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
dk y t
dt k
M
¦ bk
k 0
dkx 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
f t
is a real or complex-valued function of the (time)
variable t ! 0 and
s is a real or complex parameter. We define the
Suppose that
Laplace transform as
Chapter 1
4
F s
L^ f t `
f
³e
st
f t
dt
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
for t t 0.
iii The symbol 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
t
0 (if it exists). For instance, the function
1
diverges at t
t
0 , but the integral (1.1) exists.
v 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
constants L and M
and some
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
f
is
f
st
st
Lt st
³ f t e dt d³ f t e dt d³ M e e dt
0
M ª sL t ºf
e
¼0
Ls ¬
L^ f t `
0, f , f t e st
0
M
>0 1@
Ls
F s f
Definition: A function
f t
positive constants T and M
0
M
sL
for s ! L.
is said to be of exponential order L if
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
function f
t
a.
Solution:
Using the definition of the Laplace transform
F s
L^ f t `
f
³e
st
f t
dt
0
f
L ^a`
³e
st
a dt
0
L ^a`
a st f
e
0
s
L ^a`
a f 0
ªe e º¼
s ¬
a
>0 1@
s
provided, of course, that s ! 0 (if s is real). Thus we have
L ^a`
a
s
s!0 .
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
L^ f t `
F s
f
³e
st
f t
dt
0
^ `
L e at
f
f
at
st
³e e d t
³e
0
s a t
dt .
0
On integrating, we get
L e at
^ `
1 s a t f
e
0
sa
^ `
1
ª¬e f 1º¼
sa
L e at
Thus,
^ `
L e at
1
sa
for s ! a
Example 1. 3. Find the Laplace transform of f
Solution:
Let L
^ f t ` L >cosh at @
t
cosh at.
The Laplace Transform
and express the function
9
cosh at in its exponential form
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,
s
s a2
L > cosh at @
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 cosh at
sinh 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
for s ! a .
2
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 in its exponential form
1 iat
e e iat .
2
cosh at
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
dt
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
1
n
dx
s
f
s n 1 ³0
e x x n dx
§
Therefore, ¨ by gamma function * n
©
^ `
L tn
dx
s
f
·
x n 1
e
x
dx
¸
³0
¹
* n 1
s n 1
Thus,
^ `
L tn
* n 1
s
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
(impulse function) f
G t .
t
Solution:
An impulse is infinite at t
0 and zero elsewhere. The area under the
unit impulse is 1.
G t is defined by
The Dirac delta function
G t
­f
®
¯0
t 0
tz0
The Dirac delta function
G t a
is characterized by the following
two properties
i
­f
®
¯0
G t a
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
dt
0
L ^G t a `
f
³e
st
G ta dt
0
L ^G t a ` e st
t a.
Chapter 1
14
Thus,
L ^G t a ` e as
^ t ` 1.
and in particular L G
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
u t
­1
®
¯0
for t t 0
elsewhere
The Laplace Transform
15
u(t – a)
1
0
t
a
­1
®
¯0
u t a
for t t a
elsewhere
Figure 1.3
The step function is shown in Figure 1.2.
The Laplace transform of u t , by definition, can be written as
F s
f
L^ f t `
³e
st
st
u ta
st
dt
f t
dt
0
L^ u t a
f
` ³e
dt
0
L^ u t a
f
` ³e
a
1 st f
e
a
s
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 ` 1s
L^ u t a
Thus,
for a
0 .
Example 1.10. Find the Laplace transform of the ramp function
f t
r t .
Solution:
The ramp function r t
r t
­t
®
¯0
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
st
f t
dt
0
L^ r t
f
` ³e
st
r t
st
tdt
dt
0
f
L^ r t `
³e
0
Recall from calculus the following formula for integration by parts
b
³ u dv
a
b
b
uv a ³ v du.
a
f
L^ r t `
f
e st
1
t
³ e st dt .
s 0 s0
Integration by parts gives
L^ r t ` 0 1 st f
e
0
s2
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
1
s
L tn
^ `
n!
, n 1, 2,3.......
s n 1
^ ` s 1 a
L e at
^ `
1
sa
s
s a2
L ^1`
L e at
L ^sin at`
a
s a2
L ^cos at`
L ^sinh at`
a
s a2
L ^cosh at`
s
s a2
L ^u t a `
e as
s
L ^u t `
1
s
L ^G t ` 1
2
2
2
2
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 for every pair of constants C1 and C2.
T ^C1 f t C2 g t ` C1T ^ f t ` C2T ^ g t `.
and
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 C2 ³ g t dt .
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 `
7
s
4
1
2
2
.
s s 4 s
s4
b 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
* n 1
and
s n 1
c We have L ^t n `
* n 1
23
§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
^
t
Example 1.12. Determine L m 7 cosh 2t 4t
100
e 11t ` .
Solution:
Using Table 1.1,
^
L ^mt ` L e log m t
L ^t100 `
100!
s101
1
s
, L ^cosh 2t`
,
` s log
m
s 4
2
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 F
s
L^ f n 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
0 s n 3 f
11
0 ...... f
n 1
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
f
^
L f ' t
` ³e
st
f ' t
dt
0
Integration by parts gives
^
'
L f t
` e
st
f t
f
0
f
s ³ e st f t
dt
0
^
L f ' t
`
f
f 0 s ³ e st f t
dt
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
^
L f '' t
t must be continuous. Using Eq. (1.2), we have
f
` ³e
st
f '' t
dt
0
^
L f '' t
` e
st
f' t
f
0
f
s ³ e st f ' 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
` f 0 s ^sL ^ f t ` f 0 ` .
^
L f '' t
'
Thus,
^
` s L^ f t ` s f 0 f 0 .
L f '' t
2
'
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
0 s n 3 f
11
0 ...... f
t
sin 2t. Then f 0
0 and f ' t
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 f
n1
2 cos 2t ,
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 sa .
Proof:
The proof is immediate, since, by definition,
L ^e f t `
at
f
st at
³ e 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, e
transform of f
at
f t , if we already know the
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
s 1
F s s s 1
^
3 4t
Example 1.15. Determine L t e
2
s 1 49
`.
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 s 6
In general,
L ^e at cos bt`
L ^e at cos bt`
sa
2
s a b2
sa
sa
2
b2
and
6
s4
2
.
.
The Laplace Transform
b
L ^e at sin bt`
2
s a b2
b
L ^e at sin bt`
2
s a b2
sa
L ^e at cosh bt`
2
s a b2
2
s a b2
b
L ^e at sinh bt`
2
s a b2
b
L ^e at sinh bt`
2
s a b2
n!
sa
n 1
n!
L ^e at t n `
and
sa
L ^e at cosh bt`
L ^e at t n `
and
sa
n 1
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
L^ f t `
f
³e
st
f t
dt
0
f
½°
d n ­° st
e
f
t
d
t
®
¾
ds n ¯° ³0
°¿
dn
F s
ds n
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
dt
n
e ts
Using the nth-derivative i.e D
d
F s
ds
Thus,
f
³e
0
st
^ t f t ` d t
n
n
t e ts
`
The Laplace Transform
^
L tn 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 ¬
d § 3 ·
¨
¸
ds © s 2 9 ¹
6s
s2 9
2
.
^ e `.
Example 1.17. Determine L t
2
3t
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 F
1 §s·
F¨ ¸.
a ©a¹
L ^ f t ` , then L ^ f at `
s
Proof:
Using the definition of the Laplace transform,
F s
L^ f t `
f
st
f t
f at
dt
³e
dt
0
L ^ f at `
f
³e
st
0
let at
x Ÿ dt
dx
a
f
§s·
we get
L ^ f at `
1 ¨© a ¸¹ x
e
f x dx
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,
L ^sin at`
a
.
s a2
2
We have proved that differentiating the transform of a function
corresponds to multiplying the function by t . Similarly, integrating the
transform corresponds to dividing the function with t , as shown by the
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 dt.
f
³ F s ds.
s
Chapter 1
34
Integrating both sides of the equation,
f
ff
³ F s ds
³³ e
s
st
f t
d t ds
s 0
and reversing the order of integration, we get
f
³ F s ds
s
f
ª f st º
³0 «¬ ³s e ds »¼ f t d t
(1.3)
On integration, we get
f
f
f
ª e st º
³s F s ds ³0 «¬ t »¼ f t d t
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
³ s 1 ds
s
ª sin t º
1 f
L«
tan
s
s
¬ t »¼
1
2
s
S
tan 1 s.
2
Thus,
ª sin t º S
L«
tan 1 s.
»
¬ t ¼ 2
­ e3t e 2t ½
¾.
t
¯
¿
Example 1.20. Determine L ®
Solution:
In this case, f
^
L e3t e 2t
t
e3t e 2t , giving
` Fs
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 s2 º
­ e3t e 2t ½
L®
¾ log «
».
t
s
3
¯
¿
¬
¼
Property 1.5.7: Transforms of Integrals
If F
t
°­
°½
L ^ f t ` , then L ® ³ f W dW ¾
¯° 0
¿°
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 3t , giving
` Fs
2
1
s 4 s3
2
by the integral theorem,
­° t
½° 1
L ® ³ sin 2W e 3W dW ¾
F s
¯° 0
¿° s
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 F
s for
s or large s . In fact, it is often beneficial to determine this
small
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 F
L ^ f t ` , then f 0
s
Proof:
We have the derivative formula
^
L f' t
` f 0 sL ^ f t `
f
³e
st
f ' t dt
f 0 sF s .
0
Taking lim on both sides, we get
sof
0 f 0 lim ^s F s ` .
s of
Therefore,
lim ^s F s ` .
s of
The Laplace Transform
39
lim ^s F s ` .
f 0
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
the function f
t
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
f t
cos A cos B sin A sin B
AeD t ^cos T cos E t sin T sin E t ` u t ,
giving
L^ f t ` F s
F s
^
`
^
A cos T L eD t cos E t u t A sin T L eD t sin E t u t
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
­
½½
° °­ ª cos T s D E sin T º °°
lim ® s ® A «
»
¾¾
2
s of
s D E 2
»¼ ¿°°¿
°¯ °¯ «¬
­
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,
A cos T .
f 0
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 F
L ^ f t ` , then f f
s
lim ^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 sF s
0
Taking lim on both sides, we get
so 0
f
'
³ f t dt
0
f f f 0
f 0 lim ^s F s `
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, f
^
L 5 e 2t
t
5 e 2t , giving
` Fs
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
tion in
time
Translation
Multiplicat
ion by t
Time scale
L ^ f n t ` s n F s s n 1 f
L ^e at f t `
F sa
L ^t
dn
1
^F s `
ds n
n
f t`
L ^ f at `
n
1 §s·
F¨ ¸
a ©a¹
Integration
t
°­
°½ 1
L ® ³ f W dW ¾
L^ f t `
°¯ 0
°¿ s
Division
by t
­f t ½
L®
¾
¯ t ¿
Initial-
f 0
lim ^s F s `
f f
lim ^s F s `
value
f
³ F s ds.
s
s of
theorem
Final-value
theorem
n 1
s o0
0 s n 2 f ' 0 ......
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
s2 9
and by the first shifting theorem,
L ^e 2t t sin 3t`
F s2
F s s s 2
^
Example 1.26. Determine L t e
Solution:
Using Table 1.1,
L ^cos t` F1 s
s
s 1
2
3t
cos t` .
6 s2
s2
2
2
9
.
2
The Laplace Transform
45
using the multiplication theorem,
L ^t cos t`
F s
d
ª F1 s º¼
ds ¬
s2 1
d § s ·
¨
¸
ds © s 2 1 ¹
s2 1
2
and by the first shifting theorem,
2
L ^e t cos t` F s 3
F s s s 3
Example 1.27. Determine L
^ e t sin t` .
3t
8t
s 3 1
2
2
.
s 3 1
2
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
s2 1
and by the first shifting theorem,
2
L ^ e t sin t` F s 8
8t 2
F s s s 8
6 s 8 2
2
s 8 1
3
.
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:
t
2e 3t e 4t giving
F s
L 2e 3t e 4t
In this case, f
L^ f t `
^
`
^by direct f 0 lim 2e e
3t
t o0
F s
4 t
1 ·
§ 2
¨
¸
© 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,
f f
^by direct f f lim 2e e
3t
0.
t of
4 t
`.
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 2
22
.
125
6s 2 2
s2 1
3
Chapter 1
48
Thus,
f
³e
2 t
t 2 sin t dt
0
22
.
125
f
e t e 3t
³0 t dt .
Example 1.30. Evaluate
Solution:
We rewrite the integral as
f
et e3t
³0 t dt
^
§ 1 e 2t ·
³0 e ¨© t ¸¹ dt
t
t
§ 1 e 2t ·
¨
¸ , giving
© t ¹
` Fs
1
1
s s2
In this case f
L 1 e 2t
f
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
­1 e 2t ½
L®
¾
¯ t ¿
49
ª s2 º
log «
».
¬ s ¼
By definition of the Laplace transform,
f
2 t
·
t § 1 e
e
³0 ¨© t ¹¸ dt
­1 e 2t ½
L®
¾
¯ t ¿s 1
­° ª s 2 º ½°
®log «
»¾
¯° ¬ s ¼ ¿° s 1
log 3
thus,
f
2 t
·
t § 1 e
e
¨
³0 © t ¸¹ dt
log 3.
­ 2sinh 2t ½
¾.
t
¯
¿
Example 1.31. Determine L ®
Solution:
In this case f
^
L e 2t e 2t
t
2sinh 2t
e 2t e 2t , giving
1
1
s2 s2
` F s
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 s2 º
­ 2sinh 2t ½
L®
¾ log e «
».
s
2
¯ t
¿
¬
¼
­ 2t t
Example 1.32. Determine L ®e
¯
2W
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
s2
by the integration property,
2
9
½
³ e cos 3W dW ¾¿ .
The Laplace Transform
s2 ·
1§
¨
¸.
s ¨© s 2 2 9 ¸¹
­t
½
L ® ³ e 2W cos 3W dW ¾
¯0
¿
F s
51
Thus,
­ t
½
L ®e 2t ³ e 2W cos 3W dW ¾ F s 2
¯ 0
¿
f
Example 1.33. Evaluate
sin t
³ t
s
s 2 s2 9
dt .
0
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 1 ¸¹ ds
s
^tan s`
1
f
s
1
§
·
2
s
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 0
S
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
1
1 e sT
L^ f t `
T
st
³ e f t dt .
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
periodicity of f
L^ f t `
t .
f
³e
st
f t dt
0
We write the Laplace transform of f
T
L^ f t `
t as two integrals
f
st
st
³ e f t dt ³ e f t dt .
0
When we let t
T
u T , the last integral becomes
Chapter 1
54
f
f
st
³ e f t dt
³e
T
f
s u T
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
T
1 e sT L ^ f t `
³e
st
f t dt .
0
Thus,
L^ f t `
1
1 e sT
T
st
³ e f t dt.
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 sT , where T
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
t
2T
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
f t . The mathematical expression for f t
period is
f t
interval
K
t 0dt dT
T
with f t T
We have
L^ f t `
L^ f t `
1
1 e sT
K
T 1 e sT
T
st
³ e f t dt
0
T
st
³ e t dt .
0
f t
by
in one-
Chapter 1
56
Integrating by parts, we get
L^ f t `
K
­ 1 st T 1 st T ½
te
2 e
¾
sT ®
0
0
s
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
¿
Therefore,
L^ f t `
K
s T 1 e sT
2
^1 e
sT
sTe sT ` .
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
f t
­E 0 d t d a
®
¯ E a d t d 2a
57
with f t 2a
f t
.
We have
L^ f t `
1
1 e sT
T
st
³ e f t dt
0
L ª¬ f t º¼
2a
ª a st
º
1
e E dt ³ e st E dt »
s 2a « ³
1 e
a
¬0
¼
L^ f t `
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
s 1 e s 2a
^ e
L^ f t `
E
s 1 e s 2a
^ ª¬e
Therefore,
st a
0
sa
e st
2a
a
`
`
1º¼ ª¬e s 2 a e sa º¼ .
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
ue
sa
2
sa
2
E
s 1 e
sa
§ sa
·
E¨e 2 e 2 ¸
©
¹
sa
sa
§
·
s¨e 2 e 2 ¸
©
¹
sa
1 e
sa
1 e sa
§ 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
f t
­0 0 d t d a
®
¯1 a d t d 2a
t in a one-period is
with f t 2a
f t
2a.
2
The Laplace Transform
59
We have
L^ f t `
L^ f t `
1
1 e sT
T
st
³ e f t dt
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 f
T
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
L^ f t `
L^ f t `
1
1 e sT
T
st
³ e f t dt
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
at
³ e 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 f
T
t is called a full-wave rectifier wave and has the period
S w . The mathematical expression for f t
f t
sin wt 0 d t d
S
§ S·
with f ¨ t ¸
w
© w¹
in one period is
f t
We have
L^ f t `
L^ f t `
1
1 e sT
T
st
³ e f t dt
0
ª Sw
º
1
« st
»
e sin wt dt » .
³
S s
«
§
· 0
w
»¼
¨1 e ¸ «¬
©
¹
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
L^ f t `
L^ f t `
T
1
1 e sT
st
³ e f t dt
0
1
1 e 2S s
S
2S
°­ st
°½
sin
e
t
dt
e st sin t dt ¾ .
®³
³
°¯ 0
°¿
S
Recall from integral calculus that
eat
ª a sin bt b cos bt º¼
a 2 b2 ¬
at
³ e sin bt dt
L^ f t `
L^ f t `
1
1 e
2S s
^ªe s sin t cos t º¼ ª¬e s sin t cos t º¼ `
s 1 ¬
2S s
2S
st
S
0
1
1 e
S
st
2
s2 1
^ 1 e
sS
e s 2S e sS
`
Chapter 1
64
L^ f t `
1 e sS
1 e 2S s
2
s2 1
.
Therefore,
L^ f t `
L ª¬ f t º¼
1 e sS
1 e sS
s2 1
§Ss ·
sinh ¨ ¸
© 2 ¹
§Ss ·
s 2 1 cosh ¨ ¸
© 2 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 f
period T
f t
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
®
¯ 2a t a d t d 2a
with f t 2a
f t
We have
L ª¬ f t º¼
L ª¬ f t º¼
T
1
1 e sT
st
³ e f t dt
0
2a
ª a st
º
1
st
e
t
dt
e
a
t
dt
2
«
».
³a
1 e s 2 a ¬ ³0
¼
Integrating by parts, we get
L ª¬ f t º¼
1
ª1 e 2 as 2e as º¼ .
s 1 e s 2a ¬
2
Therefore,
L ª¬ f t º¼
L ª¬ f t º¼
1
2
s 1 e
sa
1 e sa
s 2 1 e sa
1 e
sa
1 e sa
2
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
f t
We have
­3t
®
¯6
0dt d2
2dt d4
with f t 4
f t .
The Laplace Transform
L^ f t `
L^ f t `
1
1 e sT
67
T
st
³ e f t dt
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 º
1
«®
2 ¾ ®
¾ ».
s ¿ 0 ¯ s ¿ 2 »
1 e s 4 « ¯ s
¬
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
­
°°sin T t
®
°0
°̄
function
0t of
the
S
T
S
2S
t T
T
.
Solution:
We have
L^ f t `
1
1 e sT
T
st
³ e f t dt
0
period
2S
T
defined
by
Chapter 1
68
L^ f t `
ª ST
º
1
« st
»
e sin Tt dt » .
³
2S s
«
§
· 0
T
»¼
¨1 e
¸ «¬
©
¹
On integrating, we get
L^ f t `
L^ f t `
­ st
1
®e s sin Tt T cos st
2S s
§
·
¯
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:
S
T
0
½
¾
¿
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
st
f t a u t a dt
st
f t a dt
³e
0
L^ f t a u t a `
f
³e
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 va
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
f t
­ f1 t
°
® f2 t
°
¯ f3 t
t defined by
0dt a
a dt b.
t tb
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
0;
at t
a , and, at the same time,
f3 t at t
b , and, at the same time,
f2 t
switch off the function f1 t ;
(c)
Switch on the function
switch off the function f 2 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 f
(1.6)
t is thus rewritten as follows.
f1 (t 1)
3t 3
f 2 (t 2)
replace t
t 1
replace t
6 3t
t2
f1 (t ) 3t
f 2 (t ) 3t
then its Laplace transform is
then its Laplace transform is
L ^ f1 (t )`
3
s2
Eq. (1.6) becomes
3
L ^ f 2 (t )` 2
s
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
t
3
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
­3
°
® 5 2t
°
¯1
0 d t d1
1 t d 3
t t3
In terms of the unit step function, f
t may be expressed as
Chapter 1
74
f (t )
f1 t ^ f 2 t f1 t ` u t a ^ f3 t f 2 t ` u t b
f (t ) 3 2 2t u t 1 2t 6 u t 3
Taking the Laplace transform, we have
L ^ f (t )`
L ^3` L ^ 2 2t u t 1 ` L ^ 2t 6 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 and for the second term a function of
t 3 .
We can apply the second shifting property to the result above to obtain
L ^ f (t )`
3 s
e L ^ f1 (t )` e 3 s L ^ f 2 (t )`
s
The function f
(1.7)
t is thus rewritten as follows.
f1 (t 1)
2 2t
f 2 (t 3)
replace t
t 1
replace t
2t 6
t 3
f1 (t ) 2t
f 2 (t ) 2t
then its Laplace transform is
then its Laplace transform is
L ^ f1 (t )`
2
s2
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 f (t )
­t 2 0 d t d 2
.
®
tt2
¯4t
Solution:
In terms of the unit step function, f
f (t )
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)
t2
replace t
f1 (t )
4t t 2
4 t2 t2
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
f (t )
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
then its Laplace transform is
L ^ f1 (t )`
1
s 1
2
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
f (t )
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 )`
1 eS s 2S s
e L ^ f1 (t )`
s s
The function 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 f (t )
­sin t 0 t d S 2
®
t !S 2
¯cos t
Solution:
In terms of the unit step function, f
f (t )
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 )`
S s
1
2
e
L ^ f1 (t )`
2
s 1
The function 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
1
s
L ^ f1 (t )` 2 2
s 1 s 1
(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 f (t )
­cos t 0 t d S
.
®
t !S
¯sin t
Solution:
In terms of unit step functions, f
f (t )
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 )`
s
eS s L > f1 (t )@
2
s 1
The function f
t is thus rewritten as follows.
(1.12)
The Laplace Transform
f1 (t S )
replace t
83
sin t cos 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 )`
Ss
2 s
e L ^ f1 (t )` e 2 L ^ f 2 (t )`
s
The function f
t is thus rewritten as follows.
