the Non on equivalent source model for the circuit; specify the equivalent current source and source impedance.
13.13. A small railroad locomotive is sketched in Fig. 13.28 uncoupled from a single car. The locomotive is a force source Fs(t) and has mass m 1 and viscous rolling resistance 8 1 • The coupling unit is modeled as a parallel spring K and damper 8 2 . The railroad car has ma~s m 2 and viscous rolling drag coefficient 8 3 .
Figure 13.28:
A lrain model.
(a) Form a linear graph model for the system. (b) Represent the locomotive (uncoupled) as a Thevenin equivalent model. (c) Use the Thevenin source model to draw a linear graph of the system when the car is coupled to the locomotive. (d) Use the linear graph to derive the transfer function between the car velocity motive force F...(t ).
Vm ,
-
and the loco·
Chap. 13
449
Problems
13.14. Power transmission systems are designed for maximum efficiency; that is they must deliver maximum power to the load. Impedance matching of the load to the source is an important design criterion. An electronic audio amplifier is modeled as a Thevenin source with a resistive output impedance Z0 , 1 = R,. The amplifier is connected to a loudspeaker that exhibits a resistive characteristic R. Make a plot of the power dissipated in the loudspeaker as a function of R. Prove that the power transferred to the load is a maximum when R = R,. (This is known as the maximum power transfer theorem.) 13.15. Millman's theorem states that several Thevenin or Norton sources between common nodes in a system may be combined into a single equivalent source. Figure 13.29 shows three Thevenin equivalent sources connected in parallel. Show that the three sources may be combined into a single Thevenin equivalent V. _ V1Y1 + V2Y2 + V3Y3 eq Y1 + Y2 + Y3 1
Zeq=----Yl
+ Y2 + Y3
where Y; = I I Z;, in the following steps: (a) Convert each source into its Norton equivalent. (b) Combine the parallel through variable sources into a single equivalent source. (c) Combine the three parallel admittances into a single equivalent admittance. (d) Convert the Norton equivalent to a Thevenin equivalent.
Figure 13.29: Three Thevenin sources in parallel. 13.16. A manufacturing plant uses a pair of pumps to draw chemicals into a mixing valve and deliver the mixture to a storage tank. Each pump may be modeled as a Thevenin source. Use Millman's theorem (Prob. 3.5) to combine the pumps, shown in Fig. 13.30, into a single source. Combine the pumps and delivery pipe into a Thevenin source, and consider the storage tank and outlet valve as a load impedance. Find the source and load impedances, and the equivalent Thevenin pressure source. 13.17. In a simple model of blood flow in the human body, the heart is considered as a pump and the arteries as conduits with resistance, capacitance, and inertance. A schematic model is illustrated in Fig. 13.31. (a) Form a linear graph model of the system considering the heart as a flow source, the artery as a flexible tube which can be represented by an inertance and resistance bounded at each end by a capacitance, and the veins as a resistive load. (b) Determine the impedance of the fluid system as viewed by the heart. (c) If the arteries harden, becoming less flexible, the value of each capacitance approaches zero. What is the impedance of the arterial system viewed by the heart in this condition?
450
Impedance-Based Modeling Methods
Chap. 13
Source
Figure 13.30:
A chemical mixing system.
R, I
Figure 13.31:
A simple blood flow model.
13.18. Consider the rotational system shown in Fig. I 3.32 in which an electric motor is used to control a large inertia through a gear train that provides a speed reduction of N : I . When the motor is driven by a current source, its output torque is proponional to the curre nt and may be considered as an input to the mechanical subsystem. Is (t)
+ o---{J
Motor T.s (r). l m
Figure 13.32:
r---,1
A motor drive system.
(a) Form a linear graph model for the system, considering the motor as a torque source T.., the motor inenia lm, the gear train as an ideal transformer with a ratio of N : I, and the fl ywheel inertia J . All damping in the system may be neglected. (b) Determine the effective impedance viewed by the motor; that is, reflect the flywheel inertia through the gear train and combine it with that of the motor to form an equivalent impedance driven by the motor. (c) As seen by the motor, how do the fl ywheel inertia and the motor inenia compare? How does the gear train modify the flywheel inertia viewed by the motor? 13.19. An electromechanical drive system is shown in Fig. 13.33. A motor drives a massive element through a rack and pinion drive, producing a linear velocity. The rack and pinion drive may be considered lossless and massless, while the load mass has a value m. The DC motor produces torque T K,i and has significant armature resistance R and inductance L. The motor is connected to a voltage source.
=
Chap. 13
451
Problems
Figure 13.33:
A linear motor drive system.
(a) Form a linear graph model for the system. (b) Use impedance methods to derive the transfer function relating the mass velocity to the motor input voltage. What are the system undamped natural frequency Wn and damping ratio s?
13.20. Highly tuned electronic circuits, with very little damping, are used in communications systems. These circuits use high quality inductors and capacitors in parallel and series configurations. Figure 13.34 shows two different tuned circuit configurations.
Real
inductor L, RL
(a)
Real capacitor
Real capacitor
(b)
Figure 13.34: Two tuned circuits with (a) an L-C circuit and (b) a C-G-C circuit.
(a) The circuit in Fig. 13.34(a) consists of an inductor and a capacitor in series. In such tuned circuits parasitic resistances associated with the elements dissipate energy and have a strong effect on the performance. Formulate a model of the circuit which includes resistance of the wire in the inductor and leakage resistance in the capacitor. Use impedance methods to derive the system transfer functi on relating the output voltage to the input voltage. Determine the system undamped natural frequency Wn and the damping ratio s. How do the resistances influence the damping ratio? If the residual resistance of the capacitor can be neglected, what is the damping ratio? (b) In many cases the leakage resistance of real capacitors may be ignored, but the inherent resistance of the wire of a inductor is more difficult to control. The circuit of Figure 13.34(b) has been proposed as an alternative tuned circuit that uses only capacitors coupled through an electronic gyrator. Derive the transfer function relating the output voltage to the input voltage for the circuit including the capacitor parasitic resistances. What is the circuit damping ratio? If the resistances are zero, what is the damping ratio and undamped natural frequency?
452
Impedance-Based Modeling Methods
Chap. 13
REFERENCES [ 1] Shearer, J. L., Murphy, A. T., and Richardson, H. H., Introduction to System Dynamics, AddisonWesley, Reading, MA, 1967. [2] Reid, J. G., Linear System Fundamentals, McGraw-Hill, New York, 1983. [3] Domy, C. N., Understanding Dynamic Systems, Prentice Hall, Englewood Cliffs, NJ, 1993.
14
Sinusoidal Frequency Response of Linear Systems
14.1
INTRODUCTION
Many inputs to physical systems are periodic in nature. For example, the forces exerted on marine structures by ocean waves, the acoustic and electric waveforms of music and speech, and mechanical vibrations exerted on structures due to unbalanced rotating elements are all inherently cyclic or periodic in nature and in many cases may be closely approximated by sinusoidal waveforms. Furthermore, as discussed in Chap. 15, any physical repetitive waveform may be represented by an infinite sum of hannonically related sinusoids, and therefore knowledge of the system response to a sinusoidal input provides a basis for determining the response to a broad class of periodic inputs [1-4]. In this chapter we examine in detail the response of linear systems to inputs of the form u(t) =A sin(wt
+ 1/1)
(14.1)
where A is the amplitude of the input, w is the angular frequency, and 1/1 is the phase. This waveform is periodic, with period T = 2tr1w. In Chap. 15 the analysis is extended to a much broader class of system excitation waveforms through the use of Fourier and Laplace methods.
14.2
THE STEADY-STATE FREQUENCY RESPONSE
In analyzing system response to sinusoidal inputs of the form in Eq. ( 14.1) it is generally assumed that the input u(t) has existed for all time t and a solution is sought for the steady-state periodic response after all transient terms in the response have decayed to insignificance. The response of a stable linear system to an input u(t) is the sum of a 453
Sinusoidal Frequency Response of Linear Systems
454
Chap. 14
homogeneous component and a particular solution Yp(t): n
y(t) =
L C;e}.
11
+ Yp(t)
(14.2)
i=l
where n is the system order, A; are the roots (assumed to be distinct) of the characteristic equation, and C; are n constants to be determined from the initial conditions. For a stable system all terms e'A;t will ultimately decay to zero. If the input u(t) is periodic with period T, that is, u(t) = u(t + T) for all t, the particular response Yp(t) is also periodic and therefore persists for all time. For a stable system it is convenient to consider the total response y(t) in two regions: an initial transient region in which the exponential homogeneous components e}.11 must be considered, followed by the steady-state region where the homogeneous solution components have all decayed to the point of becoming insignificant, as shown in Fig. 14.1. In the steady-state region, only the particular response component yp(t) of the response is considered. The sinusoidal system response is defined in the steady-state region: Yss(t)
= ,_00 lim y(t) = Yp(t)
(14.3)
In Chap. 12 the particular response component Yp (t) of a linear system, characterized by a transfer function H(s), to an exponential input u(t) = U(s)ef' is shown to be Yp(t)
= Y(s)e-u = H(s)U(s)es'
(14.4)
The transfer function H (s) provides the basis for development of the steady-state response to sinusoidal inputs.
F(t)
~ vm(t)
Transient
Steady-state
response
response
F<•>--~ ~o+f1H-H+++i++1H+-1+++i++1H+-1+++~
0
Input
Friction B _}
Response
Figure 14.1: Response of a linear system to sinusoidal inputs.
14.3
THE COMPLEX FREQUENCY RESPONSE
The response of a linear system to the complex exponential waveform u(t) = U(s)e'', as described in Chap. 12, may be extended to describe the response to sinusoidal inputs of the
Sec. 14.3
455
The Complex Frequency Response
form ofEq. (14.1). In order to derive the sinusoida1 response from the exponential response we initially sets to be purely imaginary, that is, s = jw, so the input takes the form of a general complex sinusoid: u(t) = U(s)e3 '1s=jw = U(jw)eiwt
(14.5)
The Euler relationships allow complex exponentials to be written as complex sine and cosine functions:
= cos wt + j sin wt
(14.6)
e-itJJI = coswt- j sinwt
(14.7)
eitJJI
With the substitutions = jw, the steady-state exponential response Yss(t) from Eq. (14.4), is also a complex sinusoid with angular frequency jw: Yss(t)
= Y(jw)eiwt =
= Yp(t), as seen
H(jw)U(jw)ejwr
(14.8)
and so the output amplitude Y(jw) is Y (jw) = H (jw) U (jw)
(14.9)
The function H (j w) is defined to be the frequency response of the system and is related directly to the system transfer function: .
Y(jw)
H(jw) = H(s)ls=jw = U(jw)
(14.10)
Equation (12.10) shows that the transfer function of a linear system is a rational function in the complex variable s, expressed as the ratio of a numerator polynomial N (s) and a denominator polynomia1 D (s): H(s) = N(s) D(s)
= bmsm + bm-JSm-l + · · · + btS + bo ansn +an-ISn-I
+ ... +aiS +ao
(14.11)
The frequency response function H(jw) is therefore a similar rational function in the · · variable jw: H(jw)
= N(jw) = bm(jw)m + bm-l (jw)m-l + · · · + bt (jw) + bo D(jw)
anUw)n +an-I (jw)n-I
+···+at (jw) + ao
(14.12)
The frequency response H (j w) is a complex function which can be expressed in terms of its real and imaginary parts: H(jw)
= ffi {H(jw)} + j~ {H(jw)}
(14.13)
456
Sinusoidal Frequency Response of Linear Systems
Chap. 14
where m{} and ~ (} are the real and imaginary operators that extract the real and imaginary parts of a complex expression. It is useful to represent H(jw) in polar form, in terms of a magnitude function and phase function, as shown in Fig. 14.2: H(jw) = IH(jw)l et/J)
(14.14)
where IH(jw)l is the magnitude of the frequency response given by IH(jw)l
= Jcm {H (jw)}) 2 + (~ {H (jw)}) 2
(14.15)
and lfJ (j w) is the phase angle of the frequency response, At.(. ) = LHU ) = tan- 1 (~ (H (jw)}) "' JW w ffl {H (jw)}
(14.16)
HUw>
(}(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)
= ~ (ei<..,+Vl _ .-J<•<+ll'l)
(14.22)
Equation (14.22) shows that the real input u(t) can be expressed as the sum of two complex exponential components, Ut(t) = (A/2j)ei and u2(t) = (-A/2j)e-i. The principle of superposition allows the sinusoidal response to be written as the sum of the responses to the two complex exponential components: Yss(t) = Yssl (t)
+ Yss2(t)
= ~ H(jw)ei
(14.23)
If H(jw) is written in its polar fonn [Eq. (14.18)] and H(- jw) is described by Eq. (14.19), Eq. (14.23) becomes Yss(t) =
~ IH(jw)l (ei(wt+T/I)ejciJ(jw)- e-j(wt+'/l>e-icfJUw>)
=A
IH(jw)l 1j (eilwt+'/l+cfJUw>J _ 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
+1
sin [{J)t
+ tf>(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 << 1/t') the phase shift is approximately zero, at a frequency w = 1/t' the phase shift is -1r /4 rad ( -45°}, and at high frequencies the phase shift approaches a maximum value of -7r/2 rad (-90°). The normalized magnitude, IH{jw}l I Ko, and phase functions are shown against a normalized frequency scale ({J)t') in Fig. 14.5. Example 14.2 The height of fluid in a mixing tank is monitored by an electric pressure transducer that generates a voltage proportional to the fluid pressure at the bottom of the tank. The transducer output is viewed on an oscilloscope, as shown in Fig. 14.6. The normal pressure variations in the tank are in the frequency range of 0-3 Hz. When the system was installed. it was found that the measured output was contaminated by a strong 60-Hz noise signal caused by electromagnetic radiation from nearby equipment. It was suggested that a first-order electric low-pass filter,
461
The Frequency Response of Ftrst- and Second-Order Systems
Sec. 14.5
IHUw)IIKo 1.2
.,u
1.0
~
0.8
= &.
r-----"""'l""----~---""T"'"----r-----....,
-8 a
·a
0.6
Ill)
ell
E
1e z0
0.4 0.2
0.0
2
4 6 Normalized frequency
8
10
WT
LHUw> 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(~ << 1), however, the gain function has a resonant peak in the region of Wn before decaying to zero at higher frequencies. In the region of this resonance the system amplifies the input; that is, the amplitude of the output sinusoid is greater than that of the input As the value of~ approaches zero, the resonant peak becomes narrower and the peak amplitude increases. The resonance associated with a lightly damped system may be used to advantage in some applications, particularly in electronic communications systems where it is often important to design a filter to emphasize sinusoidal components in a particular frequency band. In many other applications, such as mechanical systems, unwanted resonances can result in large amplitude vibrations that must be prevented by careful design or by the addition of specialized vibration suppressors. The maximum value of the magnitude function occurs at a frequency that is dependent on the damping ratio~. This frequency may be found by differentiating Eq. (14.34) with respect to w, equating the result to zero, and solving to find the frequency wp of the maximum, or peak, response, obtaining (14.36) for~ ~ lf.Ji. The resonance peak occurs at a frequency less than the undamped natural frequency Wn, with the shift increasing toward zero as the damping is increased. Substitution of wp into Eq. ( 14.34) gives the value of the amplification at resonance as a function of the damping ratio:
(14.37) Figure 14.9 shows the amplification factor and the frequency of the maximum response for underdamped second-order systems. For values of ~ ;::: 0. 707, no peak occurs in the response and the magnitude monotonically decreases with increasing frequency. The phase response of a second-order system has a phase lag characteristic that approaches zero as the input frequency approaches zero and tends toward a maximum phase shift of -1r rad ( -180°) as the frequency becomes large. At a frequency of w = Wn the phase shift is exactly -7r/2 rad (-90°). The slope and width of the transition region depend on the value of the damping ratio ~. Example 14.3
In hydroelectric generating plants a large water turbine driven by flowing water transfers energy through a shaft to an electric generator. Torsional vibrations in rotational systems consisting of a shaft supporting a large inertia, such as the turbine rotor shown in Fig. 14.10a. can cause severe damage. Any resonances caused by the inertia of the rotor and torsional compliance in the shaft can generate large-amplitude vibrations in the shaft. In order to measure any potential resonances in the turbine, the shaft-rotor system was removed from the turbine and the upper
465
The Frequency Response of First- and Second-Order Systems
Sec. 14.5
IHUw)I / CKolw~)
6.----------,----------.----------.----------. 5
1;l
g
4
~
e 0
'0
3
.E ~ "' E
2
'0
.~
-;;
E
0
z
0
2 Normalized frequency
3
4 wlwn
2
3
4 wlwn
LHUw)
0 -30 co .., ~
..,
"'c c.. "'!:!
-QO
0
-90
0
:a
..c 0..
- 120 -150 - 180 0
Normalized frequency Figure 14.8:
Magnitude and phase frequency response plots of a second-order system.
end of the support shaft was clamped, as shown in Fig. 14.1 0. The turbine rotor was subjected to a sinusoidally varying torque at many frequencies, and the angular displacement of the turbine rotor was measured at each frequency. The results of the tests showed that the system frequency response had a maximum amplitude at a frequency of 4Hz and that at that frequency the response was five times the amplitude of the response at low frequencies of excitation. It has also been estimated that when the shaft-rotor system is installed in the turbine and operated in water, the effective inertia leff of the rotor is increased by 20% and the rotational damping constant Beer is increased by 50% in comparison to the tests conducted with the
Chap. 14
Sinusoidal Frequency Response of Linear Systems
I
I i
!
-
I
!
I
~
i-......
I
I
I
"""' '
!
I
\
'
~' ~
I
I
_l_
0.2 0.3 0.4 0.5 Damping ratio
0.6
0.7 t
0
0.1
I 0.2
I
""
I I
I I
I
0.1
I
I I
I
j
'" \
I
0.3 0.4 0.5 Damping ratio
0.6
0.7
t
(b)
(a) .
Figure 14.9: The resonant peak of an underdamped second-order system frequency response. (a) Magnitude of the peak response, and (b) frequency of the peak.
Shaft clamped
Oj{t)
for testing
K
FlowQ -4-
(a)
(c)
Figure 14.10: The water turbine and shaft of a hydroelectric generator. (a) Overall ~ystem. (b) test setup for measuring resonance, and (c) a simple lumped-parameter model.
shaft-rotor system removed from the turbine. The task is to develop a model for the system and compute the natural frequency and damping ratio of the system under the test conditions. Then we wish to determine the maximum amplification ratio for the system when operating in the turbine and the frequency at which the
resonance occurs. Solution The rotor-shaft system may be represented as an equivalent inertia J (the rotor), a torsional spring K (the shaft), and a rotary damper B (the system damping). The applied test sinusoidal torque is represented by a torque source T(t). For this second-order model, shown in Fig. 14.1 Oc, the transfer function between the angular displacement of the rotor 9 (t) and the
Sec. 14.6
Logarithmic (Bode) Frequency Response PJots
467
input torque T(t) is defined by the differential equation (i)
as
1
H(s)
= J s + 8 s+ K 2
(ii)
The system's undamped natural frequency and damping ratio are (iii)
(iv) For a second-order system described by Eq. (i) with a peak amplification ratio of 5, the system damping ratio may be determined from Eq. (14.37) (or by using the plot in Fig. 14.9) to be ~ ~ 0.1. The system undamped natural frequency (l)n is related to the peak or resonant frequency by Eq. (14.36): (l)n
=
(l)p
../1-
2~ 2
=
8.11'
../1 - 2(0.1)2
= 25.4 rad/s
(v)
As stated above, when the shaft is installed in its housing and operated underwater, the damping constant is increased to 8etr = 1.58 and the inertia becomes Jeff = 1.2J. The installed undamped natural frequency (l)nl is then (vi) and the installed damping ratio is ~~=
1.58
~=
2 1.2JK
1.5 ~~=0.15 v 1.2
(vii)
The expected resonance frequency of the installed rotor is given by Eq. (14.36): (I)P
= 23../ 1 -
2~ 2 = 22.5 rad/s
(viii)
or 3.6 Hz and corresponds to a rotation critical of speed of 216 rpm. At this frequency the peak amplification ratio of the installed system is given by Eq. (14.37) as (ix)
4.6
LOGARITHMIC (BODE) FREQUENCY RESPONSE PLOTS
In system dynamic analyses, frequency response characteristics are almost always plotted using logarithmic scales. In particular, the magnitude function IH (j w) I is plotted against frequency on a log-log scale, and the phase LH(jw) is plotted on a linear-log scale. For
Sinusoidal Frequency Response of Linear Systems
468
Chap. 14
example, in Fig. 14.11 the frequency response functions of a typical first-order system T dyfdt + y = u(t), similar to those discussed in Sec. 14.5.1, are plotted on linear axes and logarithmically scaled axes. It can be seen that while two sets of plots convey the same infonnation, they have a different appearance. The logarithmic frequency scale has the effect of expanding the low-frequency region of the plots while compressing the high frequencies. The logarithmic magnitude plot can be seen to exhibit straight-line asymptotic behavior at high and low frequencies. Similarly, Fig. 14.12 compares the linear and logarithmic versions of the second-order system described in Sec. 14.5.2.1n this case the curves are a function of the damping ratio ~, but all curves on the magnitude plots can be seen to approach a pair of straight-line asymptotes as the frequency becomes very small and as the frequency becomes very large.
