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CHAPTER 3
LAPLACE TRANSFORM
Chapter Outline
3.0 Overview and Learning Outcomes
3.1 Introduction
3.2 Laplace transform
3.2.1 Definition of Laplace transform
3.2.2 Derivation of Transformations of Some Elementary Functions
3.2.3 Standard Table of Laplace transform
3.3 Properties of Laplace Transform
3.3.1 Linearity
3.3.2 First Shift Theorem
3.3.3 Multiplying by nt
3.4 Inverse Laplace transform
3.5 Properties of Inverse Laplace transform
3.5.1 Linearity Property of Inverse Laplace transform
3.5.2 First Shift Property 3.5.3 Inverse Laplace transform of Rational Functions
3.6 Discontinuous Functions and Periodic Functions
3.7 Application of Laplace transform
3.7.1 Initial Value Problems
3.7.1.1 First Order Linear Differential Equation
3.7.1.2 Second Order Linear Differential Equation
3.7.2 Electric Circuit Analysis
Practice 3 (Laplace transform)
3.0 OVERVIEW AND LEARNING OUTCOMES
Third chapter for this course is a Laplace transform. We will start the topic
with the introduction to Laplace transform and some derivation of elementary
functions. Student will be exposed to properties of Laplace transform and inverse
Laplace transform. It is very important to each student to understand deeply to both
properties studied. In order to learning smoothly, students are required to be familiar
with the provided of Laplace transform to make learning process getting easier. The
following subtopic is the application of Laplace transform in solving initial value
problems and analysis of electrics circuit.
It is expected that at the end of this course students will be able to:
• Derive the Laplace transform and Inverse Laplace transform.
• Use the definition of Laplace transform to find the Laplace transform.
• Apply the First Shift Theorem and First Shift Theorem (Inverse).
• Apply Laplace transform in solving differential equation.
• Apply Laplace transform in the daily physical problems
79
3.1 INTRODUCTION
In reality, Laplace transform has many applications in science and engineering. It has
important applications in mathematics, mechanical vibrations, electrical circuits and
control engineering. The main idea behind the Laplace transformation is that we can
solve an equation containing differential equation and integral terms by transforming
the equation in t-domain to one in s-domain with the intention that the problem easier
to solve. In mathematics, it is used for solving ordinary differential equation and
integral equations. We begin our discussion with the definition of Laplace transform.
3.2 LAPLACE TRANSFORM
3.2.1 DEFINITION OF LAPLACE TRANSFORM
Since the integral in an improper integral, by definition evaluated according to the
rule
( ) ( ) ( )dttfedttfetf
T
st
T
st ∫∫ −
∞→
∞− ==
00
limL .
Since the result of integral (3.1) depend on s, we can write
( ) ( ) ( ) ( )sFdttfedttfetf
T
st
T
st === ∫∫ −
∞→
∞−
00
limL .
In general, other examples can be written as
( ) ( )sFtf =L
( ) ( )sGtg =L
( ) ( )sYty =L .
Definition 3.1 (Definition of Laplace Transform)
Let ( )tf be a function defined over [0,∞). The integral
( )dttfest∫
∞−
0
(3.1)
is called Laplace transform of ( )tf if the integral exist. We write
( ) ( )dttfetf st∫∞
−=0
L where L interpreted as an operator.
80
3.2.2 DERIVATION OF TRANSFORMATION OF SOME ELEMENTARY
FUNCTIONS
The Laplace transform for some elementary function stated as follows.
Example 3.1
Show thats
11 =L .
Solution
( ) ( ) )1(1lim1100
−−−−== ∫∫ −
∞→
∞− dtedte
T
st
T
stL
a) If )1(then,0<s gives ∞=
+−=
−=
−
∞→
−
∞→ ss
e
s
e sT
T
Tst
T
1limlim1
0
L
b) If )1(then,0=s gives ( ) [ ] ∞===∞→∞→ ∫
T
T
T
Ttdt 0
0
lim11lim1L
c) If )1(then,0>s gives sss
e
s
e sT
T
Tst
T
11limlim1
0
=
+−=
−=
−
∞→
−
∞→L
Thus, .1
1s
=L
Example 3.2
Using the definition, determine the Laplace transforms for the following function.
a) ( ) atf = b) ( ) ttf = c) ( ) nttf = d) ( ) atetf =
Let a be constant and n non negative integer.