(1.13)
The Laplace Transform
f1 (t 1)
t2
2
2
replace t
t 1
S
t2
f 2 (t ) cos t 2
2
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 ¹
©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
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
f (t )
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 f
87
t is thus rewritten as follows.
f1 (t 1)
1 t
replace 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 .
s2
s
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
f (t )
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 > cos t @ L ^ sin t cos t u t S ` L > f (t ) @
L ^ cos 3t cos 2t 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 )`
s
eS s L ^ f1 (t )` e2S s L ^ f 2 (t )`
s 1
2
The function f
f1 (t S )
replace t
(1.15)
t is thus rewritten as follows.
cos 2t cos t
t S
f 2 (t 2S )
cos 3t cos 2t
replace t
t 2S
f1 (t ) cos 2t 2S cos t S
f 2 (t )
cos 3t cos 2t
cos 2t cos t
f 2 (t )
cos 3t cos 2t
f1 (t )
then its Laplace transform is
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
®
2dt 3
° 3t
°0
t t3
¯
.
In terms of unit step functions, f
f (t )
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 f
(1.16)
t is thus rewritten as follows.
f1 (t 1)
t 1
f 2 (t 2)
replace t
t 1
replace t
4 2t
t2
f3 (t 3)
replace t
t 3
t 3
f1 (t ) t
f 2 (t ) 2t
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
2t 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 ^ f (t )`
L ^ t 1 u t 2 ` t 3 L ^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 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 )@
The function f
(1.17)
t is thus rewritten as follows.
f1 (t 2)
t 1
f 2 (t 3)
replace t
t2
replace t
t 3
t 3
f1 (t ) t 1
f 2 (t ) t 6
then its Laplace transform is
then its Laplace transform is
L ^ f1 (t )`
1 1
s2 s
Eq. (1.17) becomes
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:
Given f (t )
sin E t u t 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
replace t
sin E 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
·
§
·½
s
sin E T ¨ 2
e Ts ®cos E T ¨ 2
2 ¸
2 ¸¾
©s E ¹
© s E ¹¿
¯
Thus,
L ^ f (t )`
^E cos E T s sin E T ` e .
Example
1.57.
f (t )
Ts
2
s E
­
°°0
®
°sin 3t 2
°̄
2
Evaluate
the
Laplace
transform
2
3
.
2
tt
3
t
Solution:
In terms of the unit step function, f
f (t )
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
2t d4
tt4
In terms of unit step functions, f
f (t )
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 )`
5
5e4 s
2 s
5
e
L
f
(
t
)
>
@
1
s2
s
The function f
t is thus rewritten as follows.
f1 (t 2)
t 1
replace t
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:
t 1 t 1 , t t 0.
Given f (t )
Clearly
­2
f (t ) ®
¯2t
0 d t d1
t !1
In terms of the unit step function, f
f (t )
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 )`
2 s
e L ^ f1 (t )`
s
The function f
t is thus rewritten as follows.
(1.21)
Chapter 1
100
f1 (t 1)
2t 2
replace 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
2t d3
t !3
In terms of the unit step function, f
f (t )
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
2 s
t is thus rewritten as follows.
f1 (t 1) 10 t 1
t 1
replace t
f1 (t )
10t
then its Laplace transform is
L > f1 (t )@
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
2t dS .
t !S
Example 1.62. Determine the Laplace transform of the Bessel
functions
a f (t )
J0 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` L ^J 0 t `
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
Clearly
L ^J 0 t `
1 § 1 § 1 · 1˜ 3 § 1 · 1˜ 3 ˜ 5 § 1 ·
·
1 ¨ 2 ¸ ¨ 4 ¸
¨ 6 ¸ ........ ¸
¨
s © 2 © s ¹ 2˜4 © s ¹ 2˜4˜6 © s ¹
¹
L ^J 0 t `
1§
1·
¨1 2 ¸
s© s ¹
1 2
.
Thus,
L ^J 0 t `
1
1 s2
b We have J 0' t
.
J1 t .
Therefore, by the derivative property,
L ^ J1 t `
Thus,
L ^ J 0' t `
^
ª
º
1»
¬ 1 s
¼.
` «
sL ^ J 0 t ` 1
s
2
The Laplace Transform
105
§
s ·
L ^ J1 t ` ¨ 1 ¸.
1 s2 ¹
©
f
Example 1.63. Evaluate
³e
2 t
1 cos 3t dt .
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
³e
2 t
1 cos 3t dt
L ^1 cos 3t` s 2
1 cos 3t dt
9
.
26
0
1 ½
­1
® 2
¾
¯s s 9¿ s 2
9
26
Thus,
f
³e
2 t
0
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
transform of F
s and write f t
is the inverse Laplace
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 F
f t
109
possesses a Laplace
s so that
f
F s
³e
st
f t dt
f
(2.1)
where s can be complex and
transform F
f t
can be recovered from its Laplace
s by means of the complex line integral
V jf
f t
1
e st F s ds.
³
2S V jf
(2.2)
where 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 dt
f
Eq. (2.3) is the Fourier transform of f
(2.3)
t
Chapter 2
110
F s s j:
F^f t `.
If s is not purely imaginary, i.e. s
V j:, Eq. (2.3) can be written as
f
F V j:
³e
V j: t
f t dt
f
(2.4)
f
F V j:
V
³ e ^e f t ` dt
j:t
t
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
From Eq. (2.5), that the Laplace transform
f t
^
V t
.
F V j: of a signal
is given by
f
F e V t f t
` ³ e ^e
j:t
V t
f t
` dt.
f
Applying the inverse Fourier transform to the above relationship, we
obtain
e V t f t
^
f
`
F 1 ¬ª F eV t f t ¼º F 1 ¬ª F V j: ¼º
1
j:t
³ e F V j: d :
2S f
(2.6)
Multiplying both sides of Eq. (2.6) by e
V t
it follows that
The Inverse Laplace Transform
111
f
f t
As s
1
e V j: t F V j: d :
³
2S f
V j: and V is a constant, ds
(2.7)
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
ª
º
a
» using the Bromwich
L1 « 2
«¬ s a 2 »¼
Example 2.1. Determine
integral formula.
Solution:
Let F
a
s a2
s
a
2
sa sa
, then
ae st
,
sa sa
e st F s
The poles F
s are s
Re s ^ a `
lim e st F s s a
s oa
Re s ^ a `
a &s
^
^
a . These are both simple poles.
lim e st F s s a
s o a
­
st
½
at
` lim ® sae a ¾ e2
s oa
`
¯
¿
­ 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
¦ sum of residues
1 at at
e e
2
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¿
L tn
^ `
n!
s n 1
t n 1
­ 1 ½
L1 ® n ¾
n 1 !
¯ s ¿
^ `
1
sa
eat
L e at
­ 1 ½
L1 ®
¾
¯s a¿
L ^sinh at`
2
a
s a2
­ a ½
sinh at L1 ® 2 2 ¾
¯s a ¿
L ^cosh at`
2
s
s a2
cosh at
­ s ½
L1 ® 2 2 ¾
¯s a ¿
L ^sin at`
2
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
Chapter 2
114
L ^u t `
1
s
u t
e as
s
L ^u t a `
­1 ½
L1 ® ¾
¯s¿
­ e as ½
L1 ®
¾
¯ s ¿
u t a
L ^G t ` 1
G t
L ^G t a ` e as
G t a
^
L e at t n
n!
`
sa
L eat cos bt
^
`
^
`
L e at sin bt
L1 ^1`
^ `
L1 e as
­° 1 ½° at t n 1
L1 ®
e
n ¾
n 1 !
°¯ s a °¿
n 1
sa
2
s a b2
b
2
sa b
2
­°
½° at
sa
L1 ®
¾ e cos bt
2
2
°¯ s a b °¿
­°
½° at
b
L1 ®
¾ e sin bt
2
2
°¯ s a b °¿
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
L1 ^C1 F s C2G s ` C1 f t C2 g t
115
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.
1
Example 2.2. Determine (i ) L
ª 2s 9 º
ª Ds E º
« 2
» (ii ) L1 « 2
».
«¬ s 9 »¼
«¬ s O 2 »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 º
»
L1 « 2
«¬ s 9 »¼
ª 2s
9 º
»
2
L1 « 2
s 9 »¼
«¬ s 9
ª s º
ª 3 º
» 3L1 « 2
».
2 L1 « 2
«¬ s 9 »¼
«¬ s 9 »¼
Applying the inverse Laplace transform to the above equation, and using
its linearity, we get the following result
Chapter 2
116
ª 2s 9 º
» 2 cos 3t 3sin 3t.
L1 « 2
«¬ s 9 »¼
(ii )
ª Ds E º
ª Ds
ª s º E ª O º
E º
L1 « 2 2 » L1 « 2 2 2 2 » D L1 « 2 2 » L1 « 2 2 »
s O »¼
«¬ s O »¼
«¬ s O
«¬ s O »¼ O «¬ s O »¼
Applying the inverse Laplace transform to the above equation, and using
its linearity, we get the following result
ª Ds E º
E
»
D
O
L1 « 2
cos
t
sin Ot.
O
«¬ s O 2 »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 f t . Expressed in the
at
inverse form, the theorem becomes
L1 ^ F s a ` e at f t .
ª
º
3
«
».
Example 2.3. Determine L
« s2 2 9 »
¬«
¼»
1
The Inverse Laplace Transform
117
Solution:
3
2
s2 9
since
3
s2 9
ª 3 º
« 2
»
«¬ s 9 »¼
sos 2
L ^sin 3t` , the shift theorem gives
­
½
3
°
°
L ®
¾
2
° s3 9 °
¯
¿
e 2t sin 3t.
Example 2.4. Determine
ª
º
s2
».
L1 « 2
«¬ s 2 s 5 »¼
1
Solution:
Observe that
s2
s 2s 5
2
s 1
2
s 1 4
since
s
s 4
2
s 1 3
s 1
2
2
s 1 4
s 1 4
ª s º
« 2
»
and
«¬ s 4 »¼
s o s 1
L ^cos 2t` and
1
2
s 1 4
1
s 4
2
3
2
s 1 4
ª 1 º
« 2
»
«¬ s 4 »¼
s o s 1
1
L ^cos 2t` .
2
Chapter 2
118
Applying the inverse Laplace transform to the above equation, and using
its linearity, we get the following result
ª
º
s2
»
L1 « 2
«¬ s 2 s 5 »¼
­
½
­
½
s 1
3
°
° 1 °
°
L1 ®
L
¾
®
¾
2
2
° s 1 4 °
° s 1 4 °
¯
¿
¯
¿
ª
s 2 º t
3
» e cos 2t e t sin 2t.
L1 « 2
2
«¬ s 2s 5 »¼
Example 2.5. Determine
ª
º
5s 2
»
L1 « 2
«¬ s 6 s 13 »¼
Solution:
First, we complete the square to get
2
s 2 6 s 13
s3 4 .
Observe that
5s 2
s 6 s 13
5 s 3 13
2
2
s3
5
s3 4
2
s3 4
13
1
2
s3 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
119
ª 5s 2 º
13
» 5e 3t cos 2t e 3t sin 2t.
L1 « 2
2
«¬ s 6s 13 »¼
2.5 Inverse Transform by Differentiation and Integration
Given the function F
s , its inverse Laplace transform f t may be
formed by evaluating the inverse transform g t
that F
sG s and f t
s
dg t
dt
of G s
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
Example 2.6. Evaluate f
t given that F s
2a 2
.
s s 2 4a 2
Solution:
Let G s
sF s
2a 2
s 2 4a 2
a
2a
2
s 2a
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
2a
0
t
1
cosh 2at 1
2
t 0
Thus,
f t
1
cosh 2at 1 .
2
Example 2.7. Evaluate f
t given that F s
§ sa·
log e ¨
¸.
© s b ¹
Solution:
Let F
log e s a log e s b
s
F' s
1
1
sa
sa
F ' s
1
1
.
sa
sa
Applying the inverse Laplace transform to the above equation, and using
its linearity, gives us the following result.
^
L1 F ' s
­
½
°¯
°¿
­
½
°¯
°¿
` L °® s 1 a °¾ L °® s 1 a °¾
1
1
The Inverse Laplace Transform
^
L1 F ' s
121
` e e .
at
bt
^
t ` We have seen that L t f
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
` L ­®¯ s 1 1½¾¿
1
2
§1·
tan 1 ¨ ¸ .
©s¹
Chapter 2
122
^
` sin t .
L1 F ' s
^
We have seen that L t f
t ` dF s
ds
F ' s
then,
tf t
sin t
Thus,
sin t
.
t
f t
Example 2.9. Evaluate f
t given that F s
§ s2 a2 ·
log e ¨
¸.
2
© s
¹
Solution:
Let F
log e s 2 a 2 log e s 2
s
F' s
2s
2
2
s
s a
F ' s
2
2s
2
s
s a
2
2
Applying the inverse Laplace transform to the above equation, and using
its linearity, we get the following result
The Inverse Laplace Transform
­° s
½°
­1 ½
2 L1 ® 2 2 ¾ 2 L1 ® ¾
¯s¿
°¯ s a °¿
^
`
^
` 2 cos at 2 .
L1 F ' s
L1 F ' s
123
^
We have seen that L t f
t ` dF s
ds
F ' s
then,
2 1 cos at
t f t
.
Thus,
2 1 cos at
f t
t
.
Example 2.10. Evaluate f
t given that F s
Solution:
Let F
log e s 2 E 2 log e s 2
s
F' s
2s
2
2
s E s
F ' s
2s
2
2
.
2
s E
s
2
§ s2 E 2 ·
log e ¨
¸.
2
© s
¹
Chapter 2
124
Taking the inverse Laplace transform, we get
^
` L ­® s 2sE ½¾ L ­®¯ 2s ¾½¿
^
` 2 cos E t 2 .
L1 F ' s
L1 F ' s
1
1
¯
^
We have seen that L t f
2
2
¿
t ` dF s
F ' s
ds
then,
tf t
2 cos E t 2 .
Thus,
f t
2 1 cos E t
t
1
Example 2.11. Determine L
ª 1 º
«
».
¬ 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
1
Using this result with L
ª 1 º
L «
»
¬ s s 1 ¼
1
1
t
­ 1 ½
®
¾ gives
¯ s 1 ¿
e W
S ³0 SW
x 2 converts this to
The change of variable W
t
ª 1 º
L «
»
¬ s s 1 ¼
2
2
e x dx,
³
S 0
1
however, the error function erf
erf x
2
dW .
x
e
S ³
0
W 2
dW
x is given by
Chapter 2
126
1
so L
ª 1 º
«
»
¬ 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 F s ` is the Laplace transform of f t a u t a . Expressed
as
in the inverse form, the theorem becomes
^
L1 e as F s
`
f t a u t a .
ª
s
1 · º»
¸
2
¨ s2 4
¸»
s
1
©
¹¼
§
1
S s
Example 2.12. Determine a L « e ¨
«
¬
ª
º
1
».
b L1 «
«¬ s 1 e s »¼
Solution:
a This may be written as L1 ª¬e S 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
The Inverse Laplace Transform
127
cos t sin t
f t
Then, using the theorem, we have
ª § s
1 · º» 1 S s
¸ L ª¬e F s º¼
L1 «eS s ¨ 2
2
¸»
« ¨ s 4
s
1
¹¼
¬ ©
b Let F s
Here,
^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 F
129
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
determining the constants A1 , A2 , A3 ,...... Ai .
In the next example, we demonstrate three such methods.
1
Example 2.13. Determine L
where F s
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
s2
s3
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 1
Hence, the given function F
2 s 2 9 s 11
s 2 s 3 s 1
3 .
s can be expanded as
1
2
3
s2
s3
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
s2
s3
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, s , and 1 gives the following system of
linear equations
A B C
2 ,
4A B C
9 ,
The Inverse Laplace Transform
3 A 2 B 6C
133
11 ,
solving this system yields A
1 , B
Hence, the given function F
s can be expanded as
2 s 2 9 s 11
s 2 s 3 s 1
2, and C
3.
1
2
3
s2
s3
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 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
s2
s3
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 Ÿ A
Next, using s
10 B
1 .
3 gives
20 Ÿ B
2.
2, 3 or -1. Using s
2
The Inverse Laplace Transform
Finally, using s
6C
18 Ÿ C
135
1 gives
3 .
As such, the given function F s can be expanded as
2 s 2 9 s 11
s 2 s 3 s 1
1
2
3
.
s2
s3
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
1
Example 2.14. Determine L
where F s
ª¬ F s º¼ ,
s2 1
.
s s 1 s 1 s 2
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
s2
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 0
1
,
2
ª
º
s2 1
B «
1, C
»
¬« s s 1 s 2 ¼» s 1
D
ª
º
s2 1
«
»
¬« s 1 s s 1 ¼» s 2
5
.
6
ª
º
s2 1
1
,
«
»
3
¬« s 1 s s 2 ¼» s 1
The Inverse Laplace Transform
137
Hence, the given function F s can be expanded as
s2 1
s s 1 s 1 s 2
11
1
1 1
5 1
2s
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
s2
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
1
, B
2
1,
C
Hence, the given function F s 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
.
2s
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
s2
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
three values for s 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
2A 1 Ÿ A
1
2
Next, using s
1 gives
2 B
2 ŸB
Next, using s
6C
1
1 gives
2 ŸC
1
3
0,1, 1 , or 2.
The Inverse Laplace Transform
Finally, using s
6D
5 ŸD
141
2 gives
5
.
6
Hence, the given function F
s2 1
s s 1 s 1 s 2
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
s2
where A and B are real numbers to be determined.
ª 3s 5 º
A «
»
¬« s 2 ¼» s 1
2 ,
Hence, the given function F
s 2 6s 5
s 2 3s 2
1
B
ª 3s 5 º
1.
«
»
¬« s 1 ¼» s 2
s can be expanded as
2
1
s 1
s2
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
s2
where A and B are real numbers to be determined.
ª s 3 º
A «
»
«¬ s 2 »¼ s 1
ª s 3 º
B «
»
«¬ s 1 »¼ s 2
2 ,
1
.
Hence, the given function F s can be expanded as
F s
s 3 5s 2 9 s 7
s 2 3s 2
s2
2
1
s 1
s2
Note that the Laplace transform of the unit-impulse function G t is a
unity and the Laplace transform of
dG t
dt
­d f t ½
i.e L ­® d G t ½¾ s. L ®
¾ sF 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
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
2
s 2s 1
s 1
s 1
«¬ 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 .
1
Example 2.17. Determine L ª¬ F s º¼ , where
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
s2
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 2
19
,D
6
ª
º
3s 2 7
«
»
«¬ s s 1 s 2 »¼ s 1
Hence, the given function F s can be expanded as
F s
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
­° 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 °¿
f t
.
Finally, using the transform pairs established in Table 2.1, we have
7
19
5
5et e 2t e t .
2
6
3
f t
ª
º
s2 6s 9
Example 2.18. Determine (a) L «
»
«¬ s 1 s 2 s 4 »¼
1
1
(b) L
ª
º
s
«
».
¬« 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
s4
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 1
C
16
,
5
ª s 2 6s 9 º
«
»
«¬ s 1 s 2 »¼ s 4
Hence, the given function F s
s 2 6s 9
s 1 s 2 s 4
B
ª s 2 6s 9 º
«
»
¬« s 1 s 4 ¼» s 2
25
,
6
1
30
can be expanded as
16 ­ 1 ½ 25 ­° 1 ½° 1 ­° 1 ½°
®
¾ ®
¾ ®
¾
5 ¯ s 1¿ 6 ¯° s 2 ¿° 30 ¯° 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
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 can be expanded as
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 ¿°
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»
1
Example 2.19. Determine L
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
«
»
«¬ 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 , which is D s
F s . The
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
A
B
C
D
s
s 1
s2
s7
s s 1 s 2 s 7
where A, B, C, and D are real numbers to be determined.
C
ª
º
4s 1
«
»
¬« s s 1 s 7 ¼» s 2
3
,
10
D
ª
º
4s 1
«
»
¬« s s 1 s 2 ¼» s 7
29
280
.
Hence, the given function F 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 each term on the right and using
the linearity property of the Laplace transform, we find that
4s1
°­
°½ § 1 · 1 ­1½ §1· 1 °­ 1 °½ § 3 · 1 °­ 1 °½ § 29 · 1 °­ 1 °½
L1 ®
¾
¾ ¨ ¸L ® ¾¨ ¸L ®
¾¨ ¸L ®
¾¨ ¸L ®
¯°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
1
Example 2.20. Determine L ª¬ F s º¼ , where
F s
ª
º
s3 5s 2 6s 7
«
»
«¬ s s 1 s 3 6 s 2 11s 6 »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
s2
s3
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 3
hence, the given function F
s 3 5s 2 6 s 7
s s 1 s 1 s 2 s 3
.
83
,
24
s can be expanded as
§ 7 · 1 § 9 · 1 § 5 · 1 § 11 · 1
§ 83 · 1
¨ ¸
¨ ¸
¨ ¸
¨ ¸ ¨ ¸
6
s
24
s
1
4
s
1
2
s
2
© ¹ © ¹
© ¹
© ¹
© 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
s2 1
F s
s 1
2
s2
A
B
C
2
s 1 s 1
s2
where A, B, and C are real numbers to be determined.
We begin by multiplying both sides by
s2 1
A s 1 s 2 B s 2 C s 1
Setting
s
s
s
1
2
0
Ÿ B 2
Ÿ C 5
Ÿ A 4
hence, the given function F
2
s 2 s 1 to obtain
s can be expanded as
2
Chapter 2
154
ª
º
s2 1
«
»
2
«¬ s 1 s 2 »¼
4
2
5
2
s 1 s 1
s2
.
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 )
ª s 2 9s 2 º
L1 «
»
2
«¬ s 3 s 1 »¼
1
(b) L
ª s 1 º
.
«
2»
¬« s 1 s ¼»
Solution:
(a) Since s 1 is a repeated linear factor with multiplicity two and
s 3 is a non-repeated linear factor, the partial fraction expansion has
the form
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
s3
where A, B, and C are real numbers to be determined.
We begin by multiplying both sides by
s 2 9s 2
2
s 3 s 1 to obtain
A s 1 s 3 B s 3 C s 1
2
Setting
Ÿ B 3
s 1
s 3 Ÿ C 1
s 0 Ÿ 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
s3
.
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
2
s 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 Ÿ C 2
s 0 Ÿ B 1 .
Equating the coefficients of
s 2 Ÿ A C 0 Ÿ A 2 .
Hence, we have
s 1
2
s 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
9 s 2 38s 55
F s
s3 s2
3
Solution:
Alternative Method: Compute partial fraction constants.