In the 1940s H. W. Bode introduced logarithmic frequency response plots as a simplified method for sketching approximate frequency response characteristics of electronic . feedback amplifiers..Bode plots, named after him, have subsequently been widely used in linear system design and analysis and in feedback control system design and analysis [5-8). The Bode $ketching method provides an effective means of approximating the frequency response of a complex system by combining the responses of simple first- and second-order systems.
IHUw>l 1A
1.2
e
1.0
= 8.
C)
"CS
0.8
·~
0.6
B
e
]
IHUw)l
I II iII ~ i II ~ ~ I ~ ·j
0.4
'if
e
0.2
~
0.01 2
4
6
10
8
I
0
t
I
f
i
-15 ~ -45
=
-60
.. llj
I
.I "!II
I
:!
~
I·
liI H
-t ill!fl\ 11Jll]l .
I
I
·~ I
I
I
I
i
I
1
3
10
30
100
(J)T
LHUw)
15
-30
l Ill I r1111 l!lli ·"'~. II I ~~~li'~
Normalized frequency
LHUw)
~
''I
~
0.01 0.03 0.1 0.3
(J)T
Normalized frequency
'Ot)a :!:!.
~
!
I
I I
r\-\.. --+· I
~,
-75
-
_j___
I i
I
-90
0
I
(a)
to
-
i
:!:!.
l-· i Li
~
o
~
!
oo -l
~~~11-~"'1""'1""!~--!"-r"""'j~~'"""1""'!"1"'1"1' i'l~! ii" ~~oi:!'T':tH!'!It-"""-11r-H-tHttt--t-+1rtHt,it-t-+!rt.IHtttUI ~ I I ·il
-
-20 ~--~~14~~1-.f-f'llid!HIHI-+-+-+++-1 1--t,-1-!-lWI*I ·lH!!I -3o .____...._,~·t...u.ul___._ ··+-1-,H·+I-+·-H·l li 111 -40 t--+-+t-t+-!HJ-t-H-irHWt--t-+HtlttJ-··-r H"1 i-th -SO I - ! !.Ill· -60 1--·~~~~--+~+#~1-'~~-++~tl~-++~tl. -70 ,..._,f-·+-11-Hol+~-+-++-tHHI-~.+H-fHI---+-IH--tt#tl 1+111
--r-
~o~-~-~~-4 ++~~--+~+~~~-+~-++#·!H• _l_
10
(I)T
-90 '---1""""'--~llo-"'-'-~l.l.lo..-""--'-..a....u.lll.ll........l.,;;;~0.01 0.03 0.1 0.3 1 3 10 30 100 carr Normalized frequency (b)
Rgure 14.11: Comparison of magnitude and phase characteristics of a first-order system l /(Ts + 1), drawn (a) using linear scales and (b) as Bode plots using the logarithmic scales.
H(s)
=
I
.
i
I I I 2 4 6" 8 Normalized frequency I.
10
i
Sec. 14.6
Logarithmic (Bode) Frequency Response Plots
469
IHUw>l
:fUw)l 1.2 ---...,.....----,-------,------.
~
J.O
,.......,I"'"T"':~-n----~~~"'"":""'"1~""'":'1"".....,...T'"'l""\'~
c:
~ e .g
1.0
=
0.8 1----+·!t-----+----t-0.6
1 - - - - f - + ! - : - - -1----+-----i
0.4
1----11--t\~r-----+----i--
0.2
~~~~·~--+----+--
•t
~...
§
0.1 0.01 0.001
z 0.0001 1
2
1--i~-tHtifl--l-+iH+H11-++HH~++-H+liHI
t-~l+#mt-~~mr~~
L.......JL....L...u.L,Uw...._...L...J....LJ.I,..IJJJ-.-J.....r...uu.JW----L.....l...L...IJ.WI
0.01 0.03 0.1 0.3
3
Normalized frequency
1
3
10
Normalized frequency
LHUw)
"Uw) o-=~-..,....---~--------,------.
o~~~~~~r.rrmm-rnm~
20
-30~~~+--
-60
~i-R~~~
·1Htt--ir-t-1H+#il-+-i-+t+i+tl
~ c:
&.
-90 -120
t--it-+1-ttf.rtt-+-t
e ~
1-----+\\-~~~ _.;;;::,o,..._~-·
-150 -180 L~..:f:~~:E:5:3::=:===1 0 2 3 4 wlwn
-loo -120
-140 -160
Normalized frequency
Normalized frequency
(a)
(b)
Figure 14.12: Comparison of magnitude and phase characteristics of a second-order system H(s) = (J);/(s 2 + 2('cuns + (J);), drawn (a) using linear scales and (b) as Bode plots using the logarithmic scales.
14.6.1 Logarithmic Amplitude and Frequency Scales Logarithmic Amplitude Scale: The Decibel Bode magnitude plots are frequently plotted using the decibel logarithmic scale to display the function IH (j w) 1. The bel, named after Alexander Graham Bell, is defined as the logarithm to base 10 of the ratio of two power levels. In practice the bel is too large a unit, and the decibel (dB), defined to be~ bel, has become the standard unit of the logarithmic power ratio. The power flow P into any element in a system may be expressed in terms of a logarithmic ratio Q to a reference power level Prer:
or
Q = I 0 log 10 ( ; , : ) decibels
(14.38)
Because the power dissipated in a D-type element is proportional to the square of the amplitude of a system variable applied to it, when the ratio of across- or through-variables
Sinusoidal Frequency Response of Linear Systems
470
Chap. 14
is computed, the definition becomes
= 10log10 (~) = 20log 10 (~) Arer Arcf 2
Q
decibels
{14.39)
where A and Arer are amplitudes ofvariables. 1 Table 14.11ists some commonly used decibel values in tenns of the power and amplitude ratios. The magnitude of the frequency response function IH U w) I is defined as the ratio of the amplitude of a sinusoidal output variable to the amplitude of a sinusoidal input variable. This ratio is expressed in decibels, that is,
.
20log 10 IHUw)l
IYUw)l . = 20log10 IUUw)l dectbels
As noted, this usage is not strictly correct because the frequency response function does not define a power ratio and the decibel is a dimensionless unit whereas IH U(J))I may have physical units. TABLE 14.1:
Decibels
Common Decibel Quantities and Their Corresponding Power and Amplitude Ratios Power/Ratio
Amplitude Ratio
-40 -20
0.0001 0.01
0.01
-10
0.1
-6 -3
0.25 0.5 1.0
0.3162 0.5 0.7071 1.0 1.414
0 3 6
10 20 40
2.0 4.0 10.0 100.0 10000.0
0.1
2.0 3.162 10.0 100.0 .
Logarithmic Frequency Scales In Bode plots the frequency axis is plotted on a logarithmic scale. Two logarithmic units of frequency ratio are commonly used: the octave which is defined to be a frequency ratio of 2: 1, and the decade which is a ratio of 10: 1. Given two frequencies c.o 1 and CtJ2, the frequency ratio W = c.o 1/CtJ2 between them may be expressed logarithmically in units of decades or This definition is strictly correct only when the two amplitude quantities are measured across a common D-type (dissipative) element Through common usage, however, the decibel has been effectively redefined to be simply a convenient logarithmic measure of amplitude ratio of any two variables. This practice is widespread in texts and references on system dynamics and control system theory. In this book we have also adopted this convention.
Sec. 14.6
Logarithmic (Bode) Frequency Response Plots
471
octaves by the relationships W = log2 ( : ) octaves
= log 10 ( : ) decades The terms above and below are commonly used to express the positive and negative values of logarithmic values of W. A frequency of tOO radls is said to be two octaves (a factor of 22) above 25 radls, while it is three decades (a factor of 10-3) below 100,000 radls. 14.6.2 Asymptotic Bode Plots of Low-Order Transfer Functions
Bode plots consist of (1) a plot of the logarithmic magnitude (gain) function, and (2) a separate linear plot of the phase shifi both plotted on a logarithmic frequency scale. In this section we develop the plots for first- and second-order terms in the transfer function. The approximate sketching methods described here are based on the fact that an approximate log-log magnitude plot can be derived from a set of simple straight-line asymptotic plots that can be easily combined graphically. Equation ( 12.11) expresses the system transfer function in terms offactored numerator and denominator polynomials: H(s) = K (s- Zt)(s- Z2) · · · (s - Zm-J)(s- Zm) (s- PJ)(s- P2) · · · (s- Pn-I)(s- Pn)
(14.40)
where the Z;, for i = 1, ... , m, are the system zeros and the p;, for i = 1, ... , n, are the system poles. In general a system may have complex conjugate pole and zero pairs, real poles and zeros, and possibly poles or zeros at the origin of the s-plane. Bode plots are constructed from a rearranged form of Eq. (14.40) in which complex conjugate poles and zeros are combined into second-order terms with real coefficients. For example, a pair of complex conjugate poles s;, Si+l = u; ± jw; is written (14.41)
w;
s.
and described by parameters Wn and The constant term 1I is absorbed into a redefinition of the gain constant K. In the following sections Bode plots are developed for the first- and second-order numerator and denominator terms. Constant-Gain Term The simplest transfer function is a constant gain, that is, H (s) IH(jw)l
=K
= K:
LH(jw) = 0
and
(14.42)
and converting to the logarithmic decibel scale, 20log 10 IH(jw)l = 20 log 10 K dB
and
LH(jw) = 0 rad
(14.43)
Sinusoidal Frequency Response of Linear Systems
472
Chap. 14
The Bode magnitude plot is a horizontal line at the appropriate gain, and the phase plot is identically zero for all frequencies.
A Pole at the Origin of the s-Piane A single pole at the origin of the s-plane, that is, H(s) = 1/s, has a frequency response IHUw)l =
_!.
LHUw)
and
(l)
= -~2
(14.44)
The value of the magnitude function in logarithmic units is (14.45)
log IH(jw)l = -log w or using the decibel scale, 201og 10 IHUw)l
= -201og 10 w dB
(14.46)
The decibel-based Bode magnitude plot is therefore a straight line with a slope of -20 dB/decade pas~ing through the 0-dB line (IHUw)l = 1) at a· frequency of 1 rad/s. The phase plot is a constant value of -1r /2 rad, or -90°, at all frequencies. The magnitude Bode plot for this system is shown in Fig. 14.13a.
i I 1
30
- ~~ -
llll'['t 1- I~2~dB/dec llh I rt,~ I
!
·i o -
i
I ll'!f'~ I I!\
·,' ]H11' ! .r•11f It
Jll..l/t II''!
~ I ·11 i Utafil 1 Jj i al 10 t-1--t-iH-ttiiil--t-H•-ti!-miii!t-1-+-:o ~l'"itlII ~i~.pe I I . -C: 0 'fifiiV!"r-;'~dB nu•Rr "'
.
II e
t-l~-H1 iHI--i,H-f-H-!tifl--i~-tH+ii~H-f-+HHI
-20
t-li-H-H-1+1fi---1,H-f--l+lf~H-ii-+Hf'lk-,-+-t+H1+fl ! .,. ...
·;
100
I
I
-W t-l~-!+iti~H-1-tttt
-30
I
30
IIIII
.I
11-,
·ji
,.
liJ!
i'
1-V+~o¥1-t!HHI--+-f-HiiiHH·!1-++++ii:iiHHI---+-1-HHHI
~~~~~~~~~~~~~
~~~~~~~~~~~~~
10
1-
o -to t-1--t-I+H-tHI-/_;,~~...-:.o-t!IHHI---+--HIH-illMt-' --+-rffiiHHI
i -30 t-li-H-tH+ifl--iH-I-!+fH.fl-ii-H-tH+ifl--if-.3''1d--H·-IHI
3
II"' i. •Liiil : HUI
Wt-i~~~-H~fl--i-H~~-H~
0-10
0.01 0.03 0.1 0.3
'!lit·, II
(I)
0.01 0.03 0.1 0.3
1
3
10
Angular frequency (rad/s)
Angular frequency (radls)
(b)
(b)
30
100
(I)
Figure 14.13: Bode magnitude plots for (a) a single pole at the origin of the s-plane, and (b) a single zero at the origin. The phase plots are not shown; they are a constant of -1r /2 for the pole and 1r /2 for the zero.
A Single Zero at the Origin A single zero at the origin of the s-plane, that is, H(s) H U w) with magnitude and phase IHUw)l = w
and
= •
s, has a frequency response 1r
LHUw) = 2
(14.47)
The logarithmic magnitude function is therefore log IH(jw)l =log w
(14.48)
Sec. 14.6
Logarithmic (Bode) Frequency Response Plots
473
or, in decibels, 20log 10 IH(jw)l = 20log 10 w dB
(14.49)
The Bode magnitude plot is a straight line with a slope of +20 dB/decade. This curve also has a gain of 0 dB (unity gain) at a frequency of 1 radls. The phase plot is a constant of 1r /2 rad, or +90°, at all frequencies. Figure 14.13b shows the magnitude Bode plot for this term.
A Single Real Pole The frequency response of a single real pole factor written in the form
I
H(s)= - -
rs + 1
(14.50)
was derived in Sec. 14.5.1: and
LH(jw) = tan- 1(-wr)
(14.51)
The logarithmic magnitude function is logiHUw)l
= -0.5log[(wr) 2 + 1]
(14.52)
or, as a decibel function, 20 log 10 IH(jw)l = -10 log 10 [ (wr) 2 + 1] decibels
(14.53)
When wr << 1, the first term may be ignored and the magnitude approximated by a lowfrequency asymptote: (14.54) which is a horizontal line on the plot at 0 dB (unity) gain. At high frequencies, for which wr >> 1, the unity term in the magnitude expression Eq. (14.53) may be ignored, and the magnitude function is approximated by a high-frequency asymptote: 20 log 10 IH(jw)l ~ -10 log 10 [(w-r) 2 ] = -20 log 10 (w)- 20 log 10 (t') decibels
(14.55)
which is a straight line when plotted against log(w), with a slope of -20 dB/decade. The high- and low-frequency asymptotes, Eqs. (14.54) and (14.55), intersect on the plot on the 0-dB line at a comer or break frequency of w = 1I r. The complete asymptotic Bode magnitude plot as defined by these two line segments is shown in Fig. 14.14a using a normalized frequency axis. The exact response (magnitude) is also shown in the figure; at thebreakfrequencyw = 1/r theactualresponseis20Jog 10 IH(jw)l = -10log 10 (2) = -3 dB.
..
Chap.14
Sinusoidal Frequency Response of Linear Systems
474
1-
=
~
·ac:
0 -10
~~
~
Slopcl..
~
-20 dB/decade
·;
' h
1-
0
-20 ~
-30
~
-=
'
~
~
0
r\" r-
I
I
'
~ ~
I I
~ -40
f
-50 -60 -70 -80 -90 0.01 0.03 O.J
\.
~
I ~
i
N.. 3
0.3
10
30
0.3
3
I
II
]()() (I)T
LHUCJ>) 90 1H - I II!! 80 I! II I !:ji :· 70 ii ii i i II 60 'CQ)D I iii !Ill 50 IIH. ~ ilil I 40 I! ~~I. ~ !I Iii .c 30 Q., 1-'l.'l T I i 20 ! 1 ~ Ti!Hi 10 ·~i 1!11 0 ~i _: ;[.~ I iii!' -10 0.01 0.03 0.1 0.3
ir
~!!'·~~
-!1111
! krill II
I IHW
! 111!11 i IIIJU ! ! IIIII I it nn I liilll
I iii II i !!! I!
p lUlU I Jill! f'! ! Ulll I II liP i ltilli ! !!UIJ
n,,
3
10
Normalized angular frequency
(a)
(b)
!!
IIIII
i i Hi i li!l! I IIIII I ~
30
100 car:
Bode magnitude and phase plots for (a) a single real pole, and (b) a single real zero.
The phase characteristic is also plotted against a normalized frequency scale in Fig. 14.14a. At low frequencies the phase shift approaches 0 rad. It passes through a phase shift of -1r /4 rad at the break frequency w = 1/T and asymptotically approaches a maximum phase lag of -1r /2 rad as the frequency becomes very large. A piecewise linear approximation may be made by assuming that the curve has a phase shift of 0 rad at frequencies more than one decade below the break frequency, a phase shift of -1r j2 rad at frequencies more than a decade above the break frequency, and a linear transition in phase between these two frequencies on the logarithmic frequency scale. This approximation is within 0.1 rad of the exact value at all frequencies. A Single Real Zero A numerator term, corresponding to a single real zero, written in the form H (s) = Ts + 1 (where T is not strictly a time constant), is handled in a manner similar to a real pole. In this case, H(jw) jwr + 1 and the magnitude and phase responses are
=
IH(jw)l
(I)'
'1~--.-~
!
Normalized angular frequency
Agure 14.14:
100
30
10
Normalized angular frequency
I
~
10
-10 0.01 0.03 0.1
100 (I)T
IIl
I II
1111
20
0
I
-40 0.01 0.03 0.1 0.3 1 3 10 30 Nonnalized angular frequency
LHUCJ>) 10 0 -10 -20 ~ -30
c:
30
= JI + (w~)2
and
LH(jw)
= tan-
1
(w~)
(14.56)
Sec. 14.6
Logarithmic (Bode) Frequency Response Plots
475
respectively. In decibels the magnitude expression is (14.57) The low-frequency asymptote is found by assuming that cur<< 1, in which case lim 20log 10 IH(jcu)l
WT-0
= 101og10 (1) = 0 dB
and the high-frequency asymptote is found by assuming that cut: 20log 10 IH(jcu)l
~
>>
(14.58) 1:
20log 10 (cur) = 201og 10 {cu)-201og 10 (r) decibels
when cu
>> 1/t: {14.59)
which is a straight line on the log-log plot with a slope of +20 dB/decade. The break frequency, defined by the intersection of these two asymptotes, is at a frequency cu = 1I 1:, and at this frequency the exact value of IH Uw) I is .J2 or +3 dB. The complete asymptotic Bode magnitude plot using a normalized frequency scale is shown in Fig. 14.14b. The phase characteristic asymptotically approaches 0 rad at low frequencies and approaches a maximum phase lead of 1r /2 rad at frequencies much greater than the break frequency. At the break frequency the phase shift is 7r/4 rad. A piecewise Jinear approximation, similar to that described for a real pole, is also shown in Fig. 14.14b.
Complex Conjugate Pole Pair The classical second-order system (14.60) described in Sec. 14.5.2 has a frequency response
IH(jw)l
= -;:::::====·===== 2 2 2 J[t -
(cu/wn) ]
+ [2~(wfwn)]
{14.61)
and LH(jcu) = tan-l -2~ (cufcun~ 1- (cufcu")
(14.62)
In logarithmic units the magnitude response is
(14.63)
476
1111111~1~
1-
·~ ·ac 0
140 dB/decade
60
0
~c
-20
·a 0
-40
I
1-
,,
40 1-
20 1-
t...,....