Solution
a) ( ) .0,
000
>=
−===
∞−∞−
∞− ∫∫ s
s
a
s
eadteadtaea
stststL
b) ( ) .0,111
2
00000
>=
−=+
−===
∞−∞−
∞−∞−
∞− ∫∫∫ s
ss
e
sdte
ss
tedttedttet
stst
stststL
c) ( ) ∫∫∫∞
−−
∞−∞−
∞− +
−===
0
1
000
dttes
n
s
etdtetdttet nst
stnstnnstnL
. 1
0
1 −∞
−−
== ∫ nnst ts
ndtte
s
nL
From result (c), we can write 11 −−
= nnt
s
nt LL
81
.1
1
11
2
1
Then
2
32
s
st
ts
t
ts
nt nn
=
=
=
−= −−
L
LL
LL
LL
M
Substitute result above into , 1−
= nn ts
nt LL
Then
21
1
3
2
1
−
−
−
−
−
=
−
=
=
n
nn
nn
ts
n
s
n
s
n
ts
n
s
nt
ts
nt
L
LL
LL
.!
1!
1!
21
1
2
+
−
=
=
=
−
−
=
n
n
n
n
s
n
ss
n
s
n
ts
n
s
n
s
n
L
L
M
d) ( ) ( )( )
( ) ( ).,
1
000
asasas
edtedteee
tastasatstat >
−=
−−===
∞−−∞−−
∞− ∫∫L
Example 3.3
Using the definition, find the Laplace transforms for the following function
( ) attf cos= where a be constant.
82
Solution
( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( ) .cos1
cos01
cossin1
sincos
sincos
coscoscos
0
2
2
0
00
00
0000
dtates
a
sdtate
s
a
s
a
s
dtates
a
s
ate
s
a
s
dtates
a
s
ate
dtatas
e
s
atedtatedtateat
stst
stst
stst
stststst
∫∫
∫
∫
∫∫∫
∞−
∞−
∞−
∞−
∞−
∞−
∞ −∞−∞−
∞−
−=
+−=
+
−−=
−
−=
−−
−
−===L
.0, cos
1 cos1
Thus
22
2
2
>+
=
=
+
sas
sat
sat
s
a
L
L
Exercise
Show that .0, sin 22
>+
= sas
aatL
Example 3.4
Using the definition, find the Laplace transforms for ( ) ( )attf sinh= where a be a
constant.
Solution
( ) ( ) ( ) ( ) ( )
( ) ( )
+
−=
+
−===
∫
∫∫∫∞
−
∞−
∞−
∞−∞−
∞−
dtates
a
s
ate
s
a
dtates
a
s
atedtatedtateat
stst
stst
stst
00
0000
sinhcosh
coshsinh
sinhsinhsinhL
( ) ( ) .sinhsinh1
0
2
2
2
0
dtates
a
s
adtate
s
a
ss
a stst ∫∫∞
−∞
− +=
+
=
83
( ) ( )
( )
( ) .0, sinh
sinh1
sinhsinh
Thus
22
22
2
0
2
2
2
>−
=
=
−
+= ∫∞
−
sas
aat
s
aat
s
a
dtates
a
s
aat st
L
L
L
]Exercise
Show that .0, cosh 22
>+
= sas
satL
Example 3.5
Using the definition of the Laplace transformation, determine ( ) tfL if
( )
>
<≤=5,1
50,5
1
t
tttf
Solution (Recall: ( ) ( )dttfetfst∫
∞−=
0
L ). Then,
( )