Multiple poles pi of order m appear as a factor of the form
m
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
mi
Cij
¦ s p
i 1
i
The residues Cij are given by
m j
Cij
1
d i
mi j ! ds mi j
therefore, given F
9 s 2 38s 55
s3 s2
A
3
^ s p
mi
i
F s
`
s pi
, i 1, 2,3....mi
s , we write
C33
C32
C31
A
3
2
s 3 s2
s2
s2
ª 9 s 2 38s 55 º
«
»
3
«¬
»¼
s2
s 3
2
j
.
Chapter 2
158
C33
2
1
d0 ­
° 9 s 38s 55 ½
°
¾
0 ®
3 3 ! ds °
s3
°s 2
¯
¿
2
­
° 9 s 38s 55 ½
°
®
¾
s3
°
°s 2
¯
¿
C32
2
1
d ­
° 9 s 38s 55 ½
°
®
¾
3 2 ! ds °
s3
°
¯
¿s 2
1
C31
1
d2 ­
° 9 s 38s 55 ½
°
¾
2 ®
3 1 ! ds °
s3
°
¯
¿s 2
2
2
Hence, the given function F
9 s 2 38s 55
s3 s2
s can be expanded as
2
3
1
2
3
2
s3 s2
s2
s2
3
Applying the inverse Laplace transform to the above equation, and using
its linearity, we get the following result
ª 9s 2 38s 55 º
3
» 2e 3t t 2 e 2t te2t 2e2t .
L «
3
2
«¬ s 3 s 2 »¼
1
Example 2.24. Evaluate the inverse transform of
F s
3s 2 7 s 6
s 1
3
3
The Inverse Laplace Transform
159
Solution:
3s 2 7 s 6
s 1
C33
3
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 7s 6 `
2
^
2
` ^ 6s 7 `
s 1
2
s 1
1
3
s 1
.
3s 2 7 s 6
s 1
C31
s 1
s 1
^ 3s 7s 6 `
Hence, the given function F
F s
2
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
Thus,
t 2 et tet 3et
.
Chapter 2
160
t 2 t 3 et .
f t
Alternative method: To compute partial fraction constants, we write
3s 2 7 s 6
s 1
A
3
s 1
3
B
s 1
2
C
s 1
from which,
2
3s 2 7 s 6
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 Ÿ C
3
Hence, the given function F
F s
3s 2 7 s 6
s 1
3
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
L1 ¬ª F s ¼º
161
­
­ 1 °
½
° 1 ½
° 1 ­
° 1 ½
°
°
3L1 ®
2 L1 ®
L
®
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 .
The differentiation procedure works particularly well with repeating linear
factors.
s5
Example 2.25. Evaluate the inverse transform of F s
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
mi
Cij
¦ s p
i 1
i
The residues Cij are given by
m j
Cij
1
d i
mi j ! ds mi j
therefore, given F
^ s p
s , we write
i
mi
F s
`
s pi
, i 1, 2,3....mi
j
m
Chapter 2
162
s5
s2 s
3
C
C
C
A
33
32
31
3
2
s2 s
s
s
ª s5 º
«
»
3
¬ s ¼s 2
A
7
8
C33
1
d 0 °­ s 5 °½
®
¾
3 3 ! ds 0 ¯° s 2 ¿°s 0
°­ s 5 °½
®
¾
¯° s 2 ¿°s 0
C32
1
d °­ s 5 °½
®
¾
3 2 ! ds °¯ s 2 °¿s 0
7
4
C31
1
d 2 ­° s 5 ½°
®
¾
3 1 ! ds 2 ¯° s 2 ¿°s 0
11
16
Hence, the given function F
s5
s2 s
3
5
2
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
n
as 2 bs c of the denominator,
A3 s B3
An s Bn
A1s B1
A2 s B2
......... n
2
3
2
as bs c as 2 bs c
as 2 bs c
as 2 bs c
where 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 2 2 s 5 is irreducible. Since
s 1 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
2
s 1 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
.
Setting
s
s
s
1
Ÿ C 1
1 Ÿ B 4
0 Ÿ A 3
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
2
s 1 2
2
1
s 1
ª
º
ª
º
s 1
2
«
»
«
»ª 1 º
2
3
« s 1 2 22 »
« s 1 2 22 » «« s 1 »»
¼
«¬
»¼
«¬
»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
165
ª
º
ª
º
ª
º
2s 2 10s
s 1
2
» 3L1 «
» 2 L1 «
» L1 ª 1 º
L1 «
«
»
« s 1 2 22 s 1 »
« s 1 2 22 »
« s 1 2 22 »
¬« s 1 ¼»
«¬
¼»
¬«
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 .
ª 4s 2 s 1 º
»
Example 2.27. Determine L «
«¬ s s 2 1 »¼
1
Solution:
We can see that the quadratic factor
s 2 1 is irreducible. Since the
linear factor s appears to the first power in the denominator of F
s
and the quadratic factor appears to the first power, the partial fraction
expansion of F
s has the form
ª 4s 2 s 1 º
«
»
«¬ s s 2 1 »¼
A Bs C
2
s
s 1
where A, B, and C are real numbers to be determined.
4s 2 s 1
A s 2 1 s Bs C
4s 2 s 1
A B s2 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,
ª 4 s 2 s 1 º 1 3s 1
«
»
«¬ s s 2 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
ª 4s 2 s 1 º
»
L «
«¬ s s 2 1 ¼»
1
­ s °
½
­ 1 °
½
°
­1 ½
1 °
L1 ® ¾ 3L1 ® 2
¾ L ® 2
¾
¯s¿
°
°
¯ s 1 °
¿
¯ s 1 °
¿
Finally, using the transform pairs established in Table 2.1, we have
ª 4s 2 s 1 º
»
L «
«¬ s s 2 1 »¼
1
1 3cos t sin t
.
1
Example 2.28. Determine L
ª
º
s 1
« 2
».
«¬ s 1 s 2 9 »¼
The Inverse Laplace Transform
167
Solution:
s 2 1 and s 2 9
are
irreducible and appear to the first power in the denominator of F
s .
We can see that both the quadratic factors
The partial fraction expansion for F
s 1
2
2
s 1 s 9
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
Chapter 2
168
s 1
s2 1 s2 9
1 ª s 1
1
1 º
s 1 º 1 ª s
s
« 2
» « 2
»
2
2
2
2
8 « s 1 s 9 » 8 « s 1 s 1 s 9 s 9 »
¬
¼ ¬
¼
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 ª ­° s ½°
­ 1 ½°
­ s ½°
­ 1 ½°º
s 1
1
1 °
1 °
1 °
«
L1 ® 2
L
L
L
L
¾
® 2 ¾
® 2 ¾
® 2
¾
® 2
¾»
2
°¯ s 1 s 9 °¿ 8 «¬ °¯ s 1 °¿
°¯ s 1 °¿
°¯ s 9 °¿
°¯ s 9 °¿»¼
Finally, using the transform pairs established in Table 2.1, we have
­°
½°
s 1
L ® 2
¾
2
°¯ s 1 s 9 °¿
1
1
>3cos t 3sin t 3cos 3t sin 3t @.
24
Example 2.29. Determine the inverse Laplace transform of
F s
4s 2 6
s 1 s 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 2 6
F s
2
s 1 s 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
1
Ÿ A
2.
Equating the coefficients of s
B
2
and s , we get
2.
2 and C
Hence, we have
F s
F s
4s 2 6
2
s 1 s 2s 2
4s 2 6
s 1 s 2 2s 2
2
2s 2
2
s 1
s 2s 2
2 s 1 4
2
2
s 1
s 1 1
.
Taking the inverse Laplace transform of each term on the right, and using
the linearity property of the Laplace transform, we find that
­
½
­
½
­°
½°
4s 2 6
­ 1 ½°
s 1 °
1
°
1 °
1 °
1 °
2
2
4
L1 ®
L
L
L
¾
®
¾
®
¾
®
¾
2
2
2
1
s
1
2
2
s
s
s
° s 1 1 °
° s 1 1 °
¯°
¿°
¯°
¿°
¯
¿
¯
¿
Finally, using the transform pairs established in Table 2.1, we have
Chapter 2
170
­°
½°
4s 2 6
t
t
t
L ®
¾ 2e 2e cos t 4e sin t.
2
°¯ s 1 s 2s 2 °¿
1
1
Example 2.30. Evaluate L
ª¬ F s º¼ where
4s 16
F s
2
s 4s 15 s 2 6s 13
Solution:
Using partial fractions, we write
4s 16
F s
2
2
s 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
4 s 16
s ,
As B s 2 6 s 13 Cs D s 2 4 s 5
Comparing the coefficients of both sides gives
s3 :
A C 0,
s2 :
6 A B 4C D 0,
s:
13 A 6 B 5C 4 D
16 :
13B 5 D 16 .
4,
.
The Inverse Laplace Transform
171
By successive elimination of variables, we obtain
A
C
0 and B
2, D
Hence, the given function F
2
F s
2
s 2 1
2 .
s can be expanded as
2
2
s3 4
.
Taking the inverse Laplace transform of each term on the right, and using
the linearity property of the Laplace transform, we find that
L ^F s `
1
ª ­
½
­
½º
1
1
° 1 °
°»
1 °
«
2 L ®
¾ L ®
¾»
2
2
«
s
2
1
s
3
4
°
°
°
°»
«¬ ¯
¿
¯
¿¼ .
Finally, using the transform pairs established in Table 2.1, we have
f t
2e 2t sin t e 3t sin 2t .
ª
º
1
» .
«¬ s 1 s 2 4 »¼
1
Example 2.31. Determine L «
Solution:
We can see that the quadratic factor
s2 4
is irreducible. Since
s 1 and s 2 4 are non-repeated factors, the partial fraction
expansion has the form
Chapter 2
172
ª
º
1
«
»
«¬ s 1 s 2 4 »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 º
« 2
»
«¬ s 4 »¼
s 1
> Bs C @s 2i
2 Bi C
1
5
ª 1 º
«
»
«¬ s 1 »¼ s 2i
1
1 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
1
5.
Hence, we have
ª
º
1
«
»
«¬ s 1 s 2 4 ¼»
1 °­ 1
s
1 ½°
s 1 °½ 1 °­ 1
®
¾
®
¾
5 ° s 1 s2 4 ° 5 ° s 1 s2 4
s 2 4 °¿
¯
¿
¯
.
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
1
»
L1 «
«¬ s 1 s 2 4 ¼»
173
1 ­° 1 § 1 · 1 § s · 1 § 1 · ½°
¸ L ¨
¸¾
®L ¨
¸ L ¨¨ 2
¸
¨ s 2 4 ¸°
5 ° © s 1 ¹
s
4
©
¹
©
¹¿ .
¯
Finally, using the transform pairs established in Table 2.1, we have
ª
º
1
»
L1 «
«¬ s 1 s 2 4 »¼
1­ t
1
½
®e cos 2t sin 2t ¾ .
5¯
2
¿
1
Example 2.32. Determine L
ª s
º
1
«e
».
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
«¬
1 t 1
e e 2 t 1 u t 1 .
3
1
Example 2.33. Determine L
F s
ª¬ F s º¼ , where
ª 3s 1 º
e2 s « 3
2
¬ s 5s 6s »¼ .
Solution:
In step 1, the denominator is in the factored form
F s
ª
º
3s 1
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 s2 s3
A
B
C
s s2
s3
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 0
C
ª 3s 1 º
«
»
«¬ s s 2 »¼ s 3
B
ª 3s 1 º
«
»
«¬ s s 3 »¼ s 2
5
,
2
8
,
3
hence, the given function F
3s 1
s s2 s3
1
,
6
175
s can be expanded as
1 1 5 1
8 1
6 s 2 s2 3 s3
.
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
«
».
«¬ s 2 s 3 »¼
F s
Find the initial value of f
finding f
t
t , by (i) the initial value theorem; (ii)
L1 ª¬ F s º¼ .
Solution:
(i) by the initial value theorem,
· °½
3s
°­ §
lim ® s ¨
¸¸ ¾
¨
s of
°¯ © s 2 s 3 ¹ °¿
f 0
lim ^s F s `
f 0
§ §
··
3s
lim ¨ s ¨
¸¸
s of ¨ ¨ s 1 2 s s 1 3 s ¸ ¸
¹¹
© ©
s of
§
·
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
s2 s3
177
A
B
s2
s3
where A and B are real numbers to be determined.
ª 3s º
A «
»
«¬ s 3 »¼ s 2
6 , B
hence, the given function F
F s
3s
s2 s3
ª 3s º
«
»
«¬ s 2 »¼ s 3
6,
s , can be expanded as
6
6
s2
s3
.
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
F s
ª
º
s2
«
».
«¬ 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,
lim ^s F s `
f 0
f 0
s of
­° ª
º ½°
s2
lim ® s «
» .¾
s of
°¯ «¬ s 1 s 2 s 3 »¼ °¿
§ §
··
§
·
s2
1
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
F s
ª
º
s2
«
»
«¬ s 1 s 2 s 3 »¼
A
B
C
s 1
s2
s3
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 1
1
2
ª
º
s2
«
»
¬« s 1 s 2 ¼» s 3
9
,
2
C
hence, the given function F
F s
B
,
179
ª
º
s2
«
»
«¬ s 1 s 3 »¼ s 2
4,
s can be expanded as
ª
º
s2
«
»
¬« s 1 s 2 s 3 ¼»
1 1
1
9 1
4
2 s 1
s2 2 s3
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
and G s
F 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
³ f u g t u du.
g t
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
f t
g t f t
iii
f t
g t
f t
g t
h t
>0, f . Then,
h t ,
h t .
Proof. To prove the equation in (i), we begin with the definition
t
f t
³ f u g t u du.
g t
0
Using the change of variables, w
f t
g t
0
t
³ f w g t w dw
³ f w g t w dw
t
,
t u , we have
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
³ f u g t u du.
g t
0
t
f t
g t
t
2
cos t
2
³ u 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
sin A cos B
1
ªsin A B sin A B º¼ .
2¬
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
t u
³ u e du.
0
Recall from trigonometry that
t
t et
et ³ u e u du.
0
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
G s
and the convolution of their transform pairs f
t
s
and
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
i
L^ f t
L ^ g t ` . Then,
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 e
L^ f t
g t`
sv
F s . Hence,
f
sv
³ g v e F s dv
0
The Inverse Laplace Transform
L^ f t
g t`
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
º
s
».
L1 « 2
«¬ s a 2 s 2 b 2 »¼
Solution:
We express
let F
f t
g t
s
ª
s
s2 a2
s 2 b2
s
s a2
and G s
2
­°
½°
s
L1 ® 2 2 ¾ cos at
¯° s a ¿°
­
½
1
°
°
L1 ® 2
2 ¾
°
¯ s b °
¿
ºª
º
s
1
»
«
»;
«¬ s 2 a 2 »¼ «¬ s 2 b 2 »¼
as «
1
so that
s b2
2
and
1
sin bt
b
185
Chapter 2
186
In case
i , if we suppose that a 2 z b 2 , then, using the convolution
integral, we have
L1 ^ F s G s `
t
f t
g t
³ f u g t u du
0
ª
º
s
L « 2 2 2 2 »
«¬ s a s b »¼
t
1
cos au sin b t u du.
b ³0
1
(2.11)
Recall from trigonometry that
cos A sin B
1
ªsin A B sin A B º¼ .
2¬
If we set A
au and B b t u , we can carry out the integration
in Eq. (2.11)
ª
º
s
L1 « 2 2 2 2 »
«¬ s a s b »¼
1
sin a b u bt sin a b u bt du
2 ³0
ª
º
s
L « 2 2 2 2 »
«¬ s a s b »¼
1 ­° cos a b u bt cos a b u bt ½°
®
¾
2 ¯°
a b
ab
¿°
1
t
ª
º
s
«
»
L
«¬ s 2 a 2 s 2 b 2 »¼
1
This simplifies to
^
cos bt cos at
.
a 2 b2
`
t
0
The Inverse Laplace Transform
ª
º
s
»
L1 « 2
«¬ s 4 s 2 9 »¼
187
1­
1
½
®cos 2t cos 3t ^cos 2t cos 3t `¾
4¯
5
¿
Thus,
ª
º
s
»
L1 « 2
«¬ s 4 s 2 9 »¼
In case
ii If a
2
ª
º
s
»
L1 «
« s2 a2 2 »
¬
¼
1
^cos 2t cos 3t ` .
5
b 2 , then
t
1
cos au sin a t u du.
a ³0
(2.12)
Recall from trigonometry that
cos A sin B
If we set A
1
ªsin A B sin A B º¼ .
2¬
au and B b t u , we can carry out the integration in
Eq. (2.12)
ª
º
s
«
»
L
« s2 a2 2 »
¬
¼
1
^sin at sin 2au at `du
b ³0
ª
º
s
«
»
L
« s2 a2 2 »
¬
¼
cos 2au at ½
1­
®sin at u ¾
b¯
2a
¿
1
1
t
t
0
Chapter 2
188
ª
º
s
»
L1 «
« s2 a2 2 »
¬
¼
t sin at
.
2a
Example 2.38. Using the convolution theorem, determine
ª
º
1
«
».
L
« s2 k 2 2 »
¬
¼
1
Solution:
We express
ª
1
s2 k 2
F s
G s
2
ºª
º
1
1
»
«
»;
«¬ s 2 k 2 »¼ «¬ s 2 k 2 »¼
as «
1
s k2
2
so that
f t
g t
½°
1 1 ­° k
L ® 2 2 ¾
k
¯° s k ¿°
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
º
1
«
»
L
« s2 k 2 2 »
¬
¼
1
t
1
sin ku sin k t u du.
k 2 ³0
(2.13)
The Inverse Laplace Transform
189
Recall from trigonometry that
sin A sin B
1
ªcos A B cos A B º¼ .
2¬
If we set A
ku and B
k t u , we can carry out the integration in
Eq. (2.13)
ª
º
1
«
»
L
« s2 k 2 2 »
¬
¼
1
(cos 2ku kt cos kt )du
2k 2 ³0
ª
º
1
«
»
L
« s2 k 2 2 »
¬
¼
1 ª1
º
sin 2ku kt u cos kt »
2 «
2k ¬ 2k
¼0
1
1
t
t
.
Thus,
ª
º
1
«
»
L
« s2 k 2 2 »
¬
¼
1
1
^sin kt kt cos kt `.
2k 3
Example 2.39. Using the convolution theorem, determine
ª
º
1
L1 « 2
» .
«¬ s s a »¼
Solution:
We express
1
s sa
2
ª 1 ºª 1 º
»;
2 «
¬ s »¼ ¬« s a ¼»
as «
Chapter 2
190
let F
f t
s
1
and G s
s2
­1 ½
L1 ® 2 ¾
¯s ¿
1
so that
sa
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
º
s
«
».
L
« s2 4 2 »
¬
¼
1
The Inverse Laplace Transform
191
Solution:
s2 4
let F
ª
1 ºª s º
»«
»;
«¬ s 2 4 »¼ «¬ s 2 4 »¼
s
We express
as «
2
1
and G s
s 4
s
s
s 4
2
2
so that
f t
g t
1 1 ­° 2 ½°
L ®
¾
2 ° s 2 22 °
¯
¿
1
sin 2t
2
­
½
s
°
°
L1 ® 2
2 ¾
° s 2 ¿
°
¯
cos 2t
and
.
Using the convolution integral, we have
L ^F s G s `
1
t
f t
g t
³ f u g t u du
0
ª
º
s
«
»
L
« s2 4 2 »
¬
¼
1
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
º
s
»
L1 «
« s2 4 2 »
¬
¼
1
^sin t sin 4u 2t ` du
4 ³0
ª
º
s
«
»
L
« s2 4 2 »
¬
¼
cos 4u 2t
1 °­
t
®sin t u 0 4°
4
¯
1
t
t
½°
¾
0°
¿
which can be simplified to
ª
º
s
»
L1 «
« s2 4 2 »
¬
¼
1
t sin 2t.
4
Example 2.41. Using the convolution theorem, determine
ª
º
s
».
L1 « 2
«¬ s 4 s 2 9 »¼
Solution:
We express
let F
s
s
s2 4 s2 9
1
and G s
s 4
2
ª
1 ºª s º
»«
»;
«¬ s 2 4 »¼ «¬ s 2 9 »¼
as «
s
s 9
2
The Inverse Laplace Transform
so that f
g t
t
193
­° 1 ½° 1
sin 2t and
L1 ® 2
¾
°¯ s 4 °¿ 2
­
½
s
°
°
L1 ® 2
2 ¾
°
¯ s 3 °
¿
cos 3t
.
Using the convolution integral, we have
L1 ^ F s G s `
t
f t
g t
³ f u g t u du
0
ª
º
s
»
L1 « 2
«¬ s 4 s 2 9 »¼
t
1
sin 2u cos 3 t u du.
2 ³0
(2.16)
Recall from trigonometry that
sin A cos B
1
ªsin A B sin A B º¼ .
2¬
If we set A
2u and B 3 t u , we can carry out the integration in
Eq. (2.16)
ª
º
s
«
»
L
«¬ s 2 4 s 2 9 »¼
1
On integration, we get
t
1
^sin 3t u sin 5u 3t `du
4 ³0
.
Chapter 2
194
t
cos 5u 3t ½°
t
1 ­°
®cos 3t u 0 ¾
4°
5
°¿
0
¯
ª
º
s
»
L « 2
«¬ s 4 s 2 9 »¼
1
which can be simplified to
ª
º
s
»
L1 « 2
«¬ s 4 s 2 9 »¼
1­
1
½
®cos 2t cos 3t ^cos 2t cos 3t `¾
4¯
5
¿
.
Thus,
ª
º
s
»
L1 « 2
«¬ s 4 s 2 9 »¼
1
^cos 2t cos 3t ` .
5
Example
the
2.42.
Using
convolution
ª
º
1
L1 « 2
».
2
«¬ s s 2 »¼
Solution:
We express
1
s2 s 2
2
ª 1 ºª 1 º
»;
2
¬ s ¼ «¬ s 2 »¼
as « 2 » «
Let
F s
So that
1
and G s
s2
1
s2
2
.
theorem,
determine
The Inverse Laplace Transform
f t
te2t
t and g t
t
L ^F s G s `
1
f t
³ f t u g u du
g t
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
º
1
».
L1 «
«¬ s s 2 1 »¼
Solution:
We express
Let
1
s s2 1
1
and G s
s
F s
so that
f t
ª1 º ª 1 º
«
»;
¬ s »¼ «¬ s 2 1 »¼
as «
1 and g t
1
s 1
2
sin t
195
Chapter 2
196
t
L ^F s G s `
1
f t
g t
³ f t u g u du
0
ª 1 º
»
L « 2
«¬ s s 1 »¼
1
t
³ sin u du.
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
­°
½°
1
L1 ® 4
¾.