0
~~~
0.0) 0.03 0.1 0.3 1 3 10 30 Normalized angular frequency
100 WT
LHUw)
to ~~sr::~s:::rn!TI'~ .. -::-r-·nl_ml_l_ml.llrTiiTIT'i·,nm il ii'tl. -20 -;-.. 1',{),5 pj' t = 0.1 II -40
-20
1-Lo-L-II.UJ.I.LL....-..1......1.....&.
1--t-1++1+1i'ICI-N \I'
~~'
Ill!"
I !Ill
!!1111
I
1i
.·111
~
~~
~0.5 I
i/
40 !--·
20 0
~If'·
1-,...
: ..... ~
rr VII/ rr~
-
120
~100
Normalized angular frequency
:....""·
t=D.lA [2[C .s
. .._,;
140
~
l/
Ill',.. ,
I
,If!.
1
0.01 0.03 0.1 0.3 3 10 30 Normalized angular frequency
LHUw) 180 160
i
.
um
uvr lTIIII
r
on ~CL
1-
-80
Chap. 14
Sinusoidal Frequency Response of Linear Systems
~
t= sV
:.,....
ll
7-.s
1
II
~,;. ~~~!
-
~)
~ ~ '-'l...oo'·
!1..,
l
[I'.I
t=O.I
10 30 0.01 0.03 0.1 0.3 3 Normalized angular frequency
100
(b)
(a)
Figure 14.15: Bode magnimde and phase plots for (a) a complex. conjugate pole pair. and (b) a complex. conjugate zero pair.
The Bode fonns of the magnitude and phase responses are plotted in Fig. 14.15~ with the damping ratio ~ as a parameter. The low-frequency asymptote is found by assuming that w / Wn << 1 and so lim (CLI/CII,a)-0
{20log 10 IH(jw)l) = -10log 10 (1) = 0 dB
(14.64)
The high-frequency response can be found by retaining only the dominant term when W/Wn
>> 1: 2o1og 10 tHUw>t"" -IOiog•o
(:Y
(I)'
!:f
/"'
~~
WI l7 .lj.
100
(14.65)
= -40 log 10 (w) + 40 log 10 (wn) decibels which is a linear function of log 10 w with a slope of -40 dB/decade. The two asymptotes intersect at a break frequency of w = Wn, as shown in Fig. 14.15a. The straight-line asymptotic form does not account in any way for the damping ratio. The resonance peak (for values of ~ < 0. 707) must be sketched in after the asymptotes have been drawn. Figure 14.16,
(1)'1
Sec. 14.6
477
Logarithmic (Bode) Frequency Response Plots
which contains the same data as Fig. 14.9, plots the logarithmic magnitude correction and frequency of the resonant peak as a function of~ [from Eqs. (14.36) and (14.37)]; it is a simple matter to sketch in the resonant peak from these values.
I iI
![
l\
-
I
:\ l \I
~
1-
i
!1-
0
I 0.1
~
~
I I
I
-
""""-!
I
I
0.2
_l I
" ""!"
I
~
! I
i
I
-
r;--
i'ooo...
0.3 0.4 0.5 Damping ratio
I
0.6
0.7 t
0
-
I
I
0.1
(a)
0.2
0.3 0.4 Damping ratio
i J
!
!
!
I
!
I 0.5
0.6
~
~ 0.7 t
(b)
Figure 14.16: (a) Second-order resonant peak value in decibels, and (b) the frequency at which the peak occurs. These curves may be used to estimate corrections to the asymptotic Bode plots for lightly damped pole and zero pairs.
The phase characteristic asymptotically approaches 0 rad at low frequencies, has a phase lag of -1r /2 at the break frequency Wn, and approaches -tr rad at high frequencies. The steepness of the transition is a function of the damping ratio and so must be sketched using the information contained in Fig. 14.15a.
r
Complex Conjugate Zero Pair Bode plots for a pair of complex conjugate zeros can be derived in a manner similar to the conjugate pole pair described above. In this case the block is assumed to have a transfer function (14.66)
and a frequency response IH(jw)l
= J[l- (w/wn) 2] 2 + [2~(w/wn)] 2
LH(jw) = tan-1 2~ (w/wn) 1 - (w/wn) 2
(14.67) (14.68)
The logarithmic magnitude response is (14.69) The asymptotic responses are derived in a similar manner to the complex pole pair; the low-frequency asymptote is lim
(Q)/Cd,)-0
{20log10 IH(jw)l)
= 10log10 (1) = 0 dB
(14.70)
478
Sinusoidal Frequency Response of Linear Systems
Chap. 14
and the high-frequency asymptote is 20Iogl0 IHUco)l"" lOioglO
c:.r
(14.71)
= 40log10 (w)- 40log 10 (wn)
decibels
for w >> Wn
The magnitude response is plotted in Fig. 14.15b. This is effectively an inverse of the characteristic in Fig. 14.15a. There is a "notch" in the response in the region of the frequency Wn, and the depth is a function of the parameter~. The plot has a low-frequency asymptote of 0 dB, a break frequency of w = Wn, and a high-frequency asymptote is a straight line with a slope of +40 dB/decade. The phase characteristic approaches 0 rad at low frequencies, passes through -1r /2 at the break frequency, and shows a maximum phase lead of 1r rad at high frequencies. As above, the slope of the curve in the transition region is dependent on the value of ~. Summary The essential features of the asymptotic forms of the seven components of the magnitude plot are summarized in Table 14.2.. TABLE 14.2:
Summary of Asymptotic Magnitude Bode Plot Parameters for the Seven Basic Blocks
(rad/s)
High-Frequency Slope (dB/decade)
1/T 1/T
0 -20 20 -20 20
cu,.
-40
cu,.
40
Break Frequency Description
Transfer Function
Constant gain Pole at the origin Zero at the origin _Real pole
K 1/s
s
1/(TS + 1) n+ 1
Real zero
w!
Conjugate poles Conjugate zeros
s2 + 2rcu,.s + ~ 1
~ (s
2
+ 2rcu,s + cu~)
n
14.6.3 Bode Plots of Higher-Order Systems
If a system with transfer function H (s) terms in Table 14.2, that is, H(s)
= K N (s) I D(s) is expressed as a product of the
= KN(s) D(s)
(14.72)
= K'[Nt (s) · · · Nm(s)]/[Dt (s) · · · Dn(s)]
where the factors N; (s) are first- or second-order zero terms, the D; (s) are pole terms, and K' is a modified constant factor. For example, lO(s + 3) H(s) = (s +O.S)(s +S)
= K'Nt(s)/[D 1(s)D2 (s)]
479
Logarithmic (Bode) Frequency Response Plots
Sec. 14.6
1 1 1 = 12(3s + 1) x 2s + 1 x 0.2s + 1 When complex numbers are represented in polar form, the magnitude of a product is the product of the component magnitudes and the angle of a product is the sum of the component angles, and so Eq. (14. 72) may be expressed in terms of its magnitude and phase functions: IHUw)l = K' x IN1 (jcu)l x ..• x INmUw)l
X
I
DJ
:jw) Ix · · · x ID.:jw) I (14.73)
LH(j(J)) = LNt (j(J))
+ ... +
1 1 LNm(j(J))- L . ) - ... + L D . Dt (j(J) n (j(J))
(14.74)
The logarithm of a product is the sum of the logarithms of its factors, and so Eq. (14.73) may be written log IH(j(J))I =log K' +log INt (j(J))I + ... +log INm(j(J))I +
I
log D1
~co) I+···+ log ID.~w) I
(14.75)
Equations ( 14.75) and ( 14.74) express the overall magnitude and phase responses as a sum of component responses of first- and second-order elementary "building blocks." In practice Bode plots are constructed by graphically adding the individual response components. Given the transfer function H (s) of a linear system, the following steps are used to construct the Bode magnitude plot:
1. Factor the numerator and denominator of the transfer function into the constant, firstorder, and quadratic terms in the form described in the previous section. 2. Identify the break frequency associated with each factor. 3. Plot the asymptotic form of each of the factors on log-log axes. 4. Graphically add the component asymptotic plots to form the overall plot in straightline form. S. "Round out" the comers in the straight-line approximate curve by hand using the known values of the responses at the break frequencies (±3 dB for first-order sections, and dependent upon ~ for quadratic factors). The phase plot is constructed by graphically adding the component phase responses. The individual plots are drawn, either as the piecewise linear approximation for the firstorder poles or in a smooth form from the exact plot, and then these are added to find the total phase shift at any frequency. Example 14.4 Plot the Bode magnitude and phase plots of a third-order system described by the transfer function
40s+4
H(s)
= s 3 + 2s2 + 2s
480
Sinusoidal Frequency Response of Linear Systems
Chap. 14
Solution The transfer function is rewritten H(s)
=
4(10s + 1) s(s2 +2s
+2)
= 2 (lOs + 1) (;) ( sl +
~ +2)
(i)
with the second-order poles and the first-order zero scaled to be in the standard form of Table 14.2. The four contp9nent tenns are:
=2 A single real zero tenn H 2 (s) = 1Os + 1 with a break frequency of w = 0.1 radls. A single real pole at the origin H3 (s) = 1Is A complex conjugate pole pair H 4 (s) = 2j(s2 + 2s + 2), characterized by Wn = ..J2
1. A constant gain term of H 1(s) 2. 3. 4.
rad/s and a damping ratio of ..J212
The component tenns are plotted and are added together to detennine the total response for a frequency range ofO.Ol to 100 rad/s in the magnitude and phase plots in Fig. 14.17.
Example 14.5 In many types of rotating equipment (engines, motors, pumps) unwanted vibrations are generated. as a result of imbalances in rotating elements. Equipment is often mounted on rubber shock mounts to reduce the transmission of vibratory forces to the support structure. In Fig. 14.18 a pump, of total mass m 100 kg, is shown supported on resilient shock mounts. In this particular pump an imbalance in the rotor generates a sinusoidal vertical force
=
Fu = Fo sin wt where Fo is the force amplitude and w is the shaft angular velocity. The pump normally operates at a constant speed of 900 rpm, however, it is periodically shut down· for maintenance and restarted, with the speed gradually increasing to the nominal speed. A set of shock mounts has been reconimended for the pump installation; we wish to determine the frequency response of the pump and shock mount system in terms of the force transmitted through the mounts to the supporting floor at the operating speed and as the motor is started up or slowed down. Solution Shock mounts have both stiffness and damping characteristics that may be determined by measuring the deflection y(t) when the mount is subjected to a sinusoidal force Fs F sin wt. A set of measurements were made on the mounts, and a frequency response magnitude plot was made as shown in Fig. 14.18c. This measured response is typical of a linear ~st-order system consisting of a parallel spring and dashpot, with a transfer function relating displacement to force, 1 1 1 H(s) Bs + K K -r:s + 1
=
=- - = - - -
which generates the frequency response IH( 'w)l J
= .!_
K J1
1
+ [(B/ K)w]2
(i)
481
Logarithmic (Bode) Frequency Response Plots
Sec. 14.6
20 log 10!H(iw)l
60 40
20
i
II)
0
"0
a
·aco
:E= -20 -40
-60 -80 0.01
0.03
0.3 I 3 Angular frequency (radls)
0.1
10
30
100
LHUoo) 90
I i -r·----r---·· I ' I
45
0
bO 4)
:s 0.1
I'll
=
-45
.c ll.o
-90
I
-1-r-----
-135 -180
0.01
0.03
Figure 14.17:
0.3 1 3 Angular frequency (rad/s)
0.1
10
30
100
Bode plots for the third-order system in Example 14.3.
where K is the equivalent stiffness, B is the damping constant, and the time constant is B I K. From Eq. (i),
t'
=
lim IH(jw)l
w-o
= _!_ K
(ii)
and so the value of the stiffness may be found directly from the low-frequency asymptote on
482
Sinusoidal Frequency Response of Linear Systems
Chap. 14
Motorm= tOO kg Rotor with imbalance
iii' -8 ~
·a~
~ ~
Resilient shock mounts
t
(a)
-110
t
y(t)
I
I.
~
~
-130
1
I I
I
~
1\ I 'i\
J
t-
::s
J:
':~
-115 -120
~-{ I !
i
t-
GJ
Displacement
l ..:;.....
g -125
F(t)
(b)
20 log 101Y(iw}IF(iw}f -105 I
I. so )()() I
I
I
5
2
10 20
\
'I
200 500 1000
(J)
Angular frequency (radls) (c)
Mounts: stiffness K dampingB
Figure 14.18: (a) A rotating pump with a rotor imbalance, (b) the vibration isolation mounts, and (c) a frequency response magnitude plot for the mounts relating · displacement y(t) to input force F(t).
the measured frequency response. From Fig. 14.18c, lim(201og 10 IH(jw)l)
w-+0
~
-108 dB
=4x
w-6 miN
Therefore, K = 25 x 1Q4 N/m. The time constant B 1K is found from the break frequency of the measured frequency response estimated from the curve as 100 rad/s; therefore, T = 11100 = 0.01 s. The corresponding value of the damping constant i.s B = KT = 25 x 1o2 N-slln. A simple second-order model for the vertical motion of the complete pump assembly, in which the pump mass m is supported by the parallel spring-damper of the shock mounts, is shown in Fig. 14.19a and its linear graph in Fig. 14.19b. The transfer function relating the force transmitted through the shock mounts Fs = FK + F8 to the input force F, is Bs+K
H(s)
= ms 2 + B s+ K (Ts
where a>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) << w, it can be used for measuring the base acceleration. 14.6. An electronic amplifier has the following transfer function relating the output to the input voltage:
H s ( ) -
2
s2
X
6
10
+ 20s + 106
(a) Determine the frequency response of the amplifier and sketch the magnitude and phase response functions. (b) The amplifier is used to amplify a signal centered on a frequency of 250 rad/s. For this signal what amplification does the amplifier provide, and what phase shift results? 14.7. A nonminimum phase linear system is defined as one with one or more zeros in the right half of the s-plane. (a) Does a nonminimum phase zero affect system stability? (b) Construct pole-zero plots and Bode plots for the following two systems: H2(s)
s-a = --s+b
(c) By comparing the magnitude and phase plots for the two systems, detennine the effect of having a zero in the right-half s-plane rather than the left s-plane. Comment on the terminology nonminimum phase.
Chap. 14
Problems
491
(d) Consider a system with transfer function H2 (s), with b =a. Compute the magnitude and phase responses, and discuss why this might be called an all-pass filter. 14.8. Construct Bode plots for the frequency responses of the four following systems:
(b)
5s+ 1
= s + 10 H2(s) = s + 10
(a) H.(s)
Ss
+1
Ss + 1
(c) H3(s)
= .2
(d) H4 (s)
= (s + 2)(s2 +lOs+ 100)
s
.
+ 3s+ 1 s+ 10
14.9. Four systems have the pole-zero plots shown in Fig. 14.28. For each system determine from the pole-zero plot
(a) the highest frequency break point in the frequency response and the slope of the high frequency magnitude asymptote (b) the asymptotic high frequency phase response
(c) the low frequency asymptotic magnitude behavior (d) the low frequency phase shift j(J)
j(J)
X
-5
-3
u
-1
-100
jl20
u
0
X
-jl20 (b)
(a)
j(J)
X
j3
-3 -2 -1 0
u
X
-j3 (c)
(d)
Figure 14.28: Four pole-zero plots.
1./
492
Sinusoidal Frequency Response of Linear Systems
Chap. 14
14.10. The Butterworth low-pass filter is commonly used in the design of communication systems. The Nth-order Butterworth filter has its N poles equally spaced on a circle in the left-half s-plane,
n
= 1,2, ... , N
where Pn is the nth pole location and We is the half-power (-3dB) cut-off frequency of the filter. (a) Sketch the pole-zero plot of a second-order Butterworth system. Determine the transfer function, assuming unity gain at low frequencies, and find the damping ratio and the undamped namral frequency (in terms of We). (b) Derive the differential equation representing the third-order Butterworth filter with a -3dB cutoff frequency of 100 Hz, and unity gain at low frequencies. Sketch the pole locations and the Bode magnitude and phase responses for this filter. 14.11. One way to design a high-pass filter is to first design a low-pass system with a cut-off frequency of We, and then to make the substitution
in the transfer function H(s). The resulting high-pass system H'(s) will have the same cut-off frequency. (a) Consider the low-pass system H(s)
=K
s- z (s- Pa>
Apply the transformation to this system to produce the high-pass transfer function H' (s). Determine and plot the poles and zeros of the new system. How many poles and zeros are there? Why is this now a high-pass system? (b) Generalize the approach to a system with n poles and m zeros (m !: n). What does the transformation do to the number of poles and zeros? Where are any additional zeros located? (c) Use this methOd to transform a unity gain second-order system with Wn = 100 rad/s and~ = 0. 707 into a high-pass system with the same cut-off frequency. 14.12. The circuit illustrated in Fig. 14.29 has two R-C filters isolated by an electronic amplifier. The 10, and acts as an ideal voltage source to amplifier draws no input current, has a voltage gain K R2 = R3 10,000 Q, Ct = 1.0 JLF, and C2 = 0.1 JLF. the second filter. For the system R1
=
=
=
___________H_~~.J Figure 14.29: Signal processing circuit
=
(a) Detennine the overall system transfer function H(s) H 1 (s)H2 (s)H3 (s). (b) Plot the magnitude and phase frequency response plots for the system. (c) What is the gain of the system for low frequencies and for high frequencies?
Chap. 14
Problems
493
14.13. A thermocouple is a temperature sensor that generates a voltage proportional to the measured temperature difference from a reference temperature. The thermal capacitance of the probe tip limits its ability to track rapidly changing temperatures. The response is described by a first-order transfer function relating output voltage VT(t) to tip temperature T(t)
K
H(s)=-T:S +]
where K is a constant and 1: = I s. An electronic amplifier is used to amplify and process the output voltage as shown in Fig. 14.30.
T(t)
=======1. . ____,~U~~t)
G:"
Thennocouple
....____Am_p_lifi-•er __
__,~~:
Figure 14.30: Thennocouple temperature sensor and amplifier.
(a) Detennine the frequency response of the thermocouple sensor. At what frequency has the amplitude response rolled off to 0. 707 (- 3 dB) of the low-frequency response? (b) It bas been suggested that the dynamic response of the amplifier can be modified to extend the useful bandwidth of the thermocouple system. A prototype amplifier has a transfer function
where Ka is a constant If T:t is adjusted to be 1 s, what value of 1:2 will double the cut-off frequency of the system? Sketch the combined thermocouple-amplifier frequency response. What range of frequencies will have a response less than 10% ( -20 dB) of the low-frequency response? 14.14. Automotive designers must mount the engine to the car frame in a manner that minimizes the transmission of vibrations. Usually rubber mounts are used to provide some filtering between the engine forces and the frame. A simple model for vibration transmission uses a lumped mass m supported by engine mounts that have both stiffness K and damping B, as shown in Fig. 14.31. The engine disturbance may be modeled as a sinusoidal force, with a frequency related to engine speed, acting at the center of gravity of the engine. F(t)
0m
Engine mass
Engine mount
Figure 14.31: Engine mount model.
494
Sinusoidal Frequency Response of Linear Systems
Chap.14
(a) Derive the transfer function that relates the force transmitted through the engine mounts to the car frame to the disturbance force generated by the engine. (b) With an engine with a mass of 1000 kg, it is noted that the most severe vibration occurs at a frequency of 20 rad/s. The mount specificationS state that the mounts provide a damping ratio of 0.2 for this engine. What is the effective stiffness of the mounts?
(c) Sketch the magnitude and phase of the frequency response relating the transmitted force to input force. What is the maximum force ratio transmitted? At what frequency does this occur? (d) The manufacturer has two other prototype sets of engine mounts available with the following properties: MountB: Ks 4K, Bs O.SB MountC: Kc 2K, Be= 2B Sketch the transmissibility force ratio H U co) as a function of frequency for these mounts. Is either of the new mounts superior in performance to the original mounts?
