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MODELING AND ANALYSIS OF DYNAMIC SYSTEMS Third Edition

EDITOR EDITORIAL ASSISTANT Steve Peterson MARKETING MANAGER Katherine Hnm11;cn SENIOR PRODUCTION EDITOR SENIOR DESIGNER Kevin

Cover

courtesy of NASA ancl bouncl

This book was set in Times Roman

Hamilton

Press. This book was

on acid-free paper. @

Copyright 2002 © John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 605 Third Avenue, New York, NY 10158-0012, (212) 850-6011, fax (212) 850-6008, E-Mail: [email protected]. To order books please call 1(800) 225-5945. Library of Congress Cataloging in Publication Data: Close, Charles M. Modeling and analysis of dynamic systems I Charles M. Close and Dean K. Frederick and Jonathan C. Newell-3rd ed. p. cm. Includes bibliographical references. ISBN 0-471-39442-4 (cloth: alk. paper) 1. System analysis. I. Frederick, Dean K., 1934- H. Newell, Jonathan C. HI. Title. QA402.C53 2001 003-dc21 2001033010

Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

To my wife, Margo, and to our children and grandchildren CMC

To my mother, Elizabeth Dean Frederick, and to the memory of my father, Charles Elder Frederick DKF

To my wife, Sigrin, and my sons, Andrew and Raymond JCN

PREFACE The primary purpose of this edition remains the same as in the previous editions: to provide an introductory treatment of dynamic systems suitable for all engineering students regardless of discipline. We have, however, made significant changes as a result of experiences with our many students, comments from numerous professors around the country, and the increasing educational use of computer packages. We have maintained flexibility in the selection and ordering of material. The book can be adapted to several types of courses. One such use is for students who need a detailed treatment of modeling mechanical and electrical systems and of obtaining analytical and computer solutions before proceeding to more advanced levels. Such courses can serve as a foundation for subsequent courses in vehicular dynamics, vibrations, circuits and electronics, chemical process control, linear systems, feedback systems, nuclear reactor control, and biocontrol systems. The book also covers such general topics as transfer functions, state variables, the linearization of nonlinear models, block diagrams, and feedback systems. Hence it is suitable for a general dynamic systems course for students who have completed a disciplinary course such as machine dynamics, electrical circuits, or chemical process dynamics. This text can also be used for students with significant modeling and analysis experience who wish to emphasize computer techniques and feedback control systems. Topics include computer solutions for both linear and nonlinear models, as well as root-locus diagrams, Bode plots, block diagrams, and operational amplifiers. We explain some of the practical design criteria for control systems and illustrate the use of analytical and computer methods to meet those criteria. Finally, the book can provide a general introduction to dynamic systems for students in broad-based engineering programs or in programs such as biomedical and materials engineering who may have limited time for this subject. We assume that the reader has had differential and integral calculus and basic college physics, including mechanics and electrical phenomena. Many students will have had a course in differential equations at least concurrently. We have been careful to present the mathematical results precisely (although without the rigorous proofs required for a mathematics book), so that the concepts learned will remain valid in subsequent courses. For example, the impulse has been treated in a manner that is consistent with distribution theory but is no more difficult to grasp than the usual approach taken in introductory engineering books.

Approach The book reflects the approach we have used for many years in teaching basic courses in dynamic systems. Whether for a particular discipline or for a general engineering

Vil

viii flJ> Preface

course, we have found it valuable to include systems from at least two disciplines in some depth. This illustrates the commonality of the modeling and analysis techniques, encourages students to avoid compartmentalizing their knowledge, and prepares them to work on projects as part of an interdisciplinary team. Mechanical systems are examined first because they are easily visualized and because most students have had previous experience with them. The basic procedures for obtaining models for analytical and computer solutions are developed in terms of translational systems. Models include those in state-variable, input-output, and matrix form, as well as block diagrams. The techniques are quickly extended to rotational systems, and there are also chapters on electrical, electromechanical, thermal, and fluid systems. Each type of system is modeled in terms of its own fundamental laws and nomenclature. After introducing block diagrams in Chapter 4, we show how to use Simulink and MATLAB to obtain responses of simple systems. We introduce Laplace transforms fairly early and use them as the primary means of finding analytical solutions. We emphasize the transfer function as a unifying theme. We treat both linear and nonlinear models, although the book allows nonlinear systems to be deemphasized if desired. Students should realize that inherent nonlinearities generally cannot be ignored in the formulation of an accurate model. Techniques are introduced for approximating nonlinear systems by linear models, as well as for obtaining computer solutions for nonlinear models. We believe that obtaining and interpreting computer solutions of both linear and nonlinear models constitute an important part of any course on dynamic systems. We introduce, early on, MATLAB and Simulink, two computer packages that are widely used in both educational and industrial settings. Although it is possible to use the book without detailed computer work, the inclusion of such methods enhances the understanding of important concepts, permits more interesting examples, allows the early use of computer projects, and prepares the students for real-life work. An important feature is providing motivation and guidance for the reader. Each chapter except Chapter 1 has an introduction and summary. There are approximately 200 examples to reinforce new concepts as soon as they are introduced. Before the examples, there are explicit statements about the points to be illustrated. Where appropriate, comments about the significance of the results follow the examples. There are over 400 end-of-chapter problems. The answers to selected problems are contained in Appendix G.

Organization The majority of the material can be covered in a one semester course, but the book can also be used as the basis for a two quarter or year long course. A number of chapters (including Chapters 10, 11, 12, 14, and 15) and a number of individual sections (including Sections 3.3, 6.7, 8.6, and 9.4) can be omitted or abbreviated without any loss of continuity. The chapters can be grouped into the following four blocks:

1. Modeling of mechanical and electrical systems: Chapters 1 through 6. The sections of Chapter 4 on Simulink and MATLAB can be deferred. However, we believe it is beneficial to encourage students to prepare simulations and to note typical features of the responses even before complete analytical explanations are presented.

Preface

2.

'"''"'"'''"'' ..":ir

ix

no:ndinear models: tool for

3,

4. A core sequence for students without a nrc'""'"Q include 1 9

to oilier we suggest including one or more of 11, and 12. 13 15 can be used as an introduction to the""'~~'""'"' and design of feedback control In '-'"'"'-'"~' of MATLAB's Control '"Q""'rn Rather than provide lengthy ae~;cnpuons we them in an iterative provH>IJ"'""·"'"'-' interpreting crnmpute:rcess to meet some design criteria. in material from The following table of prerequisites should be the later chapters:

matelial Linearization

1

9

Chapter 5 1

Electromechanical systems

10

Chapters 5-8

Thermal systems

11

Chapters 8, 9

Fluid systems

u

8,9

Block

13

8

Feedback systems

14

13

Section 9.2 also requires Chapter 6.

to l,

methods are introduced in a vV,cHj.'''"'''-''.J the rest of the book. the many the instructor's to add features to an eadier the simulation in 4 of ~U l s. a. b. c. d.

with is zero for

an t ~ 1 s and is

"'~HY"'·""'·"'·"''" x for 0 ~ t ~ 8 when M = 4

whenM=20 How does the value of M affect the trial and error, determine the range of values of M for which the x never overshoots the steady-state value.

4.19. Change the value of to 2 and repeat Problem 4.12. Give an explanation in physical terms for the change in the response from the response of Problem 4.12. 4.20. Problem 4.11 wi.th the same values, = 20 and = 20 Give an explanation in nnvQ'"'" in the response compared to that of Problem 4.1 L

for

= 30 N-s/m,

5 ROTATIONAL MECHANICAL SYSTEMS In Chapter 2 we presented the laws governing translational systems and introduced the use of free-body diagrams as an aid in writing equations describing the motion. In Chapter 3 we showed how to rearrange the equations and develop state-variable and input-output models. Extending these procedures to rotational systems requires little in the way of new concepts. We first introduce the three rotational elements that are analogs of mass, friction, and stiffness in translational systems. Two other elements, levers and gears, are characterized in a somewhat different way. The use of interconnection laws and free-body diagrams for rotational systems is very similar to their use for translational systems. In the examples, we seek models consisting of sets of state-variable and output equations, or input-output equations that contain only a single unknown variable. We include examples of combined translational and rotational systems, followed by a computer simulation. ~

5.1 VARIABLES For rotational mechanical systems, the symbols used for the variables are (), angular displacement in radians (rad)

w, angular velocity in radians per second (rad/s) o:, angular acceleration in radians per second per second (rad/s2 )

r, torque in newton-meters (N·m) all of which are functions of time. Angular displacements are measured with respect to some specified reference angle, often the equilibrium orientation of the body or point in question. We shall always choose the reference arrows for the angular displacement, velocity, and acceleration of a body to be in the same direction so that the relationships

w = ()

o:=w=B hold. The conventions used are illustrated in Figure 5 .1, where r denotes an external torque applied to the rotating body by means of some unspecified mechanism, such as by a gear on the supporting shaft. Because of the convention that the assumed positive directions for (), w, and o: are the same, it is not necessary to show all three reference arrows explicitly. The power supplied to the rotating body in Figure 5 .1 is

p=rw

(1)

95

96

Rotational Mechanical

The power is the derivative of the energy w, and the energy t is

to the

up to time

+

devices in rotational are moment and gears. We shall restrict our consideration to elements that rotate about fixed axes in an inertial reference frame.

When Newton's second law is to the differential mass element dm in Figure 5.1 and the result is over the entire body, we obtain d

dt(Jw)=r

(2)

where J w is the angular momentum of the body and where T denotes the net torque applied about the fixed axis of rotation. The J denotes the moment of inertia in kilogrammeters2 We can obtain it out the of r2 dm over the entire body. whose mass M can be considered to concentrated The moment of inertia for a is , where L is the distance from the point to the axis of rotation. Figure 5 .2 at a shows a slender bar and a solid each of which has a total mass M that is uniformly distributed the and of the for J are the center of mass. The results for the case where the axis of rotation passes for other common can be found in basic the center of mass, we can use the theorem. axis that does not pass Let denote the moment of inertia about the axis that passes the center of mass, and let a be the distance between the two axes. Then the desired moment of inertia is l= +

2

l=

+M 2

(4)

502 Element Laws

97

l= (a)

(b)

(c)

(cl)

5.2 Moments of inertia" (a) Slender bar" (b) Disk Slender bar where axis of rotation does not pass through the center of mas so

For two or more components rotating about the same we can find the total moment of inertia by the individual contributions" ff the uniform bar shown in Figure then the masses for the sections of and will be has a total mass (d1 + and + Using (4), we see that the total moment of inertia is J= l M d2 - M(df +dD + 3 2 2 - 3(d1 +dz) We consider nonrelativistic so and constant moments of reduces to fW=T

where wis the angular acceleration" As is the case for a mass translational """"''"' a rotating body can store energy in both kinetic and forms" The kinetic energy is

Wp

=

where M is the mass, g the gravitational constant, and h the height of the center of mass above its reference If the fixed axis of rotation is vertical or passes the center of mass, there is no change in the energy as the rotates, and (8) is not needed" To find the complete response of a containing a we must know its initial angular velocity If its potential energy can vary or if we want to find e(t), then we must also know

Friction A irotatfonal frktiion element is one for which there is an the tmque and the relative friction arises when two two concentric 'v"'""'"' of the T

= B!iw

98

Rotational Mechanical Oil film, B

B

B

(a)

(c)

(b)

5.3 Rotational devices characterized

viscous friction.

win be exerted on each cylinder, in directions that tend to reduce the relative angular velocity Aw. Hence the sense of the frictional must be counterclockwise on the The friction coefficient B has units of inner cylinder and clockwise on the outer newton-meter-seconds. Note that the same is used for translational viscous friction, where it has units of newton-seconds per meter. Equation also to a rotational ua.>UIJU• drive system, such as that shown in Figure or 5 .3 (c). The inertia of the parts is assumed to be negligible or else is accounted for in the mathematical model by separate moments of inertia. If the rotational friction element is assumed to have no inertia, then when a T is applied to one side, a torque of equal magnitude but opposite direction the or some other component), as shown must be exerted on the other side (by a in Figure 5.4(a), where T = B(w2 . Thus in Figure the torquer passes through the first friction element and is exerted directly on the moment of inertia J. Other types of friction, such as the damping vanes shown in Figure 5.5(a), may exert a torque that is not directly proportional to the angular velocity but that may be as shown in Figure 5.5(b). For a linear element, the described by a curve of r versus curve must be a straight line passing through the The power to the friction "'""""""' r is immediately lost from the mechanical in the form of peat.

Rotational stiffness is associated with a torsional such as the ""~U.JI-'• of a or with a relatively thin shaft. It is an element for which there is an

B

B

(b)

(a)

5.41 (a) Rotational

with u'-l';ui;.w'" inertia. element

Torque transmitted through a friction

5.2 Element Laws

e1' so the must be labeled By the law of reaction torques, the effect of the connecting shaft on disk 1 is a with its sense in the clockwise direction. We can reach the same conclusion first selecting a clockwise sense for the arrow in Figure that on disk 1, and then noting that disk 2 wm tend to drive disk 1 in the ~~''""'a if > 81. Thus the correct ex1ore:ss1on is Of course, if we had selected a counterclockwise arrow in we would have labeled the arrow either - 81) or

5.4 Obtaining the System Model

~

107

For each of the free-body diagrams, the algebraic sum of the torques may be set equal to zero by D' Alembert's law, giving the pair of equations fiw1 +B1w1+K181 - K2(82 -81)= 0 fzw2 +B2w2 +K2(82 - 81) - Ta (t)= 0

(~?)

Two of the state-variable equations are B1 = w1 and 02 = wz, and we can find the other two by solving the two equations in (26) for w1 and w2, respectively. Thus

01= w1 1

w1 = -[-(K1 +K2)81 - B1w1+K282] 11

(27)

B2= w2 W2 = ; 2 [K281 - K282 -B2w2 + Ta(t )] The output equations are

Kz(82 -fh) mr = fi w1+ fzw2

TK2

=

where mr is the total angular momentum. To obtain an input-output equation, we rewrite (26) in te1ms of the angular displacements 81 and 82. A slight rearrangement of tenns yields

liB1 + B1B1 + (K1 + K2)B1 - K2B2 = 0

(28a) (28b)

Neither of the equations in (28) can be solved separately, but we want to combine them into a single differential equation that does not contain B1. Because Bi appears in (28b) but none of its derivatives do, we rearrange that equation to solve for 81 as .. . 1 81 = - [JzB2 +B2B2 + K2B2 -Ta(t )J

Kz

Substituting this result into (28a) gives

lihBiiv) + (liB2 + JzB1)eiiii} + (liK2 + hK1 + JzK2 + B1B2)B2 + (B1K2 + B2K1 +B2K2 )f)z +K1K202

= l 1fa +Bda + (K1 + K2) Ta(t)

(29)

which is the desired result. Equation (29) is a fourth-order differential equation relating 02 and Ta(t ), in agreement with the fact that four state variables appear in (27).

In the last example, note the signs when like terms are gathered together in the torque equations corresponding to the free-body diagrams. In (28a) for Ii , all the terms involving 81 and its derivatives have the same sign. Similarly in (28b) for ]z, the signs of all the terms with B2 and its derivatives are the same. This is consistent with the comments made after Example 2.2, and can be used as a check on the work. Some insight into the reason for this can be obtained from the discussion of stability in Section 8.2. ~· EXAMPLE 5.3

Find state-variable and input-output models for the system shown in Figure 5.14(a) and studied in Example 5.2, but with the shaft connecting disk 1 to the wall removed.

108

II!> Rotational Mechanical Systems

-;IB2W2 (b)

(a)

Figure 5.15 Free-body diagrams for Example 5.3.

SOLUTION This example is analogous to Example 3.5. Because there are now only three energy-storing elements, we expect to need only three state variables. Two of these are chosen to be wi and wz, which are related to the kinetic energy stored in the disks. The relative displacement of the ends of the connecting shaft is B2 - B1, which is related to the potential energy in that element. Hence we select as the third state variable

BR = Bz - B1 although B1 - Bz would have been an equally good choice. The free-body diagrams for each of the disks, with torques labeled in terms of the state variables and input, are shown in Figure 5.15. By D' Alembert's law, Jiw1 + B1wi - KzBR = 0 ]zwz + B2w2 + Kz{}R = Ta (t)

(30)

We obtain one of the state-variable equations by noting that ()R = Bz - Bi = wz -w1, and we find the other two by rearranging the last two equations. Thus the third-order state-variable model is

()R = wz-wi 1

(3la)

wi = -(-Biwi +K2BR)

Ii

1

wz= ]z[-K2BR-B2w2+Ta(t)]

(3lb) (3lc)

The output equations for TK2 and mr are

TK2 =Kz{}R

mr = l1wi +fzwz If one of the outputs were the angular displacement Bz, it would not be possible to write an output equation for B2 as an algebraic function of ()R, w1, wz, and Ta (t). In that case we would need four state variables. For the state-variable equations, we could either use (27) with Ki = 0 or add to (31) the equation B2 = wz. The input-output differential equation relating B2 and Ta(t) can be obtained by letting Ki = 0 in (29):

Ji]z{}rv)

+ (liBz + JzB1)B~iii) + (J1K2 +JzKz + BiB2)iJ2 +(Bi +B2)K2B2 = lifa +Bi fa +K2Ta(t)

Although this could be viewed as a third-order differential equation in the variable () 2 , which is the angular velocity wz, we would need four initial conditions if we wanted to

5.4 Obtaining the System Model

·4'1

109

e,w Tll(t)

Figure 5.16 Rotational system for Example 5.4.

determine 82 rather than w2 . This is consistent with the fact that four state variables are needed when the output is fh We would expect the input-output equation relating BR and Ta(t) to be strictly third order, because only three state variables are needed for the output BR . We see from (3 l a) that w1 = w 2 - !JR. Substituting this expression into (30), we obtain

Jiw2 + B1w2 - (liBR + B 1iJR+K28R ) = 0 hw2 + B2w2 +Kz{}R = ra(t ) We can eliminate the variable w2 from this pair of equations by using the p-operator method described in Section 3.2. When this is done, we find that

JihBR + (J1B2 +hBi )i}R + (f1K2 + hK2 + B1B2)iJR +(B1 + B2)K2BR = ft fa +B1ra(t )

~-

EXAMPLE 5.4 The shaft supporting the disk in the system shown in Figure 5.16 is composed of two sections that have spring constants Ki and Kz. Show how to replace the two sections by an equivalent stiffness element, and derive the state-variable model. The outputs of interest are the angular displacements () and 8A.

