Engineering Applications of the Laplace Transform 1527573737, 9781527573734
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Contents
Preface
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Bibliography
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Engineering Applications of the Laplace Transform
Engineering Applications of the Laplace Transform By
Y.H. Gangadharaiah and N. Sandeep
Engineering Applications of the Laplace Transform By Y.H. Gangadharaiah and N. Sandeep This book first published 2021 Cambridge Scholars Publishing Lady Stephenson Library, Newcastle upon Tyne, NE6 2PA, UK British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Copyright © 2021 by Y.H. Gangadharaiah and N. Sandeep All rights for this book reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner. ISBN (10): 1-5275-7373-7 ISBN (13): 978-1-5275-7373-4
CONTENTS
Preface ......................................................................................................... x
Chapter 1 ..................................................................................................... 1 The Laplace Transform 1.1 Introduction ...................................................................................... 1 1.2 Definition ......................................................................................... 3 1.3 Existence of Laplace Transforms ..................................................... 5 1.4 Laplace Transforms of Some Standard Functions ........................... 7 1.5 Operational Properties of the Laplace Transform .......................... 19 Property 1.5.1: Linearity of the Transform..................................... 20 Property 1.5.2: Transform of the Derivative .................................. 24 Property 1.5.3: First Translation Theorem ..................................... 27 Property 1.5.4: Multiplication of
f t by t n .............................. 29
Property 1.5.5: Time-scaling .......................................................... 32 Property 1.5.6: Division by t ......................................................... 33 Property 1.5.7: Transforms of Integrals ......................................... 36 Property 1.5.8: The Initial-value Theorem ..................................... 38 Property 1.5.9: The Final-value Theorem ...................................... 41 1.6 Periodic Function ........................................................................... 52 1.6.1 Transform of a Periodic Function.......................................... 53
Contents
vi
1.7 Second Shifting Theorem or Second translation Theorem ............ 68 1.7.1 Expression for Piecewise Continuous Function
f t in
Terms of the Unit Step Function .................................................... 70 Summary ............................................................................................ 105
Chapter 2 ................................................................................................. 107 The Inverse Laplace Transform 2.1 Introduction .................................................................................. 107 2.2 The Inversion Integral for the Laplace Transform ....................... 108 2.2.1 Relationship Between Laplace Transforms and Fourier Transforms ................................................................................... 109 2.2.2 Inversion Using the Bromwich Integral Formula ................ 112 2.3 The Inverse Laplace Transform is a Linear Transform ................ 114 2.4 Inversion Using the First Shift Theorem ...................................... 116 2.5 Inverse Transform by Differentiation and Integration ................. 119 2.6 Inversion Using the Second Shift Theorem ................................. 126 2.7 The Inverse Laplace Transform Using Partial Fractions.............. 128 2.7.1 Partial Fractions: Distinct Linear Factors ............................ 129 2.7.2 Partial Fractions: Repeated linear factors ............................ 153 2.7.3 Partial Fractions: Quadratic polynomials ............................ 163 2.8 Convolutions ................................................................................ 180 2.8.1 Definition ............................................................................ 180 2.8.2 Properties of Convolution ................................................... 180 2.9 Convolution Theorem .................................................................. 183 2.10 Integral and Integro-differential Equations ................................ 201 Summary ............................................................................................ 219
Engineering Applications of the Laplace Transform
vii
Chapter 3 ................................................................................................. 221 Application of the Laplace Transform to LTI Differential Systems: Transfer Functions 3.1 Introduction .................................................................................. 221 3.2 Transfer Functions ....................................................................... 223 3.3 Transfer Function of the Linear Time-invariant System .............. 225 3.3.1 Definitions: Linear and Nonlinear nth-order Ordinary Differential Equations............................................................. 226 3.4 Electrical Network Transfer Functions ........................................ 259 3.4.1 Kirchhoff’s Current Laws ................................................... 260 3.5 Mechanical System Transfer Functions ....................................... 280 Summary ............................................................................................ 290
Chapter 4 ................................................................................................. 292 Application of the Laplace Transform to LTI Differential Systems: Solving IVPS 4.1 Introduction .................................................................................. 292 4.2 The Scheme for Solving IVPs ...................................................... 294 4.2.1 Definitions: Homogeneous and Non-homogeneous Linear Differential Equations............................................................. 294 4.3 Differential Equations with Variable Coefficients ....................... 347 4.4 Total Response of the System Using the Laplace Transform ...... 361 4.4.1 Impulse Response and Transfer Function ........................... 362 4.5 Systems of Linear Differential Equations .................................... 370 Summary ............................................................................................ 389
Contents
viii
Chapter 5 ................................................................................................. 390 Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems 5.1 Introduction ................................................................................. 390 5.2 Application to Electrical Circuits ................................................. 390 5.2.1 The Scheme for Solving Electrical Circuits ........................ 390 5.3 Application to Mass-spring-damper Mechanical System ............ 430 Summary ............................................................................................ 453
Chapter 6 ................................................................................................. 454 Application of the Laplace Transform to LTI Differential Systems: State-space Analysis 6.1 State-space Representation of Continuous-time LTI Systems ..... 454 6.2 State Model Representations........................................................ 455 6.2.1 Procedure to Find the Response of a Second-order Nonhomogenous System ............................................................... 457 6.3 Matrix Exponential (State-transition Matrix)............................... 458 6.3.1 Converting from State-space to Transfer Function ............. 459 Summary ...................................................................................... 489
Chapter 7 ................................................................................................. 490 Laplace Transform Methods for Partial Differential Equations (PDEs) 7.1 Introduction .................................................................................. 490 7.2 The Laplace Transforms of
u x, t and its Partial
Derivatives ................................................................................... 491 7.2.1 Steps in the Solution of a PDE by Laplace Transform ........ 494 Summary ............................................................................................ 521
Engineering Applications of the Laplace Transform
ix
Chapter 8 ................................................................................................. 522 Exercises and Answers
Bibliography ............................................................................................ 537
PREFACE
The Laplace transform, a technique of transforming a function from one domain to another, has a vital role to play in engineering and science. Laplace transformation methods offer simple and efficient strategies for solving many science and engineering problems, including: control system analysis; heat conduction; analyzing signal transport; mechanical networks; electrical networks; communications systems; and analog and digital filters. This book is aimed explicitly at undergraduates and graduates in applied mathematics, electrical and electronic engineering, physics, and computer science. The reader can follow the step by step problem solving and derivations presented with minimal instructor assistance. The first two chapters give a straightforward introduction to the Laplace transform, including its functional properties, finding inverse Laplace transforms by different methods, and the operating properties of inverse Laplace transform. Chapter 3 describes transfer function applications for mechanical and electrical networks to develop the input and output relationships. Chapters 4 and 5 demonstrate applications in problem solving, such as the solution of LTI differential equations arising in electrical and mechanical engineering fields, along with the initial conditions. The state-variables approach is discussed in Chapter 6 and explanations of boundary value problems connected with the heat
Engineering Applications of the Laplace Transform
xi
conduction, waves, and vibrations in elastic solids are presented in Chapter 7.
CHAPTER 1 THE LAPLACE TRANSFORM
1.1 Introduction Named in honor of the French mathematician Pierre Simon Laplace, the Laplace transform plays a vital role in technical approaches to studying and designing engineering problems. The significance of the Laplace transform is its application in many different functions. For example, the Laplace transform enables us to deal efficiently with linear constantcoefficient differential equations with discontinuous forcing functions— these discontinuities comprise simple jumps that replicate the action of a switch. Using Laplace transforms, we can also design a meaningful mathematical model of the impulse force provided by, for example, a hammer blow or an explosion. It is certainly not a lazy assumption to suggest that differential equations comprise the most important and significant mathematical entity in engineering and technology. The linear, time-invariant differential equation, Eq. (1), is one such design: N
¦ ak k 0
d k y t dt k
M
¦ bk k 0
d k x t dt k
(1)
2
Chapter 1
With the system parameters ܽ and ܾ , several systems are represented by this equation, which relates the output )ݐ(ݕ, to the input )ݐ(ݔ. However, we want to design a system where the state variables vary with time and/or space. The most natural way to describe this behavior is through differential equations. The development of a differential equation model requires a detailed understanding of the system we wish to depict. It is not enough to set up a differential equation model; we also have to solve the equations. Therefore, an essential mathematical method for modeling and analyzing linear systems is the Laplace transform. In terms of the mathematical representation of a physical system, the Laplace transform can simplify the study of its behavior considerably. The Laplace transform offers tremendous benefits. We model physical systems of continuous-time (linear time-invariant) when appropriate, with linear differential equations having constant coefficients. A clear explanation of the characteristics of the equations and physical structure is given by the Laplace transform of the LTI system. Once transformed, however, these differential equations are algebraic and are thus easier to solve. The solutions are functions of the Laplace transform variable ݏ rather than the time variable ݐwhen we use the Laplace transform to solve differential equations. Consequently, we will need a procedure to translate functions from the frequency domain to the time domain, which is called the inverse Laplace transform. In Section 1.3, we discuss the conditions under which a function’s Laplace transform occurs. In Section 1.4, we derive the Laplace transforms of standard signals as examples and present Laplace transform pairs in Table 1.1. In Section 1.5, various operational properties, i.e. linearity, timedomain shift rules, multiplication, division, and scaling practices, are
The Laplace Transform
3
described. The regulations on differentiation and integration are presented in this section also. These are more difficult to demonstrate, but are of great importance in application. The time-domain differentiation rule is essential for applying the differential equations set out in chapters 3 and 4. We also deal with two theorems in this section, i.e. the initial and final value theorems for the Laplace transform. We can see how to evaluate the Laplace transform of a periodic function in Section 1.6. Section 1.7 explains how you can determine the Laplace transform of a piecewise continuous function that uses the unit step function. LEARNING OBJECTIVES On reaching the end of this chapter, we expect you to have understood and be able to apply: x The definition of the Laplace transform. x The concept of the existence of the Laplace transform. x The standard examples of the Laplace transform. x The properties of linearity, shifting, and scaling. x The rules of differentiation and integration. x The initial and final value theorems. x The Laplace transform of a periodic function. x How to express the piecewise continuous function in terms of the unit step function.
1.2 Definition Suppose that variable
f t is a real or complex-valued function of the (time)
t ! 0 and s is a real or complex parameter. We define the
Laplace transform as
Chapter 1
4
F s
L ^ f t `
f
³e
st
f t d t
0
where the limit exists as a finite number.
(1.1)
s
is a fixed parameter
(real/complex) when evaluating the integral (1.1). However, the reader should understand that, in advanced applications of Laplace transforms primarily to solve partial differential equations in digital signal processing, it is essential to consider
s
as a complex number. Before we proceed
further, it is worth making a few observations relating to the definition in (1.1).
i The Laplace transform is an integral transform.
ii The Laplace transform only uses values of f t iii The symbol
for t t 0.
L denotes the Laplace transform operator. When it
operates on the function
f t , it transforms it into the function F s
of the complex variable
s . That is, the operator transforms the function
f t in the t domain (time-domain) into the function F s in the s domain (complex frequency domain, or frequency domain). This relationship is depicted graphically in Figure 1.1.
The Laplace Transform
5
Figure 1.1. The Laplace transform operator
iv
For the integral (1.1) to exist, any discontinuity of the integrand
inside the interval
0, f
must be a finite jump so that there are right and
left-hand limits at those discontinuous points. An exception is a discontinuity at
f t
v
t
0 (if it exists). For instance, the function
1 diverges at t t
0 , but the integral (1.1) exists.
The Laplace transform is a mathematical toolbox for solving linear
ODEs and related initial value problems, PDEs, and boundary value problems. It is used extensively in electrical engineering, control theory, and the stability of algorithms.
1.3 Existence of Laplace Transforms A Laplace transform should exist if the magnitude of the transform is finite, that is,
F s f.
Piecewise continuous: A function
f t is piecewise continuous on a
finite interval A d t d B if f is continuous on
> A, B @ , except possibly
Chapter 1
6
at many finite points. Each of these finite points
f has a finite limit on
both sides. Sufficient condition: The sufficient condition for a Laplace transform to exist if
f t is that it is piecewise continuous on 0, f and some
constants L and M
f t M e Lt , then F s
exist such that
exists for s ! L . Proof: As
f t is piecewise continuous on
integrable on
0, f .
L ^ f t `
f
³ 0
f
is
f
f t e st d t d ³ f t e st d t d ³ M e Lt e st d t 0
M ª s L t º f e ¼0 Ls ¬ L ^ f t `
0, f , f t e st
M >0 1@ Ls
F s f
Definition: A function positive constants
0
M sL
for s ! L.
f t is said to be of exponential order L if
T and M exist such that f t M e Lt , for all
t t T. The physical significance of the Laplace transform: A Laplace transform has no physical meaning except that it transforms the time domain function to a frequency domain
s .
The Laplace transform is
The Laplace Transform
7
applied to simplify mathematical computations and allow the effortless analysis of linear time-invariant systems.
1.4 Laplace Transforms of Some Standard Functions In this section, we illustrate the procedure to find the Laplace transform of the function f t . In all expressions, it is assumed that
f t satisfies the
conditions of Laplace transformability. Example 1.1. Determine the Laplace transform of the constant
f t a.
function Solution:
Using the definition of the Laplace transform
F s
L ^ f t `
f
³e
st
f t d t
0
f
L ^a`
³e
st
a dt
0
L ^a`
a st e s
L ^a`
a f 0 ªe e º¼ s ¬
f 0
provided, of course, that
L ^a`
a s
s ! 0 .
a >0 1@ s
s ! 0 (if s is real). Thus we have
Chapter 1
8
Example 1.2. Find the Laplace transform of
f t e at , where a is
constant. Solution: By definition of the Laplace transform
F s
f
L ^ f t `
³e
f t d t
st
0
^ `
L e at
f
f
at st ³e e d t
³e
0
s a t
dt .
0
On integrating, we get
^ ` s1a e
L e at
s a t
^ ` s1a ª¬e
L e at
f
f 0
1º¼
Thus,
^ ` s 1 a for s ! a
L e at
Example 1. 3. Find the Laplace transform of Solution: Let
L ^ f t `
L > cosh at @
f t cosh at.
The Laplace Transform
9
cosh at in its exponential form
and express the function
1 at at e e . 2
cosh at
The Laplace transform becomes
L ^ f t `
1 L ªe at e at ¼º 2 ¬
1ª L e at L e at º¼ 2¬
^ ` ^ `
1ª 1 1 º « » 2 ¬« s a s a ¼»
1 ªs a s aº « ». 2 « s2 a2 » ¬ ¼
L > cosh at @
Thus,
L > cosh at @
s s a2 2
for s ! a .
Example 1.4. Find the Laplace transform of
f t sinh at.
Solution: Let
L ^ f t `
L >sinh at @
and express the function
sinh at
cosh at
1 at e e at . 2
The Laplace transform becomes
in its exponential form
Chapter 1
10
1 L ¬ªe at e at ¼º 2
L >sinh at @ L >sinh at @
1ª L e at L e at ¼º ¬ 2
^ ` ^ `
1ª 1 1 º « » 2 «¬ s a s a »¼
1 ªs a s aº « » 2 « s2 a2 » ¬ ¼
Thus,
L >sinh at @
a s a2 2
for s ! a
.
Example 1.5. Find the Laplace transform of
f t cos at.
Solution: Let
L ^ f t `
L > cos at @
and express the function
cosh at
cosh at in its exponential form
1 iat e e iat . 2
The Laplace transform becomes
L ^ f t `
1 L ª¬eiat eiat º¼ 2
L > cosh at @ Thus,
1 ª iat L e L e iat º¼ ¬ 2
^ ` ^ `
1 ª s ai s ai º « ». 2 « s2 a2 » ¬ ¼
1ª 1 1 º « » 2 «¬ s ia s ia »¼
The Laplace Transform
L > cosh at @
s s a2 2
11
for s ! 0 .
Example 1.6. Find the Laplace transform of
f t sin at.
Solution: Let
L ^ f t `
L >sin at @
and express the function sin at in its exponential form
sin at
1 iat iat . e e 2i
The Laplace transform becomes
L >sinh at @
1 L ª¬eiat e iat ¼º 2i
L >sinh at @
1 ª s ia s ia º « » 2i « s 2 a 2 » ¬ ¼
1ª 1 1 º « » 2 «¬ s ia s ia »¼
a . s a2
Thus,
L >sinh at @
a s a2 2
for s ! 0 .
2
Chapter 1
12
Example 1.7. Find the Laplace transform of
f t t n , where n is a
positive integer. Solution: Using the definition of the Laplace transform
F s
f
L ^ f t `
³e
st
f t d t
0
f
^ ` ³e
L tn
st
tn d t
Setting st
x & dt
0
^ `
L t
n
^ `
L tn
f
§x· ³0 e ¨© s ¸¹ x
n
dx s
dx s
f
1 s n 1
³e
x
x n dx
0
§ Therefore, ¨ by gamma function * n ©
f
· x n 1 e x dx ¸ ³0 ¹
* n 1 ^ ` s
L tn
n 1
Thus,
* n 1 ^ ` s
L tn
n 1
n! s n 1
* n 1
n!
The Laplace Transform
13
Example 1.8. Find the Laplace transform of the Dirac delta function
f t G t .
(impulse function) Solution:
t
An impulse is infinite at
0 and zero elsewhere. The area under the
unit impulse is 1.
G t is defined by
The Dirac delta function
f G t ® ¯0
t 0 tz0
The Dirac delta function
G t a
is characterized by the following
two properties
i
f G t a ® ¯0
t a tza
f
ii
³ G t a f t
dt
f a
f
Using the definition of the Laplace transform
F s
f
L ^ f t `
³e
st
f t d t
0
L ^G t a `
f
³e
st
G t a d t
0
L ^G t a ` e st
t a
.
Chapter 1
14
Thus,
L ^G t a ` e as and in particular
L ^G t ` 1.
Example 1.9. Find the Laplace transform of the unit step function
f t u t . Solution: The unit step signal is a typical “engineering signal” in made to measure engineering applications, which often involve functions (mechanical or electrical driving forces). The unit step function u t is defined by
u(t)
1
0
Figure 1.2.
t
1 u t ® ¯0
for t t 0 elsewhere
The Laplace Transform
15
u(t – a)
1
0
t
a
for t t a
1 u t a ® ¯0
elsewhere
Figure 1.3 The step function is shown in Figure 1.2. The Laplace transform of
F s
u t , by definition, can be written as
f
L ^ f t `
³e
st
f t d t
0
L ^ u t a
f
` ³e
st
u t a
st
dt
dt
0
L ^ u t a
f
` ³e a
1 st e s
f a
Hence, the Laplace transform of the unit step function exists only if the real part of
s
is greater than zero. We denote this by
Chapter 1
16
`
1 ª0 e as º¼ s ¬ .
L ^ u t a
`
e a s , for s ! 0 s
In particular L
^ u t `
L ^ u t a Thus,
1 s
for
a
0 .
Example 1.10. Find the Laplace transform of the ramp function
f t r t . Solution: The ramp function
r t
t ® ¯0
r t is defined by
for t ! 0
f(t)
for t 0
t
Figure 1.4
The Laplace Transform
17
Using the definition of the Laplace transform, we have
F s
L ^ f t `
f
³e
f t d t
st
0
L ^ r t
f
` ³e
st
r t d t
st
tdt
0
f
L ^ r t `
³e 0
Recall from calculus the following formula for integration by parts b
³ u dv
b
uv
a
L ^ r t `
b a
³ v du. a
e st t s
f
f
0
1 st e dt . s ³0
Integration by parts gives
L ^ r t ` 0
1 st e s2
f 0
1 . s2
Thus,
L ^ r t `
1 . s2
To further progress with the Laplace transform, it is necessary to use a table of Laplace transform pairs for the most commonly occurring functions. Table 1.1 provides a list of the most useful Laplace transform pairs involving elementary functions.
Chapter 1
18
Table 1.1 Laplace Transform Pairs
L ^1`
1 s
^ `
L e at
^ `
L tn 1 sa
^ `
L e at
n! , n 1, 2,3....... s n 1 1 sa
L ^sin at`
a s a2
L ^cos at`
L ^sinh at`
a s a2
L ^cosh at`
L ^u t `
1 s
L ^G t ` 1
2
2
L ^u t a `
s s a2 2
s s a2 2
e as s
L ^G t a ` e as
Table 1.1 lists the Laplace transforms of some elementary functions. It would be helpful to familiarize yourself with these expressions. You will frequently encounter problems, solve linear differential equations with constant coefficients, find transfer functions, investigate mechanical systems, and analyze electrical circuits. The Laplace transform function can be derived directly by performing integration. However, it is much simpler to derive the Laplace transform using Laplace transform properties than through direct integration, as shown in the previous examples. The Laplace transform has many interesting operational properties. These properties are why the Laplace transform is considered such a powerful
The Laplace Transform
19
tool of mathematical analysis. We will now derive these properties one after the other and apply them to generate more transforms, as illustrated in the next section.
1.5 Operational Properties of the Laplace Transform In Table 1.1 in the previous section, we presented an essential list of commonly occurring functions
f t as Laplace transform pairs. It is
necessary to establish several fundamental properties of the transform, known as its operational properties, to solve initial value problems of linear differential equations and deal with electrical and mechanical systems. The operational properties are those properties that directly relate to the way the transform operates on any function
f t that is
transformed, rather than the effect the properties have on specific functions. We will state and explain the various properties of Laplace transforms, including linearity, first translation, the multiplication of
f t by t n , time-scaling, time integration, differentiation, initial value theorem, and final value theorem. Let us now consider their properties. Linear Operators An operator
T is linear for every pair of functions f t and g t ,
and for every pair of constants C1 and C2.
T ^C1 f t C2 g t ` C1T ^ f t ` C2T ^ g t `.
Chapter 1
20
Differentiation and integration are both linear operations. For any two differentiable functions,
f t and g t , and any two constants, C1
and C2, then
d d d C1 f t C2 g t ` C1 ^ f t ` C2 ^ g t `. ^ dt dt dt Likewise, for any two integrable functions
f t and g t , and any
two constants, C1 and C2,
³ ^C f t C g t ` dt 1
2
C1
^³ f t dt` C ^³ g t dt`. 2
We now prove that the Laplace transform is a linear operator.
Property 1.5.1: Linearity of the Transform The Laplace transform of a linear combination of two (or more) functions is equal to the linear combination of the respective Laplace transforms. Mathematically speaking,
L ^C1 f t C2 g t ` C1 L ^ f t ` C2 L ^ g t ` . This linear property easily follows from the linearity property of integrals. Proof: The proof is simple and follows directly from the fact that the integration is a linear operation; as such, the integral of a sum of functions is the sum of their integrals. Thus,
The Laplace Transform
21
f
³ e ^C f t C g t ` dt
L ^C1 f t C2 g t `
st
1
2
0
f st ½ f st ½ C1 ® ³ e f t dt ¾ C2 ® ³ e g t dt ¾ ¯0 ¿ ¯0 ¿
C1 L ^ f t ` C2 L ^ g t ` and
L ^C1 f t C2 g t ` C1 L ^ f t ` C2 L ^ g t `. The physical significance of the property of linearity: If the function can be decomposed into a linear combination of two or more component functions, then the Laplace transform of the function is the linear combination of the component functions. This simplifies the mathematical computations. Similarly, if the function can be decomposed into a linear combination of two or more component functions in the s domain, then the inverse Laplace transform of the signal is the linear combination of the component inverse Laplace transform functions. This property is used in the calculation of inverse Laplace transforms using partial fraction expansion. Example 1.11. Compute the Laplace transform of
a 7 cos 2t 4t e4t c 7
t
4 . t
b e2t 4e3t
Chapter 1
22
Solution:
a
Using the values given in Table 1.1,
L ^7`
7 , L ^cos 2t` s
s , L ^t` s 4 2
1 & L ^e 4t ` 2 s
1 . s4
Using the linearity property of the Laplace transform, we find
L ^7 cos 2t 4t e 4t ` L ^7` L ^cos 2t` 4 L ^t` L ^e 4t ` . Thus,
L ^7 cos 2t 4t e 4t `
b
7 s 4 1 2 2 . s s 4 s s4
Using the values given in Table 1.1,
L ^e 2t `
1 & L ^e 3t ` s2
1 s3
Using the linearity property of the Laplace transform, we find
L ^e 2t 4e 3t ` L ^e 2t ` 4 L ^e 4t ` L ^e 2t 4e 3t `
1 1 . 4 s2 s3
Thus,
L ^e 2t 4e 3t `
3s 11 . s2 s 6
The Laplace Transform
c
We have
23
* n 1 and s n 1
L ^t n `
§1· * n 1 n* n and * ¨ ¸ ©2¹
S
(recurrence relation of gamma function)
1½ L ®t 2 ¾ ¯ ¿
L ®t ¯
1 2
½ ¾ ¿
§3· §1 · * ¨ ¸ * ¨ 1¸ ©2¹ ©2 ¹ 32 s s3 2
§1· *¨ ¸ ©2¹ s1 2
§1· *¨ ¸ 1 ©2¹ 2 s3 2
S
S
12
s
s
1 S 2 s3 2
Using the linearity property of the Laplace transform, we find
4½ L ®7 t ¾ 7 L t¿ ¯ 4½ L ®7 t ¾ t¿ ¯
^ t ` 4L ®¯ 1t ½¾¿
7 S S 4 32 2s s
Thus,
4½ L ®7 t ¾ t¿ ¯
7 S S 4 . 32 2s s
Chapter 1
24
Example 1.12. Determine
L ^mt 7 cosh 2t 4t100 e 11t ` .
Solution: Using Table 1.1,
^
L ^mt ` L e log m t L ^t100 `
100! s101
`
1 s , L ^cosh 2t` 2 , s log m s 4 L ^e 11t `
&
1 s 11
Using the linearity property of the Laplace transform, we find
L ^mt 7 cosh 2t 4t100 e 11t ` L ^mt ` 7 L ^cosh 2t` 4 L ^t100 ` L ^e 11t ` .
Thus,
L ^mt 7 cosh 2t 4t100 e11t `
1 s 1 § 100! · . 7 2 4 ¨ 101 ¸ s log m s 4 © s ¹ s 11
Property 1.5.2: Transform of the Derivative If L^ f
n
F s t `
L ^ f t ` and f t is continuous for t t 0 then s n L ^ f t ` s n 1 f 0 s n 2 f
1
11
0 s n 3 f 0 ......
f
n1
0 .
Proof: This property sheds light on why the Laplace transform is such a useful tool in solving initial value problems. Roughly speaking, using the Laplace transform, we can replace differentiation with respect to
t and
The Laplace Transform
25
multiplication by s, thereby converting a differential equation into an algebraic one. This idea is explored further in Chapter 4. For now, we show how this property can help to compute a Laplace transform. Let us start with the definition of the Laplace transform
^
L f
f
'
t ` ³ e st
f
'
t
dt
0
Integration by parts gives
^
`
L f t '
e
st
f
f
f t s ³ e st f t d t 0
0
^ t `
L f
'
f
f 0 s ³ e st f t d t 0
^
`
L f ' t
f 0 sL ^ f t `
(1.2)
This result can be easily extended to the second-order derivative; however, the first derivative
f t must be continuous. Using Eq. (1.2), we have
f
^ t ` ³ e
L f
''
st
f
''
t
dt
0
^ t `
L f
''
f
f
e st f ' t s ³ e st f 0
'
t
0
^
L f
''
t `
f
f ' 0 s ³ e st f
'
t
0
^
`
L f '' t
^
`
f ' 0 s L f ' t .
dt
dt
Chapter 1
26
^ t `
L f
''
^
`
f ' 0 s sL ^ f t ` f 0 .
Thus,
^ t `
L f
''
s 2 L ^ f t ` s f 0 f ' 0 .
According to the recursive nature of the Laplace transform of the nthderivative L ^ f n t `
s n L ^ f t ` s n 1 f 0 s n 2 f
Example 1.13. Given that L ^sin 2t`
1
11
0 s n3 f 0 ...... f
f t sin 2t. Then f 0 0 and f ' t 2 cos 2t ,
using the derivative property
^
`
L f ' t
f 0 s L ^ f t `
L ^2 cos 2t` s L ^sin 2t` 2 L ^cos 2t` s
2 . s 4 2
Dividing by 2 gives
L ^ cos 2t`
s . s 4 2
0 .
2 , determine L ^cos 2t` . s 4 2
Solution: Let
n1
The Laplace Transform
27
Property 1.5.3: First Translation Theorem (First Shifting Theorem) If
F s
L ^e at f t `
L ^ f t ` and
a
is
any
real
number,
then
F s a .
Proof: The proof is immediate, since, by definition,
L ^e f t ` at
f
³e
st at
e dt
0
L ^e at f t `
f
³e
s a t
dt
0
L ^e at f t ` F s a and L ^e at f t ` F s a . The above property states that if we know the Laplace transform of any function, the transform of that function multiplied by an exponential can immediately be obtained by a simple shift in the
s
variable. As the
following example illustrates, this property allows us to easily calculate the Laplace transform of the function, transform of
e at f t , if we already know the
f t .
Example 1.14. Determine
L ^e t cos 7t` .
Solution: Using the values given in Table 1.1,
Chapter 1
28
L ^cos 7t`
s , s 49
F s
2
using the first shift theorem
L ^et cos 7t` F s 1 F s s
s 1
s 1 . 2 s 1 49
L ^t 3e 4t ` .
Example 1.15. Determine Solution:
Using the values given in Table 1.1,
L ^t 3 `
F s
6 , s4
by the first shift theorem
L ^t 3e 4t ` F s 4
F s s
6 s 6
In general,
L ^e at cos bt`
L ^e at cos bt`
sa
s a
2
b2
sa
s a
2
b2
and
s 4
2
.
The Laplace Transform
b
L ^e at sin bt`
s a
2
s a
2
b2
sa
L ^e at cosh bt`
s a
L ^e at cosh bt`
2
b2
s a s a
L ^e at sinh bt`
and
sa 2
b2
b
L ^e at sinh bt`
2
b2
b
s a n!
s a
n 1
n!
L ^e at t n `
and
b
L ^e at sin bt`
L ^e at t n `
b2
s a
n 1
2
b2
and and and
.
Property 1.5.4: Multiplication of f t by t n If
F s
L ^ f t ` and n 1, 2,3, 4..... , then
L ^t n f t `
1
n
dn ª F s º¼ . ds n ¬
29
Chapter 1
30
Proof: This property sheds light on why the Laplace transform is such a useful tool in solving initial value problems. Roughly speaking, we can use a Laplace transform to convert a function multiplied with t n into the
s
domain, thereby converting a differential equation with variable coefficients into an algebraic one. This idea is explored in Chapter 4. For now, we show how this property can help to compute a Laplace transform. Let us start with the definition of the Laplace transform
F s
f
L ^ f t `
³e
st
f t d t
0
dn F s ds n
dn ds n
°f st ½° ®³ e f t d t ¾ °¿ ¯° 0
Building on these assumptions, we can apply a theorem from advanced calculus (Leibniz’s rule) to interchange the order of integration and differentiation f
dn F s ds n
d n st ³0 ds n e
f t
^
Using the nth-derivative i.e D
d F s ds Thus,
f
³e 0
st
^ t
n
n
dt
e t
`
ts
f t d t
n
e ts
`
The Laplace Transform
^
`
L t n f t
1
n
31
dn F s ds n .
Example 1.16. Determine
L ^t sin 3t` .
Solution: Using the values given in Table 1.1,
L ^sin 3t`
F s
3 s 9 2
using the multiplication theorem,
L ^t sin 3t`
d ª F s º¼ ds ¬
Example 1.17. Determine
d § 3 · ¨ ¸ ds © s 2 9 ¹
6s
s2 9
2
.
L ^t 2 e3t ` .
Solution: Using Table 1.1,
L ^e3t `
1 s 3
F s
using the multiplication theorem,
L ^t 2 e3t `
1
2
d2 ª F s º¼ ds 2 ¬
1
d § 1 · ¨ ¸ ds ¨© s 3 2 ¸¹
2
s 3
3
.
Chapter 1
32
Property 1.5.5: Time-scaling If
1 §s· F¨ ¸. a ©a¹
L ^ f t ` , then L ^ f at `
F s
Proof: Using the definition of the Laplace transform,
F s
L ^ f t `
f
³e
st
f t d t
0
L ^ f at `
f
³e
st
f at d t
0
let at
x dt
dx a
f
§s·
we get
L ^ f at `
1 ¨© a ¸¹ x e f x d x a ³0
.
Thus,
L ^ f at `
1 §s· F ¨ ¸. a ©a¹
Example 1.18. Given that L ^sin t`
L ^sin at` .
F s
1 , determine s 1 2
The Laplace Transform
33
Solution: Let
f t sin t and f at sin at ,
by the time-scaling property,
L ^ f at `
1 §s· F ¨ ¸. a ©a¹
L ^sin at`
1 1 a § s ·2 ¨ ¸ 1 ©a¹
Thus,
a . s a2
L ^sin at`
2
We have proved that differentiating the transform of a function
t . Similarly, integrating the transform corresponds to dividing the function with t , as shown by the
corresponds to multiplying the function by
following theorem.
Property 1.5.6: Division by t If
F s
f t ½ L ^ f t ` , then L ® ¾ ¯ t ¿
Proof: f
By definition
F s
³e 0
st
f t d t .
f
³ F s ds. s
Chapter 1
34
Integrating both sides of the equation, f
ff
³ F s ds
³³
s
e st f t d t ds
s 0
and reversing the order of integration, we get f
f
³ F s ds
³
s
0
ª f st º « ³ e ds » f t d t ¬s ¼
(1.3)
On integration, we get f
f
³ F s ds ³ s
0
f
ª e st º « » f t d t ¬ t ¼ s
(1.4)
Thus,
f t ½ L® ¾ ¯ t ¿
f
³ F s ds. s
As before, the conditions on f (t ) and f (t ) / t have been introduced to guarantee uniform convergence. Only in this way can we be sure that the interchange of the order of integration in equation (1.3) and the operation of taking the limit under the integral sign in equation (1.4) are valid.
sin t ½ ¾. ¯ t ¿
Example 1.19. Determine L ® Solution: In this case,
f t sin t , giving
The Laplace Transform
L ^sin t` F s
1 s 1 2
by the division theorem,
ª f t º L« » ¬ t ¼
f
f
³ F s ds
³s
s
ª sin t º 1 f L« tan s s ¬ t »¼
s
2
1 ds 1
S tan 1 s. 2
Thus,
ª sin t º S L« tan 1 s. » ¬ t ¼ 2 Example 1.20. Determine
e3t e 2t ½ L® ¾. t ¯ ¿
Solution: In this case,
^
L e3t e 2t
f t e3t e 2t , giving
`
F s
1 1 s 3 s 2
by the division theorem,
ª f t º L« » ¬ t ¼
f
f
³ F s ds
³ ¨© s 3 s 2 ¸¹ ds
s
s
§ 1
1 ·
35
Chapter 1
36
e3t e 2t ½ L® ¾ t ¯ ¿
f
^log s 3 log s 2 `
f s
½ ª s 3 º ° ° ®log « »¾ ° ¬ s 2 ¼ ° ¯ ¿s .
Thus,
ª s 2 º e3t e 2t ½ L® ¾ log « ». t s 3 ¯ ¿ ¬ ¼ Property 1.5.7: Transforms of Integrals If
t ° °½ L ^ f t ` , then L ® ³ f W dW ¾ ¯° 0 ¿°
F s
1 L ^ f t ` s
Proof: t
g t
³ f W dW 0
We have
dg t dt
f t and g 0 0
Taking Laplace transforms,
dg t ½ L® ¾ ¯ dt ¿
L ^ f t `
sL ^ g t ` g 0 Hence,
(using the derivative formula)
L ^ f t ` .
F s s
.
The Laplace Transform
L ^ g t `
37
1 L ^ f t ` s
giving the result
° t ½° 1 L ® ³ f W dW ¾ L ^ f t ` ¯° 0 ¿° s
F s .
Division of the transform of a function by
s
corresponds to integration of
the function between the limits 0 and t .
° t ½° 3W dW ¾ . Example 1.21. Determine L ® ³ sin 2W e °¯ 0 °¿
Solution: In this case,
^
f t
L sin 2t e 3t
sin 2t e , giving 3t
`
F s
2 1 s 4 s3 2
by the integral theorem,
° t ½° 1 L ® ³ sin 2W e 3W dW ¾ F s °¿ s ¯° 0
1§ 2 1 · ¨ 2 ¸. s © s 4 s3¹
In the analysis of a linear time-invariant (LTI) system, it is necessary to solve differential equations using Laplace transform techniques. Unfortunately, it is sometimes the case that it is impossible to invert
F s to find the desired solution to the original problem. Numerical inversion techniques are possible and can be found in some software
Chapter 1
38
packages, especially those used by the control system. Insight into the behavior of the solution can be deduced without actually solving the differential equation by examining the asymptotic character of
s
small
s.
or large
F s for
In fact, it is often beneficial to determine this
asymptotic behavior without solving the equation, even when exact solutions are available, as these solutions are often complex and challenging to obtain, let alone interpret. In the next section, two properties help us to find the asymptotic behavior of the function.
Property 1.5.8: The Initial-value Theorem The initial value theorem allows us to find the function’s initial value without finding the inverse Laplace transform. If
L ^ f t ` , then f 0 lim ^s F s ` .
F s
s of
Proof: We have the derivative formula
^
`
L f ' t
f 0 sL ^ f t `
f
³e
st
f ' t dt
f 0 s F s .
0
Taking
lim on both sides, we get sof
0 f 0 lim ^s F s ` . s of
Therefore,
The Laplace Transform
f 0
39
lim ^s F s ` . s of
The initial value theorem’s physical significance: The initial value of the function can be found from its Laplace transform by taking the limit of
s F s , as s tends to infinity. We need not take
the inverse Laplace transform. Example 1.22. Determine the initial value of
f t
5 sinh t .
Solution: In this case
f t
L ^ 5 sinh t `
5 sinh t ,
giving
5 s 2 s s 1
F s
by the initial value theorem,
§5 s ·½ lim ® s ¨ 2 ¸ ¾ s of ¯ © s s 1 ¹¿
f 0
lim ^s F s `
f 0
s · § lim ¨ 5 2 2 ¸ s of s 1 ¹ ©
s of
Thus,
f 0 5.
§ s lim ¨ 5 2 s of ¨ s 1 1 s2 ©
· ¸ ¸ ¹
Chapter 1
40
Example 1.23. Using the Laplace transform, find the initial value of
f t
the function
AeD t cos E t T u t .
Solution: In this case,
f t
AeD t cos E t T u t .
Using the trigonometric formula
cos A B cos A cos B sin A sin B f t
AeD t ^cos T cos E t sin T sin E t ` u t ,
giving
L ^ f t ` F s
F s
^
`
^
ª º ª º E s D T A cos T « A sin » « » 2 2 «¬ s D E 2 »¼ «¬ s D E 2 »¼
by the initial value theorem,
f 0
lim ^s F s `
f 0
lim ^s F s `
s of
s of
`
A cos T L eD t cos E t u t A sin T L eD t sin E t u t
½½ ° ° ª cos T s D E sin T º °° lim ® s ® A « » ¾¾ 2 s of »¼ ¿°°¿ s D E 2 °¯ °¯ «¬ ½ § D· E ° cos T ¨1 ¸ sin T ° ° ° s¹ s © A lim ® ¾. 2 2 s of § D· E ° ° ¨1 ¸ 2 °¯ °¿ s s © ¹
The Laplace Transform
41
Thus,
f 0 A cos T . Property 1.5.9: The Final-value Theorem The final value theorem allows us to find the function’s final value without finding the inverse Laplace transform. If
L ^ f t ` , then f f lim ^s F s ` .
F s
s o0
Proof: We have the derivative formula
^
`
L f ' t
f 0 sL ^ f t `
f
³e
st
f ' t dt
f 0 s F s
0
Taking
lim on both sides, we get so 0
f
³ f t dt '
f 0 lim ^s F s `
0
f f f 0
s o0
f 0 lim ^s F s ` .
Therefore,
f f
lim ^s F s ` . s o0
s o0
Chapter 1
42
The physical significance of the final value theorem: The final value of the function can be found from its Laplace transform by taking the limit of
s F s , as s tends to zero. We need not take the inverse Laplace transform. Example 1.24. Determine the final value of
f t
5 e . 2t
Solution: In this case,
^
L 5 e 2t
f t
`
5 e , giving
F s
2t
5 1 s s2
by the final value theorem,
§5 1 ·½ lim ® s ¨ ¸¾ s o0 ¯ © s s 2 ¹¿
f f
lim ^s F s `
f f
s · § lim ¨ s ¸ s o0 © s2¹.
s o0
Thus,
f f
0 · § lim ¨ 5 ¸ 5. s o0 © 02¹
Several operational properties of the Laplace transform have been developed. These properties are applied to generate Laplace transform tables and use the Laplace transform in the solution of linear differential equations with constant coefficients. These properties are applicable in
The Laplace Transform
43
both the analysis and design of linear time-invariant physical systems. Table 1.2 gives the derived operational properties for the Laplace transform. Table 1.2 Basic Properties of the Laplace transform Linearity
L ^C1 f t C2 g t ` C1 L ^ f t ` C2 L ^ g t `
Differentia
L ^ f n t ` s n F s s n 1 f
tion in time Translation Multiplicat ion by
t
Time scale
F s a
L ^t
dn 1 n ^F s ` ds
n
f t `
L ^ f at `
n
1 §s· F¨ ¸ a ©a¹
t ° °½ 1 L ® ³ f W dW ¾ L ^ f t ` °¯ 0 °¿ s
Division
f t ½ L® ¾ ¯ t ¿
t
Initial-
f
³ F s ds. s
f 0
lim ^s F s `
f f
lim ^s F s `
value
s of
theorem Final-value theorem
0 s n2 f ' 0 ......
L ^e at f t `
Integration
by
n 1
s o0
Chapter 1
44
Table 1.2 lists the operational properties of Laplace transforms. You should familiarize yourself with them, since they are frequently encountered in problems for solving initial value problems, finding transfer functions, and in the analysis of mechanical systems and electrical circuits. Example 1.25. Determine
L ^t e 2t sin 3t` .
Solution: Using Table 1.1,
L ^sin 3t`
F1 s
3 s 9 2
using the multiplication theorem,
F s
L ^t sin 3t`
d ª F1 s º¼ ds ¬
d § 3 · ¨ ¸ ds © s 2 9 ¹
6s
s
and by the first shifting theorem,
L ^e 2t t sin 3t`
F s 2
Example 1.26. Determine
Using Table 1.1,
s s 1 2
6 s 2 s 2
L ^t e 3t cos t` .
Solution:
L ^cos t` F1 s
F s s
s 2
2
9
2
.
2
9
2
The Laplace Transform
45
using the multiplication theorem,
F s
L ^t cos t`
d ª F1 s º¼ ds ¬
d § s · ¨ ¸ ds © s 2 1 ¹
s2 1
s
2
1
2
and by the first shifting theorem, 2
L ^e t cos t` F s 3 F s s 3t
Example 1.27. Determine
s 3
s 3 1 . 2 2 s 3 1
L ^ e 8t t 2 sin t` .
Solution: Using Table 1.1,
L ^sin t`
F1 s
1 s 1 2
using the multiplication theorem,
F s
L ^t sin t` 2
d d ½ ® ª¬ F1 s º¼ ¾ ds ¯ ds ¿
§ · d ¨ s ¸ ds ¨ s 2 1 2 ¸ © ¹
6s 2 2
s
and by the first shifting theorem, 2
L ^ e t sin t` F s 8 F s s
6 s 8 2
8t 2
s 8
s 8 1 2
3
.
2
1
3
Chapter 1
46
Example 1.28. Using the Laplace transform, find the initial and final values of the function
f t
2e
3t
e 4t .
Solution: In this case,
f t
L ^ f t `
F s
^by direct
2e
3t
e 4t giving
^
L 2e 3t e 4t
`
f 0 lim 2e3t e4t t o0
1 · § 2 F s ¨ ¸ © s3 s4¹ by the initial value theorem,
§ 2 1 ·½ lim ® s ¨ ¸¾ s of ¯ © s 3 s 4 ¹¿
f 0
lim ^s F s `
f 0
§ ·½ 1 ¸ °° °° ¨ 2 lim ® ¨ ¸ ¾ 1. s of ° ¨ 1 3 s 4 ¸° s s ¹ ¿° ¯° ©
s of
Thus,
f 0 1. By the final value theorem,
`
1
The Laplace Transform
47
§ 2 1 ·½ lim ® s ¨ ¸¾ s o0 ¯ © s 3 s 4 ¹¿
f f
lim ^s F s `
f f
§ · ¨ 2 1 ¸ lim ¨ ¸ s o0 ¨ 1 3 s 4 ¸ s s¹ ©
s o0
0.
Thus,
^by direct
f f 0.
f f lim 2e 3t e4t t of
`.
0
f
Example 1.29. Evaluate
³e
2 t
t 2 sin t dt .
0
Solution: Using Table 1.1,
L ^sin t`
1 s 1
F1 s
2
using the multiplication theorem,
F s
L ^t 2 sin t`
d d ½ ® ª F1 s º¼ ¾ ds ¯ ds ¬ ¿
§ · d ¨ s ¸ ds ¨ s 2 1 2 ¸ © ¹
and by definition of the Laplace transform, f 2 t 2 ³ e t sin t dt 0
L ^t 2 sin t`
s 2
2 ½ ° 6s 2 ° ® 2 3¾ °¯ s 1 °¿ s
22 . 125 2
6s 2 2
s
2
1
3
Chapter 1
48
Thus, f
³e
2 t
t 2 sin t dt
0
22 . 125 f
³
Example 1.30. Evaluate
0
e t e 3t dt . t
Solution: We rewrite the integral as f
³ 0
et e3t dt t
^
³ 0
§ 1 e 2t · e ¨ ¸ dt © t ¹ t
§ 1 e 2t f t ¨ © t
In this case
L 1 e 2t
f
`
F s
· ¸ , giving ¹
1 1 s s2
by the division theorem,
ª f t º L« » ¬ t ¼
1 e 2t ½ L® ¾ ¯ t ¿ Thus,
f
f
³ F s ds
³ ¨© s s 2 ¸¹ ds
s
§1
1 ·
s
f
^log s log s 2 `
f s
° ª s º ½° ®log « »¾ . ¯° ¬ s 2 ¼ ¿° s
The Laplace Transform
49
ª s 2 º log « ». ¬ s ¼
1 e 2t ½ L® ¾ ¯ t ¿
By definition of the Laplace transform, f
³ 0
§ 1 e 2t · et ¨ ¸ dt © t ¹
1 e 2t ½ L® ¾ ¯ t ¿s
1
° ª s 2 º ½° ®log « »¾ ¯° ¬ s ¼ ¿° s
log 3 1
thus, f
³ 0
§ 1 e 2t · et ¨ ¸ dt © t ¹
log 3.
2sinh 2t ½ ¾. t ¯ ¿
Example 1.31. Determine L ® Solution: In this case
^
f t 2sinh 2t
L e 2t e 2t
`
F s
e 2t e 2t , giving
1 1 s2 s2
by the division theorem,
ª f t º L« » ¬ t ¼
f
f
³ F s ds
³ ¨© s 2 s 2 ¸¹ ds
s
e 2t e 2t ½ L® ¾ t ¯ ¿
§ 1
1 ·
s
f
^log s 2 log s 2 ` e
e
f s
° ª s 2 º ½° ®log e « »¾ . ¬ s 2 ¼ ¿° s ¯°
Chapter 1
50
Thus,
ª s 2 º 2sinh 2t ½ L® ¾ log e « ». s 2 ¯ t ¿ ¬ ¼ Example 1.32. Determine
t ½ L ®e 2t ³ e 2W cos 3W dW ¾ . 0 ¯ ¿
Solution: Using the first shifting theorem,
2t t 2W ½ L ®e ³ e cos 3W dW ¾ F s 2 ¯ 0 ¿ where
F s
t 2W ½ L ® ³ e cos 3W dW ¾ ¯0 ¿
Using Table 1.1,
L ^cos 3t`
F2 s
s s 9 2
by the first shifting theorem,
L ^e 2t cos 3t `
s2
s 2
by the integration property,
2
9
The Laplace Transform
F s
t ½ L ® ³ e 2W cos 3W dW ¾ ¯0 ¿
1 s
51
§ s2 · ¨ ¸. ¨ s 2 2 9 ¸ © ¹
Thus,
t ½ L ®e 2t ³ e 2W cos 3W dW ¾ F s 2 ¯ 0 ¿
s
s 2 s2 9
f
Example 1.33. Evaluate
sin t dt . t 0
³
Solution: Using Table 1.1,
L ^sin t`
F s
1 s 1 2
by the division theorem,
ª f t º L« » ¬ t ¼
sin t ½ L® ¾ ¯ t ¿
f
f
³ F s ds
³ ¨© s
s
^tan s` 1
f s
§
s
2
1 · ¸ ds 1 ¹
S tan 1 s 2
and by definition of the Laplace transform, f
³e
0t
0
Thus,
sin t dt t
sin t ½ L® ¾ ¯ t ¿s
0
S 1 ½ ® tan s ¾ ¯2 ¿s
S 0
2
.
.
Chapter 1
52 f
³e
2 t
t 2 sin t dt
0
22 . 125
1.6 Periodic Function Intuitively, a function is periodic when it repeats itself. This intuition is captured in the following definition: a continuous-time function
f t is
periodic if there exists a positive real T for which
f t T
f t ,
t .
The smallest such T is called the fundamental period of the function. i The square wave function in Figure 1.5 is periodic. The fundamental period of this square wave is T
4, but 8, 12, and 16 are also periods of
the function.
Figure 1.5. A periodic square wave function
i
A function that has period T will also have period 2T , 3T , etc. For example, the sine function has periods 2S , 4S , 6S etc. Some authors refer to the smallest period as the fundamental period or just the period of the function.
The Laplace Transform
53
Periodic functions play a significant role in many branches of engineering and applied science, particularly control systems and circuit analysis. One only has to think of springs or alternating current in household electricity to understand their prevalence. Here, we introduce a theorem on the Laplace transform of periodic functions. We prove the theorem with a few illustrative examples.
1.6.1 Transform of a Periodic Function If f (t) is piecewise continuous on
>0, f of the exponential order and
periodic with a period T , then T
L ^ f t `
1 e st f t dt . sT ³ 1 e 0
Proof: Like many proofs of the operational properties of Laplace transforms, this one begins with its definition then evaluates the integral by using the
f t .
periodicity of
L ^ f t `
f
³e
st
f t dt
0
We write the Laplace transform of T
L ^ f t `
f
st st ³ e f t dt ³ e f t dt . 0
When we let
f t as two integrals
t
T
u T , the last integral becomes
Chapter 1
54 f
f
st ³ e f t dt
³e
T
s u T
f
f u T du
0
e sT ³ e su f u du
e sT L ^ f t `
0
. Therefore, T
L ^ f t `
³e
st
f t dt e sT L ^ f t `
0
1 e sT L ^ f t `
T
³e
st
f t dt .
0
Thus, T
L ^ f t `
1 e st f t dt. sT ³ 1 e 0
The physical significance of the property of periodicity The Laplace transform of a periodic function can be found by taking the Laplace transform of one period and dividing it by is a period of the function.
1 e , where T sT
The Laplace Transform
55
Example 1.34. Find the transform of the periodic function
f t of
the period T shown in Figure 1.6.
f(t)
K
–2T
0
–T
2T
T
t
Figure 1.6 Solution: The function
f t is called a sawtooth wave and has a period T . For
0 t T . f t
f t T
can
be
defined
outside
the
interval
f t . The mathematical expression for f t in one-
period is
f t
by
K t 0dt dT T
with f t T
We have T
L ^ f t `
1 e st f t dt sT ³ 1 e 0
L ^ f t `
K e st t dt . sT ³ T 1 e 0
T
f t
Chapter 1
56
Integrating by parts, we get
L ^ f t `
K 1 st te sT ® T 1 e ¯ s
L ^ f t `
K 1 1 ½ ª¬Te sT º¼ 2 ª¬e sT 1º¼ ¾ . sT ® s T 1 e ¯ s ¿
T 0
1 st T ½ e ¾ 0 s2 ¿
Therefore,
L ^ f t `
K 1 e sT sTe sT ` . sT ^ s T 1 e 2
Example 1.35. Find the transform of the periodic function
f t of
the period 2a shown in figure 1. 7. f(t) E
t
O a
2a
3a
4a
–E
Figure 1.7. A periodic square wave function Solution: The function
f t is called a square wave and has a period T
The mathematical expression for
f t in one-period is
2a.
The Laplace Transform
E 0 d t d a f t ® ¯ E a d t d 2a
with
57
f t 2a
f t
.
We have T
L ^ f t `
L ª¬ f t º¼
L ^ f t `
1 e st f t dt sT ³ 1 e 0 2a ª a st º 1 e E dt ³ e st E dt » s 2a « ³ 1 e ¬ 0 a ¼ 2a ª a st º E st e dt e dt « ». ³ ³ 1 e s 2a ¬ 0 a ¼
After performing integration,
^
`
L ^ f t `
E st a e e st s 2a 0 s 1 e
L ^ f t `
E ª¬e sa 1º¼ ª¬e s 2 a e sa º¼ . s 2a s 1 e
Therefore,
^
2a a
`
Chapter 1
58
E e s 2 a 2e sa 1` s 2a ^ s 1 e
L ^ f t `
E 1 e sa u e
L ^ f t `
s 1 e
sa
u e
E s 1 e
sa § sa · E¨e 2 e 2 ¸ © ¹ sa sa § · s¨e 2 e 2 ¸ © ¹
sa 2
sa 2
sa
sa 2
1 e sa
1 e
§ as · E sinh ¨ ¸ © 2¹. § as · s cosh ¨ ¸ © 2¹
Thus,
E § as · tanh ¨ ¸ . s ©2¹
L ^ f t `
Example 1.36. Determine the Laplace transform of the periodic function shown in Figure 1.8. f(t) 1
0
a
2a
3a
4a
t 5a
Figure 1.8. A periodic square wave function Solution: The function
f t is called a square wave and has a period T
The mathematical expression for
f t
0 0 d t d a ® ¯1 a d t d 2a
f t in a one-period is
with
f t 2a
f t
2a.
The Laplace Transform
59
We have T
L ^ f t `
L ^ f t `
1 e st f t dt sT ³ 1 e 0 2a ª a st º 1 st e dt e dt 0 1 « ». ³ ³ 1 e s 2a ¬ 0 a ¼
On integrating, we get
L ^ f t `
2a 1 st ª º e ¼a s 1 e 2 as ¬
1 ªe 2 as e as º¼ s 2a ¬ s 1 e
. Therefore,
L ^ f t `
1 ªe as e 2 as º¼ 2 as ¬ s 1 e
Thus,
L ^ f t `
e as . s 1 e sa
e as s 1 e
sa
1 e sa
ª¬1 e as º¼ .
Chapter 1
60
Example 1.37. Determine the Laplace transform of the half-wave rectifier shown in Figure 1.9.