°¯ s 16 °¿
Solution:
ª
1 ºª 1 º
»«
»;
«¬ s 2 4 »¼ «¬ s 2 4 »¼
We express
1
4
s 16
let F
1
and G s
s 4
then,
s
2
as «
1
s 4
2
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
ª
1 ºª 1 º
»«
»;
«¬ 2 s 3 »¼ «¬ 2 s 3 »¼
as «
Chapter 2
198
let F
s
1
and G s
2s 3
1
2s 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
t
L1 ^ F s G s `
f t
³ f u g t u du
g t
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 f
t
integral equation f
t
2t 2 ³ f t u e u dt.
0
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
2e t 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
s2 s 1
.
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
F s
2 s2 1
2
s s 1
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 g t
integral equation
t
g t
3t 2 e t ³ e t u g u du
0
.
in the
The Inverse Laplace Transform
201
Solution:
We recognize the integral on the right as the convolution of g t
with
et .
Therefore,
form
g t
3t 2 et g t
the
integral
et
equation
has
the
.
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 . becomes
G s
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 t ,
Y s
4e2e s
s2 .
Applying the Laplace transform to both sides of the integro-differential
equation, we get
t
­° df 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
4e2 e s
s2
3F s
s2
4e2 e s
s 2 2s 3 s 2 2s 3 .
4 s2
Then,
4e 2 e s
s 1 s 3
s 1 s 3
4 s2
F s
Let
4 s2
F1 s
.
s 1 s 3
Using partial fractions, we get
3
1
.
s 1 s 3
F1 s
Since
Let F2
f1 t
L1 ª¬ F s ¼º
4
s
s 1 s 3
3et e 3t ,
.
Using partial fractions, we get
F2 s
since
1
1
s 1 s 3
f2 t
et e 3t ,
then finally we invert to obtain
203
Chapter 2
204
ª 4 s2
4e 2 e s º
L1 «
»
s 1 s 3 ¼»
¬« s 1 s 3
3et e 3t e 2 e t 1 e 3 t 1 u t 1
3et e 3t e 2 e t 1 e 3 t 1 u t 1 .
f t
Example 2.50. Compute L
^ f t * g t ` for the signals shown in
figures 2.2a and b.
f(t)
g(t)
1
1
0
1
(a)
t
0
1
(b)
t
Figure 2.2
Solution:
The mathematical expression for the functions is given below
f t
­1 0 d t d 1
u t u t 1
®
¯0 t ! 1
and
g t
­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
F s
t and g t , which is
1 e s
and F s
s
1 e s
s2
.
Eq. (2.19) becomes
L^ f t
§ 1 e s · § 1
s ·
¨
¸¨ 2 1 e ¸
¹
© s ¹© s
g t`
F s G s
g t`
2
1
1 e s .
3
s
Thus,
L^ f t
Example 2.51. Compute f t
F s
i
1
sa
sb
by
Partial fractions.
ii Convolution theorem.
iii Bromwich integral.
Solution:
i By partial fractions.
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
sa sb
A
B
sa
sb
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
sa sb
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 «
»
»
ba
«¬ 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
207
ii Convolution theorem.
We express F
let F1
s
s
ª
1 ºª 1 º
»«
»;
¬« s a ¼» ¬« s b ¼»
1
as «
sa sb
1
and F2 s
sa
1
.
sb
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
L ^ F1 s F2 s `
1
t
f1 t
f2 t
³ f u f t u du
1
2
0
ª
º
1
L «
»
¬« s a s b ¼»
1
t
au b t u
³e e
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
au b t u
³e e
0
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 F
1
s
sa sb
,
Then,
e st
,
sa sb
e st F s
a & s
b . Both are simple poles
the poles in F
s are s
Re s ^ a `
lim e st F s s a
s o a
Re s ^ b `
lim e st F s s b
s o b
st
­
½
at
^
` lim ® se b ¾ be a
^
` lim ® se a ¾ ae b
s o a
­
s o b
¯
¿
st
¯
½
bt
¿
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
F s
1
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
1
s 2 s 3
A
B
s2
s3
where A and B are real numbers to be determined
ª 1 º
A «
»
¬« s 3 ¼» s 2
Hence,
1 , B
ª 1 º
1
«
»
¬« s 2 ¼» s 3
Chapter 2
210
1
F s
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
e 3t e 2 t .
ii Convolution theorem.
F s
let F1
1
and F2 s
s2
s
ª
1
We express
s 2 s 3
1 ºª 1 º
»«
»;
«¬ s 2 »¼ «¬ s 3 »¼
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¿
.
and
The Inverse Laplace Transform
211
We have the convolution theorem
L1 ^ F1 s F2 s `
t
f1 t
³ f u f t u du
f2 t
1
2
0
ª
º
1
L1 «
»
«¬ s 2 s 3 »¼
t
2 u 3 t u
³e e
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 2 t .
iii Bromwich Integral.
Let F
s
1
s 2 s 3
,
then,
e F s
e st
,
s2 s3
the poles of F
s are s
st
2 &s
3 . Both are simple poles
Chapter 2
212
^
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 st ½
lim ®
¾
s o2 s 3
¯
¿
­
st
½
e 2 t
` lim ® se 2 ¾ e
s o3
¯
3t
¿
.
Using the Bromwich Integral
V jf
f t
1
e st F s ds
³
2S j V jf
Example 2.53. Compute f t
F s
¦ sum of residues
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
e 3t e 2 t .
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
B
1
and C
4
0.
Hence, we have
F s
1
s s2 4
1 ª1
s º
« 2
»
4 «s s 4 »
¬
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 º
»
L «
L
4 «¬ s »¼ 4 « s 2 4 »
¬
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
1
s s 4
F s
1
s s 2i s 2i
2
A
B
B
s
s 2i
s 2i
Where A and B are real numbers to be determined
A
B
ª 1
º
« 2
»
«¬ s 4 »¼
s 0
ª
º
1
«
»
«¬ s s 2i »¼ s 2i
1
, B
4
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
f t
­°
½° 1 ­ 1 ½ 1 ­° 1 ½° 1 ­° 1 ½°
1
1
L1 ® 2
L1 ® ¾ L1 ®
¾
¾ L ®
¾
°¯ s s 4 °¿ 4 ¯ s ¿ 8 ¯° s 2i ¿° 8 ¯° s 2i ¿°
Finally, using the transform pairs established in Table 2.1, we have
The Inverse Laplace Transform
f t
­° 1
½° 1 1
1
ei 2 t e i 2 t
L1 ® 2
¾
8
°¯ s s 4 °¿ 4 8
Thus,
f t
­°
½° 1 1 ei 2t e i 2t
1
L ® 2
¾
4
4
2
s
s
4
°¯
°¿
f t
­° 1
½° 1
L ® 2
1 cos 2t .
¾
°¯ s s 4 °¿ 4
1
1
ii Convolution theorem.
We express
º
1
1 ª
1
as ª º «
«¬ s »¼ s 2 4 » ;
s s2 4
«¬
»¼
Let F s
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
º 1t
1
»
L1 « 2
³ sin 2u du.
«¬ s s 4 »¼ 2 0
Carrying out integration gives
ª
º 1
t
1
»
cos 2u 0
L1 « 2
«¬ s s 4 »¼ 4
ª
º
1
«
»
L
2
s
s
4
«¬
»¼
1
1
1 cos 2t .
4
Example 2.54. Compute f t
F s
s
2
L1 ^ F s ` where
1
by
s4
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
s4 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
1
s 4 s2
C
C
A
22
21
2
s4 s
s
1
ª1º
A « 2»
¬ s ¼ s 4 16
C22
d 0 ­° 1 ½°
®
¾
ds 0 ¯° s 4 ¿°s 0
­° 1 ½°
®
¾
¯° s 4 ¿°s 0
C21
1
d °­ 1 °½
®
¾
3 2 ! ds °¯ s 4 °¿s 0
1
16
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
219
ii Convolution theorem.
We express
1
and G s
s4
Let F s
so that f
s2
ª 1 ºª 1 º
1
as «
» « 2 »;
s4
«¬ s 4 ¼» ¬ s ¼
t
e 4t and g t
1
s2
t
t
L ^F s G s `
1
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 H
is a rational function of
s
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):
dk y t
ak
¦
dt k
k 0
dkx t
bk
¦
dt k
k 0
N
M
(3.1)
Many systems can be approximately described by this equation, which
y t
relates the output
ak
parameters
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 y t
of the system to an arbitrary
input
x t
is
x t
i.e y t
h t * x t . In order to find the response y t
a given input
transform
given
by
the
convolution
h t
with
for
x t , it is often easier to first determine the Laplace
X s of x t
and subsequently to transform X
back to the time domain. This is because
Y s
of
s H s
X s , and hence
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
ªY s º
«
»
«¬ X s »¼
all initial conditions zero
all initial conditions zero
H s
.
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
dk y t
dt k
M
¦ bk
k 0
dkx t
dt k
(3.2)
where 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 s X s .
k
k
k 0
Y s Laplace transform of y t
transform of x t
y t .
Therefore,
N
T .F
H s
Y s
X s
¦a s
k
k
k 0
M
¦b s
k
k
k 0
.
and
X s Laplace
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
dk y t
dt k
f t
.
An nth-order ordinary differential equation is linear if it can be written in
the form
aN t
dN y t
d N 1 y t
d2y t
dy t
a
t
a
t
......
a1 t
a0 t y t
N 1
2
N
N 1
2
dt
dt
dt
dt
aN t , aN 1 t ,......a1 t and a0 t
f 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
(ii) The independent variable y t
and all of its derivatives
227
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
variable
and any of its derivatives or two or more of its
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
d3y t
2
dt 2
d2y t
dt 2
dy t
2y t
dt
d 2x t
dt 2
2
dx t
dt
x t .
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
s3 2s 2 s 2 Y s
s 2 2s 1 X s
so that the system transfer function is given by
Y s
s 2 2s 1
X s
s3 2s 2 s 2
.
Thus, the transfer function H
T .F H s
s is
Y s
s 2 2s 1
X s
s3 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
d3y t
dt 2
8
d2y t
5
dt 2
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
3
s 8s 2 5s 14
.
Chapter 3
230
Example 3.3. Consider the initial value problem
d2y t
dt
2
dy t
dt
y t
x t
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 H
(b) Determine
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
s2 s 1 Y s
X s
.
Thus, the transfer function H
H s
Y s
X s
The function
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
f 2 (t 3S ) sin t
t S
replace t
(3.3)
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
then its Laplace transform is
L ^ f1 (t )`
1
s2 1
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
233
X s
Y s
s2 s 1
1
e 2S s e S s .
2
2
s 1 s s 1
Y s
Example 3.4. Determine the system response if the transfer function of
the system is given by
ª
º
s2
« 3
»
«¬ s 6 s 2 11s 6 »¼
H s
and if the unit step function is applied as an input.
Solution:
s2
s 3 6s 2 11s 6
H s
.
Given the input
x t
u t ŸX s
1
s
we have
1
s2
s s 3 6s 2 11s 6
Y s
X s H s
Y s
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
Y s
s becomes
s
s 1 s 2 s 3
A
B
C
s 1
s2
s3
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
Y s
s can be expanded as
s
s 1 s 2 s 3
1 1
1
3 1
2
2 s 1
s2 2 s3
.
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
235
1 t
3
e 2e 2 t e 3 t .
2
2
y t
Example 3.6. Determine the system response if the transfer function of
the system given by
4s 1
H s
3
s 8s 2 5s 4
if the unit step function is applied as an input.
Solution:
4s 1
H s
s 3 8s 2 5s 4
Given the input
x t
u t ŸX s
1
s
we have
Y s
Y s
X s H s
4s 1
3
s s 8s 2 5s 14
4s 1
3
s s 8s 2 5s 14
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
Y s
s becomes
4s 1
A
B
C
D
s
s 1
s2
s7
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 0
ª
º
4s 1
«
»
«¬ s s 1 s 7 »¼ s 2
C
B
ª
º
4s 1
«
»
«¬ s s 2 s 7 »¼ s 1
1
,
8
D
ª
º
4s 1
«
»
¬« s s 1 s 2 ¼» s 7
29
280
1
,
14
3
,
10
.
Hence, the given function Y
Y s
4s 1
s s 1 s 2 s 7
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 Y
s and using Table 2.1 give
us
y t
§ 1 · § 1 · t § 3 · 2t § 29 · 7 t
¨ ¸ ¨ ¸e ¨ ¸e ¨
¸e .
© 14 ¹ © 8 ¹
© 10 ¹
© 280 ¹
Example 3.7. The impulse response h t of an LTI system is given by
h t
1 et .
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
H s
ª1 1 º
«¬ s s 1 »¼
1
.
s s 1
Given the input
e 2t Ÿ X s
1
s2
Y s
X s H s
1
s s 1 s 2
Y s
1
s s 1 s 2
x t
we have
.
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
Y s
s becomes
1
s s 1 s 2
A
B
C
s
s 1
s2
where A, B, and C are real numbers to be determined.
ª
º
1
A «
»
¬« s 1 s 2 ¼» s 0
C
ª 1 º
«
»
«¬ s s 1 »¼ s 2
ª 1 º
«
»
¬« s s 2 ¼» s 1
1
,
1
2
Hence, the given function Y
Y s
1
,B
2
1
s s 1 s 2
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
d2y t
dt 2
5
dy t
dt
d 2x t
6y 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
H s
Y s
s 2 8s 13
X s
s 2 5s 6
.
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
s2 s3
1
A
B
s2
s3
.
240
Chapter 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
H s
1
s
becomes
1
2
s2
s3
Taking the inverse Laplace transform of H
s
and using Table 2.1
gives the result
G t e 2t 2e 3t
h t
Example 3.9. Consider the causal LTI system described by the second
differential equation
d3y t
dt
3
6
d2y t
dt
2
11
dy t
dt
6y 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
d2y t
dy t
°­ d y t
L®
6
11
6y 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
H s
H s
X s
Y s
X s
Y s
X s
s
1
s 6s 11s 6
3
2
1
s 1 s 2 s 3
b The impulse response h t .
X s .
242
Chapter 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
H s
s becomes
1
s 1 s 2 s 3
A
B
C
s 1
s2
s3
where A, B, and C are real numbers to be determined.
ª
º
1
A «
»
«¬ s 2 s 3 »¼ s 1
C
ª
º
1
«
»
¬« s 1 s 2 ¼» s 3
Hence, the given function H
H s
1
,
2
ª
º
1
B «
»
«¬ s 1 s 3 »¼ s 2
1,
1
2
s can be expanded as
1
s 1 s 2 s 3
1 1
1
1 1
2 s 1
s2 2 s3
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 ŸX 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
s2
s3
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 0
ª
º
1
B «
»
¬« s s 2 s 3 ¼» s 1
1
2
C
ª
º
1
«
»
¬« s s 1 s 3 ¼» s 2
1
,
2
D
ª
º
1
«
»
¬« s s 1 s 2 ¼» s 3
1
6
Hence, the given function
Y s
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
xt
G t ŸX 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
s2
s3
where A, B, and C are real numbers to be determined.
246
Chapter 3
ª
º
1
A «
»
«¬ s 2 s 3 »¼ s 1
1
,
2
ª
º
1
«
»
¬« s 1 s 2 ¼» s 3
C
Hence, the given function H
1
2
1,
.
s can be expanded as
1
s 1 s 2 s 3
Y s
ª
º
1
B «
»
«¬ s 1 s 3 »¼ s 2
1 1
1
1 1
2 s 1
s2 2 s3
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 1 ­° 1 ½° 1 ­° 1 ½° 1 1 ­° 1 ½°
L ®
¾ L ®
¾ L ®
¾
2 ¯° s 1 ¿°
¯° s 2 ¿° 2 ¯° s 3 ¿°
y t
.
Finally, using the transform pairs established in Table 2.1, we have
1 t 2t 1 3t
e e e .
2
2
y t
Example 3.10. Consider the initial value problem.
d3y t
dt
3
7
d2y t
dt
2
e2t 4t
along with the initial conditions
Application of the Laplace Transform to LTI Differential Systems:
Transfer Functions
y 0
0 y' 0
247
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
d2y 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
s3 7s 2 Y s
X s
.
Thus, the transfer function H
H s
Y s
X s
1
s s7
2
s is
X s
Chapter 3
248
e 2t 4t
b Let x t
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
s
2
1
s7
X s
Y s
s2 s 7
s 2 4s 8
.
s4 s 2 s 7
Y s
Example 3.11. Consider the causal LTI system described by the
second differential equation
d2y t
dt
2
5
dy t
dt
6y 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
6y 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
X s
1
s 2 5s 6
The transfer function H
H s
H s
Y s
.
s is given by
X s
1
s 2 5s 6
Y s
1
X s
s2 s3
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
H s
s becomes
1
s2 s3
A
B
s2
s3
where A and B are real numbers to be determined.
ª 1 º
1, B
A «
»
«¬ s 3 »¼ s 2
Hence, the given function H
H s
ª 1 º
«
»
«¬ s 2 »¼ s 3
1
.
s can be expanded as
1
1
s2
s3
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
h t
­ 1 ½ 1 °­ 1 °½
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
h t
251
e2t e3t .
c Output response for the input x t
u t .
We have
Y s
1
X s
s2 s3
1
Y s
X s
s2 s3
Given the input
x t
1
s
u t ŸX s
therefore
Y s
1
s s2 s3
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
Y s
s becomes
1
s s2 s3
A
B
C
s
s2
s3
where A, B, and C are real numbers to be determined.
Chapter 3
252
ª
º
1
A «
»
«¬ s 2 s 3 »¼ s 0
C
ª 1 º
«
»
«¬ s s 2 »¼ s 3
1
3
1
,
6
1
2
,
.
Hence, the given function Y
Y s
ª 1 º
B «
»
«¬ s s 3 »¼ s 2
s can be expanded as
11 1 1
1 1
6 s 2 s2 3 s3
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
s2 s3
G t .
Application of the Laplace Transform to LTI Differential Systems:
Transfer Functions
1
Y s
253
X s
s2 s3
.
Given the input
x t
G t ŸX s
1
therefore
Y s
1
s2 s3
.
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
Y s
s becomes
1
s2 s3
A
B
s2
s3
where A and B are real numbers to be determined.
ª 1 º
ª 1 º
1, B «
A «
»
»
«¬ s 2 ¼» s 3
¬« s 3 ¼» s 2
Hence, the function Y
Y s
s can be expanded as
1
1
s2
s3
1
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°
L1 ®
¾ L ®
¾
¯s 2¿
°¯ s 3 °¿
H t
.
Finally, using the transform pairs established in Table 2.1, we have
e2t e3t .
y t
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
d3y t
dt
2
6
d2y t
dt
2
11
dy t
dt
6y 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
d2y t
dy t
°­ d y t
L®
6
11
6y 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 6s 11s 6 `Y s
X s
Thus, the transfer function H
s is
3
H s
2
Y s
X s
.
1
s 6 s 11s 6
3
255
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
H s
s becomes
1
s 1 s 2 s 3
A
B
C
s 1 s 2
s3
where A, B, and C are real numbers to be determined.
ª
º
ª
º
ª
º
1
1
1
6
1
, B «
.
A «
1, C «
»
»
»
«¬ s 1 s 3 ¼» s 2
¬« s 3 s 2 ¼» s 1 2
¬« s 1 s 2 ¼» s 3 2
H s
1 1
1
1 1
2 s 1 s 1 2 s 2
.
Taking the inverse Laplace transform, we get
Chapter 3
256
1 t 2t 1 3t
e e e .
2
2
h t
Example 3.13. Transform the transfer function
s 2 8s 13
H s
3s 3 7 s 2 5s 6
into a differential equation.
Solution:
Let
H s
T .F
s 2 8s 13
3s 3 7 s 2 5s 6
H s
Y s
s 2 8s 13
X s
3s 3 7 s 2 5 s 6
3s 3 7 s 2 5s 6 Y s
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
d3y t
d2y t
dy t
5
6y t
7
3
2
dt
dt
dt
d 2x t
dx t
8
13x t .
2
dt
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
Y s
s2 s 1
s 2 2s 1
T .F
H s
Y s
s2 s 1
X s
s 2 2s 1
.
X s
If the input is an impulse G
Thus we have Y
s
t , then its Laplace transform is X s
1.
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
2
s 2s 1
s 1
s 1
«¬ 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
H s
1
.
s
Its inverse Laplace transform yields the step response in the time domain.
Let us carry out partial fraction expansion of Y
Y s
s2 s 1
s s 2 2s 1
s2 s 1
s s 1
2
.
s as
Application of the Laplace Transform to LTI Differential Systems:
Transfer Functions
It is necessary to invert Y
259
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
Y s
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) w in
joules (J) required to move charge q through the device v
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
Zs
R
i t
v t
Ri t
1
v t
R
v t
R
dq t
dt
R
d 2q t
Ls
Resistor
L
di t
v t
L
v t
1
i W dW
C ³0
dt
t
i t
1
v W dW
L ³0
i t
C
v t
L
dt 2
Inductor
C
t
Capacitor
dv t
dt
v t
1
q t
C
1
Cs
Vs
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
Ri 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 1
I s
sC
1 ·
§
¨R L s ¸I s
sC ¹
©
1
E s
1 ·
§
¨R L s ¸
sC ¹ .
©
Thus, the transfer function H
H s
E s
E s
I s
T .F
263
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 s is Z1
the impedance of the parallel connection of 2 and
1
s
1
2
s
2u
Z2 s
2
2s 1
which is a voltage divider. Thus we have
Y s
Z2 s
^Z s Z s `
1
Y s
Y s
U s
2
2
2s 1
2
U S
6s 9s 5
2
2
2 3s
2s 1
2
U S
6s 9s 5
2
Thus, the transfer function H
T .F
H s
Y s
U s
s is
2
.
6s 9s 5
2
1
is
s
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 Ÿ
^
^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 Ÿ 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 Ÿ I1 s
s 1 2 s 1 I2 s
I2 s
1
s 2 3s 1
V s
but
V s
VL s
sI 2 s
VL s
s
s 3s 1
2
V s
Thus, the transfer function is
T .F
G s
VL s
V s
s
s 2 3s 1
2 s I2 s
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
x t
267
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, 2s and
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
1
s
Chapter 3
268
s
Y s
2
s 1
2
X s
2
.
If we apply a step input u t
Y s
1
.
s
1 and U s
1
2
s2 1
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
dt 2
dt
x t ,y 0
0 y' 0
where
x t
­ 1 0 d t d1
x t2
®
¯1 1 t 2
x t
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
s2 s 1 Y s
X s
.