SOLUTION Free-body diagrams for each of the sections of the support shaft and for the inertia element J are shown in Figure 5.17. No inertial torques are included on the shafts, because their moments of inertia are assumed to be negligible. The quantity r,. is the reaction torque applied by the support on the top of the shaft. Summing the torques on each of the free-body diagrams gives (32a)

K1BA - Tr=O Kz({} - 8A ) - K1 BA = 0

(32b)

Jw +Bw +K2(B - BA) - ra(t) = 0

(32c)

We need the first of these equations only if we wish to find r,., which is not normally the case. Equations (32b) and (32c) can also be obtained by considering the free-body

110

Rotational Mechanical

::;,17

Ki(e - e"i

Figure 5.18 Alternative free-body diagrams for Example 5.4.

diagrams for just the disk and the massless junction of the two shafts. The corresponding diagrams are shown in Figure 5.18. Applying D' Alembert's law to them yields and (32c). We see from that and () are proportional to each other. Specifically,

into

Jw+Bw+ where

can be reii:an1eo as an stiffness constant for the series comThe bination of the two shafts. () and was the state variables and (34), we can write the state-variable model as

B=w 1

w = J[-KeqB-Bw+ The

() i.s one of the state variables. The

(35) for

is given by (33).

5.4 Obtaining the System Model

'4

111

Figure 5.19 Rotational system for Example 5.5.

Iii- EXAMPLE 5.5 The system shown in Figure 5 .19 consists of a moment of inertia Ii corresponding to the rotor of a motor or a turbine, which is coupled to the moment of inertia Ji representing a propeller. Power is transmitted through a fluid coupling with viscous-friction coefficient B and a shaft with stiffness constant K. A driving torque Ta(t) is exerted on 11, and a load torque 1L (t) is exerted on ]z. If the output is the angular velocity w2, find the state-variable model and also the input-output differential equation. SOLUTION There are three independent energy-storing elements, so we select as state variables w1, wz, and the relative displacement ()R of the two ends of the shaft, where (36)

Note that the equation (37)

is not yet a state-variable equation because of the symbol WA on the right side. Next we draw the free-body diagrams for the two inertia elements and for the shaft, as shown in Figure 5.20. Note that the moment of inertia of the right side of the fluid coupling element is assumed to be negligible. The directions of the arrows associated with the torque B (w1 - WA) are consistent with the law of reaction torques and also indicate that the frictional torque tends to retard the relative motion.

(a)

(b)

Figure 5.20 Free-body diagrams for Example 5.5.

(c)

112 ~- Rotational Mechanical Systems

Setting the algebraic sum of the torques on each diagram equal to zero yields the three equations

fiwi + B(wi - wA) - Ta (t) = 0 B(wi - WA) - K(()A - 82) = 0

(38b)

=0

(38c)

fzw2 - K(BA - 82) + rL(t)

(38a)

Using (36), we can rewrite (38) as (39a)

J1w1+B(0J1 - wA)-ra(t) = 0

B(w1 -wA) =KBR

(39b) (39c)

Jzw2-KBR+TL(t) = 0 Substituting (39b) into (39a) and repeating (39c) give J1w1+KBR-Ta(t)= O hw2 -KBR+ rL(t)= o

(40)

Also from (39b),

WA= WI

K

-

(41)

- BR

B

We substitute (41) into (37) and rearrange (40) to obtain the three state-variable equations (42a) (42b)

(42c) Because the specified output is one of the state variables, a separate output equation is not needed as part of the state-variable model. To obtain the input-output equation, we first rewrite (38) in terms of the angular velocities w1 , w2, and WA and the torques TA(t) and rL(t) . Differentiating (38b) and (38c) and noting that 02 W2 and OA =WA, We have

=

fiw1 +B(w1 -WA) = Ta(t ) B(w1 - WA) - K(wA - w2) = 0

hw2 - K (wA - w2) +TL= 0 By using the p -operator technique to eliminate WA and obtain the input-output equation

w2+

~w2+K ()1 +

JJ

w1

from these equations, we can

w2

K 1 K K = Ta(t) - - TL- -f"L - -TL(t) (43) l1fz ]z B]z Ji]z Although this result can be viewed as a second-order differential equation in w2, we will need three initial conditions if we are to determine w2 rather than the acceleration w2. Note that if the load torque TL(t ) were given as an algebraic function of vJ2 , as it would be in practice, w2 would appear in (43). Then the input-output equation would be strictly third order.

5.4 Obtaining the System Model

- K,c,,

+"'·I

M

·~

113

~---:::

(b)

M

(a)

(c)

Figure 5.21 (a) Translational system containing a lever. (b), (c) Free-body diagrams.

The next three examples contain either a lever or a pendulum that does not rotate about its midpoint. In Example 5.6 the mass of the lever is assumed to be negligible. The pendulum in the next example is approximated by a point mass at the end of a rigid bar. The final example includes a lever whose mass is uniformly distributed along the bar. ~

EXAMPLE 5.6 Find the state-variable equations for the system shown in Figure 5.2l(a). Also find the output equation when the output is defined to be the force exerted on the pivot by the lever. The input is the displacementx4(t) of the right end of the spring K2; it affects the mass M through the lever. The lever has a fixed pivot and is assumed to be massless yet rigid. Its angular rotation is small so that only horizontal motion need be considered. In a practical situation, the springs K1 and K2 might represent the stiffness of the lever and of associated linkages that have a certain degree of flexibility.

e

e

SOLUTION The displacements x2 and X3 are directly proportional to the angle and hence to one another. Furthermore, the two springs appear to form a series combination somewhat similar to the one shown in Figure 2.24(a), because the lever has no mass. Hence we can express x2,x3, and as algebraic functions of x 1 andx4 (t). Thus we will select only x1 and v 1 as state variables, with x4(t) being the input. By inspection, we determine that one of the required state-variable equations is .X1 = vi . The next step is to draw free-body diagrams for the mass M and the lever, as shown in Figure 5.2l(b) and Figure 5.2l(c). We must pay particular attention to the signs of the force arrows and to the expressions for the elongations of the springs. For example, the elongation of spring Ki is -(x1 + x2) because of the manner in which the displacements have been defined. Summing the forces on the mass M yields

e

(44) The forces on the lever are those exerted by the springs and the reaction force J,. of the pivot. Because the lever's angle of rotation is small, the motion of the lever ends can be considered to be translational, obeying the relationships = x2 Id2 = X3 IdI and

e

e

(45)

114

Rotational Mechanical

To obtain a second lever""'''"""" that involves xi and x2 but not the

+

=0 In order to solve we must first express x2 in we obtain the

can also be obtained

x2 =

.

1 M

7

+ (di/d2)-K2

We find the second of the two state-variable terms so as to solve for iii.

iii=--

we sum moments about

+

v .. ·~m.•v••~

into we find that the state-variable

+a

where (49)

cu~~vu••a,.v

To

function of the state varito obtain

which has the desired form.

case where di = From constant for a series connection of the two that (48) reduces

as in to the equations describing the to that shown in Figure tion A the lever. In tum, this the series where a single with the coefficient connection. vUUUJ~v.:'l the effective direction of the an with mass M rotates about an axis that the center of mass. In order to obtain a of an we would have to consider the motion of its center of to the direction of moto it, directed toward the The force the D 'Alembert there is a LAjiU 0. Because the range of in positive values oft, the transform of f(t) is not affected its values before

(1) defines the one-sided Laplace transfoim, There is a more general, two-sided Laplace transform, which is useful for theoretical work bul is seldom used for solving for system responses,

190

Trnnsfonn Solutions of Linear Models t

= 0. Then it follows from

that

5£[A] = = A.:-st

-s

It-+= t=O

to converge, I must I,the wm converge,"''"'""'~"''"' Because for the transform of this function converges for an values of sin the and becomes A

In order for the

s process and that the result is Note that the variable t has 'u""""'""' a function of s. For all functions of time with which we shaU be -..,uJm-'""''°u, the transform definition wm converge for an values of s to the of some vertical Hne in the complex s-plane. Although of the region of convergence is in some advanced applicawe do not need such for the in this book. When '"'"""'"JUL'"' Laplace we shall assume that (J, the real part of s, is to ensure convergence. If you are interested in of convergence, you should consult one of the more advanced references on system cited in E The unit denoted by U(t), is an important function that will be defined and discussed in detail in Section 7.5. Note that a letter U is used, in contrast to the lowercase letter in the for a general input. The unit step function is defined to be zero fort < 0 and unity fort > 0. Its transform is given by (3) with A replaced by 1. Thus, 1 5£[U(t)] = (4)

s

uepe1aa1ng on the value of the parameter a, the function = cat represents an expodecaying function, a constant, or an exponentially function for t > 0, as shown in 7 .1. In any case, the transform of the exponential function is

"v""""UA,

E-(s+a)t

I-+=

I -(s+a)

t=O

The upper limit vanishes when a, the real

of s, is

than -a, so (5)

7.2 Transforms of Functions

191

f(t)

0

Note that if a = 0, the _,.,,~A·-·A•A-· function reduces to the constant value of l. Likewise, reduces to as it must for this value of a.

Function The unit ramp function is defined to be have

for

= t fort> 0.

in (1), we

9-?[t] = To evaluate the integral in

we use the formula for integration

udv = uv[

parts:

vdu

where the limits a and b apply to the variable t. Making the identifications u v = E-st /(-s), a= 0, and b -too, we can rewrite as tE-st

1=

1

9-?[t] = (-s) o - (-s)

r

Jo

E-st

= t and

dt

(8)

It can be shown that (9)

Because the integral in to

is 9-?[U(t)], which from

has the value

1 s

9-?[t] = 2

Functiions When

sin wt or cos wt in the definition of the ~~.,~~,,~ h·~,,~ 0 '"n'~~' by using the identities in Table l in

(11)

192

Transform Solutions of Linear Models

an

or

o'.£[si.nwt] =

si.nwt E-st

-w

(-s)2+w 2 w - s2+w2 a similar prclce~Jure. we can show that

s s2+w2

The rPr'1rC1r1rrn transform is

showni.n

7.2has a

F(s) =

Ac-st

dt+

(14) When A = 1 and when the pulse duration L becomes infinite, the pulse approaches the unit function U(t). the term csL in (14) approaches zero for an values of shaving a positive real Under these conditions, F(s) given (14) approaches l/s, which is o'.£ [U (t)] as expected.

PROPERTIES The Laplace transform has a number of that are useful in finding the transforms of functions in terms of known transforms and in solving for the rnsponses of dynamic models. We shall state, and in most cases derive those properties that wiU be useful in later work. are tabulated in E. f(t)

A

0

7.2 The

l

of

A and duration L.

7.3 Transform Properties

0 the is angular

SOLUTION

The

is described

is at its the

the differential

or

The time constant is J

With

= A for all positive

(w1)ss =

+ B2). Because w1 (0) = 0,

As should be expected, the time needed to approach steady-state conditions is to the inertia of the disk. However, once the disk reaches a constant speed, on it are those from the friction elements. Thus the steady-state and the relative sizes

EXAMPLE7.3 Find an "'""""" 0.

for aH t > 0 for the system shown in Figi.s zero for aU t < 0 but has a constant value A for aH

SOLUTION

directly, we encounter a to evaluwe shall go back and apply the general three-step

If we ate the iniHal condition

7.4 First-Order Systems

~

203

i

R e;(t)

Figure 7.7 RL circuit for Example 7.3.

transform procedure to the modeling equations. Because all the elements are in series, we could sum voltages around the closed path to get a differential equation in which the only unknown is the current i(t). However, the output is specified as e0 , so we choose instead to write a node equation at the upper right corner:

~ [e

0 -

ei(t)] + [i(O) + ~

l

eo dP-] = 0

(39)

where the quantity in brackets is the element law for the current i(t) through the inductor. The initial current i(O) must be zero since there was no excitation fort< 0. The input ei(t) is a constant A for all t > 0, so its transform is A/ s. Transforming the node equation gives

Solving algebraically for the transform of the output gives A E0 (s) = ~

s+L

Taking the inverse transform, we have

e0 (t) = Ac-Rt/L

for

t >0

(40)

When checking this result, we note that e0 (t) goes to zero as t approaches infinity. In Section 6.5, we saw that when the input is a constant, a steady-state circuit can be drawn by replacing any capacitor by an open circuit, and any inductor by a short circuit. When this is done for our example, the output is the voltage across a short circuit, so it should indeed be zero in the steady state. Also notice that the expression for e0 (t) becomes equal to A when t is replaced by zero, even though e0 (t) = 0 fort < 0 because there was no excitation until t = 0. The voltage across an inductor can suddenly jump to a new value at t = 0, because it is not directly related to the energy stored within the device. In our Laplace transform solution, however, the only inital condition encountered was i(O), which cannot change instantaneously. We shall discuss the initial condition problem in general terms next.

Initial Conditions As in the last example, it is possible for the output to undergo, at t = 0, an instantaneous change, caused by the sudden application of an input or by some other event. When we must consider functions that have a discontinuity at t = 0, it is customary to use the notations y(O-) andy(O+) for the limiting values ofy(t) as t approaches zero through negative

204 ... Transform Solutions of Linear Models y y(O+) y(O-) Yss - - - - - - - - - - - -

0

Figure 7.8 Typical response of a first-order circuit.

and positive values, respectively. The function y(t) shown in Figure 7.8 has a discontinuity at t = 0, where it jumps instantaneously from y(O-) to y(O+ ). This is typical in the response of many electrical systems. To account for the possibility of discontinuous behavior, we modify (36) slightly to read

y(t) = Yss + [y(O+) -Yss]€-t/r

(41)

Using y(O+ ), the value of y(t) immediately following any discontinuity, is consistent with the fact that the equation should be valid for all t > 0. To understand which variables might suddenly jump to new values, recall that the energy stored in any passive element cannot change instantaneously unless there is an infinitely large power fl.ow. A translational mechanical system may have potential energy stored in a stretched spring or kinetic energy stored in a moving mass. The amount of this initial stored energy is determined by XK or VM, respectively. Thus, as long as all variables remain finite, XK and VM cannot change instantly. Similarly, in a rotational system, BK and w1 cannot change instantaneously. An electrical system may have energy stored within a capacitor or an inductor. Such energy is determined by ec or iL, respectively. Thus, as long as all variables remain finite, ec and iL cannot instantly jump to new values. In any mechanical or electrical system,the initial stored energy can be accounted for by knowing the values of XK,VM,BK,WJ,ec, and iL at t = 0. When the Laplace transform is applied to the original modeling equations, these are the only initial conditions that should be expected, even in the case of higher-order systems. Because these quantities cannot normally change instantaneously, they do not suddenly jump to new values at t = 0 even if the input suddenly changes. The output e0 (t) in Example 7.3 had the form shown in Figure 7.8 with y(O-) = 0, y(O+) =A, and Yss = 0. In that example, recall that the only initial condition that was encountered was the current through the inductor, which could not change instantly. Thus there was no confusion about the proper value to use for i(O). The Laplace transform solution then automatically gave the correct value of A for e0 ( 0+). Avoiding problems with difficult initial conditions is one of the advantages of the transform method as compared with a classical solution. If we choose not to transform the original modeling equations immediately, then we may have to deal directly with a variable that has a discontinuity at t = 0. That situation will be discussed in Section 7.5 and again in Section 8.1. .... EXAMPLE 7.4 The circuit shown in Figure 7.9(a) contains one energy-storing element and three dissipative elements. The voltage across the capacitor is the node voltage eA. Fort< 0, the switch is open and no energy is stored in the capacitor. The switch then closes at t = 0. Find an expression for e0 (t) for all t > 0.

7.4 First-Order

1 +~:r o"

+

3!1

24 V _-=-

eA

205

6 f!

2F~

7.9 (a) Circuit for bx:imJJles 7.4 and 7.5. (b)

SOLUTION we have l

+3 1 l -e +6 ° 3

-e0 ) = 0

l

-eA) + 6

=0

-Ea(s)]=O l l -Ea(s) + -[Eo(s) -EA(s)] + -1 [Ea(s) - -24] 6 3 6 s We set eA(O)

=0

to zero and then solve algebraically for Ea(s),

6

1

(s+ 112) + (s+ 112) s

Using

and (34) to take the inverse ea(t) =

6E-t/l 2

+ 12(1 -

we get E -t/ 12 ) =

12-

for

t

>0

Note from this expression that e0 (0+) = 6 V, even though ea was zero for though the voltage across the capacitor must remain zero at t = 0+, there is to the voltage across a resistor from jumping to a new value. For very large values oft, after the transient term has died out, our answer reduces to (e 0 )ss = 12 V. In order to check this result, we can redraw the circuit with the switch closed and with the capacitor replaced by an open as shown in The two 6-Q resistors are in because no current can flow down the 3-Q rethe rule, = [6 / (6 + 6)] (24) = 12 V. Also note that sistor. Then (eA)ss = = 12 V, since there is no across the 3-Q resistor in the state. Finally, note that the transform of the output has a at s = -1 / T and that the transient term in ea(t) contains the factor c 1/ 7 , where T = 12 s. Pfots of both eA and e0 are shown in Figure The time constant r is 12 s. When t = r, the response is approximately 63 percent of the way from the initial to the value. Thus ea(T)::::::6+ (12-6)=9.8 whHeeA(r):=:::: (12)=7.6V.

EXAMPLE7.5 Example 7.4 when the switch in Figure

has been closed for

time for = 0.

< 0, so that steady-state conditions have been reached. The switch then opens at t Again find ea (t) fort > 0.

t

206 ""' Transform Solutions of Linear Models

12 ;../.

/

··;,·

10 So

/.

8

,•,

I /.

:1e

f

©

~ ~

.

A

(a)

6 ··········r:· I

4 /. ..

2

0'----'---'----'---'------'---'------'---'---~

-10 12

10

0

20

50

60

70

80

'-----~

10 ....... \' ... ·...... . \

(b)

4

.

\" ....

•'•

...

2 Ol___i.~_j_~_i_~_i_~.i.___i==:::::t:::::::::S::::==::J-~~ -10 10 20 0 30 40 50 60 70 80 Time (s)

Figure 7.10 Responses for the circuit in Figure 7.9. (a) Switch closes at t t=O.

= 0. (b) Switch opens at

SOLUTION We need to know the voltage across the capacitor at t = 0- in order to include the effect of the circuit's history for t < 0. The steady-state circuit for t < 0 is shown in Figure 7.9(b), which gives 12 V for eA in the steady state. This voltage cannot change instantaneously, so eA(O+) = 12 V. The two left-hand elements in Figure 7.9(a) are disconnected for all t > 0, so 1 2eA+ 3(eA-eo) =0

1

1

6eo + 3 (eo - eA)

=0

7.5 The

Function and

207

1

-eA(O)J+3[EA(s)-E0 (s)] =0

1 1 6Eo(s) + 3[E0 (s) -

= 0

8

s+18

Hence, we can write for all

t

>0

We see that e0 (0+) = 8 V, so once the output voltage 0. Notice that c0 (t) goes to zero for values oft. This is since there is no external source of energy for t > 0. The factor indicates that the time constant r is 18 s. Plots of eA and c 0 for this case are shown in of Figure 7.10.

t =

The last two show that the time constant r for the circuit in Figure 7.9(a) is 12 s when the switch is closed and 18 s when the switch is open. The time constant is a measure of how quickly any initial stored energy would be when the external input is zero fort > 0. With the switch dosed, current from the discharging capacitor can flow through the 3-0 resistor and then two different 6-0 With the switch open, one of the two 6-0 has been removed for t > 0, so we would expect it would take longer to dissipate any initial energy stored in the capacitor.