Figure 1.9. A periodic half-wave rectifier function Solution: The function
T
f t is called a half-wave rectifier wave and has a period
2S w . The mathematical expression for f t in one period is
f t
S °°sin wt 0 d t d w ® 2S S °0 dt d °¯ w w
§ 2S · with f ¨ t ¸ w ¹ ©
We have T
L ^ f t `
L ^ f t `
1 e st f t dt sT ³ 1 e 0 Sw ½ 1 ° st ° ® ³ e sin wt dt ¾ . 2S s § ·°0 ° w ¨1 e ¸¯ ¿ © ¹
Recall from integral calculus that
f t .
The Laplace Transform
³e
at
sin bt dt
L ^ f t `
L ^ f t `
61
eat ª a sin bt b cos bt º¼ a 2 b2 ¬
1 e st s sin wt w cos wt 2S s § · 2 2 w ¨1 e ¸ s w © ¹ S s § · w w e 1 ¨ ¸ 2S s § · 2 ¹ 2 © w e s w 1 ¨ ¸ © ¹
S w 0
S s § · w ¨1 e w ¸ © ¹ S s S s § ·§ · 2 2 w w ¨1 e ¸¨1 e ¸ s w © ¹© ¹
Therefore,
L ^ f t `
§ ¨1 e ©
S s w
w . · 2 2 ¸ s w ¹
Example 1.38. Determine the Laplace transform of the full-wave rectifier shown in Figure 1.10.
Figure 1.10. A periodic full-wave rectifier function
Chapter 1
62
Solution: The function
T
S w.
f t is called a full-wave rectifier wave and has the period f t in one period is
The mathematical expression for
f t sin wt 0 d t d
S § S· with f ¨ t ¸ w © w¹
f t
We have T
L ^ f t `
L ^ f t `
1 e st f t dt sT ³ 1 e 0 ª Sw 1 « st e sin wt dt S s § · « ³0 w ¨1 e ¸ «¬ © ¹
º ». » »¼
On integrating, we get
L ^ f t `
L ^ f t `
S
1 e st s sin wt w cos wt ` w ^ S s 0 § · 2 2 w ¨1 e ¸ s w © ¹
§ ¨1 e ©
S s w
§ w ¨1 e · 2 2 © ¸ s w ¹
S s w
· ¸ ¹
§ Ss · w cosh ¨ ¸ © 2w ¹ . § Ss · 2 2 sinh ¨ ¸ s w © 2w ¹
The Laplace Transform
63
Therefore,
L ^ f t `
w § Ss · coth ¨ . 2 s w © 2w ¸¹ 2
Example 1.39. Determine the Laplace transform of the rectified sine wave
f t
sin t 0 t S with f t 2S ® ¯ sin t S t 2S
f t .
Solution: We have T
L ^ f t `
L ^ f t `
1 e st f t dt sT ³ 1 e 0 S 2S 1 ° st °½ sin e t dt e st sin t dt ¾ . ³ 2S s ® ³ 1 e °¯ 0 °¿ S
Recall from integral calculus that
³e
at
eat ª a sin bt b cos bt º¼ a 2 b2 ¬
sin bt dt
L ^ f t `
L ^ f t `
ªe s 1 ^¬
1
1 e
2S s
st
2
1
1 e
2S s
s 2 1
s sin t cos t º¼
^ 1 e e sS
S 0
ª¬e st s sin t cos t º¼
s 2S
`
e sS
2S
S
`
Chapter 1
64 2
1 e . L ^ f t ` 1 e s 1 sS
2S s
2
Therefore,
1 e L ^ f t ` 1 e s 1 sS
sS
L ª¬ f t º¼
2
§Ss · sinh ¨ ¸ © 2 ¹ s 2 1 cosh §¨© S2s ·¸¹
1 §Ss · tanh ¨ ¸ . s 1 © 2 ¹ 2
Thus,
L ª¬ f t º¼
1 §Ss · tanh ¨ . s 1 © 2 ¸¹ 2
Example 1.40. Determine the Laplace transform triangular wave shown in Figure 1.11.
Figure 1.11. A periodic triangular-wave function
The Laplace Transform
65
Solution: The function period
T
f t is called a triangular-wave rectifier wave and has the
2a. The mathematical expression for f t in one period is
0dt da t f t ® ¯ 2a t a d t d 2a
with
f t 2a
f t
We have T
L ª¬ f t º¼
1 e st f t dt sT ³ 1 e 0
L ª¬ f t º¼
2a ª a st º 1 st e t dt e a t dt 2 « ». ³a 1 e s 2a ¬ ³0 ¼
Integrating by parts, we get
L ª¬ f t º¼
1 ª1 e 2 as 2e as º¼ . s 1 e s 2 a ¬ 2
Therefore,
L ª¬ f t º¼
1 s 1 e 2
sa
1 e
1 e L ª¬ f t º¼ s 1 e sa
2
sa 2
sa
sa
1 e
Chapter 1
66
L ª¬ f t º¼
§ as · sinh ¨ ¸ © 2¹ § as · s 2 cosh ¨ ¸ ©2¹
1 § as · tanh ¨ ¸ . 2 s ©2¹
Thus,
L ª¬ f t º¼
1 § as · tanh ¨ ¸ . 2 s ©2¹
Example 1.41. Determine the Laplace transform periodic function shown in Figure 1.12. f(t)
6
6 3t
t 2
4
6
Figure 1.12 Solution: The mathematical expression for a given periodic function is
3t f t ® ¯6 We have
0dt d2 2dt d4
with f t 4
f t .
The Laplace Transform
67
T
L ^ f t `
L ^ f t `
1 e st f t dt sT ³ 1 e 0 4 ª 2 st º 1 st 3 6 e t dt e dt « ». ³2 1 e s 4 ¬ ³0 ¼
Integrating by parts, we get
L ^ f t `
ª 3te st 3e st ½ 2 6e st ½ 4 1 «® ¾ ® ¾ 1 e s 4 «¬¯ s s 2 ¿ 0 ¯ s ¿ 2
º ». »¼
Thus,
L ^ f t `
3 1 e 2 s 2 se 4 s s 2 1 e s 4
.
Example 1.42. Determine the Laplace transform of the half-wave rectifier
f t
periodic
function
of
S °°sin T t 0 t T ® S 2S °0 t °¯ T T
.
the
Solution: We have T
L ^ f t `
1 e st f t dt sT ³ 1 e 0
period
2S T
defined
by
Chapter 1
68
L ^ f t `
ª ST 1 « st e sin Tt dt 2S s § · « ³0 T ¨1 e ¸ «¬ © ¹
º ». » »¼
On integrating, we get
L ^ f t `
L ^ f t `
S st ½ 1 T sin cos e s Tt T st ® ¾ 2S s 0 § · ¯ ¿ 2 2 T ¨1 e ¸ s T © ¹
S s § · T T e 1 ¨ ¸. 2S s § · © ¹ 2 2 T ¨1 e ¸ s T © ¹
Therefore,
L ^ f t `
§ ¨1 e ©
S s T
T . · 2 2 ¸ s T ¹
1.7 Second Shifting Theorem or Second Translation Theorem If
F s
L ^ f t ` and a ! 0, then
L ^ f t a u t a ` e as L ^ f t ` . Proof:
The Laplace Transform
69
The proof follows immediately from the definition of the unit step function. According to the definition of the Laplace transform, we have
L ^ f t `
f
F s
³e
st
f t dt
0
L ^ f t a u t a `
f
³e
st
f t a u t a dt
0
L ^ f t a u t a `
f
³e
st
f t a dt
a
(1.5)
In the last equation, we used the fact that
u t a is zero for t a and
equals 1 for t t a. Making the change of variable
v t a , we have dv dt , and Eq.
(1.5) becomes
L ^ f t a u t a `
f
³e
s v a
f v dv
a
f
L ^ f t a u t a ` e sa ³ e sv f v dv . a
Therefore,
L ^ f t a u t a ` e as F s . This establishes the theorem.
Chapter 1
70
The only condition on
f t is that it is a function that is of exponential
order, which means that it is free from singularities for
t ! a. The
principal use of this theorem is that it enables us to determine the Laplace transform of a function that is switched on at time
t
a.
1.7.1. Expression for Piecewise Continuous Function f t in Terms of the Unit Step Function. Consider a piecewise continuous function
f t
f t defined by
f1 t 0 d t a ° ® f2 t a d t b . ° t tb ¯ f3 t
To construct the function
f t , we can use the following ‘switching’
operations: (a)
Switch on the function
f1 t at t
(b)
Switch on the function
f 2 t at t
a , and, at the same time,
f3 t at t
b , and, at the same time,
switch off the function (c)
f1 t ;
Switch on the function
switch off the function
0;
f2 t .
In terms of the unit step function,
f t may thus be expressed as
The Laplace Transform
f (t )
71
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
Example 1.43. Obtain the Laplace transform of the piecewisecontinuous function shown in Figure 1.13 by expressing it in the unit step function.
f(t)
3
1
2
t
Figure 1.13 Solution: The function is continuous, but is defined differently on the intervals. As such, the mathematical expression for the function is
f t
0 d t d1 0 ° ®3 t 1 1 t d 2 °3 tt2 ¯
Unit step functions for
f t may be expressed as
f (t )
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
f (t )
3t 3 u t 1 6 3t u t 2
Chapter 1
72
Taking the Laplace transform, we have
L ^ f (t )` L ^ 3t 3 u t 1 ` L ^ 6 3t u t 2 ` We cannot apply the second translation theorem with
f t in this form
since, according to the second translation theorem, for the first term we must have a function of
t 1
and for the second term a function of
t 2 . We can apply the second shifting property to the result above and obtain
L ^ f (t )` e s L ^ f1 (t )` e2 s L ^ f 2 (t )` The function
f1 (t 1) replace
t
f t is thus rewritten as follows.
3t 3 t 1
6 3t
f 2 (t 2) replace
t
t2
f 2 (t ) 3t
f1 (t ) 3t then its Laplace transform is
L ^ f1 (t )`
(1.6)
3 s2
Eq. (1.6) becomes
then its Laplace transform is
L ^ f 2 (t )`
3 s2
The Laplace Transform
§3· §3· L ^ f (t )` e s ¨ 2 ¸ e2 s ¨ 2 ¸ ©s ¹ ©s ¹
73
3 s 2 s e e . s2
Example 1.44. Obtain the Laplace transform of the piecewisecontinuous function shown in Figure 1.14 by expressing it in the unit step function.
f(t) 3
0 –1
3
t
Figure 1.14 Solution: The function is continuous, but it is defined differently on the intervals. As such, the mathematical expression for the function is
f t
0 d t d1 3 ° ® 5 2t 1 t d 3 ° t t3 ¯1
In terms of the unit step function,
f t may be expressed as
Chapter 1
74
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
f (t )
f (t ) 3 2 2t u t 1 2t 6 u t 3 Taking the Laplace transform, we have
L ^3` L ^ 2 2t u t 1 ` L ^ 2t 6 u t 3 ` .
L ^ f (t )`
We cannot apply the second translation theorem with
f t in this form,
since, according to the second translation theorem, for the first term we must have a function of
t 1
and for the second term a function of
t 3 . We can apply the second shifting property to the result above to obtain
3 s e L ^ f1 (t )` e 3 s L ^ f 2 (t )` s
L ^ f (t )` The function
f1 (t 1) replace
t
f t is thus rewritten as follows.
2 2t t 1
f1 (t ) 2t
f 2 (t 3) replace
2t 6
t t 3
f 2 (t ) 2t
then its Laplace transform is
L ^ f1 (t )`
(1.7)
2 s2
then its Laplace transform is
L ^ f 2 (t )`
2 s2
The Laplace Transform
75
Eq. (1.7) becomes
L ^ f (t )`
3 s § 2 · 3s § 2 · 3 2 3s s e ¨ 2 ¸e ¨ 2 ¸ 2 e e . s ©s ¹ ©s ¹ s s
Example 1.46. Express in terms of the Heaviside unit step function and find the Laplace transform of
t 2 0 d t d 2 f (t ) ® . tt2 ¯4t
Solution: In terms of the unit step function,
f (t )
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a
f (t ) t 2 ª¬ 4t t 2 º¼ u t 2 Taking the Laplace transform, we have
L ^ f (t )`
^
`
L ^t 2 ` L 4t t 2 u t 2
which, using the result in (1.7), gives
L ^ f (t )`
2 2 s e L ^ f1 (t )` s3
(1.8)
Chapter 1
76
f1 (t 2) replace
f1 (t )
t
4t t 2
t2
4 t 2 t 2 2
f1 (t ) 4t 8 t 2 4t 4 f1 (t ) 4 t 2 then its Laplace transform is
L ^ f1 (t )`
4 2 s s3
Eq. (1.8) becomes
L ^ f (t )`
2 §4 2 · e 2 s ¨ 3 ¸ . 3 s ©s s ¹
Example 1.47. Find the Laplace transform for the function in Figure 1.15.
Figure 1.15
The Laplace Transform
77
Solution: The function is continuous, but it is defined differently on the intervals. As such, the mathematical expression for the function is
f (t )
0 ° ®sin t °0 ¯
0dt dS S t d 3S t t 3S
In terms of the unit step function,
f (t )
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
f (t ) 0 sin t u t S sin t u t 3S . Taking the Laplace transform, we have
L ^ f (t )`
L ^sin t u t S ` L ^sin t u t 3S ` .
We cannot apply the second translation theorem with
f t in this form,
since, according to the second translation theorem, for the first term we must have a function of
t S
and for the second term a function of
t 3S . We can apply the second shifting property to the result above to obtain
L ^ f (t )` eS s L ^ f1 (t )` e2S s L ^ f 2 (t )` The function
f t is thus rewritten as follows.
(1.9)
Chapter 1
78
f1 (t S ) sin t replace
t t S
f 2 (t 3S ) sin t replace
t
t 3S
f1 (t ) sin t S
f1 (t ) sin t 3S
f1 (t ) sin t
f1 (t ) sin t
then its Laplace transform is
L ^ f1 (t )`
1 s 1 2
then its Laplace transform is
L ^ f 2 (t )`
1 s2 1
Eq. (1.9) becomes
§ 1 · 2S s § 1 · L ^ f (t )` e S s ¨ 2 ¸e ¨ 2 ¸. © s 1 ¹ © s 1 ¹ Thus,
L ^ f (t )`
1 e 2S s e S s . s 1 2
Example 1.48. Express in terms of the Heaviside unit step function
1 ° and find the Laplace transform of f (t ) ®0 °sin t ¯
0dt dS S t d 2S . t ! 2S
The Laplace Transform
79
Solution: In terms of the unit step function,
f (t )
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
f (t ) 1 u t S sin t u t 2S . Taking the Laplace transform, we have
L ^ f (t )` L ^1` L ^u t S ` L ^sin t u t 2S ` We cannot apply the second translation theorem with
f t in this form,
since, according to the second translation theorem, for the first term we must have a function of
t S
and for the second term a function of
t 2S . We can apply the second shifting property to the above result to obtain
L ^ f (t )` The function
1 eS s 2S s e L ^ f1 (t )` s s f t is thus rewritten as follows.
(1.10)
Chapter 1
80
f1 (t 2S ) sin t replace
t
t 2S
f1 (t ) sin t 2S f1 (t ) sin t then its Laplace transform is
L ^ f (t )`
1 s2 1
Eq. (1.10) becomes
L ^ f (t )`
1 e S s 2S s § 1 · e ¨ 2 ¸. s s © s 1¹
Example 1.49. Express in terms of the Heaviside unit step function and find the Laplace transform of
sin t 0 t d S 2 f (t ) ® t !S 2 ¯cos t
Solution: In terms of the unit step function,
f (t )
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a
The Laplace Transform
81
§ S· f (t ) sin t > cos t sin t @ u ¨ t ¸ . © 2¹ Taking the Laplace transform, we have
L ^ f (t )`
§ S ·½ L ^sin t` L ® cos t sin t u ¨ t ¸ ¾ © 2 ¹¿ ¯
which, using the result in (1.7), gives
L ^ f (t )` The function
S s 1 2 e L ^ f1 (t )` 2 s 1
f t is thus rewritten as follows.
§ S· f1 ¨ t ¸ © 2¹ Replace t
cos t sin t t
S 2
§ S· § S· f1 (t ) cos ¨ t ¸ sin ¨ t ¸ © 2¹ © 2¹
f1 (t ) sin t cos t then its Laplace transform is
L ^ f1 (t )`
1 s 2 s 1 s 1 2
(1.11)
Chapter 1
82
Eq. (1.11) becomes
L ^ f (t )`
S s 1 s · § 1 e 2 ¨ 2 2 ¸. 2 s 1 © s 1 s 1 ¹
Example 1.50. Express in terms of the Heaviside unit step function and find the Laplace transform of
cos t 0 t d S f (t ) ® . t !S ¯sin t
Solution: In terms of unit step functions,
f (t )
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a
f (t ) cos t >sin t cos t @ u t S . Taking the Laplace transform, we have
L ^ f (t )` L > cos t @ L ^ sin t cos t u t S ` which, using the result in (1.7), gives
L ^ f (t )` The function
s eS s L > f1 (t )@ 2 s 1 f t is thus rewritten as follows.
(1.12)
The Laplace Transform
f1 (t S ) replace
83
sin t cos t
t t S
f1 (t ) sin t S cos t S
f1 (t ) sin t cos t then its Laplace transform is
L ^ f1 (t )`
s 1 2 s 1 s 1 2
Then, Eq. (1.12) becomes
L ^ f (t )`
s 1 · § s e S s ¨ 2 2 ¸ . s 1 © s 1 s 1¹ 2
Example 1.51. Express
f t in terms of the Heaviside unit step
function and find the Laplace transform
f (t )
0 t 1 °2 ° 2 S °t 1 t ® 2 °2 S ° °¯cos t t ! 2
Chapter 1
84
Solution: In terms of unit step functions,
f t can be expressed as
f (t )
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
f (t )
ªt2 º ª t2 º § S · 2 « 2 » u t 1 «cos t » u ¨ t ¸ . 2¼ © 2¹ ¬2 ¼ ¬
Taking the Laplace transform, we have
° ª t 2 º ½° ° ª t 2 º § S · ½° L ^ f (t )` L ^2` L ® « 2 » u t 1 ¾ L ® «cos t » u ¨ t ¸ ¾ 2 ¼ © 2 ¹ °¿ °¯ ¬ 2 ¼ °¿ °¯ ¬ We cannot apply the second translation theorem with
f t in this form,
since, according to the second translation theorem, for the first term we must have a function of
t 1
and for the second term a function of
§ S· ¨ t ¸. 2¹ © We can apply the second shifting property to the result above to obtain
L ^ f (t )` The function
Ss 2 s e L ^ f1 (t )` e 2 L ^ f 2 (t )` s
f t is thus rewritten as follows.
(1.13)
The Laplace Transform
t2 2 2
f1 (t 1) replace
S t2 f 2 (t ) cos t 2 2
t 1
t
replace t
2
t 3 t 2 2
f1 (t )
t
S 2
t2 S 2 S f 2 (t ) sin t t 2 8 2
then its Laplace transform is
then its Laplace transform is
1 1 3s s3 s 2 2
L ^ f1 (t )`
85
L ^ f 2 (t )`
1 1 sS 2 S s 2 1 s 3 8 2s 2
Eq. (1.13) becomes
L ^ f (t )`
2 1 3s · S s § 1 1 sS 2 S §1 e s ¨ 3 2 ¸ e 2 ¨ 2 3 2 s s 2 ¹ 8 2s ©s © s 1 s
Example 1.52. Obtain the Laplace transform of the piecewisecontinuous function shown in Figure 1.16 by expressing it in the unit step function
f(t) 1
t 1 Figure 1.16
2
3
4
· ¸. ¹
Chapter 1
86
Solution: The function is continuous, but it is defined differently on the intervals. As such, the mathematical expression for the function is
t ° f (t ) ®1 °0 ¯
0 d t d1 1 t d 4 tt4
In terms of the unit step function,
f (t )
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
f (t ) t 1 t u t 1 u t 4 . Taking the Laplace transform, we have
L ^ f (t )`
L ^t` L ^ 1 t u t 1 ` L ^u t 4 ` .
We cannot apply the second translation theorem with
f t in this form,
since, according to the second translation theorem, for the first term we must have a function of
t 1
and for the second term a function of
t 4 . We can apply the second shifting property to the above result to obtain
L ^ f (t )`
1 s e4 s e L ^ f1 (t )` s2 s
(1.14)
The Laplace Transform
The function
87
f t is thus rewritten as follows.
f1 (t 1) replace
t
1 t t 1
f1 (t ) t then its Laplace transform is
L ^ f1 (t )`
1 s2
Eq. (1.14) becomes
L ^ f (t )`
1 e4 s s 1 e s . s2
Example 1.53. Express in terms of the Heaviside unit step function
and find the Laplace transform of
f (t )
cos t 0 t d S ° ®cos 2t S t 2S . °cos 3t t t 2S ¯
Solution: In terms of unit step functions,
f (t )
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
Chapter 1
88
f (t ) cos t > cos 2t cos t @ u t S > cos 3t cos 2t @ u t 2S . Taking the Laplace transform, we have
L > f (t ) @
L > cos t @ L ^ sin t cos t u t S ` L ^ cos 3t cos 2t u t 2S `
f t in this form
We cannot apply the second translation theorem with
since, according to the second translation theorem, for the first term we must have a function of
t S
and for the second term a function of
t 2S . We can apply the second shifting property to the above result to obtain
L ^ f (t )` The function
f1 (t S ) replace
s eS s L ^ f1 (t )` e2S s L ^ f 2 (t )` s 1 2
(1.15)
f t is thus rewritten as follows.
cos 2t cos t
t t S
f 2 (t 2S ) replace
f1 (t ) cos 2t 2S cos t S f 2 (t ) f1 (t ) cos 2t cos t f (t ) 2
then its Laplace transform is
t
cos 3t cos 2t
t 2S
cos 3t cos 2t cos 3t cos 2t
then its Laplace transform is
The Laplace Transform
s s 2 s 4 s 1
L ^ f1 (t )`
2
L ^ f 2 (t )`
89
s s 2 s 9 s 4 2
Eq. (1.15) becomes
s 1 · s · § s § s e S s ¨ 2 2 ¸ e 2S s ¨ 2 2 ¸. s 1 © s 1 s 1¹ © s 9 s 4¹
L ^ f (t )`
2
Example 1.54. Express the signal shown in Figure 1.17 in terms of the Heaviside unit step function and find the Laplace transform.
f(t)
1
t–1
3 –t t
1
2
3
4
Figure 1.17 Solution: The mathematical expression for the above function is
Chapter 1
90
f t
0 d t 1 0 ° ° t 1 1 d t 2 ® ° 3 t 2 d t 3 °0 t t3 ¯
In terms of unit step functions,
f (t )
.
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
^ f t f t ` u t c 4
f (t )
3
t 1 u t 1 4 2t u t 2 t 3 u t 3 .
Then, taking Laplace transforms,
L ^ f (t )` L ^ t 1 u t 1 ` L ^ 4 2t u t 2 ` L ^ t 3 u t 3 `
We cannot apply the second translation theorem with
f t in this form,
since, according to the second translation theorem, for the first term we must have a function of
t 1 , for the second term a function of t 2 ,
and for the third term a function of
t 3 .
The Laplace Transform
91
We can apply the second shifting property to the above result to obtain
L ^ f (t )` e s L ^ f1 (t )` e2 s L ^ f 2 (t )` e3s L ^ f3 (t )` The function
f1 (t 1) replace
t
(1.16)
f t is thus rewritten as follows.
t 1 t 1
f 2 (t 2) replace
t
4 2t t2
f 2 (t ) 2t
f1 (t ) t
f3 (t 3) replace
t 3
t t 3
f 2 (t ) t
then its Laplace
then its Laplace
then its Laplace
transform is
transform is
transform is
L ^ f1 (t )`
1 s2
L ^ f 2 (t )`
2 s2
Eq. (1.16) becomes
L ^ f (t )`
1 s e 2e2 s e3s . 2 s
L ^ f3 (t )`
1 s2
Chapter 1
92
Example 1.55. Express the function shown in Figure 1.18 in terms of the Heaviside unit step function and find the Laplace transform.
Figure 1.17 Solution: The function is continuous, but is defined differently on the intervals. As such, the mathematical expression for the function is
0 ° f (t ) ® t 1 °4 ¯
0dt d2 2 t d3 t t3
In terms of the unit step function,
f t can be expressed as
f (t )
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
f (t )
t 1 u t 2 t 3 u t 3 .
The Laplace Transform
93
Taking the Laplace transform, we have
L ^ t 1 u t 2 ` t 3 L ^u t 3 ` .
L ^ f (t )`
We cannot apply the second translation theorem with
f t in this form,
since, according to the second translation theorem, for the first term we must have a function of
t 2
and for the second term a function of
t 3 . We can apply the second shifting property to the above result to obtain
L ^ f (t )` e2 s L > f1 (t )@ e3s L > f 2 (t )@ f t is thus rewritten as follows.
The function
f1 (t 2) replace
t
t 1 t2
f1 (t ) t 1 then its Laplace transform is
L ^ f1 (t )`
(1.17)
1 1 s2 s
Eq. (1.17) becomes
f 2 (t 3) replace
t 3
t t 3
f 2 (t ) t 6 then its Laplace transform is
L ^ f 2 (t )`
1 6 s2 s
Chapter 1
94
§ 1 1· § 1 6· L ^ f (t )` e2 s ¨ 2 ¸ e3s ¨ 2 ¸ . ©s s¹ ©s s¹ Example
1.56.
Evaluate
the
Laplace
transform
of
f (t ) sin E t u t T . Solution: sin E t u t T
Given f (t )
,
taking the Laplace transform,
L ^ f (t )` L ^sin E t u t T ` which, using the result in (1.7), gives
L ^ f (t )` eTs L ^ f1 (t )` f1 t T sin E t replace
t
t T
f1 (t ) sin E t E T f1 (t ) sin E t cos E T cos E t sin E T then its Laplace transform is
§ E · § s · L ^ f1 (t )` cos E T ¨ 2 sin E T ¨ 2 2 ¸ 2 ¸ ©s E ¹ ©s E ¹
(1.18)
The Laplace Transform
95
Eq. (1.18) becomes
L ^ f (t )`
§ E e Ts ®cos E T ¨ 2 2 ©s E ¯
· § ·½ s ¸ sin E T ¨ 2 2 ¸¾ ¹ © s E ¹¿
Thus,
L ^ f (t )` Example
f (t )
^E cos E T s sin E T ` e
s
1.57.
°°0 ® °sin 3t 2 °¯
2
E
2
Ts
Evaluate
the
.
Laplace
transform
2 3 . 2 tt 3 t
Solution: In terms of the unit step function,
f (t )
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a
§ 2· f (t ) sin 3t 2 u ¨ t ¸ . © 3¹ Taking the Laplace transform, we have
L ^ f (t )`
§ 2 ·½ L ®sin 3t 2 u ¨ t ¸ ¾ © 3 ¹¿ ¯
which, using the result in (1.7), gives
of
Chapter 1
96
L ^ f (t )` e
2 s 3
L > f1 (t )@
(1.19)
§ 2· f1 ¨ t ¸ sin 3t 2 © 3¹ replace t
t
2 3
f1 (t ) sin 3t then its Laplace transform is
L ^ f1 (t )`
3 s2 9
Eq. (1.19) becomes 2 s§ 3 · L ^ f (t )` e 3 ¨ 2 ¸. © s 9¹
Example 1.58. Consider the waveform given in Figure 1.19.
(a ) Write a mathematical expression for f (t). (b) Find the Laplace transform by expressing it in the Heaviside unit step function.
The Laplace Transform
97
f(t) 10 5 0
2
4
t
–5 Figure 1.19 Solution: The function is continuous, but is defined differently on the intervals. As such, the mathematical expression for the function is
5t ° f (t ) ®5 °0 ¯
0dt d2 2 t d4 tt4
In terms of unit step functions,
f (t )
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
f (t ) 5t 5 t 1 u t 2 5u t 4 Taking the Laplace transform, we have
L ^ f (t )` L ^5t` 5L ^ t 1 u t 2 ` 5L ^u t 4 `
Chapter 1
98
We cannot apply the second translation theorem with
f t in this form,
since, according to the second translation theorem, for the first term we must have a function of
t 2
and for the second term a function of
t 4 . We can apply the second shifting property to the above result to obtain
L ^ f (t )` The function
5 5e4 s 2 s 5 e L f ( t ) > @ 1 s2 s f t is thus rewritten as follows.
f1 (t 2) replace
t
t 1 t2
f1 (t ) t 3 then its Laplace transform is
L ^ f1 (t )`
1 3 s2 s
Eq. (1.20) becomes
L ^ f (t )`
5 3· e 4 s 2 s § 1 5 e 5 . ¨ 2 ¸ s2 s ©s s¹
(1.20)
The Laplace Transform
99
Example 1.59. Express in terms of the Heaviside unit step function and find the Laplace transform of
f (t )
t 1 t 1 , t t 0.
Solution: Given
Clearly
f (t )
t 1 t 1 , t t 0.
2 f (t ) ® ¯2t
0 d t d1 t !1
In terms of the unit step function,
f (t )
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a
f (t ) 2 > 2t 2@ u t 1 Taking the Laplace transform, we have
L ^ f (t )` L > 2@ L ^ 2t 2 u t 1 ` which, using the result in (1.7), gives
L ^ f (t )` The function
2 s e L ^ f1 (t )` s f t is thus rewritten as follows.
(1.21)
Chapter 1
100
f1 (t 1) replace
2t 2
t t 1
f1 (t ) 2t then its Laplace transform is
L ^ f1 (t )`
2 s2
Eq. (1.21) becomes
L ^ f (t )`
2 s § 2 · e ¨ 2 ¸. s ©s ¹
Thus,
L ^ f (t )`
2 s e s . 2 s
Example 1.60. Consider the waveform shown in Figure 1.20.
(a ) Write a mathematical expression for f (t). (b) Find the Laplace transform by expressing it in the Heaviside unit step function.
The Laplace Transform
101
f(t) 10
7
0
1
2
3
t
Figure 1.20 Solution: The function is continuous, but is defined differently on the intervals. As such, the mathematical expression for the function is
0 ° °10 t 1 f (t ) ® °7 °0 ¯
0 d t d1 1 t d 2 2 t d3 t !3
In terms of the unit step function,
f (t )
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
^ f t f t ` u t c 4
3
f (t ) 10 t 1 u t 1 17 10t u t 2 7u t 3 . Taking the Laplace transform, we have
Chapter 1
102
L ^10 t 1 u t 1 ` L ^ 17 10t u t 2 ` 7 L ^u t 3 `.
L > f (t ) @
We cannot apply the second translation theorem with
f t in this form,
since, according to the second translation theorem, for the first term we must have a function of
t 1 , for the second term a function of t 2
, and for the third term a function of
t 3 .
We can apply the second shifting property to the above result to obtain
7e3s e L ^ f1 (t )` e L ^ f 2 (t )` s
L ^ f (t )`
s
The function
f t is thus rewritten as follows.
f1 (t 1) 10 t 1 replace
f1 (t )
t 1
t
10t
then its Laplace transform is
L > f1 (t )@
2 s
10 s2
Eq. (1.22) becomes
f 2 (t 2) replace
t
17 10t t2
f 2 (t ) 10t 3 then its Laplace transform is
L > f 2 (t )@
10 3 s2 s
(1.22)
The Laplace Transform
L ^ f (t )`
103
10e s 2 s § 10 3 · 7e 3s . e ¨ 2 ¸ s2 s ©s s¹
Thus,
L ^ f (t )`
10 s 1 e e 2 s 3e 2 s 7e 3 s . 2 s s
Example 1.61. Transform
F s
1 e s 2e 2 s 2 se S s 2 into s2 s2 s s 1
the t-domain. Solution: Application of the second shifting theorem gives
f t L1 ^F s ` f t t t 2 u t 2 2u t 2 2cos t S u t S f t t t u t 2 2 cos t S u t S . Thus, the time domain function is
t ° f (t ) ®0 °2 cos t ¯
0dt d2 2 t dS . t !S
Example 1.62. Determine the Laplace transform of the Bessel functions
a f (t )
J 0 t b f (t )
J1 t . Solution:
Chapter 1
104
a We know that J0 t
t2 t4 t6 1 2 2 2 2 2 2 ........ 2 2 4 2 4 6
L ^ J 0 t ` L ^1`
1 1 1 L ^t 2 ` 2 2 L ^t 4 ` 2 2 2 L ^t 6 ` ........ 22 2 4 2 4 6
1 1 2! 1 4! 1 6! 2 3 2 2 5 2 2 2 7 ........ s 2 s 2 4 s 2 4 6 s
L ^J 0 t ` Clearly
L ^J 0 t `
1 § 1 § 1 · 1 3 § 1 · 1 3 5 § 1 · · 1 ¨ 2 ¸ ¨ 4 ¸ ¨ 6 ¸ ........ ¸ ¨ s © 2 © s ¹ 24 © s ¹ 246 © s ¹ ¹
L ^J 0 t `
1§ 1· ¨1 2 ¸ s© s ¹
1 2
.
Thus,
1
L ^J 0 t `
1 s2
b We have
.
J 0' t J1 t .
Therefore, by the derivative property,
L ^ J1 t ` Thus,
L ^ J 0' t `
^
`
sL ^ J 0 t ` 1
ª s º « 1» 2 ¬ 1 s ¼.
The Laplace Transform
§ s L ^ J1 t ` ¨1 1 s2 ©
105
· ¸. ¹
f
Example 1.63. Evaluate
³ e 1 cos 3t dt . 2 t
0
Solution: Using Table 1.1,
L ^ 1 cos 3t `
F s
1 1 . 2 s s 9
By definition of the Laplace transform, f
³ e 1 cos 3t dt 2 t
L ^1 cos 3t` s
2
0
1 ½ 1 ® 2 ¾ ¯s s 9¿ s
2
9 26
Thus, f
³ e 1 cos 3t dt 2 t
0
9 . 26
Summary In this chapter, we have described and explained the properties of Laplace transforms. x We started with a definition and description of Laplace transforms, including some of their standard functions: the constant function; the exponential function; the sine function; the cosine function; the hyperbolic function; the unit step; the delta function; and the ramp function. Some standard functions of the Laplace transform have
Chapter 1
106
been tabulated and several numerical examples have been solved based on the standard functions. x The properties of Laplace transforms have been stated and proved. These include linearity, first shifting, multiplication, and timescaling. Properties, such as time differentiation, time integration, and initial and final value theorem, have also been stated and explained. The Laplace transform has been evaluated in terms of its properties and illustrated using several solved examples. The property of periodicity has been proved and has been shown to be useful for finding the Laplace transform of some standard periodic functions. We have demonstrated the use of the property of periodicity to find the Laplace transform of any periodic function given its waveform. The physical significance of all these properties has been explained. The second shifting property has been proved, which is useful for finding the Laplace transform of discontinuous functions. We have illustrated various waveforms to find the Laplace transform of discontinuous functions using the second shifting property.
CHAPTER 2 THE INVERSE LAPLACE TRANSFORM
2.1 Introduction Virtually all operations have their inverses; for example, subtraction is the inverse of addition, division is the inverse of multiplication, and integration is the inverse of differentiation. The Laplace transform is no exception. We can define the inverse of the Laplace transform as follows: If
F s represents the Laplace transform of a function f t , that is,
L ^ f t ` F s , we then say f t is the inverse Laplace transform of
F s and write f t
L1 ^ F s ` .
In this chapter, we will mainly restrict ourselves to describing different techniques for finding the inverse Laplace transform. In Section 2.2, we derive the relation between Laplace and Fourier transforms. The properties of the inverse Laplace transform, such as linearity, shifting, differentiation, and integration, are treated in sections 2.3 to 2.6. In Section 2.7, we consider the procedure for finding the inverse Laplace transform by the partial fractions method. In Section 2.8, we define the convolution between two functions to find the inverse Laplace transform of the product of two functions in the
s
domain by applying the convolution theorem.
Chapter 2
108
LEARNING OBJECTIVES After studying this chapter, it is expected that you: x Have understood and are able to apply the definition of the inverse Laplace transform. x Can find the inverse Laplace transform of complex functions by using the table to apply the transform’s properties and are able to apply the method of partial fraction expansion. x Can find the inverse Laplace transform using the Bromwich integral formula. x Know and can apply the convolution between two functions and the convolution theorem. x Know and can apply the convolution to solve the integral equation.
2.2 The Inversion Integral for the Laplace Transform When applying the Laplace transform to most practical problems and obtaining the transform
F s of the required result, it is usually possible
to find the required inverse transform
f t by using Table 2.1. This
table presents Laplace transform pairs together with the operational properties listed in Table 1.2 (Chapter 1). As tables of transform pairs do not always contain the required inverse Laplace transform and
s
must be
allowed to be complex, some other method must be found by which to determine
f t
L1 ^ F s ` .
The Inverse Laplace Transform
The method we derive here shows that transform
109
f t possesses a Laplace
F s so that f
F s
³e
st
f t d t
f
where
s
(2.1)
can be complex and
transform
f t can be recovered from its Laplace
F s by means of the complex line integral V jf
1 e st F s ds. ³ 2S V jf
f t where
(2.2)
c ! 0 is a suitable real constant. The formula in (2.2) is called the
inversion integral of the Laplace transform
F s . To establish the
result in (2.2), we derive the relationship between the Fourier transform and the Laplace transform.
2.2.1 Relationship Between Laplace Transforms and Fourier Transforms When the complex variable s is purely imaginary, i.e. s
j: , Eq. (2.1)
becomes
F j:
F ^ f t `
f
³e
j:t
f t d t
f
Eq. (2.3) is the Fourier transform of
(2.3)
f t
Chapter 2
110
F s s
F ^ f t ` .
j:
If s is not purely imaginary, i.e. s f
F V j :
³e
V j : t
V j:, Eq. (2.3) can be written as
f t d t
f
(2.4)
f
V ³ e ^ e f t `
F V j :
j:t
t
dt
f
(2.5)
The right-hand side of Eq. (2.5) is the Fourier transform of
e V t f t .
Thus, the Laplace transform can be interpreted as the Fourier transform of
f t after multiplication by a real exponential signal e V t . From Eq. (2.5), that the Laplace transform
F V j: of a signal
f t is given by
^
f
` ³ e ^e
F e V t f t
j:t
V t
`
f t d t .
f
Applying the inverse Fourier transform to the above relationship, we obtain
^
`
eV t f t F 1 ¬ª F eV t f t ¼º F 1 ¬ª F V j: ¼º
1 2S
f
³e
j:t
F V j : d :
f
(2.6) Multiplying both sides of Eq. (2.6) by e
V t
it follows that
The Inverse Laplace Transform
f t As s
1 2S
f
³e
V j : t
111
F V j : d :
f
(2.7)
V j: and V is a constant, ds
jd : . Substituting these
values in Eq. (2.7) and changing the variable of integration from
s
to :,
we arrive at the following inverse Laplace transform V jf
f t
1 e st F s ds. ³ 2S j V jf
(2.8)
Equation (2.8), our inverse transformation, is usually known as the Bromwich integral, although it is also referred to the Fourier–Mellin theorem or the Fourier Mellin integral. This integral is evaluated by applying the standard methods of contour integration. If t ! 0, the contour may be closed by an infinite semicircle in the left half-plane. Then, by the residual theorem, V jf
f t
1 e st F s ds ³ 2S j V jf
¦ sum of residues .
In the next section, we will examine a direct method for finding inverse Laplace transforms based on the theory of functions of a complex variable. In the meantime, we shall be content with using tables, which is the simplest method for most of the applications. However, to use the tables, we require a mathematical technique to resolve transforms into the listed forms.
Chapter 2
112
2.2.2 Inversion Using the Bromwich Integral Formula
Example 2.1. Determine
ª a L1 « 2 «¬ s a 2
º » using the Bromwich »¼
integral formula. Solution: Let
F s
e st F s The poles
a s a2 2
a
s a s a
, then
ae st , s a s a
F s are s
Re s ^ a `
a &s
^
a . These are both simple poles. ae st ½ lim ® ¾ s oa ¯s a¿
`
lim e st F s s a s oa
Re s ^ a `
^
`
lim e st F s s a
s o a
e at 2
e st a ½ lim ® ¾ s o a s a ¯ ¿
e at . 2
Using the Bromwich integral V jf
f t
1 e st F s ds ³ 2S j V jf
1
¦ sum of residues 2 e
at
e at
sinh at.
To further progress with the inverse Laplace transform, it is necessary to have a table of inverse Laplace transform pairs for the most commonly
The Inverse Laplace Transform
113
occurring functions. Table 2.1 provides a list of the most useful inverse Laplace transform pairs involving elementary functions. Table 2.1 Inverse Laplace Transform Pairs
F ( s)
L^ f (t )`
Transform
L ^1`
f (t ) L1 ^F ( s)` Inverse Transform
1 s
1 ½ 1 L1 ® ¾ ¯s¿
^ `
n! s n 1
t n 1 1 ½ L1 ® n ¾ n 1 ! ¯ s ¿
^ `
1 sa
eat
L tn
L e at
1 ½ L1 ® ¾ ¯s a¿
L ^sinh at`
a s a2
a ½ sinh at L1 ® 2 2 ¾ ¯s a ¿
L ^cosh at`
s s a2
cosh at
2
2
s ½ L1 ® 2 2 ¾ ¯s a ¿
L ^sin at`
a s a2
sin at
a ½ L1 ® 2 2 ¾ ¯s a ¿
L ^cos at`
s s a2
cos at
s ½ L1 ® 2 2 ¾ ¯s a ¿
2
2
Chapter 2
114
L ^u t `
1 s
1 ½ u t L1 ® ¾ ¯s¿
e as s
L ^u t a `
u t a
e as ½ L1 ® ¾ ¯ s ¿
L ^G t ` 1
G t L1 ^1`
L ^G t a ` e as
G t a
^
L e at t n
`
n!
s a
^
`
^
`
L eat cos bt
L e at sin bt
^ `
L1 e as
° 1 ½° at t n 1 L1 ® e n ¾ n 1 ! °¯ s a °¿
n 1
s a 2 s a b2
° s a ½° at L1 ® ¾ e cos bt 2 2 °¯ s a b °¿
b
° ½° at b L1 ® ¾ e sin bt 2 2 °¯ s a b °¿
s a
2
b
2
2.3 The Inverse Laplace Transform is a Linear Transform The linearity property for the Laplace transform (Property 1.5.1) states that if C1 and C2 are any constants, then
L ^C1 f t C2 g t ` C1 L ^ f t ` C2 L ^ g t ` C1 F s C2G s . It follows from the above definition that
The Inverse Laplace Transform
115
L1 ^C1 F s C2G s ` C1 f t C2 g t C1 L1 ^F s ` C2 L1 ^G s `
L1 is also a linear operator. For example, if we consider an arbitrary function of s , there is no guarantee that we find a function of t , i.e. an inverse Laplace transform of t . One important condition is that the function of s must tend to zero as s o f. When we are sure that a function of s has arisen so that the inverse Laplace transform operator
from a Laplace transform, we can use certain techniques and theorems to help us invert it. Partial fractions simplify rational functions and can help identify standard forms, for example, the exponential, hyperbolic, and trigonometric functions. The second shift of the differentiation and integration theorems that we have come across previously further extend the repertoire of standard forms.
Example 2.2. Determine
ª 2s 9 (i ) L1 « 2 «¬ s 9
º ª Ds E » (ii ) L1 « 2 »¼ «¬ s O 2
º ». »¼
Solution:
(i ) We first rewrite the given function of s as two expressions by means of term by term division and get ª 2s 9 L1 « 2 «¬ s 9
º » »¼
ª 2s 9 2 L1 « 2 s 9 «¬ s 9
º » »¼
ª s 2 L1 « 2 «¬ s 9
º ª 3 » 3L1 « 2 »¼ «¬ s 9
Applying the inverse Laplace transform to the above equation, and using its linearity, we get the following result
º ». »¼
Chapter 2
116
ª 2s 9 L1 « 2 «¬ s 9
ª Ds E L1 « 2 2 «¬ s O
º » 2 cos 3t 3sin 3t. »¼
º ª Ds E » L1 « 2 2 2 2 s O »¼ «¬ s O
(ii )
º ª s » D L1 « 2 2 »¼ «¬ s O
º E ª O » L1 « 2 2 »¼ O «¬ s O
Applying the inverse Laplace transform to the above equation, and using its linearity, we get the following result
ª Ds E L1 « 2 «¬ s O 2
º E » D cos Ot sin Ot. O »¼
2.4 Inversion Using the First Shift Theorem Property 1.5.3 shows saw that, if
F s is the Laplace transform of
f t , then, for a scalar a,
^F s a `
is the Laplace transform of
e at f t . Expressed in the
inverse form, the theorem becomes
L1 ^ F s a ` e at f t . ª 3 Example 2.3. Determine L « « s 2 2 9 ¬« 1
º ». » ¼»
º » »¼
The Inverse Laplace Transform
117
Solution:
ª 3 « 2 «¬ s 9
3
s 2 9 2
since
3 s2 9
º » »¼ sos 2
L ^sin 3t` , the shift theorem gives
3 ° L ® 2 ° s 3 9 ¯ 1
½ ° ¾ ° ¿
e 2t sin 3t.
ª s2 L1 « 2 «¬ s 2 s 5
Example 2.4. Determine
º ». »¼
Solution: Observe that
s2 s 2s 5 2
ª s « 2 «¬ s 4
s 1 s 1 2 4 since
s s 4 2
s 1 3 s 1 2 4
s 1 3 2 s 1 4 s 1 2 4
º » and »¼ s o s 1
L ^cos 2t` and
1
ª 1 « 2 «¬ s 4
s 1 4 2
1 s 4 2
1 L ^cos 2t` . 2
º » »¼ s o s 1
Chapter 2
118
Applying the inverse Laplace transform to the above equation, and using its linearity, we get the following result
ª s2 L1 « 2 «¬ s 2 s 5
ª s2 L1 « 2 «¬ s 2s 5
º » »¼
° s 1 L1 ® 2 ° s 1 4 ¯
½ 3 ° 1 ° L ¾ ® 2 ° ° s 1 4 ¿ ¯
½ ° ¾ ° ¿
º 3 » e t cos 2t e t sin 2t. 2 »¼
Example 2.5. Determine
ª 5s 2 L1 « 2 «¬ s 6 s 13
º » »¼
Solution: First, we complete the square to get
s
2
6 s 13
s 3
2
4 .
Observe that
5s 2 s 6 s 13 2
5 s 3 13
s 3
2
4
5
s 3 13 1 2 s 3 4 s 3 2 4
Applying the inverse Laplace transform to the above equation, and using its
linearity,
we
now
use
the
L1 ^ F s a ` e at f t , to conclude
frequency
shift
theorem
The Inverse Laplace Transform
ª 5s 2 L1 « 2 «¬ s 6s 13
119
º 13 » 5e 3t cos 2t e 3t sin 2t. 2 »¼
2.5 Inverse Transform by Differentiation and Integration Given the function
F s , its inverse Laplace transform f t may be g t of G s
formed by evaluating the inverse transform
that
F s
dg t
sG s and f t
dt
F s s
, so
. Similarly, given the function
F s , its inverse transform may be found by evaluating the inverse g t
of
G s sF s ,
so
that
F s
G s s
t
f t
³ g t dt. 0
f t given that F s
Example 2.6. Evaluate
2a 2 . s s 2 4a 2
Solution:
Let
G s
sF s
2a 2 s 2 4a 2
a
2a
s 2a 2
Taking the inverse Laplace transform, we get
2
.
and
Chapter 2
120
g t a sinh 2at then, t
f t
t
³ g t dt ³ a sinh 2at dt 0
a cosh 2at
0
2a
t
t 0
1 cosh 2at 1 2
Thus,
f t
1 cosh 2at 1 . 2
Example 2.7. Evaluate
§ sa· f t given that F s log e ¨ ¸. © s b ¹
Solution: Let
F s log e s a log e s b
F ' s
1 1 s a s a 1 1 . s a s a
F ' s
Applying the inverse Laplace transform to the above equation, and using its linearity, gives us the following result.
^
`
L1 F ' s
° 1 °½ 1 ° 1 °½ L1 ® ¾ L ® ¾ °¯ s a °¿ °¯ s a °¿
The Inverse Laplace Transform
^
`
L1 F ' s
We have seen that
121
e at ebt .
L ^t f t `
dF s ds
F ' s
then,
tf t e at ebt . Thus,
f t
eat ebt . t
Example 2.8. Evaluate
f t given that F s
Solution:
§1· tan 1 ¨ ¸ ©s¹
Let F s
F ' s
1 s 1
F ' s
1 . s 1
2
2
Taking the inverse Laplace transform, we get
^
`
L1 F ' s
1 ½ L1 ® 2 ¾ ¯ s 1¿
§1· tan 1 ¨ ¸ . ©s¹
Chapter 2
122
^
`
L1 F ' s
sin t .
We have seen that
L ^t f t `
dF s ds
F ' s
then,
tf t sin t Thus,
f t
sin t . t
Example 2.9. Evaluate
§ s2 a2 · f t given that F s log e ¨ ¸. 2 © s ¹
Solution: Let
F s log e s 2 a 2 log e s 2
F ' s
F ' s
2s 2 2 s s a
2
2 2s 2 s s a 2
Applying the inverse Laplace transform to the above equation, and using its linearity, we get the following result
The Inverse Laplace Transform
^
`
° s 2 L1 ® 2 2 °¯ s a
^
`
2 cos at 2 .
L1 F ' s
L1 F ' s
L ^t f t `
We have seen that
123
½° 1 1 ½ ¾ 2L ® ¾ ¯s¿ °¿
dF s ds
F ' s
then,
t f t 2 1 cos at . Thus,
f t
2 1 cos at t
Example 2.10. Evaluate
.
f t given that F s
Solution: Let
F s log e s 2 E 2 log e s 2
F ' s
2s 2 2 s E s 2
F ' s
2s 2 . 2 s E s 2
§ s2 E 2 · log e ¨ ¸. 2 © s ¹
Chapter 2
124
Taking the inverse Laplace transform, we get
^
`
2 s ½ 1 2 ½ L1 ® 2 L ® ¾ 2¾ ¯s¿ ¯s E ¿
^
`
2 cos E t 2 .
L1 F ' s
L1 F ' s
L ^t f t `
We have seen that
dF s ds
F ' s
then,
tf t 2 cos E t 2 . Thus,
f t
2 1 cos E t t
Example 2.11. Determine
ª 1 º L1 « ». ¬ s s 1 ¼
Solution: We have
1 ½ L® ¾ ¯ t¿
1 2
½ L ®t ¾ ¯ ¿
Its inversion is
§1· *¨ ¸ ©2¹ s1 2
S
S
12
s
s
.
The Inverse Laplace Transform
1 ½ L1 ® ¾ ¯ s¿
125
1 . St
As such, the shift theorem gives
et
1 ½ L1 ® ¾ ¯ s 1 ¿
.
St
To complete the inversion process, we now make use of the Laplace transform of an integral
F s ½ L ® ¾ ¯ s ¿ 1
t
³ f W dW . 0
Using this result with
ª 1 º L « » ¬ s s 1 ¼
1 ½ L1 ® ¾ gives ¯ s 1 ¿
1
1
S
t
e W
³
SW
0
x 2 converts this to
The change of variable W
ª 1 º L « » ¬ s s 1 ¼
2 S
1
t
³e
erf x
S
x
³e 0
W 2
x2
dx,
0
however, the error function
2
dW .
dW
erf x is given by
Chapter 2
126
so
ª 1 º L1 « » ¬ s s 1 ¼
erf
t .
2.6 Inversion Using the Second Shift Theorem This theorem plays a vital role in determining inverse transforms. As indicated earlier, shifting is inherent in most practical systems and engineers are interested in knowing how they respond. In theorem 1.7, we saw that if
F s is the Laplace transform of f t , then, for a scalar a,
^e
`
as
F s is the Laplace transform of f t a u t a . Expressed
in the inverse form, the theorem becomes
^
`
L1 e as F s
f t a u t a . ª
§
s 1 2 2 ¨ s 1 © s 4
1 S s Example 2.12. Determine a L « e ¨
« ¬
ª
1 «¬ s 1 e s
b L1 «
º ». »¼
Solution:
a This may be written as L1 ª¬eS s F s º¼ , where F s
s 1 2 . s 4 s 1 2
First, we obtain the inverse transform
f t of F s
·º ¸» ¸» ¹¼
The Inverse Laplace Transform
127
f t cos t sin t Then, using the theorem, we have
ª § s 1 L1 «eS s ¨ 2 2 « ¨ s 4 s 1 ¬ ©
b Let F s Here,
·º ¸ » L1 ª¬eS s F s º¼ ¸» ¹¼
^cos t S sin t S ` u t S .
1 . s 1 e s
1 can be interpreted as the sum of a geometric series with 1 e s
the common ratio e
1 1 e s
s
, so that we may write
1 e s e 2 s e 3 s e 4 s ........
F s
1 ª¬1 e s e2 s e3s e4 s ........º¼ s
F s
1 e s e2 s e3s e4 s ...... s s s s s
Applying the inverse Laplace transform to the above equation, and using its linearity, we get the following result
f t 1 u t 1 u t 2 u t 3 u t 4 ...... Thus, the function can be written as follows (its graph is shown in Figure 2.1
Chapter 2
128
f(t)
f t
1 0 t d 1 °2 1 t d 2 3 ° °3 2 t d 3 ® 2 °4 3 t d 4 °..................... 1 ° ¯....................... 0
1
2
3
Figure 2.1
2.7 The Inverse Laplace Transform Using Partial Fractions Integral calculus shows us how to integrate rational functions by using partial fraction decomposition. In order to efficiently derive inverse Laplace transforms of rational functions using a table of Laplace transforms, we must be conversant with a variety of partial fraction decompositions. Partial fractions play an important role in deriving inverse Laplace transforms. Partial fractions provide a method for the decomposition of a rational expression into a sum of simple rational functions. In this section, we will review some of the basic algebra for the
The Inverse Laplace Transform
important cases in which the denominator of a Laplace transform
129
F s
contains distinct linear factors, repeated linear factors, and quadratic polynomials. In many cases,
F s is the quotient of two polynomials
with real coefficients. If the numerator polynomial is of the same or higher degree than the denominator polynomial, then we divide the numerator polynomial by the denominator polynomial. We continue the division until the numerator polynomial of the remainder is one degree less than the denominator. This results in a polynomial in
s
and a proper fraction. The
resulting proper fraction can be expanded into a partial fraction. The procedure for finding a Laplace transform, given the function in the frequency domain, can be stated as follows: Step 1: Factorize the denominator to get the roots of the function. Step 2: Use partial fraction expansion. Step 3: Find the inverse Laplace transform of each term. The following examples illustrate partial fraction decomposition when the denominator of
F s is factorable into: (i) distinct linear factors; (ii)
repeated linear factors; and (iii) quadratic polynomials.