Thus,
H s
Y s
1
s s 1
2
X s
b But x t is a periodic function with period T
We have
X s
1
1 e sT
L ª¬ x t º¼
T
st
³ e x t dt
0
X s
2
­ 1 st
½
1
st
e
dt
e
dt
®
¾
³1
1 e 2 s ¯ ³0
¿
X s
1
1 e 2 s
X s
therefore
1
2
­
e st
° e st
®
s 1
s
°
0
¯
1 e s
2
s 1 e s 1 e s
½
°
¾
°
¿
1 e s
s 1 e s
2.
Chapter 3
270
Y s
1
ŸY s
2
s s 1
X s
X s
s2 s 1
.
Thus,
ŸY s
1 e s
s 1 e s
s2 s 1
.
Example 3.20. The output y t of a system is
2 3e t e 3t
y t
for an input, it is
2 4e 3t
x t
Determine the corresponding input for an output
2 e t te t .
y1 t
Solution:
For the input x t
X s
2 4
s s3
2 4e 3t , the Laplace transform is
6 s 1
s s3
.
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
x1 t
u t .
Example 3.21. Compute the output
impulse response
and x t
s gives
y t for an LTI system whose
h t and input x t are given by h t
t 2u t .
u t
Chapter 3
272
Solution:
t 2u t , the Laplace transform is
For the input x t
2
s3 .
X s
The corresponding impulse response h t
H s
has the Laplace transform
1
s
Hence,
Y s
H s X s
The output y t
2
s4
is obtained by taking the inverse Laplace transform of
Y s .
Thus,
y t
t3
u t .
3
Example 3.22. The impulse response h t
by h t
of an LTI system is given
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:
Given the impulse response h t
and the transfer function H
H s
s º
ª1
2
«
»
¬ s s 1¼
Given the input x t
we have Y
Y s
s
of the LTI system h t
1 cos t
L ª¬ h t º¼
s
1
s s2 1
.
G t ŸX s
X s H s
1
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
Y s
1
2
s s 1
s becomes
A
B
C
s
si
si
where A, B, and C are real numbers to be determined.
Chapter 3
274
A
ª 1 º
« 2
»
«¬ s 1 »¼
s 0
1 , B
ª 1 º
«
»
«¬ s s i »¼ s i
1
2
C
ª 1 º
«
»
«¬ s s i »¼ s i
1
2
,
Hence, the function Y s can be expanded as
Y s
1
2
s s 1
1 1 1
1 1
2 si
s 2 si
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 1 ­° 1 ½° 1 1 ­° 1 ½°
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
2e 3t u t when the input x t
for an LTI system is found to be
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
Solution:
a For the input x t
X s
u t , the Laplace transform is
1
s.
The corresponding output has the Laplace transform
Y s
2
s3
Hence,
H s
Y s
X s
Rewriting H
H s
2s
s3
s as
2 s3 6
s3
2
275
6
s3
t
u t .
Chapter 3
276
h t
the impulse response
transform of H
is obtained by taking inverse Laplace
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
2s
s 1 s 3
H s X s
.
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
s3
where A and B are real numbers to be determined.
ª 2s º
A «
»
«¬ s 3 »¼ s 1
1 , B
ª 2s º
«
»
«¬ s 1 »¼ s 3
3
Application of the Laplace Transform to LTI Differential Systems:
Transfer Functions
1
3
s 1
s3
Y s
The output y t
277
.
is obtained by taking the inverse Laplace transform of
Y s
y t
e t 3e 3t u t .
Example
3.24.
R
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
Ri t
dt
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
4i 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 v t
H s
I s
V s
1
0.5s 4
.
12u t . The transformed current is given by
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 0
I s
3
3
s
s 8
The output
3 , B ª 24 º
«¬ s »¼
s 8
.
i t is obtained by taking the inverse Laplace transform of
I s
i t
3,
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
F t
Kx t
F t
D
K ³ v W dW
0
K
Dv t
Ds
Spring
D
Damper
dx t
dt
F t
F s
X s
Application of the Laplace Transform to LTI Differential Systems:
Transfer Functions
281
Mass
F t
M
M
d 2x t
dt
F t
2
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 2x 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
M s2 D s K X s
X s
F s
1
M s D sK
F s
2
Thus, the transfer function H
.
s
is
F s
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 ` x1 t x2 t ^ x1 t x2 t ` F t
2^ 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 x2 t
M2
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
­ 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
s 2 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
10 s 1
s 2 10s 10
'2
X2 s
T .F
s 2 10 s 10
10 F s
10 s 1
0
ª s 2 10 s 10 2 100 s 1 2 º
¬«
¼»
100 s 1 F s
'2
'
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
285
X2 s
100 s 1
F s
ª s 2 10 s 10 2 100 s 1 2 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 1s 2 D1 D3 s k X 1 s D3 sX 2 s
^
F s
`
D3 sX 1 s M 2 s 2 D2 D3 D4 s X 2 s
Solving X 2
'
' ª
¬
s using Cramer’s rule
M 1s 2 D1 D3 s k
D3 s
D3 s
^M s D D D s`
2
2
2
T .F
Thus,
3
4
'2
'
X2 s
F s
2
2
1
1
3
2
2
X2 s
2
^ 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
0
2
3
4
3
^M s D D D s` F s
2
2
2
3
4
^M s D D D s` F s
2
2
2
3
4
'
^M s D D D s` F s ^M s D D D s`
2
2
2
2
3
' F s
4
2
2
'
3
4
Chapter 3
288
T .F
^M s D D D s` .
2
X2 s
2
2
3
4
'
F 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 2z t
dt 2
3
dz t
dt
d 2x t
2z t
dt 2
6
dx t
dt
7x 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
H s
H s
1
½
°
¾
°¿
s 2 6s 7 X s
Z s
s 2 6s 7
X s
s 2 3s 2
3s 5
s 1 s 2
1
A
B
s 1
s2
.
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
s2
where A and B are real numbers to be determined.
ª 3s 5 º
2 , B
A «
»
¬« s 2 ¼» s 1
H s
1
2
1
s 1
s2
ª 3s 5 º
1
«
»
¬« s 1 ¼» s 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 2z 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 4s 10 Z s
T .F
H s
X s
Z s
H s
½
°
¾
°¿
, we get
X s
1
s 4 s 10
2
1
2
s2 6
.
Taking the inverse Laplace transform, we get
h t
1 2t
e sin
6
6t .
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
dN y t
d N 1 y t
d2y 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 t alone.
If 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
4y t
dt
et with y 0
2.
Solution:
Let y
'
t 4y 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
L ^ y t ` , we find
.
It is now necessary to invert 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
Y s
2s 1
s 1 s 4
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
1
, B
5
Hence, the given function Y
Y s
2s 1
s 1 s 4
ª 2s 1 º
«
»
«¬ s 1 »¼ s 4
9
5
.
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
1 1 ­ 1 ½ 9 1 ­° 1 ½°
L ®
¾ L ®
¾
5 ¯ s 1 ¿ 5 ¯° s 4 ¿°
y t
297
.
Finally, using the transform pairs established in Table 2.1, we have
1 t 9 4t
e e .
5
5
y t
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
d2y t
dt
2
2
dy t
dt
4y 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
ª¬ s 2Y s sy 0 y ' 0 º¼ 2 ª¬ sY s y 0 º¼ 4Y s
It is now necessary to invert Y
2
s2 4 .
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
Y s
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
As B s 2 4 Cs D s 2 2 s 4
299
.
Hence, comparing the coefficients of both sides gives
s3 :
A C 0,
s2 :
D B 0,
s:
2 D 4C
0,
1:
4B 4D
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
2
s 2s 4
14 s
s2 4
Y s
1 °­
1
s 1
°½ 1 °­
°½ 1 °­ s °½
® 2
¾ ® 2
¾ ®
¾
4 ° s 2s 4 ° 4 ° s 2s 4 ° 4 ° s 2 4 °
¯
¿
¯
¿
¯
¿
Y s
­
½
1°
1
1 ­° s ½°
° 1 ­° s 1 ½°
®
¾
®
¾
®
¾
4 ° s 1 2 3 ° 4 ° s 1 2 3°
4 ° s2 4 °
¯
¿
¯
¿
¯
¿
.
Applying the inverse Laplace transform to the above equation, and using
its linearity, we get
Chapter 4
300
1
y t
4 3
e t sin
1
3 t e t cos
4
1
3 t cos 2t .
4
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
d3y t
dt
2
2
d2y t
dt
2
dy t
dt
2y t
0
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 Y
301
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
Y s
Y s
s becomes
6
s3 2s 2 s 2
6
3
2
s 2s s 2
6
s 1 s 1 s 2
A
B
C
s 1 s 1
s2
where A, B, and C are real numbers to be determined.
ª
º
ª
º
ª
º
6
6
6
1, B «
2.
A «
3, C «
»
»
»
«¬ s 1 s 2 ¼» s 1
¬« s 1 s 2 ¼» s 1
¬« s 1 s 1 ¼» s 2
Y s
1
3
2
s 1 s 1
s2
.
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
3y 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
1
s2 .
With the initial value of y 0 , this equation can be written as
Y s
1
1
s3 s3 s2
.
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
e 3t y t
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
d2y 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
1
s s2 1
A Bs C
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
1 cos t.
y t
Example 4.6. Obtain the solution of the second-order differential
equation
d2y t
dy t
6y t
et
with the initial conditions
y 0
dt
2
5
dt
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
6y 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 5s 6 Y s
1
2 s 11
s 1
s 2 5s 6 Y s
2s 2 13s 12
s 1
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
s2
s3
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 1
1
,
2
ª 2 s 2 13s 12 º
«
»
«¬ s 1 s 3 »¼ s 3
9
,
2
C
hence, the function Y
B
ª 2 s 2 13s 12 º
«
»
«¬ s 1 s 3 »¼ s 2
s can be expanded as
6,
Chapter 4
306
2 s 2 13s 12
s 1 s 2 s 3
Y s
1 1
6
9 1
2 s 1
s2 2 s3
.
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
d2y t
dy t
5
6y t
2
dt
dt
with y 0
0 y' 0 .
G t S G t 2S
Application of the Laplace Transform to LTI Differential Systems:
Solving IVPS
307
Solution:
In
this
instance,
the
forcing
function
is
the
step
function
^u t S u t 2S ` of the above equation. This function is
discontinuous at t
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 ` e e
2
S s
'
.
This algebraic equation can be solved for L
^ y t ` as
2S s
Chapter 4
308
L ª¬ y t º¼
e S s e 2S s
1
s2 s3
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
L ª¬ y t º¼
ª 1
1 º
e S s e 2S s «
»
s 3 »¼
«¬ s 2
L ª¬ y t º¼
e S s
e S s
e2S s
e2S s
s2
s3
s2
s3
.
Applying the inverse Laplace transform to the above equation, and using
the linearity of the inverse Laplace transform, we can write the timedomain solution y t as
y t
e 2 t S 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
d2y t
dt 2
with
4y t
y 0
sin 2t
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
4y 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 y 0 and
this expression for Y
Y s
2
s 4
2
y ' 0 in this equation and solving
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
y t
1
1
10 cos 2t sin 2t t cos 2t .
8
4
Chapter 4
310
Example 4.9. Determine the solution of the initial value problem
d2y t
dt 2
4y t
F t
with y 0
y' 0
0
where F t is the periodic function
F t
­1 0 t 1
with F t 2
®
¯0 1 t 2
F t .
Solution:
Given that
d2y t
4y t
dt 2
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
L ª¬ F t º¼
T
s 2 4 L ª¬ y t º¼
1
e st F t dt
1 e st ³0
1
s 2 4 L ª¬ y t º¼
1
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
s 1 e s
Application of the Laplace Transform to LTI Differential Systems:
Solving IVPS
311
Partial fractions give
1 ª1
s º
« 2
»
4 «s s 4 »
¬
¼.
1
2
s s 4
1
1 e s
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
L ª¬ y t º¼
^ y t ` can be expressed as an infinite series
1 ª1
s º
« 2
» 1 e s e 2 s e 3s e 4 s ........
4 «s s 4 »
¬
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
d2y t
dt 2
9y t
cos 2t
with y 0
1 & y' 0
0.
Chapter 4
312
Solution:
Let
d2y t
dt 2
9y 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
L ¬ª y t ¼º
s
2
2
s 4 s 9
1
s 9
^ y t ` as
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
­°
½°
­ 1 ½°
s
1°
L ® 2
¾ L ® 2
¾
2
°¯ s 4 s 9 °¿
°¯ s 9 °¿
.
1
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
º
s
»
L1 « 2
«¬ s 4 s 2 9 »¼
1
^cos 2t cos 3t `
5
(see Chapter 2, Example 2.26 ).
Thus,
1
1
cos 2t cos 3t ` sin 3t .
^
5
5
y t
Example 4.11. Solve the differential equation
d2y t
dt
2
3
dy t
dt
2y 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 u t
above equation. This function is discontinuous at 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 ` 10s
2
'
.
This algebraic equation can be solved for L
s 2 s 10
L ª¬ y t º¼
s s 1 s 2
^ 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
B
C
s s 1
s2
ª s 2 s 10 º
ª s 2 s 10 º
A «
5,
B
10, C
»
«
»
«¬ s 1 s 2 »¼ s 0
«¬ s s 2 »¼ s 1
L ª¬ y t º¼
315
ª s 2 s 10 º
6.
«
»
«¬ s s 1 »¼ s 2
5
10
6
.
s s 1
s2
Applying the inverse Laplace transform to the above equation, and using
its linearity, we get
5 10et 6e2t .
y t
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
d2y t
dt
2
3
with y 0
dy t
dt
2y t
u t S
0 y' 0 .
Solution:
In this instance, the forcing function is the step function u t S
above equation. This function is discontinuous at t
of the
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.
Given that
d2y t
dt
2
3
dy t
dt
2y 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
2y t
2
dt
dt
¯°
½°
¾ L^ u t S `
¿°
.
Using the derivative property and assuming all initial conditions are zero
gives
^s 3s 2 `Y s
2
e S s
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
Y s
e S s
s s 2 3s 2
s becomes
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
s2
s 3 ¿°
¯° s
where A, B, and C are real numbers to be determined.
ª
º
1
A «
»
«¬ s 1 s 2 »¼s 0
Y s
ª
º
1
1
, B «
1, C
»
2
«¬ s s 2 »¼s 1
ª 6 º
1
.
«
»
«¬ s s 1 »¼s 2 2
­° 1 1
1
1 1 ½°
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
y t
§ 1 t S 1 2 t S ·
e
¨ e
¸ 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
d2y t
dt 2
t
³ f u sin t u du with y 0
y t
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
¿°
''
using
''
2
'
°­ L ª¬ y t º¼ s L ª¬ y t º¼ sy 0 y 0 °½
®
¾
¯° and convolution theorem
¿°
^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
L ª¬ y t º¼
, we get
L ª¬ y t º¼
s2 1
^ y t ` as
2 s2 1
s2 s 1
.
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
2
s s 1
4
2 2
2
s 1 s s
.
Thus, we can write the time-domain solution y t as
4et 2 2t.
y t
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
d2y t
dt
2
4
dy t
dt
4y 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°
d2y 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 ` s 22
2
'
This algebraic equation can be solved for L
L ª¬ y t º¼
^ y t ` as
2
s2
5
.
Thus we can write the time-domain solution y t as
2e2t t 4
24
y t
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
d2y t
dt
2
5
dy t
dt
6y 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 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
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
s 1 .
ª¬ s 2Y s sy 0 y ' 0 º¼ 5 ª¬ sY s y 0 º¼ 6Y s
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
s2
s3
where A, B, and C are real numbers to be determined.
ª
º
s7
A «
»
¬« s 2 s 3 ¼» s 1
C
ª
º
s7
«
»
«¬ s 1 s 2 »¼ s 3
ª
º
s7
3 , B «
»
¬« s 1 s 3 ¼» s 2
5,
2,
Y s
3
5
2
s 1
s2
s3
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
d2y t
dt 2
y t
with y 0
u t S
0 and y ' 0
1.
Solution:
In this instance, the forcing function is the step function u t S
above equation. This function is discontinuous at t
of the
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
d2y 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
1
e S s
s2 1 s s2 1
Y s
Using partial fractions, we can write this as
ª
1
1 º
S s 1
«
»
e
s2 1
s 2 1 »¼
«¬ s
Y s
.
Applying the inverse Laplace transform to the above equation, and using
its linearity, we get
sin t 1 cos t S
y t
u t S
.
Example 4.17. Determine the solution of the initial value problem
d2y t
dt
2
6
dy t
dt
9y t
12e3t t 2 with y 0
Solution:
Given that
y '' t 6 y ' t 9 y t
12 e3t t 2
0 & y' 0
0.
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 ` s 24 3
2
'
This algebraic equation can be solved for L
L ª¬ y t º¼
^ 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
d2y t
dt
2
9y t
cos 2t with y 0
§S ·
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
^ y t ` gives
L ª¬ y t º¼
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
L ª¬ y t º¼
^ 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
a
1
4
sin 3t cos 2t cos 3t
3
5
5
y t
when
t
S
2
Ÿ 1
a 1
Ÿa
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
d4y t
K
dt 4
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
Y s
K A B
s5 s 2 s 4
329
B
Taking the inverse Laplace transform of each term on the right, and using
the linearity property of the Laplace transform, we find that
Kt 4
Bt 3
At 24
6 .
y t
The boundary conditions on the right-hand side are now satisfied
y d
Kd 4
Bd 3
Ad 24
6
y '' d
Kd 2
Kd
Bd 0 Ÿ B 2
2
and A
kd 3
.
24
0
Finally, the desired solution of the boundary value problem is
Kt 4
Kd 3
Kd 3
t
t .
24
24
12
y t
Example 4.20. Obtain the solution of the second-order differential equation
d2y t
dt 2
dy t
dt
f t ,
f t
along with the initial conditions
­1 0 t 1
®
¯0 t ! 1
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 s2
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
ª d ­ 1 ½º
ª1º
1, B « ®
A « 2»
¾»
¬ s ¼ s 1
¬ ds ¯ s 1 ¿¼ s 0
C
ª 1 º
«
»
¬« s 1 ¼» s 0
1
s 1 s2
331
1,
1,
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 ¼ s 1
1 , B
Hence, the given function Y
Y s
ª 1 º
1,
«¬ s 1 »¼
s 0
1
s 1 s
1
1
s 1 s
.
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
­
½
1 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
d2y t
dt 2
4y 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
4y 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 ` s 1 3 .
2
'
Application of the Laplace Transform to LTI Differential Systems:
Solving IVPS
It is now necessary to invert Y
333
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
Y s
6
s2 4 s 3
Y s
6
s 4 s 3
2
6
s3 s2 s 2
A
B
C
s 3 s2
s2
where A, B, and C are real numbers to be determined.
ª
º
ª
º
ª
º
6
1
6
1
6
1
, B «
, C «
A «
»
»
»
«¬ s 3 s 2 ¼» s 2 20
¬« s 2 s 2 ¼» s 2 5
¬« s 2 s 3 ¼» s 2 4
Y s
1 1
1 1
1 1
5 s 3 20 s 2 4 s 2
.
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
2y t
sin 2t
along with the initial conditions 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
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 ` s 1 4 .
2
It is now necessary to invert 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 1 s2 4
Y s
.
We first observe that the quadratic factor
s2 4
and
s 1
s 2 4 is irreducible. Since
are non-repeated factors, the partial fraction
expansion has the form
ª
º
1
«
»
«¬ s 1 s 2 4 »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 º
« 2
»
«¬ s 4 »¼
s 1
> Bs C @s 2i
2 Bi C
1
5
ª 1 º
«
»
«¬ s 1 »¼ s 2i
1
1 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
º
1
«
»
«¬ s 1 s 2 4 ¼»
1 °­ 1
s 1 °½ 1 °­ 1
s
1 ½°
2
2
2
®
¾
®
¾
5 ° s 1 s 4 ° 5 ° s 1 s 4
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
y t
1 ­° 1 § 1 · 1 § s · 1 § 1 · ½° 1 § 2 ·
¸ L ¨
¸¾ L ¨
®L ¨
¸ L ¨¨ 2
¨ s 1 ¸¸
¸
¨ s2 4 ¸°
5 ° © s 1 ¹
s
4
©
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
y t
L®
dt
¯
½ °­ t
¾ L ® ³ cos 2 t u du.
¿ °¯ 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
Y s
s becomes
1
2
2
s 1
s 1 s 4
Using partial fraction decomposition, we can write
1
s 1 s2 4
1 ­° 1
1 ½°
s
2
2
®
¾
5 ° s 1 s 4
s
4
¯
¿°
Thus,
Y s
1 ­° 1
s
1 ½° 2
®
¾
2
5 ° s 1 s2 4
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
Chapter 4
338
1 ­° 1 § 1 · 1 § s · 1 § 1 · ½°
1 ·
¸ L ¨
¸ ¾ 2 L1 ¨§
®L ¨
¸ L ¨¨ 2
¸
2
¸
¨ s 4 ¸°
s 1 ¹
5 ° © s 1 ¹
s
4
©
©
¹
©
¹¿
¯
Y s
.
Finally, using the transform pairs established in Table 2.1, we have
1­ t
1
½
®11e cos 2t sin 2t ¾ .
5¯
2
¿
y t
Example 4.24.
Use the Laplace transform technique to solve the
initial value problem
d2y t
dt
2
4
dy t
dt
3y t
e2t
0 y' 0 .
along with the initial conditions 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
2
'
¬ª s Y s sy 0 y 0 ¼º 4 ¬ª sY s y 0 º¼ 3Y s
339
1
s2 .
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
s3
where A, B, and C are real numbers to be determined.
ª
º
1
A «
»
¬« s 2 s 3 ¼» s 1
C
ª
º
1
«
»
¬« s 1 s 2 ¼» s 3
Hence, the function Y
Y s
1
,
6
B
ª
º
1
«
»
¬« s 1 s 3 ¼» s 2
1
,
15
1
10
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
1 t 1 2t 1 3t
e e e .
6
15
10
y t
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
d4y t
dt
4
2
d3y t
dt
3
d2y t
dt
2
2
dy t
dt
G t
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
d3y t d2y 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
s3 1
s 4 2s3 s 2 2s
Y s
s3 1
s s 1 s 1 s 2
1
.
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
s2
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 0
1
,
2
B
ª
º
s3 1
«
»
«¬ s s 1 s 2 »¼ s 1
1,
C
D
ª
º
s3 1
«
»
«¬ s 1 s s 1 »¼ s 2
3
2
.
ª
º
s3 1
«
»
«¬ s 1 s s 2 »¼ s 1
0,
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
d2y t
dt
2
4
dy t
dt
4y t
f t , f (t )
along with the initial conditions y 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
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
f (t )
343
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)
2t 2
replace t
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
4y 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 ` s2 s e .