THE STEP FUNCTION AND IMPULSE Two of the most important inputs we encounter in the analysis of dynamic omm''>ITIO unit step function and the unit impulse. We need to understand their nrc>nPrtu"" to find the response of a system to each of them.

The

Step One frequently encounters that are zero before some reference time and that have a nonzero constant value thereafter. To treat such mathematicaUy, we define the unit which is denoted U(t). This function is defined to be zero fort::; 0 and occurs at some later unity fort> O; it is shown in Figure 7.ll(a). 3 Uthe step time t 1, as shown in Figure 7 .11 (b ), the function is defined by = {

~

fort ::; ti fort> ti

that a capital U is used for the unit step function U(t), in contrast to the lowercase letter in the symbol for a general input. The value of the unit step function at time zero could be defined to be unity, or its value could be left undefined at this instant. Defining U(O) = 0 will be convenient for our applications. 3 Note

208 ~ Transform Solutions of Linear Models

A

U(t)

0

0

0

(c)

(b)

(a)

Figure7.11 Stepfunctions. (a)U(t). (b)U(t-t1). (c)AU(t).

This notation is consistent with the fact that when any function f(t) is plotted versus t, replacing every t in f(t) by t - t1 shifts the curve t1 units to the right. If the height of the step is A rather than unity, we simply have A times the unit step function, as shown in Figure 7.ll(c). From the definition of U (t - t1), we note that

f(t)U(t-t1)={

~(t)

fort~ ti

fort>

(43)

ti

Thus the output of a system that was at rest fort~ 0 is often written in the form f(t)U(t), where the multiplying factor U (t) is used in place of the phrase "fort > O." We define the unit step response of a system, denoted by yu(t), as the output that occurs when the input is the unit step function U (t) and when the system contains no initial stored energy-that is, yu(t) is the zero-state response to the input U(t). In Section 7.4, we found the response for several examples when the input was zero for t ~ 0 and A for t > 0. To obtain the unit step response for those systems, we merely replace A by 1. For the translational system shown in Figure 7 .3,

Yu(t)

1 = K[l-t:-KtfB]

fort

2: 0

whereas for the rotational system shown in Figure 7 .6,

We can represent any function that consists of horizontal and vertical lines as the sum of step functions. Consider, for example, the pulse shown in Figure 7 .12 (a). This function is the sum of the two functions shown in parts (b) and (c) of the figure, so

fi(t) =AU(t)-AU(t-ti)

(44)

Suppose that the pulse in (44) is the input to a linear first-order system described by the differential equation 1 (45) y+ -y = u(t) 7

and for which y(O) = 0. From Section 7.4, we know that the response to AU(t) is AT(l 2: 0, and this is also the response for 0 ~ t ~ti to the input in (44). Fort> ti, we may use superposition and sum the responses to the components AU (t) and -AU (t - ti) to obtain c1fr) fort

y(t) = AT[l -

€-tfr]

-AT[l -

= AT(-1 + €tifr)€-t/r

€-(t-ti)fr]

7.5 The Step Function and Impulse f,(t)

~

209

/AU(t)

Ai------.

A------

0

0

(b)

(a)

0

-AU(t - ti)

/

-A (c)

Figure 7.12 (a) Rectangular pulse. (b), (c) Formation of the rectangular pulse by the sum of two step functions.

Hence we may write

y(t) = { AT(l - ctlr) AT(1/il 7 - l)c 1/r

for 0::::: t::::: ti fort> ti

(46)

which is shown in Figure 7.13(a). As expected, the response for the first ti seconds is the same as in Figure 7.5(b), whereas fort >ti the output decays exponentially to zero with a time constant T. The output is the superposition of the two functions shown in Figure 7.13(b) and Figure 7.13(c). It can also be written in the alternative form

y(t) = AT(l - E-t/r)U(t) -AT(l - E-(t-ti)/r)U(t - t1) It is instructive to rewrite the pulse response fort> t1 in (46) with E ti/r replaced by its Taylor-series expansion 1 E 1

/r =1+-+ti 1 (ft -) T 2! T

2

+···

Then, for t > t1, 2 ft y (t ) =AT [ -;: + 2!1 (t1) -;: +... ] E-t/r

Suppose that the pulse width is small compared to the time constant of the system, so that t1 « T. Then we can neglect all the terms inside the brackets except the first and write

y(t):::At1E-t/r

fort>t1

(47)

where At1 is the area underneath the input pulse. We could also consider input pulses that have somewhat different shapes but have the same area underneath the curve. We would find that as long as the width t1 of any pulse that is the input to a first-order system is small compared to the system's time constant, the response fort> t1 depends on the area underneath the pulse but not on its shape.

210

Transform Solutions of Linear Models y Ar

Ar

--------------

M(l-

0 (b)

(a)

11

11

+r

OF-~~=w-~~~~~~~~-..-

-Ar (C)

7.13

shown in

7.12.

As background for another note that the response of the modeled to a function of height times the unit step with y(O) = 0, is response: (t) = 1- E-t/r) fort 2:: 0 Note that d dt

(48)

for an positive values oft. Because the right side of is identical to we see that if is the derivative of the response to a t1 « r, the response fort > t1 to a of area function of

The response of a first-order to a pulse of area appears to be of the pulse as as the width t1 is smaH compared to the time constant T, so it is reasonable to try to define an idealized function whose width is small compared to the time constant of an first-order However, in order to have t1 « T for an and to have a nonzero area, the nonzero values of T, ti must be height of the must become infinitely large. Although such an idealized and mathematical difficulties, let us reconsider the rectangular input of the is the derivative of the is shown in Note that Ji (t) in part function shown in Let us specify A= in Figure 7.14 such that the area underneath Ji (t) is unity and such that the value fort > t 1 is also unity. If we continually decrease the value of the Ji (t) increases in order to maintain unit area, while (t) rises then the to its final value more rapidly. The dashed lines in 7.15 show the changes in fi(t) and when ti is halved.

7.5 The Step Function and Impulse

0,

214 .,.. Transform Solutions of Linear Models Yu(t)

h(I)

l!B

l/M

0

0

T

(b)

(a)

Figure 7.19 Responses for the system shown in Figure 7.18 when the output is the velocity. (a) Unit step response. (b) Unit impulse response.

Note in Figure 7.19(a) that when fa(t) is a step function, the velocity of the mass does not change instantaneously at t = 0. When fa(t) is an impulse, however, the velocity of the mass does undergo an instantaneous change from zero to 1/ M. In the first case, there is no change in the kinetic energy of the mass at t = 0. In the latter case, the impulse causes an instantaneous increase in the energy of the mass. Consider further the velocity expression in (56) that is caused by a unit impulse for fa(t), and recall some of the equations in Chapter 2. According to Equation (2.6), the kinetic energy that the impulse inserts into the mass is

1

2

2Mv (0+)

1

= 2M

The applied force fa(t) returns to zero for all time after the impulse, but there is still the friction force fs = Bv. From Equation (2.1), the power dissipated in the form of heat is

B p = fsv = Bvz = M2 E-2Bt/M which gradually decays to zero as the mass slows down. As explained in Chapter 2, the energy supplied between times to and t1 is ft~1 p(t) dt. Thus the total energy dissipated by friction for all t > 0 is

-B2 looo c2Bt/Mdt = loooo p(t)dt = M o

-B . -M [-c2Bt/M ] M 2 2B o

00

which reduces to 21, . As expected, all the energy that was inserted into the system by the impulse is eventually lost in the form of heat.

Iii- EXAMPLE 7.7 LetM = 0.1 kg andB = 0.1 N·s/m for the system discussed in Example 7.6. Let the input fa(t) be a rectangular pulse having a width L and a height 1/L. Construct a Simulink diagram in which fa (t) is represented by the superposition of two step functions: a positive step starting at t = 0 and a negative step starting at t = L. Plot the output velocity when L = 3 s, L = 0.5 s, and when L = 0.1 s. SOLUTION The Simulink diagram is shown in Figure 7.20, and the M-file is given below. The resulting output plots are shown in Figure 7 .21. Although the input step occurs

705 The

Function and

215

Integrator

friction coefficient

707

7.20 Simulink model for

at t = 0, for we have at t = -1 s the Start the time to -1 in the Simulation > Parameters dialog boxo The Stop time was is 1 So The area underneath each set to 8 So Notice that the time constant T for the of the input is unity since they begin at t = 0, end at t = and have an amplitude of l / L, all set by the two Step blocks at the left of the figureo vO

Oo;

% initial velocity, m/s % kg B = 0.1; % Ns/m %----- narrow pulse L = 0.1; sim('ex7_7_model') plot(t,v, t,fa,'--'); grid; hold on %----- medium-width pulse -----L = 0.5; sim(' ex7_7_model') plot(t,v, t,fa,'--') %----- wide pulse -----=

M = 0 .1;

L

=

3o0;

sim('ex7_7_model') plot(t,v, t,fa,'--'); hold off %----- make plot more informative -------ylabel ('Velocity (m/s); Force (N) ') xlabel(' Time (s) ')

The value of the velocity is reached at the end of the input after which it decays to zeroo The form of the response for L = 3 s, which is three times the time constant, is similar to the one shown in Figure 7 13 (a) The response when L = Oo 1 s, which is of the time constant, is very dose to the unit impulse response found in Example 7060 However, instead of the velocity rising to its peak value, it takes OJ s to do soo Furthermore, this value is about 905 instead of the 10 reached for the impulse responseo The reader may wish to use the M-file to the response for other values of L As Lis made extremely small v~'"''"''~V~ with the system's time constant, the response becomes virtually identical with the response, which itself can be from the Simulink diagram the letting the initial condition be v(O) = 10, the value obtained from

to

0

0

2Hi

Transfonn Solutions of Linear Models 10

I

I

9

8

~

7

(I)

~

0

6

u..

~;::;Tl

5-

4

0

~

3

ii

l"' s:o s .. I ·-. I I -I oL_~_J[__.....'.__j_~~_L_~_;_~=::~~==,,,;;;;;;;;;;;;;;j;;;;;;;;:;::;;~

2

1

3

2

0

5

4

6

7

Time (s)

7.21 Pulse responses for Example 7.7.

the Unit Impulse

The

From (1 ), we can write the transform of the unit impulse as

This expression presents a dilemma because the impulse o(t) occurs at t = 0, which is the lower limit of integration. To avoid this problem, we shall agree that when we are working with Laplace transforms, the unit impulse o(t) occurs just after t = 0, namely at t = 0+. Then the will be contained within the Hmits of integration in the definition of the Laplace transform. With this stipulation, is applicable with ta= 0+, b = 0, and c--+ oo, and we can say that

r

Jo

O(t)E-st dt =

E-sll

=

l

t=O+

so

9?[o(t)J = 1 In books where the lower limit of the transform in is taken to be t = 0-, the impulse can be considered to occur at t = 0. This convention also It is instructive to revisit the rectangular pulse shown in Figure 7.2. U we choose A = 1/ L, the pulse has unit area, occurs immediately after t = 0, and has the ~~,,.~,,~ transform ( 1/ sL) (1 - csL). If we now take the limit as L approaches zero, the becomes a unit impulse occuring after t = 0. In the limit, the expression for its transform becomes an indeterminate because both the numerator and the dezero. To evaluate this indeterminate form, we can use L'Hopital's rule, nominator differentiate ~,,,,,,,,.,.u,1, the numerator and denominator with respect to and then finally

7.5 The

217

Function and

take the new Hmit. Thus SE-sL

9?[J(t)] =

--=l s

which agrees with If the instantly that we are as v"''u'"'"' the initial conditions the values of the variables The transform of any function that is that when the input is we were to as t = 0, in which case Hs transform would be zero. Then the initial-condition terms that appear in the solution would have to include the effect of the additional energy that had been instantly inserted into the the impulse. it is to obtain a solution this way, the evaluation of the necessary initial conditions would be difficult. We shall assume that all occur after t = 0 and that 9?[a(t)] = 1. In a transform there can be other cases where we must be careful to between t = 0- and t = 0+. For the formulas for the derivatives of functions contain terms of the form f(O) and In the event of a or j at t = 0, we shall use for the initial condition the limit of or j as t zero from the left, the value just before the discontinuity occurs. the formula for the transform of a first derivative, which was derived in (24):

9?[j] =

sF(s) - f(O)

To illustrate the application of (58) when has a discontinuity at t = 0+, consider the product of cos wt and the unit step function U (t): = [coswt]U(t) = {

~oswt

fort:::; 0 fort> 0

which is shown in Figure 7.22(a). Because differs from cos wt at the within the interval 0:::; t 0. Frrst we consider the This

= lim

s_,=

where the limit exists. If F(s) is a rational function, win a finite value the numerator polynomial is less than that of the denominator-in other that m < n. If we attempt to use when m = n, the result win be infinite. Recall from Section 7.6 that to find when m = n, we must use a preliminary of division to write F(s) as the sum of a constant and a transform for which < n. Because the inverse transform of the constant is an impulse at t = 0+, the value of f(O+) is undefined when

m=n. The final-vah1e theorem states that

f( oo) = limsF(s) s_,O

provided that F (s) has no poles in the right half of the complex plane with the IJU,,•,,uno exception of a single pole at the origin, has no poles on the imaginary axis. The f (co) denotes the limit of as t approaches infinity. To gain some insight into the effect of this restriction on the use of the final-value theorem, we recall that the forms of the terms in a partial-fraction are dictated by the locations of the poles of F(s). Suppose, for example, that

A2(s+a) A3 F(s)=-+( s s+a )2 + (32+ s- b+ where a,(3,b, and ware positive real constants. The expansion implies that the of F(s) are s1 = 0, s2 = -a+ j(3, s3 = -a s4 = b, ss = and S6 = - jw. The corresponding time function for t > 0 is

The limits of the first two terms as t approaches infinity are and zero, ,.,,QnPt't"''"" increases without whereas sin wt oscillates without aPJDfO>aCJtung a constant value. Thus, because of the poles of F(s) at s4 = b, ss = and s6 = - jw, the function does not a limit as t As another "'"''""'Hf"" double at the origin will cause the to have terms of the form

F(s)=-+-+··· 2 s

s

234

Transform Solutions of Linear Models

f(t) = a limit. which The use of the initial-value and final-value theorems is iHustrated in the exand in the next In 7 .17 we consider a transform for which neither theorem is ""''""'"'"--'''"·

EXAMPLE 7.16 Use the initial-value and final-value theorems to find f(O+) and

when

F s _ s2 +2s+4 ( ) - s3 + 3s2 + 2s

SOLUTION

From

the initial value of f(t) is

f(O+) = lim s(s2+2s+4) s3 + 3s2 + 2s

s-7°0

. s3 + +4s = 11 m - - - - s-+00 s3 + 3s2 + 2s

(102)

Because f(O+) is the Hmit of a ratio of ins ass approaches infinity, we need to consider the highest powers in s in both the numerator and denominator. Hence reduces to s3 f (0+) = lim 3 = 1 s-+oo

s

Before applying the final-value theorem, we must verify that the conditions necessary for it to be valid are satisfied. In this case, we can rewrite with its denominator in factored form as F s _ s2 +2s+4 (103) () - s(s+ l)(s+2) which has distinct poles at s = 0, -1, and -2. The function F(s) has a single pole at the origin with the remaining poles inside the left half of the s-plane, so we can apply the final-value theorem. Using (100), we find that

!(=) = lim s(s2 +2s+4) s-+O s(s2 + 3s + 2)

= lim s2+2s+4 = 2 s-+0 s 2

+ 3s + 2

task to evaluate f (t) for ant The result is

>0

U~L,,U,~u.

fort> 0 and f

(

00 )

are correct.

F(s) as given

Summary EXAMPLE 7.17

Explain why the initial-value and final-value theorems are not applicable to the transform

F(s)

= s3 +2s2 +6s+8 s 3 +4s

SOLUTION

(104)

Attempting to apply the initial-value theorem, we would write

f(O+)

= lim s(s3 + 2s2 + 6s+ 8) s---+=

s 3 +4s

.

s4 + 2s3 + 6s2 + 8s s3 +4s

=hm--~----

s---+=

which is infinite. This is expected, because m = n = 3 in (104). As for the final-value theorem, we see from (104) that we can write the denominator of F(s) as s(s2 + 4) = s(s- j2)(s+ }2). Hence F(s) has a pair of imaginary poles at s2 = }2 and s3 = - }2 which violates the requirements for the final-value theorem. If we carry out the partial-fraction expansion of F (s), we find that

2 2 F(s) = l + - + - s s2 +4 which is the transform of the time function

f(t) = '5(t) + 2 + sin2t fort> 0 The initial-value theorem is invalid because of the impulse at t = 0+, and the final-value theorem is invalid because of the constant-amplitude sinusoidal term.

_..

SUMMARY The Laplace transform can be used to convert the integral-differential equations that describe fixed, linear dynamic systems into algebraic equations. The general procedure consists of three steps: transforming the system equations, solving the resulting algebraic equations for the transform of the output, and taking the inverse transform. Any initial conditions that need to be evaluated automatically appear in the first st~p. These initial conditions are often easier to find than those that are needed for a time-dclmain solution of the input-output equation. Although the basic definition can be used to find the Laplace transform of a given function of time, transforms of the most common functions are tabulated in Appendix E. The inverse transform, which should be expressed as a real function of time, can usually be found by writing a partial-fraction expansion and by again using the table in Appendix E. The appendix also summarizes the properties and theorems developed throughout this chapter. In addition to being an important tool for finding the response of a system to specified inputs, the Laplace transform can be used to develop a number of important general concepts. This further development, which builds on the notions of stability, time constant, step response, and impulse response from this chapter, will be carried out in the next chapter.

236

Transform Solutions of Linear Models

7.1. Using the definition of the '-'"I·""'''"' transform in of the functions. a.

Ji

evaluate the transforms

= t2

b.

=

c.

f3(t) =

d.

E-at

cos wt

= sin2t for 0

< t ).

*7.3. Use the tabulated in Appendix E to find the the following functions of time. a.

!1 (t)

derive expressions for

transform of each of

= tC 21 cos3t

b.

(t) =t 2 si.nwt

c.

f3(t) = -(t 2 c

d

dt

1)

d.

7.4. a. b.

Prove that .P[f(t/a)] =aF(as). this property with f(t) =cos wt to find .P[cos2wt].

7.5. a.

For the mechanical system shown in Figure P7.5, find the input-output equation relating x0 to the displacement input

b.

What is the time constant associated with this first-order system?

c.

Let Ki = l N/m, Kz = 2 N/m, and B = l N·s/m. Find x 0 as a function of time for an t > 0 when there is no energy stored in the springs for t < 0 and when fort> 0. = 1x

~I ~

B

I-

P7.5

Problems

7.6. A P7.6.

Va

A in the mechanical

to

is

a.

Write the

b.

What is the time constant r for the

c.

Sketch the response when

d.

= 0 fort

237

shown in

2: 0 and vi (0) = 10.

(b), and

P7.6

*7.7. Assume that the circuit shown in Figure P6.l contains no stored energy fort < 0 and that the input is the unit function. The differential found in Problem 6.1, is 3. I 1 4ea + :zea = 2 dt + and determine the time constant.

a.