2.7.1 Partial Fractions: Distinct Linear Factors If
D s can be factored into a product of distinct linear factors, then
D s
s s1 s s2 s s3 ......... s si ,
Chapter 2
130
where
si , represents a distinct real number. As such, the partial fraction
expansion has the form
N s
A3 Ai A1 A2 , ......... s s1 s s2 s s3 s si
D s where
Ai , represents a real number. There are various ways of
A1 , A2 , A3 ,...... Ai .
determining the constants
In the next example, we demonstrate three such methods. Example 2.13. Determine
where F s
L1 ª¬ F s º¼ ,
2 s 2 9 s 11 . s 2 s 3 s 1
Solution: Method 1: Cover-up Method. The denominator is already in the factored form completing step 1. In step 2, we use partial fraction expansion and decompose the function into three terms. We begin by finding the partial fraction expansion for
F s . The
denominator consists of three distinct linear factors. The expansion has the form
2 s 2 9 s 11 s 2 s 3 s 1
A B C s 2 s 3 s 1
The Inverse Laplace Transform
131
where A, B, and C are real numbers to be determined.
A
B
C
ª 2 s 2 9 s 11 s 2 º « » 1 , «¬ s 2 s 3 s 1 »¼ s 2 2 ª 2 s 9 s 11 s 3 º « » 2, «¬ s 2 s 3 s 1 »¼ s 3
ª 2 s 2 9 s 11 º « » ¬« s 2 s 3 ¼» s
3 . 1
Hence, the given function
F s can be expanded as
2 s 2 9 s 11 s 2 s 3 s 1
1 2 3 s 2 s 3 s 1
In step 3, we find the inverse Laplace transform of each term. Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
° 2 s 2 9s 11 ½° 1 ° 1 ½° 1 ½° 1 ½° 1 ° 1 ° L1 ® ¾ L ® ¾ 2L ® ¾ 3L ® ¾ °¯ s 2 s 3 s 1 °¿ °¯ s 2 °¿ °¯ s 3 °¿ °¯ s 1 °¿ Finally, using the transform pairs established in Table 2.1, we have
ª 2s 2 9s 11 º 2t 3t t L « » e 2e 3e . ¬« s 2 s 3 s 1 ¼» 1
Chapter 2
132
Method 2: Classical Method. The denominator is already in the factored form, completing step 1. In step 2, we use partial fraction expansion and decompose the function into three terms. We begin by finding the partial fraction expansion for
F s . The
denominator consists of three distinct linear factors and so the expansion has the form
2 s 2 9 s 11 s 2 s 3 s 1
A B C s 2 s 3 s 1
where A, B, and C are real numbers to be determined. In this procedure, which works for all partial fraction expansions, we first multiply the expansion equation by the given rational function’s denominator. This leaves us with two identical polynomials. Equating the coefficients of s
k
leads to a system of linear equations, which we can
solve to determine the unknown constants. In this example, we multiply the above equation by
s 2 s 3 s 1
and find
2s 2 9s 11 A s 3 s 1 B s 2 s 1 C s 2 s 3 . Equating the coefficients of s2, linear equations
A B C
2 ,
4A B C
9 ,
s,
and 1 gives the following system of
The Inverse Laplace Transform
3 A 2 B 6C
133
11 ,
solving this system yields A
1 , B
2, and C
Hence, the given function
F s can be expanded as
2 s 2 9 s 11 s 2 s 3 s 1
1 2 3 s 2 s 3 s 1
3.
In step 3, we find the inverse Laplace transform of each term. Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
° 2 s 2 9 s 11 ½° 1 ° 1 ½° 1 ½° 1 ½° 1 ° 1 ° L1 ® ¾ L ® ¾ 2L ® ¾ 3L ® ¾ °¯ s 2 s 3 s 1 °¿ °¯ s 2 °¿ °¯ s 3 °¿ °¯ s 1 °¿ Finally, using the transform pairs established in Table 2.1, we have
ª 2s 2 9s 11 º 2t 3t t L1 « » e 2e 3e . «¬ s 2 s 3 s 1 »¼ Method 3: Alternative Method. The denominator is already in the factored form, completing step 1. In step 2, we will use partial fraction expansion and decompose the function into three terms. Although the method of comparing coefficients is direct, it can also be laborious. A more efficient alternative, especially for fractions having
Chapter 2
134
distinct linear factors in the denominator, is based on the fact that if two polynomials of degree
k are equal for more than k replacements of the
variable, they are identical for all values of the variable. We begin by finding the partial fraction expansion for
F s . The
denominator consists of three distinct linear factors and so the expansion has the form
2 s 2 9 s 11 s 2 s 3 s 1
A B C s 2 s 3 s 1
where A, B, and C are real numbers to be determined. We have
2s 2 9s 11 A s 3 s 1 B s 2 s 1 C s 2 s 3 In this method for finding the constants A, B, and C from the above equation, we choose three values for
s
and substitute them into the above
equation to obtain three linear equations in the three unknowns. If we are careful in our choice of values for s, the system is easy to solve. In this case, the above equation is simplified if s gives
15 A
15 A
Next, using
10 B
s
1 .
3 gives
20 B
2.
2, 3 or -1. Using s
2
The Inverse Laplace Transform
Finally, using
6C
s
18 C
135
1 gives 3 .
As such, the given function
2 s 2 9 s 11 s 2 s 3 s 1
F s
can be expanded as
1 2 3 . s 2 s 3 s 1
In step 3, we find the inverse Laplace transform of each term. Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that 2 1 °½ 1 °½ ° 2s 9s 11 °½ 1 ° 1 °½ 1 ° 1 ° L1 ® ¾ L ® ¾ 2L ® ¾ 3L ® ¾ °¯ s 2 ¿° ¯° s 2 s 3 s 1 ¿° ¯° s 3 ¿° ¯° s 1 ¿°
Finally, using the transform pairs established in Table 2.1, we have
ª 2s2 9s 11 º 3t t 2t L « » e 2e 3e . «¬ s 2 s 3 s 1 »¼ 1
Chapter 2
136
L1 ª¬ F s º¼ ,
Example 2.14. Determine
s2 1 . s s 1 s 1 s 2
where F s Solution:
Method 1: Cover-up Method. The denominator is already in the factored form, completing step 1. In step 2, we use partial fraction expansion and decompose the function in four terms. We begin by finding the partial fraction expansion for
F s . The
denominator consists of four distinct linear factors and so the expansion has the form
s2 1 s s 1 s 1 s 2
A B C D s s 1 s 1 s 2
where A, B, C, and D are real numbers to be determined.
ª º s2 1 A « » ¬« s 1 s 1 s 2 ¼» s ª º s 1 B « » ¬« s s 1 s 2 ¼» s
0
1 , 2
2
ª º s 1 « » ¬« s 1 s s 1 ¼» s
1, C 1
2
D
2
5 . 6
ª º s2 1 « » ¬« s 1 s s 2 ¼» s
1
1 , 3
The Inverse Laplace Transform
Hence, the given function
s2 1 s s 1 s 1 s 2
F s
137
can be expanded as
11 1 1 1 5 1 2 s s 1 3 s 1 6 s 2
In step 3, we find the inverse Laplace transform of each term. Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
° ½° 1 1 1½ 1 ° 1 ½° 1 1 ° 1 ½° 5 1 ° 1 ½° s2 1 L1 ® ¾ L ® ¾ L ® ¾ L ® ¾ L ® ¾ ¯°s s 1 s 1 s 2 ¿° 2 ¯s ¿ ¯° s 1 ¿° 3 ¯° s 1 ¿° 6 ¯° s 2 ¿° . Finally, using the transform pairs established in Table 2.1, we have
° ½° 1 t 1 t 5 2t s2 1 L1 ® e e e . ¾ 3 6 °¯ s s 1 s 1 s 2 °¿ 2 Method 2: Classical Method. The denominator is already in the factored form, completing step 1. In step 2, we use partial fraction expansion and decompose the function into three terms. We begin by finding the partial fraction expansion for
F s .
The
denominator consists of three distinct linear factors and so the expansion has the form
Chapter 2
138
s2 1 s s 1 s 1 s 2
A B C D s s 1 s 1 s 2
where A, B, C, and D are real numbers to be determined. In this procedure, which works for all partial fractions, we first multiply the expansion equation by the given rational function’s denominator. This leaves us with two identical polynomials. Equating the coefficients of s k leads to a system of linear equations that we can solve to determine the unknown constants. In this example, we multiply the above equation by
s s 1 s 1 s 2
and find
s2 1 A s 1 s 1 s 2 Bs s 1 s 2 Cs s 1 s 2 Ds s 1 s 1 . Equating the coefficients of s3, s2,
s, and 1 gives the following system of
linear equations
A BC D
0 ,
2 A B 3C
1 ,
A 2 B 2C D 2A
0,
1 ,
solving this system yields A
Hence, the given function
1 , B 2
F s
1,
C
can be expanded as
1 , and D 3
5 . 6
The Inverse Laplace Transform
s2 1 s s 1 s 1 s 2
139
11 1 1 1 5 1 . 2 s s 1 3 s 1 6 s 2
In step 3, we find the inverse Laplace transform of each term. Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
s2 1 ° °½ 1 1 1½ 1 ° 1 °½ 1 1 ° 1 °½ 5 1 ° 1 °½ L1 ® ¾ L ® ¾ L ® ¾ L ® ¾ L ® ¾ ¯°s s 1 s 1 s 2 ¿° 2 ¯s ¿ ¯° s 1 ¿° 3 ¯° s 1 ¿° 6 ¯° s 2 ¿° . Finally, using the transform pairs established in Table 2.1, we have
° ½° 1 t 1 t 5 2t s2 1 L ® e e e . ¾ 3 6 °¯ s s 1 s 1 s 2 °¿ 2 1
Method 3: Alternative Method. The denominator is already in the factored form, completing step 1. In step 2, we use partial fraction expansion and decompose the function into three terms. Although the method of comparing coefficients is direct, it can also be laborious. A more efficient alternative, especially for fractions having distinct linear factors in the denominator, is based on the fact that if two polynomials of degree
k are equal for more than k replacements for the
variable, they are similar for all values of the variable.
Chapter 2
140
We begin by finding the partial fraction expansion for
F s . The
denominator consists of three distinct linear factors and so the expansion has the form
s2 1 s s 1 s 1 s 2
A B C D s s 1 s 1 s 2
where A, B, C, and D are real numbers to be determined. We have
s 2 1 A s 1 s 1 s 2 Bs s 1 s 2 Cs s 1 s 2 Ds s 1 s 1 . To find the constants A, B, C, and D from the above equation, we choose
s
three values for
and substitute them into the above equation to obtain
three linear equations for the three unknowns. If we are careful in our choice of values for s, the system is easy to solve. In this case, the above equation is clearly simplified if s Using
s 0 gives 1 2
2A 1 A Next, using
2 B
2 B
Next, using
6C
s 1 gives
s
1
1 gives
2 C
1 3
0,1, 1 , or 2.
The Inverse Laplace Transform
Finally, using
6D
s
5 D
141
2 gives 5 . 6
Hence, the given function
s2 1 s s 1 s 1 s 2
F s can be expanded as 11 1 1 1 5 1 . 2 s s 1 3 s 1 6 s 2
In step 3, we find the inverse Laplace transform of each term. Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
s2 1 ° °½ 1 1 1 ½ 1 ° 1 °½ 1 1 ° 1 °½ 5 1 ° 1 °½ L1 ® ¾ L ® ¾ L ® ¾ L ® ¾ L ® ¾ ¯° s s 1 s 1 s 2 ¿° 2 ¯ s ¿ ¯° s 1 ¿° 3 ¯° s 1 ¿° 6 ¯° s 2 ¿° . Finally, using the transform pairs established in Table 2.1, we have
° ½° 1 t 1 t 5 2t s2 1 L ® e e e . ¾ 3 6 °¯ s s 1 s 1 s 2 °¿ 2 1
Chapter 2
142
Example 2.15. Find the inverse Laplace transform of
a F s
s 2 6s 5 s 2 3s 2
b F s
s 3 5s 2 9 s 7 . s 2 3s 2
Solution: All these problems are tackled in a similar fashion by decomposing the expression into partial fractions then identifying the simplified expressions with various standard forms.
(a) Step 1. The denominator is in the factored form. Note that the degree of the numerator is the same as the degree of the denominator. So, we have to bring it in the fraction form to apply partial fraction expansion. In many cases,
F s is the quotient of two polynomials with real
coefficients. If the numerator polynomial is of the same or higher degree than the denominator polynomial, we divide the numerator polynomial by the denominator polynomial. The division is carried forward until the numerator polynomial of the remainder is one degree less than the denominator. This results in a polynomial in
s
plus a proper fraction. The
proper fraction can be expanded into a partial fraction.
s 2 6s 5 3s 5 . 1 2 s 3s 2 s 1 s 2 In step 2, we use partial fraction expansion and decompose the function into three terms. The proper fraction is expanded into partial fraction form
The Inverse Laplace Transform
143
s 2 6s 5 A B 1 2 s 3s 2 s 1 s 2 where A and B are real numbers to be determined.
ª 3s 5 º A « » ¬« s 2 ¼» s
2 , 1
Hence, the given function
s 2 6s 5 s 2 3s 2
B
1
ª 3s 5 º « » ¬« s 1 ¼» s
1. 2
F s can be expanded as
2 1 s 1 s 2
In step 3, we find the inverse Laplace transform of each term. Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
s 2 6s 5 ½ L1 ® 2 ¾ ¯ s 3s 2 ¿
° 1 °½ 1 ° 1 °½ L1 ^1` 2 L1 ® ¾ L ® ¾. °¯ s 1 °¿ °¯ s 2 °¿
Finally, using the transform pairs established in Table 2.1, we have
f t G t 2et e2t . (b) Here, since the degree of the numerator polynomial is higher than that of the denominator polynomial, we must divide the numerator by the denominator
F s
s 3 5s 2 9 s 7 s 2 3s 2
s2
s3 . s 1 s 2
Chapter 2
144
We use partial fraction expansion of the second term of the right-hand side. The proper fraction is expanded into partial fraction form
F1 s
s3 s 1 s 2
A B s 1 s 2
where A and B are real numbers to be determined.
ª s 3 º A « » «¬ s 2 »¼ s
1
Hence, the given function
F s
ª s 3 º B « » «¬ s 1 »¼ s
2 ,
F s
s 3 5s 2 9 s 7 s 2 3s 2
1 2
.
can be expanded as
s2
2 1 s 1 s 2
Note that the Laplace transform of the unit-impulse function unity and the Laplace transform of
dG t dt
d f t ½ i.e L ® d G t ½¾ s. L ® ¾ s F s . ¯ dt ¿ ¯ dt ¿
Thus,
f t
dG t dt
2G t 2et e2t .
Example 2.16. Find the Inverse LT of
F s
s2 s 1 . s 2 2s 1
G t is
a
The Inverse Laplace Transform
145
Solution: Note that the degree of the numerator is the same as the degree of the denominator. We have to put it in the form of a fraction and divide the numerator polynomial by the denominator polynomial using simple long division. This allows us to apply partial fraction expansion.
s2 s 1 3s 1 s2 2s 1 s2 2s 1 ª s 1 1º 3 3 s2 s 1 . 1 3 « » 1 2 2 s 2s 1 s 1 s 1 2 «¬ s 1 »¼ Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
s2 s 1 ½ L1 ® 2 ¾ ¯ s 2s 1¿
1 °½ ° 1 ½° 1 ° . L1 ^1` 3L1 ® ¾ 3L ® 2¾ °¯ s 1 °¿ ¯° s 1 ¿°
Finally, using the transform pairs established in Table 2.1, we have
f t G t 3et 3tet . Example 2.17. Determine L ª¬ F s º¼ , where 1
F s
ª º 3s 2 7 « 4 ». 3 2 ¬ s 2s s 2s ¼
Solution: In step 1, we have the denominator in the factored form
Chapter 2
146
F s
3s 2 7 . s s 1 s 2 s 1
In step 2, we use partial fraction expansion and decompose the function into three terms. We begin by finding the partial fraction expansion for
F s .
The
denominator consists of four distinct linear factors and so the expansion has the form
F s
3s 2 7 s s 1 s 2 s 1
A B C D s s 1 s 2 s 1
where A, B, and C are the real numbers to be determined.
ª º 3s2 7 7 A « , » «¬ s 1 s 2 s 1 »¼ s 0 2 ª º 3s 2 7 5, B « » ¬« s s 2 s 1 ¼» s 1 C
ª 3s 2 7 º « » «¬ s s 1 s 1 »¼ s
Hence, the given function
F s
2
19 ,D 6
F s
ª º 3s 2 7 « » «¬ s s 1 s 2 »¼ s
1
can be expanded as
7 § 1 · § 1 · 19 § 1 · 5 § 1 · ¸ ¨ ¸ ¨ ¸ ¨ ¸ 5¨ 2 © s ¹ ¨© s 1 ¸¹ 6 ¨© s 2 ¸¹ 3 ¨© s 1 ¸¹
In step 3, we find the inverse Laplace transform of each term.
.
5 3
.
The Inverse Laplace Transform
147
Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
f t
° 1 ½° 19 1 ° 1 ½° 5 1 ° 1 ½° 7 1 1 ½ L ® ¾ 5L1 ® ¾ L ® ¾ L ® ¾ 2 ¯s¿ °¯ s 1 °¿ 6 °¯ s 2 °¿ 3 °¯ s 1 °¿
. Finally, using the transform pairs established in Table 2.1, we have
f t
7 19 5 5et e 2t e t . 2 6 3
Example 2.18. Determine (a)
(b)
ª º s2 6s 9 L « » «¬ s 1 s 2 s 4 »¼ 1
ª º s L1 « ». ¬« s 1 s 2 s 3 ¼»
Solution:
(a) We begin by finding the partial fraction expansion for F s . The denominator of
F s
, which is
D s
s 1 s 2 s 4 , has
three linear factors of multiplicity 1. As such, the partial fraction expansion for
F s
has the form
s 2 6s 9 s 1 s 2 s 4
A B C s 1 s 2 s 4
Chapter 2
148
where A, B, and C are real numbers to be determined.
ª s 2 6s 9 º A « » ¬« s 2 s 4 ¼» s ª s 6s 9 º « » «¬ s 1 s 2 »¼ s
16 , 5
1
2
C
4
B
ª s 2 6s 9 º « » ¬« s 1 s 4 ¼» s
2
25 , 6
1 30
Hence, the given function
F s
s 2 6s 9 s 1 s 2 s 4
16 1 ½ 25 ° 1 ½° 1 ° 1 ½° ® ¾ ® ¾ ® ¾ 5 ¯ s 1¿ 6 ¯° s 2 ¿° 30 ¯° s 4 ¿°
can be expanded as
. Taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
ª º s 2 6s 9 L1 « » «¬ s 1 s 2 s 4 »¼
16 t 25 2t 1 4t e e e 5 6 30
(b) We begin by finding the partial fraction expansion for F s . The denominator of
F s , which is D s
s 1 s 2 s 3 , has
three linear factors of multiplicity 1. As such, the partial fraction expansion for
F s
has the form
The Inverse Laplace Transform
s s 1 s 2 s 3
149
A B C s 1 s 2 s 3
where A, B, and C are real numbers to be determined.
ª º 1 s A « , » «¬ s 2 s 3 »¼ s 1 6 ª º s 3 C « » ¬« s 1 s 2 ¼» s 3 10
ª º s B « » «¬ s 1 s 3 »¼ s
2
2 , 15
Hence, the given function
F s
s s 1 s 2 s 3
1 1 ½ 2 ° 1 ½° 3 ° 1 ½° ® ¾ ® ¾ ® ¾ 6 ¯ s 1¿ 15 ¯° s 2 ¿° 10 ¯° s 3 ¿°
can be expanded as
Taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
ª º s L1 « » ¬« s 1 s 2 s 3 ¼»
Example 2.19. Determine
ª º 4s 1 L1 « » «¬ s s 1 s 2 s 7 »¼
1 t 2 2t 3 3t e e e . 6 15 10
ª º 4s 1 L1 « » «¬ s s 1 s 2 s 7 »¼
.
Chapter 2
150
Solution: We begin by finding the partial fraction expansion for denominator of
F s .
The
F s , which is D s s s 1 s 2 s 7 , has
three linear factors of multiplicity 1. As such, the partial fraction expansion for
F s
has the form
4s 1 s s 1 s 2 s 7
A B C D s s 1 s 2 s 7
where A, B, C, and D are real numbers to be determined.
C
ª 4s 1 º « » ¬« s s 1 s 7 ¼» s
2
3 , 10
D
ª 4s 1 º « » ¬« s s 1 s 2 ¼» s
7
29 280
. Hence, the given function
4s 1 s s 1 s 2 s 7
F s
can be expanded as
§ 1 ·1 §1· 1 §3· 1 § 29 · 1 ¨ ¸ ¨ ¨ ¸ ¨ ¸ ¸ © 14 ¹ s © 8 ¹ s 1 © 10 ¹ s 2 © 280 ¹ s 7
. Taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
4s1 °½ § 1 ·L11½§1·L1° 1 °½§ 3 ·L1° 1 °½§ 29 ·L1° 1 °½ ° L1 ® ¾ ¾ ¨ ¸ ®¾ ¨ ¸ ® ¾ ¨ ¸ ® ¾ ¨ ¸ ® ¯°s s1 s2 s7 ¿° ©14¹ ¯s¿ ©8¹ ¯° s1 ¿° ©10¹ ¯° s2 ¿° © 280¹ ¯° s7 ¿° Finally, using the transform pairs established in Table 2.1, we have
The Inverse Laplace Transform
151
ª º 1 1 t 3 2t 29 7t 4s 1 L1 « e e e . » 10 280 «¬ s s 1 s 2 s 7 »¼ 14 8 Example 2.20. Determine L ª¬ F s º¼ , where 1
F s
ª s3 5s 2 6s 7 « «¬ s s 1 s 3 6 s 2 11s 6
º » »¼ .
Solution: Step 1, the denominator is in the factored form
F s
ª º s3 5s 2 6s 7 « » «¬ s s 1 s 1 s 2 s 3 »¼ .
In step 2, we use partial fraction expansion and decompose the function into five terms. We begin by finding the partial fraction expansion for
F s .
The
denominator consists of five distinct linear factors and the expansion has the form
s 3 5s 2 6 s 7 s s 1 s 1 s 2 s 3
A B C D E s s 1 s 1 s 2 s 3
where A, B, C, D, and E are real numbers to be determined.
Chapter 2
152
ª º s 3 5s 2 6 s 7 7 , A « » «¬ s 1 s 1 s 2 s 3 »¼ s 0 6 ª s 3 5s 2 6 s 7 º 9 B « , » ¬« s s 1 s 2 s 3 ¼» s 1 24 C
D
E
ª s 3 5s 2 6 s 7 º 5 , « » «¬ s s 1 s 2 s 3 »¼ s 1 4 ª s 3 5s 2 6 s 7 º 11 , « » ¬« s s 1 s 1 s 3 ¼» s 2 2 ª s 3 5s 2 6 s 7 º « » «¬ s s 1 s 1 s 2 »¼ s
hence, the given function
3
83 , 24
F s can be expanded as
s 3 5s 2 6 s 7 § 7 · 1 § 9 · 1 § 5 · 1 § 11 · 1 § 83 · 1 ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ s s 1 s 1 s 2 s 3 © 6 ¹ s © 24 ¹ s 1 © 4 ¹ s 1 © 2 ¹ s 2 © 24 ¹ s 3 . In step 3, we find the inverse Laplace transform of each term. Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
§ 7 · 1 ½ § 9 · ° 1 ½° § 5 · 1 ° 1 ½° § 11 · 1 ° 1 ½° § 83 · 1 ° 1 ½° F s ¨ ¸ L1 ® ¾ ¨ ¸ L1 ® ¾¨ ¸L ® ¾¨ ¸L ® ¾¨ ¸L ® ¾ © 6 ¹ ¯ s ¿ © 24 ¹ °¯ s 1 °¿ © 4 ¹ °¯ s 1 °¿ © 2 ¹ °¯ s 2 °¿ © 24 ¹ °¯ s 3 °¿ Finally, using the transform pairs established in Table 2.1, we have
The Inverse Laplace Transform
153
7 9 t 5 t 11 2t 83 3t e e e e . 6 24 4 2 24
f t
2.7.2 Partial Fractions: Repeated Linear Factors
ª º s2 1 Example 2.21. Determine L « ». 2 «¬ s 2 s 1 »¼ 1
Solution: Since
s 1
is a repeated linear factor with multiplicity 2 and
s 2
is a non-repeated linear factor, the partial fraction expansion has the form
s
F s
2
1
A B C 2 s 1 s 1 s 2
2
s 1 s 2
where A, B, and C are real numbers to be determined. We begin by multiplying both sides by
s
2
s 2 s 1
A s 1 s 2 B s 2 C s 1
1
Setting
s s s
1 2 0
B 2 C 5 A 4
hence, the given function
F s can be expanded as
2
2
to obtain
Chapter 2
154
ª º s2 1 « » 2 «¬ s 1 s 2 »¼
4 2 5 2 s 1 s 1 s 2
.
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
ª º 1 °½ s2 1 ° 1 ½° 1 ½ 1 ° « » 4 L1 ® L 5L1 ® ¾ 2L ® ¾ 2 2¾ ¯ s 1¿ «¬ s 1 s 2 »¼ °¯ s 2 °¿ ¯° s 1 ¿° 1
. Finally, using the transform pairs established in Table 2.1, we have
ª º s2 1 » 4e t 2te t 5e 2t . L « 2 «¬ s 1 s 2 »¼ 1
Example 2.22. Determine
(a )
(b)
ª s 2 9s 2 º L1 « » 2 «¬ s 3 s 1 »¼
ª s 1 º L1 « . 2» ¬« s 1 s ¼»
Solution:
(a) Since s 1 is a repeated linear factor with multiplicity two and
s 3 the form
is a non-repeated linear factor, the partial fraction expansion has
The Inverse Laplace Transform
ª s 2 9s 2 º « » 2 «¬ s 1 s 3 »¼
155
A B C 2 s 1 s 1 s 3
where A, B, and C are real numbers to be determined. We begin by multiplying both sides by
s
2
9s 2
s 3 s 1
2
to obtain
A s 1 s 3 B s 3 C s 1
2
Setting
B 3 s 1 s 3 C 1 s 0 A 2 hence, we have
ª s 2 9s 2 º « » 2 «¬ s 1 s 3 »¼
2 3 1 2 s 1 s 1 s 3
.
Applying the inverse Laplace transform to the above equation, and using its linearity, we get the following result
ª s 2 9s 2 º » 2et 3tet 2e 3t L « 2 «¬ s 1 s 3 »¼ 1
Chapter 2
156
(b) Since s is a repeated linear factor with a multiplicity of two and
s 1 is a non-repeated linear factor. The partial fraction expansion has the form
s 1 s 2 s 1
A B C 2 s s s 1
.
We begin by multiplying both sides by
s 1
s 1 s 2 to obtain
As s 1 B s 1 Cs 2
Setting
s 1 C 2 s 0 B 1
.
Equating the coefficients of
s 2 A C 0 A 2 . Hence, we have
s 1 s 2 s 1
2 1 2 2 s s s 1
Applying the inverse Laplace transform to the above equation, and using its linearity, we get the following result
ª s 1 º t L1 « 2 » 2 t 2e . s s 1 »¼ «¬
The Inverse Laplace Transform
157
Example 2.23. Evaluate the inverse transform of
9s
F s
2
38s 55
s 3 s 2
3
Solution: Alternative Method: Compute partial fraction constants. Multiple poles
pi of order m appear as a factor of the form
s pi
in the denominator. These multiple poles produce
m
terms in decomposition.
Cimi Ci1 Ci 2 Ci 3 ..... 2 3 m s pi s pi s pi s pi i The residues
m j
therefore, given
9s
2
38s 55
i 1
s pi
^ s p
mi
i
`
F s
s pi
, i 1, 2,3....mi
F s , we write
s 3 s 2 A
Cij
¦
Cij are given by
1 d i mi j ! ds mi j
Cij
mi
3
C33 C32 C31 A 3 2 s 3 s 2 s 2 s 2
ª 9 s 2 38s 55 « 3 «¬ s 2
º»
»¼ s
2 3
j
.
m
Chapter 2
158
1 d0 3 3 ! ds 0
C33
2 ° 9 s 38s 55 ® s 3 ° ¯
2 1 d ° 9 s 38s 55 ® 3 2 ! ds °¯ s 3
C32
1 d2 3 1 ! ds 2
C31
2
38s 55
s 3 s 2
¾ °s ¿
2 ° 9 s 38s 55 ® s 3 ° ¯
2
½°
2 ° 9 s 38s 55 ® s 3 ° ¯
Hence, the given function
9s
½°
¾ ° ¿s
½°
¾ °s ¿
3 2
1 2
½°
¾ ° ¿s
2 2
F s can be expanded as
2 3 1 2 3 2 s 3 s 2 s 2 s 2
3
Applying the inverse Laplace transform to the above equation, and using its linearity, we get the following result
ª 9s 2 38s 55 L « 3 «¬ s 3 s 2 1
º»
3 2e 3t t 2 e 2t te2t 2e2t . 2 »¼
Example 2.24. Evaluate the inverse transform of
3s F s
2
7s 6
s 1
3
The Inverse Laplace Transform
159
Solution:
3s
2
7s 6
s 1
3
C33
s 1
3
C32
s 1
C33
1 d0 3 3 ! ds 0
C32
1 d 3s 2 7 s 6 3 2 ! ds
C31
1 d2 3 1 ! ds 2
^ 3s
2
7s 6
`
^
^ 3s
2
7s 6
s 1
2 s 1
`
7s 6
C31 s 1
s 1
`
^ 6s 7 `
s 1
1
3 s 1
.
Hence, the given function
3s F s
2
2
F s can be expanded as
2
3
s 1
3
1
s 1
2
3 s 1
.
Taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
L1 ª¬ F s º¼
1 ° ½ ° 1 ½ ° 1 ° 1 ½ ° ° 2 L1 ® L ® 3L1 ® ¾ 3¾ 2 ¾ ° ° ° ¯ s 1 ° ¿ ¯ s 1 ° ¿ ¯ s 1 ° ¿
Finally, using the transform pairs established in Table 2.1, we have
f t t 2 et tet 3et Thus,
.
Chapter 2
160
f t
t
2
t 3 et .
Alternative method: To compute partial fraction constants, we write
3s
2
7s 6
s 1
A
3
s 1
3
B
s 1
2
C s 1
from which,
3s
2
7s 6
2
A B s 1 C s 1 .
If s = 1, we obtain A = 2, but we seem to have run out of convenient values with which to make substitutions. However, remember that whenever two functions are equal, so are their derivatives. Differentiating both sides of the above equation, we get
6s 7
B 2C s 1
so that s
1 can be used again, yielding B 1. Differentiating a
second time, we obtain
6
2C C
3
Hence, the given function
3s F s
2
7s 6
s 1
3
F s can be expanded as 2
s 1
3
1
s 1
2
3 s 1
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
The Inverse Laplace Transform
161
1 ° ½ ° 1 ½ ° 1 ° 1 ½ ° ° 3L1 ® 2 L1 ® L ® ¾ 3¾ 2 ¾ ° s 1 ¿ ° ° ° ¯ ¯ s 1 ° ¿ ¯ s 1 ° ¿
L1 ¬ª F s ¼º
Finally, using the transform pairs established in Table 2.1, we have
f t t 2 et tet 3et . The differentiation procedure works particularly well with repeating linear factors. Example 2.25. Evaluate the inverse transform of F s
s 5 s 1 s3
.
Solution: Alternative method: Compute partial fraction constants. Multiple poles
pi of order m appear as a factor of the form
s pi
in the denominator. These multiple poles produce
m
terms in decomposition.
Cimi Ci1 Ci 2 Ci 3 ..... m 2 3 s pi s pi s pi s pi i The residues
Cij
i 1
s pi
¦
Cij are given by
1 d i mi j ! ds mi j m j
Cij
mi
therefore, given
^ s p
F s , we write
i
mi
`
F s
s pi
, i 1, 2,3....mi
j
m
Chapter 2
162
s 5 s 2 s3 A
C33
C32
C31
C C C A 33 32 31 3 2 s2 s s s
ª s 5 º « » 3 ¬ s ¼s
7 8
2
1 d0 3 3 ! ds 0
° s 5 °½ ® ¾ ¯° s 2 ¿°s
1 d ° s 5 °½ ® ¾ 3 2 ! ds °¯ s 2 °¿s 1 d2 3 1 ! ds 2
° s 5 ½° ® ¾ ¯° s 2 ¿°s
Hence, the given function
s 5 s 2 s3
0
0
0
° s 5 °½ ® ¾ ¯° s 2 ¿°s
0
5 2
7 4 11 16
F s can be expanded as
7 1 5 1 7 1 11 1 3 2 8 s 2 2 s 4 s 16 s
.
Applying the inverse Laplace transform to the above equation and using its linearity, we get the following result
° s 5 ½° L1 ® 3¾ °¯ s 2 s °¿
7 1 1 ½ 5 1 1 ½ 7 1 1 ½ 11 1 1 ½ L ® ¾ L ® ¾ L ® ¾ L ® ¾ 8 ¯ s 2 ¿ 2 ¯ s 3 ¿ 4 ¯ s 2 ¿ 16 ¯ s ¿
Finally, using the transform pairs established in Table 2.1, we have
The Inverse Laplace Transform
° s 5 °½ L1 ® 3¾ °¯ s 2 s °¿
163
7 2t 5 2 7 11 e t t . 8 4 4 16
2.7.3 Partial Fractions: Quadratic Polynomials Case (i). A partial fraction corresponds to every single quadratic factor
as 2 bs c of the denominator,
As B , as bs c 2
here, A and B are constants. Case (ii). A partial fraction corresponds to every repeating quadratic factor
as
2
n
bs c of the denominator,
A1s B1 A2 s B2 2 as bs c as 2 bs c
where
2
A3 s B3
as
2
bs c
3
.........
An s Bn
as
2
bs c
n
Ai and Bi are constants.
ª º 2 s 2 10 s ». Example 2.26. Determine L « 2 «¬ s 2 s 5 s 1 »¼ 1
Solution: We can see that the quadratic factor the linear factor
s 1
s
2
2 s 5 is irreducible. Since
appears to the first power in the denominator of
Chapter 2
164
F s and the quadratic factor appears to the first power, the partial fraction expansion
F s has the form
ª º 2 s 2 10 s « 2 » ¬« s 2 s 5 s 1 »¼
ª º 2 s 2 10 s « » « s 1 2 22 s 1 » ¬« ¼»
A s 1 B
s 1
2
2
2
C s 1
where A, B, and C are real numbers to be determined.
2s
2
10s
2
ª¬ A s 1 B º¼ s 1 C s 1 22
.
º »ª 1 º » «« s 1 »» ¼ »¼ ¬
Setting
1 C 1 1 B 4 0 A 3
s s s
We have
ª º 2 s 2 10 s « » « s 1 2 22 s 1 » ¬« ¼»
ª º 2 s 2 10 s « » « s 1 2 22 s 1 » «¬ »¼
3 s 1 4
s 1 2 2
ª s 1 3« « s 1 2 22 «¬
2
1 s 1
º ª 2 » 2« » « s 1 2 22 »¼ «¬
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
The Inverse Laplace Transform
ª º ª 2s 2 10s » 3L1 « s 1 L1 « « s 1 2 22 s 1 » « s 1 2 22 «¬ ¼» ¬«
º ª 2 » 2 L1 « » « s 1 2 22 ¼» ¬«
165
º » L1 ª 1 º « » » ¬« s 1 ¼» ¼»
Finally, using the transform pairs established in Table 2.1, we have
ª º 2 s 2 10 s « » L « s 1 2 22 s 1 » ¬« ¼»
1
3et cos 2t 2et sin 2t e t .
º»
ª 4s 2 s 1 Example 2.27. Determine L « «¬ s s 2 1 1
»¼
Solution: We can see that the quadratic factor
s
linear factor
s
2
1 is irreducible. Since the
appears to the first power in the denominator of
F s
and the quadratic factor appears to the first power, the partial fraction
F s has the form
expansion of
º»
ª 4s 2 s 1 « «¬ s s 2 1
A Bs C 2 s s 1
»¼
where A, B, and C are real numbers to be determined.
4s
2
s 1
4s
2
s 1
A s 2 1 s Bs C
A B s
2
C s A
Chapter 2
166
Hence, comparing the coefficients of both sides, we have
s2 :
A B 4,
s:
C
1:
1,
A 1.
By solving these equations, we obtain
A
1, B
3& C
1
Hence,
º»
ª 4s 2 s 1 « «¬ s s 2 1
1 3s 1 »¼ s s 2 1
.
Taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
º»
ª 4s 2 s 1 L « «¬ s s 2 1 1
¼»
s ° 1 ½ L1 ® ¾ 3L1 ® 2 ¯s¿ ° ¯ s 1
½ 1 ° 1 ° ¾ L ® 2 ° ° ¿ ¯ s 1
Finally, using the transform pairs established in Table 2.1, we have
º»
ª 4s 2 s 1 L « «¬ s s 2 1 1
»¼
1 3cos t sin t
Example 2.28. Determine
.
ª s 1 L1 « 2 «¬ s 1 s 2 9
º ». »¼
½ ° ¾ ° ¿
The Inverse Laplace Transform
167
Solution: We can see that both the quadratic factors
s
2
1 and s 2 9
irreducible and appear to the first power in the denominator of The partial fraction expansion for
s 1
s
2
2
1 s 9
F s has the form
As B Cs D 2 s2 1 s 9
where A, B, and C are real numbers to be determined.
s 1 As B s 2 9 Cs D s 2 1 Comparing the coefficients of both sides, we have
s3 :
A C 0,
s2 :
B D 0,
s:
9A C
1,
1:
9B D
1.
Through the successive elimination of variables, we obtain
A
B
1 and C 8
Hence, we have
D
1 . 8
are
F s .
Chapter 2
168
s 1
s 1 s 9 2
2
1 ª s 1 s 1 « 2 2 8 « s 1 s 9 ¬
º 1ª s 1 1 s » « 2 2 2 2 »¼ 8 «¬ s 1 s 1 s 9 s 9
º » »¼
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
° s 1 L1 ® 2 2 °¯ s 1 s 9
½° 1 ª ° s 1 ¾ «L ® 2 8 « °¿ ¬ °¯ s 1
½° 1 ½° s 1 ° 1 ° L L ¾ ® 2 ¾ ® 2 °¿ °¯ s 1 °¿ °¯ s 9
½° 1 1 ° L ¾ ® 2 °¿ °¯ s 9
½°º ¾» °¿»¼
Finally, using the transform pairs established in Table 2.1, we have
° s 1 L ® 2 2 °¯ s 1 s 9 1
½° ¾ °¿
1 >3cos t 3sin t 3cos 3t sin 3t @. 24
Example 2.29. Determine the inverse Laplace transform of
4s F s s 1 s
2
6
2
2s 2
.
Solution: We can see that the quadratic factor the linear factor
s
2
2 s 2 is irreducible. Since
s 1 appears to the first power in the denominator of
F s and the quadratic factor appears to the first power, the partial fraction expansion for
F s has the form
The Inverse Laplace Transform
4s F s s 1 s
2
6
2
2s 2
169
A Bs C 2 s 1 s 2s 2
where A, B, and C are real numbers to be determined.
4s
2
6
A s 2 2 s 2 s 1 Bs C
Setting
s
A
1
2.
Equating the coefficients of s
B
2
and
s , we get
2.
2 and C
Hence, we have
4s F s s 1 s
2
6
2
2s 2
2
6
2
2s 2
4s F s s 1 s
2 2s 2 2 s 1 s 2s 2
2 s 1 4 2 s 1 s 1 2 1
.
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
° 4s 2 6 L1 ® 2 ¯° s 1 s 2s 2
½° 1 ½° s 1 1 ° 1 ° ¾ 2L ® ¾ 2L ® 2 ° s 1 1 ¯° s 1 ¿° ¿° ¯
½ 1 ° 1 ° ¾ 4L ® 2 ° ° s 1 1 ¿ ¯
Finally, using the transform pairs established in Table 2.1, we have
½ ° ¾ ° ¿
Chapter 2
170
° 4s 2 6 L ® 2 °¯ s 1 s 2s 2 1
Example 2.30. Evaluate
F s
½° t t t ¾ 2e 2e cos t 4e sin t. °¿
L1 ª¬ F s º¼ where
4s 16
s
2
4s 15 s 2 6s 13
Solution: Using partial fractions, we write
F s
4s 16
s
2
2
4 s 5 s 6s 13
As B Cs D 2 s 4s 5 s 6s 13 2
where A, B, and C are the real numbers to be determined. Multiplying both sides by the denominator of
F s ,
4s 16 As B s 2 6s 13 Cs D s 2 4s 5 . Comparing the coefficients of both sides gives
s3 :
A C 0,
s2 :
6 A B 4C D 0,
s: 16 :
13 A 6 B 5C 4 D 13B 5 D 16 .
4,
The Inverse Laplace Transform
171
By successive elimination of variables, we obtain
A
C
0 and B
2, D
F s can be expanded as
Hence, the given function
2
F s
2 .
2
s 2 1 s 3 4 . 2
2
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
1
L
^F s `
ª 1 ° 2 « L1 ® 2 « «¬ °¯ s 2 1
½ 1 ° 1 ° ¾ L ® 2 ° ° s 3 4 ¿ ¯
½º °» ¾» °» ¿¼ .
Finally, using the transform pairs established in Table 2.1, we have
f t
2e
2 t
sin t e 3t sin 2t . ª
1 «¬ s 1 s 2 4
1 Example 2.31. Determine L «
º » . »¼
Solution: We can see that the quadratic factor
s 1 and s 2 4 are expansion has the form
s
2
4
is irreducible. Since
non-repeated factors, the partial fraction
Chapter 2
172
ª 1 « «¬ s 1 s 2 4
º » »¼
A Bs C 2 s 1 s 4
Alternative Method: We can use the cover-up method to find A, B, and C,
A
ª 1 « 2 «¬ s 4
º » »¼ s
1
ª 1 º « » «¬ s 1 »¼ s
> Bs C @s 2i 2 Bi C
1 5
1 1 2i
1 1 2i
2i
° 1 2i ½° 1 2 i ® ¾ ¯° 1 2i ¿° 5 5
Since the real and imaginary parts are equal on both sides, we get
B
1 &C 5
1 5.
Hence, we have
ª 1 « «¬ s 1 s 2 4
º » ¼»
1 ° 1 s 1 2 ® 5 ° s 1 s 4 ¯
s 1 °½ 1 ° 1 2 2 ¾ ® s 4 °¿ 5 °¯ s 1 s 4
. Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
½° ¾ °¿
The Inverse Laplace Transform
ª 1 L1 « «¬ s 1 s 2 4
º » ¼»
1 ° 1 § 1 · 1 § s ®L ¨ ¸ L ¨¨ 2 5 ° © s 1 ¹ © s 4 ¯
173
· § 1 ¸ L1 ¨ ¸ ¨ s2 4 ¹ ©
·½ ¸ ¾° ¸° ¹¿ .
Finally, using the transform pairs established in Table 2.1, we have
ª 1 L1 « «¬ s 1 s 2 4
º » »¼
1 t 1 ½ ®e cos 2t sin 2t ¾ . 5¯ 2 ¿
Example 2.32. Determine
ª º 1 L1 «e s ». ¬« s 1 s 2 ¼»
Solution: Let
F s
1
s 1 s 2
Using partial fractions, we get
F s
1ª 1 1 º « » 3 ¬« s 1 s 2 ¼»
.
Applying the inverse Laplace transform to the above equation, and using its linearity, we get the following result
f t then,
L1 ª¬ F s º¼
1 t e e 2t , 3
Chapter 2
174
ª º 1 L1 «e s » «¬ s 1 s 2 »¼ Example 2.33. Determine
1 t 1 2 t 1 e e u t 1 . 3
L1 ª¬ F s º¼ , where
ª 3s 1 º F s e2 s « 3 2 ¬ s 5s 6s »¼ . Solution: In step 1, the denominator is in the factored form
ª º 3s 1 F s e 2 s « » «¬ s s 2 s 3 »¼ . In step 2, we use partial fraction expansion and decompose the function into three terms. We begin by finding the partial fraction expansion for
F s . The
denominator consists of three distinct linear factors and the expansion has the form
3s 1 s s 2 s 3
A B C s s 2 s 3
where A, B, and C are real numbers to be determined.
The Inverse Laplace Transform
ª º 3s 1 A « » «¬ s 2 s 3 »¼ s C
ª 3s 1 º « » «¬ s s 2 »¼ s
3
0
B
ª 3s 1 º « » «¬ s s 3 »¼ s
2
5 , 2
8 , 3
hence, the given function
3s 1 s s 2 s 3
1 , 6
175
F s can be expanded as
1 1 5 1 8 1 6 s 2 s 2 3 s 3
.
In step 3, we find the inverse Laplace transform of each term. Having obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
° ½° 3s 1 L1 ® ¾ ¯° s s 2 s 3 ¿°
1 1 1 ½ 5 1 ° 1 ½° 8 1 ° 1 ½° L ® ¾ L ® ¾ L ® ¾ 6 ¯ s ¿ 2 ¯° s 2 ¿° 3 ¯° s 3 ¿°
. Finally, using the second shifting rule and the transform pairs established in Table 2.1, we have
° ½° § 1 5 2v 8 3 t 2 · 3s 1 L1 ®e 2 s ¾ ¨ e e ¸ u t 2 . s s 2 s 3 °¿ © 6 2 3 ¹ °¯
Chapter 2
176
Example 1.34. Given the Laplace transform
ª º 3s F s « ». «¬ s 2 s 3 »¼ Find the initial value of finding
f t
f t , by (i) the initial value theorem; (ii)
L1 ª¬ F s º¼ .
Solution:
(i) by the initial value theorem, f 0
lim ^s F s `
f 0
§ lim ¨ s s of ¨ ©
s of
° lim ® s s of °¯
§ · °½ 3s ¨¨ ¸¸ ¾ © s 2 s 3 ¹ °¿
§ ·· 3s ¨¨ ¸¸ ¸¸ © s 1 2 s s 1 3 s ¹ ¹
§ · 3 lim ¨ ¸ s of ¨ 1 2 s 1 3 s ¸ © ¹
. Thus,
f 0 3.
(ii) We use partial fraction expansion and decompose the function into three terms. We begin by finding the partial fraction expansion for
F s . The
denominator consists of two distinct linear factors and the expansion has the form
The Inverse Laplace Transform
F s
3s s 2 s 3
177
A B s 2 s 3
where A and B are real numbers to be determined.
ª 3s º A « » «¬ s 3 »¼ s
6 , B 2
hence, the given function
F s
3s s 2 s 3
ª 3s º « » «¬ s 2 »¼ s
6, 3
F s , can be expanded as
6 6 s 2 s 3
.
Having obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
1 °½ 1 °½ 3s ° °½ 1 ° 1 ° L1 ® ¾ 6 L ® ¾ 6L ® ¾ °¯ s 2 s 3 °¿ °¯ s 2 °¿ °¯ s 3 °¿ . Finally, using the second shifting rule and the transform pairs established in Table 2.1, we have
f t 6e2t 6e3t . Example 1.35. Given the Laplace transform
ª º s2 F s « ». «¬ s 1 s 2 s 3 »¼
Chapter 2
178
Find the initial value of
f t , by
(i) the initial value theorem (ii) finding f t
L1 ª¬ F s º¼ .
Solution:
(i) by the initial value theorem, f 0
lim ^s F s ` s of
° lim ® s s of °¯
ª º ½° s2 « » .¾ «¬ s 1 s 2 s 3 »¼ °¿
§ § ·· § · s2 1 f 0 lim ¨ s ¨ ¸ ¸ lim ¨ ¸ s of ¨ ¨ s 1 1 s s 1 2 s s 1 3 s ¸ ¸ s of ¨ 1 1 s 1 2 s 1 3 s ¸ ¹¹ © ¹ © © Thus,
f 0 1. (ii) We will use partial fraction expansion and decompose the function into three terms. We begin by finding the partial fraction expansion for
F s . The
denominator consists of three distinct linear factors and the expansion has the form
ª º s2 F s « » «¬ s 1 s 2 s 3 »¼
A B C s 1 s 2 s 3
where A, B, and C are real numbers to be determined.
The Inverse Laplace Transform
ª º s2 A « » «¬ s 2 s 3 »¼ s C
ª º s2 « » ¬« s 1 s 2 ¼» s
hence, the given function
F s
1 2
1
3
B ,
179
ª º s2 « » «¬ s 1 s 3 »¼ s
4, 2
9 , 2
F s can be expanded as
ª º s2 « » ¬« s 1 s 2 s 3 ¼»
1 1 1 9 1 4 2 s 1 s 2 2 s 3
Having obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
° ½° 1 1 ° 1 ½° 1 ½° 9 1 ° 1 ½° s2 1 ° L1 ® L ® ¾ ¾ 4L ® ¾ L ® ¾ °¯ s 1 s 2 s 3 °¿ 2 °¯ s 1 °¿ °¯ s 2 °¿ 2 °¯ s 3 °¿ . Finally, using the second shifting rule and the transform pairs established in Table 2.1, we have
f t
1 t 9 e 4e 2t e 3t . 2 2
2.8 Convolutions It is often necessary to find the inverse transform of the product of two functions F s G s when inverse transforms
f t
and
g t of
Chapter 2
180
F s and G s are known. We shall see shortly that the inverse of F s G s is called the convolution of f and g. 2.8.1 Definition The convolution of two functions, f and g, is defined as t
f t g t
³ f u g t u du. 0
2.8.2 Properties of Convolution Let
f t , g t and h t
be piecewise continuous on
i
f t g t g t f t ,
ii
f t g t h t
iii
f t g t h t
>0, f .
Then,
f t g t f t h t ,
f t g t h t .
Proof. To prove the equation in (i), we begin with the definition t
f t g t
³ f u g t u du. 0
Using the change of variables,
w t u , we have
0
f t g t
³ t
,
t
f w g t w dw
³ f w g t w dw 0
g t f t
The Inverse Laplace Transform
181
which proves (i). The proofs of equations (ii) and (iii) are found in the exercises. Example 2.36. Compute
a t 2 *cos t b sin t *cos t c t * et .
Solution:
a Convolution between two functions f and g is given by t
f t g t
³ f u g t u du. 0
t
f t g t t cos t 2
³u
2
cos t u du.
0
Using integration by parts, we get
t 2 cos t
^
`
u 2 sin t u 2u cos t u 2sin t u
Thus,
t 2 cos t 2 t sin t .
b Convolution between two functions f and g is given by t
f t g t
³ f u g t u du. 0
t
f t g t sin t cos t
³ sin u cos t u du. 0
(2.9)
t 0.
Chapter 2
182
Recall from trigonometry that
1 ªsin A B sin A B º¼ . 2¬
sin A cos B If we set A
u and B
t u , we can carry out the integration in Eq.
(2.9) t
t
³ sin u cos t u du.
sin t cos t
0
sin t cos t
1 ^sin t sin 2u t `du 2 ³0
1 ^t sin t 0` 2
Thus,
t sin t
sin t cos t
2
.
c The convolution between two functions f and g is given by t
f t g t
³ f u g t u du. 0
t
f t g t
g t f t t et
³ue 0
Recall from trigonometry that t
t et
et ³ u e u du. 0
t u
du.
The Inverse Laplace Transform
183
We can carry out the integration and get
^
et ue u e u
t et
`
t 0
Thus,
t et
et t 1.
The convolution operation has various uses. We introduced one of the essential theorems concerning the use of Laplace transforms and convolution—the convolution theorem. The convolution theorem enables one to deduce the inverse Laplace transform of an expression. However, it must be expressed in the form of a product of functions, meaning that each inverse Laplace transform of the product is known, i.e., it expresses the relationship between the product of two Laplace transforms F s
and
G s and the convolution of their transform pairs f t and g t . 2.9 Convolution Theorem Let
f t
and
g t
be piecewise continuous on
F s
L ^ f t ` and G s
L ^ g t ` . Then,
i
L ^ f t g t ` F s G s ,
or, equivalently,
ii
L1 ^ F s G s `
f t g t .
>0, f
and
Chapter 2
184
Proof: Starting with the left-hand side of
i ,
we use the definition of
convolution
L ^ f t g t `
f
ªt º st e ³0 «¬ ³0 f t v g v dv »¼ dt.
To simplify the evaluation of this iterated integral, we introduce the unit step function
u t v and write
L ^ f t g t `
f
³e
st
0
using the fact that u t v
ªt º « ³ u t v f t v g v dv » dt , ¬0 ¼ 0 , if v ! t.
Reversing the order of integration gives
L ^ f t g t `
f
ª f st º g v ³0 «¬ ³0 e u t v f t v dt »¼ dv. (2.10)
Recall from the first translation property in Section 1.5.3 that the integral in brackets in Eq. (2.10) equals
L ^ f t g t `
e sv F s . Hence,
f
³ g v e 0
sv
F s dv
The Inverse Laplace Transform
L ^ f t g t `
185
f
F s ³ g v e sv dv 0
.
Thus,
L ^ f t g t `
F s G s .
This proves the theorem. Example 2.37. Using the convolution theorem, determine
ª s L1 « 2 2 «¬ s a s 2 b 2
º ». »¼
Solution:
We express
let
F s
s
2
a2
s
s s a2 2
b2
½° ¾ cos at ¿°
1 ° L1 ® 2 2 ° ¯ s b
2
½ ° ¾ ° ¿
s «¬ s 2 a 2
as «
and G s
° s f t L1 ® 2 2 ¯° s a g t
ª
s
and
1 sin bt b
ºª 1 »« 2 »¼ «¬ s b 2
1 so that s b2 2
º »; »¼
Chapter 2
186
In case
i ,
2
2
if we suppose that a z b , then, using the convolution
integral, we have t
L1 ^ F s G s `
f t g t
³ f u g t u du 0
ª s L « 2 2 2 2 «¬ s a s b 1
º » »¼
t
1 cos au sin b t u du. b ³0
(2.11)
Recall from trigonometry that
1 ªsin A B sin A B º¼ . 2¬
cos A sin B If we set
A au and B b t u , we can carry out the integration
in Eq. (2.11)
ª s L1 « 2 2 2 2 «¬ s a s b
º » »¼
1 sin a b u bt sin a b u bt du 2 ³0
ª s L « 2 2 2 2 s a s b «¬
º » »¼
1 ° cos a b u bt cos a b u bt ½° ® ¾ 2 ¯° a b a b ¿°
1
ª s L « 2 2 2 2 «¬ s a s b
t
1
This simplifies to
º » »¼
^
`
cos bt cos at . a 2 b2
t
0
The Inverse Laplace Transform
ª s L1 « 2 «¬ s 4 s 2 9
º » »¼
187
1 1 ½ ®cos 2t cos 3t ^cos 2t cos 3t `¾ 4¯ 5 ¿
Thus,
ª s L1 « 2 «¬ s 4 s 2 9
º » »¼
1 ^cos 2t cos 3t ` . 5
ii If
a2
b 2 , then
ª s L1 « « s2 a2 ¬
º » 2 » ¼
1 cos au sin a t u du. a ³0
In case
t
(2.12)
Recall from trigonometry that
1 ªsin A B sin A B º¼ . 2¬
cos A sin B If we set
A au and B b t u , we can carry out the integration in
Eq. (2.12)
ª s L « « s2 a2 ¬
º » 2 » ¼
1 ^sin at sin 2au at `du b ³0
ª s L « « s2 a2 ¬
º » 2 » ¼
cos 2au at ½ 1 ®sin at u ¾ b¯ 2a ¿
1
1
t
t
0
Chapter 2
188
ª s L1 « « s2 a2 ¬
º » 2 » ¼
t sin at . 2a
Example 2.38. Using the convolution theorem, determine
ª 1 L « « s2 k 2 ¬ 1
º ». 2 » ¼
Solution:
We express
1
s
2
k2
F s G s
as
2
1 s k2 2
ª 1 « 2 «¬ s k 2
ºª 1 »« 2 »¼ «¬ s k 2
º »; »¼
so that
f t g t
1 1 ° k L ® 2 2 k ¯° s k
½° ¾ ¿°
1 sin kt k .