= =
=
14.15. The sinusoidal input impedance Z U co) of a system is found directly from its input impedance Z (s) (Chap. 13) and represents the frequency response of the input across variable to the input through variable. Figure 14.32 shows parallel and series L-C circuits. Determine the impedance Z (s) of each circuit and derive its sinusoidal impedance Z(j(J)).
i(t)
c
L
~~~c
Figure 14.32: Parallel and series L-C circuits.
(a) For each circuit determine the magnitude of the impedance when L the input frequency approaches zero
ii. the frequency approaches infinity
iii. the frequency is ro
= ..;rrr:c
(b) Explain what is happening physically to the voltages and currents associated with each element at the resonance condition. Repeat the problem with the assumption that the coil has finite resistance R as well as inductance.
14.16. A simple AM radio, using the resonance of a single L-C circuit to tune a station in the AM band, is shown in Fig. 14.33(a) together with a linear graph model of the tuning circuit in Fig. 14.33(b). The transmission of the signal from the transmitter to the radio antenna is modeled as a Thevenin source of voltage V6 (t) and source resistance R3 • The coil has both inductance L and finite resistance R. The capacitor C is varied to select the desired station. The amplifier has a very high-input impedance and does not load the circuit. (a) Derive the transfer function between the output voltage vc and the source V, (t), and determine the circuit undamped natural frequency eon and damping ratio ~. (b) The AM radio band spans a frequency of 550kHz to 1600kHz. Assume L = 200 #LH. .R .01 0, and R.r 20,000 fl. Fmd the minimum and maximum values of the variable capacitor C needed to tune the complete AM band. (Hint: For a lightly damped system, the resonant peak is very close to the undamped natural frequency.)
=
=
Chap. 14
495
Problems Antenna
I I I I
c:
Tuning cicuit
I
I I
I I
I
(b)
(a)
Figure 14.33: A simple AM radio and a model of its tuning circuit.
(c) The selectivity of a radio (its ability to select out one station from all others) depends on the bandwidth of the resonant peak in its frequency response. The quality factor Q of a circuit is defined as the bandwidth /!l(J) between the half-power (-3 dB) points on the resonant peak, divided by the peak resonant frequency (/)petJJc. For a second-order system !l(J)
1
(J)peak
2~
Q=--~-
Assume that an AM station has a frequency of 1 MHz. Find the value of capacitance required to tune the radio to this station, and compute the value of r and Q at this frequency. Determine the -3 dB bandwidth of the receiver. 14.17. Figure 14.34 shows a miniature electromechanical filter that uses rotational elements to shape the frequency response. The electrical input current ls(t) produces a torque Ts(t) = K ls(t) on the inertial element J 1, the output sensor produces a voltage proportional to the angular displacement of the inertia J 2 (which is related to the torque in spring K2 ). Each inertia bas a mass of 1 gram and a diameter of 1 em. The springs have a torsional stiffness of 100 N-m/rad and the damping elements have a coefficient B 0.001 N-m-s/rad.
=
r------------------------------------------------1
Is{t)
l
::. !
Actuator
Filter
Sensor
3 s,
1
!
u
Figure 14.34: A hybrid electromechanical filter.
(a) Generate a model of the system and represent this in state-space or transfer function form. (b) Use a computer program (such as Matlab) to make Bode magnitude and phase plots of the filter's frequency response. Determine the cut-off frequency and the high frequency attenuation slope. 14.18. A series R-L-C circuit is shown in Fig. 14.35.
496
Sinusoidal Frequency Response of Linear Systems
Chap. 14
Figure 14.35: A tuned circuit.
(a) Derive tbe transfer function between the input and output voltages. Determine the undamped natural frequency and damping ratio of the circuit. (b) Assume the circuit values are L 1.0 mH, R 20 0, and C 0.1 iJ.F. Sketch the system frequency response magnitude function. At what frequency is the response a maximum? What is the circuit maximum amplification ratio. (c) What changes are required in L and R in order to double both the peak resonant frequency and the peak amplification ratio? 14.19. A Helmholtz resonator is a fluid device that exhibits significant resonance. The resonator consists of a tube connected to a fixed wall cavity as shown in Fig. 14.36. When filled with a compressible gas, it has a resonant frequency detennined by the fluid capacitance of the cavity and the inertance of the tube. The amplitude of the resonance is controlled by the fluid resistance of the tube.
=
=
=
Figure 14.36: A Helmholtz resonator.
(a) Derive the transfer function between the pressure at the tube entrance to pressure in the cavity and show that this system is the fluid analog oftbe electrical system in Prob. 14.8. Determine the system undamped natural frequency and damping ratio in terms of the system parameters. (b) In a resonator with a cavity C = 0.01 m5IN, it is observed that the peak amplification occurs at w 100 radls and has an amplitude twice that observed at low frequencies. What are the values of the tube fluid inertance and resistance? (c) What is the new resonant frequency and peak amplification ratio if the volume of the cavity is increased by a factor of four? 14.20. The quarter-car model depicted in Fig. 14.37 is used to study vehicle ride quality. In the model, one quarter of the car body mass is supported by the suspension elements associated with one wheel. As the car travels along a road it is excited by the vertical displacement profile of the roadway, generating a time varying displacement input to the model. The vertical acceleration response of the car body is frequently used to predict passenger comfort. (a) Determine the transfer function relating the car bOdy acceleration to the roadway vertical input displacement What are the system undamped natural frequency and damping ratio? (b)· Assume that the suspension of a car with a total.mass of 2,000 kg has been tuned to give a undamped natural frequency of 1.5 Hz. and a damping ratio of 0.2. Derive and sketch the frequency response of the car body acceleration to the roadway input acceleration. (c) Repeat part (b) assuming the damping ratio~ is increased to 1.0 (critical damping). How does this respo1;1se differ from the previous case? For what range of frequencies does the critically damped suspension have an increased amplitude of response?
=
Chap. 14
497
Problems
Road profile
Figure 14.37: Quaner car model.
14.21. The electric motor drive for a printing roller (shown in Fig. 14.38) has shown eiTatic speed variations which blur the print transfer to the paper. It is suspected that a resonance in the drive system is amplifying electrical interference and causing the ink to smear. A frequency response analyzer has been used to measure the transfer function between the input voltage and the rotational speed sensor output voltage. The measured transfer function is 0.2
H(s)
= s2 + 6.4s + 256
+ Amplifier
, Tachometer
Figure 14.38: A printing press roll drive system.
(a) Plot the frequency response magnitude and phase functions. At what frequency does the operation appear to have the highest possibility of rough operation? (b) It has been suggested that a filter with a transfer function
where T2 = 0.125 s, be cascaded with the amplifier. Compute the system response with the filter in place, that is, with the total transfer function HT(s) = H(s)H1 (s). First consider the case T1 = 4T2 s, and then consider the case T1 = T2 /4 s. Which filter yields an improved frequency response? 14.22. Positive displacement fluid pumps sometimes generate undesirable periodic pressure fluctuations at the outlet Fluid accumulators are often installed at the outlet to reduce the propagation of these disturbances throughout the system. A positive displacement pump may be represented as a flow source, with the fluctuating part of the output flow as Q = Q 0 sin(wt), where the angular frequency w is directly related to the shaft speed. A pump connected to a distribution pipe terminated with a valve is depicted in Fig. 14.39. The pipe has inertance I and the combined resistance of the pipe and the valve is R.
498
Sinusoidal Frequency Response of Linear Systems
lrl L;J
Chap.l4
fluid accumulator (not attached)
Pipe----~
t
R, I
Valve
Figure 14.39: A fluid distribution system.
{a) Derive the transfer function between the pump output ftow Q(t) and the pressure at the pump outlet
=
=
{b) Detennine the system frequency response assuming I 0.2 N-s 2 /m5 , R 20 N-s/m5 • What is the behavior of the pressure amplitude variations as the pump rotational speed increases?
(c) Derive the modified system transfer function if a fluid accumulator (which is a fluid capacitor C) is installed at the pump outlet What is the order of the system with the accumulator? {d) Determine and plot the system frequency response when the accumulator has a capacitance of C 1o-s m5IN. Compare the response with that of the original system. For what range of frequencies is the accumulator effective in reducing the pressure fluctuations? 14.23. Example 1.1 describes the application of a tuned-mass damper to the reduction of wind induced motions in tall buildings. Tall structures typically have low damping, with ~ ::::: 0.02. Figure 14.40(c) shows the building of Fig. 1.2 as a lumped second-order system with an equivalent mass m on a cantilevered translational spring K with a damping element B, and excited by the wind force Fw(t). For· a typical 50 storey building, the natural period of osciUation is approximately 6 s. In the schematic of Fig. 14.40(b) the elements of the tuned-mass damper (md, Kd, Bd) have been added. These dampers have natural frequencies tuned to the building's natural frequency.
=
r
y(t)
Cantilevered support
K.B
(a)
(b)
Figure 14.40: Tall building with a tuned-mass damper.
(c)
Chap. 14
499
References
(a) Determine the transfer function between the wind force and the building displacement without the damper. What is the undamped natural frequency for a building with a natural period of 6 s and a damping ratio of 0.02? Determine and plot the frequency response. What is the maximum amplification of the system? (b) Consider the system with the tuned-mass damper instaUed. Derive the system transfer function between the wind force Fw(t) and the building displacement What is the order of the complete system? {c) Assumemd = 0.04m and Bd = 0.1. Choosethedamperstiffness Kd toyieldanaturalfrequency of the damper equal to that of the building. Determine the eigenvalues of the coupled system and from the eigenvalues determine the system natural frequencies and damping ratios. {d) Detennine the transfer function of the building with the velocity of the building mass as the output Use a computer program to plot the actual frequency response of the model.
REFERENCES [1] Shearer, J. L., Murphy, A. T., and Richardson, H. H., Introduction to System Dynamics, AddisonWesley, Reading, MA, 1967. [2] Shearer, J. L., and Kulakowski, B. T., Dynamic Modeling and Control of Engineering Systems, Macmillan, New York, 1990. [3] Reid, J. G., Linear System Fundamentals, McGraw-Hill, New York, 1983. [4] Kamopp, D. C., Margolis, D. L., and Rosenberg, R. C., System Dynamics: A Unified Approach (2nd ed.), John Wiley, New York, 1990. [5] Ogata, K., Modem Control Engineering (2nd ed.), Prentice Hall, Englewood Cliffs, NJ, 1990. [6] Dorf, R. C., Modem Control Systems (5th ed.), Addison-Wesley, Reading, MA, 1989. [7] Kuo, B. C., Automatic Control Systems (6th ed.), Prentice Hall, Englewood Cliffs, NJ, 1991. [8] Franklin, G. F., PoweU, J.D., and Emami-Naeni, A.• Feedback Control of Dynamic Systems {2nd ed.), Addison-Wesley, Reading, MA, 1991.
15
Frequency Domain Methods
15.1
INTRODUCTION In Chaps. 12 and 14 the system transfer function H(s) and the frequency response function H U w) were developed and used to find the response to an exponential form of input u(t) = In this chapter we extend the use of these two system descriptions to a much broader range of input waveforms by expressing the system input as a superposition of exponential waveforms. For example, suppose that a system input u(t) may be expressed as a sum of complex n exponentials:
r'.
n
u(t)
= L a;es;t i=l
where the complex coefficients a; and constants s; are known. Assume that each component is applied to the system alone; if at time t = 0 the system is at rest, the solution component y;(t) is of the form y;(t) = (Yh (t)); +a;H(s;)es;t
where (yh(t)); is a homogeneous solution. The principle of superposition states that the total response Yp (t) of the linear system is the sum of all component outputs: n
Yp(t)
= LY;(t) i=l
n
=L i=l
500
(yh (t)); +a;H(s;)e511
Sec. 15.1
SOl
Introduction
Example 15.1
Find the response of the first-order system with the differential equation dy - +4y =2u dt to an input u(t)
=se-t+ 3e-21 , given that at timet= 0 the response is y(O) = 0.
Solution The system transfer function is 2 s+4
H(s)= - -
(i)
The system homogeneous response is (ii)
where C is a constant, and if the system is at rest at time t input u(t) = a~' is
= 0, the response to an exponential (iii)
The principle of superposition says that the response to the input u (t) = sum of two components, each similar to Eq. (iii); that is,
se-1 + 3e-21 is the
(iv)
In this chapter we examine methods that allow a function of time f (t) to be represented as a sum of elementary sinusoidal or complex exponential functions. We then show how the system transfer function H (s) or the frequency response H Uw) defines the response to each such component and through the principle of superposition defines the total response. These methods allow computation of the response to a very broad range of input waveforms, including most of the system inputs encountered in engineering practice. The methods are known collectively as Fourier analysis methods, after Jean Baptiste Joseph Fourier, who in the early part of the nineteenth century proposed that an arbitrary repetitive function could be written as an infinite sum of sine and cosine functions [1]. The Fourier series representation of periodic functions may be extended through the Fourier transform to represent nonrepeating aperiodic (or transient) functions as a continuous distribution of sinusoidal components. A further generalization produces the Laplace transform representation of waveforms. The methods of representing and analyzing waveforms and system responses in terms of the action of the frequency response function on component sinusoidal or exponential waveforms are known collectively as the frequency domain methods. Such methods, developed in this chapter, have important theoretical and practical applications throughout engineering, especially in system dynamics and control system theory.
Frequency Domain Methods
502
15.2
Chap. 15
FOURIER ANALYSIS OF PERIODIC WAVEFORMS In Chap. 14 the steady-state responses of linear time-invariant systems to two periodic wavefonns, the real sinusoid f (t) sin wt and the complex exponential f (t) ei"", were described. Both functions are repetitive; that is, each has identical values at intervals in time of t = 21r1w seconds apart In general a periodic function is a function that satisfies the relationship (15.1) f(t) = f(t + T)
=
=
for all t, or f (t) = f (t + n T) for n = ± 1, ±2, ±3, . . . . The constant T is known as the period of the function. Figure 15.1 shows some examples of periodic functions.
w
v v v · Figure 15.1: Examples of periodic functions of time.
The fundamental angular frequency directly from the period:
wo
(radls) of a periodic waveform is defined
2n'
wo=T
(15.2)
Any periodic function with period T is also periodic at intervals of n T for any positive integer n. Similarly, any waveform with a period ofT /n is periodic at intervals of T seconds. '1\vo waveforms whose periods, or fundamental frequencies, are related by a simple integer ratio are said to be harmonically related. Consider, for example, a pair of periodic functions, the first !I (t) with a period of TI = 12 sand the second /2Ct) with a period of T2 = 4 s. H the fundamental frequency wo is defined by !I (t), that is, wo = 21r/12, then /2(t) has a frequency of 3wo. The two functions are harmonically related, and f 2 (t) is said to have a frequency that is the third harmonic of the fundamental wo. H these two functions are summed together to produce a new function, g(t) = ft (t) + /2(t), then g(t) will repeat at interva1s defined by the longest period of the two, in this case, every 12 s. In general, when harmonically related waveforms are added· together, the resulting function is also periodic with a repetition period equal to the fundamental period. Example 15.2 A family ofwavefonns gN(t) (N component functions, that is,
= 1, 2, ... , 5) is formed by adding the first N of up to five N
8N(t) =
L fn(t) n:zl
503
Fourier Analysis of Periodic Wavefonns
Sec. 15.2 where
fl(t) = 1 f2(t) = sin(21rt)
/3(t) =
~ sin(61rt)
j4(t) =
~ sin(l01rt)
fs(t) =
~ sin(14Ht)
The first term is a constant, and the four sinusoidal components are harmonically related, with a fundamental frequency of c.rJo = 211' rad/s and a fundamental period of T = 2Jr I c.rJo = 1 s. (The constant term may be considered to be periodic with any arbitrary period but is commonly considered to have a frequency of 0 radls.) Figure 15.2 shows the evolution of the function that is fonned as more of the individual terms are included in the summation. Notice that in all cases the period of the resulting 8N(t) remains constant and equal to the period of the fundamental component (1 s). In this particular case it can be seen that the sum tends toward a square wave. BN(t) (N
=1, ..., 5)
2.0 r---~~--,.-------,-----,~--,--------,
1.5
.g= u
c
cE!
1 ·v;
1.0
u
-5 c
>.
ell
0.5
0.0
0.5
1.0 Time (s)
1.5
2.0 t
Figure 15.2: Synthesis of a periodic waveform by the summation of harmonically related components.
The Fourier series [2, 3] representation of a real periodic function f(t) is based upon the summation of hannonically related sinusoidal components. If the period is T, then the harmonics are sinusoids with frequencies that are integer multiples of cuo; that is, the nth harmonic component has a frequency nllJO = 21r n 1T and can be written
+ bn sin(ncvot) =An sin(nwot + t/Jn)
fn (t) =an cos(nwot)
(15.3) (15.4)
Frequency Domain Methods
504
Chap. IS
In the first form the function f, (t) is written as a pair of sine and cosine functions with real coefficients a, and b,. The second form, in which the component is expressed as a single sinusoid with an amplitude .A, and a phase tj>, , is directly related to the first by the trigonometric relationship . .A, sin(nwot + rf>,) =.A, sin t/J, cos(nwot) +An cos t/J, sin(nwot) Equating coefficients,
= An sin tj>, b, = .A, COS tPn
a,
(15.5)
and
An= Ja~ +b~ (15.6)
~ = tan-• (::)
The Fourier series representation of an arbitrary periodic waveform f (t) (subject to some general conditions des~ribed later) is as an infinite sum of harmonically related sinusoidal components commonly written in the fo11owing two equivalent forms
1
f(t) =
2ao + L 00
[a, cos(nwot)
+ b, sin(nwot)]
(15.7)
n=l
1
00
= 2ao + l:An sin(nwot + rf>,)
(15.8)
n=l
In either representation knowledge of the fundamental frequency wo and the sets of Fourier coefficients {a,} and {b,} (or {.A,} and {t/J,}) is sufficient to completely define the waveform f(t). A third, and completely equivalent, representation of the Fourier series expresses each of the harmonic components f,(t) in terms of complex exponentials instead of real sinusoids. The Euler formulas may be used to replace each sine and cosine term in the components of Eq. (15.3) with a pair of complex exponentials: f,(t)
=a, cos(nwot) + b, sin(nwot) = a; (ei"OJO' + e-JnOJOt) + ~". (einOJOt _
. .1
.
.
1
J
e-JnCZJot) .
= 2 (a,- jb,) eJn(I)Qt + 2 (a,+ jb,) e-Jnwot
= F,einwot + F_,e-Jnwoi
.
(15.9)
Fourier Analysis of Periodic Waveforms
Sec. 15.2
505
where the new coefficients (15.10)
are now complex numbers. With this substitution the Fourier series may be written in a compact form based upon hannonically related complex exponentials: f(t)
+oo
L
=
(15.11)
Fnejn(I)(Jt
n=-oo
This form of the series requires summation over all negative and positive values of n, where the coefficients of terms for positive and negative values of n are complex conjugates, (15.12)
so knowledge of the coefficients Fn for n ~ 0 is sufficient to define the function f (t). Throughout this book we adopt the practice of using uppercase letters to represent the Fourier coefficients in the complex series notation, and so the set of coefficients {G n} represent the function g(t) and {Yn} are the coefficients of the function y(t). The lowercase coefficients {an} and {bn} are used to represent the real Fourier coefficients of any function of time. Example 15.3 A periodic function
f
(t) consists of five components:
f(t) = 2 + 3 sin( lOOt)+ 4cos(100t) + 5 sin(200t +
1r /4)
+ 3 cos(400t)
It may be expressed as a finite complex Fourier series by expanding each term through the Euler formulas:
:j :j
f(t) = 2 +
+
(eiiOOt _ e-iJOOt)
+ ~ (eJIOOr + e-JIOOt)
(ei<200r+;r/4) _ e-J(200t+;r/4)) +
=2+ (2+
~ (ei400t +
:j)eJIOOt + (2- :j)e-jiOOt
5 (2.J25 2hj
(52.J2 2.J2j 5)
+ --+-- )"200 eJ '+ - - - - + ~ei400t + ~e-i400t 2
e-J400r)
2
"200'
e-J
506
Frequency Domain Methods
Chap.l5
The fundamental frequency is c.rJo = I00 radls, and the time domain function contains harmonics n = I, 2, 3, and 4. The complex Fourier coefficients are
F-1=l+~j 5
F-2 =
.r.:;-
2v2 F-3=0 3 F-4=-
(1
+ lj)
2
The finite Fourier series may be written in the complex form using these coefficients as 5
f(t)
=L
FnelniOOt
n=-5
and plotted with real and imaginary parts as in Fig. 15.3. 9t{Fn)
2.0
~
l.