Verify the differential

b.

Find and sketch ea versus t for an t > 0 when that ea(t) has a at t = 0.

c.

Check the steady-state response by replacing

is the unit step function. Notice by an open circuit.

7.8. Repeat Problem 7.7 for the circuit shown in Figure differential equation, found in Problem 6.4, is

for which the

7.9. Assume that the circuit shown in Figure P6.2 contains no stored energy fort < 0 and that the is the unit function. The differential found in is Problem

ea+ a.

Verify the differential equation and detennine the time constant.

b.

Find and sketch ea versus t for an > 0. response

the inductor

a short circuit.

7.10. The switch shown in Figure P7.10 has been closed for a so the circuit is in the state at t = 0-. If the switch opens at t = 0, find and sketch ea versus t. Show on the same scale the curves for = 5 n and for = 10 n. What would be the circuit's response

238 •

Transform Solutions of Linear Models

Figure P7 .10

*7.11. Consider a first-order system with input u(t) and output y that is described by y + 0.5y = u(t). a.

Find and sketch the unit step response yu(t).

b.

Find and sketch the response to u(t) = 2 fort> 0 and with y(O) = -1.

c.

Find and sketch the response to u(t)

d.

Find and sketch the unit impulse response h(t).

= U(t)-U(t-2) withy(O) = 0.

7.12. The voltage source in Figure P7.12 is zero fort< 0, but has a constant value of A for

all t

> 0. > 0.

a.

Find and sketch the output voltage e0 for all t

b.

Check the steady-state response by replacing the capacitor with an open circuit.

+ e;(t)

Figure P7.12

7.13. Let the voltage source in Figure 6.18(a) have a constant value of 24 V and let C=2F.

a.

The switch has been open for all t < 0, so there is no initial stored energy in the capacitor. The switch then closes at t = 0. Transform the modeling equations for t > 0, which are given by Equations (6.26), and find e0 for all t > 0.

b.

The switch has been closed for a long time, so that the circuit is in the steady state at t = 0-. The switch then opens at t = 0. Transform the modeling equations for t > 0, which are given by Equations (6.28), and find e0 for all t > 0. As part of your solution show that the initial voltage across the capacitor is eA (0) - e0 (0) = 9V.

c.

Find the time constants for parts (a) and (b) and explain why they are not the same.

7.14. The unit impulse response for the circuit in Figure 7.7 was found in Example 7.8. From Section 6.2, the energy stored in an inductor is w = Li2 , and the power dissipated in a resistor is p = Ri2 .

!

Problems '4

239

a.

Find the energy that the unit impulse instantaneously inserts into the inductor.

b.

Find the total energy dissipated in the resistor fort> 0.

c.

Compare the two results.

For Problems 7.15 through 7.17, find f(t) for the given F(s). 7.15. a.

3 2 F (s) -_ 2s + 3s + s + 4

s3

3s2 +9s+24 F(s) - - - - - . - (s- l)(s+2)(s+5)

b

4 c.

F(s) = s2(s+ 1)

d. F(s) =

2

3s

s +2s+26

7.16. a

s

.

b.

F(s) ---- s2 +8s+ 16 F(s)

1 = s(s2 +w 2 )

8s2 +20s+74 c F (s)---. - s(s2 +s+9.25) d.

F(s)=2s2 +Hs+16 (s+ 2) 2

*7.17. F

a.

s3 +2s+4 (s)= s(s+1) 2 (s+2)

b.

F(s) = 4s2 + 10s+ 10 s3 + 2s2 +5s

c.

F(s) = 3(s3 +2s2 +4s+ 1) s(s+ 3) 2

d.

s3 -4s F(s) = (s + l)(s2 + 4s + 4)

*7.18.

a.

Find the inverse transform of

3s2 + 2s + 2 ( )- (s+2)(s2 +2s+5)

F s _

by writing a partial-fraction expansion and using Equation (84). b.

Repeat part (a) by completing the square and using Equation (86).

240

Transform Solutions of Linear Models

7 .19. Find the inverse transform of the function of s by first and C and then the square in the denominator of the term F

2 72) - 4(s3 + 8s+ 2

A - s +8s +32s - s

Bs+C +8s+32

--+~---

s2

*7.20. Use

-s+5 F(s)---- s(s+ l)(s+4) Check your answer 7.21. Use

and the results of Example 7.9 to find the inverse transform of

s(-s+ 5) (s+ l)(s+4)

7.22. a.

Use the time-delay theorem in gular pulse shown in Figure 7 .2.

to derive the ,_," 1-'""'" transform of the rectan-

b.

Find the Laplace transform of the triangular shown in Figure 7.26(a) using Equation (1) directly, rather than by decomposing the pulse into ramp and step components. Compare your resuHs to (96).

*7.23. Using the definition of the Laplace transform in Equation 0), evaluate the transforms of the following functions. a.

The function Ji (t) shown in Figure P7.23(a).

b.

The function fz (t) shown in Figure P7.23(b ).

!1 (I)

fi(I)

A

0

2

-A (b)

(a)

Figure P7 .23

the functions into step functions and ramp

Problems

a.

241

F(s) = 1

=-(1-2cs+ s2

b.

*7.26. the initial-value and final-value theorems to find of the four transforms in Problem 7 .17. If either theorem is not "'""Ju,.uun.. to a this is so.

7.27. Repeat Problem 7.26 for the a.

F (s) in Problem 7.19.

b.

The functions in

c.

F

transforms.

and

of Problem 7.16.

7.28.

a.

Using Equation

and the initial-value theorem, show that f(O+) = lim[s 2F(s)-sf(O)] s---+=

provided that the limit exists. b.

Use the derived in part (a) to find f(O+) for the transform F(s) given in Equation (101). Assume that = l fort::; 0. If this property is not explain this is so.

c.

Repeat

d.

Check your answers to part (b) and part (c) by differentiating the for f(t) in Example 7.16.

assuming that

= 0 fort< 0.

given

We in Section 7.4 the for the response of any fixed Hnear by use of the Laplace transform. We transformed the modeling for t > 0, thereby converting them into equations. We solved these transformed equations for the transform of the and then took the inverse transform in order to obtain the output as a function of time. However, we the procedure only to systems with one energy-storing element. In this chapter, we first consider the complete response of more complex systems. We then examine in some detail two important cases: the zero-input response, where the excitation consists of some initial stored energy, and then the response to a given when the initial stored energy is zero. From the second case, we develop and mustrate the concept of the transfer function. We give special attention to the responses to the unit impulse, the unit step function, and sinusoidal functions. We also show that finding the transfer function for electrical systems can be simplified introducing the concept of impedances.

8.1 THE COMPLETE SOLUTION The basic procedure for higher-order systems is the same as in Section even though the algebra may be more involved. If the is described by several modeling equations, we transform them individuaUy to obtain a set of algebraic equations. We must then solve these equations simultaneously to an expression for the transformed output, with an other unknowns eliminated. The resulting expression may be fairly complicated, but we crui use the partial-fraction techniques in Section 7.6 to find the as a real function of time. The first in this section consider a mechanical system that has three energyelements, but which is at rest fort < 0. We then examine a second-order electrical that contains some initial stored energy as a result of an that was for t

< 0.

EXAMPLES.! The translational 3.lL Let undeflected when x1 response by

= x2

was modeled in 3. 7 and in Ex and = l The are = 0. Take the to be the . Find the unit step the obtained from the free-body diagrams.

242

8.1 The

Solution

243

SOLUTION shown in

+ +

-x2) = (t) -xi)= 0

+ +

with an the element values

these

to

and with

[fa(t)] =

we have

(s) = 0 Because we assume that the initial stored energy is zero when sponse, the initial of the and the initial

(a)

---Mx 1 fa(I)

M K 1(x 1 -x 2) (b)

A

K 1(x 1 -x2 )

(c)

M

(d)

(el

8.1 (a) Translational system for (d) Free-body for the

uA,rnufJ'"

x2 + K 2x 2

B2

the unit

(1)

244

Transfer Function

zem Thus x1 (0) =

becomes

= 0 andi1(0) = 0, so

=0 from these 1 ) [ s+2 ] Xi\s = s3+3s2 +3s+1

that

=

Si?[U(t)]

=

1

and then

s+ 2

a

A13

(s) = s(s+ 1)3=--;-+(s+1)3+(s+1) 2 +s+1 where

-- (s+s+2_1 -2 1)3 =s+21 =-1 = [!!_ (s+2)] = ~ [ - (s+_3:)] s=O -

S

s=-1

ds

s

2 ds

2

= _

s=-l

s

2

= -2 s=-l

we insert these numbers into (3) and take the as a function of To find the inverse transform of each term. This gives us x1(t)=2-(

+2t+

(4)

fort> 0

This equation reveals that the steady-state response is (x1)ss = 2 m, which we can check by referring to Figure When has a constant value of 1, the mass win ~ .. -~m•.r become motionless and there wiU be no inertial or friction forces. Because the friction elements can be the two springs are then i.n series and can be replaced by a equivalent spring. Using Equation (2.37), we see that Keq = N/m. The only forces on the mass are those shown in Figure so we can again conclude that =2m. we can show that the velocity and acceleration of the mass are

1

fort> 0

+l)c1 fort>O and

we see that xi(O+)=O

i1(0+)=0 i1(0+) = 1

(6c)

The fact that x1 and i1 remain zero at = O+ is expected; the of the and the of the mass cannot Because of this, we also know that at t = O+ there is no force on the mass from or . Thus the forces on the mass at = O+ are the and inertial forces, as shown in From that figure,

8.1 The Complete Solution

~

245

we see that the initial acceleration must be .X1 (O+) = 1 m/s2, which serves as a check on (6c).

a;.

EXAMPLE 8.2

Repeat Example 8.1 by transforming the state-variable equations. SOLUTION The state-variable model for Figure 8.l(a) was found in Example 3.7. From Equations (3.14), with all the element values set equal to unity, we have

i1 = v1 i11 =-xi -v1 +x2 + fa(t)

i2 = x1 -2.x2 Transforming these equations gives sX1(s)-x1(0) = Vi(s)

(7a)

sVi (s) - v1 (0) = -X1 (s) - Vi (s) + X2(s) + Fa(s) sX2(s)-x2(0) =X1(s) -2X2(s)

(7b) (7c)

where we can again set the initial-condition terms equal to zero. Then, by substituting (7a) into (7b) and rearranging the last two equations, we obtain

(s2 + s+ l)X1 (s) -X2(s) = Fa(s) -X1(s) + (s+2)X2(s) = 0 from which

s+2 ] X1(s)= [ s3+3s2+3s+l Fa(s) which agrees with (2).

a;.

EXAMPLE 8.3

Repeat Example 8 .1 by transforming the input-output differential equation. SOLUTION The input-output equation for Figure 8 .1 (a) was derived in Example 3 .11. From Equations (3.27) with the element values set equal to unity, we have

:X'1+3.X1+3i1 +x1=ia+2fa(t) We transform this equation, getting [s3X1 (s) - s2x1 (0) - si1 (0) -.X1 (O)J + 3[s2X1 (s) - sx1 (0) -i1 (O)J + 3(sX1 (s) -x1 (O)J +X1 (s) = sFa(s) - fa(O) + 2Fa(s)

(8)

We note that (8) involves four different initial conditions: x1(0),ii(O),.X1(0), and fa(O). Because there is zero initial stored energy and because the step-function input is assumed to occur just after t = 0 (that is, at t = O+ ), each of these initial conditions is zero. Then

246 .,,. Transfer Function Analysis (8) reduces to (s 3 +3s 2 +3s+l)X1(s) = (s+2)Fa(s) which is equivalent to (2).

Although somewhat different, the methods used in the last three examples gave the same equations for the transformed output X1 (s) and for x1 (t) for all t > 0. When the equations from the free-body diagrams were transformed immediately or when the statevariable equations were transformed, the only initial conditions needed were for functions that did not change instantaneously. This was not the case, however, when we obtained the input-output equation before applying the Laplace transform. In this method, it is not unusual for one or more of the initial conditions to involve functions that have discontinuities. For the acceleration .X1 in the last example, .X1 (0-) = 0, but it was found that .X1 (O+) = 1 m/s 2 . Whenever there is a discontinuity at the time origin, we use the value at t = 0- for the initial-condition term, for the reasons explained in Section 7.5. This is in contrast to the classical solution of an input-output differential equation which requires the values of the initial conditions at t = O+. One advantage of the transform method is that evaluating the initial-condition terms is often easier. In the last three examples, we chose to denote the transform of the unit step input fa(t) by Fa(s) rather than to immediately replace it by ~[U(t)] = l/s. Thus the expression for X1 (s) in (2) was a somewhat more general result than necessary. Equation (2) holds for any input, provided only that the input has a Laplace transform and that the initial stored energy is zero. The quantity inside the brackets is called the transfer function and will become increasingly important. When finding the response to a particular input, however, we more often replace its transform immediately by the appropriate expression from Appendix E.

• EXAMPLE 8.4 After steady-state conditions have been reached, the switch in Figure 8.2(a) opens at t = 0. Find the voltage e0 across the capacitor for all t > 0. .·

SOLUTION The circuit fort> 0 is shown in Figure 8.2(b), with the switch open and with the node voltages e0 and eA shown. The current-law equations at these two nodes are

e0 -12+ ~e 0 + iL(O) + 4 1

t

l

(e 0

-

eA) df.. = 0 1

2(eA -12) - iL(O) +4 Jo (eA -e0 )df..+ 2 eA = 0 Transforming these equations, we have

iL(O) 4 12 1 Eo(s)- - + -4 [sE 0 (s) -e0 (0)] + + -[E0 (s)-EA(s)] = 0 s s s

!2

[EA(s) - 12 ] - iL(O) s s

+ ~[EA(s)-Eo(s)] + !EA(s) = s

2

0

We must find the numerical values of the initial conditions e0 (0) and iL(O) that appear in these transformed equations. Because e0 and iL are measures of the energy stored in the capacitor and inductor, respectively, they cannot change instantaneously and do not have discontinuities at the time origin.

247 20

+

12

v-=

20

(b)

8.2 Circuit for Example 8.4. (a) Original circuit. (b) Circuit valid for t > 0.

When the circuit is in the steady state with the switch closed, the and inductor may be replaced by open and short respectively, as explained at the end of Section 6.5. This is done in Figure from which we find that e0 (0) = 4 V and iL (0) = 8 A. Substituting these initial conditions into the transformed and collecting like terms, we get

1 + ~] E (s) - ~EA(s) = [ ~s+ 4 s s 0

l+

~s

14

-~E 0 (s) + [1 + ~] s s

s We want to find the capacitor voltage e0 (t), so the next step is to solve eliminating (s). Noting from that

for

(s)

= 4E (s) + 14

(s)

0

s+4 and substituting this

into (9a), we find that

Eo(s)

= 4(s2 + 8s + 72) s 3 + 8s2 + 32s

One of E0 (s) is s = 0, and the the poles are s1 = 0, s2 = -4 + and output into the form

s3

(s)--+ s

=

two are the roots of s 2 + Ss + 32 = 0. Hence -4 and we can the transformed

s+ 4-

A3 s + 4 + j4

+---

248 ~ Transfer Function Analysis

2n

2n

IQ

IQ

+

+

2n

12 V-=-

+

e,

12

2n

v-=-

(a)

+ (e,).,

(b)

\

' ' ',

9+5-VZc 4 '

'.......... .......... 9V

/

--- - -------

--~==--

~-~

9-5-VZc4 '

4V

0

0.75

0.50

0.25

1.00 t, s

(c)

Figure 8.3 (a) Equivalent circuit for Example 8.4 just before the switch opens. (b) Equivalent circuit valid as t approaches infinity. (c) Complete response.

where the coefficients are Ai= sEo(s)ls=O =

(4)(72) -----n= 9

Az = (s+4- j4)Eo(s)ls=-4+j4

4[(-4 + j4) 2 + 8(-4 + j4) + 72] ( -4 + j4) (-4 + j4 + 4 + }4) 4(40) (-4+}4)(18)

5 = -(l+Jl)

2_Ej3?r/4

V2

and A -A* - 2_E-j37f/4 3-

2-

V2

Comparing the second and third terms on the right side of (10) to Equation (7.83), we see rad. Using Equation (7.84) to write the part of that a= 4, w = 4, K = 5/V2,, and¢= the response that corresponds to the pair of complex poles, we have

i7f

e0 (t) = 9 + 5-J2E- 41 cos ( 4t +

~7f)

fort> 0

As a check on the work, note that as t approaches infinity, this expression gives a constant value of 9. We can also find the steady-state behavior with the switch open from

249 Figure

""""'"""! and inductor have ..,~~"'~'

been by open and short combination of the 2-Q and 1-Q resistors is

2

e0 ( 00 ) = - 2

= 9V

2+3 fort > 0.

which agrees with the The response is shown in The transient c::::::-::'lp.~.'l~~:',.'- has an enwith a time constant of 0.25 s and has a of

We discussed in Section 7.4 how the response of first-order could be divided into the sum of zero-input and zero-state and also into the sum of transient and 'Hr-.Qt"·'" components. To iHustrate the extension of these to systems and to examine further the role of any initial stored energy, we consider in the next vA•~u1fY1" a simple second-order mechanical Part of the solution will be canied out in literal form. In order to emphasize the concepts, the details of efficients in the partial-fraction expansions will be left for the reader. A of the chapter the electrical analog of the mechanical

EXAMPLE8.5 In the system shown in Figure 8.4(a), the input is the

force v, the velocity of the mass. The effects of the input fort < 0 are summarized conditions x( 0) and v (0), which determine the initial energy stored in the and in the mass, and which we assume are known, First find a for the transform of the in terms of the transformed input and the initial conditions. Then assume that M = l kg, B = 3 N·s/m, and K = 2 N/m and the transforms of the zero-state and zero-input of the Fort > 0, let the input be the sum of constant and sinusoidal co1mpom~nt:s, let = 1 +sin t. Find for an > 0 and the components. Q1CP"'1"T-QTCITP

SOLUTION The free-body is in Figure is v rather than x, we shall write the displacement in terms of the definite = x(O) + J~ vd'A. Then

Mv+Bv+K by

,_,,"'"'"'u"" the transform of the

+ and

v that the transform of the constant

x(O) is x(O) / s, we write (s) -v(O)] + both sides of this for V (s) yields

V(s) =

K

K

+ -x(O) + -V(s) =

s s by s and collect the terms

sF(s) + sMv(O)-Kx(O) Ms 2 +Bs+K Ms 2 -1-Bs+K

V(s) on

250 ti- Transfer Function Analysis

(a)

Bv-~1

M

Kx---1:

I

• f.Vl

.----Mv

(b)

Figure 8.4 A mass-spring-friction system for Example 8.5. (a) The system. (b) Free-body diagram.