We have the convolution theorem
L1 ^ F s G s `
t
f t g t
³ f u g t u du 0
ª 1 L « « s2 k 2 ¬ 1
º » 2 » ¼
t
1 sin ku sin k t u du. k 2 ³0 (2.13)
The Inverse Laplace Transform
189
Recall from trigonometry that
1 ªcos A B cos A B º¼ . 2¬
sin A sin B If we set
k t u , we can carry out the integration in
A ku and B
Eq. (2.13)
ª 1 L « « s2 k 2 ¬
º » 2 » ¼
1 (cos 2ku kt cos kt )du 2k 2 ³0
º » 2 » ¼
1 2k 2
º » 2 » ¼
1 ^sin kt kt cos kt `. 2k 3
1
ª 1 L « « s2 k 2 ¬ 1
t
.
Thus,
ª 1 L « 2 « s k2 ¬ 1
Example 2.39. Using the convolution theorem, determine
ª º 1 L1 « 2 » . «¬ s s a »¼ Solution:
We express
1 ª 1 ºª 1 º as « 2 » « »; s s a ¬ s ¼ ¬« s a ¼» 2
t
ª1 º «¬ 2k sin 2ku kt u cos kt »¼ 0
Chapter 2
190
let
F s
f t
1 and G s s2
1 ½ L1 ® 2 ¾ ¯s ¿
1 so that s a
t and g t
° 1 ½° at L1 ® ¾ e . ¯° s a ¿°
Using the convolution integral, we have t
L1 ^ F s G s `
f t g t
³ f u g t u du 0
ª º 1 L1 « 2 » «¬ s s a »¼
t
³u e
a t u
du.
0
(2.14)
ª º at t 1 au L « 2 » e ³ u e du. s s a »¼ «¬ 0 1
We can carry out the integration in Eq. (2.14). Therefore
ª º 1 at 1 ªe at 1º¼ . L1 « 2 » 2 ¬ «¬ s s a »¼ a Example 2.40. Using the convolution theorem, determine
ª s L « « s2 4 ¬ 1
º ». 2 » ¼
The Inverse Laplace Transform
191
Solution:
We express
let
F s
ª
s
s
2
4
1 2 «¬ s 4
as «
2
1 and G s s 4
2
ºª s »« 2 »¼ «¬ s 4
s s 4
2
º »; »¼
so that
1 1 ° 2 L ® 2 ° s 2 22 ¯
f t
g t
s ° L1 ® 2 2 ° s 2 ¯
½° ¾ °¿
1 sin 2t 2 and
½ ° ¾ ° ¿
cos 2t .
Using the convolution integral, we have 1
L
^F s G s `
t
f t g t
³ f u g t u du 0
ª s L « 2 « s 4 ¬ 1
º » 2 » ¼
t
1 sin 2u cos 2 t u du. 2 ³0
Recall from trigonometry that
sin A cos B
1 ªsin A B sin A B º¼ . 2¬
(2.15)
Chapter 2
192
If we set A
2u and B
2 t u , we can carry out the integration in
Eq. (2.15)
ª s L1 « 2 « s 4 ¬
º » 2 » ¼
1 ^sin t sin 4u 2t ` du 4 ³0
ª s « L « s2 4 ¬
º » 2 » ¼
t cos 4u 2t ½° 1 ° t ®sin t u 0 ¾ 4° 4 °¿ 0 ¯
1
t
which can be simplified to
ª s L1 « « s2 4 ¬
º » 2 » ¼
1 t sin 2t. 4
Example 2.41. Using the convolution theorem, determine
ª s L1 « 2 «¬ s 4 s 2 9
º ». »¼
Solution:
We express
let
F s
as
ª 1 « 2 «¬ s 4
1 and G s s 4
s s 9
s
s
2
2
4 s2 9
2
ºª s »« 2 »¼ «¬ s 9
º »; »¼
The Inverse Laplace Transform
so that
° 1 L1 ® 2 °¯ s 4
f t
g t
s ° L1 ® 2 2 ° ¯ s 3
193
½° 1 sin 2t and ¾ °¿ 2
½ ° ¾ ° ¿
cos 3t .
Using the convolution integral, we have t
L1 ^ F s G s `
f t g t
³ f u g t u du 0
ª s L1 « 2 «¬ s 4 s 2 9
º » »¼
t
1 sin 2u cos 3 t u du. 2 ³0 (2.16)
Recall from trigonometry that
1 ªsin A B sin A B º¼ . 2¬
sin A cos B If we set
A 2u and B 3 t u , we can carry out the integration in
Eq. (2.16)
ª s L « 2 «¬ s 4 s 2 9 1
On integration, we get
º » »¼
t
1 ^sin 3t u sin 5u 3t `du 4 ³0
.
Chapter 2
194
ª s L « 2 «¬ s 4 s 2 9 1
t cos 5u 3t ½° t 1 ° ®cos 3t u 0 ¾ 4° 5 °¿ 0 ¯
º » »¼
which can be simplified to
ª s L1 « 2 «¬ s 4 s 2 9
º » »¼
1 1 ½ ®cos 2t cos 3t ^cos 2t cos 3t `¾ 4¯ 5 ¿
. Thus,
ª s L1 « 2 «¬ s 4 s 2 9
Example
º » »¼
2.42.
Using
1 ^cos 2t cos 3t ` . 5 the
convolution
ª º 1 L1 « 2 ». 2 «¬ s s 2 »¼ Solution:
We express
1 s2 s 2
2
as
ª 1 ºª 1 º »; 2 « s2 » « ¬ ¼ «¬ s 2 »¼
Let
F s
So that
1 and G s s2
1
s 2
2
.
theorem,
determine
The Inverse Laplace Transform
f t t and g t te2t 1
L
t
^F s G s `
f t g t
³ f t u g u du 0
ª º 1 L1 « 2 » 2 «¬ s s 2 »¼
t
³ t u ue
2 u
du.
0
On integration by parts, this gives
ª º 1 L1 « 2 » 2 «¬ s s 2 »¼
1 t 1 t 1 e 2t . 4
Example 2.43. Using the convolution theorem, determine
ª 1 L1 « 2 «¬ s s 1
º ». »¼
Solution:
We express
Let
F s
so that
1
s s2 1
as
1 and G s s
ª1 º ª 1 «¬ s »¼ « s 2 1 «¬
1 s 1 2
f t 1 and g t sin t
º »; »¼
195
Chapter 2
196
1
L
t
^F s G s `
f t g t
³ f t u g u du 0
ª 1 º » L « 2 «¬ s s 1 »¼
t
sin u du. ³
1
0
Carrying out integration gives
ª 1 º » L « 2 s s 1 «¬ »¼ 1
1 cos t .
Example 2.44. Using the convolution theorem, determine
° 1 L1 ® 4 °¯ s 16
½° ¾. °¿
Solution:
We express
let
F s
then,
ª 1 « 2 «¬ s 4
1 4 s 16
1 and G s s 4 2
as
ºª 1 »« 2 »¼ «¬ s 4
1 s 4 2
º »; »¼
The Inverse Laplace Transform
197
1 ½ 1 L1 ^F s ` L1 ® 2 ¾ sin 2t ¯s 4¿ 2 and 1 ½ 1 L1 ^G s ` L1 ® 2 ¾ sinh 2t . ¯s 4¿ 2 We have the convolution theorem t
L1 ^ F s G s `
f t g t
³ f u g t u du 0
ª 1 º L « 4 ¬ s 16 »¼ 1
t
1 sinh 2u sin 2 t u du. 4 ³0
(2.17)
We can carry out the integration in Eq. (2.17) t ª 1 º 1 ª¬cosh 2u sin 2 t u sinh 2u cos 2 t u º¼ 0 L1 « 4 » ¬ s 16 ¼ 16 .
Thus,
ª 1 º 1 L1 « 4 >sinh 2t sin 2t @. ¬ s 16 »¼ 16 Example 2.45. Using the convolution theorem, determine
1 ½ L1 ® 2 ¾. ¯ 4s 9 ¿ Solution:
We express
1 4s 2 9
as
ª 1 ºª 1 º « »« »; «¬ 2 s 3 »¼ «¬ 2 s 3 »¼
Chapter 2
198
let F
s
1 and G s 2 s 3
1 2 s 3
then,
1 ½ L1 ^F s ` L1 ® ¾ ¯ 2s 3 ¿
1 23 t e 2
3 1 ½ 1 2t L1 ^G s ` L1 ® e . ¾ ¯ 2s 3 ¿ 2
We have the convolution theorem
L1 ^ F s G s `
t
f t g t
³ f u g t u du 0
ª 1 º L1 « 2 ¬ 4 s 9 »¼
3 3 u t u 1 2 2 e e du. ³ 40 t
(2.18) We can carry out the integration in Eq. (2.18)
ª 1 º L « 2 » ¬ 4s 9 ¼ 1
t
1 32 t e ³ e 3u du 4 0
Thus, 3 ª 1 º 1 2t L1 « 2 » e 1 e 3t . ¬ 4s 9 ¼ 12
Example 2.46. Using the convolution theorem, solve for t
integral equation
f t 2t 2 ³ f t u e u dt. 0
f t in the
The Inverse Laplace Transform
199
Solution: Integral equations are equations in which the unknown function appears under the integral. If the derivatives of the function are also in the equation, then they are called integro-differential equations. We recognize the integral on the right as the convolution of
f t with
e t . Therefore, the integral equation has the form
f t 2t 2 f t et
.
We apply the Laplace transform and the convolution theorem to this equation to get
4 1 F s . s3 s 1
F s
Then F s
4 4 4 , which we invert to obtain f t 3 s s
Example 2.47. Using the convolution theorem, solve for
2 2t 2 t 3 . 3
f t in the
integral equation t
f t
2e t ³ sin t u f u du 0
Solution: We recognize the integral on the right as the convolution of
e t . Therefore, the integral equation has the form
f t with
Chapter 2
200
f t 2et f t sin t
.
We apply the Laplace transform and the convolution theorem to this equation to get
F s
F s
2 1 F s 2 . s 1 s 1
2 s2 1
s s 1 2
.
It is now necessary to invert
F s and to accomplish this, some
algebraic manipulation is necessary to identify the terms on the right using the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
F s
2 s2 1
s s 1 2
F s becomes 4 2 2 2 s s s 1
which we invert to obtain
f t 2t 2 4e t .
Example 2.48. Using the convolution theorem, solve for integral equation t
g t
3t 2 e t ³ e t u g u du 0
.
g t in the
The Inverse Laplace Transform
201
Solution: We recognize the integral on the right as the convolution of
et .
Therefore,
the
integral
g t 3t 2 et g t et
equation
has
g t with the
form
.
We apply the Laplace transform and the convolution theorem to this equation to get
G s
6 1 1 G s . 3 s s 1 s 1
It is now necessary to invert
G s . to accomplish this. Some algebraic
manipulation is necessary to identify the terms on the right using the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
G s
G s . becomes
6 6 1 2 s3 s 4 s s 1
which we invert to obtain
f t 3t 2 t 3 1 2e t .
2.10 Integral and Integro-differential Equations Equations with unknown functions under the integral are termed integral equations. If the equation also contains derivatives, then it is termed an integro-differential equation. Such equations are usually difficult to solve. However, if the integrals are in the form of a convolution, then we can use Laplace transforms to solve them. The following example shows how to
Chapter 2
202
apply the Laplace transform method to arrive at the general solution of integro-differential equations using the convolution theorem. Example 2.49. Using the convolution theorem, solve for
f t in the
integro-differential equation
df t dt y t
t
3³ f t u e 2u du
y t , f 0
4
0
0 t 1 0 ® 2t t !1 ¯4e .
Solution: Equations containing an unknown function within the integral are termed integral equations. If there are also derivatives in the equation, then it is called an integro-differential equation. Using the second-shifting property and Table 2.1, we find the Laplace transform of
Y s
y t ,
4e2e s s2 .
Applying the Laplace transform to both sides of the integro-differential equation, we get
° df t t ½° 3³ f t u e 2u du ¾ Y s L® 0 ¯° dt ¿° . Applying the convolution theorem to this equation, we get
The Inverse Laplace Transform
sF s 4 F s
3F s s2
4e2 e s s2
4 s 2
4e2 e s s 2 2s 3 s 2 2s 3 .
Then,
F s
Let
4 s 2
4e 2 e s s 1 s 3 s 1 s 3
F1 s
4 s 2
.
s 1 s 3
Using partial fractions, we get
F1 s
Since Let
3 1 . s 1 s 3
f1 t
F2 s
3e e ,
L1 ª¬ F s ¼º
4
s 1 s 3
t
.
Using partial fractions, we get
F2 s since
1 1 s 1 s 3 f 2 t et e 3t ,
then finally we invert to obtain
3t
203
Chapter 2
204
ª 4 s 2 4e 2 e s º L1 « » ¬« s 1 s 3 s 1 s 3 ¼»
3e
f t
t
3e e e e t
3t
2
t 1
e 3 t 1 u t 1
e 3t e 2 e t 1 e 3 t 1 u t 1 .
Example 2.50. Compute
L ^ f t * g t ` for the signals shown in
figures 2.2a and b.
f(t)
g(t)
1
0
1 1 (a)
t 0
1 (b)
t
Figure 2.2 Solution: The mathematical expression for the functions is given below
1 0 d t d 1 f t ® u t u t 1 ¯0 t ! 1 and
t g t ® ¯1
0 d t d1 t !1
We have the convolution theorem
L ^ f t g t ` F s G s
(2.19)
The Inverse Laplace Transform
205
Using the second-shifting property and Table 2.1, we find the Laplace transform of
F s
f t and g t , which is
1 e . s
1 e s and F s s
s2
Eq. (2.19) becomes
L ^ f t g t `
§ 1 e s · § 1 s · F s G s ¨ ¸¨ 2 1 e ¸ ¹ © s ¹© s
Thus,
L ^ f t g t `
1 1 e s 3 s
Example 2.51. Compute f t
F s
1
s a s b
by
i
Partial fractions.
ii
Convolution theorem.
iii Bromwich integral. Solution:
i By partial fractions.
2
. L1 ^ F s ` where
Chapter 2
206
We begin by finding the partial fraction expansion for
F s . The
denominator consists of two distinct linear factors and the expansion has the form
F s
1
A B s a s b
s a s b
where A and B are real numbers to be determined
ª 1 º A « » «¬ s b »¼ s
a
ª 1 º 1 B « » ba «¬ s a »¼ s ,
b
1 a b
Hence,
F s
1
s a s b
1 ª 1 º 1 ª 1 º « » « » b a «¬ s a »¼ b a «¬ s b »¼
.
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find
L1 ª¬ F s º¼
ª 1 º ª 1 º 1 1 L1 « L1 « » » b a «¬ s a »¼ b a «¬ s b »¼
.
Finally, using the transform pairs established in Table 2.1, we have
ª º 1 L1 « » «¬ s a s b »¼
1 e bt e at . a b
The Inverse Laplace Transform
ii
Convolution theorem.
We express
let
207
F1 s
F s
1
as
s a s b
1 and F2 s s a
ª 1 ºª 1 º « »« »; ¬« s a ¼» ¬« s b ¼»
1 . s b
Then,
1 ½ at L1 ^F1 s ` L1 ® ¾ e ¯s a¿
and
1 ½ bt L1 ^F2 s ` L1 ® ¾ e ¯s b¿ . We have the convolution theorem 1
L
^F s F s ` 1
2
t
f1 t f 2 t
³ f u f t u du 1
2
0
ª º 1 L « » ¬« s a s b ¼» 1
t
³
e au e b t u du
0
(2.20) We can carry out the integration in Eq. (2.20)
ª º 1 L « » «¬ s a s b »¼ 1
Thus,
t
³ 0
e au e b t u du
Chapter 2
208
ª º 1 L1 « » «¬ s a s b »¼
1 e bt e at . a b
iii By Bromwich integral. Let
1
F s
s a s b
,
Then,
e st F s the poles in
e st , s a s b F s are s
a & s
b . Both are simple poles
^
`
Re s ^ a `
s o a
Re s ^ b `
lim e st F s s b
lim e st F s s a
s o b
^
`
e st ½ lim ® ¾ s o a s b ¯ ¿ e st ½ lim ® ¾ s o b s a ¯ ¿
e at ba e bt a b
Using the Bromwich integral V jf
f t
1 e st F s ds 2S j V ³jf
¦ sum of residues
1 e bt e at . a b
The Inverse Laplace Transform
Example 2.52. Compute f t
1
F s
s 2 s 3
209
L1 ^ F s ` where
by
i Partial fractions. ii
Convolution theorem.
iii Bromwich integral. Solution:
i Partial fractions. We begin by finding the partial fraction expansion for
F s . The
denominator consists of two distinct linear factors and the expansion takes the following form
F s where
1
s 2 s 3
A and B are real numbers to be determined
ª 1 º A « » ¬« s 3 ¼» s Hence,
A B s 2 s 3
1 , B 2
ª 1 º « » ¬« s 2 ¼» s
1 3
Chapter 2
210
F s
1
s 2 s 3
ª 1 º ª 1 º « »« » «¬ s 2 »¼ «¬ s 3 »¼ .
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find
ª 1 º 1 ª 1 º L1 ª¬ F s º¼ L1 « »L « » «¬ s 2 »¼ «¬ s 3 »¼ . Finally, using the transform pairs established in Table 2.1, we have
ª º 1 L1 « » «¬ s 2 s 3 »¼
ii
3t
e 2t .
Convolution theorem.
We express
let
e
F1 s
F s
1
s 2 s 3
1 and F2 s s 2
as
1 s 3
then
1 ½ 2t L1 ^F1 s ` L1 ® ¾ e ¯s 2¿
1 ½ 3t L1 ^F2 s ` L1 ® ¾ e ¯ s 3¿ .
ª 1 ºª 1 º « »« »; «¬ s 2 »¼ «¬ s 3 »¼
and
The Inverse Laplace Transform
211
We have the convolution theorem
L1 ^ F1 s F2 s `
t
f1 t f 2 t
³ f u f t u du 1
2
0
ª º 1 L1 « » «¬ s 2 s 3 »¼
t
³
e 2u e3 t u du
0
(2.21)
We can carry out the integration in Eq. (2.21)
ª º 3t t u 1 L1 « » e ³ e du «¬ s 2 s 3 »¼ 0 Thus,
ª º 1 L1 « » «¬ s 2 s 3 »¼
e
3t
e 2t .
iii Bromwich Integral. Let
F s
1
s 2 s 3
,
then,
e F s st
e st , s 2 s 3
the poles of F s are s
2 &s
3 . Both are simple poles
Chapter 2
212
^
`
e st ½ lim ® ¾ s o2 s 3 ¯ ¿
`
e st ½ 3t lim ® ¾ e s o3 s 2 ¯ ¿ .
Re s ^ 2 ` lim e F s s 2 s o2
st
^
Re s ^ 3 ` lim e st F s s 3 s o3
e 2 t
Using the Bromwich Integral V jf
1 e st F s ds ³ 2S j V jf
f t
Example 2.53. Compute f t
F s
¦ sum of residues e L1 ^ F s ` where
1 by s s 4
2
i Partial fractions. ii
Convolution theorem.
Solution:
i Partial fractions. We can take the partial fractions two ways
F s
1 s s 4
2
A Bs C s s2 4
1 A s 2 4 Bs C s
3t
e 2t .
The Inverse Laplace Transform
213
Hence, comparing the coefficients of both sides, we find
s2 :
A B 0,
s:
C
0,
1:
A
1 4.
Using these equations, we obtain
A
1 and C 4
B
0.
Hence, we have
F s
1 s s2 4
1 ª1 s « 2 4 «s s 4 ¬
º » »¼ .
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
L1 ª¬ F s º¼
1 1 ª 1 º 1 1 ª s L « L 4 «¬ s »¼ 4 « s 2 4 ¬
º » »¼
Finally, using the transform pairs established in Table 2.1, we have
f t
1 ^1 cos 2t `. 4
Alternative method:
Chapter 2
214
We begin by finding the partial fraction expansion for
F s . The
denominator consists of two complex roots, linear factors, and the expansion has the form
F s
1 s s 4
2
1 s s 2i s 2i
A B B s s 2i s 2i
Where A and B are real numbers to be determined
A
B
ª 1 « 2 ¬« s 4
ª º 1 « » «¬ s s 2i »¼ s
º » »¼ s
2 i
1 , B 4 0
ª º 1 « » ¬« s s 2i ¼» s
2i
1 8i 2
1 , 8
1 8
Hence,
F s
1 s s 4
2
1 s s 2i s 2i
11 1 1 1 1 4 s 8 s 2i 8 s 2i
. Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
° 1 f t L1 ® 2 °¯ s s 4
½° 1 1 ½ 1 ° 1 ½° 1 ° 1 ½° 1 L1 ® ¾ L1 ® ¾ ¾ L ® ¾ °¿ 4 ¯ s ¿ 8 ¯° s 2i ¿° 8 ¯° s 2i ¿°
Finally, using the transform pairs established in Table 2.1, we have
The Inverse Laplace Transform
° 1 f t L1 ® 2 °¯ s s 4
½° 1 1 1 ei 2 t e i 2 t ¾ 8 °¿ 4 8
f t
° 1 L ® 2 °¯ s s 4
½° 1 1 ei 2t e i 2t ¾ 4 4 2 °¿
f t
° 1 L ® 2 °¯ s s 4
½° 1 1 cos 2t . ¾ °¿ 4
Thus,
ii
1
1
Convolution theorem.
We express
Let F s
º 1 1 ª 1 as ª º « «¬ s »¼ s 2 4 » ; s s2 4 «¬ »¼
1 and G s s
1 s 4 2
so that
f t 1 and g t
L1 ^ F s G s `
1 sin 2t 2 t
f t g t
³ f t u g u du 0
215
Chapter 2
216
ª 1 L1 « 2 «¬ s s 4
º 1t » ³ sin 2u du. »¼ 2 0
Carrying out integration gives
ª 1 L1 « 2 «¬ s s 4
ª 1 L « 2 «¬ s s 4
º 1 t » cos 2u 0 »¼ 4
º » »¼
1
1 1 cos 2t . 4
Example 2.54. Compute f t
F s
L1 ^ F s ` where
1 by s s 4 2
i Partial fractions. ii
Convolution theorem.
Solution:
i Partial fractions. We can take the partial fractions in two ways.
F s
1 s 4 s2
A B C 2 s 4 s s
The Inverse Laplace Transform
1 As 2 B s 4 s C s 4
217
.
Hence, comparing the coefficients of both sides, we get
s2 :
A B 0,
s:
4B C
1:
4C
0,
1.
Using these equations, we obtain
A
B
1 and C 16
1 . 4
Hence,
F s
1 s 4 s2
1 1 1 1 1 1 16 s 4 16 s 4 s 2
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find
L1 ¬ª F s ¼º
1 1 ° 1 °½ 1 1 1 ½ 1 1 1 ½ L ® ¾ L ® ¾ L ® ¾ 16 °¯ s 4 °¿ 16 ¯ s ¿ 4 ¯ s 2 ¿
Finally, using the transform pairs established in Table 2.1, we have
L1 ª¬ F s º¼
1 4t 1 1 e t. 16 16 4
Another way to take partial fractions is to note that
Chapter 2
218
C C A 22 21 2 s4 s s
1 s 4 s2 ª1º A « 2» ¬ s ¼s C22
C21
d0 ds 0
4
1 16
° 1 ½° ® ¾ ¯° s 4 ¿°s
0
° 1 ½° ® ¾ ¯° s 4 ¿°s
1 d ° 1 °½ ® ¾ 3 2 ! ds °¯ s 4 °¿s
0
1 16
0
1 4
.
Hence, we have
F s
1 s 4 s2
1 1 1 1 1 1 16 s 4 16 s 4 s 2
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find
L1 ¬ª F s ¼º
1 1 ° 1 ½° 1 1 1 ½ 1 1 1 ½ L ® ¾ L ® ¾ L ® ¾ 16 ¯° s 4 ¿° 16 ¯ s ¿ 4 ¯ s 2 ¿
.
Finally, using the transform pairs established in Table 2.1, we have
ª º 1 4t 1 L1 « e 4t 1 . 2» «¬ s 4 s »¼ 16
The Inverse Laplace Transform
ii
Convolution theorem.
We express
Let F s so that
1
L
219
ª 1 ºª 1 º 1 as « » « 2 »; s2 s 4 «¬ s 4 ¼» ¬ s ¼ 1 and G s s4
1 s2
f t e 4t and g t t t
^F s G s `
f t g t
³ f t u g u du 0
ª º 1 L1 « 2» ¬« s 4 s ¼»
t
³e
4 t u
udu
0
ª º 4 t t 4u 1 L « e ³ e udu 2» 0 ¬« s 4 s ¼» . 1
Carrying out integration gives
ª º 1 4t 1 L1 « e 4t 1 . 2» ¬« s 4 s ¼» 16
Summary In this chapter, we have described and explained the properties of the inverse Laplace transform and the convolution theorem:
220
Chapter 2
x We started with a definition of the inverse Laplace transform and presented some standard functions in a table. The formula for the inverse Laplace transform uses a contour integration, which has been illustrated using simple solved examples. This contour integration is a little difficult to solve. An evaluation of the inverse Laplace transform using the partial fraction method for distinct linear factors, repeated linear factors, and quadratic factors has been given with several examples. x A definition of the convolution between two functions has been stated and the properties of convolution, including its commutative, distributive, and associative characteristics, have been stated and proved. Convolution between two functions has been illustrated using simple solved examples. Convolution of two time-domain functions results in the multiplication of their Laplace transforms in the frequency domain. The convolution theorem is useful for finding the inverse Laplace transform of the product of two functions in the frequency domain. Evaluation of the inverse Laplace transforms using the convolution theorem has also been explained with several examples.
CHAPTER 3 APPLICATION OF THE LAPLACE TRANSFORM TO LTI DIFFERENTIAL SYSTEMS: TRANSFER FUNCTIONS
3.1 Introduction This chapter describes mathematical models for certain electrical circuits. These models describe the relationship between voltage and current (Kirchhoff’s laws) in a circuit. The properties of resistance, inductance, and capacitance are assumed to be due to the devices present at specific locations in the circuit and the connecting wires are ideal conductors. Such linear elements in circuit models yield linear, time-invariant differential equations with constant coefficients. We also cover some mechanical systems that are described by similar linear ordinary differential equations with constant coefficients, which should provide physical intuition for analogous circuits and their components. These mathematical models represent the input/output characteristics of the electrical and mechanical system. We will mainly limit ourselves to systems described by ordinary linear differential equations with constant coefficients and with initial conditions all assumed to be equal to zero. The transfer function is a rational function of
s
H s
and the impulse response can thus be
determined by transforming this back to the time domain. Newton’s laws
Chapter 3
222
(for the mechanical system) and Kirchhoff’s laws (for the electrical system) lead to mathematical models that describe the relationship between dynamical system inputs and outputs. One such model is the linear time-invariant differential equation in Eq. (3.1):
d k y t ak ¦ dt k k 0 N
d k x t bk ¦ dt k k 0 M
(3.1) Many systems can be approximately described by this equation, which relates the output parameters
ak
y t and
to the input
x t by way of the system
bk . The transfer function H s can be
determined by partial fraction expansion followed by transformation back to the time domain. The response input
x t
is
given
by
y t the
of the system to an arbitrary
convolution
of
h t
with
x t i.e y t h t * x t . In order to find the response y t for a given input transform
x t , it is often easier to first determine the Laplace
X s of x t and subsequently to transform X s H s
back to the time domain. This is because
X s , and hence
Y s X s H s , is a rational function for a large class of inputs. The inverse Laplace transform
y t of Y s can then immediately be
determined by partial fraction expansion. This simple standard solution
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
223
method is yet another advantage of the Laplace transform over other transforms. The transfer function for the linear time invariant system described by the differential equations treated in Section 3.3 shows that the initial conditions are always zero. In Section 3.4, we will show that the Laplace transform can equally well be applied to an electrical circuit to find the transfer function. Finally, in Section 3.5 we briefly describe how the Laplace transform can be used to find the transfer function for the mechanical system. LEARNING OBJECTIVES On reaching the end of this chapter, we expect that you: x Have understood and can apply The definition of the transfer function. x Can find the input-output relations using the transfer function. x Can find the transfer function of the linear time-invariant system described by differential equations. x Can find the transfer function of the electrical system. x Can find the transfer function of the mechanical system.
3.2 Transfer Functions The transfer function of a linear time-invariant system (LTI) is defined as the Laplace transform of the system output to the Laplace transform of the system input, assuming that the initial conditions are equal to zero
Chapter 3
224
Input X (s)
Transfer Function H(s)
Output Y (s)
Figure 3.1 Transfer Function
T .F
H s
L > output @ L >input @
all initial conditions zero
or
ª L ^ y t ` º « » «¬ L ^ x t ` »¼
H s
all initial conditions zero
ª Y s º « » «¬ X s »¼
.
all initial conditions zero
Using this equation, we can obtain the transfer function using any inputoutput pair
x t y t . Indeed, if the input is an impulse G t , then its
Laplace transform is 1 and the transfer function simply equals the Laplace transform of the corresponding output, which by definition is the impulse response.
i
Transfer functions are frequently used in engineering to
characterize the input-output relationships of linear time-invariant systems.
ii
Transfer functions play an essential role in the analysis and
design of LTI systems.
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
225
3.3 Transfer Function of the Linear Time-invariant System Consider a linear time-invariant system characterized by the differential equation N
¦ ak k 0
where
d k y t dt k
M
¦ bk k 0
d k x t dt k
(3.2)
y t is the output to the input x t applied at time t
0.
We use the differentiation and linearity properties of the Laplace transform to obtain the transfer function T .F
H s
Y s X s
.
Taking the Laplace transform, and with all the initial conditions assumed to be zero, we get N
¦ ak s k Y s k 0
Where
M
¦b
k
sk X s .
k 0
Y s Laplace transform of y t and X s Laplace
transform of x t
y t .
Therefore, N
T .F
H s
Y s X s
¦a s
k
¦b
sk
k
k 0 M
k
k 0
.
Chapter 3
226
The following examples show how to use the Laplace transform method to obtain the transfer function of a linear time-invariant system (LTI) described by linear non-homogeneous differential equations with constant coefficients.
3.3.1 Definitions: Linear and Nonlinear nth-order Ordinary Differential Equations The general nth-order ordinary differential equation can be written symbolically as N
¦ ak t k 0
d k y t dt k
f t .
An nth-order ordinary differential equation is linear if it can be written in the form
aN t
d N y t d N 1 y t d 2 y t dy t a t a t ...... a1 t a0 t y t f t N 1 2 N N 1 2 dt dt dt dt
aN t , aN 1 t ,......a1 t and a0 t
,
which are all functions of the independent variable t alone. A nonlinear ordinary differential equation is an ordinary differential equation that is not linear. From the definition, we can see that for an ordinary differential equation to be linear it is important that:
(i) Each coefficient function ak t depends only on the dependent variable
t and not on the independent variable y t .
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
227
(ii) The independent variable y t and all of its derivatives y t occur algebraically to the first degree only. That is, the power of each term containing y and its derivatives is equal to 1.
(iii) There are no terms that involve the product of either the independent
y t and any of its derivatives or two or more of its
variable
derivatives.
(iv) Functions of y t or any of its derivatives, such as y, log y, ev or sin y (nonlinear functions cannot appear in the equation). Example 3.1. Determine the transfer function
H s of the system
represented by the differential equation
d 3 y t dt 2
d 2 y t dy t d 2 x t dx t 2 2 y t 2 x t . dt 2 dt dt 2 dt
Solution: Let
y ''' t 2 y '' t y ' t 2 y t x '' t 2 x ' t x t
.
Because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation, we have
^
`
^
` ^
`
^
`
^
`
L y ''' t 2 L y '' t L y ' t 2 L ^ y t ` L x '' t 2 L x ' t L ^ x t `
Chapter 3
228
Using the derivative property and assuming all initial conditions to be zero gives
s
3
s
2s 2 s 2 Y s
2
2s 1 X s
so that the system transfer function is given by
s
Y s X s
s
3
2
2s 1
2s 2 s 2
Thus, the transfer function
T .F H s
.
H s is
s
Y s X s
s
3
2
2s 1
2s 2 s 2
.
In this equation, we show the Laplace-transform solution of differential equations with constant coefficients, which transforms these differential equations into algebraic equations. We then solve the algebraic equations and use partial-fraction expansion to transform the solutions back into the time domain. For an equation in which the initial conditions are ignored, this method of solving the equations takes us to the transfer-function representation of LTI systems. The transfer-function approach is a standard procedure for the analysis and design of LTI systems. An important use of transfer functions is to determine the characteristics of an LTI system.
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
229
Example 3.2. Determine the transfer function of the system represented by the differential equation
d 3 y t dt 2
8
d 2 y t dt 2
5
dy t dt
14 y t 4
dx t dt
x t .
Solution: Let
y ''' t 8 y '' t 5 y ' t 14 y t 4 x ' t x t
.
Because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation, we have
^
`
^
`
^
`
^
`
L y ''' t 8 L y '' t 5 L y ' t 14 L ^ y t ` 4 L x ' t L ^ x t ` Using the derivative property, and assuming all initial conditions are zero, gives
s
3
8s 2 5s 4 Y s
4s 1 X s
so that the system transfer function is given by
Y s X s
4s 1
s
3
8s 2 5s 14
.
Thus,
T .F
H s
Y s X s
4s 1
s
3
8s 2 5s 14
.
Chapter 3
230
Example 3.3. Consider the initial value problem
d 2 y t dy t y t x t dt 2 dt along with the initial conditions
y 0 0 y' 0 , where
x(t )
0 ° ®sin t °0 ¯
0dt dS S t d 3S t t 3S
(a) Determine the system transfer function (b) Determine
H s .
Y s .
Solution: Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
° d 2 y t dy t ½° L® y t ¾ L ^ x t ` 2 dt ¯° dt ¿° L ª¬ y '' t º¼ L ª¬ y ' t º¼ L ª¬ y t º¼
X s
ª¬ s 2Y s sy 0 y ' 0 º¼ ª¬ sY s y 0 º¼ Y s
X s
.
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
231
Imposing the initial conditions, we obtain the Laplace transform of the solution as
s
2
s 1 Y s
X s
Thus, the transfer function
H s
Y s X s
The function
.
H s is
1 s s 1 2
x t is continuous, but it is defined differently on the
intervals. As such, the mathematical expression for the function is
x(t )
0 ° ®sin t °0 ¯
0dt dS S t d 3S t t 3S
In terms of unit step functions,
x t can be expressed as
x(t ) x1 t ^ x2 t x1 t ` u t a ^ x3 t x2 t ` u t b x(t ) 0 sin t u t S sin t u t 3S
.
Then, taking the Laplace transform,
L ^ x(t )` L ^sin t u t S ` L ^sin t u t 3S ` and on using the result in (1.7), we get
Chapter 3
232
L ^ x(t )` eS s L ^ f1 (t )` e2S s L ^ f 2 (t )`
f1 (t S ) sin t replace
(3.3)
f 2 (t 3S ) sin t
t t S
replace t
t 3S
f1 (t ) sin t S
f1 (t ) sin t 3S
f1 (t ) sin t
f1 (t ) sin t
then its Laplace transform is
1 s2 1
L ^ f1 (t )`
then its Laplace transform is
L ^ f 2 (t )`
As such, Eq. (3.3) becomes
§ 1 · § 1 · L ^ x(t )` eS s ¨ 2 ¸ e2S s ¨ 2 ¸ © s 1 ¹ © s 1 ¹ thus,
X s
L ^ x(t )`
1 e 2S s e S s . s 1 2
We have
Y s X s
1 s s 1 2
1 s2 1
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
Y s
Y s
233
X s
s
2
s 1
1 e 2S s e S s . 2 2 s 1 s s 1
Example 3.4. Determine the system response if the transfer function of the system is given by
ª s2 « 3 «¬ s 6 s 2 11s 6
H s
º » »¼
and if the unit step function is applied as an input. Solution:
H s
s2 s 3 6s 2 11s 6
.
Given the input
x t u t X s
1 s
we have
Y s
Y s
X s H s
1 s2 s s 3 6s 2 11s 6
s s 6 s 11s 6 3
2
s s 3 6s 2 11s 6
Chapter 3
234
It is now necessary to invert
Y s . To accomplish this some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s becomes
s s 1 s 2 s 3
A B C s 1 s 2 s 3
where A, B, and C are real numbers to be determined
ª º s 1 A « ,B » «¬ s 2 s 3 »¼ s 1 2 ª º s 3 C « » ¬« s 2 s 1 ¼» s 3 2
ª º s « » «¬ s 1 s 3 »¼ s
2 2
.
Hence, the given function
Y s
Y s can be expanded as
s s 1 s 2 s 3
1 1 1 3 1 2 2 s 1 s 2 2 s 3
.
Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find
y t
1 °½ 3 1 ° 1 °½ 1 1 ° 1 °½ 1 ° L ® ¾ 2L ® ¾ L ® ¾ 2 °¯ s 1 °¿ °¯ s 2 °¿ 2 °¯ s 3 °¿
.
Finally, using the transform pairs established in Table 2.1, we have
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
y t
235
1 t 3 e 2e 2 t e 3 t . 2 2
Example 3.6. Determine the system response if the transfer function of the system given by
H s
4s 1
s
3
8s 2 5s 4
if the unit step function is applied as an input. Solution:
H s
4s 1
s
3
8s 2 5s 4
Given the input
x t u t X s
1 s
we have
Y s
Y s
X s H s
4s 1 s s 8s 2 5s 14 3
4s 1 s s 8s 2 5s 14 3
It is now necessary to invert
.
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with
Chapter 3
236
the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s becomes
4s 1 s s 1 s 2 s 7
A B C D s s 1 s 2 s 7
where A, B, C, and D are real numbers to be determined.
ª º 4s 1 A « » «¬ s 1 s 2 s 7 »¼ s
ª 4 s 1 º « » «¬ s s 1 s 7 »¼ s
C
2
0
1 , 14
B
3 , 10
D
ª 4s 1 º « » «¬ s s 2 s 7 »¼ s
ª 4 s 1 º « » ¬« s s 1 s 2 ¼» s
1
7
1 , 8
29 280
. Hence, the given function
Y s
Y s can be expanded as
4s 1 s s 1 s 2 s 7
§ 1 ·1 §1· 1 §3· 1 § 29 · 1 ¨ ¸ ¨ ¨ ¸ ¨ ¸ ¸ © 14 ¹ s © 8 ¹ s 1 © 10 ¹ s 2 © 280 ¹ s 7
. Taking the inverse Laplace transform of
Y s and using Table 2.1 give
us
§ 1 · §1· §3· § 29 · 7 t y t ¨ ¸ ¨ ¸ e t ¨ ¸ e 2t ¨ ¸e . © 14 ¹ © 8 ¹ © 10 ¹ © 280 ¹ Example 3.7. The impulse response
h t 1 et .
h t of an LTI system is given by
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
237
a Determine the transfer function H s of the system. b Determine the response to the input x t
e 2t .
Solution: Given the impulse response
h t of an LTI system h t 1 et ,
then the transfer function is
H s L ª¬ h t º¼
ª1 1 º H s « ¬ s s 1 »¼
1 . s s 1
Given the input
x t
1 s2
e 2t X s
we have
Y s X s H s
Y s
1 s s 1 s 2
1 s s 1 s 2
.
It is now necessary to invert
Y s . . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with
Chapter 3
238
the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s becomes
1 s s 1 s 2
A B C s s 1 s 2
where A, B, and C are real numbers to be determined.
ª º 1 A « » ¬« s 1 s 2 ¼» s C
ª 1 º « » «¬ s s 1 »¼ s
2
0
ª 1 º « » ¬« s s 2 ¼» s
1 1
,
1 2
Hence, the given function
Y s
1 ,B 2
1 s s 1 s 2
Y s can be expanded as
11 1 1 1 2 s s 1 2 s 2
.
Now that we have obtained the partial fraction expansion, we can consider the inverse Laplace transform of each term on the right-hand side and apply the Laplace transform’s linearity property. We arrive at
y t
1 1 1 ½ 1 ° 1 ½° 1 1 ° 1 ½° L ® ¾ L ® ¾ L ® ¾ 2 ¯s¿ ¯° s 1 ¿° 2 ¯° s 2 ¿°
.
Finally, using the transform pairs established in Table 2.1, we have
y t
1 t 1 2t e e . 2 2
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
239
Example 3.8. Use the Laplace transform to find the transfer function and the impulse response of the system if the differential equation describes the system
d 2 y t dt 2
5
dy t dt
d 2 x t
6 y t
dt 2
8
dx t dt
13x t .
Solution: Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
^
`
^
`
^
`
^
`
L y '' t 5 L y ' t 6 L ^ y t ` L x '' t 8 L x ' t 13L ^ x t ` Using the derivative property, and assuming all initial conditions are zero, gives
s
2
5s 6 Y s
s
2
8s 13 X s
Thus, the transfer function is given by
T .F
s 8s 13 H s X s s 5s 6 . Y s
2
2
Note that the degree of the numerator is the same as the degree of the denominator. As such, we have to put it into fraction form so that we can apply partial fraction expansion
H s 1
3s 7 A B 1 s 2 s 3 s 2 s 3
.
Chapter 3
240
It is now necessary to invert
H s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
H s 1
H s becomes
1 2 s 2 s 3
Taking the inverse Laplace transform of
H s and using Table 2.1
gives the result
h t
G t e 2t 2e 3t
Example 3.9. Consider the causal LTI system described by the second differential equation
d 3 y t dt
3
6
d 2 y t dt
2
11
dy t dt
6 y t x t
a Determine the transfer function H s .
b Determine the impulse response h t .
c Determine the output response for the input x t
u t .
d Determine the output response for the input x t
G t .
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
241
Solution:
a The transfer function H s . Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write 3 d 2 y t dy t ° d y t L® 6 11 6 y t 3 2 dt dt °¯ dt
°½ ¾ L ^x t ` °¿
L ª¬ y ''' t º¼ 6 L ª¬ y '' t º¼ 11L ª¬ y ' t º¼ 6 L ª¬ y t º¼
s
3
6 s 2 11s 6 Y s
Y s X s
1 s 3 6s 2 11s 6
The transfer function
H s
H s
X s
Y s
H s
1 s 6s 11s 6
X s
Y s
1 s 1 s 2 s 3
X s
3
2
b The impulse response h t .
X s .
Chapter 3
242
It is now necessary to invert
H s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
H s
H s becomes
1 s 1 s 2 s 3
A B C s 1 s 2 s 3
where A, B, and C are real numbers to be determined.
ª º 1 A « » «¬ s 2 s 3 »¼ s C
ª º 1 « » ¬« s 1 s 2 ¼» s
Hence, the given function
H s
1
3
1 , 2
ª º 1 B « » «¬ s 1 s 3 »¼ s
1, 2
1 2
H s can be expanded as
1 s 1 s 2 s 3
1 1 1 1 1 2 s 1 s 2 2 s 3
Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find
h t
1 1 ° 1 ½° 1 ° 1 ½° 1 1 ° 1 ½° L ® ¾ L ® ¾ L ® ¾ 2 °¯ s 1 °¿ °¯ s 2 °¿ 2 °¯ s 3 °¿
Finally, using the transform pairs established in Table 2.1, we have
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
h t
243
1 t 2t 1 3t e e e 2 2 .
c Output response for the input x t
u t .
We have
Y s X s Y s
1 s 1 s 2 s 3 1 X s s 1 s 2 s 3
Given the input x t
u t X s
1 s
therefore
Y s
1 s s 1 s 2 s 3
It is now necessary to invert
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s becomes
1 s s 1 s 2 s 3
A B C D s s 1 s 2 s 3
where A, B, C, and D are real numbers to be determined.
Chapter 3
244
ª º 1 A « » «¬ s 1 s 2 s 3 »¼ s ª º 1 B « » ¬« s s 2 s 3 ¼» s C
D
1
ª º 1 « » ¬« s s 1 s 3 ¼» s ª º 1 « » ¬« s s 1 s 2 ¼» s
Hence, the given function
Y s
2
3
0
1 , 6
1 2 1 , 2 1 6
.
Y s can be expanded as
11 1 1 1 1 1 1 6 s 2 s 1 2 s 2 6 s 3
Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find
y t Y s
1 1 1 ½ 1 1 ° 1 °½ 1 1 ° 1 °½ 1 1 ° 1 °½ L ® ¾ L ® ¾ L ® ¾ L ® ¾ 6 ¯ s ¿ 2 ¯° s 1 ¿° 2 ¯° s 2 ¿° 6 ¯° s 3 ¿°
Finally, using the transform pairs established in Table 2.1, we have
y t
y t
1 1 t 1 2t 1 3t e e e . 6 2 2 6
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
d Output response for the input x t
245
G t .
We have
Y s X s Y s
1 s 1 s 2 s 3 1 X s s 1 s 2 s 3
.
Given the input
x t G t X s 1 therefore
Y s
1 . s 1 s 2 s 3
It is now necessary to invert
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s becomes
1 s 1 s 2 s 3
A B C s 1 s 2 s 3
where A, B, and C are real numbers to be determined.
Chapter 3
246
ª º 1 A « » «¬ s 2 s 3 »¼ s ª º 1 « » ¬« s 1 s 2 ¼» s
C
Hence, the given function
Y s
1
3
1 , 2 1 2
ª º 1 B « » «¬ s 1 s 3 »¼ s
1, 2
.
H s can be expanded as
1 s 1 s 2 s 3
1 1 1 1 1 2 s 1 s 2 2 s 3
Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find
y t
1 1 ° 1 ½° 1 ° 1 ½° 1 1 ° 1 ½° L ® ¾ L ® ¾ L ® ¾ 2 ¯° s 1 ¿° ¯° s 2 ¿° 2 ¯° s 3 ¿°
.
Finally, using the transform pairs established in Table 2.1, we have
y t
1 t 2t 1 3t e e e . 2 2
Example 3.10. Consider the initial value problem.
d 3 y t dt
3
7
d 2 y t dt
2
e2t 4t
along with the initial conditions
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
247
y 0 0 y ' 0 y '' 0 .
a Determine the system transfer function H s . b Determine
Y s .
Solution: (a) Assuming that the Laplace transform of the solution
Y s exists,
and using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
° d 3 y t d 2 y t L® 7 3 dt 2 ¯° dt
½° ¾ L ^ x t ` ¿°
L ª¬ y ''' t º¼ 7 L ª¬ y '' t º¼
X s
ª¬ s 3Y s s 2 y 0 sy ' 0 y '' 0 º¼ ª¬ s 2Y s sy 0 y ' 0 º¼ . Imposing the initial conditions, we obtain the Laplace transform of the solution as
s
3
7s2 Y s
X s
Thus, the transfer function
H s
Y s X s
.
H s is
1 s s 7 2
X s
Chapter 3
248
b Let x t
e 2t 4t
1 4 2 s2 s
X s
L ^ x(t )`
X s
s 2 4s 8 . s2 s 2
s 2 4s 8 s2 s 2
We have
Y s X s Y s
Y s
1 s s 7 2
X s s2 s 7 s 2 4s 8 . s 4 s 2 s 7
Example 3.11. Consider the causal LTI system described by the second differential equation
d 2 y t dt
2
5
dy t dt
6 y t x t .
a Determine the transfer function H s . b Determine the impulse response h t . c Determine the output response for the input x t
u t .
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
d Determine the output response for the input x t
249
G t .
Solution:
a The transfer function H s . Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write 2 dy t ° d y t 5 6 y t L® 2 dt °¯ dt
°½ ¾ L ^x t ` °¿
L ª¬ y '' t º¼ 5 L ª¬ y ' t º¼ 6 L ª¬ y t º¼
s
2
5s 6 Y s
Y s X s
1 s 2 5s 6
The transfer function
H s
H s
X s
Y s X s
.
H s is given by
1 s 2 5s 6
Y s
1
X s
s 2 s 3
X s .
Chapter 3
250
b The impulse response h t . It is now necessary to invert
H s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
H s
H s becomes
1
A B s 2 s 3
s 2 s 3
where A and B are real numbers to be determined.
ª 1 º A « » «¬ s 3 »¼ s
2
ª 1 º 1, B « » «¬ s 2 »¼ s
Hence, the given function
H s
1 3
.
H s can be expanded as
1 1 s 2 s 3
Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
1 ½ 1 ° 1 °½ h t L1 ® ¾ L ® ¾ ¯s 2¿ ¯° s 3 ¿°
.
Finally, using the transform pairs established in Table 2.1, we have
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
251
h t e2t e3t .
c Output response for the input x t
u t .
We have
Y s
1
X s
s 2 s 3 1
Y s
s 2 s 3
X s
Given the input
x t
u t X s
1 s
therefore
Y s
1 s s 2 s 3
It is now necessary to invert
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s becomes
1 s s 2 s 3
A B C s s 2 s 3
where A, B, and C are real numbers to be determined.
Chapter 3
252
ª º 1 A « » «¬ s 2 s 3 »¼ s C
ª 1 º « » «¬ s s 2 »¼ s
3
Y s
0
1 3
Hence, the given function
1 , 6
ª 1 º B « » «¬ s s 3 »¼ s
2
1 2
,
.
Y s can be expanded as
11 1 1 1 1 6 s 2 s 2 3 s 3
Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
y t
1 1 1 ½ 1 1 ° 1 ½° 1 1 ° 1 ½° L ® ¾ L ® ¾ L ® ¾ 6 ¯ s ¿ 2 °¯ s 2 °¿ 3 °¯ s 3 °¿
.
Finally, using the transform pairs established in Table 2.1, we have
y t
1 1 2t 1 3t e e . 6 2 3
d Output response for the input x t We have
Y s
1
X s
s 2 s 3
G t .
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
Y s
1
s 2 s 3
253
X s .
Given the input
x t G t X s 1 therefore
Y s
1
s 2 s 3 .
It is now necessary to invert
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s becomes
1
s 2 s 3
A B s 2 s 3
where A and B are real numbers to be determined.
ª 1 º A « » ¬« s 3 ¼» s
2
Hence, the function
Y s
ª 1 º 1, B « » ¬« s 2 ¼» s
1 3
Y s can be expanded as
1 1 s 2 s 3
Chapter 3
254
Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
1 ½ 1 ° 1 ½° H t L1 ® ¾ L ® ¾ ¯s 2¿ °¯ s 3 °¿
.
Finally, using the transform pairs established in Table 2.1, we have
y t e2t e3t . Example 3.12. Use the Laplace transform to find the transfer function and the impulse response of the system if the third-order differential equation describes the system
d 3 y t dt
2
6
d 2 y t dt
2
11
dy t dt
6 y t x t .
Solution: Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have 3 d 2 y t dy t ° d y t L® 6 11 6 y t 2 2 dt dt °¯ dt
°½ ¾ L ^ x t ` °¿ .
Using the derivative property and assuming all initial conditions are zero gives
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
^s
3
`
6 s 2 11s 6 Y s X s
Thus, the transfer function
H s
Y s
X s
255
.
H s is
1 s 6 s 11s 6 3
2
.
The transfer function can be factored as
H s
1 s 1 s 2 s 3
It is now necessary to invert
.
H s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
H s
H s becomes
1 s 1 s 2 s 3
A B C s 1 s 2 s 3
where A, B, and C are real numbers to be determined.
ª º 1 A « » ¬« s 3 s 2 ¼» s
H s
1
ª º 1 1 , B « » 2 «¬ s 1 s 3 ¼» s
1 1 1 1 1 2 s 1 s 1 2 s 2
2
.
Taking the inverse Laplace transform, we get
ª º 6 1, C « » ¬« s 1 s 2 ¼» s
3
1 . 2
Chapter 3
256
1 t 2t 1 3t e e e . 2 2
h t
Example 3.13. Transform the transfer function
H s
s 3s
3
2
8s 13
2
7 s 5s 6
into a differential equation. Solution: Let
H s
3s
3
2
3
8s 13
2
7 s 5s 6
H s
T .F
3s
s
s
Y s X s
3s
7 s 2 5s 6 Y s
3
2
8s 13
2
7 s 5s 6
s
2
8s 13 X s
.
Take the inverse Laplace transform of both sides and
^ ^
` `
L1 s 3 Y s y ''' t ° °° L1 s 2Y s y '' t ® 1 ' ° L ^sY s ` y t ° 1 °¯ L ^Y s ` y t using
½ ° °° ¾ ° ° °¿ , we get
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
3 y ''' t 7 y '' t 5 y ' t 6 y t x '' t 8 x ' t 13 x t
257
.
Thus,
3
d 3 y t d 2 y t dy t d 2 x t dx t 8 13x t . 7 5 6 y t dt 3 dt 2 dt dt 2 dt
Example 3.14. Compute the impulse response and step response of the transform with the transfer function
s2 s 1 s 2 2s 1 .
H s
Solution: We have
T .F
Y s
H s
Y s X s
s2 s 1 s 2 2s 1
s2 s 1 X s s 2 2s 1 .
If the input is an impulse Thus we have
G t ,
then its Laplace transform is
X s 1.
Y s X s and the impulse response is simply the
inverse Laplace transform of
H s .
Note that the degree of the numerator is the same as the degree of the denominator. As such, we have to put it in fraction form so that we can apply partial fraction expansion.
Chapter 3
258
H s
s2 s 1 3s 1 2 2 s 2s 1 s 2s 1
H s
ª s 1 1º s2 s 1 3 3 1 3 « » 1 2 2 s 2s 1 s 1 s 1 2 «¬ s 1 »¼ .
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find
y t
s2 s 1 ½ L1 ® 2 ¾ ¯ s 2s 1 ¿
1 ½° ° 1 °½ 1 ° L1 ^1` 3L1 ® 3 L ¾ ® 2¾ °¯ s 1 °¿ ¯° s 1 ¿°
. Finally, using the transform pairs established in Table 2.1, we have
y t G t 3et 3tet . Next, we compute the step response. If the input is a step function or
x t u t , then X s
1 . s
The step response in the Laplace transform domain is Y s
1 H s . s
Its inverse Laplace transform yields the step response in the time domain. Let us carry out partial fraction expansion of
Y s
s2 s 1 s s 2 2s 1
s2 s 1 s s 1
2
.
Y s as
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
It is necessary to invert
259
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s becomes
s2 s 1 1 2 3 2 2 s s 1 s s 1 s 1
.