( ).15
1
5
1
5
5
1
5
10
5
5
1
5
1
5
1
5
1
5
1
5
2
22
5
5
22
555
22
55
55
0
2
55
0
5
0
0
5
05
5
0
s
s
ssssss
sststsstst
stststst
es
ss
e
s
e
ss
e
s
e
s
e
ss
e
s
e
s
e
s
e
s
te
s
edt
s
e
s
te
s
edtedtedttetf
−
−
−−−−−−
−−−−−−
∞−−
∞−−
−=
+−=
++−−=+
−−
−−=
+
−−=+
−−
−=
−+=+=
∫
∫∫∫L
Exercises
1. By using definition, find the Laplace transform of the following function.
a) ( ) tetf 2= b) ( ) 2ttf = c) ( ) ttf 3cos= d) ( ) 3−=tf
Answer
1. a) ( )2
1
−=s
sF b) ( )3
2
ssF = c) ( )
92 +=s
ssF d) ( )
ssF
3−=
84
3.2.3 STANDARD TABLE OF LAPLACE TRANSFORM
Table of Laplace transform are shown in Figure 1. These are generally used after
analyzing some elementary function by using the definition.
Figure 1: Table of Laplace Transforms
( )tf ( ) ( )sFtf =L ( )tf ( ) ( )sFtf =L
a s
a bteat sin ( ) 22
bas
b
+−
,..3,2,1, =nt n 1
!+ns
n bteat cos ( ) 22
bas
as
+−
−
ate as −
1 atn et ( ) 1
!+− n
as
n
atsin 22 as
a
+ ( )tft n ( ) ( )[ ]sF
ds
dn
nn
1−
atcos 22 as
s
+ ( )tfeat ( )asF −
atsinh 22 as
a
− ( )ty′ ( ) ( )0yssY −
atcosh 22 as
s
− ( )ty ′′ ( ) ( ) ( )002 ysysYs ′−−
Exercises
1. By using the table of Laplace transforms, find the Laplace transforms for given
( )tf .
a) ( ) 5=tf b) ( ) tetf 3= c) ( ) ttf 3cos= d) ( ) 3ttf =
2. Using table of Laplace transforms, find ( ) tfL for given ( )tf .
a) ( ) ttf3
2sinh= b) ( ) ttf
4
5cosh= c) ( ) ttf 4sin= d) ( ) tetf 10−=
Answer
1 a) ( )s
sF5
= b) ( )3
1
−=s
sF c) ( )92 +
=s
ssF d) ( )
4
6
ssF =
2a) ( )49
62 −
=s
sF b) ( )2516
162 −
=s
ssF c) ( )
16
42 +
=s
sF d) ( )10
1
+=s
sF
85
3.3 PROPERTIES OF LAPLACE TRANSFORM
In this subtopic, we will discuss properties of Laplace transform such as
linearity property, First Shift property, multiplication by nt (Differentiation of a
transform) and Inverse Laplace transform. By using all this four properties, we would
be able to find Laplace transforms easily without doing tedious calculation. First, we
will discuss linearity property.
3.3.1 LINEARITY
Example 3.6
Find the Laplace transforms for given function.
a) ( ) tttf 72 3 +−= b) ( ) tetf t 611 3 +=
b) c) ( ) tetttf 5965cos2 −+−=
Solution
a) .712!1
7!3
272722424
33
sssstttt +−=
+
−=+−=+− LLL
b) .6
3
11!16
3
111611611
22
33
sssstete
tt +−
=
+
−
=+=+ LLL
c) 965cos2965cos2 55 tt ettett −− −+−=−+− LLLL
+
−
+
+
−=5
19
!16
252
22 sss
s
.5
96
25
222 +−+
+−=
sss
s
Example 3.7
If ( )atat eeat −+=2
1cosh , determine cosh atL .
Theorem 3.1 (Linearity Property)
If ( ) 1 tfL and ( ) 2 tfL exist, and if α and β are constants, then
( ) ( ) ( ) ( ) 2121 tftftftf LLL βαβα +=+
We say that the operator L is linear.
86
Solution
( )
22
2
2
11
2
11
2
1
2
1
2
1
2
1
2
1
2
1cosh
as
s
asas
eeeeeeat atatatatatat
−⋅=
+
+
−
=
+=
+=
+= −−− LLLLL
22 as
s
−=
Exercises
Find the ( ) tfL for given ( )tf .