Inserting numerical values gives

s ] sv(O) - 2.x(O) [ Vs= () s2+3s+2 F.s+ a() s2 +3s+2

(11)

The zero-state response corresponds to the case where the initial values of the state variables are zero, so the transform of the zero-state response is the first of the two terms in (11). With s2 + 3s+ 2 = (s+ l)(s+ 2), Vzs(s)= [(s+l;(s+ 2)]Fa(s)

(12)

Note that . 1 1 s2 +s+l Fa(s)=~[l+smt]=-+z--1 = ( 2 l) s s+ ss+ so

Vs-[ s ][ss(s2 +s+l] 2 +1) zs( )- (s+l)(s+2)

(13)

Because s2 + 1 has the complex factors (s - jl) (s + jl ), we write the partial-fraction expansion in the following form: Ai A2 A3 Bs+C Vzs(s) = --1 +--2+-+-2--1 s+ s+ s s + Using the techniques in Section 7.6, we find in a straightforward way thatA1=1/2,A2 = -3/5,A3 = 0, B = 1/10, andC = 3/10. Then fort> 0 Vzs (t ) =

1

2€

-t -

3 -2t SE +

1 3 10 cost+ 10 smt 0

Two sinusoidal terms having the same angular frequency can be combined into a single term with a phase angle. Using Table 1 in Appendix D, we can write Vzs(t) =

~E-t - ~E- 21 +0.3162 sin(t + 0.3218)

(14)

Kl The

Vzi(s)

251

Solution

= ------- = --+ - s+ l

s+2

+

and that

so

= -[v(O) + 2x(O)]E- 1 +2[v(0) + The complete response of the system is The response, which is the Vss(t)

the sum of the vr.,·

= 0.3162

The transient response, which dies out as

and is

000 ,

+

t "'"'"""".,,...

(16)

is

l

+ 2x(O)] E- 21

(17)

We conclude this section by examining further the results of the last For the case when there is no initial stored energy, we see from that the transform of the zerostate response is the product of the function of s within the brackets and the transformed input, no matter what the input function might be. The quantity inside the which is determined the system and the element values, is called the transfer function and is often represented by the symbol H(s). In this-, ...... ,,.~, s s H(s) = s2+3s+2 = (s+l)(s+2) which has poles, as defined in Section 7 .6, at s = -1 ands =

- 2. The transformed

is

s2 +s+ 1 (s) = s(s 2 +1) which has poles at s = 0, and - jl. The poles ofV28 (s), some of which come from the denominator of H(s) and some of which come from the denominator of the transformed the process of input, determined the form of the partial-fraction expansion. taking the inverse transform and finding Vzs ( t), we see that the of H (s) gave rise to the transient terms, while the poles of the transformed input gave rise to the terms. The input in the last example consisted of a constant and a sinusoidal component. We would that the steady-state would contain terms of the same kind. that does contain a sinusoidal term having the same fre(t). There is not, a constant term in the v~>JUU.~•v•u for Vzs(s), note that the value of A3 in the term was zero. This could have been anticipated, because a numerator factor in canceled In effect, the system refused to '""' "''u"" the s in the denominator of state to the constant term in the This can also be seen from force is a constant, the spring win becomes zero. that the To appreciate the role of the initial stored energy, consider the response in This consists of transient terms that have the same form as the transient terms in Vzs (t). initial stored energy does not contribute to the response at aH, 0

252 Ii> Transfer Function Analysis nor does it affect the form of the transient terms. Its only effect is to change the size of the transient terms in the complete response.

8.2 THE ZERO-INPUT RESPONSE The zero-input response is defined to be the output y(t) when the input u(t) is zero for all t > 0 and when the initial conditions are nonzero. Because it is understood that the input is zero, we shall generally omit the subscript zi throughout this entire section. The general form of the input-output differential equation for a fixed linear nth-order system was given in Equation (3.25). When the input terms are zero, this becomes

any(n) + an-lY(n-l) + · · · + al.Y + aoy = 0

(18)

where y(n) denotes dny/dtn. Using the expressions for the transforms of derivatives given in Appendix E, we can transform (18) term by term to obtain

an[s11Y(s) -s11- 1y(O) - · · · -y(n-l)(O)]

+ an-1 [s11- 1Y (s) - si- 2y(O) - .. · -y(n- 2 ) (O)] + ···+ al[sY(s) -y(O)] +aoY(s) = 0

(19)

where Y(s) = 5E[y(t)]. If we retain the terms involving Y(s) on the left-hand side and collect those involving the initial conditions on the right-hand side, (19) becomes

(ansn +an-lgi-l + · · · +a1s+ao)Y(s) = any(O)sn-l

+ [any(O) + an-1Y(O)]sn- 2 + ... + [any(n-l) (0) + an-lY(n- 2 )(0) + · · · + aly(O)]

Thus the transform of the zero-input response is

F(s) Y(s) = P(s)

(20)

where

F(s) = any(O)sn-l + [an.Y(O) +an-1Y(O)]sn- 2 + ...

+ [any(n-l)(O) + an-lY(n- 2)(0) + · · · + aly(O)] and

P(s)

= ansn + an-1Sn-l + · · · + a1s + ao

(21)

(22)

Because P (s) is of degree n and F (s) is at most of degree n - 1, Y (s) is a strictly proper rational function. The numerator polynomial F (s) depends on the initial conditions. We shall show in the next section that the denominator polynomial P(s) is identical to the denominator of the transfer function. It is a characteristic of the system and is sometimes called the characteristic polynomial. To factor P(s), we can find the roots of the characteristic equation P(s) = 0. There are explicit formulas for doing this for first- and second-order polynomials, but we may have to use a computer or hand calculator for higher-order cases. When P (s) is factored, it will have the form

P(s) = an(s- s1)(s - s2) ... (s - sn) The quantities s1,s2, ... ,sn are the poles of the transformed output (as well as the poles of the transfer function); hence they determine the form of the zero-input response. For a

8.2 The Zero-Input Response

0. The second case corresponds to the characteristic equation s2 + (3 2 = 0 and to the response y(t) =Kcos((3t+ 0, it is useful to rewrite the characteristic equation in the standard form +wn2 =0 s2+ The parameter Wn is called the undamped natural frequency and has units of radians per second. The parameter ( is dimensionless and is known as the damping ratio. For ( > 1, the roots are distinct negative numbers, and y(t) consists of two decaying exponentials. For ( = 1, there is a repeated root at s = -wn, and y(t) consists of terms having the form cwnt and tcwnt. For 0 :::; ( < l, the roots are complex and are s1

= -(wn + jwnJI=C?

Sz

= -(wn - jwnJI=C?

Then by (26), the zero-input response is

= Kt-(wnt cos(wnJI=C? t + 0 it is the inverse transform of The poles of Yu(s) consist of the poles of and a at s = 0. Provided has no pole at s = 0, the time functions that make up yu(t) wm be the mode functions of the zero-input response and a constant term from the pole at s = 0. fa fact, i.f the all the mode functions wm to zero as t approaches infinity, and the of the step response will be due of this to the pole at s = 0. The coefficient in the term, and thus win be

(yu )ss = We can also

this result

the final-value theorem to

(s) as

Yu(=)= Because the initial condi.ti.ons do not affect the response of a stable even when the initial stored energy is not zero. If has a at (s) as win have a double at s = 0, and the response win contain a ramp function in addition to a constant term. Then Yu wiH grow without bound as t

272

Transfer Function

EXAMPLE 8.14 Find the unit ti on

described

response for a second-order

the

equa-

y+ Pfot the response for several values of the uu,HµAHi'-

SOLUTION

From

yu(t) = :£

-1

We shaH omit the details of find that

Yu(t) = 1-

the inverse transform. For 0

c( ~

2)

< ( < 1, for example, we

s:in(w12 JI=(Z t + ¢)

where

¢ = tan- 1 ( (/ JI=(Z) The unit step response is shown in Figure 8.16 for several values of(, with Wn held constant. For ( > 0, the response is unity. Note that the value of ( determines to what extent, if any, the response will overshoot its steady-state value. The overshoot :is 100 for ( = 0 and decreases to zero when the damping ratio is unity. For ( > 1, the response its value monotonically. The case ( = 1 is the boundary between responses that oscillate and those that do not, and the is said to have c:ritical damping. Second-order systems for which ( > 1 have more than critical damping; those for which 0 < ( < 1 are less than critically damped. When ( = 0 the system is Figure 8.11 shows the unit impulse response of the general second-order system described by (61) for the values of (used in Figure 8.16. Note that the steady-state response is zero for ( > 0 and that the value of the damping ratio establishes the character of the response. In the design of systems, the choice of ( depends upon several factors, but consider the following two simple examples. In many electrical measuring instruments, we want the

2 w,,

4

6 w,,

8

8.16 The unit step response for a second-order system described

10 t w,,

(61).

8.3 The Zero-State Ke1monse

273

to settle down so that a may be obtained An amJropnate value of ( be a Httle less than one, 0.8. On the olher if we want a robot 3Jffi1 to move out and a flat surface as it passes we would need ( to be somewhat than one. the arm overshoot and dent the surface that is to be

EXAMPLE 8.15

Use

transforms to find the unit

iwa1Y""'"

Y" + ~~·~,,,_,,~

=

+-

2u

+

deresponse of lhe ...-.,..,· ..-··y for which we found uu1Ju"'" response in

8.13.

of the

SOLUTION to be

Hence the transform of the unit

differential

the transfer function

2s+ 1 s 3 +4s2 +3s response is 2s+ l (s)

whose

= s 2 (s 2 + 4s + 3)

expansion can be shown to be

(s) = 1/3

+ 2/9 _

s2

s

1/2 s+l

+ 5/18 s+3

Thus

1 1 -I = -t + -2 - -£ + -5£ -31

fort > 0 3 9 2 18 As shown in Figure the steady-state of the unit step response contains a ramp component with a slope of in addition to the constant of . Th.is is because H(s) has a pole at s = 0. This pole with the pole at s = 0 that is to the step-function to give a double pole ofY(s) at s = 0. As an aid in the unit step response, we note that in addition to Yu (O+) = 0, the initial slope is zero, because is the derivative of the step response and Fig= 0. ure

YU

uu,,~.,,~

In Section

response is the derivative of the unit and yu(t), we use the theorem for the

to write

Because ~!Yu(t)] =

where 'A is a

Yu(t) =

it foHows that

274 ~ Transfer Function Analysis Yu

2

0

3

4

t, s

Figure 8.17 Step response for Example 8.15.

In graphical terms, (62) states that the unit step response yu(t) is the area from 0 to t underneath the curve of the unit impulse response h(t). We can also use the theorem for a derivative to write ~[Yu(t)] =

sYu(s) -yu(O)

When we are dealing with zero-state responses, we assume that the output, the input, and all of their derivatives are zero fort::=; 0. Any discontinuities are assumed to occur at t = O+. With Yu(s) = H(s) /sand with yu(O) = 0,

~[Yu(t)] ;;_'lf(s) = ~[h(t)] Noting that Yu (t) and h(t) must be zero fort ::=; 0, we conclude that d

h(t) = dtyu(t)

(63)

which agrees with Equation (7.54). Of course, either (62) or (63) follows immediately from the other equation. For a linear system that contains no initial stored energy, the above arguments can be used to prove a more general statement. Suppose we know the response to a certain input. If we substitute a new input that is the derivative of the old input, then the new response is the derivative of the old response. This property may be useful when constructing block diagrams for certain types of system models.

8.4 FREQUENCY RESPONSE In discussing the impulse and step responses, we were interested in both the transient and steady-state components. However, when considering the sinusoidal input

u(t) = sinwt

(64)

8.4 Frequency Response

, respectively, we can write (65) Yss(t) =Asin(wt+) which is referred to as the sinusoidal steady-state response. It plays a key role in many aspects of system analysis, including electronic circuits and feedback control systems. We now show how we can express A and in terms of the transfer function H (s). The initial conditions do not affect the steady-state response of a stable system, so we start with the transform of the zero-state response as given by (56). Because ~[sinwt] = w/ (s 2 + w2 ), the transformed response to (64) is

Y(s) =H(s)

[~] s +w

(66)

Hence the poles of Y (s) will be the poles of H (s) plus the two imaginary poles of U (s) at s = jw and s = - jw. For a stable system, all the poles of H (s) will be in the left half of the complex plane, and all the terms in the response corresponding to these poles will decay to zero as time increases. Thus the steady-state response will result from the imaginary poles of U (s) and will be a sinusoid at the frequency w. To determine the steady-state component of the response, we can write (66) as

Y(s) =H(s)

[(s- jw~s+

jw)]

C1 C2 . = --.-+--.-+[terms correspondmg to poles of H(s)] S- JW s+ JW

(67)

where C1

= (s- jw)Y(s)ls=jw =H(s)

=

w.

I

(s+ JW) s=jw

H(jw) 2}

The constant C2 is the complex conjugate of C1; that is,

C __ H(-jw) 2-

2}

Because all the terms in (67) corresponding to the poles of H (s) will decay to zero if the system is stable, the transform of the steady-state response is

_ H(- jw)/2} ( ) -_ H(jw)/2} YssS . . S- JW s+ JW

(68)

In general, H (jw) is a complex quantity and may be written in polar form as

H(jw) =M(w)EjO(w)

(69)

where M (w) is the magnitude of H (jw) and () (w) is its angle. Both M and () depend on the value of w, as is emphasized by thew within the parentheses. The quantity H(- jw) is the

276

Transfer Function

and we can write it as = M(w)E-je(w)

and

M (

EJe

cJe )'

(s)=- - - - - ss+

we find that

the inverse transforms of the two terms in

= M [EjeEjwt

E

-je E-}wt]

as in Table 1 in Appendix

Using the

+e)

=M

which has the form in where A = similar derivation to show that for a stable system, the B + ¢i) is Yss(t) = + ¢1 + e) response to

we obtain

. One can use a

(72)

=Bcos(wt+¢2) is (73)

+¢2 +B)

Yss(t) =

In summary, the sinusoidal steady-state response of a stable linear system is a sinusoid having the same frequency as the input, an amplitude that is M (w) times that of the input, and a phase angle that is () (w) plus the input angle, where M (w) and () (w) are the magnitude and angle of H (iw), respectively. As a consequence, the function H (jw), which is the is known as the frequency-response function. transfer function evaluated for s = Calculating and interpreting the frequency-response function are illustrated in the following examples. In the first example, we find the steady-state response to a sinusoidal input that has a specified frequency. In the next two examples, we sketch curves of M(w) and () (w) as functions of w. Such curves indicate how the magnitude and angle of the sinusoidal steady-state response change as the frequency of the input is changed.

EXAMPLE!U6 Use and function

to find the steady-state response of the system described by the transfer

9s+ 14 3(s + 4s+ 3) 2

to the input

= 6 cos 2t, and compare the result to that from Example 8. U.

SOLUTION

We replaces

with w = 2 rad/s, the frequency of the input, to form

j18 + 14 3(-4+ +3) The urnsrnm"''"' of denominator:

14 + j18 3(-l+j8)

is the magnitude of its numerator divided M= l14+j18I =

31-1 + j&I

22.80 =·0 427 3(8.062) ·9

the magnitude of i.ts

8.4 Frequency Response

0 when the input = 0 fort> 0, but when = 1 and = -2.

SOLUTION For the = 0.5 to write Yss = (0.5) ( 1)

with

response to a unit

= 0.5. For the

() 2 s = s(s+ 2) 2

0.5 s

so fort> 0

YU(t) = 0.5-tt- 21 -

1

(s+

2) 2

0.5

s+2

282

Transfer Function

Note that the vious answer for Yss· To find the need

which agrees with our pre-

0 ""''""'"- 0 ' "

0 '"''"'''- 0 ''""'

we

response to a sinusoid with an

2 = 0.25E-jn/Z (2+ }2) 2 ) with M = 0.25 and & = -?T /2 rad Yss(t) =LS - n: /4). Finally, for the fast of the is zero for t > 0, but the initial conditions are nonzero. We cannot make use of which is valid for the zero-state response. we know that the input-output is ji + + side is zero fort > 0, we transform this to get -y(O)] +4[sY(s)-y(O)] +

= 0

the numerical values for the initial conditions and solving algebraically for we obtain = l / (s + 2), so y(t) = c 21 for an t > 0. For the tnmsfonn of the the transfer we would normally the zero-input response to contain modes of the form c 21 and , but the second of these is not excited our choice of initial conditions.

EXAMPLE 8.20 The pole-zero plot for a certain transfer function is shown in Figure 8.22. It is also known that the steady-state response to a unit step is 0.5. Find the system's input-output equation.

SOLUTION

Because the transfer function has zeros at -1 and -3, and poles at -2 and

H(s) = K(s+ l)(s+3) (s+2)(s+4) To evaluate the multiplying constant K, we note that H(O) = 3K/8, which by (82) must equal 0.5. Thus K = 4/3. Multiplying out the numerator and denominator of H(s) gives

so ji+ 6y+

H(s) = (4/3)(s 2 +4s+3) s2 +6s+8 = (4/3)(ii+4u+ 3u(t)].

EXAMPLE 8.21 The unit-step response of a certain system is yu (t) = 4+ c 21 (cost-18 the damping ratio (. Also find the steady-state response to the n:/3)+ +?T/4).

-x--o-x---o -4

-3

-2

8.22 Pole-zero

-1

for Example 8.20.

fort> 0. Find

= 4 + 3sin(t +

8.6 Impedances

SOLUTION

0. LetJ = 1 ,B = 5

shown in andK=6N·m.

a.

Find the zero-input response in terms of 8(0) and

b.

the mode functions. For each mode im11A. 0 and when the initial conditions are y(O) = 0 and y(O) = 1.

*8.36. The unit step response of a certain linear system is

yu(t) = [2+2c:- 1 cos(2t+n/4)JU(t)

294

Transfer Function

a.

Find the numerical values of the quencywn.

b.

Find the unit

response

ratio ( and the . Sketch Yu

natural fre-

and

*8.37. a.

P8.37 is

Verify that the transfer function for the circuit shown in

s2 +2s+ l s2 +4s+4 b.

Find the unit impulse response.

c.

Find the unit step response. 211

e;(t)

Figure P8.37

8.38. The mechanical system described in Problem 8.10 has two inputs, the gravitational force M2g and the force fa(t) applied to Mi. a.

Solve for xo, the constant displacement caused by gravity when the applied force fa (t) is zero.

b.

Find the transformed outputX(s) when x(O) tion.

c.

Find for the conditions in part (b) when =Bi = B2 = Kz = 1 in a consistent set of uni.ts. Check your answer evaluating Xss and x(O+ ).

= xo and fa(t)

is the unit step func-

8.39. Use the Laplace transform to solve for the unit

response of the mechanical described in Problem 8.11 when the parameter values are M = J = R = = K2 = l and B = 5 in a consistent set of units. 8.40. For the rotational system shown in Figure 5.19 and discussed in Example = 0.5, B = 1, and K = 2 in a consistent set of units.

let

Ii =

= ~[w2(t)] by

a.

Find a general expression for

b.

If the system is at rest and the inputs are t > 0, find and sketch w2 fort> 0.

= 0 and Ta(t) = c

21

for

*8.41. Find and sketch e0 versus for the circuit shown in Figure 6.17 and discussed in Example 6.4. The values are C = 0.5 F, = 2 = l and L = 0.5 H. The inputs are e1 (t) = 2 V for all t and V. conditions exist at t = 0-.