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find
y t
1 ½° 1 ½° s 2 s 1 ½ 1 1 ½ 1 ° 1 ° L ® 2 ¾ L ® ¾ 2L ® ¾ 3L ® 2¾ ¯s¿ °¯ s 1 °¿ ¯ s 2s 1 ¿ °¯ s 1 °¿ 1
. Finally, using the transform pairs established in Table 2.1, we have
y t 1 2et 3tet . 3.4 Electrical Network Transfer Functions In this section, we apply the transfer function to the mathematical modeling of electric circuits. The equivalent circuits of the electric networks that we work with initially consist of three passive linear components: resistors, capacitors, and inductors. Table 3.1 summarizes the components and the relationships between voltage and current and between voltage and charge under zero initial conditions.
Chapter 3
260
Some Basic Definitions Electric Circuit: An electric circuit is a network of electrical devices whose terminals are connected together by ideal conducting wires. The three basic linear circuit elements are resistors, capacitors, and inductors. Current: The current that flows through a circuit device is the time rate of change of charge i
dq , which has units of amperes (A) defined as dt
coulombs/second (C/s). Voltage: The voltage across a circuit device is the work (energy) joules (J) required to move charge q through the device v
w
in
dw , which dt
has units of volts (V) defined as joules/coulomb (J/C).
3.4.1 Kirchhoff’s Current Laws G. R. Kirchhoff formulated the physical principles governing electrical circuits in 1859. They are as follows. 1. Kirchhoff’s current law: The algebraic sum of the currents flowing into any junction point must be zero. 2. Kirchhoff’s voltage law: The algebraic sum of the voltage drops around any closed loop must be zero.
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
261
Table 3.1 Modeling of Electric Circuits ImpeComponent
Voltage-
Current-voltage
Voltage-charge
current
dance
Z s
R v t Ri t
i t
1 v t R
v t R
dq t dt
R
d 2q t
Ls
Resistor
L v t
L
di t dt
t
i t
1 v W dW L ³0
v t L
dt 2
Inductor
C t
v t
Capacitor
1 i W dW C ³0
i t C
dv t dt
v t
1 q t C
1 Cs
V s I s
Chapter 3
262
The following examples show how to use the Laplace transform method to obtain the transfer function of a linear time-invariant system (LTI) of an electrical system. Example 3.15. Determine the transfer function of the electrical system shown in Figure 3.2.
R
L
V I = 0/p
I/P = e (t)
C
Figure 3.2 Solution: Applying KVL to the above circuit, the integro-differential equation that characterizes the electrical system is
di t
t
1 ³ i W dW e t R i t L dt C0 Here, i
dq is the current. The three terms on the left give the voltage dt
drop across the resistor, inductor, and capacitor, respectively. Taking the Laplace transform of the governing equation and assuming all initial conditions are zero gives
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
R I s L sI s
263
1 I s E s sC
1 · § ¨ R L s ¸ I s E s sC ¹ ©
I s
1
1 · E s § ¨R L s ¸ sC ¹ . © Thus, the transfer function
H s
T .F
H s is
I s
1
E s
1 · § ¨R L s ¸ sC ¹ ©
.
Example 3.16. Compute the transfer function of the network shown in Figure 3.3.
3H + + u(t)
~
1F
y(t)
– – Figure 3.3
Chapter 3
264
Solution: The impedance of the series connection of 3 and the impedance of the parallel connection of 2 and
Z2 s
1 s 1 2 s 2u
2 2s 1
which is a voltage divider. Thus we have
Y s
Z2 s
^Z s Z s ` 1
Y s
Y s
2 2s 1 2 2 3s 2s 1
2 U S 6s 9s 5 2
2 U S 6s 9s 5 2
Thus, the transfer function
T .F
U s
2
H s
Y s U s
H s is 2 . 6s 9s 5 2
s
is
1 is s
Z1 s 2 3s;
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
VL s
Example 3.17. Determine the transfer function T .F
V s
265
of the
system shown in Figure 3.4.
1H
v (t)
+ –
+ 1H
vL(t) –
Figure 3.4 Solution: Applying KVL to the left loop, we have
L1
d d i1 t ` R1 ^i1 t i2 t ` v t ^ ^i1 t ` ^i1 t i2 t ` v t dt dt . Applying KVL to the right loop, we have
R2 ^i2 t ` L2
d d i2 t ` R2 ^i2 t i1 t ` 0 i2 t ^i2 t ` ^i2 t i1 t ` 0 ^ dt dt
Taking the Laplace transform of the governing equation and assuming all initial conditions are zero gives, using
Chapter 3
266 ' ° L ¬ªi t ¼º sI s °½ ® ¾ °¯ L ª¬i t º¼ I s °¿
s 1 I1 s I 2 s V s I1 s 2 s I 2 s 0 I1 s 2 s I 2 s
s 1 2 s 1 I s 2
I2 s V s but
V s
1 s 2 3s 1
VL s sI 2 s
VL s V s
s s 3s 1
2
Thus, the transfer function is
T .F
G s
VL s V s
s s 2 3s 1
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
y t of the network shown in
Example 3.18. Determine the output Figure 3.5 for the input
267
x t u t . +
x(t)
2H
y(t)
1F
– Figure 3.5 Solution: The input
x t is a current source; the output y t is the voltage across
the capacitor. The impedance of the inductor and capacitor are, respectively, The impedance of their parallel connection is
1 s 1 2s s 2s u
2s 2s 2 1 .
Thus, the input and output of the network are related
T .F
H s
Y s X s
2s 2s 2 1
s
s2 1
2
2
2s and
1 s
Chapter 3
268
Y s
s 2
s 1
2
If we apply a step input
Y s
X s
2
.
u t 1 and U s
1 . s
1
s2 1
2
2
then its output is
y s
^
2 sin 1
`
2 t .
Example 3.19. Solve the initial value problem
d 2 y t dy t y t x t , y 0 0 y' 0 2 dt dt where
1 0 d t d1 x t ® x t 2 x t ¯1 1 t 2
a Determine the system transfer function H s
b Determine Y s .
Y s X s
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
269
Solution:
a
Taking the Laplace transform of the governing equation and
assuming all initial conditions are zero gives
s
2
s 1 Y s
X s
.
Thus,
H s
b
But
Y s
X s
1 s s 1
2
x t is a periodic function with period T
We have T
X s L ª¬ x t º¼
1 e st x t dt sT ³ 1 e 0
X s
2 1 st ½ 1 st e dt e dt ® ¾ ³1 1 e2s ¯³0 ¿
X s
1 ° e st ® 1 e2 s ¯° s
1
e st s 0
s 2
2
1
½ ° ¾ ° ¿
1 e 1 e X s s 1 e 1 e s 1 e s
therefore
s
s
s
2.
Chapter 3
270
Y s
X s
1 Y s 2 s s 1
X s
s
2
.
s 1
Thus,
1 e Y s . s 1 e s s 1 s
s
Example 3.20. The output
y t
2 3e
2
y t of a system is
e 3t
t
for an input, it is
x t
2 4e 3t
Determine the corresponding input for an output
y1 t
2 e
t
te t .
Solution: For the input
X s
x t 2 4e 3t , the Laplace transform is
2 4 s s3
6 s 1 s s 3
.
The corresponding output has the Laplace transform
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
Y s
2 3 1 s s 1 s 3
6 s s 1 s 3
271
.
Hence,
H s
Y s
1
X s
s 1
2
.
Given
y1 t
2 e
t
te t
.
The output has the Laplace transform
Y1 s
2 1 1 s s 1 s 1 2
1 s s 1
2
Hence, the Laplace transform of the corresponding input is
X1 s
Y s H s
1 s
.
The inverse Laplace transform of
X 1 s gives
x1 t u t . Example 3.21. Compute the output impulse response and
y t for an LTI system whose
h t and input x t are given by h t u t
x t t 2u t .
Chapter 3
272
Solution: For the input
x t t 2u t , the Laplace transform is
2 s3 .
X s
The corresponding impulse response
H s
h t has the Laplace transform
1 s
Hence,
Y s
H s X s
The output
2 s4
y t is obtained by taking the inverse Laplace transform of
Y s . Thus,
y t
t3 u t . 3
Example 3.22. The impulse response by
h t of an LTI system is given
h t 1 cos t.
a Determine the transfer function H s of the system. b Determine the response to the input x t
G t .
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
273
Solution:
h t of the LTI system h t 1 cos t
Given the impulse response and the transfer function
s º ª1 H s « 2 » ¬ s s 1¼ Given the input
we have
Y s
H s
L ª¬ h t º¼
1
s s2 1
.
x t G t X s 1
Y s
X s H s
1
2
s s 1
1
2
s s 1
It is now necessary to invert
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
1
2
s s 1
Y s becomes A B C s s i s i
where A, B, and C are real numbers to be determined.
Chapter 3
274
A
ª 1 « 2 «¬ s 1
C
º » »¼ s
1 , B 0
ª 1 º « » «¬ s s i »¼ s
Y s
1
2
s s 1
i
1 2
,
1 2
i
Hence, the function
ª 1 º « » «¬ s s i »¼ s
Y s
can be expanded as
1 1 1 1 1 s 2 s i 2 s i
Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find
1 ½ 1 ° 1 ½° 1 1 ° 1 ½° y t L1 ® ¾ L1 ® ¾ L ® ¾ ¯ s ¿ 2 ¯° s i ¿° 2 ¯° s i ¿°
.
Finally, using the transform pairs established in Table 2.1, we have
y t 1
1 it 1 it e e 2 2
y t 1
1 it e e it 2
Thus,
y t 1 cos t .
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
Example 3.23. The output
y t for an LTI system is found to be
2e 3t u t when the input x t is u t .
a Find the impulse response h t
b Find the output y t
of the system.
when the input
x t is e t u t .
Solution:
a For the input x t X s
1 s
u t , the Laplace transform is
.
The corresponding output has the Laplace transform
Y s
2 s3
Hence,
H s
Y s X s
Rewriting
H s
2s s3
H s as
2 s 3 6 s3
275
2
6 s3
Chapter 3
276
h t is obtained by taking inverse Laplace
the impulse response transform of
H s .
Thus,
h t 2G t 6e3t u t .
b For given input x t X s
e t u t , the Laplace transform is
1 s 1
Thus,
Y s H s X s
2s s 1 s 3
.
We begin by finding the partial fraction expansion for
Y s . The
denominator consists of two distinct linear factors and the expansion has the form
Y s
2s s 1 s 3
A B s 1 s 3
where A and B are real numbers to be determined.
ª 2s º A « » «¬ s 3 »¼ s
1 , B 1
ª 2s º « » «¬ s 1 »¼ s
3 3
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
1 3 s 1 s 3
Y s
The output
277
.
y t is obtained by taking the inverse Laplace transform of
Y s
y t
e
Example
R
t
3e 3t u t .
3.24.
The RL circuit shown in Figure 3.6 with
4: and L
0.5 H .
R i(t) v(t ) +–
L
Figure 3.6
a
Determine the transfer function H s
I s V s
b Find the output i t when the input v t
is
.
12u t .
Chapter 3
278
Solution: Applying KVL to this circuit, we get
L
di t dt
R i t v t .
dq is the current. The two terms on the left give the voltage dt
Here, i t
drop across the inductor and resistor, respectively
0.5
di t dt
4 i t v t .
Taking the Laplace transform of the governing equation and assuming all initial conditions are zero gives
0.5s 4 I s
V s
.
We define the circuit input to be the voltage and the output to be the current; hence, the transfer function is
H s
I s V s
Now we let
H s
1 0.5s 4
.
v t 12u t . The transformed current is given by
I s V s
1 0.5s 4
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
I s
1 V s 0.5s 4
I s
12 s 0.5s 4
I s
24 s s 8
279
.
We begin by finding the partial fraction expansion for
I s . The
denominator consists of two distinct linear factors and the expansion has the form
I s
24 s s 8
A B s s 8
where A and B are real numbers to be determined.
ª 24 º A « » «¬ s 8 »¼ s
I s
0
3 3 s s 8
The output
3 , B ª 24 º «¬ s »¼ s
.
i t is obtained by taking the inverse Laplace transform of
I s
3, 8
i t 3 1 e 8t u t .
Chapter 3
280
3.5 Mechanical System Transfer Functions In this section, we formally apply the transfer function to the mathematical modeling of a translational mechanical system. Automated systems, like electrical networks, have three passive linear components. Two of them— the spring and the mass—are energy-storage elements; one of them—the viscous damper—dissipates energy. Let us take a look at these mechanical elements, which are shown in Table 3.2. In the table, K, D, and M are termed spring constant, coefficient of viscous friction, and mass, respectively. Table 3.2 Modeling of Mechanical Elements Impedance Component
Force-
Force-
displacement
velocity
K
Z s
t
F t K x t
F t K ³ v W dW 0
K
Spring
D
F t D Damper
dx t dt
F t D v t
Ds
F s X s
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
281
Mass
F t M M
d 2 x t dt
2
F t M
dv t
M s2
dt
The following examples show how to use the Laplace transform method to obtain the transfer function of a linear time-invariant system (LTI) of a mechanical system. Example 3.25. The suspension system of an automobile is shown in Figure 3.7.
a Find a differential equation to describe the system. b Determine the transfer function
X s F s
.
Displacement X
K Force F M
D Figure 3.7
Chapter 3
282
Solution:
a
The model consists of a block with mass M, which denotes the
weight of the automobile. When the automobile hits a pothole, a vertical force
F t is applied to the mass and causes the automobile to oscillate.
The suspension system consists of a spring with a spring constant K and a dashpot, which represents the shock absorber D. The spring generates force
Kx t where x t is the vertical displacement measured from
equilibrium. The dashpot is modeled to generate viscous friction as D
dx dt
The differential equation to describe the system is
M
d 2 x t
b
dt
2
D
dx t dt
K x t F t
Taking the Laplace transform of the governing equation and
assuming all initial conditions are zero gives
M s 2 X s D sX s K X s F s
M
s2 D s K X s F s
X s F s
1 M s D sK 2
Thus, the transfer function
.
H s is
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
T .F
H s
X s F s
283
1 . M s D sK
2
Example 3.26. Determine the transfer function T .F
X2 s F s
of the
translational mechanical system shown in Figure 3.8. x1(t)
x2(t) 1N/m
f(t)
0.1kg
1 N-s/m
0.1kg
Frictionless
Figure 3.8 Solution: Applying Newton’s law to the node
x1 t
d2 d M 1 2 ^ x1 t ` k x1 t x2 t D ^ x1 t x2 t ` F t dt dt 0.1
d2 d x t x t x2 t ^ x1 t x2 t ` F t 2 ^ 1 ` 1 dt dt
Chapter 3
284
d2 d x t 10 x1 t x2 t 10 ^ x1 t x2 t ` 10 F t 2 ^ 1 ` dt dt . Applying Newton’s law to the node
M2
x2 t
d2 d x t k x2 t x1 t D ^ x2 t x1 t ` 0 2 ^ 2 ` dt dt
d2 d 0.1 2 ^ x2 t ` x2 t x1 t ^ x2 t x1 t ` 0 dt dt d2 d x t 10 x2 t x1 t 10 ^ x2 t x1 t ` 0 2 ^ 2 ` dt dt
.
Taking the Laplace transform of the governing equations and assuming all initial conditions are zero gives
using
s
2
L ª x ''1 t º s 2 X 1 s ½ ¼ ° ¬ ° ° ° ' ® L ¬ª x 1 t ¼º sX 1 s ¾ ° ° °¯ L ª¬ x1 t º¼ X 1 s °¿
10s 10 X 1 s 10 s 1 X 2 s 10 F s
10 s 1 X 1 s s 2 10s 10 X 2 s 0 Solving
X 2 s using Cramer’s rule
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
s
'
2
10 s 10
10 s 1
ª s 2 10 s 10 2 100 s 1 2 º ¬« ¼»
s 10s 10 s 10s 10 10 F s
10 s 1
285
2
2
'2
10 s 1
X2 s T .F
'2 '
100 s 1 F s
0
100 s 1 F s '
X2 s
100 s 1 F s
100 s 1
F s
' F s
ª s 2 10s 10 2 100 s 1 2 º «¬ »¼
. Thus, the transfer function T .F is
T .F
X2 s F s
100 s 1 ª s 2 10 s 10 ¬«
2
2 100 s 1 º ¼»
.
286
Chapter 3
Example 3.26. Determine the transfer function T .F
X2 s F s
of the
translational mechanical system shown in Figure 3.9.
Figure 3.9 Solution: Applying Newton’s law to the node
x1 t
d2 d d M 1 2 ^ x1 t ` kx1 t D1 ^ x1 t ` D3 ^ x1 t x2 t ` F t dt dt dt . Applying Newton’s law to the node
x2 t
d2 d d d M 2 2 ^ x2 t ` D2 ^ x2 t ` D4 ^ x2 t ` D3 ^ x2 t x1 t ` 0 dt dt dt dt Taking the Laplace transform of the governing equations and assuming all initial conditions are zero gives
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
287
L ª x ''1 t º s 2 X 1 s ½ ¼ ° ¬ ° ° ° ' ® L ¬ª x 1 t ¼º sX 1 s ¾ ° ° °¯ L ª¬ x1 t º¼ X 1 s °¿
using
M s D D s k X s D sX s F s D sX s ^M s D D D s` X s 0 2
1
1
3
1
3
2
2
3
1
Solving
2
2
3
4
2
X 2 s using Cramer’s rule
M s D D s k 2
1
'
1
^M s D 2
D3 s
' ª ¬
D3 s
3
2
2
X2 s
Thus,
'2 '
X2 s F s
2
2
1
1
3
2
2
3
4
^M s D 2
2
2
T .F
`
D3 D4 s
^ M s D D s k u M s D D D s` D s º¼
D3 s F s M 2 s D2 D3 D4 s 0
'2
2
2
3
`
D3 D4 s F s
^M s D D D s` F s 2
2
2
3
4
'
^M s D 2
2
2
`
D3 D4 s F s
' F s
^M s D 2
2
2
`
D3 D4 s
'
Chapter 3
288
T .F
^M s D 2
X2 s
2
2
F s
`.
D3 D4 s
'
Example 3.27. Use the Laplace transform to find the transfer function and impulse response of the system if the differential equation describes the system
d 2 z t dt 2
3
dz t dt
d 2 x t
2z t
dt 2
6
dx t dt
7 x t .
Solution: Taking the Laplace transform of the governing equation and assuming all initial conditions are zero gives
^
`
^
`
^
`
^
`
L z '' t 3L z ' t 2 L ^ z t ` L x '' t 6 L x ' t 7 L ^ x t ` using, we get '' 2 2 ° L ª¬ z t ¼º s L ª¬ z t º¼ s Z s ® ' °¯ L ª¬ z t º¼ sL ª¬ z t º¼ sZ s
s
2
3s 2 Z s
T .F
s
2
H s 1
6s 7 X s
s H s X s s Z s
½ ° ¾ °¿
2 2
3s 2
6s 7
3s 5 A B 1 s 1 s 2 s 1 s 2
.
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
We begin by finding the partial fraction expansion for
289
H s , the
denominator consists of two distinct linear factors and so the expansion has the form
3s 5 s 1 s 2
A B s 1 s 2
where A and B are real numbers to be determined.
ª 3s 5 º A « » ¬« s 2 ¼» s
H s 1
ª 3s 5 º « » ¬« s 1 ¼» s
2 , B 1
2 1 s 1 s 2
1 2
.
Taking the inverse Laplace transform, we get
h t G t 2et e2t . Example 3.28. Use the Laplace transform to find the transfer function and the impulse response of the system if the differential equation describes the system
d 2 z t dt 2
4
dz t dt
10 z t x t .
Solution: Taking the Laplace transform of the governing equation and assuming all initial conditions are zero gives
Chapter 3
290
^
`
^
`
L z '' t 4 L z ' t 10 L ^ z t ` L ^ x t ` '' 2 2 ° L ¬ª z t ¼º s L ª¬ z t º¼ s Z s ® ' °¯ L ¬ª z t ¼º sL ¬ª z t ¼º sZ s using
s
2
½ ° ¾ °¿ , we get
4s 10 Z s X s
T .F
H s
Z s
H s
X s
1 s 4 s 10 2
1
s 2
2
6
.
Taking the inverse Laplace transform, we get
h t
1 2t e sin 6
6 t . Summary
In this chapter, we have described and explained the concept of the transfer function. x
We have developed the input-output relationship by defining the transfer function. The Laplace transform of the impulse response of the system is called the transfer function of the system in the
s
domain. The Laplace transform converts a time-domain differential equation into a simple algebraic equation in the frequency domain.
Application of the Laplace Transform to LTI Differential Systems: Transfer Functions
291
Several simple examples have been solved to illustrate the relevant concepts. x LTI systems governed by differential equations have been analyzed by finding the transfer function under the assumption of zero initial conditions. This is useful for developing the input-output relationship. The transfer function of electrical and mechanical systems have also been explained with several examples.
CHAPTER 4 THE APPLICATION OF THE LAPLACE TRANSFORM TO LTI DIFFERENTIAL SYSTEMS: SOLVING IVPS
4.1 Introduction This chapter aims to show how Laplace transforms can be used to solve initial value problems for linear differential equations i. The advantages of the transform method for linear differential equations are listed below. (i) If we compare the transform method with the classical method to solve ordinary linear differential equations with constant coefficients using homogeneous and particular solutions, the Laplace transform has the advantage. The Laplace transform takes the initial conditions into account in the calculation, which reduces the amount of calculation considerably, especially for higher-order differential equations. (ii) Other important advantages of the transform method are worth noting. For example, the technique can easily handle equations involving forcing functions having jump discontinuities (unit step and impulse functions).
i
The Laplace transform was first introduced by Pierre Laplace in 1779 in his research on probability. G. Doetsch helped develop the use of Laplace transforms in solving differential equations. His work in the 1930s served to justify the operational calculus procedures earlier used by Oliver Heaviside.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
293
(iii) Further, the method can be used to solve certain other types of equation, such as linear differential equations with variable coefficients, integro-differential equations, systems of differential equations, and partial differential equations. In Section 4.2, we will show the scheme by which the Laplace transform can equally be applied to the initial value problem. An evaluation of the total response (natural and forced) of the LTI system by the Laplace transform method is presented in Section 4.3. Even more general are the systems of several coupled ordinary linear differential equations with constant coefficients in Section 4.4. Again, these systems can be solved by applying the same method, although, in the s domain, we have a system of several equations. For convenience, we restrict ourselves to systems of only two differential equations. LEARNING OBJECTIVES On reaching the end of this chapter, we expect you to have understood and be able to apply: x The Laplace transform method to solve initial value problems. x How the impulse and step response of the LTI system is described by an ordinary linear differential equation with constant coefficients. x The use of the Laplace transform to find the total response of the LTI system. x The use of the Laplace transform to solve coupled ordinary linear differential equations with constant coefficients and under initial conditions.
Chapter 4
294
4.2 The Scheme for Solving IVPs The Laplace transform technique for solving nth-order ordinary linear differential equations with constant coefficients (homogeneous or nonhomogeneous) is a three-step process. First, linear differential equations with constant coefficients are transformed by the Laplace transform into an algebraic equation in
s and L ^ f t `
F s the Laplace
transform of the solution of the initial value problem. Next, the unknown quantity in the algebraic equation
F s is solved by algebraic
manipulation. Finally, the inverse Laplace transform is applied to the equation to derive the solution of the initial value problem. The scheme for solving a linear differential equation is outlined below. Step 1: Consider the Laplace transform on both sides of the equation. Step 2: Use the properties of the Laplace transform and the initial conditions, simplify the algebraic equation obtained for
Y s in the S
domain. Step 3: Find the inverse transform of
Y s to obtain y t , the solution
of the differential equation.
4.2.1 Definitions: Homogeneous and Non-homogeneous Linear Differential Equations An nth-order ordinary differential equation of the form is
aN t
d N y t d N 1 y t d 2 y t dy t a t a t ...... a1 t a0 t y t N 1 2 N N 1 2 dt dt dt dt
f t
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
295
aN t , aN 1 t .........a1 t and a0 t , which are all functions of the independent variable If
t alone.
f t 0, then the above linear differential equation is said to be
homogeneous. If
f t z 0 , then the above linear differential equation is said to be non-
homogeneous. The following example shows how to use the Laplace transform method to obtain a general solution for a linear differential equation with constant coefficients. Example 4.1. Determine the solution of the initial value problem
dy t dt
4 y t et with y 0 2.
Solution: Let
y ' t 4 y t et .
Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
L ª¬ y ' t º¼ 4 L ª¬ y t º¼ L ª¬et º¼
1 ª sL ª¬ y t º¼ y 0 º 4 L ª¬ y t º¼ ¬ ¼ s 1 .
Chapter 4
296
Substituting the given initial conditions into the above equation, and solving the resulting equation for
L ª¬ y t º¼
2s 1 s 1 s 4
It is now necessary to invert
L ^ y t ` , we find
.
Y s L ^ y t ` and to accomplish this
some algebraic manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
2s 1 s 1 s 4
Y s becomes
A B s 1 s 4
where A and B are real numbers to be determined.
ª 2s 1 º A « » «¬ s 4 »¼ s
1
ª 2s 1 º 1 , B « » 5 «¬ s 1 »¼ s
Hence, the given function
Y s
2s 1 s 1 s 4
4
9 5
.
Y s can be expanded as
1 1 9 1 5 s 1 5 s 4
Now that we have obtained the partial fraction expansion. Taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
y t
1 1 1 ½ 9 1 ° 1 ½° L ® ¾ L ® ¾ 5 ¯ s 1 ¿ 5 ¯° s 4 ¿°
297
.
Finally, using the transform pairs established in Table 2.1, we have
y t
1 t 9 4t e e . 5 5
As a quick check of your computational correctness, it is recommended that you verify if the computed solution meets the given initial conditions. This example illustrates a fundamental difference between the solutions of initial value problems obtained using the Laplace transform and using the classical approach (finding complementary functions and a particular integral). In the classical approach, when solving an initial value problem, first a general solution is found and then the arbitrary constants are matched to the initial conditions. However, in the Laplace transform approach, the initial conditions are incorporated when the equation is transformed and the inversion of
Y s gives the required solution of the
initial value problem immediately. Example 4.2. Obtain the solution of the differential equation
d 2 y t dt
2
2
dy t dt
4 y t sin 2t
along with the initial conditions
y 0 0
y' 0 .
Chapter 4
298
Solution: Assuming that Laplace transform of the solution
Y s exists and using
the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
° d 2 y t ½° dy t L® 2 4 y t ¾ L ^sin 2t` 2 dt °¯ dt °¿
L ª¬ y '' t º¼ 2 L ª¬ y ' t º¼ 4 L ¬ª y t ¼º
2 s 4 2
2 ª¬ s 2Y s sy 0 y ' 0 º¼ 2 ª¬ sY s y 0 º¼ 4Y s 2 s 4. It is now necessary to invert
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s becomes
2 s 2s 4 s 2 4 2
.
Using partial fractions, we can write this as
Y s
2 s 2s 4 s 2 4 2
As B Cs D 2 s 2s 4 s 4 2
where A, B, C, and D are real numbers to be determined.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
2
299
As B s 2 4 Cs D s 2 2s 4 .
Hence, comparing the coefficients of both sides gives
s3 :
A C 0,
s2 :
D B 0,
s:
2 D 4C
1:
4B 4D
0,
2.
Using these equations, we obtain
A
1 ,B 2
1 ,C 4
1 and D 4
0.
Hence, we have
Y s
1 2 s 1 4 1 4 s
Y s
1 ° 1 ® 2 4 ° s 2s 4 ¯
Y s
1° 1 ® 4 ° s 1 2 3 ¯
s
2
2s 4
s
2
4
s 1 °½ 1 ° ¾ ® 2 °¿ 4 °¯ s 2 s 4
½ 1 ° s ° 1 ° s 1 ½° ¾ ® ¾ ® 2 4 ° s2 4 ° 4 ¯° s 1 3 ¿° ¯ ¿
°½ 1 ° s ¾ ® 2 °¿ 4 °¯ s 4
°½ ¾ °¿
½° ¾ ¿°
.
Applying the inverse Laplace transform to the above equation, and using its linearity, we get
Chapter 4
300
y t
1 4 3
e t sin
3 t 14 e
t
cos
3 t 14 cos 2t .
The following example illustrates how to use the Laplace transform method to obtain the general solution of second-order linear homogeneous differential equations with constant coefficients. Example 4.3. Obtain the Solution of differential equations
d 3 y t dt 2
2
d 2 y t dy t 2 y t 0 dt 2 dt
along with the initial conditions
y 0 0 y ' 0 & y '' 0 6. Solution: Assuming that Laplace transform of the solution
Y s exists, and using
the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write 3 d 2 y t dy t ° d y t °½ L® 2 2 y t ¾ L ^0` 2 2 dt dt °¯ dt °¿
L ª¬ y ''' t º¼ 2 L ª¬ y '' t º¼ L ª¬ y ' t º¼ 2 L ª¬ y t º¼ 0 3 2 ' '' 2 ' ° ¬ª s Y s s y 0 sy 0 y 0 ¼º 2 ¬ª s Y s sy 0 y 0 ¼º °½ ® ¾ 0 ª¬ sY s y 0 º¼ 2Y s °¯ °¿
.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
It is now necessary to invert
301
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s
Y s becomes
6
s
3
2s 2 s 2 6
s
3
2
2s s 2
6 s 1 s 1 s 2
A B C s 1 s 1 s 2
where A, B, and C are real numbers to be determined.
ª º ª º 6 6 1, B « A « » » «¬ s 1 s 2 ¼» s ¬« s 1 s 2 ¼» s 1
Y s
1 3 2 s 1 s 1 s 2
1
ª º 6 3, C « » ¬« s 1 s 1 ¼» s
2. 2
.
Applying the inverse Laplace transform to the above equation, and using the linearity of the inverse Laplace transform, gives us
y t et 3et 2e2t . The following examples illustrate how to use the Laplace transform method to obtain the general solution of first and second-order linear nonhomogeneous differential equations with constant coefficients.
Chapter 4
302
Example 4.4. Use Laplace transform techniques to find the solution to the first-order differential equation
dy t dt
3 y t e 2t with y 0 1.
Solution: Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
dy t ½ L® 3 y t ¾ L e 2t ¯ dt ¿
^ ` .
Using the derivative property and assuming all initial conditions are zero gives
^sY s y 0 ` 3Y s With the initial value of
Y s
1 s2 .
y 0 , this equation can be written as
1 1 s 3 s 3 s 2
.
We expand this S domain solution into partial fractions as
Y s
1 1ª 4 1 º « » s 3 5 ¬« s 3 s 2 ¼»
.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
303
Applying the inverse Laplace transform to the above equation, and using its linearity, we get
y t e 3t
4 3t 1 2t e e 5 5
which is the required general solution of the given linear differential equation. Example 4.5. Solve the differential equation
d 2 y t dt 2
y t 1 with y 0 0
y' 0 .
Solution: Let
y '' t y t 1
Because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation, we have
L ª¬ y ''' t º¼ L ª¬ y t º¼ L >1@
1 ª s 2 L ª¬ y t º¼ sy 0 y ' 0 º L ª¬ y t º¼ ¬ ¼ s It is now necessary to invert
L ^ y t ` . To accomplish this, some
algebraic manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
L ^ y t ` becomes
Chapter 4
304
L ª¬ y t º¼
L ª¬ y t º¼
A Bs C s s2 1
1
s s2 1
s 1 2 . s s 1
Applying the inverse Laplace transform to the above equation, and using the linearity of the inverse Laplace transform, give us the results in which we can write the time-domain solution
y t as
y t 1 cos t. Example 4.6. Obtain the solution of the second-order differential equation
d 2 y t dt
2
5
dy t dt
6 y t et
with the initial conditions
y 0 2 and y ' 0 1.
Solution: Assuming that Laplace transform of the solution
Y s exists, and using
the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write 2 dy t ° d y t L® 5 6 y t 2 dt °¯ dt
°½ t ¾ L e °¿
^ `
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
L ª¬ y '' t º¼ 5L ª¬ y ' t º¼ 6 L ª¬ y t º¼
s
2
1 2 s 11 s 1
2s 2 13s 12 s 1
5s 6 Y s
s 2 5s 6 Y s
Y s
2 s 2 13s 12 s 1 s 2 s 3
305
1 s 1
.
We begin by finding the partial fraction expansion for
Y s . The
denominator consists of three distinct linear factors and so the expansion has the form
Y s
2 s 2 13s 12 s 1 s 2 s 3
A B C s 1 s 2 s 3
where A, B, and C are real numbers to be determined.
ª 2 s 2 13s 12 º A « » «¬ s 2 s 3 »¼ s C
ª 2 s 2 13s 12 º « » «¬ s 1 s 3 »¼ s
hence, the function
1
3
1 , 2
B
ª 2 s 2 13s 12 º « » «¬ s 1 s 3 »¼ s
9 , 2
Y s can be expanded as
6, 2
Chapter 4
306
Y s
2 s 2 13s 12 s 1 s 2 s 3
1 1 6 9 1 2 s 1 s 2 2 s 3
.
Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
y t
1 ½° 9 1 ° 1 ½° 1 1 ° 1 ½° 1 ° L ® ¾ 6L ® ¾ L ® ¾ 2 ¯° s 1 ¿° ¯° s 2 ¿° 2 ¯° s 3 ¿°
.
Finally, using the transform pairs established in Table 2.1, we have
y t
1 t 9 e 6e 2t e 3t . 2 2
Some of the most useful and interesting applications of the Laplace transform method are found in the solution of linear differential equations with impulsive non-homogeneous functions. Equations of this type frequently arise in the analysis of the flow of current in electric circuits or the vibrations of mechanical systems, where voltages or forces of large magnitude act over very short time intervals. We will now discuss some discontinuous non-homogeneous functions as illustrations. Example 4.7. Solve the initial value problem
d 2 y t dy t 5 6 y t G t S G t 2S dt 2 dt with
y 0 0 y' 0 .
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
307
Solution: In
this
instance,
the
^u t S u t 2S ` discontinuous at t
forcing
function
is
the
step
function
of the above equation. This function is
S and t
2S . Therefore, it is not the solution of
any linear homogeneous differential equation with constant coefficients. This is an example of an initial value problem that we cannot solve easily using the method of undetermined coefficients. The advantage of the Laplace transform method is that the solution of the above initial value problem can be obtained with one application of the given method
y '' t 5 y ' t 6 y t G t S G t 2S . Because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation, we have
L ¬ª y '' t ¼º 5 L ¬ª y ' t ¼º 6 L ª¬ y t º¼ L ^ G t S G t 2S `
L ª y '' t º s 2 L ª¬ y t º¼ sy 0 y ' 0 ½ ° ¬ ¼ ° ® ¾ ' °¯ L ¬ª y t ¼º sL ¬ª y t ¼º y 0 °¿
^¬ªs L ¬ª y t ¼º sy 0 y 0 ¼º 5 ¬ªsL ª¬ y t º¼ y 0 ¼º 6L ª¬ y t º¼ ` 2
'
. This algebraic equation can be solved for
L ^ y t ` as
eS s e2S s
Chapter 4
308
L ª¬ y t º¼
e
S s
e 2S s
s 2 1 s 3
It is now necessary to invert
.
L ^ y t ` . To accomplish this, some
algebraic manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
L ^ y t ` becomes
ª 1 1 º e 2S s « » «¬ s 2 s 3 »¼
L ª¬ y t º¼
e
L ª¬ y t º¼
e S s e S s e2S s e2S s s 2 s 3 s 2 s 3
S s
.
Applying the inverse Laplace transform to the above equation, and using the linearity of the inverse Laplace transform, we can write the timedomain solution
e
y t
2 t S
y t as
e 3 t S u t S e 2 t 2S e 3 t 2S u t 2S .
Example 4.8. A mass attached to a vertical spring undergoes forced vibration so that the motion is described by the second-order differential equation
d 2 y t dt 2 with
4 y t sin 2t
y 0 10 and y ' 0 0.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
309
y t is the displacement at time t. Use the Laplace transform to determine the removal at time
t.
Solution: Because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation, we have 2 ° d y t 4 y t L® 2 °¯ dt
°½ ¾ L ^ sin 2t` °¿
.
Using the derivative property and assuming all initial conditions are zero gives
^s Y s sy 0 y 0 ` 4Y s 2
'
With the initial values of this expression for
Y s
2 s 4 2
y 0 and y ' 0 in this equation and solving
Y s yields two rational function components
10 s 2 2 s 4 s2 4
2
.
Taking inverse the Laplace transform and applying the convolution theorem to the second term on the right-hand side, we get
1 1 y t 10 cos 2t sin 2t t cos 2t . 8 4
Chapter 4
310
Example 4.9. Determine the solution of the initial value problem
d 2 y t dt 2 where
4 y t F t with y 0 0
y' 0
F t is the periodic function
1 0 t 1 F t ® with F t 2 F t . ¯0 1 t 2 Solution:
Given that
d 2 y t dt 2
4 y t
F t .
Because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation, we have
L ª¬ y ''' t º¼ 4 L ª¬ y t º¼
s
2
L ª¬ F t º¼ T
4 L ª¬ y t º¼
1 e st F t dt 1 e st ³0 1
1 s 2 4 L ª¬ y t º¼ e st dt 2 s ³ 1 e 0
. Thus,
L ª¬ y t º¼
1
s 1 e s s 2 4
1 s 1 e 2 s
1 e s 1 1e s
s
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
311
Partial fractions give
1 2 s s 4
1 1 e s
1 ª1 s « 2 4 «s s 4 ¬
º » »¼ .
can be interpreted as the sum of a geometric series with a
s common ratio e , so we may write
1 1 e s
1 e s e 2 s e 3 s e 4 s ........
In other words,
L ª¬ y t º¼
L ^ y t ` can be expressed as an infinite series
1 ª1 s « 2 4 «s s 4 ¬
º » 1 e s e 2 s e 3s e 4 s ........ »¼ .
Applying the inverse Laplace transform to the above equation, and using the linearity of the inverse Laplace transform, results in
y t
1 1 1 >1 cos 2t @ ª¬1 cos 2 t 1 º¼ u t 1 ª¬1 cos 2 t 2 º¼ u t 2 ............... 4 4 4
Example 4.10. Use Laplace transform techniques to find the solution to the second-order differential equation
d 2 y t dt 2
9 y t cos 2t with y 0 1 & y ' 0 0.
Chapter 4
312
Solution:
Let
d 2 y t dt 2
9 y t cos 2t .
Because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation, we have
L ª¬ y ''' t º¼ 9 L ª¬ y t º¼ L ª¬cos 2t º¼
L ¬ª y '' t ¼º s 2 L ª¬ y t º¼ sy 0 y ' 0 ½° ° ® ¾ ' °¯ L ª¬ y t º¼ sL ª¬ y t º¼ y 0 °¿ using , we get
s ª s 2 L ª¬ y t º¼ sy 0 y ' 0 º 9 L ª¬ y t º¼ 2 ¬ ¼ s 4 . This algebraic equation can be solved for
L ¬ª y t ¼º
s
s
2
2
4 s 9
L ^ y t ` as
1 s 9 2
.
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
y t
° s L ® 2 2 °¯ s 4 s 9 1
½° 1 1° ¾ L ® 2 °¿ °¯ s 9
½° ¾ °¿ .
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
313
This solution is in the form of the product of two known Laplace transforms. Thus we invert using either partial fractions or the convolution theorem. Using the convolution theorem yields
ª s L1 « 2 «¬ s 4 s 2 9
º » »¼
1 ^cos 2t cos 3t ` 5
(see Chapter 2, Example 2.26 ). Thus,
y t
1 1 cos 2t cos 3t ` sin 3t . ^ 5 5
Example 4.11. Solve the differential equation
d 2 y t dt
2
3
dy t dt
2 y t 10u t
subject to the initial conditions
y 0 1 and y ' 0 2.
Solution: In this instance, the forcing function is the step function above equation. This function is discontinuous at
t
u t of the
0 . Therefore, it
cannot offer a solution to any linear homogeneous differential equation with constant coefficients. This is an example of an initial value problem that we cannot solve quickly using the method of undetermined coefficients. The advantage of the Laplace transform method is that the
Chapter 4
314
solution of above the initial value problem can be obtained with one application of the method, given that
y '' t 3 y ' t 2 y t 10u t . Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
L ª¬ y '' t º¼ 3L ª¬ y ' t º¼ 2 L ª¬ y t º¼ 10 L ^u t `
L ª y '' t º s 2 L ª¬ y t º¼ sy 0 y ' 0 ½ ° ¬ ¼ ° ® ¾ ' °¯ L ª¬ y t º¼ sL ª¬ y t º¼ y 0 °¿ using , we get
^ª¬s L ª¬ y t º¼ sy 0 y 0 º¼ 3 ª¬sL ª¬ y t º¼ y 0 º¼ 2L ª¬ y t º¼ ` 2
'
10 s
. This algebraic equation can be solved for
s 2 s 10 L ª¬ y t º¼ s s 1 s 2
L ^ y t ` as
.
It is now necessary to invert
L ^ y t ` . To accomplish this, some
algebraic manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
L ^ y t ` becomes
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
L ª¬ y t º¼
s 2 s 10 s s 1 s 2
ª s 2 s 10 º A « » «¬ s 1 s 2 »¼ s
L ª¬ y t º¼
0
315
A B C s s 1 s 2
ª s 2 s 10 º 5, B « » «¬ s s 2 »¼ s
1
ª s 2 s 10 º 10, C « » «¬ s s 1 »¼ s
6. 2
5 10 6 . s s 1 s 2
Applying the inverse Laplace transform to the above equation, and using its linearity, we get
y t 5 10et 6e2t . The above equation is the required general solution of the given linear differential equation. Example 4.12. Use the Laplace transform to find the output response of the system described by the second-order differential equation
d 2 y t dt
2
with
3
dy t dt
2 y t u t S
y 0 0 y' 0 .
Solution: In this instance, the forcing function is the step function above equation. This function is discontinuous at
u t S of the
t S and, therefore, is
not the solution of any linear homogeneous differential equation with constant coefficients. This is an example of an initial value problem that
Chapter 4
316
we cannot solve easily using the method of undetermined coefficients. The advantage of the Laplace transform method is that the solution of the above initial value problem can be obtained with one application.
d 2 y t
Given that
dt
2
3
dy t dt
2 y t u t S .
Because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation, we have
° d 2 y t dy t L® 3 2 y t 2 dt dt ¯°
½° ¾ L ^ u t S ` ¿° .
Using the derivative property and assuming all initial conditions are zero gives
^s
2
e S s s
`
3s 2 Y s
.
It is now necessary to invert
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s becomes
e S s s s 2 3s 2
e S s s s 1 s 2
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
Y s
eS s s s2 3s 2
Y s eS s
317
eS s s s 1 s 2
1 s s 1 s 2
° A B C ½° eS s ® ¾ ¯° s s 2 s 3 ¿°
where A, B, and C are real numbers to be determined.
ª º 1 A « » «¬ s 1 s 2 »¼s
0
ª º 1 1 , B « » 2 «¬ s s 2 »¼s
1
ª 6 º 1, C « » «¬ s s 1 »¼s
2
1 . 2
° 1 1 1 1 1 ½° Y s eS s ® ¾ ¯° 2 s s 1 2 s 2 ¿° . Applying the inverse Laplace transform to the above equation, and using its linearity, results in
1 §1 · y t ¨ e t S e2 t S ¸ u t S 2 ©2 ¹ . The above equation is the required general solution of the given linear differential equation. The next example is a differential equation of an unusual type because the function
y t
occurs, not only as the dependent variable in the differential
equation, but also inside the convolution integral that forms the nonhomogeneous term. Equations of this type, involving both the integral of an unknown function and its derivative, are called integro-differential equations.
Chapter 4
318
Example 4.13. Solve the integro-differential equations
d 2 y t dt 2
t
y t
³ f u sin t u du
with y 0 1& y ' 0 0.
0
Solution: Given that t
y '' t y t
³ f u sin t u du 0
Because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation we have t ° °½ L ª¬ y t º¼ L ª¬ y t º¼ L ® ³ f u sin t u du ¾ ¯° 0 ¿° ''
'' 2 ' ° L ª¬ y t º¼ s L ª¬ y t º¼ sy 0 y 0 °½ ® ¾ and convolution theorem ° ¯ ¿° , we get using
^ª¬s L ª¬ y t º¼ sy 0 y 0 º¼ L ª¬ y t º¼ ` 2
'
This algebraic equation can be solved for
L ª¬ y t º¼
L ª¬ y t º¼ s2 1
L ^ y t ` as
2 s2 1
s s 1 2
.
We expand this S domain solution into partial fractions as
.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
L ª¬ y t º¼
2 s2 1
s s 1 2
4 2 2 2 s 1 s s
Thus, we can write the time-domain solution
.
y t as
y t 4et 2 2t. The above equation is the required general solution of the given linear differential equation. Example 4.14. Determine the solution of the initial value problem
d 2 y t dt
2
4
dy t dt
4 y t e2t t 2 with y 0 0 & y ' 0 0.
Solution: The solution process is the same as in example 1. Taking the Laplace transform of the given differential equation and imposing the initial conditions, we find, successively
° d 2 y t ½° d 2 y t 2 t 2 L® y t 4 4 ¾ L e t 2 2 dt ¯° dt ¿°
^
` .
Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
^
L ª¬ y '' t º¼ 4 L ª¬ y ' t º¼ 4 L ª¬ y t º¼ L e 2t t 2
`
319
Chapter 4
320
L ª y '' t º s 2 L ª¬ y t º¼ sy 0 y ' 0 ½ ° ¬ ¼ ° ® ¾ ' °¯ L ª¬ y t º¼ sL ª¬ y t º¼ y 0 °¿ using , we get
^ª¬s L ª¬ y t º¼ sy 0 y 0 º¼ 4 ª¬sL ª¬ y t º¼ y 0 º¼ 4L ª¬ y t º¼ ` 2
2
'
This algebraic equation can be solved for
L ª¬ y t º¼
L ^ y t ` as
2
s 2
5
.
Thus we can write the time-domain solution
y t
s 2
2e2t t 4 24
y t as
e2t t 4 . 12
The above equation is the required general solution of the given linear differential equation. Example 4.15. Obtain the solution of the differential equations
d 2 y t dt
2
5
dy t dt
6 y t 2e t
along with the initial conditions
y 0 1 and y ' 0 0.
3
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
321
Solution: Assuming that the Laplace transform of the solution using the fact that
Y s exists and
L is a linear operator, we take the Laplace transform
of the above differential equation and write
° d 2 y t ½° dy t t L® y t 5 6 ¾ 2L e 2 dt ¯° dt ¿°
^ `
L ¬ª y '' t ¼º 5L ¬ª y ' t ¼º 6 L ¬ª y t ¼º
2 s 1
2 ª¬ s 2Y s sy 0 y ' 0 º¼ 5 ª¬ sY s y 0 º¼ 6Y s s 1 . Imposing the initial conditions, we obtain the Laplace transform of the solution
s
2
5s 6 Y s
2 s 5. s 1
Solving this expression for
Y s yields two rational function
components
Y s
2 s5 s 1 s 2 s 3 s 1 s 2 s 3
.
Chapter 4
322
We begin by finding the partial fraction expansion for
Y s . The
denominator consists of three distinct linear factors and so the expansion has the form
Y s
s7 s 1 s 2 s 3
A B C s 1 s 2 s 3
where A, B, and C are real numbers to be determined.
ª º s7 A « » ¬« s 2 s 3 ¼» s C
ª º s7 « » «¬ s 1 s 2 »¼ s
Y s
1
ª º s7 3 , B « » ¬« s 1 s 3 ¼» s
5, 2
2, 3
3 5 2 s 1 s 2 s 3
y t 3et 5e2t 2e3t . The above equation is the required general solution of the given linear differential equation.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
323
Example 4.16. Use the Laplace transform to find the output response of the system described by the second-order differential equation
d 2 y t dt 2 with
y t u t S
y 0 0 and y ' 0 1.
Solution: In this instance, the forcing function is the step function above equation. This function is discontinuous at
u t S of the
t S and, therefore, is
not the solution of any linear homogeneous differential equation with constant coefficients. This is an example of an initial value problem that we cannot solve easily using the method of undetermined coefficients. Given that
d 2 y t dt 2
y t u t S
because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation, we have 2 ° d y t L® y t 2 °¯ dt
°½ ¾ L ^ u t S ` °¿
.
Using the derivative property and assuming all initial conditions are zero gives
Chapter 4
324
^
`
s2 1 Y s 1
e S s s
Solving this expression for
Y s yields two rational function
components
Y s
1 e S s s2 1 s s2 1
Using partial fractions, we can write this as
Y s
ª 1 1 S s 1 « 2 e 2 s 1 s 1 «¬ s
º » »¼ .
Applying the inverse Laplace transform to the above equation, and using its linearity, we get
y t sin t 1 cos t S u t S
.
Example 4.17. Determine the solution of the initial value problem
d 2 y t dt
2
6
dy t dt
9 y t 12e3t t 2 with y 0 0 & y ' 0 0.
Solution: Given that
y '' t 6 y ' t 9 y t 12 e3t t 2
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
325
because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation, we have
^
L ¬ª y '' t º¼ 6 L ª¬ y ' t º¼ 9 L ª¬ y t º¼ 12 L e3t t 2
`
L ª y '' t º s 2 L ª¬ y t º¼ sy 0 y ' 0 ½ ° ¬ ¼ ° ® ¾ ' °¯ L ª¬ y t º¼ sL ª¬ y t º¼ y 0 °¿ using , we get
^ª¬s L ª¬ y t º¼ sy 0 y 0 º¼ 6 ª¬sL ª¬ y t º¼ y 0 º¼ 9L ª¬ y t º¼ ` 2
'
This algebraic equation can be solved for
L ª¬ y t º¼
24
s 3
L ^ y t ` as
24
s 3
5
.
Thus, we can write the time-domain solution
y t as
y t e 3t t 4 . It should be noted from the above examples, as we have noted in previous chapters, that the distinct advantage of using the Laplace transform is that it enables us to replace the operation of differentiation with an algebraic operation. Consequently, by taking the Laplace transform of each term in a differential equation, it is converted into an algebraic equation in the S domain. This may then be rearranged using algebraic rules (partial fractions) to obtain an expression for the Laplace transform of the
3
Chapter 4
326
response; the desired time response is then obtained by taking the inverse transform. Example 4.18. Determine the solution of the boundary value problem
d 2 y t dt
2
§S · 9 y t cos 2t with y 0 1, y ¨ ¸ 1. ©2¹
Solution: Since
y ' 0 is not given, we assume y ' 0 a.
Because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation, we have
L ª¬ y '' t ¼º 9 L ª¬ y t º¼ L > cos 2t @ '' 2 ' ½ ° L ¬ª y t ¼º s L ª¬ y t º¼ sy 0 y 0 ° ® ¾ ' °¯ L ª¬ y t º¼ sL ª¬ y t º¼ y 0 °¿
solving for
L ª¬ y t º¼
L ^ y t ` gives
sa s s2 9 s2 9 s2 4
It is now necessary to invert
.
L ^ y t ` . To accomplish this, some
algebraic manipulation is necessary if we are to identify the terms on the
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
327
right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
L ª¬ y t º¼
L ^ y t ` becomes
a 1 s 4 s 2 s 9 5 s 9 5 s2 9 2
.
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
y t
a 1 4 sin 3t cos 2t cos 3t 3 5 5
when
t
S 2
1
a 1 a 3 5
12 5 .
Hence, the solution is
y t
4 1 4 sin 3t cos 2t cos 3t 5 5 5
Thus,
y t
1 cos 2t 4sin 3t 4 cos 3t . 5
Chapter 4
328
Example 4.19. Obtain the solution of the fourth-order differential equation
d 4 y t dt 4
K
along with the boundary conditions
y 0 y '' 0 y d y '' d 0. Solution: Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
° d 4 y t L® 4 ¯° dt
½° ¾ L^ K ¿°
L ª¬ y '''' t º¼
K s
`
Using the derivative formula, we get
s 4Y s s 3 y 0 s 2 y ' 0 sy '' 0 y ''' 0 Using the given initial conditions, we get
s 4Y s s 2 y ' 0 y ''' 0
K s
The two unknown initial conditions are replaced with
K s .
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
y' 0
A and y '' 0 B
Y s
K A B s5 s 2 s 4
329
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
y t
Kt 4 Bt 3 At 24 6 .
The boundary conditions on the right-hand side are now satisfied
y d
y '' d and A
Kd 4 Bd 3 Ad 24 6
0
Kd 2 Kd Bd 0 B 2 2 kd 3 . 24
Finally, the desired solution of the boundary value problem is
y t
Kt 4 Kd 3 Kd 3 t t . 24 24 12
Example 4.20. Obtain the solution of the second-order differential equation
d 2 y t dy t dt 2 dt
f t ,
1 0 t 1 f t ® ¯0 t ! 1
along with the initial conditions
y ' 0 0 and y '' 0 1.
Chapter 4
330
Solution: Assuming that Laplace transform of the solution
Y s exists and using
the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
° d 2 y t dy t L® 2 dt °¯ dt
½° ¾ L ^ f t °¿
`
^
`
L ª¬ y '' t º¼ L ª¬ y ' t º¼ L f1 t f 2 t f1 t u t a
1 e s s s Y s 1 s s 2
Y s
1 e s 1 2 2 s s 1 s s 1 s s 1
Y s
1 e s 1 2 2 s s 1 s s 1 s s 1
.
We begin by finding the partial fraction expansion for
Y s .
Partial fraction expansion of the first term of the right side of the above equation gives
1 s 1 s 2
A B C 2 s 1 s s
where A, B, and C are real numbers to be determined.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
ª1º A « 2» ¬ s ¼s C
ª d 1 ½º
1
ª 1 º « » ¬« s 1 ¼» s
1 s 1 s 2
1, B « ® ¾» ¬ ds ¯ s 1 ¿¼ s
331
1, 0
1, 0
1 1 1 2 s 1 s s
.
Partial fraction expansion of the third term of the right side of the above equation gives
1 s 1 s
A B s 1 s
where A and B are real numbers to be determined.
ª1 º A « » ¬ s ¼s
1 , B 1
Hence, the given function
Y s
ª 1 º «¬ s 1 »¼ s
1, 0
1 s 1 s
1 1 s 1 s
.
Y s can be expanded as
§ 1 1 1 1 1 1· 1 1 2 e s ¨ 2 ¸ ¨ s 1 s s ¸ s 1 s s 1 s s © ¹
.
Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
Chapter 4
332
§ 1 ½ 1 1 ·° ° 1½ 2 ¸¾ L1 ® 2 ¾ L1 ®e s ¨ ¨ ¸ ¯s ¿ ° ¯ © s 1 s s ¹ ° ¿
y t
.
Finally, using the transform pairs established in Table 2.1, we have
y t t e t 1 t 2 u t 1 . Example 4.21. Obtain the solution of the second-order differential equation
d 2 y t dt 2
4 y t e 3t
along with the initial conditions
y 0 0
y' 0 .
Solution: Assuming that Laplace transform of the solution
Y s exists and using
the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
° d 2 y t 4 y t L® 2 °¯ dt
½° 3t ¾ L e °¿
^ `
L ª¬ y '' t º¼ 4 L ª¬ y t º¼
1 s 3
^ª¬s Y s sy 0 y 0 º¼ 4Y s ` 2
'
1 s 3
.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
It is now necessary to invert
333
Y s . To accomplish this, some algebraic
manipulation is necessary to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s
Y s becomes
6 s 2 4 s 3
6 s 4 s 3 2
6 s 3 s 2 s 2
A B C s 3 s 2 s 2
where A, B, and C are real numbers to be determined.