1a) ( ) ( )21 ttf −= b) ( ) 672cosh2 ++= tettf
c) ( ) tettf t 7sinh6534 92 +−+−=
2. ( ) attf sinh= if ( )atat eeat −−=2
1sinh
Answer
1a) ( )32
221
ssssF +−= b) ( )
sss
ssF
6
1
7
4
22
+−
+−
=
c) ( )49
42
9
56423 −
+−
−+−=ssss
sF 2. 22 as
a
−
3.3.2 FIRST SHIFT THEOREM
Proof
( ) ( ) ( )sFtdttfetf st === ∫∞
−
0
L
Then
( ) ( )
( ) ( )
( )
( )( ).
then,Since,
,
0
0
0
asF
asppF
aspdttfe
dttfe
dttfeetfe
pt
tas
statat
−=
−==
−==
=
=
∫
∫
∫
∞−
∞−−
∞−L
Theorem 3.2 (First Shift Theorem)
If ( ) ( )sFtf =L and a is a constant, then ( ) ( )asFtfeat −=L .
87
Example 3.8
Find the Laplace transforms of the following functions.
a) ( ) 43 tetf t= b) ( ) tetf t 2cos3−=
Solution
a) Using Table 3.1, ( ) ( )sFss
ttf ====+ 514
4 24!4LL .
Then, ( )( )5
43
3
243
−=−=s
sFte tL
b) Using Table 3.1, ( ) ( )sFs
sttf =
+==
42cos
2LL .
Then, ( ) ( )( ) 136
3
43
332cos
22
3
++
+=
++
+=+=−
ss
s
s
ssFte tL
Exercises
Find the Laplace transforms of the following functions.
a) ( ) 35 tetf t= b) ( ) tetf t 3sin6−= c) ( ) tetf t 2sinh7−=
Answer
a) ( )( )4
35
5
65
−=−=
ssFte tL
b) ( )( ) 4512
3
96
363sin
22
6
++=
++=+=−
ssssFte tL
c) ( )( ) 4514
3
47
273sin
22
7
++=
−+=+=−
ssssFte tL
3.3.3 MULTIPLYING BY nt
Example 3.9
Find the Laplace transforms for given function.
Example 3.9
a) ( ) tttf sin= b) ( ) tttf 2cos2=
Theorem 3.3 (Differentiation of a Transform)
If ( ) ( )sFtf =L and for n=1, 2, 3…then ( ) ( ) ( )[ ].1 sFds
dtft
n
nnn −=L
88
Solution
a) Recall: 1
1sin
2 +=s
tL . Then,
( ) ( )( ) ( )22222
1
2
1
21
1
11sin
+=
+
−−=
+
−=s
s
s
s
sds
dttL
b) Recall: 4
2cos2 +
=s
stL . Then,
( )( ) ( )
( )32
35
222
2
22
222
4
48122
4
1
4
2
412cos
+
−−−=
++
+
−=
+
−=
s
sss
ss
s
ds
d
s
s
ds
dttL
Exercises
a) ( ) tttf 2sinh= b) ( ) tttf sin2=
Answer
a) ( )22
4
4
−s
s b)
( )32
2
1
26
+
−
s
s
3.4 INVERSE LAPLACE TRANSFORM
.
Example 3.10
Find the inverse Laplace transform of the following functions.
a) s
6 b)
3
4
s− c)
9
5
−s d)
4
12 +s
e) 64
82 +s
s f)
4
82 −s
g) 16
42 −s
s
Solution
a) 661 =
−
sL b) 2
12
1
3
12
!2
2
44t
ss−=
−=
−
+−− LL
Theorem 3.4 (Inverse Laplace Transform)
If ( ) ( )sFtf =L , then ( )tf is called inverse Laplace transforms of ( )sF and it is
written as
( ) ( )tfsF =− 1L .
The operator 1−L is known as the operator for inverse Laplace transform.