Problems

8.42. Use the transform to solve for the unit in Problem 8.12 when the values are C = L units.

a.

=

295

response of the circuit described = 1 in a consistent set of

For the circuit shown in current law to nodes A and 0 leads to the

1

-

2

+ c.

+

+

Use tlle Laplace transfmm to solve for the unit

=0

response.

2F

2 fl

Figure P8.43

8.44. By a derivation similar to the one used to obtain Equation show that the = B cos wt is Yss ( t) = state response of a stable system to the where M and () are given by Equation 8.45. Derive the expression given in

for the steady-state response to

Bsin(wt+1). 8.46. The modeling equation for Figure was given in "-''-l"'"''v" represented by the Simulink diagram in Figure 4.13. As in ~"""""'V B=4 K = 16 N/m, and = 8si.nwt. The response when w = was plotted in the top half of Figure 4.20. Problem 4. rn asked for the for several other values of w. a.

Use the frequency-response function rad/s.

b.

Use to check the w = 27r(0.5)

urn.)',H.m."''°'

to check the

and angle of the

8.47. For each of the following transfer functions, of M (w) versus w and () (w) versus w, and comment in part (c), indude the numerical values for w"' 2 a.

rad/s

response when

draw curves on your results. For the function and 10.l

296

Transfer Function

b.

s

c.

s2

+ 0.2s + 100

Problem 8.47 for the

*8.48. a.

s-1 s+l

b.

(s+ 1) 2 s(s + 5)

transfer functions.

s

c.

- (s2 + 0.2s + 100) 2 to the = si.n5t +sin IOt +sin 15t

has the form

=A

+81) +Bsin(lOt+

+c

the transfer function in part of Problem use For a system described tion to find the values and and calculate the ratios and C / B. 8.50. The response of a system described the transfer function in Problem 8.47 to the input = 1 + sint +sin 10t

of

has the form Cakulate

=A+ B sin(t + 82 ) +c sin(lOt + 83) and C, and comment on the ratios B /A and C /A.

*8.51. a.

For the series RI£ circuit shown in Figure 6.13, find the transfer function

b.

By examining H (jw), determine the value of w such that the steady-state response to e;(t) = sinwt is = (1/R) sinwt.

8.52. a.

Find the transfer function in Figure 6.14.

RI£ circuit shown

for the

determine the value of w for which the steady-state response

b.

= Rsinwt. ""'"a'u~'"

a.

6.17 we used Simulink and MATLAB to the response of the circuit where w 0.2JT. The ·~~,, .. ~n•~n• ~ 0 •• 0 ~.a«>~ values used in the simulation.

=

andEi(s).

Problems

b.

Evaluate the calculate the Hrn.i;rn 0.

> 0 when the input is

response when

=3+5

8.56. A transfer function is known to have the form

+as+b s2 +cs+d

s2 ----Use the following facts to determine the values of the real constants a, b, c, and d. The

response to a unit

is 2.

The

response to the input

=A sin2t is zero.

8.57. Find the transfer function for the circuit shown in elements their impedances.

8.19

the

8.58. a. b.

Draw the s-domain version of the circuit shown in the circuit drawn in part

P8.37

determine the transfer function

(s).

8.59. a.

Draw the s-domain circuit for Figure by their

with the

elements characterized

298

Bit>

Transfer Function Analysis

b.

Use the result of part (a) to determine the circuit's transfer function.

c.

From the answer to part (b), write the input-output differential equation.

8.60. Repeat Problem 8.59 for the circuit shown in Figure P6.6. *8.61. Repeat Problem 8.59 for the circuit shown in Figure P6.12. 8.62.

a.

Use impedances and the results of Example 8.23 to find the transfer function for the op-amp circuit shown in Figure P8.62.

b.

Plot the pole-zero pattern.

+

Figure PS.62

8.63. Repeat Problem 8.62 for the circuit shown in Figure P6.34.

9 DEVELOPING A LINEAR MODEL In nearly all the examples used in earlier chapters, the elements were assumed to be linear. In practice, however, many elements are inherently nonlinear and may be considered linear over only a limited range of operating conditions. When confronted with a mathematical model that contains nonlinearities, the analyst has essentially three choices: (1) to attempt to solve the differential equations directly, (2) to derive a fixed linear approximation that can be analyzed, or (3) to obtain computer solutions of the response for specific numerical cases. The first alternative is possible only in specialized cases and will not be pursued. We discussed computer solutions in Chapter 4, and we now present the linearization approach in this chapter. We first develop a method for linearizing an element law where the two variables, such as force and displacement, are not directly proportional. We then show how to incorporate the linearized element law into the system model. We consider mechanical systems with nonlinear stiffness or friction elements, as well as nonlinear electrical systems. We conclude with an example that uses different models for different stages of the solution. ~

9.1 LINEARIZATION OF AN ELEMENT LAW The object of linearization is to derive a linear model whose response will agree closely with that of the nonlinear model. Although the responses of the linear and nonlinear models will not agree exactly and may differ significantly under some conditions, there will generally be a set of inputs and initial conditions for which the agreement will be satisfactory. In this section, we consider the linearization of a single element law that is a nonlinear function of a single variable. We can express such an element law as an algebraic function f(x). If x represents the total length of a nonlinear spring and f(x) the force on the spring, the function f(x) might appear as shown in Figure 9.l(a), where xo denotes the free or unstretched length. We shall carry out the linearization of the element law with respect to an operating point, which is a specific point on the nonlinear characteristic denoted by x and f. A sample operating point is shown in Figure 9.l(b). We discuss the procedure for determining the operating point in the following section; for now, we shall assume that the values of x and J are known. We can write x(t) as the sum of a constant portion, which is its value at the operating point, and a time-varying portion x(t) such that

x(t) = .x+x(t)

(1)

299

300 I

I

f ------------Operating point

(b)

(a)

9.1 (a) A nonlinear spring characteristic. (b) Nonlinear spring characteristic with point

I

f

Operating point

Figure 9.2 Nonlinear spring characteristic with linear approximation.

The constant term xis caned the nominal value of X, and the time-varying term x(t) is the incremental variable corresponding to x. Likewise, we can write f(t) as the sum of hs nominal value f and the incremental variable f(t): =

f + f(t)

where the dependence of f on time is shown explicitly. Because x and f always denote a point that lies on the curve for the nonlinear element law,

f

=

(3)

Having defined the necessary terms, we shaH develop two ~" ~H,.U methods of linearizing the element law to relate the incremental variables x and l The first uses a approach; the second is based on a expansion.

Figure 9.2 shows the nonlinear element law f(x) with the to the curve at the operating point appearing as the line. For the moment, we note that the tangent line to the nonlinear curve that the independent variable win be a good x does not deviate greatly from its nominal value x and that the curvature of the curve

9.1 LineMization of an Element Law

of the ~ ,_,,,. _,... ,, point. The

is smaH in the

dx

of the

301

line is

lx=x

k- dx

Ix

x after the vertical Hne indicates that the derivative must be evaluated

where the

the

that has the coordinates

=! +

-i)

f-1=

-x)

f

and

which can be written as Noting from (1) and

that the incremental variables are

i=x-x

l=!-f we see that

reduces to

f =ki

where k is given (4). We can represent in graphical form the nonlinear function with a coordinate system whose axes are x and and whose origin is located at the operating the point, as depicted in Figure 9 .3. The incremental variables x and f are Hneady constant of proportionality k the slope of the tangent line at the Obviously, the accuracy of the linear approximation depends on the curvature of in the vicinity of the operating (x,f) and on the extent to which x deviates from i as the system responds to its excitations. We that the linearized model should be a good approximation for those values of x for which the line the original curve. If the nonlinear spring is part of a larger system, we are to know in advance what range of values we wiH encounter for x. The is further complicated by the fact that usually we are primarily interested in how well certain responses of the overall system are approximated by cakulations made on the linearized model. For example, we can expect some of the elements in a particular system to a more critical role than others in determining the response of interest. Generally, the only way to assess

9.3 NonlineM

characteristic with incremental-variable coordinates.

302 with nonlinear model and the linearized model.

pn~se1mte,d.

we can derive the linearized about the

av11or-·!ill'n'1?s exl!l!U11sio1n

i after the vertical line indicates that the associated derivative is evaluwhere the ated at x = i. We can find the first two terms of this provided that f and its first V~•.UW•UV'U to the actual curve, so we shall sut,seqm:ont terms, which are in x - .i. The justification for truncating the series after the first two terms is that if xis sufficiently dose to .i, then the higher-order terms are negligible to the constant and linear terms. Hence-we write

+-I

dx x

(x-i)

(8)

to which reduces and k = Because J = to in (8) on the extent to which the The accuracy of the linearized ,,.__.. ~,. ~.. ~·v• terms we have omitted from the Taylor-series expansion are truly negligible. This in tum on the of x - .i, which is the incremental independent and on the values of the. higher-order derivatives of f(x) at the operating point. the li.nearization of the complete model, we shaH consider two numerical examples that iHustrate the Hnearization of a nonlinear element law.

EXAMPLE9.1 A nonlinear translational spring obeys the force-displacement relationship f(x) = lxlx, where x is the elongation of the from its unstretched length. Determine the linearized element law in numerical form for each of the operating points corresponding to i1 = -1 = O,i3 = 1, and i4 = 2. the nominal spring

SOLUTION

We can rew1ite

as for x for x

The

of the

at the operating

k-

=

is

-2.i

- { dx Ix 2.i

21.il

0

for .i < 0 for .i > 0

for an i characteristic is

f = 21.ili where the coefficient k = 21.il can be of as an effective constant whose ~A~H,,,AUV'H i, numerical value depends on the nominal value of the

9.1 Linearization of an Element Law

303

TABLE 9.1

f; -1 0 1 4

i; -1 0 1 2

1 2

3 4

k; 2 0 2 4

f

4

for

x

4

=

2

3

for

x

2

2 =

0

x -I

for

x,

9.4 Nonlinear

the four specified values of i into (9) and gives the values of f and k that appear in Table 9.1. Figure 9.4 shows the four linear on the nonlinear spring characteristic. Note that the value of the effective constant k is dependent on the location of the operating In k vanishes for i = 0, which i =0 implies that the would not appear in the linearized model of a as its operating point. It is also interesting to note that k is the same for i = -1 and i = + 1, although the values differ. Concerning the accuracy of the approximation, one say that for deviations in x up to 0.25 from the the approximation seems to be for deviait would be poor. n is difficult to make a definitive statement, the in which the element is to appear.

EXAMPLE9.2 A

exerted on a that can rotate with an TM = DsinB. Determine the linearized element law """~'H·'A Consider the five to 01 = 0, = 7r / 4, 03 = and Bs =Jr. TM

304

a Linear Model

TABLE9.2

1 2 3 4 5

Bi

fM

0

0

D

D

0

0

-D

Jr

k;

TM

03 =~ D

for 6l 1 "'0

rr

6l

2

-D

9.5 Nonlinear characteristic curve and linear approximations for Example 9.2.

SOLUTION

The Taylor-series expansion for TM is TM=

d I (8 - 8)- + ... Dsin8- + -Dsin8

dB

=DsiniJ+

e +···

Using the first two terms in this series, and noting that

= DsinB, we can write

= D(cosiJ){j =

where k = D is the linearized stiffness constant. In Table 9 .2, the values of iJ, fM, and k are Hsted for each of the five operating The values of which were found from ki = DcoslJi, are also the The to the characteristic curve drawn at the operating these are shown in Figure 9.5.

of the tan-

for the shown in one of the we used the small-angle ~"'"~""'·'~•wu In Examples 5.7 and identity to obtain a linearized element law for small variations about iJ = 0 and iJ = 1r. We found that for {} = 0 and that these same results were obtained by a more for iJ = Jr. In Example linearization method.

9.2 Linearization of the Model

~

~

305

9.2 LINEARIZATION OF THE MODEL We shall now consider the process of incorporating one or more linearized element laws into a system model. Starting with a given nonlinear model, we need to do the following: 1. Determine the operating point of the model by writing and solving the appropriate nonlinear algebraic equations. Select the proper operating-point value if extraneous solutions also appear. 2.

Rewrite all linear terms in the mathematical model as the sum of their nominal and incremental variables, noting that the derivatives of constant terms are zero.

3.

Replace all nonlinear terms by the first two terms of their Taylor-series expansionsthat is, the constant and linear terms.

4.

Using the algebraic equation(s) defining the operating point, cancel the constant terms in the differential equations, leaving only linear terms involving incremental variables.

5.

Determine the initial conditions of all incremental variables in terms of the initial conditions of the variables in the nonlinear model.

For all situations we shall consider, the operating point of the system will be a condition of equilibrium in which each variable will be constant and equal to its nominal value and in which all derivatives will be zero. Inputs will take on their nominal values, which are typically selected to be their average values. For example, if a system input were u(t) = A+ B sin wt, then the nominal value of the input would be taken as ii =A. Under these conditions, the differential equations reduce to algebraic equations that we can solve for the operating point, using a computer if necessary. Upon completion of step 4, the terms remaining in the model should involve only incremental variables, and the terms should all be linear with constant coefficients. In general, the coefficients involved in those terms that came from the expansion of nonlinear terms depend on the equilibrium conditions. Hence we must find a specific operating point before we can express the linearized model in numerical form. The entire procedure will be illustrated by several examples .

.... EXAMPLE 9.3 Derive a linearized model of the translational mechanical system shown in Figure 9.6(a), where the nonlinear spring characteristic fK(x) is given in Figure 9.6(b) and where the average value of the applied force fa (t) is zero. SOLUTION First we derive the nonlinear model by drawing the free-body diagram shown in Figure 9.6(c) and summing forces. This yields

Mx+Bi+ fk(x) = !a(t)

(11)

To find the operating point, we replace fa (t) by its average value la and x by i:

Mx+Bx+ fK(x) =la Noting that that

x= x= 0 because i is a constant and that la was specified to be zero, we see

Thus the operating point is at i = characteristic in Figure 9.6(b).

fK =fK(i) = 0 O,/K = 0, which corresponds to the origin of the spring

306

a Linear Model

Slope"' k(O)

(a)

point

(b)

(c)

Figure 9.6 (a) Nonlinear system for Example 9.3. (b) Nonlinear (c) diagram.

i =

characteristic.

The next step is to rewrite the linear terms in (11) in terms of the incremental variables x-x and Ja(t) = fa(t) - This yields

M(x+i) + B{x+i) + fK(x) =la+ f~(t)

x x= 0, we can rewrite the equation as

Because =

+ =la+ fa(t) Expanding the spring force fK (x) about x = 0 gives Mf+

fK(x) = fK(O)

(12)

+ dfK_ I x+ ... dx x=O

Substituting the first two terms into (12) yields the approximate equation

Mi+ Bi+ fK(O) +

+ fa(t)

The constant denotes the derivative evaluated at x = 0 and is the slope of the tangent to the characteristic at the operating point, as indicated in Figure 9.6(b). The spring force at the operating point is fK(O) =la= 0, so the linearized model is

(t) which is a fixed linear differential in the incremental variable x whh the incremental input . The coefficients are the constants and To solve we must know the initial values x(O) and which we find from the initial values and Because - i and - for an values of t,

x

-x

x

In this x = = 0, so = x(O) and Once we have solved the linearized we find the solution of the nonlinear model by adding the nominal value x to the incremental solution . Remember that the sum of the terms, x + , is only an approximation to the actual solution of the nonlinear model.

9.2 Linearization of the Model

Developing a Linear Model

where k(.i) = [dfK/dx]l.x and is the slope of the straight line in Figure 9.7(b). Note that the form of the model given by (17) with la =f 0 is the same as that with la = 0, which is given by (13). The only difference between the two equations is the value of the effective spring constant. The value of k(.i) depends on the value of .i at which the slope of fK(x) is measured. Hence the responses of the two linearized models could be rather different, even for the same incremental applied force la (t).

'1> EXAMPLE 9.5 Derive a linear model for the mechanical system and spring characteristic shown in Figure 9.8, where x = 0 corresponds to an unstretched spring. SOLUTION We obtain the nonlinear model of the system by drawing the free-body diagram shown in Figure 9.9(a) and setting the sum of the vertical forces equal to zero. Because the mass is constrained to move vertically, we must include its weight Mg in the free-body diagram. The resulting nonlinear model is

Mx+Bi+ fK(x) =!a(t)+Mg Note that its form is similar to that given by ( 11 ), the nonlinear model for the two preceding examples. By setting x = .i and fa (t) =la and by noting that = = 0, we find the algebraic equation for the operating point to be

x x

fK(.i) =la+ Mg which is the same as ( 14) except for inclusion of the force Mg on the right side. The nonlinear spring characteristic in Figure 9.8(b) is repeated in Figure 9.9(c). Figure 9.9(b) shows fa(t) and la as in ~l;lffiple 9.4, but it also shows the total nominal force la +Mg that must be projected ont6 the characteristic curve in Figure 9.9(c) to establish the operating point Ai. The point Az, which is obtained by projecting the force la from Figure 9 .9(b), would be the operating point if the motion of the mass were horizontal, as it is in Example 9.4. For applied forces having the same average values, the two systems will have different linearized spring characteristics if the curve of fK (x) does not have the same slope at points Ai and Az. If the spring were linear, however, the slope of the characteristic in Figure 9 .9(c) would be constant and the presence of the weight would have no influence on the effective spring constant, as was observed in Example 2.5. With the provision that

x

(a)

(b)

Figure 9.8 (a) Mechanical system for Example 9.5. (b) Nonlinear spring characteristic.

309

9 .2 Linearization of the Model

/,,(/) Mg (a)

/.(1)

J.

+Mg

0

(b)

Figure 9.9 (a) Free-body

(c)

for Example 9.5. with two operating

k(x) is the slope of fK(x) measured at point

characteristic

rather than at point

earized model is again given by

EXAMPLE9.6 A high-speed vehicle of mass M moves to a linear retarding force Bv caused viscous friction associated with the and a nonlinear retarding force caused air Obtain a linear model that is vali.d when the driving force undergoes variations about a nominal value SOLUTION

The nonlinear differential

v =ii and

and

that

the vehicle's

is

v= 0, we have

Bv+Dlvlv = and we

+Bv-

=0

310 ~ Developing a Linear Model

By inspection we see that (19) will have two real roots, one positive and the other negative. However, we are interested only in the positive root, which is

_ -B+JB2 +4laD v= 2D

(20)

The negative root was introduced when we replaced Iii Iii by (ii) 2 and is not a root of the original operating-point equation. Provided that v always remains positive, we can replace DI ii Iii in (18) by Dv 2 • Then, using v =ii+ v and fa(t) =la+ fa(t) and noting that = 0, we can rewrite (18) as

v

MB+ B(ii + v) + Dv 2 =la+ fa(t) (21) To linearize the term v 2 , we replace it by the constant and linear terms in its Taylor series: (ii)2 +

~(v 2 ) I dv

v

(v -ii)= (ii) 2 + 2iiv

(22)

Substituting (22) for v 2 into (21) and regrouping, we have

MB+ (B + 2Dii)v + Bii + D(ii) 2 =la+ fa(t) The constant terms cancel because of (19), the operating-point equation. Thus the desired linearized model, which holds for la> 0 and ii> 0, is MB+bv=fa(t) where b denotes the effective damping coefficient

b=B+2Dii with ii given by (20) as a function of the average driving force la· It is worthwhile to observe that in this particular case the Taylor series of the nonlinearity has only three terms, and )Ve) could have obtained it without differentiation by writing v 2 =(ii+ v) 2 = (ii) 2 + 2iiv + (v) 2 In this example, we can see that (v ) 2 is the error introduced by replacing v 2 by (ii) 2 + 2iiv. Provided that ii » Iv j, the error will be small compared to the two terms that are retained.