2
3
l.
4
5
n
n
-2.0 Figure _15.3: Spectral representation of the waveform discussed in Example 15.3.
The values of the Fourier coefficients in any of the above three forms are effectively measures of the amplitude and phase of the harmonic component at a frequency of nwa. The spectrum of a periodic wavefonn is the set of all the_Fourier coefficients, for example, {A,} and {q,,.}, expressed as a function of frequency. Because the harmonic components exist at discrete frequencies, periodic functions are said to exhibit line spectra, and it is common to express the spectrum graphically with frequency w as the independent axis and the Fourier coefficients plotted as lines at intervals of wo. The first two forms of the Fourier series, based upon Eqs. (15.3) and (15.4), generate uone-sided" spectra because they are
Sec. 15.2
507
Fourier Analysis of Periodic Wavefonns
defined for only positive values of n, whereas the complex form defined by Eq. (15.11) generates a ''two-sided" spectrum because its summation requires positive and negative values of n. Figure 15.3 shows the complex spectrum for the finite series discussed in Example 15.3. 15.2.1 Computation of the Fourier Coefficients
The derivation of the expressions for computing the coefficients in a Fourier series is beyond the scope of this book, and we simply state without proof that if f(t) is periodic with period T and fundamental frequency coo, in the complex exponential form the coefficients Fn may be computed from the equation
ll't+T j(t)e-jnOJQt dt Fn =T ''
(15.13)
where the initial time t1 for the integration is arbitrary. The integral may be evaluated over any interval that is one period T in duration. The corresponding formulas for the sinusoidal forms of the series may be derived directly from Eq. (15.13). From Eq. (15.10) it can be seen that an= Fn
+ F-n
= -T1 l',,'+T f(t). (einOJOt + e-inwot) dt
(15.14)
= -21''+T f(t) cos(nwot) dt T ''
and similarly,
= T21''+T ,, f(t) sin(nwot) dt
bn
TABLE 15.1:
(15.15)
Summary of Analysis and Synthesis Equations for Fourier Analysis and Synthesis
Sinusoidal Fonnulation
1
Synthesis j(t)
Exponential Formulation
~
= 20o + L [an cos(nQJot) + bn sin(nWot)]
+~
j(t)
=
21't+T j(t) cos(nWot) dt
an = -
T ,,
bn
= T21't+T f(t) sin(nWot) dt II
FneinOJOt
n ..-~
nal
Analysis
L
11'1+T f(t)e-inOJOt dt
Fn = -
T ,,
508
Frequency Domain Methods
Chap. IS
The calculation of the coefficients for a given periodic time function /(t) is known as Fourier analysis or decomposition because it implies that the waveform can be "decomposed" into its spectral components. On the other hand, the equations that express f (t) as a Fourier series summation [Eqs. (15.7), (15.8), and (15.11)] are termed Fourier synthesis equations because they imply that /(t) can be created (synthesized) from an infinite set of harmonically related oscillators. Table 15.1 (see previous page) summarizes the analysis and synthesis equations for the sinusoidal and complex exponential formulations of the Fourier series. Example 15.4 Fmd the complex and real Fourier series representations of the periodic square wave f (t) with period T, /(t)
= {I
0
0::; t < T /2 T/2:=;t
as shown in Fig. 15.4.
0.5
f(t)
- -
-1.0 1--
-9-8-7~-5-4-3-2-1
0.5
I
-T
f-
0
0 1 2 3 4 5 6 7 8 9
n
.S{Fnl
T
2T
0.
i
f
0.
3T
-9 -8 -7 ~ -5 -4 -3 -2 -1
(a)
n
-0.
-0.
(b)
Figure 15.4: (a) A periodic square wave. and (b) its spectrum.
Solution The complex Fourier series is defined by the synthesis equation f(t)
=
+oo
L
Fnelnwot
(i)
n:::-00
In this case the function is nonzero for only half of the period, and the integration limits can be restricted to this range. The zero frequency coefficient Fo must be computed separately:
Fo
1 = -T1 1T/2 ei0 dt = o 2
(ii)
Sec. 15.2
509
Fourier Analysis of Periodic Wavefonns
and all the other coefficients are
(iii)
since CrJo T
= 27r. Because e-imr = -1 when n is odd and +1 when n is even, ifn is odd ifn is even
The square wave can then be written as the complex Fourier series f(t)
=
! + f:, 2
m:;;:J
(e-i<2m-J)QI()t _ ei<2m-J)QI()t)
j
(2m- 1)31'
(iv)
where the terms F,eiWn' and F_,e-iWnt have been combined in the summation. If the Euler formulas are used to expand the complex exponentials. the cosine terms cancel and the resulting series involves only sine terms: 00
1
f(t)
= 2+ L
2 (2m _ 1)1r sin [(2m - 1) CrJol]
nuzJ
=
!2 + ~3l' [sin(CrJot) +!3 sin(3coot) +!5 sin(5CrJot) + -7 sin(7euot) + ···]
(v)
1
Comparison of the terms in this series with the components of the waveform synthesized in Example 15.2, and shown in Fig. 15.2, shows how a square wave may be progressively approximated by a finite series.
15.2.2 Properties of the Fourier Series
A full discussion of the properties of the Fourier series is beyond the scope of this boo~ and the interested reader is referred to the references [1-10]. Some of the more important properties are summarized as follows. 1. Existence of the Fourier series: For the series to exist. the integral of Eq. (15.13) must converge. A set of three sufficient conditions. known as the Dirichelet conditions, guarantees the existence of a Fourier series for a given periodic wavefonn f (t):
• The function f(t) must be integrable in the absolute sense over any period, that is, t,+T
1 It
for any 11.
1/(t)l dt
<
00
(15.16)
510
Frequency Domain Methods
Chap. 15
• There must be at most a finite number of maxima and minima in the function f(t) within any period. • There must be at most a finite number of discontinuities in the function f(t) within any period, and all such discontinuities must be finite in magnitude. These requirements are satisfied by almost all waveforms found in engineering practice. The Dirichelet conditions are a sufficient set of conditions to guarantee the existence of a Fourier series representation. They are not necessary conditions, and there are some functions that have a Fourier series representation without satisfying all three conditions. 2. Linearity of the Fourier series representation: The Fourier analysis and synthesis operations are linear. Consider two periodic functions g(t) and h(t) with identical periods T and their complex Fourier coefficients,
Gn
= ..!_ {T g(t)e-i"OJO' dt T
lo
Hn = ..!_ {T h(t)e-jnOJOt dt T lo and a third function defined as a weighted sum of g(t) and h(t), /(t)
= ag(t) + bh(t)
where a and b are constants. The linearity property, which may be shown by direct substitution into the integral, states that the Fourier coefficients of f (t) are
Fn =aGn +bHn That is, the Fourier series of a weighted sum of two time domain functions is the weighted sum of the individual series.
= 0 axis, the Fourier series representation may be simplified. If f(t) is an even function of time, that is, f(-t) = f(t), the complex Fourier series has coefficients Fn that are purely real, with the result that the real series contains only cosine terms, so Eq. (15.7) simplifies to l 00 (15.17) /(t) = 2ao + an cos(nwat)
3. Even and odd functions: If f(t) exhibits symmetry about the t
L
n=l
.
Similarly, if f (t) is an odd function of time, that is, f (-t) = - f (t), the coefficients Fn are imaginary and the one-sided series consists of only sine terms: 00
f(t)
=L
bn sin(nwat)
n=l
Notice that an odd function requires that f(t) have a zero average value.
(15.18)
Sec. 15.2
Fourier Analysis of Periodic Waveforms
511
4. The Fourier series of a time-shifted function: If the periodic function f(t) has a Fourier series with complex coefficients Fn, the series representing a ''time-shifted" version g(t) = f(t + r) has coefficients einOJOT Fn. If
11T
=f(t)e-JnOJOt dt T o
Fn
then Gn =
~ LT f(t + T)e-jn..,t dt
Changing the variable ofintegration v = t
+ t' gives
}1T+T f(v)e-inOJO(v-T) dv
Gn = T
r
. }1T+T f(v)e-JnOJOvt . dv = eJnOJOTT
T
= einOJOT Fn If the nth spectral component is written in tenns of its magnitude and phase fn(t) =An sin(na>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<
T
2
shown in Fig. 15.5. f(t)
Figure 15.5:
A periodic sawtooth waveform.
Solution The complex Fourier coefficients are
Fn = -1
lT/2
2 . -te-Jnwor dl T -T/2 T
(i)
and integrating by parts, Fn
= .....!:.}__2 [te-JnworiT/2 ner>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 - -
= 4 + r' { 5
8 8 T 3T - -
2
-+-t 4 T
shown in Fig. 15.6. /(I)
Figure 15.6: A modified sawtooth function having a shift in amplitude and time origin.
Solution If f(t) is rewritten as 5T T - -
8 -
8 T 3T - -
(i)
then f(t) =
/J
(t
+ ~) + h
(t
+
i)
(ii)
where / 1(t) is the square wave function analyzed in Example 15.4: f(t) = { 1
0
0 ~ t < T/2 T/2 ~ t < T
(iii)
and /2(t) is the sawtooth function in Example 15.5. f(t)
2 = -t T
ltl
T
<
2
(iv)
Therefore, the function f (t) is a time-shifted version of the sum of two functions for which we already know the Fourier series. The Fourier series for f(t) is the sum Oinearity property) of phase-shifted versions (time-shifting property) of the pair of Fourier series derived in Examples
514
. Frequency Domain Methods 15.4 and 15.5. The time shift ofT/8 seconds adds a phase shift of n(I)()T/8 each component:
Chap. 15
= me j4 radians to
(v)
. ( 5Wot + 51r) + 1 sm . ( 11r) ] + S1 sm 4 1 7a>ot + 4 + · · ·
/2 (t +
i) = ~ [sin (a>ot + ~) - ~sin 2wot + I) + ~ (
-
~ sin(4a>ot + 1r) + j sin (5ClJQt + 5: ) -
3
sin ( 3Wot + : )
(vi)
•• ·]
The sum of these two series is
f = /1 (t) + h
15.3
(vii)
THE RESPONSE OF LINEAR SYSTEMS TO PERIODIC INPUTS
Consider a linear single-input single-output system with a frequency response function HU(JJ) as defined in Chap. 14. Let the input u(t) be a periodic function with period T and assume that all initial condition transient components in the output have decayed to zero. Because the input is a periodic function, it can be written in terms of its complex or real Fourier series: 00
u(t)
=L
UneinOJOt
(15.19)
n=-oo
1
00
= 2ao + L An sin(nWot + t/>n)
(15.20)
n=l
The nth real hannonic input component, u,(t) =.A, sin(nwot + tf>n), generates an output sinusoidal component Yn(t) with a magnitude and a phase determined by the system's frequency response function H U(JJ): Yn(t)
= IHUn(JJQ)I.A, sin [nCLJot + tPn + LHUncuo)]
(15.21)
The Response ·of Linear Systems to Periodic Inputs
Sec. 15.3
515
as described in Chap. 14. The principle of superposition states that the total output y(t) is the sum of all such component outputs, or (X)
y(t)
= LYn(t) n::::O
= 4aoHUO) + "f.A, IHUnwa)lsin[nwot +tPn + LHUnWo)]
(15.22)
n=l
which is itself a Fourier series with the same fundamental and harmonic frequencies as the input. The output y(t) is therefore also a periodic function with the same period T as the input, but because the system frequency response function has modified the relative magnitudes and the phases of the components, the waveform of the output y(t) differs in form and appearance from the input u(t). In the complex formulation the input waveform is decomposed into a set of complex exponentials u,.(t) = Unei" 010'. Each such component is modified by the system frequency response, and so the output component is y,.(t)
= HUnwa)U,.ei"
010
(15.23)
'
and the complete output Fourier series is (X)
y(t)
=
(X)
L
y,.(t)
n=-CXl
=
L
HUnwa)U,.einOJOt
(15.24)
n=-CXl
Example 15.7
The first-order electric network shown in Fig. 15.7 is excited with the sawtooth function discussed in Example 15.5. Fmd an expression for the series representing the output Vout(t). vin(t)
1.0
R
~ ~
Vin(t)
0.5 0.0
= .s -0.5 Q,
-1.0 (a)
Rgure 15.7:
(b)
(a) A first-order electric system (b) driven by a sawtooth input waveform.
516
Frequency Domain Methods
Chap. 15
Solution .The electric network has a transfer function 1
H(s)
= RCs + 1
(i)
and therefore has a frequency response function IH(ja>)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 ~~~
I
-~ ~
-20
c.
~
n
20
Ie
J
e ~ ~ 10
u..
I
0
I
2
I
I
4
~
v~ ~ ~ I
l
_l_
6
~
uu
I
I
8
10
Time (s) (b)
Figure 15.10: (a) The frequency response magnitude function of the mechanical system in Example 15.8, and (b) an input square wave function and its response.
12 t
Sec. 15.4
519
Fourier Analysis of Transient Wavefonns
The resonance in IH U (J)) I at the undamped natural frequency (J)n = 5 rad/s has a large effect on the relative amplitude of the fifth hannonic in the output y(t). Figure 15.10b shows the system input and output wavefonns. The effect of the resonance can be clearly seen, for the output appears to be almost sinusoidal at a frequency of 5 radls. In fact the output is still a periodic waveform with a period of 21r seconds, but the fifth harmonic component dominates the response and makes it appear to be sinusoidal at its own frequency.
15.4
FOURIER ANALYSIS OF TRANSIENT WAVEFORMS
Many waveforms found in practice are not periodic and therefore cannot be analyzed directly using Fourier series methods. A large class of system excitation functions can be characterized as aperiodic, or transient, in nature. These functions are limited in time; they occur only once and decay to zero as time becomes large [2, 4, 5]. Consider a function f(t) of duration fl. that exists only within a defined interval t 1 < t !:: t1 + A and is identically zero outside this interval. We begin by making a simple assumption, namely, that in observing the transient phenomenon f(t) within any finite interval that encompasses it, we have observed a fraction of a single period of a periodic function with a very large period, much larger than the observation interval. Although we do not know what the duration of this hypothetical period is, it is assumed that f(t) will repeat itself at some time in the distant future, and in the meantime it is assumed that this periodic function remains identically zero for the rest of its period outside the observation interval. f(t)
.~'
l
::
i\ p
n
1\
.l i:
i : i
: i :
. . -2T
II
i i:
:
l
-T
:
/(t)
/
! J, H
H/
:.
.j \:
I\ I \
I\ T
I
!
;:
fl
1\
1\
i\
Ii
l \
I\ I \ l \
:
2T
l
3T
Figure 15.11: Periodic extension of a transient waveform.
The analysis thus conjectures a new function /p(t), known as a periodic extension of f(t), that repeats every T seconds (T > fl.), but at our discretion we can letT become very large. Figure 15.11 shows the hypothetical periodic extension /p(t) created from the observed f(t). As observers of the function /p(t) we need not be concerned with its pseudoperiodicity because we will never be given the opportunity to experience it outside the first period. Furthermore, we can assume that if /p(t) is the input to a linear system, T is so large that the system response decays to zero before the arrival of the second period. Therefore, we assume that the responses of the system to f (t) and /p (t) are identical within our chosen observation interval. The important difference between the two functions is that /p(t) is periodic and therefore has a Fourier series description.
F~uency
·
520
Domain Methods
Chap. IS
The development of Fourier analysis methods for transient phenomena is based on the limiting behavior of the Fourier series describing /p(t) as the period T approaches infinity. Consider the behavior of the Fourier series of a simple periodic function as its period Tis varied, for example, an even periodic pulse function f(t) of fixed width L\: ltl < L\12 L\12 ~ ltl ~ T - L\12
1
f(t)
={0
(15.25)
as shown in Fig. 15.12. Assume that the pulse width L\ remains constant as the period T varies. The· Fourier coefficients in complex fonn are Fn
= -1 1/i/2 e-jnOJQt dt T -li/2
= _j_ (e-inOJOii/2 _
einOJOii/2)
2n1r
(15.26)
1 . ( n7rL\) =-SID n1r T L\ sin (n1r L\ IT) = T (n1r L\IT) and 1
1/i/2
n;=o
L\
(15.27)
Fo =1dt =T -li/2 T
and the spectral Jines are spaced along the frequency axis at intervals of coo = 21r IT radians/second To investigate the behavior of the spectrum as the period T is altered, we define a continuous function of frequency F( )
= sin(wL\ 12)
w
wL\12
f(t)
,...
II
-
-
1.0
o.s
.! i
1.1
,,
-2T
-T
,, 0
T
It 2T·
I
It
3T
4T
Figure 15.12: A periodic rectangular pulse function of fixed duration 1:1 but varying period T.
Sec. 15.4
521
Fourier Analysis of Transient Waveforms
and note that the Fourier coefficients may be computed directly from F((J)):
Fn = e,. sin (we,./2) T we,./2 e,. = T F(w)lw=nOJO
I
(15.28)
w=21rn/T
(15.29)
The function F((J)} depends on only the pulse width fl. and is independent of the period T. As T is changed, apart from the amplitude scaling factor e,. 1T, the frequency dependence of the Fourier coefficients .is defined by F ((J)); the relative strength of the nth complex harmonic component, at a frequency nwo, is defined by F(nwo}. The function F((J)) is therefore an envelope function that depends on only f(t} and not on the length of the assumed period. Figure 15.13 shows examples of the line spectra for the periodic pulse train as the period T is changed. The following general observations on the behavior of the Fourier coefficients as T varies can be made:
1. As the repetition period T increases, the fundamental frequency Wo decreases and the spacing between adjacent Jines in the spectrum decreases. TFII
f(t)
T=4
I
1.0 ""'"~
I
I
0
f()t 1.0
T=2
I
T= 1
1
;.L
I
0
I
I
-
I
I
I
I
n
4
J
2
0
1.0 - -
I
3
.:...!...
r--
f()t
I
2
=-rr/2 rad/s
I
~
=-rrradls
I
3
4
r---
~
I
I
I
2
3
4
Figure 15.13: Line spectra of periodic extensions of an even rectangular pulse function.
n
n
522
Frequency Domain Methods
Chap. 15
2. As the repetition period T increases, the scaling factor .6. 1T decreases, causing the magnitude of all the spectral lines to be diminished. In the limit as T approaches infinity, the amplitude of the individual lines becomes infinitesimal.