ª º 6 A « » ¬« s 2 s 2 ¼» s
Y s
2
ª º 1 6 , B « » 5 ¬« s 3 s 2 ¼» s
2
ª º 1 6 , C « » 20 ¬« s 2 s 3 ¼» s
1 1 1 1 1 1 5 s 3 20 s 2 4 s 2
2
1 4
.
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
y t
1 3t 1 2t 1 2t e e e . 5 20 4
Chapter 4
334
Example 4.22. Obtain the solution of the first-order differential equation
2
dy t dt
2 y t sin 2t
along with the initial conditions
y 0 2.
Solution: Assuming that the Laplace transform of the solution using the fact that
Y s exists, and
L is a linear operator, we take the Laplace transform
of the above differential equation and write
dy t ½ L ®2 2 y t ¾ L ^ sin 2t ¯ dt ¿
L ª¬ y ' t º¼ L ª¬ y t º¼
`
1 s2 4
^¬ªsY s y 0 ¼º Y s ` It is now necessary to invert
1 s2 4
.
Y s . To accomplish this, some algebraic
manipulation is necessary to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s becomes
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
1 2 s 1 s 2 4 s 1
Y s
.
We first observe that the quadratic factor
s
2
4
and
s 1
s
2
4 is irreducible. Since
are non-repeated factors, the partial fraction
expansion has the form
ª 1 « «¬ s 1 s 2 4
º » »¼
A Bs C 2 s 1 s 4
.
We can use the cover-up method to find A, B, and C, as follows
A
ª 1 « 2 «¬ s 4
> Bs C @s 2i 2 Bi C
º » »¼ s
1 5 1
ª 1 º « » «¬ s 1 »¼ s
1 1 2i
2i
1 1 2i
° 1 2i ½° 1 2 i ® ¾ ¯° 1 2i ¿° 5 5
Since the real and imaginary parts are equal on both sides, we get
B
1 &C 5
Hence, we have
1 5.
335
Chapter 4
336
ª 1 « «¬ s 1 s 2 4
º » »¼
1 ° 1 s 1 2 ® 5 ° s 1 s 4 ¯
s 1 °½ 1 ° 1 2 2 ¾ ® s 4 °¿ 5 °¯ s 1 s 4
½° ¾ ¿°
. Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
1 ° 1 § 1 · 1 § s ®L ¨ ¸ L ¨¨ 2 5 ° © s 1 ¹ © s 4 ¯
y t
· § 1 ¸ L1 ¨ ¸ ¨ s2 4 ¹ ©
·½ § 2 · ¸ ¾° L1 ¨ ¨ s 1 ¸¸ ¸° © ¹ ¹¿
Finally, using the transform pairs established in Table 2.1, we have
y t
1 t 1 ½ t ®e cos 2t sin 2t ¾ 2e . 5¯ 2 ¿
Example 4.23. Obtain the solution of the integro-differential equation
dy t dt
t
y t
³ cos 2 t u du. 0
along with the initial condition y 0
2.
Solution: Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
dy t ½ ° t y t ¾ L ® ³ cos 2 t u du. L® ¯ dt ¿ °¯ 0
°½ ¾ °¿
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
337
L ª¬ y ' t º¼ L ª¬ y t º¼ L ^1 cos t`
^ª¬sY s y 0 º¼ Y s ` L >1@u L >cos 2t @ s 1 Y s
1 2 s 4
2
It is now necessary to invert
.
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s becomes
1 2 2 s 1 s 4 s 1
Y s
Using partial fraction decomposition, we can write
1 s 1 s 2 4
1 ° 1 1 s 2 2 ® 5 ° s 1 s 4 s 4 ¯
½° ¾ ¿°
Thus,
Y s
1 ° 1 s 1 2 2 ® 5 ° s 1 s 4 s 4 ¯
½° 2 ¾ s 1 ¿°
.
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
Chapter 4
338
Y s
1 ° 1 § 1 · 1 § s ®L ¨ ¸ L ¨¨ 2 5 ° © s 1 ¹ © s 4 ¯
· § 1 ¸ L1 ¨ ¸ ¨ s2 4 ¹ ©
·½ 1 · ¸ ¾° 2 L1 ¨§ ¸ ¸° s 1 ¹ © ¹¿
. Finally, using the transform pairs established in Table 2.1, we have
y t
1 t 1 ½ ®11e cos 2t sin 2t ¾ . 5¯ 2 ¿
Example 4.24.
Use the Laplace transform technique to solve the
initial value problem
d 2 y t dt
2
4
dy t dt
3 y t e2t
along with the initial conditions y 0
0 y' 0 .
Solution: Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation to write
° d 2 y t ½° dy t 2 t L® y t 4 3 ¾ L e 2 dt ¯° dt ¿°
^ `
L ª¬ y '' t º¼ 4 L ª¬ y '' t º¼ 3L ª¬ y t º¼
1 s2
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
339
1 2 ' ¬ª s Y s sy 0 y 0 ¼º 4 ¬ª sY s y 0 º¼ 3Y s s 2 . Imposing the initial conditions, we obtain the Laplace transform of the solution as
s
2
4s 3 Y s
1 s2
Y s
1 s 1 s 2 s 3
Y s
1 s 1 s 2 s 3
A B C s 1 s 2 s 3
where A, B, and C are real numbers to be determined.
ª º 1 A « » ¬« s 2 s 3 ¼» s
C
Y s
1
ª º 1 « » ¬« s 1 s 2 ¼» s
Hence, the function
1 , 6
3
B
ª º 1 « » ¬« s 1 s 3 ¼» s
2
1 , 15
1 10
Y s can be expanded as
1 1 ½ 1 ° 1 ½° 1 ° 1 ½° ® ¾ ® ¾ ® ¾ 6 ¯ s 1 ¿ 15 ¯° s 2 ¿° 10 ¯° s 3 ¿°
.
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
Chapter 4
340
y t
1 t 1 2t 1 3t e e e . 6 15 10
The Laplace transform method is useful for solving non-homogeneous linear differential equations and initial value problems with constant coefficients when the forcing function is a discontinuous function or an impulse function with a Laplace transform. Forcing functions, which are discontinuous functions or impulse functions, frequently occur in electrical and mechanical systems. The following example shows how to use the Laplace transform method to obtain a general solution for a linear differential equation with constant coefficients when the forcing function is an impulse. Example 4.25. Obtain the solution of the fourth-order differential equation
d 4 y t dt 4
2
d 3 y t d 2 y t dy t 2 G t 3 2 dt dt dt
along with the initial condition
y 0 1 and y ' 0 y '' 0 y ''' 0 0. Solution: Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
° d 4 y t d 3 y t d 2 y t dy t L® 2 2 4 3 2 dt dt dt ¯° dt
½° ¾ L ^ G t ` ¿°
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
341
L ª¬ y '''' t º¼ 2 L ª¬ y ''' t º¼ L ª¬ y '' t º¼ 2 L ª¬ y ' t º¼ 1
s
4
2s3 s 2 2s Y s s 3
Y s
Y s
s3 1 s 4 2s3 s 2 2s
1
s3 1 s s 1 s 1 s 2
.
We begin by finding the partial fraction expansion for
Y s . The
denominator consists of four distinct linear factors and so the expansion has the form
s3 1 s s 1 s 1 s 2
A B C D s s 1 s 1 s 2
where A, B, C, and D are real numbers to be determined.
ª º s3 1 A « » ¬« s 1 s 1 s 2 ¼» s B
D
ª º s3 1 « » «¬ s s 1 s 2 »¼ s ª º s3 1 « » «¬ s 1 s s 1 »¼ s
1 , 2
0
1, 1
2
3 2
.
C
ª º s3 1 « » «¬ s 1 s s 2 »¼ s
0, 1
Chapter 4
342
Hence, the function
Y s can be expanded as
11 1 3 1 2 s s 1 2 s 2
Y s
.
Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we find that
1 1 1 ½ 1 ° 1 ½° 3 1 ° 1 ½° L ® ¾ L ® ¾ L ® ¾ 2 ¯s¿ ¯° s 1 ¿° 2 ¯° s 2 ¿°
y t
.
Finally, using the transform pairs established in Table 2.1, we have
y t
1 t 3 2t e e . 2 2
Example 4.26. Obtain the solution of the second-order differential equation
d 2 y t dt
2
4
dy t dt
4 y t f t , f (t )
along with the initial conditions
y 0 0
Solution: First, we find the Laplace transform of
f (t )
t 1 t 1 , t t 0.
t 1 t 1 , t t 0
y' 0 .
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
0 d t d1 t !1
2 f (t ) ® ¯2t
In terms of the unit step function,
f (t )
343
f t can be expressed as
f1 t ^ f 2 t f1 t ` u t a
f (t ) 2 > 2t 2@ u t 1
.
Then, taking the Laplace transform
L ^ f (t )` L > 2@ L ^ 2t 2 u t 1 ` which, on using the result in (1.7), gives
L ^ f (t )`
2 s e L ^ f1 (t )` s
f1 (t 1) replace t
2t 2 t 1
f1 (t ) 2t then its Laplace transform is
L ^ f1 (t )`
Eq. (4.1) becomes
2 s2
(4.1)
Chapter 4
344
L ^ f (t )`
2 s § 2 · e ¨ 2 ¸ s ©s ¹
thus,
L ^ f (t )`
2 s e s . s2
Now we start solving the given initial value problem. Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
° d 2 y t dy t L® 4 4 y t 2 dt ¯° dt
½° ¾ L ^ f t ` ¿°
L ª¬ y '' t º¼ 4 L ª¬ y ' t º¼ 4 L ª¬ y t º¼
2 s e s . 2 s
^ª¬s Y s sy 0 y 0 º¼ ª¬sY s y 0 º¼ 4Y s ` 2
'
It is now necessary to invert
2 s e s . 2 s
Y s . To accomplish this, some algebraic
manipulation is necessary to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s becomes
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
Y s
2
s e s
s s 2 2
345
2
ª º 1 1 e s 2 Y s 2 « » 2 2 s s 2 »¼ «¬ s s 2 . Applying the inverse Laplace transform to the above equation, and using the linearity of the inverse Laplace transform, results in
y t
ª ° ½° 1 ° s ½°º 1 1 1 « » L ®e 2 2 L ® 2¾ 2¾ «¬ °¯ s s 2 °¿ °¯ s s 2 °¿»¼ .
Using the convolution theorem and the second shifting theorem, we have
1 t ½ 1 y t 2 ® e2t e2t 1 ¾ t 2 te2 t 1 u t 1 . 2 ¯2 ¿ 2
^
`
Example 4.27. Obtain the solution of the integro-differential equation
dy t dt
t
y t
³e
t u
u du.
0
along with the initial condition y 0
1.
Solution: Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
Chapter 4
346
dy t ½ ° t °½ L® y t ¾ L ® ³ eu t u du. ¾ ¯ dt ¿ °¯ 0 ¿°
^
L ª¬ y ' t º¼ L ª¬ y t º¼ L et t
^ª¬sY s y 0 º¼ Y s `
L ª¬et º¼ u L >t @
1 1 s s 1
s 1 Y s
`
2
It is now necessary to invert
.
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
1 s s 1 2
2
Y s becomes
1 s 1
.
Using partial fraction decomposition, we can write
1 s 2 s 1
2
2 1 2 1 2 s s s 1 s 1 2
Thus,
Y s
2 1 3 1 2 s s s 1 s 1 2
.
.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
347
Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
§2· § 1 y t L1 ¨ ¸ L1 ¨ 2 ©s¹ ©s
· 1 § 3 · 1 § 1 ¸¸ L ¨ ¸ L ¨¨ 2 ¨ s 1 ¹ © ¹ © s 1
· ¸ ¸ ¹.
Finally, using the transform pairs established in Table 2.1, we have
y t 2 t 3et tet . 4.3 Differential Equations with Variable Coefficients We have the multiplication by
L ^t f t `
t
formula
d L ^ f t ` ds
which is taken to be the nth-derivative of
^
`
L t f n t
f t , then
d L f n t ds .
^
`
Using the derivative formula
^
`
L t f n t
d n s L ^ f t ` s n 1 f 0 s n 2 f ' 0 ...... f n 1 0 . ds
^
This equation can be used to transform a linear differential equation with variable coefficients into a differential equation involving the transform. The following examples show how to use the Laplace transform method to obtain the general solution of linear differential equations with variable coefficients.
`
Chapter 4
348
Example 4.28. Determine the solution of the initial value problem
t
d 2 y t dt
2
2
dy t dt
ty t cos t with y 0 1.
Solution: Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
L ª¬ty '' t º¼ 2 L ª¬ y ' t º¼ L ª¬ty t º¼ L ^ cos t ` d L ª y t ¼º ® L ¬ªty t ¼º ds ¬ Using ¯
^
`
.
d ½ Y s ` ¾ ^ ds ¿ , we get
d 2 d s s Y s sy 0 y ' 0 2 ^sY s y 0 ` ^Y s ` 2 ds ds s 1
^
`
d s d ½ ® s 2 Y s 2sY s ¾ y 0 0 2sY s 2 y 0 ^Y s ` 2 ds s 1 ¯ ds ¿
s 1 dsd Y s 1 2
s s2 1
d s s Y s 2 ds s 1 s2 1
2
.
Taking the inverse Laplace transform and using
1 d ½ ½ ® L ® ^Y s `¾ ty t ¾ ¯ ds ¿ ¯ ¿ , we get
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
349
1 sin t t sin t 2 .
ty t Thus,
1§ 2· ¨1 ¸ sin t. 2© t ¹
y t
Example 4.29. Determine the solution of the initial value problem
t
d 2 y t dy t t y t 0 dt 2 dt
with
y 0 0 and y ' 0 2.
Solution: Because of the linearity of the equations and the Laplace transform operation, we have 2 ° d y t dy t °½ L ®t t y t ¾ 0 2 dt °¯ dt °¿ .
Using,
L ^ty t ` Y ' s ° d 2 y t ½° 2 ' L ®t ¾ 2sY s s Y s 2 ¯° dt ¿°
Chapter 4
350
dy t ½ L ®t ¾ ¯ dt ¿
Y s sY ' s .
Therefore 2 ° d y t dy t °½ L ®t t y t ¾ 2 sY s s s 1 Y ' s 0 2 dt ¯° dt ¿°
Y ' s Y s
2ds s 1
.
On integration, we get 2
log Y s log s 1 log C ° C ½° log ^Y s ` log ® 2¾ ¯° s 1 ¿°
Y s
C
s 1
2
.
Taking the inverse Laplace on both sides of the above equation, we get
y t Ctet
.
Since
y ' t Ctet Cet
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
351
y ' 0 C 2, therefore
y t 2tet
.
Example 4.30. Determine the solution of the initial value problem
d 2 y t dy t t ty t 0 with y 0 2 and y ' 0 0. 2 dt dt Solution: Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
L ª¬ty '' t º¼ L ª¬ y ' t º¼ L ª¬ty t º¼ 0
d L ª y t º¼ ® L ª¬ty t º¼ ds ¬ ¯
^
Using
`
.
d ½ Y s ` ¾ ^ ds ¿ , we get
d 2 d s Y s sy 0 y ' 0 ^sY s y 0 ` ^Y s ` 0 ds ds
^
`
d d ½ ® s 2 Y s 2sY s ¾ y 0 0 sY s y 0 ^Y s ` 0 ds ¯ ds ¿ d s Y s 2 Y s 0 ds s 1
Chapter 4
352
implying
d s Y s 2 Y s ds s 1 dY s Y s
s ds 0 s 1 2
.
On integration, we get
1 ln Y s ln s 2 1 2
ln c
ln c
ln Y s
s
2
1
c
Y s
s
2
.
1
Taking the inverse Laplace transform on both sides of the above equation, we get
° cL ® ° ¯
y t
½ ° ¾ c J0 t . 2 s 1 ° ¿ 1
1
Example 4.31. (Bessel’s equation) Determine the solution of the initial value problem
x2
d 2 y x dx
2
x
dy x dx
x 2 y x 0 with y 0 1 and y ' 0 0.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
353
Solution: Dividing the given differential equation by x and replacing
x t , we
get
t y '' t y ' t ty t 0
.
Take the Laplace transform of both sides of the above differential equation
L ª¬ty '' t º¼ L ª¬ y ' t º¼ L ª¬ty t º¼ 0
d L ª¬ y t º¼ ® L ª¬ty t º¼ ds ¯ Using
^
`
.
d ½ Y s ` ¾ ^ ds ¿ , we get
d 2 d s Y s sy 0 y ' 0 ^sY s y 0 ` ^Y s ` 0 ds ds
^
`
d d ½ ® s 2 Y s 2sY s ¾ y 0 0 sY s y 0 ^Y s ` 0 ds ¯ ds ¿ d s Y s 2 Y s 0 ds s 1
implying
d s Y s 2 Y s ds s 1
Chapter 4
354
dY s Y s
s ds 0 s 1
2
.
On integration, we get
1 ln Y s ln s 2 1 2
ln c
ln c
ln Y s
s
2
1
c
Y s
s
2
1
Taking the inverse Laplace on both sides of above equation, we get
y t
° cL ® ° ¯
½ ° ¾ c J0 t . 2 s 1 ° ¿ 1
1
Example 4.32. Use the Laplace transform to find the output the system described by d 2 y t dt
2
3
dy t dt
2 y t
dx t dt
3x t
with input given by
x t e 5t u t with y 0 1 and y ' 0 2. Solution:
y t of
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
355
Given that
y '' t 3 y ' t 2 y t x ' t 3x t . Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
^
`
^
`
^
`
L y '' t 3L y ' t 2 L ^ y t ` L x ' t 3L ^ x t ` '' 2 ' ½ ° L ¬ª y t ¼º s L ª¬ y t º¼ sy 0 y 0 ° ® ¾ ' ° L ª¬ y t º¼ sL ª¬ y t º¼ y 0 °¿ using ¯ , we get
^ª¬s L ª¬ y t º¼ sy 0 y 0 º¼ 3 ª¬sL ª¬ y t º¼ y 0 º¼ 2L ª¬ y t º¼ ` 2
'
L ^ y t `
s 3 L ^ x t `
s
2
3s 2
ª sL ª¬ x t º¼ x 0 º 3L ª¬ x t º¼ ¬ ¼
s 5
s
2
3s 2
we have,
L ^ x t `
1 , s 5
It is now necessary to invert
L ^ y t ` . To accomplish this, some
algebraic manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
L ^ y t ` becomes
Chapter 4
356
L ^ y t `
L ^ y t `
s 5 s 3 s 2 3s 2 s 5 s 2 3s 2
s s
2
2
11s 28
3s 2 s 5
.
Using partial fractions, we have
L ^ y t `
s 2 11s 28
s 1 s 2 s 5
10 1 9 3 6 2 s 1 s 2 s 5
.
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
9 t 10 2t 1 5t e e e . 2 3 6
y t
Example 4.33. Obtain the solution of the second differential equation
d 2 y t dt
2
4
dy t dt
5 y t 8sin t
along with the initial conditions y 0
0 y' 0 .
Solution: Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
357
2 dy t ° d y t °½ 4 5 y t ¾ 8L ^sin t` L® 2 dt ¯° dt ¿°
L ª¬ y '' t º¼ 4 L ª¬ y ' t º¼ 5L ª¬ y t º¼
8 s 1 2
8 ª¬ s 2Y s sy 0 y ' 0 º¼ 4 ª¬ sY s y 0 º¼ 5Y s 2 s 1 . Imposing the initial conditions, we obtain the Laplace transform of the solution as
s
2
4s 5 Y s
Y s
8 s2 1
8 s 1 s 2 4s 5 2
.
Using partial fractions, we can write this as
Y s
As B Cs D 2 2 s 1 s 4s 5
.
We expand this S domain solution into partial fractions as
Y s
s 1 s3 s2 1 s 2 4s 5
Chapter 4
358
Y s
s 1 s2 1 s2 1 s2 1 s 2 4s 5 s 2 4s 5
.
Applying the inverse Laplace transform to the above equation, and using the linearity of the inverse Laplace transform, results in
y t cos t sin t e2t cos t e2t sin t
.
Example 4.34. Use the Laplace transform to find the output
y t of
the system described by
dy t dt
5 y t x t
with input given by
x t 3e 2t u t with y 0 2.
Solution: Given that
y' t 5 y t x t because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
^
`
L y ' t 5 L ^ y t ` L ^ x t `
^ L ª¬ y t º¼ '
Using
.
` , we get
sL ª¬ y t º¼ y 0
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
359
ª sL ª¬ y t º¼ y 0 º 5L ª¬ y t º¼ L ª¬ x t º¼ ¬ ¼ . We have
L ^ x t `
3 , s 2
and thus
L ^ y t `
3
s 2 s 5
2 s 5
It is now necessary to invert
L ^ y t ` . To accomplish this, some
algebraic manipulation is necessary to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
L ^ y t ` becomes
L ^ y t `
1 1 2 s 2 s 5 s 5
L ^ y t `
1 3 s 2 s 5
.
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
y t e2t 3e5t .
Chapter 4
360
Example 4.35. Determine the solution of the initial value problem
d 2 y t dt
2
3
dy t dt
2 y t 4t with y 0 0 y ' 0 .
Solution: Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
^
`
^
`
L y '' t 3L y ' t 2 L ^ y t ` 4 L ^t`
4 ª s 2 L ^ y t ` sy 0 y ' 0 º 3 ª sL ^ y t ` y 0 º 2 L ^ y t ` 2 ¬ ¼ ¬ ¼ s Find the expression
L ^ y t ` in the form of an algebraic function.
Substituting the values for
y 0 0 y ' 0 ; rearranging the above
equation gives
s
2
3s 2 L ^ y t `
4 s2 .
This algebraic equation can be solved for
L ^ y t `
.
4 s s 1 s 3 2
.
L ^ y t ` as
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
361
This solution is in the form of the product of two known Laplace transforms; we can invert either using partial fractions. Using the partial fractions yields
4 s ( s 1)( s 2) 2
4
A B C D 2 s s s 1 s 2
As( s 1)( s 2) B( s 1)( s 2) Cs 2 ( s 2) Ds 2 ( s 1)
Let s 0
B
2
s
-1
C
4
s
-2
D -1 A 3 3 2 4 1 Thus, F ( s ) 2 s s s 1 s 2 Applying the inverse Laplace transform to the above equation, and using its linearity, results in
y t 3 2t 4et e2t . 4.4 Total Response of the System Using the Laplace Transform LTI systems can be explained by a constant-coefficient differential equation. The response of the system is derived by solving the differential equation. The Laplace transform converts the differential equation in the time domain to an algebraic equation in the Laplace domain. These algebraic equations can be solved to find the solution. The total response of the system can be expressed as
Chapter 4
362
Total response
Natural response Forced response
The usual response of the system is solely due to the initial conditions. This is also the zero input response, i.e. when the externally-applied input is zero. The forced response is due to the externally-applied input function. The forced response consists of the steady-state response and the transient response. The forced response is also a zero state response, i.e. when the initial conditions are zero.
4.4.1 Impulse Response and Transfer Function For a linear ODE with constant coefficients, the output is derived as a convolution of its impulse response function
y t h t * x t
h t and the input x t :
x t * h t , assuming zero initial states. This has
been demonstrated in Chapter 3 for second and higher-order systems in the time domain, and it also holds for higher-order LTI systems. Taking the convolution property of the Laplace transform, the output is the product of
Y s
transforms
H s
X s H s
H s X s ,
L ^h t ` , is the transfer function, and X s
where
L ^ x t ` , is
the input. From this expression, we find that the impulse response function of a system is generated when
x t G t .
Example 4.36. Find the forced response of the system with the differential equation given by
d 2 y t dt
2
5
dy t dt
6 y t x t
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
to the input given by
363
x t et u t .
Solution: We first find the transfer function of the system. The solution process is the same as for example 1. Taking the Laplace transform of the given differential equation and imposing the initial conditions, we find, successively
° d 2 y t ½° d 2 y t 2 t 2 4 4 L® y t ¾ L e t 2 2 dt ¯° dt ¿°
^
^
`
^
`
`
L y '' t 5 L y ' t 6 L ^ y t ` L ^ x t ` '' 2 2 ° L ¬ª y t ¼º s L ª¬ y t º¼ s Y s ® ' ° L ª¬ y t º¼ sL ª¬ y t º¼ sY s ¯ using
s
2
5s 6 Y s X s
T .F
H s
Y s X s
1 s 2 5s 6
Thus,
H s
Y s X s
1 s 5s 6 2
.
½ ° ¾ ¿°
Chapter 4
364
Y s
1
X s
s 2 s 3
we have
X s
1 . s 1
Y s
1 s 1 s 2 s 3
.
We begin by finding the partial fraction expansion for Y s . The denominator consists of three distinct linear factors and so the expansion has the form
Y s
1 s 1 s 2 s 3
where
A,
B,
and
C
ª º 1 A « » «¬ s 2 s 3 »¼ s C
ª º 1 « » «¬ s 1 s 2 »¼ s
Y s
1
3
are
A B C s 1 s 2 s 3 real
1 , 2
numbers
to
be
determined.
ª º 1 B « » «¬ s 1 s 3 »¼ s
1 , 2
1 1 1 1 1 2 s 1 s 2 2 s 3
.
1, 2
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
365
We take the inverse Laplace transform to find the solution. The response due to the input is the steady-state response and the response due to the poles of the system is the transient response
1 §1 · h t ¨ et e 2t e3t ¸ u t . 2 ©2 ¹ Thus,
1 §1 · § · h t ¨ et u t ¸ ¨ e2t e3t ¸ u t . 2 ©2 ¹ © ¹ Forced response = steady state response + transient response. Example 4.37. Find the natural response, forced response, and the total response of the system with the differential equation given by
d 2 y t dt
2
5
dy t dt
6 y t x t
to the input given by
x t u t . The initial conditions are
y 0 0 and y ' 0 2. Solution: In this instance, the forcing function is the step function
u t of the
above equation. Therefore, this function is discontinuous and is not the solution of any linear homogeneous differential equation with constant coefficients. This example of an initial value problem that we cannot solve quickly uses undetermined coefficients. The advantage of the Laplace
Chapter 4
366
transform method is that the above initial value problem’s solution can be obtained with one application of the method. Given that
d 2 y t dt
2
5
dy t dt
6 y t x t
We first find the transfer function of the system. The solution process is the same as for example 1. Taking the Laplace transform of the given differential equation and imposing the initial conditions, we find, successively
d 2 y t ½ dy t ° ° L® 5 6 y t ¾ 2 dt ° dt ° ¯ ¿
L ^ x t ` .
We take the Laplace transform of each term of the left side of the given differential equation and apply the initial conditions
^
`
^
`
L y '' t 5 L y ' t 6 L ^ y t ` L ^ x t ` '' 2 ' ° L ¬ª y t ¼º s L ª¬ y t º¼ sy 0 y 0 ® ' ° L ª¬ y t º¼ sL ª¬ y t º¼ sY s y 0 using ¯
s
2
5s 6 Y s s 7 X s
Solving this expression for components
½ ° ¾ °¿
.
Y s yields two rational function
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
Y s
Y s
X s
s
2
5s 6
s 7
s
2
5s 6
s 7 s 2 s 3 s 2 s 3 X s
367
.
The response due to the first term is due to the initial conditions and hence is the natural response
YN s
s 7 s 2 s 3
.
Using partial fraction expansion, we have
YN s
5 4 s 2 s 3
.
Applying the inverse Laplace transform to the above equation, and using the linearity of the inverse Laplace transform, results in
yN t
5e
2 t
4e 3t u t .
The response due to the second term is the forced response. To find the forced response, we apply the input x t
YF s
X s
s 2 s 3
u t .
Chapter 4
368
YF s
1 s s 2 s 3
.
It is now necessary to invert
YF s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
YF s
YF s becomes
1§1· 1§ 1 · 1§ 1 · ¸ ¨ ¸ ¨ ¸ ¨ 6 © s ¹ 2 ¨© s 2 ¸¹ 3 ¨© s 3 ¸¹
.
Applying the inverse Laplace transform to the above equation, and using the linearity of the inverse Laplace transform, results in
1 §1 · § 1 · yF t ¨ u t ¸ ¨ e2t u t e3t u t ¸ . 3 ©6 ¹ © 2 ¹ The total response was given by the addition of the natural response and the forced response
9 11 §1 · yT t yN t yF t ¨ u t e2t u t e3t u t ¸ . 2 3 ©6 ¹ The natural response is the complementary function, and the forced response is the particular integral. This is easily checked as to whether it is the correct solution. Now, it is up to you whether our approach to solving this differential equation is easier than the alternatives, but we suggest that the Laplace transform method provides a straightforward solution to the differential equation.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
369
Example 4.38. Find the zero-input, zero-state, transient, steady-state, and complete responses of the system governed by the differential equation
d 2 y t dt 2
4
dy t dt
4 y t
d 2 x t dx t 2x t dt 2 dt
with the initial conditions
y 0 0 and y ' 0 4 and the input x t u t , the unit-step function. Solution: We first find the transfer function of the system. Take the Laplace transform of the given differential equation and apply the initial conditions
^
`
^
`
^
` ^
`
L y '' t 4 L y ' t 4 L ^ y t ` L x '' t L x ' t 2 L ^ x t ` '' 2 ' ° L ¬ª y t ¼º s L ª¬ y t º¼ sy 0 y 0 ® ' °¯ L ª¬ y t º¼ sL ª¬ y t º¼ sY s y 0 using
s
2
4s 4 Y s 2s 8 s 1
Solving this expression for
2 s
.
Y s yields two rational function
components
Y s
s2 s 2 2s 11 2 2 s s 4s 4 s 4s 4
½ ° ¾ °¿
.
Chapter 4
370
The first term on the right-hand side is
X s H s and corresponds to
the zero-state response. The second term is due to the initial conditions and corresponds to the zero-input response. Expanding into partial fractions, we get
2
Y s
s s2 s s 2
2
2s 11 s s 2
2
1 1 2 2 2 2 7 s s 2 s 2 2 s 2 s 2 2
. Taking the inverse Laplace transform, we get the complete response
yT t yN t yF t
zero-state
zero-input 1 1 § · 2 t 2 t 2 t 2 t ¨ u t e u t 2te u t ¸ 2e u t 7te u t 2 ©2 ¹
5 §1 · yT t ¨ u e 2t 5te 2t ¸ u t . 2 ©2 ¹ 4.5 Systems of Linear Differential Equations Similar to how we can convert a single differential equation into a single algebraic equation using a Laplace transform, so we can convert a pair of differential equations into simultaneous algebraic equations. The differential equations we solve are linear and so a couple of linear differential equations will convert into simultaneous linear algebraic equations familiar from school. We first solve algebraic equations for each of the transformed functions and then find the inverse Laplace transforms in the usual way. The following example illustrates how to use the Laplace
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
371
transform method to obtain the general solution of a system of linear differential equations with constant coefficients. Example 4.39. Determine solution of initial value problem
d 2 x t dt
2
y t
and
dx t dy y t 4 4x t dt dt
with y 0 1, x 0 0 and x ' 0 1. Solution: Because of the linearity of the equations and of the Laplace transform operation, taking the Laplace to transform for both differential equations, we have
s2 X s Y s 1
4 4s X s s 1 Y s Solving for
X s
X s H s 1 1 1 s 1 2
s 1 4 4s s 1
1
.
gives
1 ; s s 4s 4 3
2
Chapter 4
372
s2 1 4 4s 1
Y s
1 s2 4 4s s 1
s 2 4s 4 s3 s 2 4s 4 .
It is now necessary to invert
X s and Y s . To accomplish this,
some algebraic manipulation is necessary to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
1 s s 4s 4
X s
3
2
2
Y s
s 4s 4 s s 2 4s 4 3
X s and Y s becomes
A B C s 1 s 2 s 2
1 1 1 3 2 6 s 1 s 2 s 2
A B C s 1 s 2 s 2
1 2 3 2 3 s 1 s 2 s 2
. Applying the inverse Laplace transform to the above equations, and using the linearity of the inverse Laplace transform, results in
x t
1 t 1 2t 1 2t e e e . 3 2 6
y t
1 t 2 e 2e 2t e 2t . 3 3
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
373
Example 4.40. Determine the solution of the initial value problem
dx t
2 x t y t 2e5t and
dt
dy t dt
x t 2 y t 3e2t y 0 0 x 0 .
Solution: Because of the linearity of the equations and the Laplace transform operation, taking the Laplace to transform to both differential equations, we have
s 2 X s Y s
2 s 5 3 s 2
X s s 2 Y s
It is now necessary to invert
.
X s and Y s . To accomplish this,
some algebraic manipulation is necessary to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
X s
2s
2
X s and Y s becomes
5s 7
s 1 s 2 s 3 s 5
5 3 3 1 4 4 s 1 s 2 s 3 s 5
5
1
3s 13 4 1 4 Y s s 1 s 3 s 5 s 1 s 3 s 5 .
Chapter 4
374
Applying the inverse Laplace transform to the above equations and using the linearity of the inverse Laplace transform results in
x t
5 t 3 e 3e 2t e3t e5t . 4 4
y t
5 1 e t e 3t e 5 t . 4 4
Example 4.41. Determine the solution of the initial value problem
dy t dt
dz t
2 y t z t 0 and
subject to the initial conditions
dt
2z t y t 0
y 0 1 and z 0 0.
Solution: Because of the linearity of the equations and the Laplace transform operation, taking the Laplace to transform to both differential equations, we have
dy t ½ L® 2 y t z t ¾ 0 ¯ dt ¿ dz t ½ L ® 2z t y t ¾ 0 ¯ dt ¿ . The method for solving single differential equations carries over into systems of equations. After transforming each equation and making use of the initial conditions, we obtain
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
s 2 Y s Z s
375
1
Y s s 2 Z s 0 Solving for
Y s and Z s gives
1 1 0 s 2
Y s
s2 1
1 s 2
s2 1 1 0 s2 1 s 2 1
Z s
s2
s 2
2
1
2
1
;
1
s 2
.
Applying the inverse Laplace transform to the above equations, and using the linearity of the inverse Laplace transform, results in
y t e2t cos t
z t e2t sin t . Example 4.42. Determine the solution of the initial value problem
d 2 x t dt 2 with
10 x t 4 y t 0 and
y 0 0
x 0 , x' 0
d 2 y t
y t 4x t 0 dt 2 1 and y ' 0 1.
Chapter 4
376
Solution: Because of the linearity of the equations and the Laplace transform operation, taking the Laplace to transform to both differential equations, we have
° d 2 x t L® 10 x t 4 y t 2 ¯° dt ° d 2 y t L® y t 4x t 2 ¯° dt
½° ¾ 0 ¿°
½° ¾ 0 ¿° .
The method for solving single differential equations carries over into systems of equations. After transforming each equation and making use of the initial conditions, we obtain
s
2
10 X s 4Y s 1
4 X s s 2 1 Y s 1 Solving for
X s
.
X s and Y s gives
4 1 1 s 2 1 s 2 10 4 4 s2 1
s2 s 2 2 s 2 12
;
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
Y s
s 2 10 1 4 1
s 2 10 4 4 s2 1
It is now necessary to invert
s2 6 s 2 2 s 2 12
377
X s and Y s . To accomplish this,
some algebraic manipulation is necessary to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
X s
s
s
2
X s and Y s becomes
2
2 s 2 12
s 6 s 2 s 2 12 2
2 3 5 5 2 2 s 2 s 12
2
Y s
6 1 5 5 2 2 s 2 s 12
Applying the inverse Laplace transform to the above equations, and using the linearity of the inverse Laplace transform, results in
x t
1 sin 5 2
2t 5 612 sin 12t .
y t
2 sin 5 2
2t 5 312 sin 12t .
Chapter 4
378
Example 4.43. Determine the solution of the initial value problem
dy t dt
2 x t sin 2t and
dx t dt
2 y t cos 2t with
y 0 0, x 0 1. Solution: Because of the linearity of the equations and the Laplace transform operation, taking the Laplace to transform to both differential equations, we have
dy t ½ L® 2 x t ¾ L ^sin 2t ¯ dt ¿
`
dx t ½ L® 2 x t ¾ L ^cos 2t ¯ dt ¿
` .
The method for solving single differential equations carries over into systems of equations. After transforming each equation and making use of the initial conditions, we obtain
2 X s sY s
sX s 2Y s
Solving for
2 s2 4
s 1 s 1 2
.
X s and Y s gives
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
Y s
2 s2 4
1ª 2 2s « 2 2 2« s 4 s 4 ¬
X s
379
º » »¼ .
Applying the inverse Laplace transform to the above equation, and using the linearity of the inverse Laplace transform, results in
x t
1 sin 2t 2 cos 2t . 2
y t sin 2t . Example 4.44. Determine the solution of the initial value problem
d 2 x t dy t dx t d 2 y t sin t and cos t dt 2 dt dt dt 2 with y 0 1, x 0 0, y ' 0 0 and x ' 0 1. Solution: Because of the linearity of the equations and the Laplace transform operation, taking the Laplace to transform to both differential equations, we have 2 ° d x t dy t L® 2 dt °¯ dt
°½ ¾ L ^ sin t °¿
`
Chapter 4
380 2 ° d y t dx t L ® 2 dt °¯ dt
°½ ¾ L ^ cos t °¿
` .
The method for solving single differential equations carries over into systems of equations. After transforming each equation and making use of the initial conditions, we obtain
2s 2 1 s2 1
s X s sY s 2
(4.2)
§ s2 2 · X s sY s ¨ 2 ¸ © s 1 ¹
(4.3)
Adding (4.2) and (4.3), we get
X s
s2 1
s
2
1
2
(4.4)
Using (4.4) in (4.3), we get
Y s
1 2s s s2 1
2
.
Applying the inverse Laplace transform to the above equations, and using the linearity of the inverse Laplace transform, results in
x t t cos t .
2 ½ ° s 1 ° t cos t L ® 2 ¾ 2 s 1 ° ° ¯ ¿ 1
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
y t
° 2s L1 ® 2 ° ¯ s 1
1 t sin t .
½ ° 2 ¾ ° ¿
381
t sin t .
Example 4.45. Find the solution of the set of the differential equations
dy t dt
2 y t z t 0 and
dz t dt
subject to the initial conditions y 0
2z t y t 0
1 and z 0 0.
Solution: Because of the linearity of the equations and the Laplace transform operation, taking the Laplace transform, we have
dy t ½ L® 2 y t z t ¾ 0 ¯ dt ¿ dz t ½ L® 2z t y t ¾ 0 ¯ dt ¿ . The method for solving single differential equations carries over into systems of equations. After transforming each equation and making use of the initial conditions, we obtain
s 2 Y s Z s
1
Y s s 2 Z s 0 Solving for
.
Y s and Z s gives
Chapter 4
382
1 1 0 s 2
Y s
s2
s2 1 1 s 2
2
1
2
1
s 2
;
s2 1 1
0 s2 1 1 s 2
Z s
1
s 2
.
Applying the inverse Laplace transform to the above equations, and using the linearity of the inverse Laplace transform, results in
y t e2t cos t z t e2t sin t . Example 4.46 Determine solution of initial value problem
d 2 x t dt 2
2x t y t 0
and
d 2 y t dt 2
2 y t x t 0
with y 0 0, x 0 0, x ' 0 1 and y ' 0 1. Solution: Because of the linearity of the equations and the Laplace transform operation, taking the Laplace transform, we have
s
2
2 X s Y s
1
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
X s s2 2 Y s 1
X s and Y s gives
Solving for
1
Y s
1
1 s 2 2
X s
.
2
s 2
1
1
s2 2
s 2
;
1
2
s 2
1
s 2 2 1 1 1 2
1 s 3 2
1 s 3 2
Thus,
X s
Y s
1 s 3 2
1 s2 3
.
Taking the inverse Laplace transform, we get
x t
1 sin 3
3t .
383
Chapter 4
384
y t
1 sin 3
3t .
Example 4.47. Determine the solution of the initial value problem
dy t y t dt where F
t
F t , y 0
0
0 0 d t d S . ® ¯sin t t ! S
Solution: Given that
dy t y t dt
F t
because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform, we have
sL ª¬ y t º¼ L ª¬ y t º¼
s 1 L ª¬ y t º¼ L ª¬ y t º¼ eS s
L ª¬ F t º¼
1 S s e s 1 2
1
s 1 s 2 1
° 1 y t L1 ®eS s s 1 s 2 1 ¯°
½° ¾ ¿°
.
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
385
Refer to example 2. 24
ª 1 L1 « «¬ s 1 s2 1
º 1 1 ½ t » ®e cos 2t sin 2t ¾ 2 ¿ »¼ 5 ¯
then,
° ½° 1 1 1 ½ tS y t L1 ®eSs ¾ ®e cos2 t S sin2 t S ¾u t S . 2 5¯ 2 ¿ ¯° s 1 s 1 ¿°
Thus,
y t
1 t S 1 ½ cos 2 t S sin 2 t S ¾ u t S ®e 5¯ 2 ¿ .
Example 4.48. Obtain the solution of the differential equation
d 2 y t dt 2
y t 2u t 2 u t 4 ,
along with the initial conditions
y 0 0 y' 0
.
Solution: Assuming that the Laplace transform of the solution using the fact that
L
exists and
is a linear operator, we take the Laplace transform of
the above differential equation and write
° d 2 y t L® y t 2 ¯° dt
Y s
½° ¾ L ^2u t 2 u t 4 ` ¿°
Chapter 4
386
L ª¬ y '' t º¼ L ª¬ y t º¼
^
2e2 s e4 s . s s
ª¬ s 2Y s sy 0 y ' 0 º¼ Y s
It is now necessary to invert
`
2e2 s e4 s s s
.
Y s . To accomplish this, some algebraic
manipulation is necessary to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s
Y s becomes
2e2 s e4 s 1 s s s2 1
2e 2 s e 4 s s s s2 1
.
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
° ½° ½° 1 1 1 1 ° 4 s 1 ° y t L1 ®2e 2 s L e L ¾ ® ¾ ® 2 2 2 1 1 s s s s °¿ °¿ ¯° ¯° ¯° s 1
Using the convolution theorem
½° ¾ ¿°
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
387
° 1 ½° L1 ® 2 ¾ 1 cos t s s 1 °¯ °¿ .
Finally,
y t 2 1 cos t 2 u t 2 1 cos t 4 u t 3 sin t. Hence, the solution of the given initial value problem is
y t 2 1 cos t 2 u t 2 1 cos t 4 u t 3 sin t. Example 4.49. Obtain the solution of the second-order differential equation
d 2 y t dt
2
5
dy t dt
6 y t 3G t 2 4G t 4
along with the initial conditions y 0
0 y' 0 .
Solution: Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
° d 2 y t dy t 5 6 y t L® 2 dt ¯° dt
½° ¾ L ^ 3G t 2 4G t 4 ¿°
L ª¬ y '' t º¼ 5 L ª¬ y ' t º¼ 6 L ª¬ y t º¼
3e 2 s 4e 4 s
`
Chapter 4
388
s
2
5s 6 Y s 3e 2 s 4e 4 s
Y s 3
Y s
e 2 s e 4 s 4 s 2 5s 6 s 2 5s 6
e 2 s e 4 s 3 4 s 2 s 3 s 2 s 3
.
We begin by finding the partial fraction expansion for
Y s . The
denominator consists of two distinct linear factors and so the expansion has the form
1
A B s 2 s 3
s 2 s 3
where A and B are real numbers to be determined.
ª 1 º A « » ¬« s 3 ¼» s
1, B 2
hence, the given function
ª 1 º « » ¬« s 2 ¼» s
1, 3
Y s can be expanded as
ª 1 ª 1 1 º 1 º 4 s Y s 3e 2 s « » 4e « » «¬ s 2 s 3 »¼ «¬ s 2 s 3 »¼ . Now that we have obtained the partial fraction expansion, taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
Application of the Laplace Transform to LTI Differential Systems: Solving IVPS
389
y t 3 e 2 t 2 e 3 t 2 u t 2 4 e 2 t 4 e 3 t 4 u t 4 .
Summary In this chapter, we have described and explained the solution of initial value problems. x LTI systems governed by differential equations, along with their initial conditions, have been analyzed by finding complete solutions. The advantage of the Laplace transform method when forcing functions, such as step functions and impulse functions, appear as inputs have been discussed with several examples. LTI systems governed by simultaneous differential equations and their initial conditions have also been analyzed by finding solutions. x The analysis of LTI systems has been given using Laplace transforms. Terms, such as impulse response, natural response, steady-state response, transient response, and forced response, have been explained. Simple examples have been solved to illustrate the relevant concepts. The zero state response and zero input response have also been described.
CHAPTER 5 APPLICATION OF THE LAPLACE TRANSFORM TO LTI DIFFERENTIAL SYSTEMS: ELECTRICAL CIRCUITS AND MECHANICAL SYSTEMS
5.1 Introduction The most familiar application of Laplace transformations in the physical sciences is in analyzing mechanical system problems and electrical circuits. The Laplace transform is a handy tool for analyzing linear timeinvariant (LTI) electric circuits. We can use Laplace transforms to solve differential equations relating an input voltage/current signal to another output signal in the circuit. Laplace transforms can also be used to analyze a circuit directly in the Laplace domain, where the circuit components are replaced by their impedances observed as transfer functions. The Laplace transform has paramount importance in studying the displacement in various dynamical systems (mechanical vibration).
5.2 Application to Electrical Circuits 5.2.1 The Scheme for Solving Electrical Circuits The scheme for solving electrical circuits is outlined below.
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
391
Step 1: Apply Kirchhoff’s laws to the given circuit, which yields linear ordinary differential equations with constant coefficients. Step 2: Take the Laplace equation to transform both sides of the equation. A differential equation with constant coefficients is transformed into an algebraic equation. Step 3: Use the properties of the Laplace transform and the initial conditions and simplify the algebraic equation obtained for
L ^i t ` in the S domain.
I s
Step 4: A table of transforms, rather than the inversion integral of (2.2), is used to find the inverse transform. Step 5: In general, a partial-fraction expansion is required to expand complicated functions of s into the simpler functions available in Laplace transform tables. Step 6: Find the inverse transform of
I s to obtain i t , the solution
of the differential equation is the required output of the circuit (voltage and current). LEARNING OBJECTIVES On reaching the end of the chapter, we expect you to have understood and be able to apply the following: x To explore the procedure for solving a complete electrical circuit using the Laplace transform technique. x The use of the Laplace transform technique to obtain the desired response of an electrical system.
Chapter 5
392
x How to determine the transfer function and the impulse and step response of an electrical system. x To know the relation between the input and the output in the S domain using the transfer function and can use it to calculate the displacement of a mechanical system. Example 5.1. Use the Laplace transform technique to find the current in the RC-circuit shown in Figure 5.1a. Suppose an impulse voltage
G t
is applied as an input shown in Figure 5.1b. The circuit is
assumed to be quiescent before the input is applied.
C
t R Figure 5.1a
Figure 5.1b
Solution: Applying KVL to loop Figure 5.1a, we get t
1 R i t ³ i W dW C0
G t
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
393
dq is the current. The two terms on the left give the voltage dt
Here, i
drop across the resistor and capacitor, respectively. We take Laplace equations to transform both sides of the equation where
I s
L ^i t ` ,
R I s
1 I s 1 C s
1 · § ¨ R ¸ I s 1 sC ¹ © . This algebraic equation can be solved for
1 s 1 · R§ ¨s ¸ RC ¹ ©
I s
I s
I s as
ª§ 1 · 1 º s ¨ ¸ « 1 © RC ¹ RC » « » 1 · » R« § s ¸ «¬ ¨© RC ¹ »¼
½ 1 ° °° 1° RC ®1 ¾ 1 ·° R° § s °¯ ¨© RC ¸¹ °¿
Taking the inverse Laplace transform of the result
I s and using Table 2.1 gives
Chapter 5
394
i t
1ª 1 RC1 t º G t e » « R¬ RC ¼
which is the required current in the RC-circuit. Example 5.2 The L C R a capacitor C
circuit consists of a resistor R
1F , and an inductor L 1H connected in series
together with a voltage source V at
t
2: ,
1volts . Before closing the switch
0 , both the charge on the capacitor and current is zero.
Determine the charge and current. Solution: We now demonstrate the use of the Laplace transform in solving for the current in an electric circuit. Consider the L C R Figure 5.2, where V
series circuit in
1volts R
L
i + V
C
–
Figure 5.2 Applying KVL to the above circuit, the integro-differential equation that characterizes the electrical system is
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
395
t
di 1 Ri t L ³ i W d W V dt C 0
d 2q dq q L 2 R V dt dt C . Here, q represents the electrical charge and i
dq represents the dt
current. The three terms on the left give the voltage drop across the inductor, resistor, and capacitor, respectively. Substituting the given values for the resistance, inductance, and capacitance gives
d 2q dq 2 q 1 2 dt dt . Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
s
2
1 s
2 s 1 L ^q t `
.
This algebraic equation can be solved for
L ^q t `
1 s s 1
2
.
L ^q t ` as
Chapter 5
396
The inverse Laplace transform of However,
I s is the desired current i t , t ! 0
L ^q t ` is not given in Table 2.1. In cases such as this, we
use partial fraction expansions to express a Laplace transform as a sum of simpler terms in the table. We can express
L ª¬ q t º¼
A B C s s 1 s 1 2
L ^q t `
1 1 1 s s 1 s 1 2
L ^q t ` as
.
Applying the inverse Laplace transform to the above equation, and using the linearity of the inverse Laplace transform, results in
q t 1 e t te t Thus, the current
i t
.
i t , t ! 0 is
dq et tet et tet dt .
Example 5.3. An inductor of L
1 henrys and a capacitor of C
farads are connected in series with a generator of the charge q as a function of time if
V t
0 0 d t 1 . ® ¯V0 t t 1
1
V t volts. Find
q 0 0 i 0 , and
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
397
Solution: The differential equation that characterizes the electrical system is
L
d 2q t dt
2
1 q t V0 u t 1 C .
Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
° d 2 q t L® q t 2 dt ¯°
½° ¾ V0 L ^ u t 1 ` ¿°
.
Using the derivative property and all initial conditions being zero gives
s
2
1 Q s V0
Q s
e s s
V0 e s
.
s s2 1
We expand this S domain solution into partial fractions as
ª1 s Q s V0 e s « 2 s s 1 «¬
º » »¼ .
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
q t V0 1 cos t 1 u t 1 .
Chapter 5
398
Example 5.4. The differential equation for the current
i t in an LR
circuit is
Ri t L
di t
E t .
dt
Find the current in an LR circuit if the initial current is i t given that
L
4: , and E t
2H , R
0A
2e 5t with time t
measured in seconds. Solution:
L
di t dt
Here, i
R i t
E t
.
dq is the current. The two terms on the left give the voltage dt
drop across the inductor and resistor, respectively. Substituting the given values for the resistance and inductance gives
di t dt
2 i t
e 5t .
Taking the Laplace transform in the usual way and using gives
s 2 L ^i t `
1 s5 .
This algebraic equation can be solved for
L ^i t ` as
i 0 0 A
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
399
1
L ^i t `
s 2 s 5 .
Resolving into partial fractions gives
L ^i t `
A B s 2 s 5
L ^i t `
1§ 1 1 · ¨¨ ¸ 3 © s 2 s 5 ¸¹
.
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
i t
1 2t e e 5t . 3
Example 5.5. The LR circuit consists of a resistor R and an inductor L connected in series together with a voltage source Prior to closing the switch at t
E t .
0, the currents are zero. Determine
the current. Solution: We now demonstrate the use of the Laplace transform in solving for the current in an electric circuit. Consider the LR circuit in Figure 5.3.
Chapter 5
400
Figure 5.3 Applying KVL to the above circuit, the differential equation that characterizes the electrical system is
Ri t L Here, i
di t dt
E t .
dq is the current. The two terms on the left give the voltage dt
drop across the resistor and inductor, respectively
di t R E t i t dt L L . Taking the Laplace transform in the usual way and using gives
R· § ¨ s ¸ L ^i t ` L¹ ©
E Ls
i 0 0 A
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
This algebraic equation can be solved for
L ^i t `
401
L ^i t ` as
E R· § Ls ¨ s ¸ L¹. ©
The inverse Laplace transform of
I s is the desired current
i t , t ! 0 However, L ^i t ` is not given in Table 2.1. In cases such as this, we use partial fraction expansion to express a Laplace transform as a sum of simpler terms in the table. We can express
L ^i t `
L ^i t `
L ^i t ` as
A B R· s § ¨s ¸ L¹ © ª º « » E 1 1 « » R ·» R «s § s ¨ ¸ «¬ L ¹ »¼ . ©
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
i t
E ª R L t º 1 e »¼ R «¬ .
Chapter 5
402
Example 5.6. The LR circuit consists of a resistor R and an inductor
L connected in series together with a voltage source Ee at . Prior to closing the switch at current
t
0 , the currents are zero. Determine the
i t .
Solution: We now demonstrate the use of the Laplace transforms in solving for the current in an electric circuit. Consider the LR circuit in Figure 5.4 with the voltage source Ee
at
.
Figure 5.4 By applying KVL we arrive at the following differential equation
Ri t L Here, i
di t dt
Ee at .
dq is the current. The two terms on the left give the voltage dt
drop across the resistor and inductor, respectively
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
403
di t R E at i t e dt L L . Taking the Laplace transform in the usual way, and using
i 0 0 A ,
gives
L ^i t `
E R· § L s a ¨ s ¸ L¹. ©
The inverse Laplace transform of
I s is the desired current
I t , t ! 0. However, the transform for L ^i t ` is not given in Table 2.1. In cases such as this, we use partial fraction expansion to express a Laplace transform as a sum of simpler terms in the table. We can express
L ^i t ` as
L ^i t `
L ^i t `
A B s a § s R · ¨ ¸ L¹ © ª º « » E 1 1 « » R aL « s a § s R · » ¨ ¸ «¬ L ¹ »¼ . ©
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
Chapter 5
404
i t
R tº ª at E e e L » R aL «¬ ¼
Example 5.7. The series
C
.
LRC circuit shown in Figure 5.5 consists of
5: L 1H .
0.25 F R
Use
the
Laplace
transform
technique. R
+ Vm
C
~ –
VC
i(t)
L
Figure 5.5
a Determine the differential equation relating Vin b Obtain
Vc for Vin
to
Vc .
e t with Vc 0 1 and Vc' 0 2.
Solution:
a By applying KVL we arrive at the following differential equation di t
t
1 ³ i W dW Vin Ri t L dt C 0 It is known that i rewritten as
C
.
dVc , hence the above differential equation can be dt
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
405
d 2VC dV LC 2 R C C VC Vin dt dt . Substituting the given values for the resistance, inductance, and capacitance gives
d 2VC dV 5 C 4VC 4Vin 2 dt dt
b Given d 2VC dV Vin , 2 5 C 4VC 4e t u t dt dt . In this instance, the forcing function is the step function above equation. This function is discontinuous at
t
u t of the
0 and, therefore, is
not the solution of any linear homogeneous differential equation with constant coefficients. This example of an initial value problem that we cannot solve easily uses undetermined coefficients. The advantage of the Laplace transform method is that the solution of the above initial value problem can be obtained with one application of the method. Taking the Laplace equation to transform both sides and using the initial conditions
Vc 0 1 and Vc' 0 2.
s
2
5s 4 L ª¬VC t º¼ s 2 5
s
2
5s 4 L ª¬VC t º¼
4 s 1
4 s7 s 1
Chapter 5
406
s
2
5s 4 L ª¬VC t º¼
s 2 8s 11 s 1 .