~ EXAMPLE 9.7

A nonlinear system obeys the state-variable equations i=y

(23a)

Y = -lxlx- 2.x- 2y3- 3 +Asint

(23b)

Find the operating point and develop the linearized model in numerical form.

SOLUTION

The operating point, described by

x and y,

must satisfy the conditions

x= y = 0 with the incremental portion of the input set to zero. Hence the operating-point equations reduce to

y=O

(24) lxlx+2X+3 = o which have the solution x = -1, y = 0. We replace the two nonlinear elements in (23b) by the first two terms in their respective Taylor-series expansions. For jxjx, we write

1.x1.x+21.x1.x

9.2 Linearization of the Model

x=x+xandy= x=y+.9 y=Withi = -1andji=0,

+

+i)-

311

we obtain

- 3 +Asint

reduces to

x=y

y = -4i+Asint which is the linearized model in state-variable form. (26) with that all the constant terms have been the coefficient in the second is zero because y = 0, and the coefficient of reflects the combined effects of the linear and nonlinear terms.

x

We have defined capacitors, and inductors as elements for which there is an braic relationship between the voltage and current, and and current and flux linkage, respectively. If the two variables involved in the algebraic are proportional to one another, then the element is Hnear, as was the case in an the examples in Chapter 6. We now consider circuits with nonlinear resistors. The general for vv,....,,.u,"' a linearized model is the same as that used for the mechanical examples. As in Section we express the variables as the sum of a constant portion and a portion. For example, we write a voltage e0 as

eo = eo +eo where the constant term e0 is the nominal value, to a point, and where eO is the incremental nrr>P-''1'UV1TI we indicate the fact that a resistor is nonlinear a curved Hne The: procedure is illustrated two examples of increasing ~~,,,., .. w,,.

EXAMPLE9.8 The circuit shown in Figure 9.lO(a) contains a nonlinear resistor that obeys the element law i0 = 2e~. Write the differential equation relating e0 and If = 18 +Acoswt, find the operating point and derive the linearized input-output differential Also determine the time constant of the linearized modeL SOLUTION

,,_,,.

,,,-,,~

resistor is nonlinear because the current i0 is not directly the currents the node at the upper

~~.urnrnrn'"'

gives l

2eo + [eo Thus the unn.H·-vuc 0 andjl = 0.5 9.6. f(z)

=

{-JIZT vz

for z for z

0 and the flux-density vector 'B points to the right, toward the south pole. Thus the positive sense of em is directed toward the viewer, and the corresponding current direction is out of the paper, opposite to the reference arrow for i in the figure. The circuit shown in Figure 10.8(b) was drawn with this assumed polarity for em, so we have

ae

em= (2N'Bfa)(}

(12)

Summing torques on the coil and using (11) give

JO+ B() +KB = (2N'Bfa )i

(13)

Using (12) in a voltage-law equation for the single loop that makes up the electrical part of the system gives di dt

.

L- +Ri + (2N'Bfa)B = ei(t)

(14)

344

Electromechanical

third-ordermodeL From these two "''Y''"'"'UU" the transfer and the inductance of the coil is often u~ ~ ~ This is in for the response to an because the model becomes second-order. When this term is we can solve (14) for the current, l i= R ...

•.

. . ··•]

where a = 2N '13£a is the electromechanical coupling coefficient Co1m¥mrmg the left side of with the left side of Equation l ), we find that the m"~""·'"~·~ and the ratio ( of the are

Wn

Wn=f§ l (=2m

urnjan11pe;d natural frequency on the mechanical parameters K and J, ratio on both the mechanical and electrical parameters and the electromechanitcal coupling coefficient a. We can find the sensitivity of the in radians per volt solving for the steady-state response to the constant excitation e;(t) =A. The steady-state value of(:) can be found setting the derivatives in (15) zero, so ~~.,,,,,.,,~

Thus a higher flux density will make the The galvanometer sensitivity is Bss/A = device more sensitive, and increasing either the stiffness or the electrical resistance will reduce its sensitivity. We can arrive at the same result for the sensitivity by the argument: Because no voltage will be induced in the coils. Thus in the steady state the coils win be the steady-state current will be The torque exerted on the coils through the action of the magnetic field will be (re)ss = which must be balanced entirely by the torsional that exerts the steady-state Equating these two gives (16). If we wish to characterize the by a set of state-variable equations, we can choose and w as the state variables and use to write

e

w=

K

-~B­

J

Had we retained the inductance of the variable.

+~) w+~e (t) JR JR' the current i would have become the third state

10.3 Devices Coupled by Magnetic Fields

._

345

Figure 10.9 Representation of a microphone.

The Microphone The microphone shown in cross section in Figure 10.9 consists of a diaphragm attached to a circular coil of wire that moves back and forth through a magnetic field when sound waves impinge on the diaphragm. The magnetic field is supplied by a cylindrical permanent magnet having concentric north and south poles, which result in radial lines of flux directed inward toward the axis of the magnet. The coil has N turns with a radius of a and is connected in series with an external resistor R, across which the output voltage is measured. The positive direction for the current i is assumed to be counterclockwise from the perspective of viewing the magnet from the diaphragm. In the figure, the resistance of the coil has been neglected, but the inductance of the coil is represented by the inductor L located externally to the coil. If the coil resistance were not negligible, another resistor could be added in series with the external resistor in order to account for it. A single stiffness element K has been used to represent the stiffness of the entire diaphragm, and the viscous-friction element B has been used to account for the energy dissipation due to air resistance. The net force of the impinging sound waves is represented by la(t), which is the input to the system. Although the forces acting on the diaphragm are certainly distributed in nature, it is a justifiable simplification to consider them as point forces associated with the lumped elements K, B, and M. The first step in modeling the microphone is to construct idealized diagrams of the electrical and mechanical portions. In developing the equations for the galvanometer, we assumed at the outset positive senses for Te and em. Then we wrote expressions for these quantities, determining their signs by using the right-hand rules. This time, we shall first determine the positive directions of le, the electrically induced force on the diaphragm, and the mechanically induced voltage em when the other variables are positive. After that, we shall label the diagram with senses that agree with the positive directions of le and em. When we use this approach, we know in advance that the expressions for le and em will have plus signs. Figure 10 .10( a) shows the upper portion of a single tum of the coil as viewed from the diaphragm looking toward the magnet. Because the flux arrow points downward from the north to the south pole and the current arrow points to the left, the right-hand rule shown in Figure 10.S(b) indicates that the positive sense of le is toward the diaphragm. Because the velocity arrow points away from the diaphragm, the right-hand rule shown in Figure 10. 6(b) indicates that the positive sense of the induced voltage is the same as that of the current. We can reach similar conclusions by examining any other part of the coil. Thus we can draw the complete diagram of the system, as shown in Figure 10.1 O(b).

346

Electromechanical

(a)

l (b)

lo.IO

Relationships for a portion of a single coiL for analysis.

used

Because of the radial ummP,trr 0. behavior of the

the

364

Electromechanical

a

~,I_

L

R

12 V

Pl0.7

*H>.8. The conductor of mass M shown in Figure Pl0.8 can move vertically through a uniform magnetic field of flux 'B whose positive sense is into the page. There is no is a source. friction, the effective of the conductor in the field is d, and The elements and R are outside the magnetic field.

a.

Write a set of state-variable equations.

b.

For what constant voltage e; win the conductor remain stationary?

c.

What will be the steady-state velocity of the conductor if e; (t) is

x x x x x x x x x x x x x x x x x x x x x x x x x >< x x x

zero?

fv

r

l-d-1 Figure Pl0.8

10.9. A plunger is made to move horizontaHy the center of a fixed cylindrical the of a Attached to the plunger is a coil N turns and radius a. a crosssectional view of the paths for the magnetic flux by dashed lines. The field between the plunger and the south its assumed to have a constant flux density '13. The resistance and inductance of the coil are represented by the lumped elements Rand L, respectively, and the plunger has mass M.

Problems """'

365

Figure Pl0.9

a.

Verify that the following is a valid set of state-variable equations: i=v iJ

1 = M[-(K1 +K2)x-Bv+ai]

1 d;d" = L[-av-Ri+e;(t)] where a= 27raN'B. b.

Determine the direction in which the plunger will move if there is no initial stored energy and if e;(t) = U(t).

c.

Find the steady-state displacement of the plunger for the input in part (b ).

10.10. For the electric motor shown in Figure 10.14(b), let ip(t) have the constant value lp. a.

Write the differential equation describing the system and identify the time constant when iA(t) is the input and w the output.

b.

If iA (t) = iA 1 for all t < 0 and iA (t) = iA 2 for all t that steady-state conditions exist at t = 0-.

c.

Repeat part (b) when iA 2 is replaced by -iA 1 • Find the value oft for which w = 0.

> 0, sketch w versus t.

Assume

*10.11. a.

Write the differential equation describing the motor shown in Figure 10.14(b) when iA(t) and ip(t) are separate inputs and when= kq,ip(t).

b.

Find an expression for w at the operating point corresponding to lA and lF.

c.

If iA(t) = lA and ip(t) = lF + fp(t), find a linearized model that is valid in the vicinity of the operating point you found in part (b ).

366

Electromechanical

10.12. Let the shaft in Figure have a stiffness constant K rather than being rigid. Also the current source IA a voltage source with its sense Assume that

Oa and the capacitance is being heated. If ii.i < 0, then 0 < Oa and the capacitance is being cooled. If ii.i = 0, then the nominal value of the temperature is 0 = Oa and f}i(t) = qi(t). Because the system is linear, the incremental model given by (15) has the same coefficients regardless of the operating point. Recall from Chapter 8 that the transfer function is H(s) = Y(s)/U(s), where U(s) is the transformed input and Y (s) is the transform of the zero-state response. Thus we transform (15) with B(O) = 0 to obtain 1 1 s0(s) +RC e(s) = CQi(s) A

A

A

We can rearrange this transformed equation to give the transferfunctionH(s) = e(s)/Qi(s) as 1

H(s)=~ s+RC

which has a single pole at s = -1 /RC. The response of {} to a unit step function for qi (t) will be {j = R(l - €-t/RC) fort > 0 which approaches a steady-state value of R with the time constant RC. Note that we can also find the steady-state value of{} by evaluating H(s) at s = 0. This is an application of the result given in Equation (82) in Chapter 8, where the steady-state response to a constant input was shown to be the input times H(O). To find the response in terms of the actual temperature(}, we merely add 0 to B, getting

0=0+R(l-€-t/RC)

fort>O

which is shown in Figure 11.5, along with the input qi(t) .

.._ EXAMPLE 11.4 A system with two thermal capacitances is shown in Figure 11.6. Heat is supplied to the left capacitance at the rate qi(t) by a heater, and heat is lost at the right end to the environment,

376

r»

Thermal Systems

e lJ + R

- - - - - - - - - - - - - - - - - -:_:--:_:-,;;.,;--;..::--=--- - . . . , - -

t

R

A

----"-l'- - - - - - - -

~- - - - - - - - - - - - - - - - - - - - - _ _l _-

RC

0

(a)

q,(I)

___ q...,,

_l±,

0 (b)

Figure 11.5 (a) Temperature response for Example 11.3. (b) Heat input rate.

Figure 11.6 Thermal system with two capacitances.

which has the constant ambient temperature Ba. Except for the thermal resistances R1 and Rz, the enclosure is assumed to be perfectly insulated. Find the transfer function relating the transforms of the incremental variables tJi (t) and

ez.

SOLUTION Taking 01 and Bz as the two state variables and using (2) for each thermal capacitance, we can write the differential equations

1 01. = -1 [qi(t) - -(01 - Bz) ] C1

Ri

. = - 1 [ -(01 1 1 Bz -Oz) - -(Oz -Ba) ] Cz Ri Rz

(16)

11.3 Dynamic Models of Thermal Systems

~

377

At the operating point corresponding to the nominal input qi, (16) reduces to 1 Rl

-

ii.i - -(B1 - B2) = 0 1 1 -(B1 -B2)- -(B2-Ba) = 0

Ri

(17)

R2

from which

IJ2

= Ba+ R2ii.i

(18a)

iJ1 =Ba+ (R1 + R2)ii.i

(18b)

At the operating point, where equilibrium conditions exist and the temperatures are constant, the heat flow rates through Rl and R2 must both equal ii.i· Because the rates of heat flow are the same, we can regard the two resistances as being in series. The equivalent resistance is Req = Rl + R2, as in (6), so IJ1 must be the temperature difference (R1 + R2)ii.i plus the ambient temperature Ba, which agrees with (18b). We define the incremental temperatures 01 = Bl - 01 and 02 = B2 - 02 and substitute these expressions into (16). After canceling the constant terms by using (17), we obtain ~ 111

}..

(

1

(} 2 + RiC2

1 R1C1

1 R1C1

1 C1

+ - - B 1 = --B2 + -t}i(t) 1

+ R2C2

A

A

)A

1 A (} 2 = R1C2 (} 1

(19)

where the ambient temperature Ba no longer appears. To find the transfer functionH(s) = 02(s)/Q;(s), we transform (19) with 01(0) = 02(0) = 0, getting

[s+ ( - 1 R1C2

+1- )] E>2(s) = A

R2C2

1 -E>1(s) R1C2 A

Combining these equations to eliminate 01 (s) and rearranging to form the ratio 02 (s) /Qi (s), we get 1

H() s

= s2+

(-1-+ R1C1

R1C1C2 _1_ _1_) s+ 1 R1C2 R2C2 R1C1R2C2

+

(20)

Observe thatH(O) = R2. This reflects the fact that in the steady-state, all the heat supplied must leave the region at temperature B2 by passing through the barrier with resistance R2· Note that the denominator of H(s) is a quadratic function of s, which implies that it will have two poles. In fact, for any combination of numerical values for Rl, R2, C1, and C2, these poles will be real, negative, and distinct. As a consequence, the transient response of the system consists of two decaying exponential functions .

._ EXAMPLE 11.5 Consider a bar of length L and cross-sectional area A that is perfectly insulated on all its boundaries except at the left end, as shown in Figure 11.7. The temperature at the left end

378 ._ Thermal Systems

Figure 11.7 Insulated bar considered in Example 11.5.

Figure 11.8 Single-capacitance approximation to the insulated bar shown in Figure 11.7.

of the baris ei(t), a known function of time that is the system's input. The interior of the bar is initially at the ambient temperature ea. The specific heat of the material is O" with units of joules per kilogram-kelvin, its density is p expressed in kilograms per cubic meter, and its thermal conductivity is a expressed in watts per meter-kelvin. Although the system is distributed and can be modeled exactly only by a partial differential equation, develop a lumped model consisting of a single thermal capacitance and resistance and then find the step response.

e

SOLUTION Assume that all points within the bar have the same temperature except the left end, which has the prescribed temperature ei(t). Then the thermal capacitance is C = ClpAL

(21)

with units of joules per kelvin. To complete the single-capacitance approximation, we assume that the thermal resistance of the entire bar, which from (4) is

R=~ Aa

(22)

with units of kelvins per watt, is all lumped together at the left end of the bar. The lumped approximation is shown in Figure 11.8, with C and R given by (21) and (22), respectively. From (2), with q 0 ut = 0 because of the perfect insulation, and with

1

qm = "R[ei(t)-e]

it follows that the single-capacitance model is

. 1 e =RC [ei(t) - e] or (23)

where e(O) =ea.

11.3 Dynamic Models of Thermal Systems

0

which indicates that the temperature of the bar will rise from the ambient temperature to that of its left end with the time constant

RC= opL2 a

• EXAMPLE 11.6 Analyze the step response of the insulated bar shown fn Figure 11.7, using the twocapacitance approximation shown in Figure 11.9.

SOLUTION

As indicated in Figure 11.9, we take the value of each thermal capacitance as 0.5C, where C is given by (21). Likewise, we take the value of each thermal resistance

0.5 R

Figure 11.9 A two-capacitance approximation to the insulated bar shown in Figure 11.7.

380 .,_ Thermal Systems

as 0.5R, where R is given by (22). Applying (2) to the left capacitance with qin =

1

o. 5R [ei(t) -ei]

and

gives the state-variable equation . 1 el = 0. 25RC [ei(t) - 281 + e2]

(25)

Applying (2) to the right capacitance with

and q0 ut = 0 gives the

seco~d state-variable equation (26)

To evaluate the responses of el and e2 to the input ei(t) =ea+ BU(t), we can define the relative temperatures 81 =el - ea. 82 = e2 -ea. and oi(t) = ei(t) - ea and then derive the transferfunctionsH1(s) = B1(s)/Bi(s) andH2(s) = B2(s)/Bi(s). KnowingH1(s) and H2(s), we can write, for the step responses,

B] el(t) =ft -1 [ Hi(s)s A

Rewriting (25) and (26) in terms of the relative temperatures, we have ;. 8A 4A 4A el+ Rcei = Rce2 + Rcei(t) (27)

4 4 e2 + RCe2 =RC el A

A

A

Transforming (27) with 01 (0) = 02(0) = 0 gives the pair of algebraic equations (28a) 4)A 4 A ( s+02(s) = -01(s) RC RC

(28b)

Solving (28b) for B2(s), substituting the result into (28a), and rearranging terms, we find that [ (s+

R~) (s+ R~ ) - (R~

rJ

B1(s) = [

R~s+ (R~

rJ

ei(s)

11.3

factor on the left side and

the

381

Models of Thermal

(s) =

for

we

obtain

s+-

4

(s) = RC [

4 16 12 RC

2

+

s

- RC

+ (RC)2 s+RC 4

4

-

1

(s+ 1}~8) (s+ 1~~72)

[

The transfer function has response has two terms that

and RC L528

s2

T!

=

T2

= 10.472 = 0.0955RC

l

= -

and the free

= 0.6545RC

RC

Because (0) = l from For this input,

the steady-state value of B1

B when

You can verify that the numerical values of the coefficients in the are

expansion

Ai =B

= -0.7235B = -0.2764B Substituting these values into (30) and

the inverse transform, we find that fort

B1(t)=B(1From

4

A

---81

RCs+4 16

[ s ( s+-l.528)B( s + 10.472) -RC RC

whose inverse transform can be shown to be

=B

-1.1708E-l.528t/RC+0.

l

> 0,

382 Ill> Thermal Systems

0.5

1.0

1.5

2.5

2.0

3.0

t/RC

Figure 11.10 Response to a step function in Oi(t) for the two-capacitance approximation to the insulated bar shown in Figure 11.9.