89
c) tess
911 59
15
9
5=
−
=
−
−− LL d) tss
2sin2
1
2
2
2
1
4
122
1
2
1 =
+
=
+−− LL
e) ts
s
s
s8cos8
88
64
822
1
2
1 =
+
=
+−− LL
f) tss
2sinh42
2
2
8
4
822
1
2
1 =
−
=
−−− LL
g) ts
s
s
s4cosh3
43
16
322
1
2
1 −=
−−=
−− −− LL
Exercises
Find the inverse Laplace transform of the following functions.
a) 94 2 +s
s b)
49
422 −s
c) 4
6
s d)
52
1
+s
Answer
a) t2
3cos
4
1 b) t7sinh6 c) 3t d)
t
e 2
5
2
1 −
3.5 PROPERTIES OF INVERSE LAPLACE TRANSFORM
3.5.1 LINEARITY PROPERTY OF INVERSE LAPLACE TRANSFORM
Example 3.11
Find the inverse Laplace transform of the following functions.
a) 4
1252 ++
s
s b)
25
562 −+
s
s
Solution
a)
+
+
+=
+
+ −−−22
1
22
1
2
1
2
2
2
112
25
4
125
ss
s
s
sLLL
.2sin62cos52
26
25
22
1
22
1tt
ss
s+=
+
+
+= −− LL
b) .5sinh5cosh65
5
56
25
5622
1
22
1
2
1 ttss
s
s
s+=
−+
−=
−+ −−− LLL
Theorem 3.5 (Linear Property of Inverse Laplace Transform)
If ( ) ( )tfsF =−
1L and ( ) ( )tgsG =−
1L exist, and if α and β are constants, then
( ) ( ) ( ) ( ) ( ) ( ). 111 tgtfsGsFsGsF βαβαβα +=+=+ −−− LLL
90
Exercises
Find the inverse Laplace transform of the following functions.
a) 16
742 −+−
s
s b)
94
62 −−s
s c)
169
562 ++
s
s
Answer
a) tt 4sinh4
74cosh4 +− b) tt
2
3sinh
2
3cosh
4
1−
c) tt3
4sin
12
5
3
4cos
3
2+
3.5.2 FIRST SHIFT PROPERTY
.
Example 3.12
Find the inverse Laplace transform of the following functions.
a) 3
2
−s b)
34
1
−s c)
( )42
10
+s d)
( )22
5
−s
s e)
( ) 93
52 ++
−
s f)
106
32 ++
+ss
s
Solution
a) ( ) .2121
23
2 33131 ttt ees
es
==
=
−
−− LL
b) .4
11
4
1
4
3
1
4
1
4
34
1
34
14
3
14
3
111tt
es
e
sss
=
=
−=
−=
−−−−− LLLL
c) ( )
.3
5!3
6
10!3
!3
1010
2
10 32
4
12
4
12
4
12
4
1te
se
se
se
s
tttt −−−−−−−− =
=
=
=
+LLLL
d) Given function can be summarized as follows:
( )( )[ ]( )
( )( ) ( ) ( )2222
2
10
2
5
2
1025
2
225
2
5
−+
−=
−
+−=
−
+−=
− sss
s
s
s
s
s.
Theorem 3.6 (First Shift of Inverse Laplace Transform)
If ( ) ( )tfsF =− 1L and a is a constant, then ( ) ( )tfeasF at=−− 1L or can be
written as ( ) ( ) sFeasF at 11 −− =− LL .
91
Then
( ) ( ) ( ) ( ) ( )
( ) ( ) .10510151
101
5
110
15
2
10
2
5
2
10
2
5
2
5
2222
2
1212
2
1212
2
11
2
1
2
1
teetees
es
e
se
se
sssss
s
tttttt
tt
+=+=
+
=
+
=
−+
−=
−+
−=
−
−−
−−
−−−−
LL
LL
LLLL
e) ( )
.3
3
3
5
3
5
93
522
13
22
13
2
1
+
−=
+
−=
++
− −−−−−
se
se
s
tt LLL
.3sin3
5
3
3
3
5 3
22
13te
se
tt −−− −=
+
−= L
f) ( )
.cos113
3
106
3 3
22
13
2
13
2
1te
s
se
s
se
ss
s ttt −−−−−− =
+=
++
+=
++
+LLL
Exercises
Find the inverse Laplace transform of the following functions.
a) 1
1
+−s
b) 52
1
+s c)
( )21−ss
d) ( )22
72
+
+
s
s e)
12
32 +−−ss
s
Answer
a) te−− b) t
e 2
5
2
1 − c) tt tee + d) tt tee 22 32 −− + e) tt tee 2−
3.5.3 INVERSE LAPLACE TRANSFORM OF RATIONAL FUNCTIONS
A rational function is basically a division of two polynomial functions. It is a
polynomial divided by another polynomial. In general, the rational function of s can
be written as ( )( )sQ
sP when the degree of ( )sQ is higher than ( )sP . The several examples
inverse Laplace transform involving the rational function which needs rules of partial
fraction in order to solve it.