This algebraic equation can be solved for
L ª¬VC t º¼
L ª¬VC t º¼
s 2 8s 11 s 1 s 2 5s 4
L ^VC t ` as
s 2 8s 11
s 4 s 1
2
.
The inverse Laplace transform of
L ^VC t ` is the desired voltage
VC t . However, the transform for L ^VC t ` is not given in Table 2.1. In cases such as this, we use partial fraction expansion to express a Laplace transform as a sum of simpler terms in the table. Expanding
L ª¬VC t º¼
L ^Vc t ` using partial fraction expansion
s 2 8s 11
s 4 s 1
A B C s 4 s 1 s 1 2
2
Solving for A, B, and C, we obtain
A Thus,
5 ,B 9
14 and C 9
2 . 3
.
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
L ª¬VC t º¼
407
5 ° 1 ½ ° 14 ° 1 ½ ° 2 ° 1 ½ ° ® ¾ ® ¾ ® 2 ¾ 9 ¯ 4 9 1 3 s s ¿° ¿° °¯ s 1 °¿ ° ° ¯ .
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
VC t
5 4t 14 t 2 t e e te . 9 9 3
Example 5.8. An electromagnetic field (EMF) of applied to an
V
EG t is
LCR series circuit. Determine the charge on the
capacitor and the resulting current in the circuit at time t; initially there is no current in the circuit and no charge on the condenser. Solution: We now demonstrate the use of Laplace transforms in solving for the current in an electric circuit. Consider the 5.6, where
V
LCR series circuit in Figure
EG t . R
L
i + V –
Figure 5.6
C
Chapter 5
408
Applying KVL to the above circuit, the integro-differential equation that characterizes the electrical system is
Ri t L
L
di t dt
t
1 i W dW V C ³0
d 2q dq q R V 2 dt dt C
Here,
§ ¨ i ©
dq · ¸ dt ¹ .
q represents the electrical charge and i
dq dt
represents the
current. The three terms on the left give the voltage drop across the inductor, resistor, and capacitor, respectively.
d 2q dq q L 2 R E G t dt dt C d 2 q R dq q dt 2 L dt LC
E G t L .
Because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation, we have
1 § 2 R ¨s s L LC ©
· ¸ L ^q t ` ¹
E L
.
The algebraic equation can be solved for
L ^q t ` as
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
L ^q t `
L ^q t `
409
E 1 1 · L§ 2 R ¨s s ¸ L LC ¹ ©
ª E« 1 L « s O 2 p 2 «¬
º » » »¼
where O
§ 1 R R2 · 2¸ & p2 ¨ 2L © LC 4 L ¹ .
Taking the inverse Laplace transform, we get
q t
E Ot ªe sin pt º¼ Lp ¬ .
Thus,
i t
dq dt
E Ot ªe p cos pt O sin pt º¼ . Lp ¬
Example 5.9. An inductor of L
1 henrys and a capacitor of C 1
farads are connected in series with a generator of the charge q as a function of time if
V t volts. Find
q 0 0 i 0 and
0 0 d t 1 V t ® . ¯1 t t 1 Solution: The differential equation that characterizes the electrical system is
d 2q t 1 L q t V t dt 2 C
.
Chapter 5
410
The two terms on the left give the voltage drop across the inductor, and capacitor, respectively.
d 2q t dt 2
q t u t 1 .
Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write 2 ° d q t L® q t 2 °¯ dt
°½ ¾ L ^ u t 1 °¿
e s s
L ª¬ q '' t º¼ L ª¬ q t º¼ e s s
s 1 Q s 2
Q s
`
e s s s2 1
§1 s Q s e s ¨ 2 ¨s s 1 ©
· ¸ ¸ ¹.
Now that we have obtained the partial fraction expansion. Taking the inverse Laplace transform of each term on the right, and using the linearity property of the Laplace transform, we find that
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
§1 s ° q t L1 ®e s ¨ 2 °¯ ¨© s s 1
411
· ½° ¸¾ ¸° ¹¿
Finally, using the transform pairs established in table 2.1, we have
q t
1 cos t 1 u t 1 .
Example 5.10 A 20-volt battery is applied to an circuit with an inductance of L and resistance of
L C R series
1H capacitance of C 104 F ,
R 160:. Determine the charge on the capacitor
and the resulting current in the circuit at time t; initially there is no current in the circuit and no charge on the condenser. Solution: We now demonstrate the use of Laplace transforms in solving for the current in an electric circuit. Consider the Figure 5.6, where V
L C R series circuit in
20 Volts. R
L
i + V –
Figure 5.7
C
Chapter 5
412
Applying KVL to the above circuit, the integro-differential equation that characterizes the electrical system is
Ri t L
di t dt
t
1 i W dW V C ³0
.
The charge q on the LCR circuit is determined by
L
d 2q dq q R V dt 2 dt C
§ ¨ i ©
dq · ¸ dt ¹ .
Here, q represents the electrical charge and i
dq represents the dt
current. The three terms on the left give the voltage drop across the inductor, resistor, and capacitor, respectively. Substituting the given values for the resistance, inductance, and capacitance gives
d 2q dq 160 104 q 20 2 dt dt
D
2
160 D 104 q
20
.
Because of the linearity of the equation and the Laplace transform operation, taking the Laplace transform of the differential equation, we have
s
2
160 s 104 L ^q t `
20 s .
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
This algebraic equation can be solved for
20 s s 160 s 104
L ª¬ q t º¼
2
The inverse Laplace transform of However,
413
L ^q t ` as
. I s is the desired current i t , t ! 0.
L ^q t ` is not given in Table 2.1. In cases such as this, we
use partial fraction expansion to express a Laplace transform as a sum of simpler terms in the table. We can express
L ^q t `
A Bs c 2 s s 160 s 104
L ^q t `
1 ª1 s 160 « 2 500 « s s 160s 104 ¬
L ^q t `
1 ª1 s 80 80 « 2 500 « s s 160s 104 ¬
ª 1 «1 500 « s ¬«
s 80
s 80
2
602
L ^q t ` as
º » »¼
º » »¼
º » 80 2 2 » s 80 60 ¼» . 1
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
Chapter 5
414
1 ª 4 § ·º 1 e 80t ¨ cos 60 t sin 60 t ¸ » « 500 ¬ 3 © ¹¼ .
q t
The resulting current
i t in the circuit is given by
1 80t ªe sin 60 t º¼ 3¬ .
i t
Example 5.11. The differential equation for the current
i t in an
LR circuit is
Ri t L
dR dt
V t with L 1H , R
4: and a voltage source
(in volts)
V t
2 ® ¯0
0 t 1 t t1
.
a Find the current i t
if
b Compute the current at
i 0 0.
t 1.5 seconds, i.e. compute i 1.5 .
c Evaluate lim ^i t `. t of Solution:
a
Since
circuit is
V t 2 2u t 1 , the differential equation of the LR
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
di t dt
415
4i t 2 2u t 1 .
Taking the Laplace transform in the usual way gives
s 4s L ^i t ` Solving for
L ª¬i t º¼
2 2e s s s .
L ^i t ` gives
2 2e s . s s 4 s s 4
Let
G s
2 s s 4
,
using partial fractions, we obtain
G s
º 1 ª1 1 « » 2 ¬ s s 4 ¼
.
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
g t
1 1 e 4t 2
since
L ª¬i t º¼
G s e sG s ,
Chapter 5
416
we can take the inverse Laplace to transform to get the current
1 1 ª¬1 e 4t º¼ ª1 e 4 t 1 º u t 1 . ¼ 2 2¬
i t
b We can express i t as follows 1 1 e 4t ° °2 ® ° 1 e 4 t 1 e 4t ° ¯2
i t
For t
0 t 1
t t1
1.5, we get
i 1.5
1 4 1.51 e e 4 1.5 2
0.0664.
c For t t 1, we have i t
1 4 t 1 e e 4t 2 .
Therefore,
ª1 º lim ª¬i t º¼ lim « e4 t 1 e4t » 0. t of t of 2 ¬ ¼
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
417
Example 5.12. Consider the following parallel RL circuit shown in Figure 5.8 with L
1H , R 1: .
a Determine the differential equation relating I s
b
Obtain the zero-state response for
transform for
c Obtain
IL.
I L t using the Laplace
I s t e 5t .
the zero-input response for
transform with
and
I L t using the Laplace
I L 0 1.
Figure 5.8 Solution:
a
I s and I L are related by the following differential equation IL IR.
IS
dI L t dt
I L t Is t .
Chapter 5
418
b For the given I s , dI L t dt
I L t e5t .
Taking the Laplace transform in the usual way gives
1 s5
sL ^ I L t ` I L 0 L ^ I L t ` since
I L 0 0 for zero states,
s 1 L ^I L t ` L ^ I L t `
1 s5
1
s 1 s 5 . L ^ I L t ` is the desired current
The inverse Laplace transform of
I L t . However, L ^ I L t ` is not given in Table 2.1. In cases such as this, we use partial fraction expansion to express a Laplace transform as a sum of simpler terms in the table. We can express
L ^I L t `
1ª 1 1 º « » 4 ¬« s 1 s 5 ¼»
.
L ^ I L t ` as
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
419
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
I L t
1 t e e 5t . 4
c For the zero-input response I s 0
0, and given that I L 0 0,
we have to find the solution of the following differential equation for the zero-input response
dI L t dt
I L t 0, I L 0 1 .
Taking the Laplace transform in the usual way gives
sL ^ I L t ` 1 L ^ I L t ` 0
s 1 L ^I L t `
1
.
Thus,
L ^ I L t `
1 s 1 .
The inverse transform of
I L t et u t .
L ^ I L t ` is the zero-input response given by
Chapter 5
420
Example 5.12. Use the Laplace transform technique to find the current in the RC circuit in figures 5.9a and 5.9b, if a single rectangular wave with voltage is applied as an input. The circuit is assumed to be quiescent before the wave is applied.
Figure 5.9a
Figure 5.9b
Solution: The input in terms of a unit step function is given by
v t V0 ª¬u t a u t b º¼ with a b. Applying KVL to the above loop, we get t
1 R i t ³ i W dW V0 ª¬u t a u t b º¼ C0 Taking the Laplace transform in the usual way gives
.
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
R I s
421
ª e as ebs º 1 I s V0 « » C s s s ¼ ¬
ª e as ebs º 1 · § R I s V 0« ¨ ¸ » sC ¹ s s ¼ © ¬
I s
ª º as bs « » V0 e e « » 1 · § 1 ·» R «§ s s «¬ ¨© RC ¸¹ ¨© RC ¸¹ »¼ .
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
V0 R
i t
1 t b u ª RC1 t a º RC e u t a e u t b « » ¬ ¼.
Example 5.13. Use the Laplace transform technique to find the current through the resistance to the circuit shown in Figure 5.10 with
L
2H , C
50 P F , R 100: and V t sin 100t . There is
no current flowing in either loop prior to closing the switch at time
t
0.
Chapter 5
422
C
i2 t=0
+
i1 L
V
R
–
Figure 5.10 Solution: Applying Kirchhoff’s second law to each of the two loops in turn gives Loop-1:
1 ª di di º i1 dt L « 1 2 » V ³ C ¬ dt dt ¼ . Substituting the given values for the inductance and capacitance gives
1 ª di di º i dt 2 « 1 2 » sin 100t 6 ³ 1 50 u10 ¬ dt dt ¼
.
Taking the Laplace equation to transform both sides of the equation where
I1 s
L ^i1 t ` & I 2 s
L ^i2 t ` ,
2 u 106 I1 s 2 sI1 s 2 sI 2 s s
100 s 1002 2
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
s
2
104 I1 s s 2 I 2 s
50
423
s s 1002 2
(5.1) Loop-2:
ª di di º R2i2 L « 2 1 » 0 ¬ dt dt ¼ . Substituting the given values for the inductance and capacitance gives
ª di di º 100i2 L « 2 1 » 0 ¬ dt dt ¼
.
Taking the Laplace equation to transform both sides of the equation where
I1 s
L ^i1 t ` & I 2 s
L ^i2 t ` ,
100 I 2 s 2sI 2 s 2sI1 s 0 sI1 s s 50 I 2 s 0 (5.2) Solving (5.1) and (5.2), we get
I1 s
ª s s 50 « 2 « s 100 s 2 104 ¬
I2 s
ª s2 « 2 « s 100 s 2 104 ¬
º » » ¼ .
º » » ¼
Chapter 5
424
We need to find the inverse Laplace transform of this function of s and we expand using partial fractions, giving
I2 s
ª s2 « 2 2 4 « ¬ s 100 s 10
I2 s
ª§ 1 · 1 1 s §1· § 1 · «¨ ¨ ¸ ¨ ¸ ¸ 2 2 © 200 ¹ s 104 «¬© 200 ¹ s 100 © 2 ¹ s 100
º » » ¼
ª A B Cs D « 2 2 s 100 s 100 s 104 ¬«
º » »¼
. The current through the resistance
i2 t is
ª§ 1 · 100t § t · 100t § 1 · 4 ¨ ¸e ¨ i2 t «¨ ¸e ¸ cos 10 t ©2¹ © 200 ¹ ¬© 200 ¹
º» . ¼
Example 5.14. Use the Laplace transform technique to find the current through the capacitance to the circuit shown in Figure 5.11 with
R1
1:
R2 , L1
1H
L2 , C
1F & V
10V .
There is no charge on the capacitors and no current flowing in the inductances prior to closing the switch at time
Figure 5.11
t
0.
º » ¼»
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
425
Solution: Applying Kirchhoff’s second law to each of the two loops in turn gives Loop-1:
R i1 i2 L
d 1 i1 i2 ³ i1 dt dt C
V .
Substituting the given values for the inductance and capacitance gives
i1 i2
d i1 i2 ³ i1 dt dt
10 .
Taking the Laplace transform, where
I1 s
L ^i1 t ` & I 2 s
L ^i2 t ` ,
1 I1 s I 2 s s I1 s I 2 s I1 s s
s 1 I1 s s 1 I 2 s
I1 s s
10 s
10 s (5.3)
Loop-2:
R2i2 L
di2 1 i1 t dt C ³
0 .
Substituting the given values for the inductances and capacitance gives
i2
di2 i1 t dt ³
0
Taking the Laplace transform, where
Chapter 5
426
I1 s
L ^i2 t ` ,
L ^i1 t ` & I 2 s
I 2 s sI 2 s
I1 s s
0 (5.4)
Solving (5.3) and (5.4), we get
I1 s
ª 10 « 2 «¬ s s 2
ª 10 I1 s « 2 ¬« s s 2
º » & I2 s »¼
ª 10 « «¬ s s 1 s 2 s 2
º » I1 s »¼
ª º « » « » 10 « » « § § 1 ·2 § 7 ·2 · » «¨¨ s ¸ ¨ ¸ ¸» ¨ 2 2 © ¹ « © ¹ ¸¹ »¼ ¬© .
The current through the capacitance
i1 t
º » »¼
i1 t is
ª§ 20 · §¨ 12 ·¸t § 7 ·º © ¹ t ¸¸ » . sin ¨¨ «¨ ¸e 2 7 © ¹ © ¹ »¼ ¬«
Example 5.15. Use the Laplace transform technique to find the loop currents shown in Figure 5.12 with
R1 1: , R1 1.4:, L1 1H and L2 § § 1 ·· V t 100 ¨ u t u ¨ t ¸ ¸ . © 2 ¹¹ ©
0.8 H and
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
L1
427
L2 i1
i2
R2
R1 V Figure 5.12 Applying KVL to loop-1 with current
i1 t , and substituting the given
values for the resistance and inductance, gives
0.8
ª di1 § 1 ·º 1 i1 i2 1.4 i1 100 «u t u ¨ t ¸ » dt © 2 ¹¼ ¬
ª di1 § 1 ·º 3 i1 1.25i2 125 «u t u ¨ t ¸ » dt © 2 ¹¼ . ¬ Applying KVL to loop-2 with current
i2 t , and substituting the given
values for the resistance and inductance, gives
1
di2 1 i2 i1 dt
di2 i1 i2 dt
0 .
0
Chapter 5
428
Taking
the
^i 0
0 i2 0 ` , we get
1
Laplace
transform,
s 3 I1 s 1.25I 2 s
assuming
s º ª 2 1 e » 125 « «s s » ¬« ¼»
I1 s s 1 I 2 s 0 Solving for
I1 s
I2 s
I1 s and I 2 s
s 125 s 1 § · ¨1 e 2 ¸ § 1 ·§ 7 · ¹ s ¨ s ¸¨ s ¸ © © 2 ¹© 2 ¹ s § · 125 2 1 e ¨ ¸ § 1 ·§ 7 · ¹ s ¨ s ¸¨ s ¸ © 2 2 © ¹© ¹ .
Using partial fractions, we get
125 s 1 1 ·§ 7· § s¨ s ¸¨ s ¸ 2 ¹© 2¹ ©
A B C 1· § 7· s § ¨s ¸ ¨s ¸ 2¹ © 2¹ ©
the
initial
conditions
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
ª º « 125 s 1 » » A « « § s 1 ·§ s 7 · » ¸ «¬ ¨© 2 ¸¨ ¹© 2 ¹ »¼ s ,B
ª º «125 s 1 » « » « s§s 7 · » ¸ «¬ ¨© 2 ¹ »¼ s
500 7 0
429
125 ,C 3
1 2
ª º «125 s 1 » « » « s§s 1 · » ¸ «¬ ¨© 2 ¹ »¼ s
625 21 7 2
s s s · · · 500 § 250 § 625 § 2 2 ¨1 e ¸ ¨1 e ¸ ¨1 e 2 ¸ 7s © ¹ 3§ s 1 · © ¹ 21§ s 7 · © ¹ ¨ ¸ ¨ ¸ 2¹ 2¹ © ©
I1 s . Similarly
I2 s
s s s · · · 500 § 250 § 250 § 2 2 2 1 1 1 e e e ¨ ¸ ¨ ¸ ¨ ¸ 7s © ¹ 3§ s 1 · © ¹ 21§ s 7 · © ¹ ¨ ¸ ¨ ¸ 2¹ 2¹ © ©
. Applying the inverse Laplace transform to the above equations, and using its linearity, results in
i1 t i2 t
125 2t 625 72t 500 e e 3 21 7 250 2t 250 72t 500 e e 3 21 7
Chapter 5
430
5.3 Application to Mass-spring-damper Mechanical System Suppose a mass
m
is hung on an idealized spring whose upper end is
supported rigidly. This is an idealized spring with negligible mass and with its restoring force proportional to its extension. Let 0, the origin, be the equilibrium position and the
y t coordinate with downward
displacement being positive and upward negative. Suppose that the load is pulled downward at a distance
y t by force F t .
By Hooke’s law
F ky t where
k is a constant called the force constant of the spring.
By Newton’s second law
m
d 2 y t
k y t
dt 2
.
Finally, assume that a damping force proportional to the velocity is present, where the damping force is
D
dy t dt
,
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
431
which is a fairly accurate assumption for small velocities. This is the fundamental equation of damped simple harmonic motion
m
d 2 y t dt 2
D
dy t dt
k y t 0.
In every oscillatory system, the oscillations cannot be maintained indefinitely due to the gradual dissipation of mechanical energy, unless energy is supplied to the system. If an external force
F t is applied, the
equation becomes
m
d 2 y t dt 2
D
dy t dt
k y t F t .
Such cases are called forced oscillations. The following example shows how to use the Laplace transform method to obtain the displacement of mechanical systems described by a linear differential equation with constant coefficients.
Chapter 5
432
Example 5.16. Use the Laplace transform technique to find the displacement of mechanical systems shown in Figure 5.13 with spring constant
k
25 , damping constant D 6,
external force
F t
mass M
1 and
4sin 2t.
K=4
D=2
M=1
y(t) F(t) = sin 2t
Figure 5.13 Solution: A mass–spring–damper system can be modeled using Newton’s law and Hooke’s law (using Table 3.2). The governing equation representing the above system is given by
d 2 y t dt
2
2
dy t dt
4 y t sin 2t .
Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
433
2 dy t ° d y t °½ L® 2 4 y t ¾ L ^sin 2t` 2 dt ¯° dt ¿°
L ª¬ y '' t º¼ 2 L ª¬ y ' t º¼ 4 L ª¬ y t º¼
2 s 4 2
ª¬ s 2Y s sy 0 y ' 0 º¼ 2 ª¬ sY s y 0 º¼ 4Y s It is now necessary to invert
2 s 4 2
Y s . To accomplish this, some algebraic
manipulation is necessary if we are to identify the terms on the right with the entries in Table 2.1. When expressed in terms of partial fractions, after a little manipulation,
Y s
Y s becomes
2 s 2s 4 s 2 4 2
.
Using partial fractions, we can write this as
Y s
2
2 s 2s 4 s 2 4 2
As B Cs D 2 s 2s 4 s 4 2
As B s 2 4 Cs D s 2 2s 4
Hence, comparing the coefficients of both sides we get
s3 :
A C 0,
s2 :
D B 0,
Chapter 5
434
2 D 4C
s:
0,
4B 4D 2 .
1:
By these equations, we obtain
A
1 ,B 2
1 and D 4
1 ,C 4
0.
Hence, we have
Y s
1 2 s 1 4 1 4 s
Y s
1 ° 1 ® 2 4 ° s 2s 4 ¯
Y s
1° 1 ® 4 ° s 1 2 3 ¯
s
2
2s 4
s
2
4
s 1 °½ 1 ° ¾ ® 2 4 ° s 2s 4 ¿° ¯
½ 1 ° s ° 1 ° s 1 ½° ¾ ® ¾ ® 2 4 ° s2 4 ° 4 ¯° s 1 3 ¿° ¯ ¿
°½ 1 ° s ¾ ® 2 4° s 4 ¿° ¯
°½ ¾ ¿°
½° ¾ ¿°
.
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
1
y t
4 3
e t sin
3 t 14 e
t
cos
3 t 14 cos 2t .
Example 5.17. The differential equation for a mass-spring system is
m
d 2 x t dt
2
D
dx t dt
K x t f t .
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
Consider a mass-spring system with mass m a spring with constant
K
1kg that is attached to
5 N / m. The medium offers a damping
force six times the instantaneous velocity, i.e.
D
6 N . s / m.
a
Determine the position of the mass
initial conditions:
b
external force
D 6 N .s / m.
x t if it is released with
x 0 3, x ' 0 1. There is no external force.
Determine the position of the mass
initial conditions:
435
x 0 0
x t if it is released with the
x ' 0 , and the system is driven by an
f t 30 sin 2t in Newtons with time t measured in
seconds. Solution:
a
The initial-value problem for the position of the mass is
d 2 x t dt 2 We have
6
dx t dt
5 x t
f t .
f t 0 the initial conditions are x 0 3, x ' 0 1.
Taking the Laplace transform on both sides of the differential equation, we get 2 dx t ½ ° d x t °½ L® ¾ 6L ® ¾ 5 L ^ x t ` 0 2 dt °¯ dt °¿ ¯ ¿
Chapter 5
436
s
2
6 s 5 ^ x t ` 3s 19
.
This algebraic equation can be solved for
L ^ x t `
L ^ x t ` as
3s 19 . s 1 s 5
The inverse Laplace transform of
L ^ x t ` is the desired displacement
x t , t ! 0 However, the transform for L ^ x t ` is not given in Table 2.1. In cases such as this, we use partial fraction expansion to express a Laplace transform as the sum of simpler terms in the table. We can express
L ^ x t ` as
L ^ x t `
4 1 s 1 s 5
.
Taking the inverse Laplace transform of gives the result
b
L ^ x t ` and using Table 2.1
x t 4e t e 5t .
We have
f t 30 sin 2t and the initial conditions are
x 0 0 and x ' 0 0.
d 2 x t dt
2
6
dx t dt
5 x t 30sin 2t .
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
437
Taking the Laplace transform on both sides of the differential equation, we get
° d 2 x t ½° dx t ½ L® ¾ 6L ® ¾ 5 L ^ x t ` 30 L ^sin 2t` 2 dt °¯ dt °¿ ¯ ¿
s
2
6 s 5 ^ x t `
60 s 4 2
This algebraic equation can be solved for
L ^ x t `
60 s 1 s 5 s 2 4
The inverse Laplace transform of
L ^ x t ` as
.
L ^ x t ` is the desired displacement
x t , t ! 0 However, the transform for L ^ x t ` is not given in Table 2.1. In cases such as this, we use partial fraction expansion to express a Laplace transform as the sum of simpler terms in the table. We can express
L ^ x t ` as L ^ x t `
1 ° 1 ½ ° 15 « 1 » 12 ° 3® « » ¾ ® 2 ° ° s 4 ¯ s 1 ° ¿ 29 ¬« s 5 ¼» 29 ¯
½ 72 ° s ° ¾ ® 2 ° 29 ¯ ° s 4 ¿
. Applying the inverse Laplace transform to the above equation, and using its linearity, results in
½ ° ¾ ° ¿
Chapter 5
438
x t
3et
15 5t 12 72 e sin 2t cos 2t 29 29 u 2 29
thus,
x t
3et
15 5t 6 72 e sin 2t cos 2t. 29 29 29
Example 5.18. An 8 lb weight is attached to a spring with a spring constant equal to 4lb / ft. Neglecting damping, the weight is released from rest at 4 ft below the equilibrium position. At t struck with a hammer, providing an impulse of the displacement function
2sec, it is
5lb sec. Determine
y t of the weight.
Solution: This situation mass-spring system can be modeled using Newton’s law and Hooke’s law (using Table 3.2). The governing equation representing to the above system is given by 2 8 d y t 4 y t 5G t 2 32 dt 2
along with the initial conditions
d 2 y t dt 2
y 0 4, y ' 0 0.
16 y t 20G t 2
Assuming that the Laplace transform of the solution
Y s exists, and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems 2 ° d y t 16 y t L® 2 °¯ dt
439
°½ ¾ 20 L ^G t 2 ` °¿
L ª¬ y '' t º¼ 16 L ª¬ y t º¼
20e 2 s
ª¬ s 2Y s sy 0 y ' 0 º¼ 16Y s 20e 2 s . Imposing the initial conditions, we obtain the Laplace transform of the solution as
s
2
16 Y s
Y s
20e 2 s 3s
e 2 s s 20 2 3 2 s 16 s 16
.
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
y t
5sin 4 t 2 u t 2 3cos 4t .
Example 5.19 A 2 kg mass is suspended from a spring with constant
512 N / m. It is set in motion by pulling it 10 cm above its equilibrium position and then releasing it. A sinusoidal force mass, but only for
A sin 8t acts on the
t ! 1. Find the position of the mass as a function of
time with negligible damping. Solution: The initial-value problem for displacement is
Chapter 5
440
2
d 2 x t dt
2
512 x t A sin 8t u t 1
x 0
1 ' , x 0 0. 10
If we take the Laplace transform, 2 ° d x t °½ 2L ® ¾ 512 L ^ x t ` A L ^sin 8t u t 1 ` 2 °¯ dt °¿
s · § 2 ¨ s 2 L ª¬ x t º¼ ¸ 512 L ª¬ x t º¼ 10 ¹ ©
AL ª¬sin 8t u t 1 º¼
s · § 2 ¨ s 2 L ª¬ x t º¼ ¸ 512 L ª¬ x t º¼ 10 ¹ ©
AL ª¬sin 8t u t 1 º¼
.
Using the second shifting theorem
s
2
256 L ª¬ x t º¼
s
2
256 L ª¬ x t º¼
s
2
256 L ª¬ x t º¼
s A s e L ª¬sin 8 t 1 º¼ 10 2
s A s e ^cos8 L >sin 8t @ sin 8 L > cos8t 10 2
@`
s A 8 s ° e s ®cos 8 2 sin 8 2 10 2 s 64 s 64 ° ¯
½ ° ¾ ° ¿
. Hence,
L ª¬ x t º¼
s A s ° 8 s sin 8 2 e ®cos8 2 2 2 s 256 s 64 s 256 s 2 64 10 s 256 2 ¯°
½° ¾ ¿°
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
x t
1 1 ª s L « 10 « s 2 256 ¬
º A ° ª 1 » ®8cos8 L1 «e s 2 »¼ 2 ° «¬ s 256 s 2 64 ¯
Taking the inverse Laplace transform of
441
º ª s » sin8L1 « 2 »¼ «¬ s 256 s 2 64
º ½° »¾ »¼ ° ¿
L ^ x t ` and using Table 2.1
gives us the result x t
1 A 1 ½ cos16t ®cos8 sin 8 t 1 sin 8cos8 t 1 cos8 sin16 t 1 sin 8cos16 t 1 ¾ u t 1 10 384 ¯ 2 ¿
Example 5.20 A 100 g mass is suspended from a spring with constant
50 N / min . It is set in motion by raising it 10 cm above its equilibrium position and giving it a velocity of 1 m/s downward. During the subsequent motion, a damping force acts on the mass and the magnitude of this force is twice the velocity of the mass. If an impulse force of magnitude mass at t
2 N . is applied vertically upward to the
3sec, find the position of the mass for all time.
Solution: The initial-value problem for the position of the mass is 2 dx t 1 d x t 1 ' 2 50 x t 2G t 3 , x 0 , x 0 1. 2 10 dt dt 10
If we multiply the differential equation by 10, and take the Laplace transform
° d 2 x t ½° dx t ½ L® ¾ 20 L ® ¾ 500 L ^ x t ` 20 L ^G t 3 ` 2 dt °¯ dt °¿ ¯ ¿
Chapter 5
442
s
2
20 s 500 L ª¬ x t º¼
L ª¬ x t º¼
L ª¬ x t º¼
20e 3 s s 10 1
s 10 1 20e 3s s 2 20 s 500 s 2 20 s 500
20e 3 s
s 10
2
20
2
1 10
s 10
s 10
Taking the inverse Laplace transform of
2
202
.
L ^ x t ` and using Table 2.1
gives us the result
ª s 10 L « « s 10 2 202 «¬ 1
x t
º » » »¼
§ 1 s e 10t L ¨ 2 ¨ s 202 ©
e 10 t 3 sin 20 t 3 u t 3
· ¸ ¸ ¹
e 10t cos 20t ,
1 10t e cos 20t. 10
Example 5.21. A mass attached to a spring is released from rest 1 m below the system’s equilibrium position and begins to vibrate. After
2S seconds, the mass is struck by a hammer exerting an impulse on the mass. The system is governed by the differential equation for a mass-spring system, which is
d 2 x t dt
2
16 x t 5G t 2S and x 0 1 & x ' 0 0.
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
Where
443
x t denotes the displacement from equilibrium at time t.
Determine
x t .
Solution: Taking the Laplace transform on both sides of the differential equation, we get
° d 2 x t ½° L® ¾ 16 L ^ x t ` 5L ^G t 2S ` 2 ¯° dt ¿°
s
2
16 L ^ x t ` 5e 2S s s
.
Solving for
x t gives
L ^ x t `
e 2S s s 2 5 2 s 16 s 16
.
Taking the inverse Laplace transform and using the second shifting theorem, we get
x t
5 sin 4 t 2S u t 2S cos 4t 4 .
Chapter 5
444
Example 5.22. A mass-spring system with mass 4, damping 8, and spring constant 20 is subject to a hammer blow at time
t
0. The
blow imparts a total impulse of 1 to the system, which was initially at rest. Find the response
y t of the system.
Solution: This situation of the mass–spring-damper system can be modeled using Newton’s law and Hooke’s law (using Table 3.2). The governing equation representing the above system is given by
4
d 2 y t dt
2
8
dy t dt
20 y t G t
along with the initial conditions
y 0 0 y' 0 .
Assuming that the Laplace transform of the solution
Y s exists and
using the fact that L is a linear operator, we take the Laplace transform of the above differential equation and write
° d 2 y t dy t L ®4 8 20 y t 2 dt dt ¯°
½° ¾ L ^G t ` ¿°
4 L ¬ª y '' t ¼º 8 L ¬ª y '' t ¼º 20 L ª¬ y t º¼ 1 4 ª¬ s 2Y s sy 0 y ' 0 º¼ 8 ª¬ sY s y 0 º¼ 20Y s 1
.
Imposing the initial conditions, we obtain the Laplace transform of the solution as
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
4s
2
445
8s 20 Y s 1
Y s
1 4 s 2s 5
Y s
1
2
2
4 s 1 4
.
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
y t
1 t e sin 2t. 8
Example 5.23. Motion of a falling body (falling bodies and air resistance). An object falls through the air towards Earth. Assuming that gravity and air resistance are the only forces applied to the object, determine its velocity as a function of time. Newton’s second law states that force is equal to mass times acceleration. We can express this by the equation
m
dV dt
F .
Near Earth’s surface, the force of gravity is the same as the object’s weight and is also directed downward. This force can be expressed in mg, where g is the acceleration due to gravity. Air resistance acting on the object is represented by
cV , where c is a positive constant depending on the
Chapter 5
446
density of the air and the shape of the object. We use the negative sign because air resistance is a force that opposes motion. Applying Newton’s law, we obtain the first-order differential equation
m
dV dt
mg cV with V 0 V0
D Velocity V mg Parachutist Figure 5.14
dV t dt
c V t m
g with V 0
V0
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
447
Velocity of a parachute We take the Laplace transform equation and get the velocity of a parachute
s L ^V t ` V0
c L ^V t ` m
c · § ¨ s ¸ L ^V t ` m ¹ © L ^V t `
g s
g V0 s
g c · § s¨ s ¸ m ¹ ©
V0 c · § ¨ s ¸ m ¹ . ©
Resolving into partial fractions gives
L ª¬V t º¼
L ª¬V t º¼
V0 A B c · § c · s § ¨ s ¸ ¨ s ¸ m¹ © m¹ © ª g «1 1 « c mc « s § ¨ s «¬ m ©
º » V0 » c · ·» § ¸» ¨ s ¸ m ¹. ¹¼ ©
Taking the inverse Laplace transform, we get
V t
c c tº t g ª m m «1 e » V0 e . mc ¬ ¼
Chapter 5
448
Example 5.24. The motion of a body falling in a resisting medium may be described by
d 2 y t m dt 2
mg D
dy t dt
when the retarding force is proportional to the velocity. Find for the initial conditions
y t
y 0 0 y' 0 .
Solution: Let
y '' t
D ' y t m
g .
Taking the Laplace transform of both sides of the given differential equation
L ª¬ y '' t º¼
D L ª¬ y ' t º¼ m
gL >1@
^s L ª¬ y t º¼` mD sL ¬ª y t ¼º 2
g s
This algebraic equation can be solved for
L ª¬ y t º¼
g D· § s2 ¨ s ¸ m¹. ©
L ^ y t ` as
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
The inverse Laplace transform of body
449
L ^ y t ` is the desired motion of a
y t , t ! 0 However, the transform for L ^ y t ` is not given in
Table 2.1. In cases such as this, we use partial fraction expansion to express a Laplace transform as a sum of simpler terms in the table. We can express
L ^ y t ` as
L ª¬ y t º¼
A B C 2 D· s s § ¨s ¸ m¹ © ½ ° ° m g° 1 ° mg 1 ½ m 2 g 1 ½ ® ¾ ® ¾ ® ¾ D · ° D ¯ s ¿ D2 ¯ s2 ¿ D2 ° § s ¨ ¸ ° m ¹° ¯© ¿ 2
L ª¬ y t º¼
Applying the inverse Laplace transform to the above equation, and using its linearity, results in
y t
m 2 g mD t mg m 2 g e t. D2 D D2
Another area to which the unit function and impulse function can be applied is studying the deflection of beams, where discontinuous and concentrated loads are encountered. The following example shows how to use the Laplace transform method to obtain the deflection of a beam.
Chapter 5
450
Example 5.25. Use the Laplace transform technique to find the lateral deflection
y t of the beam shown in Figure 5.15.
w
W
EI, L
P
P
l1
l2
y(t) Figure 5.15 If the loading is non-uniform, the use of the Laplace transform method, in which we make use of the Heaviside unit function and the impulse function, has a distinct advantage. The figure illustrates a uniform beam of length l, freely supported at both ends and bending under a uniformly distributed weight
W.
Our aim is to determine the transverse deflection
y t of the beam.
From the elementary theory of beams, we have
d 4 y t d 2 y t EI P W t dt 4 dt 2 . Where
W t is the transverse force per unit length, with a downwards
force taken to be positive, and EI is the flexural rigidity of the beam. It is
t
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
451
assumed that the beam has uniform elastic properties and a uniform crosssection over its length so that both Young’s modulus of elasticity and the moment of inertia
I
E
of the beam about its central axis are
constants. Using the Heaviside step function and the Dirac delta function, the force function
W t can be expressed as
W t W ª¬1 u t l1 º¼ W ª¬G t l2 º¼ . Therefore, equation (1) becomes
d 4 y t d 2 y t EI P dt 4 dt 2
w ª¬1 u t l1 º¼ W ª¬G t l2 º¼ .
d 4 y t 2 d 2 y t wˆ ª¬1 u t l1 º¼ Wˆ ª¬G t l2 º¼ . a 4 2 dt dt Where
a2
P , wˆ EI
w and Wˆ EI
W . EI
Since the left end is hinge support and the right end is sliding support, the boundary conditions are
at t 0
deflection 0 y 0 0.
bending moment 0 y '' 0 0.
Chapter 5
452
at t
slope 0 y ' L 0.
L
0 y ''' L 0.
shear force
Applying the Laplace transform, we get
L ª¬ y '''' t º¼ a 2 L ª¬ y '' t º¼
wˆ L ª¬1 u t l1 º¼ Wˆ L ª¬G t l2 º¼ .
Incorporating the properties and Laplace transforms of the impulse and step functions, we get
^s L ª¬ y t º¼ sy 0 y 0 ` a ^s L ª¬ y t º¼ y 0 ` 4
'
s s a L ª¬ y t º¼ 2
2
2
L ¬ª y t ¼º
''''
2
2
'
ª¬1 el1s º¼ Wˆ el2 s wˆ s
ª¬1 e l1 s º¼ s y 0 y 0 a y 0 wˆ Wˆ e l2 s s 2
'
''''
2
'
ª¬1 e l2 s º¼ ª¬ y ''' 0 a 2 y ' 0 ¼º e l1s ˆ ˆ W 2 2 w 3 2 s2 s2 a2 s s a2 s s a2 s2 a2 y' 0
Taking inverse transforms and making use of the second shift theorem gives the deflection
y x
y t as
y' 0 ½ 1§ 1 · · § t l 1 sin at ¬ª y ''' 0 a 2 y ' 0 ¼º 2 ¨ t sin at ¸ Wˆ ¨ 2 1 3 sin a t l1 ¸ u t l1 ° ° a © a ¹ © a a ¹ ° a ° ® ¾ ° wˆ ª 1 a 2t 2 2 cos at 1 º wˆ ª 1 a 2 t l 2 2 cos a t l 1 º ½ u t l ° »¾ 2 ° »¼ ® «¬ 2a4 2 2 ° ¬« 2a 4 ¼¿ ¯ ¯ ¿
Application of the Laplace Transform to LTI Differential Systems: Electrical Circuits and Mechanical Systems
To obtain the value of the undetermined constants, we
employ
the
unused
boundary
453
y ' 0 and y '' 0 ,
conditions
at
x
L, , i.e.
y ' L 0 and y '' L 0. Summary In this chapter, we have described and explained the responses of electrical and mechanical systems using the Laplace transform method. x LTI systems governed by differential equations via electrical circuits with initial conditions have been analyzed by finding the system response. The Laplace transform method’s advantage when forcing functions, such as step and impulse functions, are inputs have been discussed with several examples. x LTI systems governed by differential equations via mechanical systems with initial conditions have been analyzed by finding the system response. The Laplace transform method’s advantage in the solution of simultaneous differential equations obtained via electrical and mechanical systems have also been explained with several examples.
CHAPTER 6 APPLICATION OF THE LAPLACE TRANSFORM TO LTI DIFFERENTIAL SYSTEMS: STATE-SPACE ANALYSIS
6.1 State-space Representation of Continuous-time LTI Systems Definition: The state of a system at time
t0 is the minimal information
required that is sufficient to determine the state and the output of the system for all times,
t t t0 , for the known system input at all times,
t t t0 . The variables that contain this information are called state variables. LEARNING OBJECTIVES After studying this chapter, it is expected that you will: x Understand the relationship between the input and the output of LTI systems through state-space model representation. x Be able to apply and explore the procedure for solving state model representation using the Laplace transform technique. x Be able to apply the Laplace transform technique to obtain the state transition matrix of the state model.
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
455
x Be able to determine the state model and its response of an electrical system.
6.2 State Model Representations Consider a single-input single-output continuous-time LTI system described by the following second-order differential equation
d 2 y t dt 2
a1
dy t dt
a2 y t bu t .
y t is the system output and u t is the system input
y t a1 y t a2 y t b u t
(6.1)
Define the following useful set of state variables
y t x1 t y t x1 t x2 t
(6.2)
y t x2 t
y t a1 y t a2 y t bu t
x2 t a1 x2 t a2 y t bu t Putting (6.2) and (6.3) in matrix form, we get
(6.3)
Chapter 6
456
ª º « x1 t » « » «¬ x2 t »¼
y t
ª 0 « ¬ a2
1 º a1 »¼
ª x1 º ª0 º « x » «b » u t ¬ 1¼ ¬ ¼
ª x t º » ¬ 2 t ¼
>1 0@ « x1
(6.3a)
(6.3b)
Eqs. (6.3a) and (6.3b) can be written more compactly as
X t
A X t B u t
(6.4)
y t C X t In general, the state equations of a system can be described by
X t
A X t B u t
(6.5)
y t C X t D u t The vector
X t is called the state vector or response of the system.
If
u t 0, then equation (6.5) is called a homogeneous system.
If
u t z 0, then equation (6.5) is called a non-homogeneous system.
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
457
6.2.1 Procedure to Find the Response of a Second-order Nonhomogeneous System Consider a second-order non-homogeneous system
X t
A X t B u t
Multiplying both sides by
e At , we get
ª º e At « X t A X t » ¬ ¼
e At B u t
d ª e At X t º¼ dt ¬
e At B u t .
Integrating both sides between 0 and t, we get t
ª¬ e
At
X t X 0 º¼
³e
AO
B u O dO
0 t
ª¬ e At X t º¼
X 0 ³ e AO B u O d O 0
.
Thus, t
X t e At X 0 ³ e A t O B u O d O 0 hom ogenous solution Non hom ogenous solution
(6.6) Eq. (6.6), represents the response of a second order system.
Chapter 6
458
6.3 Matrix Exponential (State-transition matrix) The matrix exponential is denoted by
e At
I t
and is computed using
the Laplace transform as follows
ª Aº s «1 » ¬ s¼
1
> sI A@
1
> sI A@
1
1
· 1§ A A2 A3 2 3 ............ ¸ 1 ¨ s© s s s ¹
§ I A A2 A3 · ¨ 2 3 4 ............ ¸ s s ©s s ¹
Taking the inverse Laplace transform we get
1
§ At A2t 2 A3t 3 · ............ ¸ ¨1 1! 2! 3! © ¹
1
e At
L1 > sI A@
L1 > sI A@ Thus,
The matrix exponential is e
At
^
I t L1 sI A
1
`.
In applying the matrix Laplace transform method, it is straightforward (although possibly tedious) to compute
sI A
1
, but the computation
of the inverse transform may require the use of some of the special techniques (such as partial fractions) discussed in Chapter 2.
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
459
6.3.1 Converting from State-space to Transfer Function Given the state and output equations
X t
A X t B u t
y t C X t D u t
,
taking the Laplace transform, and assuming zero initial conditions, we can deduce the transfer function as
Y s X s
H S
C
^ > sI A@ ` B D 1
Example 6.1. Obtain the state-space representation of a system described by the following differential equation
d 3 y t dt
3
2
d 2 y t dt
2
3
dy t dt
4 y t u t .
Solution: The order of the differential equation is three. Hence, the three-state variables are
x1 t
y t , x1 t
dy t , x3 t dt
d 2 y t dt 2
The first derivatives of the state variables are
Chapter 6
460
x1 t x2 t
(6.7)
x2 t x3 t
(6.8)
x3 t 4 x1 t 3x2 t 2 x3 t u t
(6.9)
The state-space representation in matrix form (putting (6.6), (6.7), and (6.9) in matrix form) is
ª º « x1 t » « » « x2 t » « » « x3 t » ¬ ¼
ª 0 1 0 º ª x1 t º ª0 º « 0 0 1 » « x t » «0 » u t . « » « 2 » « » «¬ 4 3 2 »¼ «¬ x3 t »¼ «¬1 »¼
The system output
y t
y t becomes
ª x1 t º >1 0 0@ «« x2 t »» « x2 t » ¬ ¼
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
461
Example 6.2. Obtain the state-space representation for the electrical circuit shown in Figure 6.1, considering variables and
Vc1 , i1 and Vc 2
as state
Vc1 as the output y t .
Figure 6.1
Solution: The state variables for the circuit are
x1 t Vc1 , x2 t i1 , x3 t Vc 2 From the relationship between the voltage
dx1 t dt
Vc1 and current i1 , we obtain
x2 t
Kirchhoff’s voltage equation around the closed-loop gives
Chapter 6
462
Vs x1 t
dx2 t x3 t 0 dt
This equation can be rewritten as
dx2 t x1 t x3 t Vs dt with current,
i3 t
x3 t , and current, i3 t
dx3 t dt
By Kirchhoff’s current law
i1
i2 i3
implying that
x2 t
dx3 t x3 t dt .
Hence,
dx3 t dt
x2 t x3 t
The voltage
VC1 is taken as the output y t ,
y t x1 t The state-space representation of the circuit is given by
.
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
ª º « x1 t » « » « x2 t » « » « x3 t » ¬ ¼
ª 0 1 0 º ª x1 t º ª0 º « 1 1 1» « x t » «1 » V « » « 2 » « » s «¬ 0 1 1»¼ «¬ x3 t »¼ «¬0 »¼
and the system output
y t
463
y t becomes
ª x1 t º >1 0 0@ «« x2 t »» «¬ x2 t »¼
Example 6.3. Compute e At for
ª4 1º A « ». ¬0 4¼
Solution: To find the state transition matrix, we first calculate the matrix
> sI A@
ª1 0 º ª 4 1 º s« »« » ¬0 1 ¼ ¬0 4 ¼
The determinant of
sI A
ª s 0º ª4 1 º «0 s » «0 4 » ¬ ¼ ¬ ¼
sI A is given by
s 4 1 0 s4
s 4
2
.
We next calculate the adjoint of this matrix
sI A .
ª s 4 1 º « 0 s 4 »¼ ¬
Chapter 6
464
1 º ªs 4 Adj sI A « . s 4 »¼ ¬ 0 The inverse matrix is the adjoint matrix divided by the determinant
1
> sI A@
ª 1 «s 4 « « « 0 ¬
adj sI A det sI A
1
º s 4 »» 1 » » s4 ¼. 2
Taking the inverse of the elements of this matrix, we find that
I t e At
1
L1 > sI A@
ªe 4t « ¬0
te 4t º » e 4t ¼
.
Thus,
e
At
ªe 4t « ¬0
te 4t º ». e 4t ¼
Example 6.4. Find the state transition matrix of the non-homogeneous system
ª º « x1 t » « » «¬ x2 t »¼
ª 1 1 º ª x1 « 1 1» « x ¬ ¼¬ 2
º » ¼
ª0 º «1 » u t with X 0 ¬ ¼
and also find the response of the system.
ª0 º «1 » ¬ ¼
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
465
Solution: To find the state transition matrix, we first calculate the matrix
ª1 0 º ª 1 1 º s« »« » ¬0 1 ¼ ¬ 1 1¼
> sI A@
The determinant of
ª s 1 1 º « 1 s 1» ¬ ¼
sI A is given by
s 1 1 1 s 1
sI A
ª s 0 º ª 1 1 º «0 s » « 1 1» ¬ ¼ ¬ ¼
sI A ,
s 1
2
1 .
We next calculate the adjoint of this matrix
ªs 1 1 º Adj sI A « ». ¬ 1 s 1¼ The inverse matrix is the adjoint matrix divided by the determinant 1
> sI A@
adj sI A det sI A
ªs 1 1 º « » s 1 1 ¬ 1 s 1¼ 1
2
.
The state transition matrix is the inverse Laplace transform of this matrix. Finally, taking the inverse Laplace transform of each term, we obtain a state transition matrix
I t e At
1
L1 > sI A@
ª cos t sin t º et « ». ¬ sin t cos t ¼
The solution of the homogeneous part is
Chapter 6
466
X h t e At X 0 0 (since the circuit is initially relaxed) The solution of the non-homogeneous part is t
X nh t
³e
A t O
B Vs O d O
with Vs O u t 1
0
t
X nh t
³ 0
³ t
X nh t
t O cos t O °ªe ® « t O sin t O ° ¯ «¬ e
0
ª « « « « «¬
X nh t
»º °½ d O »¾ »¼ °¿
ª t O sin t O °« e ® « t O cos t O °« e ¯¬ t
º sin t O d O » » » e t O cos t O d O » »¼
³ e
t O
0
t
³ 0
t
VC t
e t O sin t O º ª0 º ½ ° » « » ¾ dO t O cos t O »¼ ¬1 ¼ ° e ¿
x2 t
³ e
t O
cos t O d O
0
t
VC t
³ e
t O
ª¬cos t cos O sin t sin O º¼ d O
0
t
t
VC t e cos t ³ e cos O d O e sin t ³ eO sin O d O O
t
0
t
0
.
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
467
Integration by parts gives t
t
VC t e t cos t ³ eO cos O d O e t sin t ³ eO sin O d O 0
0
Thus, the complete solution is
VC t
1 1 e t sin t e t cos t . 2
Hence,
VC t
x2 t
1 1 e t sin t e t cos t . 2
Example 6.5. Obtain the state-space representation for the electrical circuit shown in Figure 6.2 with
, L 1H and V zero at t
Figure 6.2
R1
4:, R2
6: , C
0.25 F
12Volts . Assume all currents and charges to be
0, the instant when the switch is closed.
Chapter 6
468
Solution: Appling KVL to the above loops, we get Loop-1:
L
di1 R1 i1 i2 V dt
di1 4 i1 i2 12 i1' 4i1 4i2 12 dt (6.10) Loop-2:
1 i2 dt R2i2 R1 i2 i1 0 C³ 1 i2 dt 6i2 4 i2 i1 0 0.25 ³
4 ³ i2 dt 10i2 4i1
4i2 10i2' 4i11 i2'
i2'
0
differentiating , we get
0 i2'
0.4i1' 0.4i2
0.4 4i1 4i2 12 0.4i2
1.6i1 1.2i2 4.8 (6.11)
Putting (6.10) and (6.11) in matrix form, we get
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
ªi1' t º «' » ¬«i2 t ¼»
I t
4 º ªi1 ª 4 « 1.6 1.2 » «i ¬ ¼¬ 2
469
º ª12 º » « 4.8 » u t ¼ ¬ ¼
A I t B u t
Example 6.6. Find the state transition matrix of the homogeneous system
ª º « x1 t » « » ¬« x2 t ¼»
ª 0 1 º ª x1 º « 2 3» « x » with X 0 ¬ ¼¬ 2 ¼
ª0 º «1 » ¬ ¼
and find the response of the system. Solution: To find the state transition matrix, we first calculate the matrix
> sI A@
ª1 0 º ª 0 1 º s« »« » ¬0 1 ¼ ¬ 2 3¼
ª s 0º ª 0 1 º «0 s » « 2 3» ¬ ¼ ¬ ¼
. The determinant of
sI A
sI A is given by
s 1 2 s3
s 1 s 2 .
We next calculate the adjoint of this matrix
ª s 3 1º Adj sI A « ». ¬ 2 s ¼
sI A , ª s 1 º « » ¬ 2 s 3¼
Chapter 6
470
The inverse matrix is the adjoint matrix divided by the determinant 1
> sI A@
1
> sI A@
adj sI A
ª s 3 1º s 1 s 2 ¬« 2 s ¼» 1
det sI A
ª s 3 « « s 1 s 2 « 2 « ¬« s 1 s 2
º s 1 s 2 »» » s » s 1 s 2 ¼» 1
.
Expressing each element of this matrix in terms of partial fractions, we get
1
> sI A@
1 ª 2 « s 1 s 2 « « 2 2 « ¬ s 1 s 2
1 1 º s 1 s 2 »» 1 2 » s 1 s 2 »¼
The state transition matrix is the inverse Laplace transform of this matrix. Finally, taking the inverse Laplace transform of each term, we obtain a state transition matrix
I t e At
1
L1 > sI A@
ª 2e t e 2t « t 2 t ¬ 2e 2e
e t e 2t º » e t 2e 2t ¼
The response of the system is
X t e At X 0
ª 2e t e 2t « t 2 t ¬ 2e 2e
e t e 2t º ª1 º »« » e t 2e 2t ¼ ¬0 ¼
.
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
X t
471
ª 2e t e 2t º « t 2 t » ¬ 2e 2e ¼ .
Example 6.7. Obtain the state-space representation of the circuit shown in Figure 6.3 considering the current through the inductor and voltage across the capacitor as state variables and the voltage across the capacitor as the output
y t .
Figure 6.3 Solution: The state variables for the circuit are
x1 t I L t , x2 t Vc1 t since the voltage across the capacitor is equal to the voltage across the series inductor and resistor branch, we obtain
x2 t
dx1 t x1 t dt .
This equation can be rewritten as
Chapter 6
472
dx1 t x1 t x2 t dt . Kirchhoff’s voltage equation around the closed-loop gives
Vs t i1 t x2 t 0 Hence,
i1 t x2 t Vs t
.
By Kirchhoff’s current law
i1 t x1 t
dx2 t dt
Therefore,
x2 t Vs t x1 t
dx2 t dt
This equation can be rewritten as
dx2 t x1 t x2 t Vs t dt The voltage
.
Vc t is taken as the output y t ,
y t x2 t The state-space representation of the circuit is given by
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
ª º « x1 t » « » «¬ x2 t »¼
ª 1 1 º ª x1 t º ª0 º » « » Vs t . « »« ¬ 1 1¼ ¬ x2 t ¼ ¬1 ¼
and the system output
y t
473
y t becomes
ª x t º ». ¬ 2 t ¼
>0 1@ « x1
Example 6.8. Find the matrix exponential of
ª1 1º A « » . ¬ 2 2 ¼
Solution: To find the state transition matrix, we first calculate the matrix
> sI A@
ª1 0 º ª1 1º s« »« » ¬0 1 ¼ ¬ 2 2 ¼
ª s 0 º ª1 1º «0 s » « 2 2 » ¬ ¼ ¬ ¼
sI A ªs 1 1 º « 2 s 2 » ¬ ¼
. The determinant of
sI A
sI A is given by
s 1 1 2 s 2
s s 1 .