3. The "shape" of the spectrum is defined by the function F((J)) and is independent of
T. Assume that we have an aperiodic function f(t) that is nonzero only for a defined time interval d and without loss of generality assUme that the interval is centered around the time origin (t 0). Then asSUJ!le a periodic extension /p(t) of f(t) with period T that fully encompasses the interval .6.. The Fourier series description of /p(t) is contained in the analysis and synthesis equations
=
1 1T/2 /p(t)e- jnOJQt dt
Fn = T /p(t)
=
(15.30)
-T/2 00
L
FneinOJQt
(15.31)
n=-oo
These two equations may be combined by substituting for Fn in the synthesis equation
Loo
/p(t) =
n=-oo
[ CUO 27r
1T/2 /p(t)e-jnOJQt dt ] einOJOt - T/2
(15.32)
where the substitution coo/21l' = 1/T has also been made. The period T is now allowed to become arbitrarily large, with the result that the fundamental frequency coo becomes very small, and we write cuo = 8(1). We define f(t) as the limiting case of /p(t) as T approaches infinity, that is, f(t) = lim /p(t)
T-.oo
=
f: - [1T/-T/2 T-.oo n=-oo 2
1 -
lim
/p(t)e-jn&(J)t dt] ein&(J)I 8(J)
27r
=
(15.33)
oo -1 [1-oo f(t)e-J(l)f . dt ] e'(J)' . d(J) 1-oo 21l' -oo
where in the limit the summation has been replaced by an integral. If the function inside the braces is defined to be F(j(J)), Eq. (15.33) may be expanded into a pair of equations known as the Fourier transform pair:
FUw) = /(t)
L:
= -21
7r
which are the equations we seek.
f(t)e-i"" dt
1
(15.34)
00
-oo
FUw)ei(J)t dw
(15.35)
Sec. 15.4
523
Fourier Analysis of Transient Waveforms
Equation (15.34) is known as the forward Fourier transform and is analogous to the analysis equation of the Fourier series representation. It expresses the time domain function f (t) as a function of frequency, but unlike the Fourier series representation, it is a continuous function of frequency. Whereas the Fourier series coefficients have units of amplitude, for example, volts or newtons, the function F(jw) has units of amplitude density; that is, the total "amplitude" contained within a small increment of frequency is F (j w) ~w /21r. Equation (15.35) defines the inverse Fourier transform. It allows computation of the time domain function from the frequency domain representation F (j w) and is therefore analogous to the Fourier series synthesis equation. Each of the two functions f (t) and F(jw) is a complete description of the function, and Eqs. (15.34) and (15.35) allow the transformation between the domains. We adopt the convention of using lowercase letters to designate time domain functions, and the same uppercase letters to designate the frequency domain functions. We also adopt the nomenclature
~ F(jw)
f(t)
as denoting the bidirectional Fourier transform relationship between the time and frequency domain representations, and we also frequently write F(jw)
= F{f(t)}
f(t) =
.r- 1{F(jw)}
as denoting the operation of taking the forward .11} and inverse F respectively.
1 {}
Fourier transforms,
15.4.1 Fourier Transform Examples
In this section we present five illustrative examples of Fourier transforms of common time domain functions. Example 15.9 Find the Fourier transform of the pulse function f(t)
=
{a 0
It! < ~/2 otherw1se
shown in Fig. 15.14.
Solution From the definition of the forward Fourier transform, (i)
(ii) (iii)
(iv)
524
Frequency Domain Methods
/(t)
Chap. 15
FUw>
a Fourier transfonn
-T/2 0 T/2
Figure 15.14:
An even aperiodic pulse function and its Fourier transform.
_
-a
T sin(wT12) wTI2
(v)
The Fourier transform of the rectangular pulse is a real function of the form (sin x) I x centered around the j w = 0 axis. Because the function is real, it is sufficient to plot a single graph showing only IFUw)l, as in Fig. 15.14. Notice that while F(jw) is a generally decreasing function of w, it never becomes identically zero, indicating that the rectangular pulse function contains frequency components at all frequencies. The function (sinx)lx = 0 when the argument X= mr for any integer n (n :f:. 0). The main peak or "lobe" of the spectrum FUw) is therefore contained within the frequency band defined by the first two zero crossings lwT121 < 1r and lwl < 27r IT. Thus as the pulse duration T is decreased. the spectral bandwidth of the pulse increases as shown in Fig. 15.15, indicating that short-duration pulses have a relatively larger high-frequency content. /(t)
FUw) aT
a
Fourier transfonn
~
T I
-T/2
0
I
I
T/2
/(t)
FUw> Fourier transfonn
a
T/4
~ aT/4
-T/8 0 T/8 Figure 15.15:
Dependence of the bandwidth of a pulse on its duration.
Sec. 15.4
Fourier Analysis of Transient Waveforms
525
Example 15.10 Find the Fourier transform of the Dirac delta function B(t). Solution The delta impulse function, defined in Chap. 8, is an important theoretical function in system dynamics, and is defined as 8(t)
0
= { undefined
t ~0
t
=0
(i)
with the additional defining property that /_: B(t) dt
=1
The delta function exhibits a "sifting" property when included in an integrand: /_: B(t - T)f(t) dt
= /(T)
(ii)
When substituted into the forward Fourier transform, F(j(l))
= /_: 8(t)e-JC11 dt (iii)
=1
by the sifting property. The spectrum of the delta function is therefore constant over all frequencies. It is this property that makes the impulse a very useful test input for linear systems. Example 15.11 Find the Fourier transform of the finite-duration sinusoidal "tone burst" f(t)
= {sinl.c1Qt 0
ltl < ~/2
otherwise
shown in Fig. 15.16. S{FU(I)))
j(t)
1.0
Fourier transform
-1.0
Figure 15.16: A sinusoidal tone burst and its Fourier transform.
Frequency Domain Methods
526
Chap. 15
Solution The forward Fourier transform is FU(J))
=
=
L:T/2
(i)
f(t)e-iwt dt
l
sin(cr>ot)e-iCIII dt
(ii)
-T/2
The sinusoid is expanded using the Euler fonnula: FU(J)>
= -:--1 lT/2 (e-i
e-it)
dt
(iii)
}2 -T/2
=
1
(e-j(OJ-OIO)t,T/2 _
2((J)- a>o)
= _
-T/2
1
2((J) + CtJo)
[e-i
/!.. {sin [((J)- CtJo) T /2] _ sin [((J) + CtJo) T /2]} ((J) - OJo) T /2
2
-T/2
(w + cr>o) T /2
(iv) (v)
which is a purely imaginary odd function that is the sum of a pair of shifted imaginary (sin x) I x functions centered on frequencies ±(J)Q, as shown in Fig. 15.16. The zero crossings of the main lobe of each function are at cr>o ± 27r /T, indicating again that as the duration of a transient waveform is d~ its spectral width increases. In this case notice that as the duration T is increased, the spectrum becomes narrower, and in the limit, as T ~ oo, it achieves zero · width and becomes a simple line spectrum. Example 15.12
Fmd the Fourier transfonn of the one-sided real exponential function 0
f(t) = {
e-'
t < 0 t ;::: 0
for a> 0, as shown in Fig. 15.17.
lFUw>l
f(t}
1.0
Fourier transform 0.5
0
2
3
4
at
Figura 15.17: The one-sided real exponential function and its spectrum.
Sec. 15.4
527
Fourier Analysis of Transient Wavefonns
Solution From the definition of the forward Fourier transfonn, FUw)
=
L:
f(t)e-iOJt dt
(i)
= looo e-at e-iOJt dt
(ii)
=
[~e-
o
1
(iii) (iv)
=--
a+ jw
which is complex, and in terms of a magnitude and phase function is . lFUw)l=
1
~
va2 + w2
LFUw) = tan- 1 (
~w)
(v) (vi)
Example 15.13 Fmd the Fourier transform of a damped one-sided sinusoidal function 0 f(t) = { e-at sin(l)Qt
t < 0 t ~0
for u > 0, as shown in Fig. 15.18.
/(t)
1.0
Fourier transform
~
0.5
0 -0.5 -1.0
Figure 15.18:
A damped sinusoidal function and its specttum.
Solution From the definition of the forward Fourier transform, FUw)
=
L:
f(t)e-iOJt dt
(i)
528
Frequency Domain Methods
= ~1
1
00 • e-a' (eiwot- e-Jwo') e-l(l)f dt
}2 0
= 2._ [ j2
1
a+ j(OJo- w) _!_ [
(ii)
e-[cr+J(a~-wo)}rloo-
1
a+ j(wo + w)
j2
(iii)
0
e-[cr+J<~>Jtloo
ClJQ
o
(iv) (v)
=----~
(a
Chap. IS
+ jw)2 + ~
The magnitude and phase of this complex quantity are IF(jw)l •
=
wo
J
LFUw) =tan
-2crw
-1
2 2 a 2 +Wo-w
(vi)
(vii)
15.4.2 Properties of the Fourier Transform
A full description of the properties of the Fourier transform is beyond the scope and intent of this boo~ and the interested reader is referred to the many texts devoted to the Fourier transform [2, 4, 5]. The properties listed below are presented because of their importance in the study of system dynamics.
1. Existence of the Fourier transform: A modified fonn of the three Dirichelet conditions presented for the Fourier series guarantees the existence of the Fourier transform. These are sufficient conditions but are not strictly necessary. For the Fourier transform the conditions are • The function f(t) must be integrable in the absolute sense over all time; that is,
L:
1/(t)l dt < oo
• There must be at most a finite number of maxima and minima in the function f(t). Notice that periodic functions are excluded by this and the previous condition. • There must be at most a finite number of discontinuities in the function f(t), and all such discontinuities must be finite in magnitude. 2. Linearity of the Fourier transform: Like the Fourier series, the Fourier transform is a linear operation. If two functions of time g(t) and h(t) have Fourier transforms G (j w) and H (j w), that is,
k h(t) k
g(t)
G(jw) H(jw)
Sec. 15.4
Fourier AnaJysis of Transient Waveforms
529
and a third function f(t) = ag(t) + bh(t), where a and b are constants, then the Fourier transform of f(t) is F(jw) = aG(jw)
+ bH(jw)
(15.36)
3. Even and odd functions: The Fourier transform of an even function of time is a purely real function, and the transform of an odd function is an imaginary function. The Fourier transform shows conjugate symmetry, that is, F(jw) = F(- jw}
(15.37)
or 9l [F(jw}] = 9l [F(- jw)] ~
[F(jw)] = -t) [F(- jw)]
(15.38) (15.39)
therefore the Fourier transform of an even function is both real and even, while the transform of an odd function is both imaginary and odd. 4. Time-shifting: Let /(t) be a waveform with a Fourier transform F(jw}, that is, F{f(t)} = F(jw)
then the Fourier transform of f(t F{f(t
+ t), a time-shifted version of f(t), is
+ t)} = ejw-c F(jw)
This result can be shown easily since by definition
L: =i:
:F{f(t + T)} = If the variable v = t
+t
f(t
+ T)e-i"" dt
is substituted in the integral,
:F{f(t + T)}
f(v)e-j.,(u-r) dv
1
00
= ejw-c
f(v)e-jwv dv
(15.40)
-oo
= eiw-c F(jw)
If F (j w) is expressed in polar form, having a magnitude and phase angle, this relationship may be rewritten as F{f(t
+ t)} =
IF(jw)l ei[ LF(jw)+w-c]
(15.41)
which indicates that the Fourier transform of a time-shifted waveform has the same magnitude function as the original waveform but has an additional phase-shift term that is directly proportional to frequency.
530
Freque~cy Domain Methods
Chap. IS
S. Waveform energy: We have asserted that the time domain repres~ntation f(t) and the frequency domain representation F Uw) are both complete descriptions of the function related through the Fourier transform f(t)
~ F(jw)
If we consider the function f(t) to be a system through- or across-variable, the instantaneous power that is dissipated in a D-type element with a value of unity is equal to the square of its instantaneous value. For example, the power dissipated when voltage v(t) is applied to an electric resistance of 1 Cis v2 (t). The power associated with a complex variable v(t) is lv(t)l 2 • The "energy" of an aperiodic function in the time domain is defined as the integral of this hypothetical instantaneous power over an time E
=
L:
2
(15.42)
1/(1)1 dl
Parseval's theorem [3] asserts the equivalence of the total waveform energy in the time and frequency domains by the relationship
L:
2
1/(1)1 dl = =
~ ~
L: L:
2
IFUwll da> (15.43)
FUwlFUw) dw
In other words, the quantity IF(jw)l 2 is a measure of the energy of the function per unit bandwidth. The energy !J.E contained between two frequencies w 1 and C02 is !J.E
= -21 1CI>l F(jw)F(jw)dw 7r
(15.44)
CI.IJ
Notice that this is a one-sided energy content and that because the Fourier transform is a two-sided spectral representation, the total energy in a real function includes contributions from both positive and negative frequencies. The function 4>(jw)
= IF(jw)l 2
is a very important quantity in experimental system dynamics and is known as the
energy ·density spectrum. It is a real function, with units of energy per unit bandwidth, and shows how the.energy of a waveform f(t) is distributed across the spectrum.
6. Relationship between the Fourier transform and the Fourier series of a periodic extension of an aperiodic function: Let f (t) be a function, with Fourier transform F(j{J)), that exists only in a defined interval It I < 8/2 centered on the time origin. Then if /p(t) is a periodic extension of f(t), formed by repeating f(t) at intervals
Sec. 15.5
Fourier Transform-Based Properties of Linear Systems
531
T > A, each period contains f(t). Then the Fourier coefficients describing /p(t) are
1 fT/2 /p(t)e-lnOJot dt T -T/2
Cn = -
= -1 [ T
00
f(t)e-}nOJot dt
(15.45)
-oo
= ]_F(jnwa) T
The Fourier coefficients of a periodic function are scaled samples of the Fourier transform of the function contained within a single period. The transform thus forms the envelope function for the definition of the Fourier series as discussed in Sec. 15.2. 7. The Fourier transform of the derivative of a function: H a function f(t) has a Fourier transform F (j w), then
.r{d,} = j"'FU"') which is easily shown using integration by parts:
:F{ df} = foo df e-l(J)t dt dt dt -00
= lf(t)e-i(J)t l~oo- [ = 0 + jwFUw)
00
f(t)(- jw)e-j1JJ1 dt
-oo
since by definition f(t) = 0 at t = ±oo. This result can be applied repetitively to show that the Fourier transform of the nth derivative of f(t) is (15.46)
5.5
FOURIER TRANSFORM-BASED PROPERTIES OF LINEAR SYSTEMS 15.5.1 Response of Linear Systems to Aperiodic Inputs
The Fourier transform provides an alternative method for computing the response of a linear system to a transient input. Assume that a linear system with frequency response H U w) is initially at rest and is subsequently subjected to an aperiodic input u(t) having a Fourier transform U(jw). The task is to compute the response y(t). Assume that the input function u(t) is a periodic function up(t) with an arbitrarily large period T and that since up(t) is periodic, it can be described by a set of complex Fourier coefficients {Unl· The output Yp(t) is periodic with period T and is described
532
Frequency Domain Methods
Chap. IS
by its own. set of Fourier coefficients {Yn}. In .Sec. 15.3.2 it was shown that the output coefficients are
Yn
= H(jn(IJQ)Un
(15.47)
and that the Fourier synthesis equation for the output is co
Yp(t) =
E
Yne)nOJOt
(15.48)
E
H Un(IJQ)UneinOJOt
(15.49)
n;:::-co co
=
n=-co
=
Eco
HUn(IJQ) [ -W()
1T/2 Up(t)e-JnOJOt . dt]eJnOJOt .
21r
n=-co
As in the development of the Fourier transform, we let T the limit we write the summation as an integral:
y(t)
(15.50)
-T/2 ~
oo, so uP (t)
~
u (t). And in
= T-+co lim Yp(t)
1co HUw) [-21r1 1co u(t)e-J(I)t. dt]eJ(I)f. dw = 2._ 1co HUw)U(jw)ei(l)t dw 21r
=
-co
-co
(15.51)
-co
This equation expresses the system output y(t) in the form of the inverse Fourier transform of the product H (jw )U (jw) or
y(t)
=F
1
{HUw)U(jw)}
(15.52)
We can therefore write
Y(jw)
= HUw)U(jw)
(15.53)
which is the fundamental frequency domain input-output relationship for a linear system. The output spectrum is therefore the product of the input spectrum and the system frequency response function: Given a relaxed linear system with a frequency response H U"') and an input that possesses a Fourier transfonn, the response may be found using the following three steps: 1. Compute the Fourier transfonn of the input
uuCI)> = .1"{u(t)} 2. Form the output spectrum as the product Y(j"')
= HUw>UUw)
3. Compute the inverse Fourier transfonn y(t)
=F
1
{YUw)}
Sec. 15.5
533
Fourier Transform-Based Properties of Linear Systems
Figure 15.19 illustrates the steps involved in computing the system response using the
Fourier transfonn method.
u(t)
n~ uu(J)>
·I ~
Linear system
y(t)
HU(J)>
1]"~' ~
Multiplication
YU(J)) = UU(J))HU(J)>
Figure 15.19: Frequency domain computation of system response.
Example 15.14 Use the Fourier transform method to find the response of a linear first-order system with a transfer function 1
H(s)= - -
-rs+ 1
to a one-sided decaying exponential input u(t)
0
t <
= { e-at
0
t~O
where a> 0. Solution The frequency response of the system is H(
JW 0
)
= 11T1/T + JW o
(i)
and from Example 15.12 the Fourier transform of a decaying exponential input is UUw)
= F{u(t)} =J='{e-at}
(ii)
The output spectrum is the product of the transfer function and the frequency response: YUw>
= 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"'(•+
= v + 't',
HUwlUUw) =
=
f(t- T)g(T)dT] e-i"" dt
(t) * g(t)] e-'"" dt
(15.64)
= :F{f(t) * g(t)} A dual property holds: If any two functions, f(t) and g(t), are multiplied together in the time domain, then the Fourier transfonn of their product is a convolution of their spectra. The dual convolution-multiplication properties are /(t) * g(t)
k
:F
/(t)g(t)
<===:}
F(j(I))G(jw) 1 . 23r F(jw) * G(j(l))
(15.65) (15.66)
15.5.5 The Frequency Response of Interconnected Systems
w)
If two linear systems Ht (j(I)) and H2 U are connected in cascade or series, as shown in Fig. 15.22, so that the output variable of the first is the input to the second, then provided
Sec. 15.6
The Laplace Transform
539
the interconnection does not affect Yl (t), the overall frequency response is Y2(jw) = H2(jw)Y1 (jw)
= H2(jw) {H1(jw)U(jw)}
(15.67)
= {H2(jw)Ht(jw)} UUw)
u(t)-8--B----
y(t}
~ u(t} ~y(t}
y(t}
~ u(t}~y(t}
u(t)
Figure 16.22: Frequency response of cascaded and parallel linear systems.
The overall frequency response is therefore the product of the two cascaded frequency responses: (15.68)
Similarly, if two linear systems are connected in parallel so that their outputs are summed together, then Y(jw)
= Ht (jw)U (jw) + H2(jw)U (jw)
= {Ht Uw) + H2(jw)} U(jw)
(15.69)
and so the overall frequency response is the sum of the two component frequency responses: (15.70)
5.6
THE LAPLACE TRANSFORM While the Fourier transform is an important theoretical and practical tool for the analysis and design of linear systems, there are classes of waveforms for which the integral defining the transfonn does not converge. Two important functions that do not have simple Fourier transforms are the unit step function, u,(t) = {
~
and the ramp function, r(t)
= {~
t::50 t>O
Frequency Domain Methods
540
Chap. 15
Neither of these functions is integrable in the absolute sense; for example,
L:
lu,(t)l dt = oo
and the forward Fourier transform
FUw) =
L:
f(t)ti-J"" dt
does not converge for either function. The Laplace transform is a generalized form of the Fourier ttansfonn that exists for a much broader range of functions [8--1 0]. The development of the Fourier transform, described in Sec. 15.3, requires that the time function /(t) be limited in duration and can be described by a Fourier series of a periodic extension of the waveform. Neither the step nor the ramp function satisfies this condition; they are representative of a broad range of functions that are unlimited in extent The Laplace transform of f(t) is the Fourier transform of a modified function, formed by multiplying f(t) by a weighting function w(t) that forces the product f(t)w(t) to zero as time t becomes large. In particular, the Laplace transform uses an exponential weighting function w(t) = e-trt (15.71) where u· is real. Figure 15.23 shows how this function forces the product w(t)/(t) to zero for large values of t. Then_ for a given value of u, provided
L:
!w(t)f(t)! dt < oo
the Fourier transform of f(t)e-tr', (15.72) will exist. The modified transform is not a function of angular frequency w alone but also of the value of the weighting constant u. The Laplace transform combines both w and a into a single complex variable s,
s =a+ jw
(15.73)
and defines the two-sided transform as a function of the complex variable s:
F(s)
=
1
00
(f(t)e-tr')e-i(l)t dt
-oo
=
1
(15.74)
00
-oo
f(t)e-st dt
Sec. 15.6
541
The Laplace Transfonn
Figure 15.23: Modification of a function by a multiplicative weighting function.