1. A linear factor ( )as + gives partial fraction ( )as
A
+, where A is a constant
2. A repeated linear factor ( )2as + gives ( ) ( )2as
B
as
A
++
+
3. A repeated linear factor ( )nas + gives ( ) ( ) ( ) ( )n
n
as
A
as
A
as
A
as
A
+++
++
++
+..
3
3
2
21
92
4. A quadratic factor ( )cbss ++2 gives ( )cbss
BAs
+++
2
5. A repeated quadratic factor
( )ncbss ++2 gives ( ) ( ) ( )nnn
cbss
BsA
cbss
BsA
cbss
BsA
++
+++
++
++
++
+222
22
2
11 ..
where all niBA ii ,..,3,2,1,and = are constants need to be determined.
Example 3.13
Find the inverse Laplace transform of the following functions.
a) ( )11
+ss b)
( )( )32
1
+− ss c)
( )11
2 +ss d) ( )1
12 ++ss
s e)
16
432 −−
s
s
f) ( )( )245
9
+− ss g)
( )( )11 2
2
+− ss
s
Solution
a) By using partial fraction, given function can be written as a sum of two functions as shown below
( ) ( )11
1
++=
+ s
B
s
A
ss
where 1and1 −== BA . Then
( ) ( ).
1
11
1
1
+−=
+ ssss
Thus ( )
tessss
−−−− −=
+
−
=
+1
1
11
1
1 111 LLL .
b) The function can be written as
( )( ) ( ) ( )3232
1
++
−=
+− s
B
s
A
ss
where 5
1and
5
1−== BA . Then
( )( ).
3
1
5
1
2
1
5
1
32
1
+
−
−
=+− ssss
Thus ( )( )
ttee
ssss
32111
5
1
5
1
3
1
5
1
2
1
5
1
32
1 −−−− −=
+
−
−
=
+−LLL .
c) The function can be written as
( ) ( )11
122 +++=
+ s
C
s
B
s
A
ss
where 1and1,1 ==−= CBA . Then
( ) ( ).
1
111
1
122 +++−=
+ sssss
Thus ( )
tetsssss
−−−−− ++−=
+
+
+
−=
+1
1
111
1
1 1
2
11
2
1 LLLL .
d) The function can be written as
93
( ) ( )11
122 ++
+=++
s
CBs
s
A
ss
s
where 1and1,1 =−== CBA . Then
( ) ( ) ( ).11
1
1
1
1222 +
++
−=+
+
ss
s
sss
s
Thus ( ) ttss
s
sss
ssincos1
1
1
1
1
1
12
1
2
11
2
1 ++=
++
+−
=
+
+ −−−− LLLL .
e) The function can be written as
( )( ) ( ) ( )4444
43
16
432 +
+−
=+−
−=
−−
s
B
s
A
ss
s
s
s
where 2and1 == BA . Then
( )( ) ( ) ( ).
4
2
4
1
44
43
16
432 +
+−
=+−
−=
−−
ssss
s
s
s
Thus ttee
sss
s 4411
2
12
4
12
4
1
16
43 −−−− +=
+
+
−
=
−
−LLL .
f) The function can be written as
( )( ) ( ) ( ) ( )2244545
9
++
++
−=
+− s
C
s
B
s
A
ss
where 1and9
1,
9
1−=−== CBA . Then
( )( ) ( ).