We next calculate the adjoint of this matrix
ª s 1 1 º Adj sI A « ». ¬ 2 s 2¼ The inverse matrix is the adjoint matrix divided by the determinant
Chapter 6
474
1
> sI A@
adj sI A
1 ª s 2 1 º s 1¼» s s 1 ¬« 2
det sI A
1
> sI A@
ª s 2 « « s s 1 « 2 « «¬ s s 1
1 º » s s 1 » s 1 » » s s 1 »¼
Expanding each term in the matrix on the right by partial fractions yields
1
> sI A@
1 ª2 « s s 1 « «2 2 « ¬ s s 1
1 1º » s 1 s » 1 2 » » s s 1 ¼
.
Finally, taking the inverse Laplace transform of each term, we obtain a state transition matrix is
I t e At
1
L1 > sI A@
ª 2 et « t ¬ 2 2e
et 1 º » 2e t 1¼
Example 6.9. Find the state transition matrix of the non-homogeneous system
ª º « x1 t » « » «¬ x2 t »¼
ª 0 1 º ª x1 « 6 5» « x ¬ ¼¬ 2
º » ¼
ª0 º «1 » u t with X 0 ¬ ¼
and find the response of the system.
ª0 º «1 » ¬ ¼
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
475
Solution: To find the state transition matrix, we first calculate the matrix
ª1 0 º ª 0 1 º s« »« » ¬0 1 ¼ ¬ 6 5¼
> sI A@
The determinant of
ª s 1 º «6 s 5» ¬ ¼
sI A is given by
s 1 2 s3
sI A
ª s 0º ª 0 1 º «0 s » « 6 5» ¬ ¼ ¬ ¼
sI A ,
s 2 s 3 .
We next calculate the adjoint of this matrix
ª s 5 1º Adj sI A « ». ¬ 6 s ¼ The inverse matrix is the adjoint matrix divided by the determinant 1
> sI A@
1
> sI A@
adj sI A det sI A
ª s 5 « « s 2 s 3 « 6 « «¬ s 2 s 3
ª s 5 1º s 1 s 2 «¬ 6 s »¼ 1
º s 2 s 3 »» » s » s 2 s 3 »¼ 1
.
Expanding each term in the matrix on the right by partial fractions yields
Chapter 6
476
2 ª 3 « s 2 s 3 « « 6 6 « ¬ s 2 s 3
1
> sI A@
2 3 º s 2 s 3 »» 2 3 » s 2 s 3 »¼
.
The state transition matrix is the inverse Laplace transform of this matrix. Finally, taking the inverse Laplace transform of each term, we obtain a state transition matrix
I t e
1
At
L
> sI A@ 1
e 2e
ª 3e 2t 2e 3t « « 6e 2t 6e 3t ¬
2 t
e 3t
2 t
3e 3t
. The solution of the homogeneous part is
X h t e
X h t
At
X 0
e 2e
ª 3e 2t 2e 3t « « 6e 2t 6e 3t ¬
ª e 2t e 3t « « 2e 2t 3e 3t ¬
2 t 2 t
e 3t
3e 3t
º » » ¼.
The solution of the non-homogeneous part is t
X nh t
³e 0
A t O
B u O dO
with u O 1
º ª0 º »« » » ¬1 ¼ ¼
º » » ¼
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
³ t
X nh t
0
t
X nh t
³ 0
X nh t
X nh t
e 2e
2 t O ª 2e 3 t O ° « 3e ®« 2 t O 6e 3 t O ° « 6e ¯¬
ª « « « « «¬
2 t O
2 t O
2 t O ª e 3 t O °« e ®« 2 t O 3e 3 t O ° « 2e ¯¬
t
³ e
2 t O
e 3 t O d O
0
t
³ 2e
2 t O
3e 3 t O
0
ª 1 1 2t 1 3t º «2 2 e 3e » « » e 2t e 2t ¬ ¼
e3 t O
3e3 t O
477
½ º » ª0 º ° d O » «¬1 »¼ ¾ ° »¼ ¿
º½ » ° dO »¾ »¼ °¿
º » » » dO » »¼
.
Thus, the complete solution is
X t
X h t X nh t
X t
ª 1 1 2t 2 3t º «2 2 e 3 e » « » «¬ e 2t 2e 3t »¼ .
ª e 2t e 3t « « 2e 2t 3e 3t ¬
º ª 1 1 e 2t 1 e 3t º » » «2 2 3 » » « 2 t 3t e e ¼ ¬ ¼
Chapter 6
478
Example 6.10. Using the Laplace transform approach, obtain an expression for the state
X t of the system characterized by the state
equation
ª º « x1 t » « » «¬ x2 t »¼
ª 1 0 º ª x1 « 1 3» « x ¬ ¼¬ 2
º » ¼
ª1 º «1 » u t with X 0 ¬ ¼
ª1 º «1 » ¬ ¼
Solution: To find the state transition matrix, we first calculate the matrix
ª1 0 º ª 1 0 º s« »« » ¬0 1 ¼ ¬ 1 3¼
> sI A@
The determinant of
ªs 1 0 º « 1 s 3» ¬ ¼
sI A is given by
s 1 0 1 s 3
sI A
ª s 0 º ª 1 0 º «0 s » « 1 3» ¬ ¼ ¬ ¼
sI A ,
s 1 s 3 .
We next calculate the adjoint of this matrix
ªs 3 0 º Adj sI A « ». ¬ 1 s 1¼ The inverse matrix is the adjoint matrix divided by the determinant 1
> sI A@
adj sI A det sI A
ªs 3 0 º s 1 s 3 «¬ 1 s 1»¼ 1
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
1
> sI A@
1 ª « s 1 « « 1 « ¬ s 1 s 3
479
º » » 1 » s 3 »¼ . 0
Expanding each term in the matrix on the right by partial fractions yields
1
> sI A@
1 ª « s 1 « «1 ª 1 1 º « « » «¬ 2 ¬ s 1 s 3 ¼
º » » 1 » » s 3 »¼ . 0
The state transition matrix is the inverse Laplace transform of this matrix. Finally, taking the inverse Laplace transform of each term, we obtain a state transition matrix
I t e
At
1
L
^> sI A@ ` 1
ª et 0 º « » « 1 e t e 3t e 3t » ¬« 2 ¼» .
The solution of the homogeneous part is
ª et 0 º « » ª1º 1 « e t e 3t e 3t » ¬«1¼» ¬« 2 ¼»
Xh
e At X 0
Xh
ª º et « » « 1 e t e 3t » ¬« 2 ¼» .
Chapter 6
480
The solution of non-homogeneous part is (alternating approach)
X Nh
^
1· § ¨U s L u t ¸ s¹ ©
`
1
L1 > sI A@ b U s
1
> sI A@
1
> sI A@
bU s
ª 1 °« s 1 °« ®« °« 1 ª 1 1 º ° « 2 « s 1 s 3 » ¼ ¯¬ ¬
º ½ » ° » ª1º ° 1 « »¾ 1 » ¬1¼ ° s » s 3 »¼ °¿
bU s
1 ª º « » s s 1 « » «1 ª 1 » 1 º « « »» «¬ 2 ¬ s s 1 s s 3 ¼ »¼ .
0
Expanding each term in the matrix on the right by partial fractions yields
1
> sI A@
X nh
X t
1
L
1 1 ª º « » s s 1 « » «§ 5 1 1 1 1 1 ·» «¨ ¸» «¬¨© 6 s 2 s 1 6 s 3 ¸¹ »¼
bU s
^> sI A@
1
`
bU s
X h t X nh t
ª º 1 et « » « 5 1 e t 1 e 3t » 6 ¬« 6 2 ¼»
º ª º ª 1 et et » « »« « 1 e t e 3t » « 5 1 e t 1 e 3t » «¬ 2 »¼ «¬ 6 2 »¼ . 6
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
481
Thus, the complete solution is
X t
X h t X nh t
ª 1 º «1 » « e 3t » ¬3 ¼.
Example 6.11. Find the matrix exponential of
ª0 1º A « ». ¬ 8 6 ¼
Solution: To find the state transition matrix, we first calculate the matrix
ª1 0 º ª 0 1 º s« »« » ¬0 1 ¼ ¬ 8 6 ¼
> sI A@
The determinant of
ª s 1 º «8 s 6 » ¬ ¼
sI A is given by
s 1 8 s6
sI A
ª s 0º ª 0 1 º «0 s » « 8 6 » ¬ ¼ ¬ ¼
sI A ,
s 2 6s 8 .
We next calculate the adjoint of this matrix
Adj sI A
ª s 6 1º « 8 s » . ¬ ¼
The inverse matrix is the adjoint matrix divided by the determinant 1
> sI A@
adj sI A det sI A
ª s 6 1º s 2 s 4 «¬ 8 s »¼ 1
Chapter 6
482
1
> sI A@
s6 ª « s 2 s 4 « « 8 « ¬ s 2 s 4
1
º s s 2 4 »» » s s 2 s 4 »¼
.
Expanding each term in the matrix on the right by partial fractions yields
1
> sI A@
ª 2 1 « « s 2 s 4 « 4 4 « «¬ s 2 s 4
1§ 1 1 ·º ¨¨ ¸» 2 © s 2 s 4 ¹¸ » » 2 1 » s 2 s 4 »¼ .
The state transition matrix is the inverse Laplace transform of this matrix. Finally, taking the inverse Laplace transform of each term, we obtain the state transition matrix
I t e
At
1
L
^> sI A@ ` 1
ª 2 t 4 t « 2e e « 2 t 4 t ¬ 4e 4e
1 2t º e e 4t » 2 » e 2t 2e 4t ¼ .
Example 6.12. Find the state transition matrix of the homogeneous system
ª º « x1 t » « » «¬ x2 t »¼
ª 0 2 º ª x1 º « 2 5» « x » with X 0 ¬ ¼¬ 2 ¼
and find the response of the system.
ª0 º «1 » ¬ ¼
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
483
Solution: To find the state transition matrix, we first calculate the matrix
ª1 0 º ª 0 2 º s« »« » ¬0 1 ¼ ¬ 2 5¼
> sI A@
ª s 0º ª 0 2 º « »« » ¬0 s ¼ ¬ 2 5¼
sI A , ª s 2 º « » ¬ 2 s 5¼
. The determinant of
sI A is given by
s 2 2 s5
sI A
s 1 s 4 .
Next, we calculate the adjoint of this matrix
Adj sI A
ª s 5 2º « 2 s » . ¬ ¼
The inverse matrix is the adjoint matrix divided by the determinant 1
> sI A@
1
> sI A@
adj sI A det sI A s5 ª « s 1 s 4 « « 2 « ¬ s 1 s 4
ª s 5 2º s 1 s 4 «¬ 2 s »¼ 1
2
º s 1 s 4 »» » s s 1 s 4 »¼
Expanding each term in the matrix on the right by partial fractions yields
Chapter 6
484
1
> sI A@
ª§ 4 1 1 1 · «¨¨ ¸¸ «© 3 s 1 3 s 4 ¹ « « 2 § 1 1 · « 3 ¨¨ s 1 s 4 ¸¸ ¹ ¬ ©
2§ 1 1 · º ¨¨ ¸ » 3 © s 1 s 4 ¹¸ » » § 1 1 4 1 ·» ¨¨ ¸¸ » © 3 s 1 3 s 4 ¹ ¼ .
The state transition matrix is the inverse Laplace transform of this matrix Finally, taking the inverse Laplace transform of each term, we obtain the state transition matrix
I t e At
^
1
L1 > sI A@
`
ª§ 4 t 1 4t · 2 t º e e 4t » «¨ 3 e 3 e ¸ 3 ¹ «© » «2 § 1 t 4 4t · » t 4 t « e e ¨ e e ¸ » 3 © 3 ¹¼ ¬3
. The solution of the homogeneous part is
ª§ 4 t 1 4t · 2 t º e e 4t » «¨ 3 e 3 e ¸ 0 3 ¹ «© » ª« º» «2 § 1 t 4 4t · » ¬1 ¼ t 4 t « e e ¨ e e ¸ » 3 © 3 ¹¼ ¬3
Xh
e At X 0
Xh
ª 2 t º 4 t « 3 e e » « ». «§ 1 e t 4 e 4t · » ¸» «¬¨© 3 3 ¹¼
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
485
Example 6.13. Convert the state and output equations to a transfer function
ª º « x1 t » « » « x2 t » « » « x3 t » ¬ ¼
ª 0 1 0 º ª x1 t º ª1 º « 0 0 1 » « x t » «0 » u t « » « 2 » « » «¬ 1 2 3»¼ «¬ x3 t »¼ «¬0 »¼
with the system output
y t
y t ,
ª x1 t º >1 0 0@ «« x2 t »» « x2 t » ¬ ¼.
Solution: The general state and output equations are
X t
A X t B u t
y t C X t D u t and the transfer function is given by
H S
Y s X s
We first find
C
^ > sI A@ ` B D
sI A as
1
(6.14)
Chapter 6
486
ª1 0 0 º ª 0 1 0 º s ««0 1 0»» «« 0 0 1 »» ¬«0 0 1 ¼» ¬« 1 2 3¼»
> sI A@
The determinant of
ª s 1 0 º «0 s 1 » « » ¬«1 2 s 3¼»
sI A is given by
s 1 0 0 s 1 1 2 s3
sI A
ª s 0 0º ª 0 1 0 º « » « » «0 s 0 » « 0 0 1 » ¬«0 0 s ¼» ¬« 1 2 3¼»
s
3
3s 2 2s 1
.
The inverse matrix is the adjoint matrix divided by the determinant
adj sI A
1
> sI A@
det sI A
Substituting
sI A
B
ª1 º «0 » « » «¬0 »¼
C
>1
D
0
1
ª s 2 3s 2 s 3 1º 1 « » 2 s 3s s » 1 « 3 2 s 3s 2 s 1 « s 2 s 1 s 2 »¼ ¬
, B, C and D into Eq. (6.14), where
0 0@
we obtain the final result for the transfer function
H S
Y s X s
C
^ > sI A@ ` B D 1
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
s
H S
s
3
2
3s 2
3s 2 2 s 1
487
.
Example 6.14. Convert the state and output equations to a transfer function
ª º « x1 t » « » ¬« x2 t ¼»
ª 1 1 º ª x1 « 1 1» « x ¬ ¼¬ 2
with the system output
y t
º » ¼
ª1 º «0 » u t ¬ ¼
y t ,
ª x t º ». t 2 ¬ ¼
>1 0@ « x1
Solution: We have the general state and output equations are
X t
A X t B u t
y t C X t D u t and the transfer function is given by
H S
Y s X s
C
^ > sI A@ ` B D 1
To find the state transition matrix, we first calculate the matrix
(6.15)
sI A ,
Chapter 6
488
ª1 0 º ª 1 1 º s« »« » ¬0 1 ¼ ¬ 1 1¼
> sI A@
ª s 0 º ª 1 1 º «0 s » « 1 1» ¬ ¼ ¬ ¼
ª s 1 1 º « 1 s 1» ¬ ¼
. The determinant of
sI A is given by
s 1 1 1 s 1
sI A
s
2
2s 2
.
Next, we calculate the adjoint of this matrix
ªs 1 1 º Adj sI A « ». ¬ 1 s 1¼ The inverse matrix is the adjoint matrix divided by the determinant 1
> sI A@
adj sI A det sI A
Substituting
sI A
B
ª1 º «0 » ¬ ¼,
and
D 0
1
C
1 2 s 2s 2
ªs 1 1 º « 1 s 1» ¬ ¼
, B, C and D into Eq. (6.15), where
>1 0@
we obtain the final result for the transfer function
H S
Y s X s
C
.
^ > sI A@ ` B D 1
Application of the Laplace Transform to LTI Differential Systems: State-space Analysis
H S
489
s 1 . s 2s 2 2
Summary In chapters 3 and 4, we specified a continuous-time LTI system model using a transfer function or differential equation. For both cases, the system input-output characteristics have been given. In this chapter, another model — the state-variable model — has been developed. We have described and explained the concept of state-space representation of continuous-time LTI systems and also explained the state vector and state transition matrix using the Laplace transform method. x We started with a definition of state variables. State model representations of LTI systems were developed and analyzed by finding the system response. The solution of state model equations, for both homogeneous and non-homogeneous LTI systems, has been illustrated using several solved examples. x The state transition matrix (matrix exponential) has been defined using the Laplace transform method. The exponential matrix evaluation using the Laplace transform method has been illustrated using several solved examples.
CHAPTER 7 LAPLACE TRANSFORM METHODS FOR PARTIAL DIFFERENTIAL EQUATIONS (PDES)
7.1 Introduction In sections 4.2 and 4.3, we illustrated the effective use of Laplace transforms in solving ordinary differential equations. The transform replaces a differential equation in transform
y t with an algebraic equation in its
Y s . It is then a matter of finding the inverse transform of
Y s , either by the partial fraction method or by using Table 2.1. The Laplace transform can also be used to solve certain types of partial differential equations involving two or more independent variables. When the transform is applied to the variable t in a partial differential equation for a function
u x, t , the result is an ordinary differential equation for
the transform
U x, s . The ordinary differential equation is solved for
U x, s and the function is inverted to yield u x, t . LEARNING OBJECTIVES On reaching the end of this chapter, we expect that you will have understood and be able to apply the following:
Laplace Transform Methods for Partial Differential Equations (PDEs)
491
x Determine the Laplace transform of partial derivatives. x The use of the Laplace transform in solving partial differential equations with initial and boundary conditions. x The use of the Laplace transform in solving special partial differential equations like heat and wave equations.
7.2 The Laplace transforms of u x, t and its partial derivatives First, we obtain the Laplace transforms of the partial derivatives
wu wu w 2u w 2u , , , wt wx wt 2 wx 2
u x, t , t ! 0 Using the same procedure as that used to
of the function
obtain the Laplace transform of standard derivatives in property 1. 5.2 (Chapter 1), we have the following
L ^u x, t `
f
³e
st
u x, t d t U x, s
0
a
wu x, t ½ L® ¾ ¯ wt ¿
(7.1)
L ^ut x, t `
f
³e
st
ut x, t d t
0
Using integration by parts, we get f
f
L ^ut x, t ` e st u x, t s ³ e st u x, t d t 0
0
Chapter 7
492
so that
L ^ut x, t ` sU x, s u x, 0 (7.2)
b Writing v x, t
wu , and repeated application of (7.2), gives wt
wv ½ L ® ¾ sV x, s v x,0 ¯ wt ¿ wv ½ L ® ¾ s ¬ª sU x, s u x,0 ¼º v x,0 ¯ wt ¿ ° w 2u x, t ½° 2 L® ¾ L ^utt x, t ` s U x, s su x, 0 ut x, 0 . 2 ¯° wt ¿°
c we have wu x, t ½ L® ¾ ¯ wx ¿
L ^u x x, t `
wu x, t ½ L® ¾ ¯ wx ¿
f º d ª st « ³ e u x, t d t » dx ¬ 0 ¼
f
³e 0
st
w ^u x, t ` d t wx
d ^U x, s ` U x x, s dx
so that
wu x, t ½ L® ¾ ¯ wx ¿
d ^U x, s ` dx
(7.3)
Laplace Transform Methods for Partial Differential Equations (PDEs)
d Writing v x, t
wu , and repeated application of (7.3), gives wx
wv ½ d L® ¾ L ^v x, t ` ¯ wx ¿ dx
^
`
d §d · ¨ U x, s ¸ dx © dx ¹
so that 2 ° w u x, t °½ L® ¾ 2 °¯ wx °¿
d2 ^U x, s ` U xx x, s dx 2
Table 7.1 Laplace Transform Partial Derivative Pairs S.N
Transform Partial Derivative Pairs
1
wu ½ L ® ¾ sU x, s u x,0 ¯ wt ¿
2
w 2u ½ L ® 2 ¾ s 2U x, s su x, 0 ut x, 0 ¯ wt ¿
3
wu ½ d L® ¾ ^U x, s ` ¯ wx ¿ dx
4
w 2u ½ L® 2 ¾ ¯ wx ¿
d2 ^U x, s ` dx 2
5
w nu ½ L® n ¾ ¯ wx ¿
dn ^U x, s ` , n 1, 2,3,..... dx n
493
Chapter 7
494
7.2.1 Steps in the Solution of a PDE by Laplace Transform The procedure for solving partial differential equations by Laplace transform may be summarized as follows: Step 1: Apply the Laplace transform (using Table 7.1) to convert the given PDE into an ordinary differential equation (ODE) using the terms
U x, s . Step 2: Find the general solution of the ODE and use the boundary and/or initial conditions of the original problem to determine the precise form of the transform U x, s . Step 3: Invert the transform
U x, s to find the required solution
u x, t . Example 7.1 Solve the initial-boundary value problem
wu x, t wx
2
wu x, t wt
u x, t , u x,0 6e3 x
bounded for all x ! 0, t ! 0. Solution: Taking the Laplace transforms of the given partial differential equation with respect to
dU x, s dx
t , we obtain
2 ^sU x, s u x, 0 ` U x, s
in which we have also used the given initial condition.
Laplace Transform Methods for Partial Differential Equations (PDEs)
Because only derivatives with respect to
x
495
remain, we replace the partial
derivative with an ordinary derivative,
dU x, s dx
2 s 1 U x, s
12e 3 x .
This is a linear first-order ODE with constant coefficients. We solve it by finding the complementary function integral
PF , where
C.F U h x, s c1 e 2 s 1 x and the particular integral is
12
P.I1 6
and particular
C.F is the general solution of the homogeneous
differential equation
P.I1
CF
1 e 3 x replace D a D 2s 1 D 3
1 e 3 x 2 s
Now we have the general solution
U x, s U h x, s U Nh x, s U x, s c1 e 2 s 1 x 6
1 e 3 x . 2 s
Chapter 7
496
where
c1 is an arbitrary constant. Since u x, t must be bounded as
x o f, we must have U x, s also bounded as x o f. . As such, we must choose
c1
0.
Hence,
U x, s 6
1 e 3 x 2 s
.
Taking the inverse Laplace transform, we get
u x, t 6e2t e 3 x
6 e 2 t 3 x .
Example 7.2. Solve the initial-boundary value problem
wu x, t wu x, t x, x ! 0, t ! 0 wx wt with the boundary and initial conditions
u x,0 0, x ! 0, and u 0, t 0, t ! 0. Solution: Apply the Laplace transform with respect to time to the PDE equation to obtain
dU x, s dx dU x, s dx
sU x, s u x, 0 sU x, s
x s
x s
Laplace Transform Methods for Partial Differential Equations (PDEs)
This is an ordinary differential equation if
x
function of
alone,
s
U x, s is regarded as a
being a parameter. The transformed differential
equation is a linear first-order ODE. Comparing with the standard form
dU PU dx here P
Q
x s
s and Q
and we solve it by finding the integrating factor
I .F
e³
sdx
e sx .
The formula for the solution,
U I .F
³ Q I .F . dx C
U x, s e sx U x, s
³e e sx s
sx
x dx C s
³ e
sx
x dx C e sx
We can use integration by parts to evaluate the integral
U x, s
497
x 1 3 C e sx 2 s s
We can evaluate the constant C using the boundary conditions
Chapter 7
498
0 U 0, s
1 C C s3
1 s3
so we have
x 1 1 e sx 3 3 2 s s s
U x, s
Taking the inverse Laplace transform, we have
u x, t
t x xt
2
2
t2 u t x . 2
Example 7.3. Solve the initial-boundary value problem
w 2 u x, t
w 2 u x, t
wt 2
wx 2
, 0 x 1, t ! 0
with boundary and initial conditions here
u x,0 0,
wu x,0 wt
u 0, t 0 u 1, t
2sin S x 4sin 3S x , 0 x 1.
Solution: Taking the Laplace transform and applying the initial condition, the transformed equation becomes
d 2U x, s dx
2
s 2U x, s 2sin S x 4sin 3S x
Laplace Transform Methods for Partial Differential Equations (PDEs)
This is an ordinary differential equation, if
x
function of here
alone,
s
499
U x, s is regarded as a
being a parameter. The transformed
differential equation is a linear second-order ODE
D
2
s 2 U x, s 2sin S x 4sin 3S x
This is a differential form of linear second-order ODE with constant coefficients. We solve it by finding the complementary function and particular integral where
C .F
C.F is the general solution of the homogeneous problem U h x, s
c1 e s x c2 e s x
and the particular integral is
P.I1
2
P.I1
2
P.I 2
4
P.I1
4
1 replace D 2 a 2 sin x S D2 s2 D 2 S 2
1 sin S x S s2
2
1 replace D 2 a 2 sin 3 x S D2 s2 D 2 9S 2
1 sin 3S x 9S s 2 2
Now, we have the general solution
U x, s U h x, s U Nh x, s
Chapter 7
500
U x, s c1 e s x c2 e s x
2 4 sin S x sin 3S x 2 2 9S s 2 S s
2
We can evaluate the constants c1 and c2 using the boundary condition
u 0, t 0 U 0, s 0 and u 1, t 0 U 1, s 0 so we have
U 0, s 0 c1 c2 and U 1, s 0 c1 e s c2 e s Solving for c1 and c2, we get
c1 0
.
c2
therefore,
U x, s
2 4 sin S x sin 3S x 2 2 9S s 2 S s
2
.
Taking the inverse Laplace transform, we get
u x, t
2
S
sin S t sin S x
4 sin 3S t sin 3S x . 3S
Example 7.4. A very long taut string is supported from below so that it lies motionless on the positive x axis. At time t
0, the support is
removed and gravity is permitted to act on the string. If the end
x 0 is fixed at origin, the initial boundary-value problem describing displacements
w 2 u x, t wt 2
c2
w 2 u x, t wx 2
u x, t of points in the string is
g,
x ! 0, t ! 0
with boundary and initial conditions
Laplace Transform Methods for Partial Differential Equations (PDEs)
501
u 0, t 0, t ! 0, u x,0 0, x ! 0, ut x,0 0, x ! 0, 9.81, and c ! 0 is a constant depending on the material
where g
and tension of the string. Use Laplace transforms to solve this problem. Solution: Take the Laplace transform and apply the initial condition
c
2
d 2U x, s dx
2
s 2U x, s s u x, 0 ut x, 0
g s
d 2U x, s s 2 g , U x s dx 2 c2 c2 s This is an ordinary differential equation if
x
function of
alone,
s
U x, s is regarded as a
being a parameter. The transformed differential
equation is a linear second-order ODE
§ 2 s2 ¨D 2 c ©
· ¸ U x, s ¹
g c2 s
This is a differential form of a linear second-order ODE with constant coefficients.
Chapter 7
502
We solve it by finding the complementary function and particular integral, where
C.F is the general solution of the homogeneous problem
C.F U h x, s c1 e sx c c2 e sx c and the particular integral is
P.I
U Nh x, s
g 1 e0 x 2 2 c s § 2 §s· · ¨¨ D ¨ ¸ ¸¸ replace D 0 ©c¹ ¹ ©
P.I
U Nh x, s
g s3
Now we have the general solution
U x, s U h x, s U Nh x, s U x, s
c1 e sx c c2 e sx c
g s3 (7.4)
We can evaluate the constants c1 and c2 using the boundary condition
u 0, t 0 U 0, s 0 and u 1, t 0 U 1, s 0 For this function to remain bounded as x o f, we must set which case condition (7.4) implies that
c2 Thus,
g s3
c1
0, in
Laplace Transform Methods for Partial Differential Equations (PDEs)
U x, s
503
g sx c 1 e s3 .
Taking the inverse Laplace transform, we have
u x, t
2 g § 2 § x · § x ·· ¨ t ¨ t ¸ u ¨ t ¸ ¸¸ . 2 ¨© © c ¹ © c ¹¹
Example 7.5. Solve the initial-boundary value problem
wu x, t wu x, t 2u x, t 0, x ! 0, t ! 0 wx wt with boundary and initial conditions
u x,0 sin x,
x ! 0, and u 0, t 0, t ! 0.
Solution: Applying the Laplace transform with respect to time to the PDE equation, we obtain
dU x, s dx dU x, s dx
sU x, s u x, 0 2U x, s s 2 U x, s
sin x
This is an ordinary differential equation if function of
x
alone,
s
0
U x, s is regarded as a
being a parameter. The transformed differential
equation is a linear first-order ODE. Compared to the standard form
Chapter 7
504
dU PU dx here,
P
Q
s 2 and Q
sin x .
We solve it by finding the integrating factor
I .F
e³
s 2 dx
e s 2 x .
The formula for a solution is
U I .F
³ Q I .F . dx C
U x, s e s 2 x U x, s e s 2 x
³ e
³e s 2 x
s 2 x
sin x dx C
sin x dx C e s 2 x
by integration, we get
U x, s
1 ª s 2 sin x cos x º¼ C e s 2 x s 4s 5 ¬ 2
We can evaluate the constant C using the boundary condition
1 C C s 4s 5
0 U 0, s
2
1 s 4s 5 2
so we have
U x, s
1 1 ª¬ s 2 sin x cos x º¼ 2 e s 2 x s 4s 5 s 4s 5 2
Laplace Transform Methods for Partial Differential Equations (PDEs)
U x, s
505
½ ½ 1 1 ° s 2 ½ ° ° ° 2 x ° sx ° sin x ® ¾ cos x ® ¾ e ®e ¾ 2 2 2 s 2 1 °¿ ° ° ° ¯ s 2 1° ¿ ¯ s 2 1° ¿ ¯
Taking the inverse Laplace transform, we have
^
`
sin x e 2t cos t cos x e 2t sin t e 2 x e 2 t x sin t x u t x .
u x, t
u x, t e 2t ª¬sin x t sin t x u t x º¼ . Example 7.6. Solve the initial-boundary value problem
w 2 u x, t
w 2 u x, t
wt 2
wx 2
, 0 x 1, t ! 0
with boundary and initial conditions
u 0, t 0, u 1, t 0 u x,0 0,
wu x,0
sin 2S x , 0 x 1.
wt
Solution: Taking the Laplace transform and applying the initial condition, we have
d 2U x, s dx 2
s 2U x, s sin 2S x .
This is an ordinary differential equation if function of
x
alone,
s
U x, s is regarded as a
being a parameter. The transformed differential
equation is a linear second-order ODE
Chapter 7
506
D
2
s 2 U x, s sin 2S x
This is a differential form of a linear second-order ODE with constant coefficients. We solve it by finding the complementary function and particular integral where
C.F
C.F is the general solution of the homogeneous problem U h x, s
c1 e s x c2 e s x
and the particular integral is
P.I
U Nh x, s
P.I
U Nh x, s
2 2 1 sin 2S x replace D a 2 D s D 2 4S 2
2
1 sin 2S x 4S s 2
2
Now we have the general solution
U x, s U h x, s U Nh x, s U x, s c1 e s x c2 e s x
1 sin 2S x 4S s 2 2
We can evaluate the constants c1 and c2 using the boundary condition
u 0, t 0 U 0, s 0 and u 1, t 0 U 1, s 0 so we have
Laplace Transform Methods for Partial Differential Equations (PDEs)
U 0, s 0 c1 c2 and U 2, s 0 c1 e s c2 e s Solving for c1 and c2 we get
c1
507
.
0 c2
therefore,
U x, s
1 sin 2S x 4S s 2
2
.
Taking the inverse Laplace transform, we have 2
u x, t 3e 4S t sin 2S x . Example 7.7. A very long taut string is supported from below so that it lies motionless on a positive x axis. At time t
0, the support is
removed and gravity is permitted to act on the string. If the end
x 0 is fixed at origin, the initial boundary-value problem describing displacements
w 2 u x, t
w 2 u x, t
wt 2
wx 2
,
u x, t of points in the string is
x ! 0, t ! 0
with boundary and initial conditions
u 0, t 1, t ! 0, u x,0 0, x ! 0, lim u x, t 0, t ! 0. x of
Chapter 7
508
Here,
u x, t represents the temperature of a very long rod that,
initially, is at a temperature of 0 ºC and for which, at time t
0, the
one end that is nearest to us is raised to, and held thereafter at 1 ºC. Note that we require
u x, t o f as x o f.
Solution: Taking the Laplace transform and applying the initial condition
d 2U x, s
sU x, s u x,0
dx 2 d 2U x, s dx 2
sU x, s 0 .
This is an ordinary differential equation if function of
x
alone,
s
U x, s is regarded as a
being a parameter. The transformed differential
equation is a linear second-order homogeneous ODE
D
2
s U x, s
0
with the general solution
C.F U x, s c1 e
sx
c2 e
sx
.
Now we have the general solution
U x, s c1 e
sx
c2 e
sx
.
We can evaluate the constants c1 and c2 using the boundary condition
Laplace Transform Methods for Partial Differential Equations (PDEs)
509
1 s
u 0, t 1 U 0, s so we have
U 0, s
u0 s
c1 c2 .
For this function to remain bounded as ݔ՜
c1
0, c2
U x, s
1 e s
҄,
we must therefore set
1 s sx
Taking the inverse Laplace transform, we have
§ x · u x, t erfc ¨ ¸. © 2 kt ¹ where erfc (x) is the complementary error function
erfc x 1 erf x 1
2
S
x
³e
W 2
dW
0
2
S
f
³e
W 2
x
Example 7.8. Solve the initial-boundary value problem
w 2 u x, t wt 2
c
2
w 2 u x, t wx 2
, 0 x L, t ! 0
with boundary and initial conditions
u 0, t 0, u L, t 0
dW .
Chapter 7
510
u x, 0
§ S · wu x, 0 A sin ¨ x ¸ , wt ©L ¹
0, 0 x 1.
Solution: Taking the Laplace transform and applying the initial condition
d 2U x, s § s · s §S · ¨ ¸ U x, s 2 A sin ¨ x ¸ 2 dx c ©c¹ ©L ¹ 2
This is an ordinary differential equation if function of
x
alone,
s
U x, s is regarded as a
being a parameter. The transformed differential
equation is a linear second-order ODE
§ 2 § s ·2 · s §S · ¨¨ D ¨ ¸ ¸¸ U x, s 2 A sin ¨ x ¸ c ©c¹ ¹ ©L ¹ © This is a differential form linear second order ODE with constant coefficients. We solve it by finding the complementary function and particular integral where
C .F
C.F is the general solution of the homogeneous problem U h x, s
c1 e s
c x
and the particular integral is
c2 e s
c x
Laplace Transform Methods for Partial Differential Equations (PDEs)
P.I
U Nh x, s
replace D 2 D2
P.I
511
s 1 §S · A sin ¨ x¸ 2 c § 2 § s ·2 · © L ¹ ¨¨ D ¨ ¸ ¸¸ ©c¹ ¹ ©
a 2
S2 L2
U Nh x, s
As §S sin ¨ 2 2 2 s c S L ©L
2
· x¸ ¹
.
Now we have the general solution
U x, s U h x, s U Nh x, s U x, s
c1 e s
c x
c2 e s
c x
As §S · sin ¨ x ¸ 2 2 2 s c S L ©L ¹
2
.
We can evaluate the constants c1 and c2 using the boundary conditions
u 0, t 0 U 0, s 0 and u 1, t 0 U 1, s 0 give in turn c1 = 0 = c2. Therefore,
U x, s
As §S · sin ¨ x ¸ 2 2 2 s c S L ©L ¹ 2
.
Chapter 7
512
Taking the inverse Laplace transform, we have
§Sc · §S A cos ¨ t ¸ sin ¨ © L ¹ ©L
u x, t
· x ¸. ¹
Example 7.9. Solve the initial-boundary value problem
wu x, t
w 2 u x, t
wt
wx 2
, 0 x 2, t ! 0
with boundary and initial conditions
u 0, t 0, u 2, t 0 u x,0 3sin 2S x . Solution: Taking the Laplace transform and applying the initial condition
d 2U x, s
sU x, s u x,0
dx 2 d 2U x, s dx 2
sU x, s 3sin 2S x
This is an ordinary differential equation if function of
x
alone,
s
being a parameter. The transformed differential
equation is a linear second-order ODE
D
2
U x, s is regarded as a
s U x, s 3sin 2S x
Laplace Transform Methods for Partial Differential Equations (PDEs)
513
This is a differential form of a linear second-order ODE with constant coefficients. We solve it by finding the complementary function and particular integral where
C.F is the general solution of the homogeneous problem
C.F U h x, s c1 e
c2 e
sx
sx
and the particular integral is
P.I
U Nh x, s 3
P.I
U Nh x, s 3
1 replace D 2 a 2 sin 2 S x D2 s D 2 4S 2
1
4S
2
s
sin 2S x
Now we have the general solution
U x, s U h x, s U Nh x, s U x, s
c1 e
sx
c2 e
sx
3
1
4S
2
s
sin 2S x .
We can evaluate the constants c1 and c2 using the boundary condition
u 0, t 0 U 0, s 0 and u 2, t 0 U 2, s 0 so we have
U 0, s 0 c1 c2 and
U 2, s 0 c1 e 2
s
c2 e 2
s
.
Chapter 7
514
Solving for c1 and c2 we get c1 = 0 = c2. Therefore,
U x, s 3
1
4S
2
s
sin 2S x .
Taking the inverse Laplace transform, we have 2
u x, t 3e 4S t sin 2S x . Example 7.10. Solve the initial-boundary value problem
wu x, t
w 2 u x, t
wt
wx 2
sin S x , 0 x 1, t ! 0
with boundary and initial conditions
u 0, t 0, u 1, t 0 u x,0 0, ut x,0 0. Solution: Taking the Laplace transform and applying the initial condition
d 2U x, s dx
2
d 2U x, s dx
2
s 2U x, s s u x,0 ut x,0
s 2U x, s
sin S x s
.
sin S x s
Laplace Transform Methods for Partial Differential Equations (PDEs)
This is an ordinary differential equation if function of
x
alone,
s
515
U x, s is regarded as a
being a parameter. The transformed differential
equation is a linear second-order ODE
D
2
s 2 U x, s
sin S x s
.
This is a differential form of a linear second order ODE with constant coefficients. We solve it by finding the complementary function and particular integral, where
C.F is the general solution of the homogeneous problem
C.F U h x, s c1 e s x c2 e s x and the particular integral is
P.I
U Nh x, s
P.I
U Nh x, s
2 2 1 1 sin S x replace D a 2 2 s D s D 2 S 2
1 1 sin S x s S 2 s2
Now we have the general solution
U x, s U h x, s U Nh x, s U x, s
c1 e s x c2 e s x
1 1 sin S x 2 s S s2
.
Chapter 7
516
We can evaluate the constants c1 and c2 using the boundary condition
u 0, t 0 U 0, s 0 and u 1, t 0 U 1, s 0 so we have
U 0, s 0 c1 c2 and U 1, s 0 c1 e s c2 e s Solving for c1 and c2, we get c1 = 0 = c2. Therefore,
U x, s
1 1 sin S x 2 s S s2
Using partial fractions, we get
U x, s
1 ª1 1 « 2 2 S « s S s2 ¬
º » sin S x »¼
Taking the inverse Laplace transform, we have
u x, t
1 ª 1 º 1 sin S t » sin S x 2 « S ¬ S ¼
Thus,
u x, t
1
ªS sin S t º¼ sin S x . S3 ¬
.
Laplace Transform Methods for Partial Differential Equations (PDEs)
517
Example 7.11. A very long cylindrical rod is placed along a positive
x axis with one end at x
0. The rod is so long that any effect
from its right end may be neglected. Its sides are covered with perfect insulation so that no heat can enter or escape there through. At time
t
0 0, the temperature of the rod is 0 c throughout. Suddenly, the
left end of the rod has its temperature raised to u0 and maintained at this temperature thereafter. The initial boundary-value problem describing temperature
wu x, t wt
k
w 2 u x, t wx 2
u x, t at points in the rod is
, 0 x 2, t ! 0
with boundary and initial conditions
u 0, t u0 ; u x,0 0,
k is a constant described as the thermal diffusivity of the material in the rod. Use Laplace transforms on the variable t to find where
u x, t . Solution: Taking the Laplace transform and applying the initial condition
k
k
d 2U x, s dx 2 d 2U x, s dx 2
sU x, s u x,0 sU x, s 0 .
Chapter 7
518
This is an ordinary differential equation if function of
x
alone,
s
U x, s is regarded as a
being a parameter. The transformed differential
equation is a linear second-order homogeneous ODE
§ 2 s· ¨ D ¸ U x, s 0 k¹ © with the general solution
U x, s c1 e
s x k
c2 e
s x k
.
We can evaluate the constants c1 and c2 using the boundary condition
u 0, t u0 U 0, s
u0 s
so we have
U 0, s
u0 s
c1 c2
and require c1 = 0 to have
U x, s o f as ݔ՜ ҄.
Further,
c1
0, c2
U x, s
u0 s
u0 e s
s x k
Taking the inverse Laplace transform, we have
Laplace Transform Methods for Partial Differential Equations (PDEs)
519
§ x · u x, t u0 erfc ¨ ¸. © 2 kt ¹ where
erfc x is the complementary error function x
2
erfc x 1 erf x 1
2
2
W ³ e dW
S
S
0
f
³e
W 2
dW .
x
Example 7.12. Use the Laplace transform method to find the solution of the modified wave equation
w 2 u x, t wx
2
c2
w 2 u x, t wt
2
2cO
that remains finite for
u x, 0 0
and
wu x, t wt
O 2 u x, t
t ! 0 and satisfies the initial conditions
ut x, 0 0
and
the
boundary
u 0, t sin t for t ! 0. Solution: Taking the Laplace transform and applying the initial condition
d 2U x, s dx
2
d 2U x, s dx
2
c 2 s 2U x, s 2cO sU x, s O 2U x, s 2
cs O U x, s 0
condition
Chapter 7
520
This is an ordinary differential equation if function of
x
alone,
s
U x, s is regarded as a
being a parameter. The transformed differential
equation is a linear second-order homogeneous ODE
D cs O U x, s 2
2
0 .
This is a differential form of a linear second-order homogeneous ODE with constant coefficients. We solve it by finding the complementary function and particular integral where
C.F is the general solution of the homogeneous problem
C.F U h x, s c1 e cs O x c2 e cs O x U x, s c1 e cs O x c2 e cs O x (7.5) We can evaluate the constants c1 and c2 using the boundary condition
u 0, t
sin t U 0, s
and require c1 = 0 to have Further,
c2 Thus,
1 s 1 2
1 c1 c2 s 1 2
U x, s o f as ݔ՜ ҄.
Laplace Transform Methods for Partial Differential Equations (PDEs)
U x, s
521
1 e cs O x s 1 2
1 ½ U x, s e O x ® e cs x 2 ¾ s 1 ¿ . ¯ Taking the inverse Laplace transform, we have
u x, t e O x sin t cx u t cx . Summary In this chapter, we have described and explained the solutions of partial differential equations using the Laplace transform method: x
The Laplace transforms of partial derivatives have been established and tabulated. The solutions of partial differential equations with various initial and boundary conditions have been explained with several examples.
x The solutions of heat and wave equations have also been discussed with several examples.
f t
4cos 3t 5sin 3t e 2 t
f t 3cosh 6t 8sinh 3t
4
5
f t te 3t cos 2t
3
iwt
3t 2
1 e
f t e
f t
Ans
Ans
Ans
Ans
Ans
F s
F s
F s
F s
F s
2
6s 13
2
.
3s 24 . 2 s 36 s 9 4 s 23 . 2 s 4s 13 2
s
s 2 6s 5
1 2 1 . s s s6 s iw . s 2 w2
Find the Laplace transforms of the following functions.
2
季 1
EXERCISES AND ANSWERS
CHAPTER 8
12
11
10
9
8
7
6
W
sin W
sin at t
0
³
eW
dW
f t sin t
§ S· f t cos ¨ t ¸ © 6¹
f t sin 4t 5
f t t 2 cos at
f t
f t
t
f t t cos wt
Ans
Ans
Ans
Ans
Ans
Ans
Ans
2
w2
2
F s
F s
F s
F s
.
cot 1 s 1 . s
s
s 2 w2
2
a2
3
.
S 1 4 s e . 2 s3 2
1 3s 1 . 2 s2 1
4 cos 5 s sin 5 . s 2 16
s
2 s s 2 3a 2
§a· F s tan 1 ¨ ¸ . ©s¹
F s
F s
Exercises and Answers 523
17
16
f t
f t with
with
f t
f t
f t 4
f t .
f t 2 f t .
a t with f t b b
2 0 d t 1 ° ®1 1 t d 3 °0 3 t d 4 ¯
3 0 d t 1 f t ® ¯1 1 t d 2
f t sin t with f t 2S
14
0dt da ; f t 2a f t a d t d 2a
K f t ® ¯ K
15
Chapter 8
Ans
Ans
Ans
Ans
Ans
aª1 1 º « bs ». s « bs e 1 » ¬ ¼
1 2 e s e3s 4 s s 1 e
1 . s 1 2
K § as · tanh ¨ ¸ . s © 2¹
1 3 2e s e2 s 2 s s 1 e
F s
F s
F s
F s
F s
Find the Laplace transforms of the following periodic functions.
13
524
22
21
20
f t
with
Ans
Ans
Ans
F s
F s
F s
2 tan Ts . 4 Ts 2
1 . s 1 1 eS s 2
1 coth S s . 2 s 1 2
t ® ¯0
0 d t d1 t !1
Ans
Ans
F s
F s
1 e s s 1 e s . s2
2e S s . s2 4
Find the Laplace transforms of the following functions by expressing in to unit step function.
f t .
T 2
T dt dT 2
0dt
f t sin 2t u t S
f t T
f t
f t .
2t °° T ® °2 ¨§1 t ¸· °¯ © T ¹
f t 2S
I sin t 0 d t S f t ® m S d t d 2S ¯0
19
f t
with
f t sin t with f t S
18
Exercises and Answers 525
27
26
25
24
23
526
f t
f t
f t
f t
f t
2 t ° ®6 °t 4 ¯
0 d t d1 1 t d 2 t!2
1 0 d t 1 ° ®2 1 t d 4 °1 t!4 ¯
0 0 d t 1 ° ®1 1 t d 3 °2 t !3 ¯
°e t 0 d t d1 ® 2t t !1 °¯e sin t 0 t 2S ® t ! 2S ¯0
Chapter 8
Ans
Ans
Ans
Ans
F s
1 1 e s e 4 s s
1 s e e 3 s s
1 1 e 2S s . s 1 2
2 1 s § 3 1 · 3s § 1 1 · e ¨ ¸ e ¨ ¸. s s2 © s s2 ¹ © s s2 ¹
F s
F s
F s
s 1 e s 2 °½ °1 e F s ® ¾. s s 1 2 ° °¿ ¯ Ans
34
33
32
31
30
29
28
F s
2
3s . 12s 48s 36
3
1 s s 1
F s
s6 s 4 s 1 s 2
2
36 s s 1 s2 9
2
s 1 s 1
4s
1 2s . 2 s 4s 5
e 2S s . s s 2 1
F s
F s
F s
F s
F s
1 cos t u t 2S .
f t et e t 2tet .
f t e2t 5sin t 2cos t .
f t
Ans
f t
f t
Ans
Ans
f t
3 3t 1 t e e. 8 8
et 1 t
1 2 t . 2
1 4t 7 t 4 2t e e e . 15 5 3
9 1 f t 4 cos t cos3t. 2 2 Ans
Ans
Ans
Ans
Find the inverse Laplace transforms of the following functions.
Exercises and Answers 527
§ s 1 · F s log ¨ ¸ © s 1 ¹ 6 s 2 50 F s s 3 s 2 4
40
41
F s
1
Ans
Ans
f t
f t
1 t sin t. 2
2sinh t . t
5 1 sin t sin 2t. 3 3
cos t
f t 8e3t 2cos 2t 3sin 2t.
f t
Ans
2
s2 6 s2 1 s2 4
s
f t 2sin t 2sin 2t.
f t e 2 t 2e 3 t e 6 t .
f t e2t et tet . Ans
Ans
Ans
Ans
Chapter 8
2
s3
s 1 s 2 3s 2
39
1
F s
2
1 s 4
6
38
s
2
2s s 2 s 3 s 6
F s
F s
F s
37
36
35
528
45
45
44
43
42
F s
F s
F s
F s
F s
1 . s s 7
s2 . s 2s 1
2
Ans
Ans
f t 9t 2 2 2cosh 3t.
f t 2 cos t t 2 2.
Ans
Ans
Ans
Initial value
Initial value
Initial value
0, final value
1, final value
0, final value
Find the initial and final values of the functions using the initial and final value theorem.
s2 . s s 2
2
162 3 s s2 8
2 s3 s 2 1
Exercises and Answers
1 7
0
f
529
52
51
50
49
48
47
530
t
1
O
1
the
solution
2 2
2 2
s s2 a
s
2
1 s 2 s 2 1
of
s 1 s 3
1
1 s s 1
the
integro-differential
dy y W cos t W dW 0, y 0 1 dt ³0
Find
F s
F s
F s
F s
F s
equation Ans
Ans
Ans
Ans
Ans
Ans
1 2 t . 2
2a
4
.
. sin at
cos Ot
2 2cos at at y t 1
f t
2O
3
sin Ot Ot
t sin t .
f t
f t
1 3t t e e . 4
f t
f t et 1.
Evaluate the inverse Laplace transforms of the following functions using the convolution theorem.
Chapter 8
58
57
56
55
54
53
W
4 y t x t
0
³ t W e dW
t
2
7
dt
dc t
6c t dt
du t 11u t .
Find the DE corresponding to the transfer function
dt
2
d c t
2
Ans
Ans
C § 1 sin ¨ L © LC
Ans
Ans
Ans
Ans
dt
dt
dc t 6
d 2c t 2
C s U s
s
2
s 2 2
.
dt
du t
. u t .
7s 6
s 11
s
s 1
2c t 2
Y s X s
Y s X s
H s
H s
H s
· t¸ ¹
1 . s 4
t t y t 2 t 3e te .
i t V
Find the transfer function of the following differential equations
0, y 0 1
d y t dy t dx t 2 y t x t . 2 dt dt dt
dt
dy t
dy y dt
Find the solution of the integro-differential equation
t
di 1 L ³ i W dW V , i 0 1 dt C 0
Find the solution of the integro-differential equation
Exercises and Answers 531
62
61
60
59
532
s
.
with
y 0 1 & y ' 0 1
with
y 0 y ' 0 , y '' 0 1
y '' t 2 y ' t 5 y t 8et sin 2t
y 5 y '' 2 y ' 8 y sin t
with '''
y 0 1 & y' 0 4
y '' t 4 y t 8sin 2t
y 0 3 & y ' 0 6
with
y t 2t 1 et cos 2t.
1 cos t 13sin t 170 Ans
1 t 2 2t 8 4t e e e 3 15 85
y t 2t 1 cos 2t 3sin 2t
y t et 3t 2 e 2t .
y t
Ans
Ans
Solve the following differential equations using the Laplace transform.
6s 2
2s 1
2
y '' t y ' t 2 y t 9e2t
H s
Chapter 8
67
66
65
64
63
L
di Ri dt
AG t t0 , i 0
y 0 1 & y' 0 2 0
y '' t 2 y ' t y t 4et 2et
y 0 1 & y ' 0 1
y '' t 5 y ' t 6 y t G t 2
y 0 1 & y' 0 2
y '' t 3 y ' t 2 y t et u t 2
y 0 6 & y' 0 0
y '' t y ' t 2 y t 54te2t
with
with
with
with
Exercises and Answers
Ans
Ans
Ans
Ans
Ans
2 t 1
u t 2 .
i t
A e L
L
R t t0
u t t0 .
y t et t 2 5t 2 et .
2 t 2 3 t 2 3t 2t y t 3e 4e e e u t 2 .
y t e2t 1 t et e
y t 6et 9t 2 6t 1 e2t .
533
72
71
70
69
68
534
x 0
of the free oscillator, assuming
2
§ w · f0 sin w0t sin wt ¸ . 2 ¨ m w w0 © w0 ¹
y 0 0 & x 0 4
2x 2 y
dx dy 2 x y 2e t ; 4 z 2 y 4e 2 t dt dt and dz x z; y 0 3, x 0 9& z 0 1 dt
with
dx dy 2 x 2 y 16tet ; dt dt
y 0 0 & x 0 0
dx dy 2 x 5 y 5et cos t ; x 2 y 10et sin t dt dt with
t y t 5e 1 cos t x t 5et 1 cos t sin t
Ans
z t 3t 1 et e 2t 2e3t .
3t t 3t y t 6t 1 4e 8e 8e x t 3t 2 2et 3e2t 8e3t .
t 3t 2t y t 2e 4t 3 2e 8e x t et 12t 13 e3t 16e2t .
Ans
Ans
. Solve the following simultaneous differential equations using the method of Laplace transform.
x 0 0
'
w0
x t
displacement as a function of time. Notice that it is a linear combination of two simple harmonic motions, one with the frequency of the driving
force and one with the frequency
Ans
f 0 sin wt . Find the
An undamped oscillator is driven by force
Chapter 8
75
74
73
2 y 3z,
dz dt
t, x 1 with
d2y dy 4 10 y x 2 dt dt
Find the impulse response of the system described by the differential equation
y 0 8 & z 0 3.
dx dy 2 2y dt dt
with
dy y dt
2 y z
y 0 8 & z 0 3.
dy dt
Exercises and Answers
y t
1 2t e sin 6
6t
3 3 2t 1 e t 4 4 2 3 3 2t 1 e t 4 4 2
x t
z t 5et 2e4t .
t 4t y t 3e 5e
h t
Ans
Ans
535
78
77
76
536
2
wu x, t 0, x ! 0, t ! 0
0
g,
wx 2 ut x, 0 and
w 2 u x, t
wx 2
wt
0, x ! 0, t ! 0 x of
u x,0 0, ux 0, t 1 and lim ux x, t 0.
w 2 u x, t
wu x, t
x of
u 0, t 0 lim u x x, t 0.
wt 2 u x, 0
w 2 u x, t
wt wx u x, 0 3, u 0, t 5.
wu x, t
with
with
Solve following initial-boundary value problem
Chapter 8
Ans
u x, t
Ans
Ans
t
0
u x, t ³
2
x ct
x t ct
1 4xW e dW . SW
gt 2 °° 2 ® ° g x 2 2cxt °¯ 2k 2
§ x· u x, t 3 2 H ¨ t ¸ . © 2¹
BIBLIOGRAPHY
1. Shaila Dinkar Apte. (2016) Signals and systems: principles and applications, Cambridge University Press, ISBN 978-1-107-14624-2 2. Won Y. Yang et al. (2009) Signals and Systems with MATLAB. Springer-Verlag, Berlin, Heidelberg, ISBN 978-3-540-92953-6. 3. Edsberg, Lennart. (2016) Introduction to computation and modeling for differential equations. John Wiley & Sons, Inc., Hoboken, New Jersey, ISBN 978-1-119-01844-5 4. Shima H., Nakayama T. (2009) Laplace Transformation. In Higher Mathematics for Physics and Engineering. Springer, Berlin, Heidelberg. https://doi.org/10.1007/b138494_13 5. Michael Corinthios. (2009) Signals, systems, transforms, and digital signal processing with MATLAB CRC Press, 978-1-4200-9048-2. 6. Chi-Tsong Chen. (2004) Signals and systems. Oxford University Press, ISBN 0-19-515661-7 7. Paul Blanchard. (2012) Differential Equations, Cengage Learning, 1133109039. 8. Lennart Edsberg. (2015) Introduction to Computation and Modeling for Differential Equations, John Wiley & Sons, 1119018463.