The ratios {jl (t)/B and {j2(t)/B are shown in Figure 11.10 versus the normalized time variable t /RC. Other lumped-element approximations that lead to more accurate results for the insulated bar are investigated in some of the problems at the end of the chapter.

ti> EXAMPLE 11.7 The insulated vessel shown in Figure 11.11 is filled with liquid at a temperature (), which is kept uniform throughout the vessel by perfect mixing. Liquid enters at a constant volumetric flow rate of w, expressed in units of cubic meters per second, and at a temperature ()i(t). It leaves at the same rate and at the temperature 00 • Because of the perfect mixing, the exit temperature () 0 is the same as the liquid temperature 0. The thermal resistance of the vessel and its insulation is R, and the ambient temperature is Oa, a constant. Heat is added to the liquid in the vessel by a heater at a rate qh (t). The volume of the vessel is V, and the liquid has a density of p (with units of kilograms per cubic meter) and a specific heat of a (with units of joules per kilogram-kelvin). Derive the system model and find the appropriate transfer functions. SOLUTION The thermal capacitance of the liquid is the product of the liquid's volume, density, and specific heat. The heat entering the vessel is the sum of that due to the heater and that contained in the incoming stream. The heat leaving the vessel is the sum of that taken out by the outgoing stream and that lost to the ambient through the vessel walls and insulation. Mixer

R

-6,a,p, V

-- -~=====i -----Figure 11.11 Insulated vessel with liquid flowing through it.

11.3

From

383

Models of Thermal

. 1 () = -

(31)

c

where

+ 1

B+-(BR

C= we obtain the first~order differen-

Substituting these extJre:ssHms itnto and tital =m•M••~~ 1 ) w +RC () = V();(t) The titme constant of the

+1

1

+-e RC a

is 1

1

T= W

V+ RC which approaches RC as w approaches zero (no liquid As R approaches infinity (perfect insulation), the time constant becomes which is the time required to the total tank volume at the volumetric flow rate w. Because the initial values of the state variables must be zero when we calculate the transfer functions, we first rewrite the model in terms of incremental variables defined with respect to the operating point. Setting (J equal to zero in the relationship between the nominal values iJh, iJ;, B, and the ambient temperature Ba. It is iih +

v (B; - B)- = R1 (e- -

Cw -

in terms of the incremental variables (} = () - iJ, e; (t) qh (t) - iJh and using we obtain for the incremental model Rewriting

~

lA

(34)

=

and

l

WA

e+ -:;:e = ve;(t) + cqh(t) The equilibrium condition, or operating point, corresponds to the initial condition e(O) 0 and to incremental IJ.1i(t) = e;(t) = 0. To find the transfer function Hi (s) e(s)/Qh(s), we transform (35) with 0(0) = 0 and = 0, obtaining

(s+ ~)

=

~Qh(s)

Solving for the ratio 1

c --1

s+T

which has a T l (0)=-=--~

c

+R

=

=

384

Thermal

either a flow rate w or a low thermal resistance R tends to reduce the state effect of a :in the heater In simHar u11>'.1Jluu.u, is that the transfer function

v s+-

--1 T

The

response of{) to a unit WT

v 1 1+-Thus a in the inlet temperature with constant heater affects the steadystate value of the liquid temperature only some fraction of the change. Either a high fllow rate or a high thermal resistance will tend to make that fraction approach In an actual process for which () must be maintained constant in spite of variations in , a feedback control system may be used to sense changes in e and make corresponding in qh. The of mathematical modeling and transfer-function analysis that we have demonstrated here an in the design of such control

11.4

SYSTEM

Producing chemicals almost always requires control of the temperature of liquids contained in vessels. In a continuous process, a vessel within which a reaction is taldng place typically has liquid into and out of :it continuously, and a control system is needed to maintain the at a constant temperature. Such a system was modeled in Example 11. 7. In a batch process, a vessel would typically be fiHed with liquid, sealed, and then heated to a prescribed temperature. In the design and operation of batch processes, it is important to be able to calculate in advance the time for the liquid to reach the desired temperature. Such a batch is modeled and analyzed in the following case

Figure 11.12 shows a insulated filled with liquid, that contains an electrical heater immersed in the The heating element is contained within a metal jacket that The thermal resistance of the vessel and its :insulation is has a thermal resistance of The heater has a thermal and the has a thermal of The heater is of the is which is assumed to be uniform because of the mixer in the vessel. The rate at which energy is to the element is The heater and the liquid are at the ambient with the heater turned off. At time t = 0, the heater is connected to an electrical source that ~"iJ'"u'o" energy at a constant rate. We wish to determine the response of the temperature BL

385

1L4 A Thermal

Mixer

11.12 Vessel with heater.

denoted by

and to calculate the time

= 1.0 x

Liquid capacitance:

= 1.0 x 10- 3

Heater-liquid resistance:

= 5.0 x 10- 3 s·K/J

Liquid-ambient resistance:

Ambient temperature: Ba= 300 K Desired temperature: ()d = 365 K We win derive the system model for an arbitrary input q; (t) and for an arbitrary ambient temperature using ()Land ()Has the state variables. Then we will define a set of variables relative to the ambient conditions and find the transfer function relating the transforms of the relative liquid temperature and the input FinaHy, we wiH cakulate the time required for the liquid to reach the desired temperature.

System Model Because Bn and model as

represent the energy stored in the system, we can write the state-variable . ()H

. ()L

where qnL = and qLa and the state-variable equations

qHL

l = -[q;(t) -

CH 1

= -CL

and qLa = these expressions for numerical parameter values into (36) leads to the of x 10- 4]q;(t)

+ = -[l.20

x 10- 3]BL +

where the initial conditions are At this we could transform Then we could solve for

+

>< 10- 3]Ba

for a specific ambient temperature and a and take its inverse transform to find

386

Thermal

where the last that ij; = 0. model in terms of the relative variables is

to rewrite

we find that the

x

where the initial conditions are When transformed and

of

equations

x

(s+ (s+ L20 x Combining the two equations in is

we find that the transfer function

to eliminate 0.50 x 10- 4

1000s2 + 51.20s + 0.010 0.50 x 10- 1 +0.05120s+ 0.50 x (s + 0.0510)(s + 0.000196)

Having found the transfer function f:h(s)/Q;(s) given (41), we can now solve for the response to various inputs. SpecificaHy, we shaH find for the ambient temperature = 300 K (approximately 27 °C or 80 °F) and fora step-function input qi(t) = 1.50 x 104 Wfort > 0. To.find we Q;(s)=[L50x (l/s)by as given by 0.750 x 10- 3

s(s + 0.0510)(s + 0.000196) Next we

in partial fractions to obtain 75.0 = -s-

Thus fort

0.289

+ s + 0.0510 -

75.29

-s-{---0-.0-0_01_9_6

> 0, the relative (43) t .. ,,,,,.,,,.,.,,tm·p

We obtain the actual = 375.0+

which has a

value of 375 Kand is shown in

of 300 K to

11.4 A Thermal System

"111

387

365 K

t,

0

5 x 10 3

104

= 10,300 s 1.5 x 104

I, S

Figure 11.13 Responses of liquid and heater temperatures.

From inspection of either (43) or (44), we see that the transient response of the system has two exponentially decaying modes with time constants of 7i

1 = 0.0510 = 19.61 s

72

1 = 0.000196 = 5102 s

Because 72 exceeds 7i by several orders of magnitude, the transient having the longer time constant dominates the response. The principal effect of the shorter time constant is to give BL a zero slope at t = 0+, which it would not have if the transient response were a single decaying exponential. Finally, we must calculate the value of ti, the time required for the liquid to reach the desired temperature Ba = 365 K. Because two exponential terms are present, an explicit solution of (44) is not possible. However by writing (44) in terms of the time constants Ti and 72 as BL = 375.0 + 0.289€-t/ri - 75.29E-t/rz and noting that the solution for ti must be much greater than 7i, we can see that the term 0.289c 1/r1 is negligible when t =ti. Hence we can use the simpler approximate expression

BL= 375.0 - 75.29E-t/5102

(45)

where the value of 5102 s has been substituted for 72. Because (45) contains only one exponential function, we can solve it explicitly for t1. Replacing the left side of (45) by the value 365.0 and replacing t by ti on the right side, we have 365.0 = 375.0- 75.29E-ti/Sl02 which leads to ti

75.29 ) ( = 5102 ln 375.0- 365.0 = 10,300 s

which is approximately 2.86 hours.

388 ..,. Thermal Systems

Though we have completed our original task, we can use the preceding modeling and analysis results to carry out a variety of additional tasks. For instance, we can find the response of the heater temperature On for the specified qi(t) by eliminating eL(s) from (40). The heater temperature On is shown in Figure 11.13, and you are encouraged to evaluate the analytical expression for it. In practice, it might be important to evaluate the time ti for constant values of tli(t) other than the value 1.50 x 104 W that we used. For example, if tli(t) is doubled to 3.0 x 104 W, the desired liquid temperature Od = 365 K will be reached in only 2916 s (48.6 minutes). On the other hand, if the constant energy-input rate is less than 1.30 x 104 W, the liquid will never reach 365 K. If we want to investigate the effects of replacing the constant power source by a variable source, we can multiply the transfer functionH(s) by Qi(s) to give the transform of the relative temperature. After we find the inverse transform, we can add it to the ambient temperature to get 0£.

llt-

EXAMPLE 11.8

Consider the situation that might arise if this heated chamber needed to be maintained at a temperature near 365 K, but the only power source was a heater that was twice as large as was needed. In order to keep the desired temperature while using this heater, it is decided to switch it off half of the time. The question we want to address is: How fast does the liquid temperature approach 365 K, and how much does it change around its final average value if a heater delivers 26 kW for 25 minutes, followed by 25 minutes off, and this on-off cycle continues indefinitely?

SOLUTION A Simulink diagram to simulate the system in Figure 11.12 and solve (36) is shown in Figure 11.14. It uses th11flame parameter values given above. The heater power qi(t) is zero for 25 minutes, then ig~reases to 26 kW for 25 minutes, and then returns to

tt time

0 solution time

----~ theta_H

Figure 11.14 A Simulink diagram to solve (36).

389

390

380

I

370

100

200

300

400

500

Time (min)

11.15 Heater and liquid temperatures.

zero and repeats this This is achieved a level of ij; to the output of the Signal Generator, which varies from to +ij;. The result is a signal that varies between zero and 2ij;. This diagram uses several new Simulink features: Goto, From, and Display blocks. The Goto and From blocks are used together to avoid the need to draw long lines across Simulink diagrams. A signal which is introduced to any Goto block with a particular name appears automatically at all From blocks that have that same name. This feature is used in Figure U.14 for the variables Ba and qHL Because of the lengthy solution time required by the 30,000 s stopping time (500 minutes), we include a Display block whose input is the time variable t t. This allows the user to monitor the problem time as the simulation progresses, without having to be concerned if something has gone wrong. Di sp 1 a y blocks could also be connected to the outputs of the two integrators to allow of the heater and liquid temperatures. This also uses the integrator block which shows the initial condition, Ba, explicitly. and that of the Figure 11.15 contains plots of the temperature of the Both start at the ambient temperature of 300 K, and increase while the heater is on. It is during the fourth "on" cycle of the heater that the liquid first reaches 365 K. The average Ii.quid temperature reaches a steady state only after about 400 minutes. Thereafter, the temperature varies between about 373 Kand 357 K. If this range were unacceptable, it would be relatively easy to change the frequency of the signal to switch more rapidly, which would reduce the size of the ten1perature uuvrncuuvu,o, the cost of more demands on the heater controller.

SUMMARY We first introduced the basic and heat flow rate-and then presented the element laws. In contrast to the three types of elements in mechanical

390

Thermal

there are thermal often necessary in order to The of each Because there is only one ofthe transfer functions are real and the mode functions that characterize the free response are ~,.,,~"''~".,~'rather than sinusoidal terms. Of course, mechanical or other elements to the thermal of a the nature of this response. We which culminated in the case a state-variable an ~ -·-······, or a transfer function are similar to those used for mechanical and electrical examthe methods of and introduced in earlier "rn~"'"'" of we shall them to

11.1. Find the

thermal resistance for the three resistances shown in Figure Pl 1.1. and at the interfaces in terms of 81 and

/

Figure Pll.l

*11.2. The hollow enclosure shown in Figure Pl 1.2 has four sides, each with resistance and two each with resistance Find the thermal resistance between the interior and the exterior of the enclosure. Also cakulate the total heat flow rate from the interior to the exterior when the interior of the enclosure is at a constant temperature 01 and the ambient temperature is R, (each side)

PH.2

is assumed to be uniform and is

Problems

391

denoted (). The heat ~"''"""''""' a. Write the differential equation b.

c.

(a) for () in terms of the

you found in Sketch () versus t when

= 1 for 10
re~rnKm

the coefficients a and

532

Feedback

with MATLAB h, (I)

hs

Pa

x

_--:----= p,

Ah

- --pl

- L'.p

+ P15.15

b.

Write the linearized in terms of the incremental variables h, and fh. Then draw the block of the with as the inputs and as the output. Evaluate the transfer funcand

c.

Taking = 0, draw a block when the controHer is described X and the sensor is described Find the closed-loop transfer function Explain why should be negative. the controller

d.

Repeat part

using a dynamic model of the sensor such that Hs(s)/H(s)

=

1/(rs+ 1). 15.16. Find the phase margin for the solution given for Example 15.4. 15.17. Find the unit step response to a disturbance ple 15.4 when = 16.

directly to the plant in Exam-

15.18. Find the value of Kc required to produce a closed-loop having a damping ratio ( = 0. 9.

in Example 15.4

IX

A UNITS We use the International System of Units, abbreviated as SI for Systeme International d'Unites. In Table A.1, we list the seven basic SI units, of which we use only the first five in this book. A supplementary unit for plane angles is the radian (rad). In Table A.2, we give those derived SI units that we use. In addition to the physical quantity, the unit, and its symbol, there is a fourth column that expresses the unit in terms of units previously given. For example, 1 N = 1 kg · m/ s2 .

Basic Units TABLEA.1

Names and Symbols of the Basic Units

Physical quantity Length Mass Time Electrical current Thermodynamic temperature Luminous intensity Amount of substance

Name meter kilogram second ampere kelvin candela mole

Symbol m kg s A K cd mol

Derived Units TABLE A.2

Names, Symbols, and Equivalents of the Derived Units

Physical quantity Force Energy Power Electrical charge Voltage Electrical resistance Electrical capacitance Inductance Magnetic flux

Name newton joule watt coulomb volt ohm farad henry weber

Symbol N J

w c v

Q

F H Wb

In terms of other units kg·m/s kg·m2/s 2 J/s As W/A V/A A·sN V·s/A Y.s

533

534

'"'"t""'Q for decimal multiples and of a unit are given in eleones used in this book are kHo mi.Hi (m), and mi.cro appear are as foHows: 1 kO = 1 m V = 10- 3

The problems.

three constants are used in the numerical solution of many examples and

At1nospJt1encpressure at sea level: Pa= L013 x 105 N/m 2 Base of natural logarithms:

E

= 2.718

Gravitational constant at the surface of the earth: g = 9. 807

IX

AP

This appendix is intended as a refresher for the reader who has had an course in linear and is those sections of the book that use matrix methods. n is not a suitable introduction for someone who has not had formal exposure to matrices. those of the that are used in Sections 3.3 and 14.1 are""''!'"''"""'"''"'· References to more complete treatments appear in Appendix F.

A matrix is a rectangular array of elements that are either constants or functions of time. We refer to a matrix having m rows and n columns as being of order m x n. The element in the ith row and the }th column of the matrix A is denoted by aiJ, such that

A=

A matrix having the same number of rows as columns, in which case m = n, is a squan~ matrix of order n. A matrix with a single column is a col.umn vector. Examples of column vectors are

b= [

·

l

r

x(t) =

x1.(t)

l

A matrix with a row is called a row vector. A matrix that consists of a single element-that is, one that has one row and one column-is referred to as a scalar. A square matrix of n rows and n columns that has for each of the elements on its main and zero for the remaining elements is the matrix of order n, denoted

and

[~ ~ ~ I 0

are the matrices of order 2 and 3, irP

Appendix G Answers to Selected Problems

9.10

a. .x = 4,i+ 2i+x = Bsin3t b. For A = 4, i = 1, k = 2; for A = -4, .i = -1, k = 2 c. .X(O) = 0.5, i(O) = 0.5

9.12

a. b.

9.18

a. b. c.

y= -1, y+2y+4y=Bcost y=3, y+2y+8y=Bcost ~=2.

0+ 20 + 120 = 0(0)

fa(t).

= -1.5 rad,

0(0)

= -0.5 rad/s

9.21

.i = ffe,19,Mi+Bi+ 3(Mg)( 2/ 3) x = fa(t) c. .i = 1.260ffe,19,Mi+Bi+ 4.762(Mg)( 2 / 3) x = fa(t)

b.

9.23

a. .i= -4,y= -2 b.

i = -2X+ 12y,Y = x+ cost

c. i+2i-12X= 12cost 9.27 b.

ZL

C.

T

b. c.

e

9.28

= 0.3622 A; eo = 0.3976 S

= 0.3936 V; dtL/dt = -2.515iL + 0.05714cost; eo = 2.113iL

0 = 4 V, l0 = 16 A ~o + 1€0 + eo = 2€i + e;(t)

Chapter 10-Electromechanical Systems 10.1

a. b.

Lreo + Rreo = L2(x)ei + R2(x)e;(t) Lre0 +Rreo = [x(t)/xmaxHLre;+Rrei(t)]

10.6 E0(s)/Fa(s) = dBR 0/P(s) where P(s) =MLs2 + (BL+MR+MR 0)s+B(R+R0) + (dB) 2 10.8

di/dt=[-Ri+Bdv+e;(t)]/L, e; =RMg/Bd c. fJ =RMg/(Bd) 2

a. b.

v=(-Bdi+Mg)/M

10.11

(JR +h)w +(BR+ BL)w = "fkm

= 28.95° Bode Diagrams

Grn=11.054 dB (at 3.8942 rad/sec), Prn=28.946 deg. (at 1.8644 rad/sec) 100 50

0 -50

CJ)

"'C

.a ·2

-100'---~~~........._,_~~~~_._,_,_~~~~._.__,_~~~....._._,_._,

:::2:

-50~~~~~~~~~~~~~~~~~~~

g>

Cl

~ -100 CJ)

gj -150 .c:

a. -200 0

-250

........ .

-300~~~~~~~~~~~~~~~~~~~

10-2

10-1

10° Frequency (rad/sec)

b. K=45.16 c. K* = 142.8

15.7 a. H(s) = -(R2Cs+ l)/(R1Cs) b. Pole at s = 0, zero at s = -l/(R2C) c. PI compensator

569

10 1

570

15.9

= -0.384rnd

c. d.

1.: ~---~--~~---~---.-]

0.5 ~---------~·

"

' "

"

' "

0.5 -

I j

1.5

...........

j

·-· .

2~~--~--~~---~--~

3

2

0 Real Axis

e.