2
s
'
2
It is now necessary to invert 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
Y s
2
2
s s2
2
345
s e s
ª
º
1
1
e s 2
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
y t
1
­ t
½ 1
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
t u
³ e 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
½ °­ t u
°½
¾ L ® ³ e t u du. ¾
¿ °¯ 0
¿°
­ dy t
L®
y t
¯ dt
^
L ª¬ y ' t º¼ L ª¬ y t º¼ L et t
`
^ª¬sY s y 0 º¼ Y s ` L ª¬e º¼ u L >t @
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
1
Y s
2
s s 1
2
s becomes
1
s 1
.
Using partial fraction decomposition, we can write
1
s2 s 1
2
2 1
2
1
2
2
s s
s 1
s 1
Thus,
Y s
2 1
3
1
2
2
s s
s 1
s 1
.
.
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
y t
§ 3 · 1 § 1 ·
§2·
§ 1·
¸
L1 ¨ ¸ L1 ¨ 2 ¸ L1 ¨
L ¨
2
¨ s 1 ¸¸
¨
¸
©s¹
©s ¹
s
1
©
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 t formula
L ^t f t `
d
L^ f t `
ds
which is taken to be the nth-derivative of f
^
L t fn t
` dsd L ^ f
n
t
`
t , then
.
Using the derivative formula
^
L t fn t
` dsd ^s L ^ f t ` s
n
n 1
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.
`
f 0 s n 2 f ' 0 ...... f n 1 0 .
Chapter 4
348
Example 4.28. Determine the solution of the initial value problem
t
d2y 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 ¯
` dsd ^Y s ` ¾¿½
^
.
, 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
¿
s2 1
d
s
Y s 1 2
ds
s 1
d
Y s
ds
s
s
2
2
s 1
s2 1
.
Taking the inverse Laplace transform and using
­ 1 ­ d
½
® L ® ^Y s `¾ ty t
¯ ds
¿
¯
½
¾
¿ , we get
Application of the Laplace Transform to LTI Differential Systems:
Solving IVPS
349
1
sin t t sin t
2
.
ty t
Thus,
y t
1§ 2·
¨1 ¸ sin t.
2© t ¹
Example 4.29. Determine the solution of the initial value problem
t
d2y t
dy t
t
y t
2
dt
dt
with y 0
0
0 and y ' 0
2.
Solution:
Because of the linearity of the equations and the Laplace transform
operation, we have
2
dy t
°­ d y 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
dy t
°­ d y t
°½
L ®t
t
y t ¾ 2 sY s s s 1 Y ' s
2
dt
¯° dt
¿°
Y' s
Y s
2ds
s 1
0
.
On integration, we get
log s 1
log Y s
2
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
y' 0
351
C 2,
therefore
2tet
y t
.
Example 4.30. Determine the solution of the initial value problem
t
d2y t
dt
2
dy t
dt
ty t
2 and y ' 0
0 with y 0
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 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
` dsd ^Y s ` ½¾¿
^
.
, 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
ds
s 1
0
Chapter 4
352
implying
d
Y s
ds
s
2 Y s
s 1
dY s
s
2 ds 0
Y s
s 1
.
On integration, we get
1
ln Y s ln s 2 1
2
ln Y s
s2 1
ln c
ln c
c
Y s
s2 1
.
Taking the inverse Laplace transform on both sides of the above equation,
we get
­
°
cL ®
°
¯
½
°
¾ c J0 t .
2
s 1 °
¿
1
1
y t
Example 4.31. (Bessel’s equation) Determine the solution of the initial
value problem
x2
d2y x
dx
2
x
dy x
dx
x2 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
^
.
` dsd ^Y s ` ½¾¿
, 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
ds
s 1
implying
d
Y s
ds
s
2 Y s
s 1
0
Chapter 4
354
dY s
s
2 ds 0
Y s
s 1
.
On integration, we get
1
ln Y s ln s 2 1
2
ln c
s2 1
ln Y s
ln c
c
Y s
s2 1
Taking the inverse Laplace on both sides of above equation, we get
y t
­
°
cL ®
°
¯
1
½
°
¾ c J0 t .
2
s 1 °
¿
1
Example 4.32. Use the Laplace transform to find the output y t
the system described by
d2y t
dt
2
3
dy t
dt
2y t
dx t
dt
3x t
with input given by
x t
e 5t u t with y 0
Solution:
1 and y ' 0
2.
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 ` ª¬sL ª¬ x t º¼ x 0 º¼ 3L ª¬ x t º¼
2
'
L^ y t `
s 3 L ^x t `
2
s 3s 2
s5
2
s 3s 2
we have,
L ^x t `
1
,
s5
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 `
s3
s 2 3s 2 s 5
s5
s 2 3s 2
s 2 11s 28
s 2 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
s5
.
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
d2y t
dt
2
4
dy t
dt
5y 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
ª¬ s 2Y s sy 0 y ' 0 º¼ 4 ª¬ sY s y 0 º¼ 5Y s
8
s2 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
358
Chapter 4
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
5y 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 5L ^ y t ` L ^ x t `
^ L ª¬ y t º¼ sL ª¬ y t º¼ y 0 ` , we get
'
Using
.
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
,
s2
and thus
L^ y t `
3
s2 s5
2
s5
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
s2
s5
s5
L^ y t `
1
3
s2
s5
.
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
d2y t
dt
2
3
dy t
dt
2y 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
L^ y t `
4
s s 1 s 3
2
.
^ 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
Thus, F ( s )
3 2
4
1
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 h t
y t
h t *x 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
d2y t
dt
2
5
dy t
dt
6y t
x t
Application of the Laplace Transform to LTI Differential Systems:
Solving IVPS
to the input given by x t
363
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°
d2y 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
T .F
H s
X 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
1
Y s
s2 s3
X s
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
1
s 1 s 2 s 3
Y s
where
A,
B,
and
C
are
A
B
C
s 1
s2
s3
real
ª
º
1
A «
»
«¬ s 2 s 3 »¼ s 1
1
,
2
ª
º
1
«
»
«¬ s 1 s 2 »¼ s 3
1
,
2
C
Y s
numbers
to
be
determined.
ª
º
1
B «
»
«¬ s 1 s 3 »¼ s 2
1 1
1
1 1
2 s 1
s2 2 s3
.
1,
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 t 2t 1 3t ·
¨ e e e ¸u t .
2
©2
¹
h t
Thus,
§ 1 t
· § 2t 1 3t ·
¨ e u t ¸ ¨ e e ¸ u t .
2
©2
¹ ©
¹
h t
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
d2y t
x t
to the input given by
x t
y 0
5
dy t
6y t
dt
2
dt
0 and y ' 0
u t . The initial conditions are
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
d2y t
dt
2
5
dy t
dt
6y 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
­d2y t
½
dy t
°
°
L®
5
6y 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
Y s
½
°
¾
°¿
.
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
X s
s2 s3
367
s7
s 2 5s 6
s7
s2 s3
.
The response due to the first term is due to the initial conditions and hence
is the natural response
YN s
s7
s2 s3
.
Using partial fraction expansion, we have
YN s
5
4
s2
s3
.
Applying the inverse Laplace transform to the above equation, and using
the linearity of the inverse Laplace transform, results in
yN t
5e 2t 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
s2 s3
u t .
Chapter 4
368
YF s
1
s s2 s3
.
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
YF s
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
yF t
1 3t
§1
· § 1 2t
·
¨ u t ¸ ¨ e u t e u t ¸.
3
©6
¹ © 2
¹
The total response was given by the addition of the natural response and
the forced response
yT t
y N t yF t
9 2t
11 3t
§1
·
¨ u t e u t e 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
d2y t
dt
2
4
dy t
4y t
dt
d 2x t
dt
2
dx t
2x t
dt
with the initial conditions
0 and y ' 0
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
½
°
¾
°¿
s2 s 2
2s 11
2
2
s s 4s 4
s 4s 4
.
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
s s2
Y s
s s2
2
2s 11
s s2
2
1
1
2 2 2 2 7
2
2
s s2
s2
s2
s2
.
Taking the inverse Laplace transform, we get the complete response
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
zero-state
yT t
y N t yF t
yT t
5 2t
§1
2 t ·
¨ u e 5te ¸ 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 2x t
dt
2
with y 0
y t
1, x 0
dx t
dy
y t
dt
4
0 and x ' 0
1.
and
dt
4x t
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
X s
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
.
X s
It is now necessary to invert
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
X s
1
s s 4s 4
3
2
2
Y s
s 4s 4
s s 2 4s 4
3
s
and Y s becomes
A
B
C
s 1
s2
s2
1
1
1
3 2 6
s 1
s2
s2
A
B
C
s 1
s2
s2
1
2
3 2 3
s 1
s2
s2
.
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
dt
2 x t y t 2e5t and
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
X s s 2 Y s
3
s2
.
X s
It is now necessary to invert
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
X s
Y s
s
and Y s becomes
5
3
3
1
4 4
s 1
s2
s 3
s 5
2 s 2 5s 7
s 1 s 2 s 3 s 5
3s 13
s 1 s 3 s 5
5
1
4 1 4
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
2y t z t
dz t
0 and
subject to the initial conditions
dt
y 0
2z t y t
1 and z 0
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®
2y 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
s2 Y s Z s
1
Y s s2 Z s
0
Solving for Y
s and Z s gives
1
1
0 s2
Y s
s2
s2
2
1
s2
1
s 2 1
s2 1
1
0
s2
1
s2
1
Z s
375
;
1
2
s 2 1
.
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 2x t
dt 2
with
10 x t 4 y t
y 0
0
0 and
x 0 , x' 0
d2y t
dt 2
1 and y ' 0
y t 4x t
1.
0
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
½°
¾ 0
¿°
­° d 2 y t
L®
y t 4x t
2
¯° dt
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
X s
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
377
s2 6
s 2 2 s 2 12
s 2 10 4
4
s2 1
X s
It is now necessary to invert
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
X s
s
2
s 2 2 s 2 12
2
Y s
s 6
s 2 s 2 12
2
s
and Y s becomes
6
1
5 5
2
2
s 2
s 12
2
3
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 y t
2
sin
5 2
2t 6
sin 12t .
5 12
3
sin 12t .
5 12
Chapter 4
378
Example 4.43. Determine the solution of the initial value problem
dy t
dt
2x t
sin 2t and
0, x 0
1.
y 0
dx t
dt
2y t
cos 2t with
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
2
s2 4
sX s 2Y s
s
1
s 1
Solving for X
s
2
.
and Y s gives
Application of the Laplace Transform to LTI Differential Systems:
Solving IVPS
Y s
2
s2 4
X s
1ª 2
2s º
« 2
»
2
2« s 4
4
s
»¼
¬
.
379
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 2x t
dy t
sin t
dt 2
dt
with y 0 1, x 0
and
0, y ' 0
dx t
d2y t
dt
dt 2
0 and x ' 0 1.
cos t
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
(4.2)
§ s2 2 ·
¨ 2
¸
© s 1 ¹
(4.3)
2
s X s sY s
X s sY s
Adding (4.2) and (4.3), we get
X s
s2 1
s2 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
­
½
° 2s °
L1 ®
2 ¾
2
s
1
°
°
¯
¿
1 t sin t .
381
t sin t
.
Example 4.45. Find the solution of the set of the differential equations
dy t
dt
2y t z t
0 and
dz t
subject to the initial conditions y 0
dt
2z t y t
1 and z 0
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®
2y 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
s2 Y s Z s
1
Y s s2 Z s
0
Solving for Y
.
s and Z s gives
Chapter 4
382
1
1
0 s2
Y s
s2
2
s2
1
1
s2
s 2 1
;
s2 1
1
0
s2
1
1
s2
Z s
1
2
s 2 1
.
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 2x t
dt 2
2x t y t
with y 0
0, x 0
0
d2y t
and
0, x ' 0
dt 2
2y t x t
1 and y ' 0
0
1.
Solution:
Because of the linearity of the equations and the Laplace transform
operation, taking the Laplace transform, we have
s2 2 X s Y s
1
Application of the Laplace Transform to LTI Differential Systems:
Solving IVPS
X s s2 2 Y s
Solving for X
Y s
.
s and Y s gives
1
X s
1
1
1 s 2 2
2
s 2
1
1
s2 2
s 2 2 1
1
1
2
s 2
1
1
s2 2
1
s 3
2
;
1
s 3
2
Thus,
X s
1
s 3
Y s
1
s2 3
2
.
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
F t , y 0
where F t
­0 0 d t d S
.
®
¯sin t t ! S
0
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
y t
L ª¬ F t º¼
1 S s
e
s 1
2
1
s 1 s2 1
­°
½°
1
L1 ®eS s
¾
2
1
1
s
s
¯°
¿°
.
Application of the Laplace Transform to LTI Differential Systems:
Solving IVPS
385
Refer to example 2. 24
ª
º 1­
1
1
½
t
»
L1 «
®e cos 2t sin 2t ¾
2
2
¿
«¬ s 1 s 1 »¼ 5 ¯
then,
yt
­°
½° 1­
1
1
½
tS
L1 ®eSs
¾
®e cos2 t S sin2 t S ¾u t S .
2
2
s 1 s 1 ¿° 5 ¯
¿
¯°
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
d2 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 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
L®
y t
2
¯° dt
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 Y
`
2e2 s e4 s
s
s .
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
Y s
Y s
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
y t
­°
½°
­
½°
­ 1 ½°
1
1
1 ° 4 s
1 °
L1 ®2e 2 s
L
e
L
¾
®
¾
® 2
¾
2
2
1
1
1
s
s
s
s
s
¯°
¿°
¯°
¿°
¯°
¿°
Using the convolution theorem
Application of the Laplace Transform to LTI Differential Systems:
Solving IVPS
387
­° 1 ½°
L1 ® 2
¾ 1 cos t
s
s
1
°¯
°¿
.
Finally,
2 1 cos t 2 u t 2 1 cos t 4 u t 3 sin t.
y t
Hence, the solution of the given initial value problem is
2 1 cos t 2 u t 2 1 cos t 4 u t 3 sin t.
y t
Example 4.49. Obtain the solution of the second-order differential
equation
d2y t
dt
2
5
dy t
dt
6y 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
6y 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
e 2 s
e 4 s
4
s 2 5s 6
s 2 5s 6
Y s
3
Y s
e 2 s
e 4 s
3
4
s2 s3
s2 s3
.
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
s2 s3
A
B
s2
s3
where A and B are real numbers to be determined.
ª 1 º
1, B
A «
»
¬« s 3 ¼» s 2
hence, the given function Y
Y s
ª 1 º
«
»
¬« s 2 ¼» s 3
1,
s can be expanded as
ª 1
ª 1
1 º
1 º
4 s
3e 2 s «
» 4e «
»
s 3 »¼
s 3 »¼
«¬ s 2
«¬ s 2
.
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
y t
389
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 ` ,
RI s 1 I s
C s
1 ·
§
¨R ¸I s
sC ¹
©
1
1
.
This algebraic equation can be solved for I
1
s
1 ·
R§
¨s
¸
RC ¹
©
I s
I s
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 I
the result
s and using Table 2.1 gives
Chapter 5
394
1ª
1 RC1 t º
G
t
e »
«
R¬
RC
¼
i t
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
dt C 0
V
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
1
s
s 2 2 s 1 L ^q t `
.
^
This algebraic equation can be solved for L q t
L ^q t `
1
s s 1
2
.
` as
Chapter 5
396
s is the desired current i t , t ! 0
The inverse Laplace transform of I
^
However, 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 q t
L ª¬ q t º¼
A
B
C
2
s
s 1
s 1
L ^q t `
1
1
1
2
s s 1
s 1
` 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 , t ! 0 is
i t
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 V
the charge q as a function of time if
V t
­0 0 d t 1
.
®
¯V0 t t 1
q 0
1
t volts. Find
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
C
V0 u t 1
.
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
s2 1 Q s
Q s
V0
e s
s
V0 e s
s s2 1
.
We expand this S domain solution into partial fractions as
Q s
ª1
º
s
»
V0 e s « 2
s 1 »¼
«¬ s
.
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
Ri t
E t
.
Here, i
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
2i t
e 5t
.
Taking the Laplace transform in the usual way and using i 0
gives
s 2 L ^i t `
1
s5 .
^ ` as
This algebraic equation can be solved for L i t
0A
Application of the Laplace Transform to LTI Differential Systems:
Electrical Circuits and Mechanical Systems
L ^i t `
399
1
s2 s5
.
Resolving into partial fractions gives
L ^i t `
A
B
s2
s5
L ^i t `
1§ 1
1 ·
¨¨
¸
3© s2
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 E t .
Prior to closing the switch at 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
di t
dt
E t
.
dq
is the current. The two terms on the left give the voltage
dt
Here, i
drop across the resistor and inductor, respectively
di t
dt
R
i t
L
E t
L
.
Taking the Laplace transform in the usual way and using i 0
gives
R·
§
¨ s ¸ L ^i t `
L¹
©
E
Ls
0A
Application of the Laplace Transform to LTI Differential Systems:
Electrical Circuits and Mechanical Systems
401
^ ` as
This algebraic equation can be solved for L i t
L ^i t `
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
as a sum of simpler terms in the table. We can express L i t
L ^i t `
L ^i t `
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 RL 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 t
0 , the currents are zero. Determine the
current 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
di t
R
i t
dt
L
403
E at
e
L
.
Taking the Laplace transform in the usual way, and using i 0
0A ,
gives
L ^i t `
E
R·
§
L sa ¨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
R·
sa §
¨s ¸
L¹
©
ª
º
«
»
E
1
1
«
»
R ·»
R aL « s a §
¨s ¸»
«¬
L ¹¼ .
©
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
L
«e e »
R aL ¬
¼
.
Example 5.7. The series LRC circuit shown in Figure 5.5 consists of
C
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 to Vc .
b Obtain Vc for Vin
e t with Vc 0
1 and Vc' 0
2.
Solution:
a By applying KVL we arrive at the following differential equation
t
di 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 u t
above equation. This function is discontinuous at 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 VC
t ` as
s 2 8s 11
s 1 s 2 5s 4
L ª¬VC t º¼
s 2 8s 11
L ª¬VC t º¼
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 Vc
L ª¬VC t º¼
t ` using partial fraction expansion
s 2 8s 11
s 4 s 1
A
B
C
2
s4
s 1
s 1
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 LCR series circuit in Figure
5.6, where V
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 EG 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 ^q t `
L
LC 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
E
1
1 ·
L§ 2 R
¨s s
¸
L
LC ¹
©
L ^q t `
L ^q t `
409
ª
E«
1
L « s O 2 p2
«¬
º
»
»
»¼
where O
§ 1
R
R2 ·
2¸
& p2 ¨
2L
© LC 4 L ¹
.
Taking the inverse Laplace transform, we get
E Ot
ªe sin pt º¼
Lp ¬
.
q t
Thus,
dq
dt
i t
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 V
the charge q as a function of time if
q 0
t volts. Find
0 i 0
­0 0 d t 1
.
®
¯1 t t 1
V t
Solution:
The differential equation that characterizes the electrical system is
L
d 2q t
dt
2
1
q t
C
V t
.
and
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
°¿
L ª¬ q '' t º¼ L ª¬ q t º¼
2
s 1 Q s
`
e s
s
e s
s
Q s
e s
s s2 1
Q s
§1
·
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
q t
411
­ §1
s · ½°
°
¸¾
L1 ®e s ¨ 2
¨
¸
s
s
1
°¯ ©
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 L C R series
circuit with an inductance of L
and resistance of R
1H capacitance of C 104 F ,
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 L C R series circuit in
Figure 5.6, where V
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 L q t
20
s s 160 s 104
L ª¬ q t º¼
413
` as
2
.
The inverse Laplace transform of I
^
However, L q t
s is the desired current i t , t ! 0.
` 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
L ^q t `
A
Bs c
2
s
s 160 s 104
L ^q t `
º
1 ª1
s 160
« 2
»
500 « s s 160s 104 »
¬
¼
L ^q t `
1 ª1
s 80 80 º
« 2
»
500 « s s 160s 104 »
¬
¼
ª
1 «1
500 « s
¬«
s 80
s 80
2
602
80
` as
º
»
2
s 80 602 »»
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
i t
1 80t
ªe sin 60 t º¼
3¬
.
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 i 0
0.
b Compute the current at t 1.5 seconds, i.e. compute i 1.5 .
c Evaluate lim ^i t ` .
t of
Solution:
a Since V t
circuit is
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
4i t
415
2 2u t 1
.
Taking the Laplace transform in the usual way gives
s 4 s L ^i t `
2 2e s
s
s .
^ ` gives
Solving for L i t
L ª¬i t º¼
2
2e s
.
s s4 s s4
Let
G s
2
s s4
,
using partial fractions, we obtain
G s
1 ª1
1
« 2 ¬s
s4
º
»
¼ .
Applying the inverse Laplace transform to the above equation, and using
its linearity, results in
Ÿg 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
1H , R 1: .
Figure 5.8 with L
a Determine the differential equation relating I s and I L .
b
IL t
using the Laplace
c Obtain the zero-input response for I L t
using the Laplace
Obtain the zero-state response for
transform for
e 5t .
Is t
transform with 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
IL t
Is t
.
Chapter 5
418
b For the given I s ,
dI L t
dt
IL 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 `
1
s5
L ^I L t `
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
L ^I L t `
1ª 1
1 º
«
»
s 5 ¼»
4 ¬« s 1
.
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
IL 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
IL 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 L I L
IL t
et u t .
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
RI s 1I s
C s
421
ª e as ebs º
V0 «
»
s
s ¼
¬
1 ·
§
¨R ¸I s
sC ¹
©
ª e as ebs º
V0 «
»
s
s ¼
¬
ª
º
as
bs
«
»
V0
e
e
«
»
1 · §
1 ·»
R Ǥ
s
s
«¬ ¨© RC ¸¹ ¨© RC ¸¹ »¼
.
I s
Applying the inverse Laplace transform to the above equation, and using
its linearity, results in
1
t b u
º
V0 ª RC1 t a
RC
e
u
t
a
e
u
t
b
«
»
R¬
¼
i t
.
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
423
s
50 2
s 1002
s 2 104 I1 s s 2 I 2 s
(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
sI1 s s 50 I 2 s
0
0
(5.2)
Solving (5.1) and (5.2), we get
I1 s
ª
º
s s 50
«
»
2
« s 100 s 2 104 »
¬
¼
I2 s
ª
º
s2
«
»
2
« s 100 s 2 104 »
¬
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
º
s2
«
»
2
« s 100 s 2 104 »
¬
¼
I2 s
ª§ 1 ·
º
1
1
s
§1·
§ 1 ·
Ǭ
»
¨ ¸
¸
¨
¸
2
© 200 ¹ s 2 104 »¼
«¬© 200 ¹ s 100 © 2 ¹ s 100
ª
A
«
s
100
¬«
B
s 100
2
Cs D º
»
s 2 104 ¼»
.