For a given f(t) the integral may converge for some values of a but not others. The region of convergence (ROC) of the integral in the complex s-plane is an important qualification that should be specified for each transfonn F(s). Notice that when a= 0, so w(t) = 1, the Laplace transform reverts to the Fourier transform. Thus, if f(t) has a Fourier transform FUw)
= F(s) ls==jw
(15.75)
Stated another way, a function f(t) has a Fourier transform if the region of convergence of the Laplace transfonn in the s-plane includes the imaginary aXis. In engineering analyses it is usual to restrict application of the Laplace transform to functions for which f (t) = 0 for t < 0. Under this restriction the integrand is zero for all negative time and the limits on the integral may be changed: F(s) =
fo"" f(t)e-" dt
(15.76)
which is commonly known as the one-sided Laplace transform. In this book we discuss only the properties and use of this one-sided transform and refer to it as the Laplace transform. It should be kept clearly in mind that the requirement f(t)
=0
fort< 0
must be met in order to satisfy the definition of the Laplace transform. The inverse Laplace transform may be defined from the Fourier transform. Since F(s)
= F(a + jw) = .1" {f(t)e-ar}
the inverse Fourier transform of F(s) is f(t)e-at
= .1"{F(s)} = -1
21Z"
00
/
F(u -oo
+ jw)e1wr dw •
(15.77)
Frequency Domain Methods
542
Chap. IS
If each side of the equation is multiplied by etr',
= 2n1 100 -oo F(s)t!' d(J)
/(t)
(15.78)
The variable of integration may be changed from (J) to s =a+ j(J), and sods= j d(J) and with the corresponding change in the limits the inverse Laplace transform is f(t)
= -2 I . 1tr+joo F(s)est ds 1CJ a-joo
(15.79)
The evaluation of this integral requires integration along a path parallel to the j (I)-axis in the complex s-plane. As will be shown below, it is rarely necessary to compute the inverse Laplace transform in practice. The one-sided Laplace transform pair is defined as
fo"" f(t)e-" dt
(15.80)
1 1a+joo f(t) = - . F(s)t!1 ds 21CJ tr-joo
(15.81)
F(s) =
The equations are a transform pair in the sense that it is possible to move uniquely between the two representations. The Laplace transform retains many of the properties of the Fourier transform and is widely used throughout engineering systems analysis. We adopt a nomenclature similar to that used for the Fourier transform to indicate Laplace transform relationships between variables. Tune domain functions are designated by a lowercase letter, such as y(t), and for the frequency domain function use the same uppercase letter, Y (s). For one-sided waveforms we differentiate between the Laplace and Fourier transforms by the argument F(s) or FU(J)) on the basis that F(j(J))
= F(s)ls=jtl>
A bidirectional Laplace transform relationship between a pair of variables is indicated by f(t)
~
F(s)
and the operations of the forward and inverse Laplace transforms are written £ {/(t)} = F(s)
.c-• {F(s)} =
f(t)
15.6.1 Laplace Transform Examples Example 15.16
Find the Laplace transform of the unit step function 0
u,(t)
={1
t < 0 t ~ 0
Sec. 15.6
543
The Laplace Transform
Solution From the definition of the Laplace transform, F(s)
= Loo f(t)e-'' dt
(i)
= Loo e_,, dt = [-~e-''1~ 1
(ii)
=s
provided u > 0. Notice that the integral does not converge for u unit step does not have a Fourier transform.
= 0, and therefore that the
Example 15.17
Find the Laplace transform of the one-sided real exponential function t <0
0 { ~,
I (t) =
t ;:
o
Solution In Example 15.12 the Fourier transform of a real exponential waveform with a negative exponent was found. In this example we let the exponent be positive or negative: F(s)
1
=
00
f(t)e-'' dt
(i)
= Loo e-
= [--~-e-
0
1
=-s-a
(ii)
The integral will converge only if u >a, and therefore the region of convergence is all of the s-plane to the right of u =a. Example 15.18
Find the Laplace transform of the one-sided ramp function f(t) = { 0 t
t < 0 0
t:::
Solution The ramp function does not possess a Fourier transform, but its Laplace transform
is F(s)
= Loo te-
11
dt
(i)
and integrating by parts, F(s)
00 11 1
1 = [ --te-
s
0
+-1 100 e-st dt s
0
(ii)
544
Frequency Domain Methods
Chap. 15
(iii)
The region of convergence is all of the s-plane to the right of u
= 0, that is, the right half-plane.
Example 15.19
Find the Laplace transform of the Dirac delta function cS(t). In Example 15.10 it was shown that cS(t) has the important property that F(cS(t)) = 1. Solution When substituted into the Laplace transfo~
1
~(s) =
00
&(t)e-11 dt
(i) (ii)
=1
by the sifting property of the impulse function. Thus, cS(t) has a similar property in the Fourier and Laplace domains; its transform is unity, and it converges everywhere. Example 15.20
Find the Laplace transform of a one-sided sinUsoidal function
/(t)
= { sin~t
tSO t>O
Solution The Laplace transform is
1
00
F(s) =
(i)
sin(wor)e-s' dt
and the sine may be expanded as a pair of complex exponentials using the Euler formula: F(s)
= ~ roo (e-[CT+j(CII-GlQ))t J2
=[ -
=
lo
I
u
+ j(CIJ- wo)
e-[CT+}(CIIi-CIIo)1t)
dt
1
e-(CT+}(CII-GlQ)Jtloo- [ -
o
(ii)
u
+ j(w + wo)
e-(CT+j(CII+ct~D>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-<1 dt
and if the variable of integration is changed to v C.{f(t + T)) =
= t + T,
fo"" f(v)e-•<•-•J dv (15.84)
= e" fo"" f(v)e-•• dv =esT F(s)
4. The Laplace transform of the derivative of a function: If a function f(t) has a Laplace transfonn F(s), the Laplace transform of the derivative of f(t) is C. {
i,} =
(15.85)
sF(s)- /(0)
Using integration by parts, £ { df } dt
=
roo df e-st dt
)0
dt
= [/(t)e-"1:;" + fooo sf(t)e_., dt = sF(s)- /(0) This procedure may be repeated to find the Laplace transform of higher-order derivatives. For example, the Laplace transform of the second derivative is
. £
dt2 = s [s£ {f(t)}- f(O)] - Ml dt { ~/}
=s
2
dfl F(s)- sf(O)-dt
t=O
t=O
(15.86)
Sec. 15.6
547
The Laplace Transfonn
which may be generalized to (15.87)
for the nth derivative of f(t). 5. The Laplace transform of the integral of a function: Iff (t) is a one-sided function of time with a Laplace transform F(s), the Laplace transform of the integral of f(t) is C
{f
f(r)dr} = ;F(s)
(15.88)
If a new function g(t), which is the integral of f(t), is defined as g(t) =
f
/(r)dr
then the derivative property shows that £ {/(t)} = sG(s)- g(O)
and since g(O) = 0, we obtain the desired result G(s)
1 = -F(s) s
6. The Laplace transform of a periodic function: The Laplace transform of a onesided periodic continuous function with period T ( > 0) is F(s) =
1 1- e-s
T
f.T f(t)e-st dt 0
Define a new function /1 (t) over one period of the waveform
/I (t) = {/ (t) 0 < t ~ T 0
otherwise
so that /(t) may be written . f(t)
= /t(t) + /I(I- T) + ft(l- 2T) + /1(t- 3T) + · · ·
Then using the time-shifting property above, F(s)
=
F1 (s) +e-sT Ft (s)
+ e-s2T F1 (s) + e-s3T F1 (s) + · · ·
= (l +e-sT+ e-s2T + e-s3T + .. ·) Ft (s)
(15.89)
.
Frequency Domain Methods
548
Chap.l5
The quantity in parentheses is a geometric series. whose sum is 1/(1 - e-"T), with the desired result F(s)
1
= 1 _e-sT Ft (s) =
1 1
-e
_T s
(T f(t)e-"
Jo
dt
7. The final value theorem: The final value theorem relates the steady-state behavior of a time domain function l(t) to its Laplace transform. It applies only if l(t) does in fact settle down to a steady {constant) value as t ~ oo. For example, a sinusoidal function sin wt does not have a steady-state value, and the final value theorem does not apply. If I (t) and its first derivative both have Laplace transforms, and if lim,_oo I (t) exists, then (15.90) lim l(t) = .J-0 lim s F(s) ,_00 To prove the theorem, consider the limit ass approaches zero in the Laplace transform ·of the derivative lim foo [dd f(t)] e-st dt = lim [sF(s)- 1(0)]
s-olo
s-o
t
from the derivative property above. Since lims-o e-st
f' [:,J] dt
= /(1)
= 1,
ig"
= l(oo)- 1(0) =lim sF(s)- 1(0)
s-o
from which we conclude that
l(oo) = lim s F(s)
s-o
15.6.3 Computation of the Inverse Laplace Transform
Evaluation of the inverse Laplace transform integral, defined in Eq. ( 15.81 ), involves contour integration in the region of convergence in the complex plane along a path parallel to the imaginary axis. In practice this integral is rarely solved, and the inverse transform is found by recourse to tables of transform pairs, such as Table 15.2. In systems analysis Laplace transforms usually appear as rational functions of the complex variable s, that is, .
F() s
= N(s)
D(s)
where the degree of the numerator polynomial N (s) is at most equal to the degree of the denominator polynomial D(s). The method of partial fractions, described in App. C, may
Sec. 15.6
The Laplace Transform
549
be used to express F (s) as a sum of much simpler rational functions, all of which have well-known inverse transforms. For example, suppose that F(s) may be written in factored
form as F(s) = K(s + bJ)(s + b,.) · • · (s + bm) (s + a1)(s + a2) · • · (s +an)
where n ~ m, and a1, a2, •.. , an and bt, b,., ... , bm are all either real or appear in complex conjugate pairs, if all the a; are distinct, then the transform may be written as a sum of first-order tenns: A1
F(s) = - s + a1
A2
An
+- + ··· + s + a2 s +an
where the partial fraction coefficients At, A2, ... , An are found from the residues,
N(s)]
A;= [ (s +a;)-D(s) s=-ai
as described in App. C. From Table 15.2 each first-order term corresponds to an exponential time function, -at J:, 1 e <==> - s+a
so the complete inverse transform is
Example 15.21 Find the inverse Laplace transform of
F(s)
=
6s + 14 s 2 +4s + 3
(i)
Solution The partial fraction expansion is
6s
F(s)
+ 14
= (s + 3)(s + 1) At A2 =--+-s+3 s+1
(ii)
where A1 and A2 are found from the residues,
At
= [(s + 3)F(s)]s=-3 =[ =2
4 6 ;:
~ lc-3
(iii)
550
Frequency Domain Methods and similarly, A2
Chap. 15
= 4. Then from Table 15.2, f(t}
= .c-• {F(s)}
{-2 }+.c-• {-4 } s+l
(iv}
=£-] . s+3 =
2e-3t
+ 4e-'
for t > 0
As described in App. C, ifthe denominator polynomial D (s) contains repeated factors, the partial fraction expansion of F (s) contains additional terms involving higher powers of the factor. Example 15.22 F~d
the inverse Laplace transform of F(s}
= 5s2 + 3s + 1 = 5s2 + 3s + I s2 (s + 1}
sl +s2
Solution In this case there is a repeated factor s 2 in the denominator, and the partial fraction
expansion contains an additional term: AI
F(s)
=7 2
A2
A3
+ "$2 + s + I
I =; + s2
. (i)
3 + s+ 1
The inverse transform of the three components can be found in Table 15.2, and the total solution is therefore "f(t)
= .c-1 {F(s}}
= 2£-1 { ~} + .c-• { s12 } + 3£-J { s = 2+t +3e-' 15.7
for
t
!
I}
(ii}
> 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 <
558
Frequency Domain Methods
Chap. 15
15.3. Determine the exponential and sinusoidal forms of the Fourier series and plot the spectrum for the waveform shown in Fig. 15.24. Show how closely the first four terms of the series represents the function. j(t)
tiT
Figure 15.24: A periodic triangular waveform.
15.4. Figure 15.25 shows a half-\vave diode rectifier and the output voltage waveform. Compute the spectrum of this waveform in both exponential and sinusoidal forms. Detennine the average or de value of the waveform.
cos(SOm)
-0.04 -0.02
0
0.02
0.04
t (s)
Figure 15.25: A half-wave rectified sinusoidal waveform.
15.5. A periodic waveform /(t), with period T, is described by a complex Fourier series with coefficients Fn, that is /(t)
=
L 00
Fnejn(21r/T)t
n...-oo
Determine the corresponding coefficients for the waveforms (a) /(-t)
(b) df dt (c) f(t- T)
f(t)dt (Assume Fo
(d) [
= 0)
00
15.6. Let the wavefonn in Fig. 15.4 be designated f(t) and the waveform in Fig. 15.5 be designated g(t), each with period T.
=
(a) Sketch the waveform r(t) f(t - T /4) + 0.5g(t) (b) Determine the Fourier series for r(t) from the Fourier series for /(t) and g(t).
Chap. 15
559
Problems
15.7. A periodic signal /(t) = lsin(wt)l is the input to a system with transfer function 1 H(s) = 0.2s + 1
(a) Determine the Fourier series for f(t) and plot the spectrum. (b) Determine the steady-state system response using the Fourier series of the input
15.8. Consider the waveform in Fig. 15.26. ft.t)
-2
-1
2
0
tiT
Figure 15.26: A square waveform.
(a) Use the time shift and linearity properties to determine the Fourier series of the waveform using
the results from Ex. 15.4. (b) Determine the Fourier series of a waveform similar to that of Fig. 15.26, the difference being
that it spans an amplitude range of -1.0 to 1.0 instead of from 0.0 to 2.0 as in the figure. 15.9. A rotating cam and follower, shown in Fig. 15.27, is used to drive a valve in a fluid delivery system. The cam's profile is specified as its radius r(9) as a function of angular rotation 9 in the range -1r /2 ::: 9 < 1r r(9)
4 = 5- -191 1r
191 ::: 1r /2
=3
Ir/2 < 191 < 3Ir/
The cam rotates at a constant angular velocity of 100 rpm and the follower stays in contact with the cam at all times.
Figure 15.27: A cam and follower.
(a) Express the linear displacement y(t) of the follower as a Fourier series. (b) Express the linear velocity v(t) of the follower as a Fourier series.
Frequency Domain Methods
560
Chap. 15
15.10. An electric generator is driven by a multicylinder diesel engine. The generator rotor-flexible drive-bearing system has a transfer function relating the rotor angular velocity Q,(t) to the diesel engine shaft angular velocity C(t) of 40000
H(s)
= s2 + 20s + 40000
The diesel engine output angular velocity has a number of harmonic variations because of the multicylinder firings. Measurements have shown the engine output angular velocity may be represented as C(t) =Co [1 + ~ sin(2Cot) + ~sin(4Cor)] where Co is the nominal angular velocity. (a) If the engine operates at Co = 50 radls, determine the rotor speed. Do any of the harmonics in
the motor output excite the resonance in the shaft-rotor system? (b) What is the effect of doubling the nominal operating speed to 100 rad/s?
15.1L Detennine the Fourier transform of the transient functions (a) A single sine pulse: f(t)
= sin(21rt/T) =0
for - T /2 ~ t < T /2 for ltl > T/2
(b) A half cosine pulse: f(t)
= cos(27rt/T) =0
for - T /4 ~ t < T /4 for ltl > T/4
(c) A decaying cosine:
e-21 cos(21rt) =0
f(t) =
fort 2:: 0 fort< 0
15.U. Detennine the Fourier transfonns of the transient signals in Fig. 15.28. How are the signals and their transforms related? j(t)
j(t)
T ··-·~
0 ~1
(a)
Figure 15.28:
(b)
Two ttansient signals.
Chap. 15
561
Problems
15.13. Use the time shift and linearity properties of the Fourier transform and the results of Ex. 15.9 to detennine the Fourier transfonn of a single pulse f(t)
= 20 =0
forO~ t ~ 4s fort 4s
15.14. By computing the inverse Fourier transfonn, show that the transfonns of sinusoidal wavefonns are impulses in the frequency domain
F{cos (CtJot)}
=1r [cS(w- Wo) + cS(w + Wo)]
F{sin (Wot)} = ~ [cS(w- Wo)- cS(w + Wo)] J
[Hint: Use the sifting property of cS(t).]
15.15. Parseval's theorem (Sec. 15.4.2) relates the tota1 energy in a wavefonn in the time and frequency domains
Prove this relationship. 15.16. A real waveform f(t) has a Fourier transform F(jw) with a magnitude
IFU{I)>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
= E
x(xT)
(A.25)
i=l
A.4.2 The Determinant
The determinant of an n x n square matrix, written det A, is an important scalar quantity that is a function of the elements of the matrix. When written using the elements of the matrix, the determinant is enclosed between vertical bars, for example, det A =
a11
a12
a131
a21
a22
a23
D31
a32
a33
I
(A.26)
The determinant of a matrix of size n x n is defined recursively in terms of lower-order determinants [(n- l)(n- 1)] as follows. The minor of an element a;i, written M;1 , is another determinant of order n - 1 that is formed by deleting the i th row and the jth column from the original determinant. In the above example, (A.27)
The cofactor a;1 of an element aiJ in a determinant is simply its minor M;1 multiplied by either 1 or -1, depending upon its position, that is, cof a;J
= aiJ = (-1)''+'1 M;J
(A.28)
In the case of the 3 x 3 example above, a23
=
5
(-1) M23
=-I
au
a31
(A.29)
At the limit, the determinant of a 1 x 1 matrix (a scalar) is the value of the scalar itself. The determinant of a high-order matrix is found by expanding in terms of the elements of any selected row or column and the cofactors of these elements. If the expansion is done by selecting the i th row, the determinant is defined as a sum of order n - 1 determinants as n
det A =
L
a;iaiJ
(A.30)
J=l
that is, the sum of products of the elements of the row and their cofactors. Similarly, the expansion in terms of the elements of the j th column of the determinant is n
det A =
L a;Jaii i=l
(A.31)
Introduction to Matrix Algebra
570
App.A
H a 2 x 2 determinant is expanded· by the top row, the result is detA
= I au a21
a
12
an
1
= auau + a12a12 =au an- a12a21
(A.32)
If A is a 3 x 3 matrix and the expansion is done by the first column, then
au
det A = a21
a12
a131
a22
a23
a31
a32
a33
I
=
I
au an
a32
I
a131 a33
a231- a2t at2 a33 a32
+ a 31 Ia12
= au (ana33 - a32a23) - a21 {a12a33
+ a31 (a12a23 -
a131
an
a23
{A.33)
- a32a13)
ana13)
A.4.3 The Matrix Inverse Matrix division is not defined. For a square matrix, multiplication by an inverse may be thought of as an analogous operation to the process of multiplication of a scalar quantity by its reciprocal. For a square n x n matrix A, its-inverse A -I is defined as the matrix (if it exists) that pre- or posbnultiplies A to give the identity matrix I,
A- 1A=I
(A.34)
The importance of the inverse matrix can be seen from the solution of a set of algebraic linear equations such as
(A.35)
Ax=b H the inverse A- 1 exists, then premultiplying both sides gives
A- 1Ax = A- 1b
(A.36)
Ix = A- b
(A.37)
1
and since premultiplying a column vector of length n by the nth-order identity matrix I does not affect its value, this process gives an explicit solution to the set of equations X=
A-lb
(A.38)
The inverse of a square matrix does not always exist. Hit does exist, the matrix is nonsingular, if it does not exist, the matrix is singular. The adjoint matrix adj A of a square matrix A is defined as the transpose of the matrix of cofactors of the elements of A, that is,
adjA =
.r
[~II ~~~
a22
~a2n
[~~~
a12
.. .
an
Unl
an2
O!nn
aln
a2n
a12
...