The current through the resistance i2
i2 t
t is
ª§ 1 · 100t § t · 100t § 1 ·
4 º
¨ ¸e
¨
¸ cos 10 t » .
«¨ 200 ¸ e
¹
©2¹
© 200 ¹
©
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 t
Figure 5.11
0.
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
I s
s 1 I1 s s 1 I 2 s 1
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 s
I 2 s sI 2 s 1
s
0
(5.4)
Solving (5.3) and (5.4), we get
I1 s
I1 s
ª
º
10
« 2
» & I2 s
«¬ s s 2 »¼
ª
º
10
«
»
«¬ s s 1 s 2 s 2 »¼
ª
º
10
« 2
» Ÿ I1 s
«¬ s s 2 ¼»
ª
º
«
»
«
»
10
«
»
« § § 1 ·2 § 7 ·2 · »
«¨¨ s ¸ ¨
¸ ¸»
¨
2
2
©
¹
«
©
¹ ¸¹ »¼
©
.
The current through the capacitance i1 t is
i1 t
ª§ 20 · §¨ 12 ·¸t
§ 7 ·º
© ¹
t ¸¸ » .
sin ¨¨
Ǭ
¸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
V t
§
§ 1 ··
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
L2
i2
i1
427
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
º
ª
2
1 e »
125 « «s s »
¬«
¼»
s 3 I1 s 1.25I 2 s
I1 s s 1 I 2 s
Solving for I1
I1 s
I2 s
s
assuming
0
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 0
,B
ª
º
«125 s 1 »
«
»
« s§s 7 · »
¸
«¬ ¨©
2 ¹ »¼ s 1
429
500
7
125
,C
3
2
ª
º
«125 s 1 »
«
»
« s§s 1 · »
¸
«¬ ¨©
2 ¹ »¼ s 7
625
21
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
125 2t 625 72t 500
e e 3
21
7
i2 t
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
d2y 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
d2y 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
d2y 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
d2y t
dt
2
2
dy t
dt
4y 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 Y
2
s 4
2
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
Y s
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 2 s 4
Hence, comparing the coefficients of both sides we get
s3 :
A C 0,
s2 :
D B 0,
Chapter 5
434
s:
2 D 4C
1:
4B 4D 2 .
0,
By these equations, we obtain
1
,B
2
A
1
and D
4
1
,C
4
0.
Hence, we have
1 2 s 1 4
Y s
2
s 2s 4
14 s
s2 4
Y s
1 °­
1
s 1
°½ 1 °­
°½ 1 °­ s °½
® 2
¾ ® 2
¾ ®
¾
4 ° s 2s 4 ° 4 ° s 2s 4 ° 4 ° s 2 4 °
¯
¿
¯
¿
¯
¿
Y s
­
½
1°
1
1 ­° s ½°
° 1 ­° s 1 ½°
®
¾
®
¾
®
¾
4 ° s 1 2 3 ° 4 ° s 1 2 3°
4 ° s2 4 °
¯
¿
¯
¿
¯
¿
.
Applying the inverse Laplace transform to the above equation, and using
its linearity, results in
1
y t
4 3
e t sin
1
3 t e t cos
4
1
3 t cos 2t .
4
Example 5.17. The differential equation for a mass-spring system is
m
d 2x 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 x t
initial conditions: x 0
435
3, x ' 0
D 6 N .s / m.
if it is released with
1. There is no external force.
b Determine the position of the mass x t if it is released with the
initial conditions: x 0
external force
f t
0
x ' 0 , and the system is driven by an
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 2x t
dt 2
6
We have f
dx t
t
dt
5x 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
L ^x t `
^ 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
s5
.
Taking the inverse Laplace transform of L
4e t e 5t .
gives the result x t
b
We have
f t
x 0
0 and x ' 0
0.
d 2x t
dt
2
6
dx t
dt
^ x t ` and using Table 2.1
5x t
30 sin 2t and the initial conditions are
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 6s 5 ^ x t `
60
s 4
2
This algebraic equation can be solved for L
60
s 1 s 5 s2 4
L ^x t `
The inverse Laplace transform of L
^ x t ` as
.
^ 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
½ 72 °
­
½
­
s
° 1 ½
° 15 « 1 » 12 °
°
°
3®
«
»
¾
® 2
¾
® 2
¾
°
° s 4 ¿
° 29 ¯
° s 4 ¿
°
¯ s 1 °
¿ 29 ¬« s 5 ¼» 29 ¯
.
Applying the inverse Laplace transform to the above equation, and using
its linearity, results in
Chapter 5
438
3et x t
15 5t
12
72
e sin 2t cos 2t
29
29 u 2
29
thus,
3et x t
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
2sec, it is
struck with a hammer, providing an impulse of 5lb sec. Determine
the displacement function 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
4y t
32 dt 2
5G t 2
along with the initial conditions y 0
d2y t
dt 2
16 y t
4, y ' 0
0.
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 A sin 8t acts on the
mass, but only for 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 2x t
dt
2
512 x t
A sin 8t u t 1
x 0
1 '
,x 0
10
0.
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 A s
e L ª¬sin 8 t 1 º¼
10 2
s 2 256 L ª¬ x t º¼
s A s
e ^cos8 L >sin 8t @ sin 8 L > cos8t @`
10 2
s 2 256 L ª¬ x t º¼
­
½
s
A
8
s
°
°
e s ®cos 8 2
sin 8 2
¾
10 2
s 64
s 64 ¿
°
°
¯
.
Hence,
L ª¬ x t º¼
½°
s
A s ­°
8
s
e
cos8
sin
8
®
¾
s 2 256 s 2 64
s 2 256 s 2 64 ¿°
10 s 2 256 2 ¯°
Application of the Laplace Transform to LTI Differential Systems:
Electrical Circuits and Mechanical Systems
xt
441
ª
º
ª
º ½°
1 1 ª s º A ­°
1
s
1
» ®8cos8 L1 «e s 2
»
«
»¾
L « 2
sin8
L
2
2
2
10 « s 256 » 2 °
s
256
s
64
s
256
s
64
«
»
«
»¼ °
¬
¼ ¯
¬
¼
¬
¿
Taking the inverse Laplace transform of 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 2 N . is applied vertically upward to the
mass at t
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
2
50 x t
2
10 dt
dt
2G t 3 , x 0
1 '
,x 0
10
1.
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
20e 3 s s 10 1
s 2 20 s 500 L ª¬ x t º¼
L ª¬ x t º¼
L ª¬ x t º¼
s 10 1
20e 3s
s 2 20 s 500
s 2 20 s 500
20e 3 s
2
s 10 20
2
1
10
s 10
2
s 10 202
Taking the inverse Laplace transform of L
.
^ x t ` and using Table 2.1
gives us the result
ª
º
s 10
«
»
L
« s 10 2 202 »
«¬
»¼
1
§
1
s
e 10t L ¨ 2
¨ s 202
©
e 10 t 3 sin 20 t 3 u t 3 x t
·
¸
¸
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 2x 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 x t
443
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
4
u t 2S
cos 4t
.
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
d2y t
dt
2
8
dy t
dt
20 y t
G t
0 y' 0 .
along with the initial conditions 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
4 s 2 8s 20 Y s
Y s
445
1
1
4 s 2s 5
2
1
Y s
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
º
«
»
V0
g 1
1
« »
c ·» §
c ·
mc « s §
¨ s ¸» ¨ s ¸
«¬
m ¹¼ ©
m ¹.
©
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
d2y t
m
dt 2
mg D
dy t
dt
when the retarding force is proportional to the velocity. Find y t
0 y' 0 .
for the initial conditions 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
L ª¬ y t º¼
g
D·
§
s2 ¨ s ¸
m¹.
©
^ y t ` as
Application of the Laplace Transform to LTI Differential Systems:
Electrical Circuits and Mechanical Systems
The inverse Laplace transform of L
449
^ y t ` is the desired motion of a
body 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
l1
P
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
d4y t
d2y t
EI
P
dt 4
dt 2
Where W
W t
.
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 ª¬1 u t l1 º¼ W ª¬G t l2 º¼ .
W t
Therefore, equation (1) becomes
d4y t
d2y t
EI
P
dt 4
dt 2
2
d4y t
2 d y t
a
dt 4
dt 2
w ª¬1 u t l1 º¼ W ª¬G t l2 º¼ .
wˆ ª¬1 u t l1 º¼ Wˆ ª¬G t l2 º¼ .
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
deflection 0 Ÿ y 0
0.
bending moment 0 Ÿ y '' 0
0.
at t 0
Chapter 5
452
at t
slope 0 Ÿ y ' L
L
0 Ÿ y ''' L
shear force
0.
0.
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
'
2
2
2
s s a L ª¬ y t º¼
L ¬ª y t ¼º
y' 0
s2 a2
''''
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
Taking inverse transforms and making use of the second shift theorem
gives the deflection y t as
y x
­ 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 °
2
2
2 °
»¼ ® «¬ 2a 4
»¼ ¾
° ¬« 2a 4
¯
¿
¯
¿
Application of the Laplace Transform to LTI Differential Systems:
Electrical Circuits and Mechanical Systems
To obtain the value of the undetermined constants,
we
employ
y' L
the
unused
0 and y '' L
boundary
453
y ' 0 and y '' 0 ,
conditions
at
x
L, , i.e.
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
d2y 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
bu t
(6.1)
Define the following useful set of state variables
y t
x1 t y t
x1 t
x2 t
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
(6.2)
(6.3)
Chapter 6
456
ª ˜ º
« x1 t »
« ˜ »
«¬ x2 t »¼
y t
ª 0
«
¬ a2
1 º ª x1 º ª0 º
u t
a1 »¼ «¬ x1 »¼ «¬b »¼
(6.3a)
ªx t º
>1 0@ « x1 t »
¬ 2
¼
(6.3b)
Eqs. (6.3a) and (6.3b) can be written more compactly as
X t
A X t Bu t
y t
CX t
(6.4)
In general, the state equations of a system can be described by
X t
A X t Bu t
y t
C X t Du t
The vector X
(6.5)
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 Bu t
Multiplying both sides by e
ª
º
e At « X t A X t »
¬
¼
d
ª e At X t º¼
dt ¬
At
, we get
e At B u t
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
hom ogenous
solution
0
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
L1 > sI A@
1
L1 > sI A@
§ At A2t 2 A3t 3
·
............ ¸
¨1 1!
2!
3!
©
¹
e At
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 Bu t
y t
C X t Du 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
d3y t
dt
3
2
d2y t
dt
2
3
dy t
dt
4y 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
d2y 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 becomes
y t
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 Vc1 , i1 and Vc 2
as state
variables and 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 Vc1 and current i1 , we obtain
dx1 t
dt
x2 t
Kirchhoff’s voltage equation around the closed-loop gives
Chapter 6
462
Vs x1 t dx2 t
x3 t
dt
0
This equation can be rewritten as
dx2 t
dt
x1 t x3 t Vs
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
463
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 becomes
y t
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
s4
2
.
We next calculate the adjoint of this matrix
sI A .
ª s 4 1 º
« 0
s 4 »¼
¬
Chapter 6
464
Adj sI A
1 º
ªs 4
.
« 0
s 4 »¼
¬
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
»
s4 »
1 »
»
s4 ¼.
2
Taking the inverse of the elements of this matrix, we find that
e At
I t
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 º ª0 º
« 1 1» « x » «1 » u t with X 0
¬
¼¬ 2 ¼ ¬ 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
ª s 0 º ª 1 1 º
«0 s » « 1 1»
¬
¼ ¬
¼
1º
> sI A@ s «0 1 » « 1 1»
¬
¼ ¬
The determinant of
¼
ª s 1 1 º
« 1 s 1»
¬
¼
sI A is given by
s 1 1
1
s 1
sI A
sI A ,
2
s 1 1
.
We next calculate the adjoint of this matrix
Adj sI A
ªs 1 1 º
« 1 s 1» .
¬
¼
The inverse matrix is the adjoint matrix divided by the determinant
1
> sI A@
ªs 1 1 º
«
»
s 1 1 ¬ 1 s 1¼
adj sI A
det sI A
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
Xh 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
X nh t
t O
­
cos t O
°ªe
³0 ® «« e t O sin t O
°
¯¬
X nh t
­ ª t O sin t O
°« e
³0 ®« e t O cos t O
°«
¯¬
X nh t
ª t t O
º
sin t O d O »
« ³ e
« 0
»
« t
»
« ³ e t O cos t O d O »
«¬ 0
»¼
VC t
x2 t
t
t
e t O sin t O º ª0 º ½
°
» « » ¾ dO
t O
cos t O »¼ ¬1 ¼ °
e
¿
º½
» ° dO
»¾
»¼ °¿
t
³e
t O
cos t O d O
0
t
VC t
³e
t O
ª¬cos t cos O sin t sin O º¼ d O
0
t
VC t
t
O
e cos t ³ e cos O d O e sin t ³ eO sin O d O
t
0
t
0
.
Application of the Laplace Transform to LTI Differential Systems:
State-space Analysis
467
Integration by parts gives
t
VC t
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
dt
di1
4 i1 i2
dt
V
12 Ÿ i1' 4i1 4i2 12
(6.10)
Loop-2:
1
i2 dt R2i2 R1 i2 i1
C³
0
1
i2 dt 6i2 4 i2 i1
0.25 ³
0
4 ³ i2 dt 10i2 4i1
0
4i2 10i2' 4i11
0 Ÿ i2'
differentiating , we get
0.4i1' 0.4i2
Ÿ i2'
0.4 4i1 4i2 12 0.4i2
Ÿ i2'
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
469
4 º ªi1 º ª12 º
ª 4
« 1.6 1.2 » «i » « 4.8 » u t
¬
¼¬ 2 ¼ ¬ ¼
AI t Bu 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
ª1 0 º ª 0
1º
> sI A@ s «0 1 » « 2 3»
¬
¼ ¬
¼
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
Adj sI A
ª s 3 1º
« 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@
det sI A
ª s 3 1º
s 1 s 2 ¬« 2 s ¼»
ª
s3
«
« s 1 s 2
«
2
«
¬« s 1 s 2
º
»
s 1 s 2 »
»
s
»
s 1 s 2 ¼»
.
adj sI A
1
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
«
s2
¬ s 1
1
1 º
s 1
s2 »
»
1
2 »
»
s 1
s2 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
471
ª 2e t e 2t º
«
t
2 t »
¬ 2e 2e ¼ .
X t
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
dt
x1 t x2 t
.
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
dt
x1 t x2 t Vs t
.
The voltage Vc
y t
t is taken as the output 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
473
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 becomes
y t
ªx t º
>0 1@ « x1 t » .
¬ 2
¼
Example 6.8. Find the matrix exponential of A
ª1 1º
« 2 2 » .
¬
¼
Solution:
To find the state transition matrix, we first calculate the matrix
ª1 0 º ª1 1º
ª s 0 º ª1 1º
«0 s » « 2 2 »
¬
¼ ¬
¼
> sI A@ s «0 1 » « 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
Adj sI A
ª s 1 1 º
« 2 s 2» .
¬
¼
The inverse matrix is the adjoint matrix divided by the determinant
Chapter 6
474
1
> sI A@
1 ª s 2 1 º
s 1¼»
s s 1 ¬« 2
adj sI A
det sI A
1
> sI A@
ª s2
«
« 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 1
¬s
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 º ª0 º
« 6 5» « x » «1 » u t with X 0
¬
¼¬ 2 ¼ ¬ 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
1º
> sI A@ s «0 1 » « 6 5»
¬
¼ ¬
The determinant of
ª s 1 º
«6 s 5»
¬
¼
sI A is given by
s 1
2 s3
sI A
¼
ª s 0º ª 0 1 º
«0 s » « 6 5»
¬
¼ ¬
¼
sI A ,
s2 s3
.
We next calculate the adjoint of this matrix
Adj sI A
ª s 5 1º
« 6 s » .
¬
¼
The inverse matrix is the adjoint matrix divided by the determinant
1
> sI A@
1
> sI A@
det sI A
ª s 5 1º
s 1 s 2 «¬ 6 s »¼
ª
s5
«
« s2 s3
«
6
«
«¬ s 2 s 3
º
»
s2 s3 »
»
s
»
s 2 s 3 »¼
.
adj sI A
1
1
Expanding each term in the matrix on the right by partial fractions yields
Chapter 6
476
2
ª 3
« s2 s3
«
« 6
6
«
s3
¬ s2
1
> sI A@
2
3 º
s2
s3 »
»
2
3 »
»
s2
s3 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
1
> sI A@
ª 3e 2t 2e 3t
«
« 6e 2t 6e 3t
¬
º
»
2 t
3t »
2e 3e ¼
e 2t e 3t
.
The solution of the homogeneous part is
At
ª 3e 2t 2e 3t
«
« 6e 2t 6e 3t
¬
Xh t
e X 0
Xh t
ª e 2t e 3t º
«
»
« 2e 2t 3e 3t »
¬
¼.
º ª0 º
»
2 t
3t » « 1 »
2e 3e ¼ ¬ ¼
e 2t e 3t
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
Application of the Laplace Transform to LTI Differential Systems:
State-space Analysis
2 t O
­ª
2e 3 t O
° « 3e
³0 ®« 6e2 t O 6e3 t O
°«
¯¬
e 2 t O e3 t O
t
X nh t
2e2 t O 3e3 t O
X nh t
2 t O
­ª
e 3 t O
°« e
³0 ®« 2e2 t O 3e3 t O
°«
¯¬
X nh t
ª
«
«
«
«
«¬
t
X nh t
t
³e
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
477
½
º
» ª0 º ° d O
» «¬1 »¼ ¾
°
»¼
¿
º½
» ° dO
»¾
»¼ °¿
º
»
»
»
dO »
»¼
.
Thus, the complete solution is
ª e 2t e 3t º ª 1 1 e 2t 1 e 3t º
»
«
» «2 2
3
»
« 2e 2t 3e 3t » «
2 t
3t
e e
¬
¼ ¬
¼
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
.
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 º ª1 º
« 1 3» « x » «1 » u t with X 0
¬
¼¬ 2 ¼ ¬ ¼
ª1 º
«1 »
¬ ¼
Solution:
To find the state transition matrix, we first calculate the matrix
ª1 0 º ª 1
> sI A@ s «0 1 » « 1
¬
¼ ¬
The determinant of
ª s 0 º ª 1 0 º
«0 s » « 1 3»
¬
¼ ¬
¼
ªs 1 0 º
« 1 s 3»
¬
¼
sI A is given by
s 1 0
1 s 3
sI A
0º
3»¼
sI A ,
s 1 s 3
.
We next calculate the adjoint of this matrix
Adj sI A
ªs 3 0 º
« 1 s 1» .
¬
¼
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»¼
s 1 s 3 «¬ 1
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 »
»
s3 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
« «
»
s3 ¼
«¬ 2 ¬ s 1
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
«
« 1 e t e 3t
¬« 2
The solution of the homogeneous part is
ª
et
«
« 1 e t e 3t
¬« 2
Xh
e At X 0
Xh
ª
º
et
«
»
« 1 e t e 3t »
¬« 2
¼» .
0 º
» ª1º
« »
e 3t » ¬1¼
»¼
0 º
»
e 3t »
¼» .
Chapter 6
480
The solution of non-homogeneous part is (alternating approach)
X Nh
^
1
L1 > sI A@ b U s
1
> sI A@ bU s
1
> sI A@ bU s
§
¨U s
©
`
­ª
1
°«
s 1
°«
®«
°« 1 ª 1 1 º
»
°« 2 « s 1
s3 ¼
¯¬ ¬
L u t
1·
¸
s¹
º ½
» °
» ª1º ° 1
« »¾
1 » ¬1¼ ° s
»
s 3 »¼ °¿
0
1
ª
º
«
»
s s 1
«
»
«1 ª 1
»
1 º
« «
»»
«¬ 2 ¬ s s 1 s s 3 ¼ »¼
.
Expanding each term in the matrix on the right by partial fractions yields
1
> sI A@ bU s
X nh
X t
1
L
1
1
ª
º
«
»
s s 1
«
»
Ǥ 5 1 1 1
1 1 ·»
Ǭ
¸»
«¬¨© 6 s 2 s 1 6 s 3 ¸¹ »¼
^> sI A@ bU s `
1
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 h t X nh t
X t
ª 1 º
«1
»
« e 3t »
¬3
¼.
Example 6.11. Find the matrix exponential of A
ª0 1º
« 8 6 » .
¬
¼
Solution:
To find the state transition matrix, we first calculate the matrix
ª1 0 º ª 0
1º
ª s 0º ª 0 1 º
«0 s » « 8 6 »
¬
¼ ¬
¼
> sI A@ s «0 1 » « 8 6»
¬
¼ ¬
The determinant of
ª s 1 º
«8 s 6 »
¬
¼
sI A is given by
s 1
8 s6
sI A
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
« s2 s4
«
«
8
«
¬ s2 s4
1
º
s2 s4 »
»
»
s
»
s2 s4 ¼
.
Expanding each term in the matrix on the right by partial fractions yields
1
> sI A@
ª 2
1
«
s4
« s2
« 4
4
«
s4
«¬ s 2
1§ 1
1 ·º
¨¨
¸»
2© s2
s 4 ¹¸ »
»
2
1
»
s2
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
2º
> sI A@ s «0 1 » « 2 5»
¬
¼ ¬
¼
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 ¹¼ .
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 Bu t
y t
C X t Du 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 0º ª 0 1 0 º ª s 0 0º ª 0 1 0 º ª s 1 0 º
s ««0 1 0»» «« 0 0 1 »» ««0 s 0 »» «« 0 0 1 »» ««0 s 1 »»
¬«0 0 1 ¼» ¬« 1 2 3¼» ¬«0 0 s ¼» ¬« 1 2 3¼» ¬«1 2 s 3¼»
> sI A@
The determinant of
sI A is given by
s 1 0
0 s
1
1 2 s3
sI A
s 3 3s 2 2s 1
.
The inverse matrix is the adjoint matrix divided by the determinant
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
¬
adj sI A
1
> sI A@
det sI A
Substituting
sI A
1
, B, C and D into Eq. (6.14), where
B
ª1 º
«0 »
« »
«¬0 »¼
C
>1 0 0@
D
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 2 3s 2
H S
s 3 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 º ª1 º
« 1 1» « x » «0 » u t
¬
¼¬ 2 ¼ ¬ ¼
with the system output y t ,
y t
ªx t º
>1 0@ « x1 t » .
¬ 2
¼
Solution:
We have the general state and output equations are
X t
A X t Bu t
y t
C X t Du 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
1º
> sI A@ s «0 1 » « 1 1»
¬
¼ ¬
¼
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
« 1 s 1» .