.. ]
~
a21
an2
...
ann
(A.39)
Sec.A.S
571
Eigenvalues and Eigenvectors
The inverse of A is found from the determinant and the adjoint of A, A_ 1 = adjA detA
(A.40)
Notice that the condition for the inverse to exist, that is, for A to be nonsingular, is that detA ::j:. 0. For a 2 x 2 matrix, (A.41) For higher-order matrices the elements in the adjoint matrix must be found by expanding the cofactors of each element For numerical matrices 'of order 4 and greater it is usually expedient to use one of the many computer matrix manipulation packages to compute the inverse.
tS
EIGENVALUES AND EIGENVECTORS In the study of dynamic systems we frequently consider a set of linear algebraic equations ofthefonn (A.42) AX=Ax
where xis a column vector of length n, A is an n x n square matrix, and A is a scalar quantity. The problem to be solved is to find the values of A satisfying this equation and the corresponding vectors x. The values of A are known as the eigenvalues of the matrix A, and the corresponding vectors are the eigenvectors. Equation (A.42) may be written as a set of homogeneous equations, AX -Ax
= (AI -A) X = 0
(A.43)
where 0 is a null column vector and A- au -a21
(Al-A)=
-a12
A-
a22
:
(A.44)
[ -aml
-am2
A theorem of linear algebra states that for a set of homogeneous linear equations a nontrivial set of solutions exists only if the detenninant of the matrix of coefficients is zero. In this case we must have det (A.I - A) = 0 (A.45) The detenninant det (AI - A) is known as the characteristic determinant of the matrix A. Expansion of the determinant results in an nth-order polynomial in A known as the characteristic polynomial of A, and the n eigenvalues of A are the n roots of the characteristic equation fonned by equating the characteristic polynomial to zero as above.
Introduction to Matrix Algebra
572
App.A
ExampleA.l F'md the eigenvalues.of the matrix
The characteristic equation is given by
and has roots (and hence eigenvalues) A1 = 6 and A2 = 1. The eigenvectors corresponding to each of these values may be found by substituting back into the original equations; for example, if A = 6, the equations become -x1 +4x2 =0
(i)
XJ-4X2=0
(ii)
which has the nonunique solution x 1 = 4x2 • Any vector maintaining this ratio between its two elements is therefore an eigenvector coiTeSponding to the eigenvalue A = 6. In general. if X is an eigenvector of A, then so is kx for any scalar value k.
A.6
CRAMER'S METHOD
Cramer's method is a convenient method for manually solving low-order nonhomogeneous sets of linear equations. If the equations are written in matrix form, (A.46)
then the i th element of the vector x may be found directly from a ratio of determinants, detA<1> Xi=--detA
(A.47)
where det A
b. ExampleA.2 Use Cramer's rule to solve the system of equations 2x1 -X2 +2x3
=2
Xt + 10x2 - 3x3 = 5 -x1 +x2 +x3 = -3
Then
2 detA= -1
-1 10
2 -3 =46
Sec.A.6
573
Cramer's Method
and XI=
1 46
-1 10 -3 1
2 -3 1
2 -3
-1
2 5 -3
2 1 -1
-1 10 1
2 5 -3
2 5
=2
X2
1 = 46
2
1
=0
1 X3= 46 =-1
B
Complex Numbers
8.1
INTRODUCTION There are many equations, such as 2x 2 + 2x + 1 = 0, that do not have any solutions that are real numbers. The quadratic fonnula applied to the above equation gives
-2±.J=4
x=----
2
= -1
±.J=I
which is not a real number. However, if we define the symbol j =
.r-1
(B.l)
and write the solution to the above quadratic equation as x=-l±jl
substitution of either one of this pair of complex numbers into the original quadratic equation will show that it does indeed satisfy the equation. Complex numbers, written as either a + j b or a + bj, are defined as having two parts: a is known as the real part, and b is the imaginary part. The symbol j (in some texts written i) is the imaginary unit. For each pair of real numbers a, b there is a corresponding complex number a + jb. Complex numbers are often represented as points in a complex plane by defining the horizontal x-axis as the real axis and the vertical y-axis as the imaginary axis. The resulting xy-coordinate system for representing the value of a complex number may be considered a generalization of the use of the number line for representing the value of real numbers. For 574
Sec. B.2
Complex Number Arithmetic
575
the complex number z = x + j y there is a unique point P in the complex plane with coordinates (x, y). The complex plane upon which complex numbers are represented is known as an Argand diagram. Figure B.I shows this cartesian form of complex number representation. The operators 9t{z) and ~{z) extract the real and imaginary parts of 2, respectively. 5{t} z=x+jy y ---------.. I I I
I I
I I I I
x
9t{zJ
Figure 8.1: Cartesian representation of a complex number on an Argand diagram.
1\vo complex numbers are defined to be equal if they are coincident on the complex plane, that is, if their real parts and their imaginary parts are equal: ZI
= z2
if and only if
Xt
= x2
and
YI
= Y2
Inequalities between complex numbers, such as z1 ::: z2, have no meaning.
1.2
COMPLEX NUMBER ARITHMETIC The arithmetic rules for manipulating complex numbers are defined as follows. Addition: The sum of a pair of complex numbers is defined as the complex number obtained by adding the real and imaginary parts of the two numbers: (B.2)
Subtraction: The difference between a pair of complex numbers is the inverse of addition and is defined as the difference between the real and imaginary parts of two numbers: (B.3)
576
. Complex Numbers
App.B
MultipUcation: The product of a pair of complex numbers is defined as the complex number obtained by applying the ordinmy rules of arithmetic and replacing j 2 by -1: Z1Z2
= (XJ + iYI)(X2 + iY2) = (XtX2- YIYl) + j(XIY2 + X2Y1)
(B.4)
= =
Division: The quotient of a pair of complex numbers, z zi!z2, is defined as the inverse of the product; that is, z is the number that satisfies ZI ZZ2· The easiest way to do this is to multiply both numerator and denominator by x2 - iYl= Zl
XI+ iYl
Z2
X2 + iYl (XI+ iYI)(X2- jy2)
-=---
= (X2 + jy2)(x2- iYl) = XtX2 + YIY2 x~ + y~
(B.S)
.X2Yl - XtY2
+J-~-=--
x~ + y~
Complex numbers obey the same commutative, associative, and distributive laws as real numbers do. Example B.l Let Z1
= 3 + 4j and z2 = 1 -
2j. Then
Zl +z2
=4+j3
Zl -Z2
=2+ j6
ZlZ2
= (3 + j4)(1- 2j) =3-6j+4j+8
= Il-2j Zl
3+4j 1-2j (3 + 4j)(l + 2j) = (1 - 2j)(l + 2j)
i; =
= -1 +2j
The complex conjugate of a number x + j y is defined as the complex number in which the sign of the imaginary part has been reversed, in the above case x - j y. For example, the conjugate of 3 - 4 j is 3 + 4 j. We will denote the conjugate by the bar notation In addition we will denote the real part of a complex number as 9l{z} and the imaginary part as ~{z}. Then, if x = x + jy,
z.
ffi{z}
= x = 21 (z + Z)
~{z}
= y
1
= 2j
(z - Z)
(B.6) (B.7)
Sec. B.3
577
Polar Representation of Complex Numbers
It is easy to show that (ZJ
+ Z2) = Zl + Zl
(B.8)
= Zl - Zt
(B.9)
(Zt - Z2)
= ZtZl Zt/Z2 = "fi/fi ZJZ2
i.3
(B.IO) (B.ll)
POLAR REPRESENTATION OF COMPLEX NUMBERS The complex number z may be considered to be a point P (x, y) in the complex plane, or the directed line segment from the origin to the point P as shown in Fig. B.2. Both interpretations are useful at different times. The magnitude of the complex number may be thought of as the length of the directed line segment, written lz I, and from Fig. B.2 is seen to be (B.12)
while its angle, written Lz, may be defined as a counterclockwise rotation from the real axis by an angle 8 (B.l3)
The magnitude and angle are sufficient to define the location of the point P in the complex plane and are therefore a complete description of the complex number. When written in terms of its magnitude and angle, z = lzl Lz, the complex number is said to be expressed in polar form as opposed to the rectangular or cartesian form z = x + j y. It can be seen from Fig. B.2 that
= lzlcos8 y = lzlsin8
x
(B.l4) (B.l5)
The polar fonn is particularly useful for the operations of multiplication and division of complex numbers. If the pair of complex numbers z1 and z2 are written
= lztlcosOI + j lzd sin01 Z2 = lz2l cos fh + j lz2l sin fh Zl
then the product z1Z2 is ZtZ2
= lztllz2l [(cos Ot cos 82 -sin 9t sin (h) + j (sin 81 cos 82 +cos 81 sin (h)] = lz11lz21 [cos (81 + 82) + j sin (81 + 82)]
578
Complex Numbers
App.B
giving the important results
lztZ21 = lztllz2l LztZ2
(B.l6)
= L1.1 + L1.2
(B.17)
9{t}
z. =x+jy
y
0
9t{z.}
Figure 8.2: Polar representation of a complex number.
Similarly, from the definition of complex division, it follows that
lz2l I Z2ZII=~ Zt
L- =
LZI-
1.2
(B.18)
LZ2
(B.19)
The polar form allows the operations of addition and subtraction to be viewed in terms of vector arithmetic, using the normal parallelogram rules for vector addition and subtraction as shown in Fig. B.3. 9t{z.}
Figure 8.3:
Vector addition and subtraction of complex quantities in polar form.
In polar form the complex conjugate
z becomes
'f = x- jy
= lzl L(-Lz)
and so the magnitude is unchanged but the angle is negated.
Sec. 8.4
1.4
579
Euler's Theorem
EULER'S THEOREM
Euler's theorem allows complex exponentials to be expressed in terms of simple complex numbers. The power series expansions for sin 8 and cos 8 are sin8 =8-
93
9S
91
82
84
86
-+--+··· 3! 5! 7!
cos9=1--+---+··· 2! 4! 6! which can be combined and written
·e3
·82
·84
cos8+ jsin8 = 1 + (j9) + L + L + !.._ +··· 2! 3! 4! The power series expansion for the exponential ex is x2
x3
x4
2!
3!
4!
~=l+x+-+-+-+
...
and comparison of these two series gives Euler's theorem:
cos8 + j sin9
= ei0
(B.20)
A similar argument will show that the conjugate is
cos8- j sin8
= e-i9
(B.21)
These two formulas may be used to express sines and cosines in terms of complex exponentials since by adding and subtracting the two equations,
cos8
=!2 (ei6 + e-i8 )
sin 9 =
2~ (ei
6
-
e-16 )
(B.22) (B.23)
Notice that ei8 is a complex quantity with a magnitude of unity and an angle of 8. Then a complex number z with magnitude R and angle 8 may be written as a complex exponential:
In this form it is easy to show that (B.24)
or that (cos (J + j sin 8)" = cos n6 + j sin n(J which is the well·known theorem of De Moivre.
(B.25)
c Partial Fraction Expansion ofRational Functions
C.1 · INTRODUCTION Rational functions of the fonn F(s)
= bmsm + bm-tSm-l + ·· · + bts + bo ansn
+ an-tSn-l + •· · + a1s + ao
N(s)
(C.l)
= D(s) occur frequently in the field of system dynamics. If the function F (s) is a proper rational function of s, that is, if it has a numerator polynomial N (s) of degree m and a denominator polynomial D(s) of degree n with m < n and with no factors in common, then the method of partial fraction decomposition will allow F (s) to be expressed as a sum of simpler rational functions, F(s)
= Ft (s) + F2(s) + ·· · + Fn(s)
where the component functions F; (s) are of the form
A
or
Bs+C (s 2
+ bs + c)k
(C.2)
The nature and number of terms in the partial fraction expansion are detennined by the factors of the denominator polynomial D(s). It is well known that the linear factors of any polynomial with real coefficients, such as D(s), are either real or appear in complex conjugate pairs. Each pair of complex conjugate factors may be considered to represent two distinct complex factors, or they 580
Sec.C.l
581
Introduction
may be combined to generate a quadratic factor with real coefficients. For example, the polynomial D(s) = s 3 - s2 - s - 15 may be written in tenns of one real linear and one real quadratic factor or in terms of a real linear factor and a pair of complex linear factors: D(s)
= (s = (s -
3)(s2 + 2s + 5) 3)[s + (1
+ 2j)][s + (1 -
2j)]
Each representation will produce a different but equivalent set of partial fractions, and the choice of the representation will depend upon the particular application. The use of the quadratic form will, however, avoid the necessity for complex arithmetic in the expansion. Assume that F (s) is written so that D (s) has a leading coefficient of unity and that D(s) has been factored completely into its linear and quadratic factors. Then, let all repeated factors be collected together so that D(s) may be written as a product of distinct factors of the fonn (s +a;)Pi or where p; is the multiplicity of the factor and where the coefficients a; in the linear factors may be either real or complex, but where the coefficients b; and c; in any quadratic factors must be real. Once this is done, the partial fraction representation of F (s) is composed of the following terms.
Linear Factors: of p terms
For each factor of D(s) of the fonn (s + a; )P, there will be a set A·1
A,·2
A;p
--'+ (s + a; )2 +···+ ---=-s + a; (s + a; )P where Ail, A;2, ... , A;p are constants to be determined.
Quadratic Factors: will be a set of p terms
For each factor of D(s) of the fonn (s 2 + b;s
+ c;)P, there
B;ps + C;p -Bns -+-Cn- + B;2s + C;2 2 + ... + ~--=------:.s2 + b;s + c; (s2 + b;s + c;) (s2 + b;s + c;)P
where B;r, B; 2 , ••• , B;p and Cit, Ci2, ... , C;p are constants to be detennined. Example C.l
Detennine the form of a partial fraction expansion for the rational function F(s)
where N(s) written
=
2
s
2
+ 3s+ 2
= 2 and D(s) = s2 + 3s + 2. If the denominator is factored. the function may be 2
F(s)
= s2 + 3s + 2 =
2
(s
+ l)(s + 2)
582
Partial Fraction Expansion of Rational Functions
App.C
and according to the rules above, each of the two linear factors will introduce a single term into the partial fraction expansion
At
A2
F(s)=-+-s + 1 s +2 where A 1 and A2 have to be found.
ExampleC.l Determine the form of the terms in a partial fraction expansion for the rational function
F(s)2s + 3 - s2(s + l)(s2 + 2s + 5) This example bas a repeated factor s2 , a single real factor (s + 1), and a quadratic factor s 2 + 2s + 5 = [s + (1 + 2j)][s + (1 - 2j)]. The partial fraction expansion may be written in two possible forms: with the quadratic term, F( )
At2
An
B3s + C3
A2
s =-;-+-;r+s+1 +s2 +2s+5 or in terms of a pair of complex factors, F( s)
C.2
An
A3
A12
A4
=-+-+--+ s s2 s + 1 s + (1 + 2j)
As s + (I - 2j)
+---~
EXPANSION USING LINEAR ALGEBRA
The simplest conceptual method of determining the unknown constants A; is to multiply both the original rational function and the expansion by D(s), and then to expand and collect like tenns. Then, by equating coefficients on both sides, a set of independent algebraic linear equations in the unknown constants will be produced and any of the standard solution methods may be used to find the set of values. ExampleC.3 Detennine the partial fraction expansion for the rational function
2s+4 F(s) = s3- 2s2 2s+4 - s2 (s- 2)
The partial fraction expansion will be
F(s)
= Au + A122 + ~ s
s
s-2
Multiplying both sides by D(s) gives
2s + 4
= Aus(s -
2) + At2(s - 2) + A 2 s2
=(Au+ A2)s 2 + (-2An +·At2)s- 2At2
Sec. C.2
583
Expansion Using Linear Algebra
Equating coefficients on both sides gives the set of equations
An
+A2
-2Au
=
0
=
2
=
4
and solving these equations yields
Au= -2
A12
= -2
and so
-2 -2 2 ++s s2 s- 2
F(s) = -
Example C.4 Detennine the partial fraction expansion for the rational function F(s)
=
s2 + s - 2 3s3 - s 2 + 3s - 1 s2 +s -2 =---~(3s- l)(s 2 + 1)
using the quadratic factor s2 + 1 as one of the terms. The partial fraction expansion will be
s2 + s- 2
B2s + C2
At
(3s - 1) (s2 + 1) = 3s - 1 +
s2 + 1
and multiplying through by (3s- 1)(s2 + 1) gives
Equating coefficients gives a set of three equations in the three unknowns which may be solved using any of the methods of linear· algebra: At At
+3B2 -82
= +3C2 -C2
= =
1
-2
These equations may be solved using any of the methods of linear algebra, but if these equations are written in matrix fo~
584
Partial Fraction Expansion of Rational Functions
App.C
Cramer's rule (App. A) may be used to find the constants, for example,
det [
!
-2 At= det
[ 1 ~
=-s7 Similarly, Cramer's rule could be used to find that
4
Bz=-
s
and
giving the final result
s +s - 2 7 1 1 4s + 3 ---..,.........= ---+ --(3s- l)(s2 + 1) S 3s- I 5 s2 + 1 2
C.3
DIRECT COMPUTATION OF THE COEFFICIENTS
If the partial fraction expansion is done in terms of linear factors, the constants may be found directly. To demonstrate the method, assume initially that there are no repeated factors in the denominator polynomial, and so we can write F(s)
=
N(s)
(s
+ a1)(s + a2) · · · (s +a,)
At A2 ~ ~ =--+--+--+···+-s +at s + a2 s +a; s + an If both sides of the equation are now multiplied by the factor s
+ a;' the result is
At A2 A,. F(s)(s +a;)= - - ( s +a;)+ - - ( s +a;)+ A;+···+ - - ( s +a;) s + a1
s + a2
s +an
(C.3)
If this expression is evaluated at s = -a;, all the tenns on the right-hand side except for the A; term will be identically zero, giving an explicit formula for A;: A; = [F(s)(s
+ a;)]ls=-a1
(C.4)
This method will not work directly for all the tenns associated with a repeated factor in D (s). For example, suppose D(s) has a factor with multiplicity of 2 and we write F(s)
=
N(s) (s + at)(s + a2) 2 At A21 A22 =--+--+--"""'""'='2 s +at s +a2 (s +a2)
Sec.C.3
585
Direct Computation of the Coefficients
Then if we attempt to compute resulting expression
A21
as above by multiplying both sides by s +
F(s)(s + a2) = (s
a2,
the
N(s)
+ at)(s + a2)
At = --(s+a2)+A21 s +at
A22
+ -s +a2
cannot be evaluated at s = -a2 because of the remaining terms s + a2 in the denominators. However, if we form
(C.5)
then
and if Eq. (C.5) is differentiated with respect to s, the derivative at s A21 and so
= -a2 has the value (C.6)
In general if a factors+ a; has multiplicity p, then the constant A;1 associated with the
tenn (s
is Aik
+ a;)k
}I = (P _1 k)'· { ddP-k p-Ic [(s- a;)k F(s)] s=-a; S
which will allow direct computation of all the required constants. ExampleC.S
Repeat Example C.4 expanding the quadratic factor as a pair of complex linear factors. If the expansion is done in tenns of complex linear factors, F (s) =
+
s2 s - 2 3s 3 - s 2 3s - 1
+
s2 +s-2
=------1)(s + j)(s- j) (3s -
Then since there are no repeated factors. the partial fraction expansion is A1 A2 A3 -s-+-s --2- - = - +- +(3s - 1)(s + j)(s - j) 3s - 1 s+j s- j 2
(C.7)
586
Partial Fraction Expansion of Rational Functions Using Eq. (C.4) directly, A1
= F(s)(3s- l)ls:::l/3 2
= A2
s +s-21 s2
+1
s=I/3
7 5
=
= F(s)(s + i>ls:::-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