¬
¼
Adj sI A
The inverse matrix is the adjoint matrix divided by the determinant
1
> sI A@
adj sI A
det sI A
Substituting
B
sI A
ª1 º
«0 »
¬ ¼,
and D
1
C
ªs 1 1 º
1
«
»
s 2 2s 2 ¬ 1 s 1¼
, B, C and D into Eq. (6.15), where
>1 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
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 y t
transform Y
with an algebraic equation in its
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
L ^ut x, t ` e st u x, t
f
0
f
s ³ e st u x, t d t
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
f
­ 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
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
` dxd ©¨§ dxd U x, s 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 t , we obtain
dU x, s
dx
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)
495
Because only derivatives with respect to x remain, we replace the partial
derivative with an ordinary derivative,
dU x, s
2 s 1 U x, s
dx
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 is the general solution of the homogeneous
differential equation
C .F U h x , s
c1 e 2 s 1 x
and the particular integral is
P.I1
12
P.I1 6
CF and particular
1
e 3 x replace D a
D 2s 1
D 3
1
e 3 x
2s
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 .
2s
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
2s
.
Taking the inverse Laplace transform, we get
6 e 2t 3 x .
6e2t e 3 x
u x, t
Example 7.2. Solve the initial-boundary value problem
wu x, t
wx
wu x, t
wt
x, x ! 0, t ! 0
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 U
497
x, s is regarded as a
function of x alone, s 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
sx
x
³ e s dx C
e sx
s
sx
³ e x dx C e
sx
We can use integration by parts to evaluate the integral
U x, s
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 Ÿ 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
xt u x, t
tx
2
2
t2
u tx .
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 0, t
u x,0
0,
wu x,0
wt
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 U
function of here
x
alone,
s
499
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 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.I1
2
P.I1
2
P.I 2
4
P.I1
4
1
sin S x
2
D s2
replace D 2
S 2
1
sin S x
S s2
2
1
sin 3S x
2
D s2
1
sin 3S x
9S s 2
2
Now, we have the general solution
U x, s
D2
a 2
U h x, s U Nh x, s
replace D 2
D2
a 2
9S 2
Chapter 7
500
c1 e s x c2 e s x U x, s
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 Ÿ U 0, s
0 and u 1, t
0 Ÿ U 1, s
0
so we have
U 0, s
0 c1 c2 and U 1, s
Solving for c1 and c2, we get
c1 0
0 c1 e s c2 e s
.
c2
therefore,
2
4
sin S x sin 3S x
2
2
9S s 2
S s
U x, 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 u
w 2 u x, t
wt 2
c2
w 2 u x, t
wx 2
g,
x, t of points in the string is
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,
where g
9.81, and c ! 0 is a constant depending on the material
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
d 2U x, s
dx 2
s 2U x, s s u x, 0 ut x, 0 s2
2 U x, s
c
g
s
g
2
cs
This is an ordinary differential equation if U
x, s is regarded as a
function of x alone, s being a parameter. The transformed differential
equation is a linear second-order ODE
§ 2 s2 ·
¨ D 2 ¸ U x, s
c 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
P.I
U Nh x, s
g
1
e0 x
2
2
c s § 2 §s· ·
¨¨ D ¨ ¸ ¸¸
replace D 0
©c¹ ¹
©
g
3
s
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 Ÿ U 0, s
0 and u 1, t
0 Ÿ U 1, s
0
For this function to remain bounded as x o f, we must set c1
which case condition (7.4) implies that
Ÿ c2
Thus,
g
s3
0, in
Laplace Transform Methods for Partial Differential Equations (PDEs)
503
g sx c
1
e
s3
.
U x, s
Taking the inverse Laplace transform, we have
2
g § 2 § x · § x ··
¨ t ¨ t ¸ u ¨ t ¸ ¸¸ .
2 ¨©
© c ¹ © c ¹¹
u x, t
Example 7.5. Solve the initial-boundary value problem
wu x, t
wx
wu x, t
wt
2u x, t
0, x ! 0, t ! 0
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
0
sin x
This is an ordinary differential equation if U
x, s is regarded as a
function of x alone, s being a parameter. The transformed differential
equation is a linear first-order ODE.
Compared to the standard form
Chapter 7
504
dU
PU
dx
Q
s 2 and Q sin x .
here, P
We solve it by finding the integrating factor
I .F
e³
s 2 dx
e s2 x .
The formula for a solution is
³ Q I .F . dx C
U I .F
U x, s e s 2 x
U x, s
³e
s2 x
sin x dx C
e s 2 x ³ e s 2 x 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 Ÿ 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
° s2 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
e 2t ª¬sin x t sin t x u t x º¼ .
u x, t
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
u x,0
0,
0
wu x,0
wt
sin 2S x , 0 x 1.
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 U
x, s is regarded as a
function of x alone, s 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 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
P.I
U Nh x, s
U Nh x, s
1
sin 2S x
D s2
replace D 2
2
D2
a 2
4S 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 Ÿ U 0, s
so we have
0 and u 1, t
0 Ÿ U 1, s
0
Laplace Transform Methods for Partial Differential Equations (PDEs)
U 0, s
0 c1 e s c2 e s
0 c1 c2 and U 2, 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
u x, t
2
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 u
w 2 u x, t
w 2 u x, t
wt 2
wx 2
,
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
dx 2
d 2U x, s
dx 2
sU x, s u x,0
sU x, s
0
.
This is an ordinary differential equation if U
x, s is regarded as a
function of x alone, s 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 s x c2 e s x
.
Now we have the general solution
U x, s
c1 e s x c2 e s x
.
We can evaluate the constants c1 and c2 using the boundary condition
Laplace Transform Methods for Partial Differential Equations (PDEs)
u 0, t
509
1
s
1Ÿ U 0, s
so we have
U 0, s
u0
s
c1 c2
.
For this function to remain bounded as ‫ ݔ‬՜ ҄, we must therefore set
c1
0, Ÿ c2
U x, s
1
s
1 sx
e
s
Taking the inverse Laplace transform, we have
u x, t
§ x ·
erfc ¨
¸.
© 2 kt ¹
where erfc (x) is the complementary error function
erfc x
1 erf x
1
2
x
W 2
e dW
S ³
0
2
f
W 2
e dW .
S ³
x
Example 7.8. Solve the initial-boundary value problem
w 2 u x, t
wt 2
c
2
w 2 u x, t
, 0 x L, t ! 0
wx 2
with boundary and initial conditions
u 0, t
0, u L, t
0
Chapter 7
510
§ S · wu x, 0
A sin ¨ x ¸ ,
wt
©L ¹
u x, 0
0, 0 x 1.
Solution:
Taking the Laplace transform and applying the initial condition
d 2U x, s
dx
2
2
§s·
¨ ¸ U x, s
©c¹
s
§S ·
2 A sin ¨ x ¸
c
©L ¹
This is an ordinary differential equation if U
x, s is regarded as a
function of x alone, s being a parameter. The transformed differential
equation is a linear second-order ODE
§ 2 § s ·2 ·
¨¨ D ¨ ¸ ¸¸ U x, s
©c¹ ¹
©
s
§S ·
2 A sin ¨ x ¸
c
©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 is the general solution of the homogeneous problem
C .F
U h x, s
c1 e s c x c2 e s c x
and the particular integral is
Laplace Transform Methods for Partial Differential Equations (PDEs)
P.I
U Nh x, s
replace D 2
D2
P.I
511
s
1
§S ·
2A
sin
¨ x¸
c § 2 § s ·2 · © L ¹
¨¨ D ¨ ¸ ¸¸
©c¹ ¹
©
a 2
S2
L2
U Nh x, s
As
§S ·
sin ¨ x ¸
2 2
2
s c S L
©L ¹
2
.
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 Ÿ U 0, s
0 Ÿ U 1, s
0 and u 1, t
give in turn c1 = 0 = c2.
Therefore,
U x, s
As
§S ·
sin ¨ x ¸
2 2
2
s c S L
©L ¹
2
.
0
Chapter 7
512
Taking the inverse Laplace transform, we have
u x, t
§Sc ·
§S ·
A cos ¨
t ¸ sin ¨ x ¸ .
© L ¹
©L 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
dx 2
d 2U x, s
dx 2
sU x, s u x,0
sU x, s
3sin 2S x
This is an ordinary differential equation if U
x, s is regarded as a
function of x alone, s being a parameter. The transformed differential
equation is a linear second-order ODE
D 2 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 s x c2 e s x
and the particular integral is
P.I
P.I
U Nh x, s
3
U Nh x, s
3
1
sin 2S x
D2 s
1
2
4S s
replace D 2
D2
a 2
4S 2
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 s x c2 e s x 3
1
4S 2 s
sin 2S x
.
We can evaluate the constants c1 and c2 using the boundary condition
u 0, t
0 Ÿ U 0, s
0 and u 2, t
0 Ÿ 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
2
4S s
sin 2S x
.
Taking the inverse Laplace transform, we have
2
3e 4S t sin 2S x .
u x, t
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
u x,0
0, ut x,0
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 U
515
x, s is regarded as a
function of x alone, s 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
P.I
U Nh x, s
U Nh x, s
1
1
sin S x
2
s D s2
replace D 2
D2
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
.
a 2
S 2
Chapter 7
516
We can evaluate the constants c1 and c2 using the boundary condition
u 0, t
0 Ÿ U 0, s
0 and u 1, t
0 Ÿ U 1, s
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
º
1 ª1
1
«
» sin S x
S 2 « s S 2 s2 »
¬
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 ¬
.
0
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 u
wu x, t
k
wt
w 2 u x, t
wx 2
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 U
x, s is regarded as a
function of x alone, s being a parameter. The transformed differential
equation is a linear second-order homogeneous ODE
§ 2 s·
¨ D ¸ U x, s
k¹
©
0
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 Ÿ 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, Ÿ c2
u0
s
s
U x, s
u0 k x
e
s
Taking the inverse Laplace transform, we have
Laplace Transform Methods for Partial Differential Equations (PDEs)
519
§ x ·
u0 erfc ¨
¸.
© 2 kt ¹
u x, t
where erfc x is the complementary error function
1 erf x
erfc x
x
2
1
2
2
W
³ e dW
f
W 2
e dW .
S ³
S 0
x
Example 7.12. Use the Laplace transform method to find the solution
of the modified wave equation
w 2 u x, t
wx
c2
2
w 2 u x, t
wt
2
2cO
that remains finite for
u x, 0
0
u 0, t
sin t for t ! 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
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 U
x, s is regarded as a
function of x alone, s being a parameter. The transformed differential
equation is a linear second-order homogeneous ODE
D 2 cs O
2
U x, s
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
c1 e cs O x c2 e cs O x
U x, s
(7.5)
We can evaluate the constants c1 and c2 using the boundary condition
u 0, t
sin t Ÿ U 0, s
and require c1 = 0 to have U
Further,
Ÿ c2
Thus,
1
s 1
2
1
c1 c2
s 1
2
x, s o f as ‫ ݔ‬՜ ҄.
Laplace Transform Methods for Partial Differential Equations (PDEs)
U x, s
1
e cs O x
s 1
U x, s
1 ½
­
e O x ® e cs x 2
¾
s 1 ¿ .
¯
521
2
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
f t
f t
f t
3
4
5
f t
2
季
1
2
4cos 3t 5sin 3t e
2 t
3cosh 6t 8sinh 3t
te 3t cos 2t
e
iwt
1 e 3t
Ans Ÿ
Ans Ÿ
F s
F s
F s
ŸF s
Ans Ÿ
Ans
Ans
ŸF s
2
.
3s
24
.
2
s 36 s 9
4 s 23
.
2
s 4 s 13
2
s 2 6s 13
s 2 6s 5
1 2
1
.
s s s6
s iw
.
s 2 w2
Find the Laplace transforms of the following functions.
EXERCISES AND ANSWERS
CHAPTER 8
§ S·
cos ¨ t ¸
© 6¹
f t
f t
11
12
sin t
sin 4t 5
f t
10
dW
t 2 cos at
W
sin W
sin at
t
0
W
³e
t
t cos wt
f t
f t
f t
f t
9
8
7
6
Ans
ŸF s
Ans Ÿ
F s
F s
ŸF s
ŸF s
ŸF s
Ans Ÿ
Ans
Ans
Ans
Ans
ŸF s
Exercises and Answers
2
.
3
.
S 1 4s
e
.
2 s3 2
1 3s 1
.
2 s2 1
4 cos 5 s sin 5
.
s 2 16
s2 a2
2 s s 2 3a 2
§a·
tan 1 ¨ ¸ .
©s¹
cot 1 s 1
.
s
s 2 w2
s 2 w2
523
17
f t
f t
f t
15
16
f t
Chapter 8
0dt da
; f t 2a
a d t d 2a
with
­2 0 d t 1
°
®1 1 t d 3
°0 3 t d 4
¯
a
t with f t b
b
with
­3 0 d t 1
®
¯1 1 t d 2
f t
f t .
f t .
f t
f t4
f t2
f t
Ans
Ans
Ans
Ans
Ans
ŸF s
ŸF s
ŸF s
ŸF s
ŸF s
aª1
1 º
« bs
».
s « bs e 1 »
¬
¼
1
2 e s e3s
4 s
s 1 e
1
3 2e s e2 s
2 s
s 1 e
1
.
s 1
2
K
§ as ·
tanh ¨ ¸ .
s
© 2¹
Find the Laplace transforms of the following periodic functions.
sin t with f t 2S
­K
f t ®
¯ K
14
13
524
22
21
20
19
18
f t
f t
f t T
f t
f t
T
2
T
dt dT
2
with
0dt Ans
Ans
Ans
ŸF s
ŸF s
ŸF s
2
tan Ts .
4
Ts 2
1
.
s 1 1 e S s
2
1
coth S s .
2
s 1
2
­t
®
¯0
0 d t d1
t !1
sin 2t u t S
Ans
Ans
ŸF s
ŸF s
1
e s
s
1 e .
s2
s
2e S s
.
s2 4
Find the Laplace transforms of the following functions by expressing in to unit step function.
f t .
­ 2t
°° T
®
°2 ¨§1 t ¸·
°¯ © T ¹
f t .
­ I m sin t 0 d t S
®
S d t d 2S
¯0
with
sin t with f t S
f t 2S
f t
f t
Exercises and Answers
525
27
26
25
24
23
526
f t
f t
f t
f t
f t
0 d t d1
1 t d 2
t!2
Ans
Ans
­1 0 d t 1
°
®2 1 t d 4
°1
t!4
¯
­2 t
°
®6
°t 4
¯
Ans
Ans
Ans
­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
ŸF s
1
1 e s e 4 s
s
1 s
e e 3 s
s
2
1
1 e 2S s .
s 1
s 1
e s 2 °½
°­1 e
®
¾.
s 2 °¿
°¯ s 1
2 1 s § 3 1 · 3s § 1 1 ·
e ¨ ¸ e ¨ ¸.
s s2 © s s2 ¹ © s s2 ¹
ŸF s
ŸF s
ŸF s
ŸF s
34
33
32
31
30
29
28
F s
F s
F s
F s
F s
F s
F s
2
2
3s
.
12s 48s 36
3
1
s s 1
s6
s 4 s 1 s 2
36
s s 1 s2 9
2
s 1 s 1
4s
1 2s
.
2
s 4s 5
e 2S s
.
s s2 1
Ans
Ÿf t
f t
Ÿf t
Ÿf t
Ÿ f t
Ÿf t
Ÿf t
Ans Ÿ
Ans
Ans
Ans
Ans
Ans
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
4 cos t cos3t.
2
2
et e t 2tet .
e2t 5sin t 2cos t .
1 cos t u t 2S .
Find the inverse Laplace transforms of the following functions.
Exercises and Answers
527
F s
39
41
F s
F s
F s
38
40
F s
F s
F s
37
36
35
528
6
2
Ÿ f t
Ÿ f t
§ s 1 ·
log ¨
¸
© s 1 ¹
6 s 2 50
s 3 s2 4
Ans
Ans
f t
f t
Ÿf t
Ÿ f t
Ans Ÿ
s2 6
s2 1 s2 4
s2 1
Ans Ÿ
f t
Ans Ÿ
Ans
Ans
2
s3
s 1 s 2 3s 2
1
s 1 s 4
2
2s
s2 s3 s6
Chapter 8
1
t sin t.
2
8e3t 2cos 2t 3sin 2t.
2sinh t
.
t
5
1
sin t sin 2t.
3
3
cos t e2t et tet .
2sin t 2sin 2t.
e 2 t 2e 3 t e 6 t .
45
45
44
43
42
F s
F s
F s
F s
F s
1
.
s s7
s2
.
s 2s 1
2
s2
.
s s2
2
Ans
Ans
Ÿf t
Ÿ f t
9t 2 2 2cosh 3t.
2 cos t t 2 2.
Ans Ÿ Initial value
Ans Ÿ Initial value
Ans Ÿ 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.
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
the
solution
2 2
2 2
s s2 a
1
s O
2
1
s2 s2 1
of
s 1 s 3
1
1
s s 1
the
1
integro-differential
dy
y W cos t W dW 0, y 0
dt ³0
Find
F s
F s
F s
F s
F s
equation
Ans
Ans
Ans
Ans
Ans
Ans
Ÿ
y t
Ÿf t
Ÿf t
Ÿ f t
Ÿ f t
Ÿ f t
1
1 2
t .
2
2 2cos at at sin at
.
2a 4
sin Ot Ot cos Ot
.
2O 3
t sin t .
1 3t t
e e .
4
et 1.
Evaluate the inverse Laplace transforms of the following functions using the convolution theorem.
Chapter 8
58
57
56
55
0
7
dt
dc t
dt
dy t
4y t
W
6c t
2y t
x t
1
Ans
Ans
Ÿ y t
Ÿ
i t
C
§ 1 ·
sin ¨
t¸
L
© LC ¹
dt
du t
dt
dx t
11u t .
x t .
dt
2
d 2c t
ŸH s
ŸH s
Ans Ÿ
Ans
Ans
Ans
ŸH s
6
dt
dc t
C s
U s
Y s
X s
s 11
dt
du t
.
.
u t .
s 7s 6
2
s2 s 2
s 1
2c t 2
Y s
X s
1
.
s4
2 t 3et tet .
V
Find the transfer function of the following differential equations
³ t W e dW 0, y 0
t
Find the DE corresponding to the transfer function
dt
2
dc t
2
dt
2
d y t
2
dt
dy t
dy
y
dt
Find the solution of the integro-differential equation
1
54
t
di 1
L ³ i W dW V , i 0
dt C 0
53
Find the solution of the integro-differential equation
Exercises and Answers
531
2s 1
y 0 1 & y' 0
1
8et sin 2t
y '' t 2 y ' t 5 y t
4
with
62
y 0 1 & y' 0
9e2t
y 5 y '' 2 y ' 8 y sin t with y 0
with
'''
6
8sin 2t
3 & y' 0
y '' t 4 y t
y 0
.
Chapter 8
y ' 0 , y '' 0
with
1
2t 1 cos 2t 3sin 2t
et 3t 2 e 2t .
1 t 2 2t 8 4t
e e e
3
15
85
Ÿy t
Ÿy t
Ÿy t
2t 1 et cos 2t.
1
cos t 13sin t
170
Ans
y t
Ans
Ans
Solve the following differential equations using the Laplace transform.
s 6s 2
2
y '' t y ' t 2 y t
H s
61
60
59
532
67
66
65
64
63
6 & y' 0
1 & y' 0
2
1 & y ' 0
L
2
0
4e t 2 e t
G t 2
et u t 2
54te2t
AG t t0 , i 0
1 & y' 0
di
Ri
dt
y 0
y '' t 2 y ' t y t
y 0
y '' t 5 y ' t 6 y t
y 0
1
0
y '' t 3 y ' t 2 y t
y 0
y '' t y ' t 2 y t
with
with
with
with
Exercises and Answers
i t
Ÿy t
Ans Ÿ
Ans
u t2 .
A
e
L
L
R t t0
u t t0 .
et t 2 5t 2 et .
y t 3e3t 4e2t e2 t 2 e3 t 2 u t 2 .
2 t 1
6et 9t 2 6t 1 e2t .
Ÿ y t e 2 t 1 t et e
Ÿy t
Ans Ÿ
Ans
Ans
533
72
71
70
69
68
534
wt
. Find the
0
x' 0
x t
2
§ w
·
f0
sin w0t sin wt ¸ .
2 ¨
m w w0 © w0
¹
Ans Ÿ
0&x 0
0
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
dx
dy
2 x 5 y 5et cos t ; x 2 y 10et sin t
dt
dt
with
y t
2et 4t 3 2e3t 8e2t
5et 1 cos t sin t
5et 1 cos t
6t 1 4e3t 8et 8e3t
3t 1 et e 2t 2e3t .
3t 2 2et 3e2t 8e3t .
y t
z t
x t
Ans Ÿ
et 12t 13 e3t 16e2t .
Ans Ÿ
x t
y t
x t
Ans Ÿ
.
Solve the following simultaneous differential equations using the method of Laplace transform.
x 0
force and one with the frequency w0 of the free oscillator, assuming
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
An undamped oscillator is driven by force f 0 sin
Chapter 8
75
74
73
dz
dt
8&z 0
2 y 3z,
8&z 0
3.
t, x 3.
with
dy
y
dt
2 y z
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
dx
dy
2
2y
dt
dt
y 0
dy
dt
Exercises and Answers
h t
y t
Ans Ÿ
z t
Ans Ÿ
3e4t 5et
6t
3 3 2t 1
e t
4 4
2
3 3 2t 1
e t
4 4
2
1 2t
e sin
6
x t
5et 2e4t .
y t
535
78
77
76
536
wu x, t
0, ux 0, t
x of
1 and lim ux x, t
wx 2
wt
u x,0
w 2 u x, t
wu x, t
0.
0.
0, x ! 0, t ! 0
0 lim u x x, t
u 0, t
x of
0
g,
5.
0, x ! 0, t ! 0
wx 2
ut x, 0 and
w 2 u x, t
wx
3, u 0, t
2
wt 2
u x, 0
w 2 u x, t
wt
u x, 0
wu x, t
with
with
Solve following initial-boundary value problem
Chapter 8
Ans Ÿ
u x, t
Ans Ÿ
Ans Ÿ
u x, t
0
³
t
2
1 4xW
e dW .
SW
x ct
x t ct
§ x·
3 2H ¨ t ¸ .
© 2¹
­ gt 2
°°
2
®
° g x 2 2cxt
°̄ 2k 2
u x, t
BIBLIOGRAPHY
1. Shaila Dinkar Apte. (2016) Signals and systems: principles and
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