4
1
4
1
9
1
5
1
9
1
45
922 +
−
+
−
−
=+− sssss
Thus
( )( ) ( )
.9
1
9
1
4
1
4
1
9
1
5
1
9
1
45
9
445
2
111
2
1
ttt teee
sssss
−−
−−−−
−−=
+−
+
−
−
=
+−LLLL
g) The function can be written as
( )( ) ( ) ( )1111 22
2
++
+−
=+− s
CBs
s
A
ss
s
where 2
1and
2
1,
2
1=== CBA . Then
( )( ) .1
1
2
1
12
1
1
1
2
1
11 222
2
+
+
+
+
−
=+− ss
s
sss
s
Thus
( )( ).sin
2
1cos
2
1
2
1
1
1
2
1
12
1
1
1
2
1
11 2
1
2
11
2
21
tte
ss
s
sss
s
t ++=
++
++
−
=
+−−−−− LLLL
.
94
Exercises
Find the inverse Laplace transform of the following functions.
a) ( )21
+ss b)
( )( )21
1
++ ss c)
( )22
2 +ss
Answer
a) te 2
2
1
2
1 −− b) tt ee 2−− − c) tet 2
2
1
2
1 −++−
3.6 DISCONTINUOUS FUNCTIONS AND PERIODIC FUNCTIONS
For real number, t = a, the unit step function become
at
atatu
>
<
=−1
0)(
Now, multiplying the constant, M to the function gives
at
at
MatuM
>
<
=−0
)(
Example 3.14
Find the Laplace transform of
5
52
2
5
2
1
)(
>
<<
<
−=
t
t
t
tf
Definition 3.2 (Unit Step Function)
The unit step function u(t) is defined as
0
0
1
0)(
>
<
=t
ttu
Theorem 3.7 (Transform of Unit Step Functions)
If 0≥a , then
( )s
eatu
as−
=− L
since 0>s
95
Solution
Let express f (t) in term of unit step functions. Plot the above function:
Hence, we have
)5(7)2(31)( −+−−= tututf
Taking Laplace transform gives
( ) ( ) ( )57231 −+−−=− tutuatu LLLL
s
e
s
e
s
ss 52
731 −−
+−=
In practice, it is more common to face with the problem of computing the transform of
function as ).()( atutg − To compute ( ),)( atutg −L we identify g(t) with )( atf −
so that ).()( atgtf += the transform become:
( ) )()( atgeatutg as +=− − LL
Example 3.15
Determine )()(sin π−tutL
f(t
)
t
+ 7 units
-3 units
5
1
-2
2 5
Theorem 3.8 (Transform of Translation in t)
Let )( tfLF(s) = exist for .0≥> αs
)()()( sFeatuatfas−
=−−L
for a positive and if f(t) is continuous on [0, )∞ , we have
)()()( atuatfsFeas
−−=−1-
L
Theorem 3.8 (Transform of Translation in t)
96
Solution
Here ttg sin)( = and .π=a Hence
ttatatg sin)(sin)(sin)( −=+=+=+ π
and so the Laplace transform of )( atg + is
1
1sin)(
2 +=−=+s
tatg LL
Thus, we get
1
1)()(sin
2 +−=− −
setut sππL
Example 3.16
The square wave function in the following figure can be expressed as
21
10
0
1)(
<<
<<
=t
ttf
Example 3.17
Determine )( tfL where f is the function in example 3.16.
A function of f(t) is said to be periodic of period T if
)()( tfTtf =+
for all t in the domain of f.
Definition 3.3 (Periodic Function)
1
1 2 3 4 -1 0 -2
Theorem 3.9 (Transform of Periodic Function)
If f has period T and is piecewise continuous on [0,T], then
sT
T st
sT
T
e
dttfe
e
sFtf
−
−
−−
∫=
−=
1
)(
1
)()(
0L
97
Solution
Given 2=T . Here we introduce the notation for the “windowed” version of periodic
function f(t) as
otherwise
0
0
)()(
TttftfT
<<
=
In order to transform this function, we have to write the unit step function of )(tfT ,
)1(1)( −−= tutfT
So
)1(111
)1(1 ss es
ess
tu −− −=−=−−L
Therefore
)1(
1
1
1)1(
1)(
2 ss
s
esee
stf
−−−
+=
−−=L
3.7 APPLICATION OF LAPLACE TRANSFORM
3.7.1 INITIAL VALUE PROBLEMS
The Laplace transform has many applications, mostly involving solving differential
equations. Previously, methods discussed in Chapter 2 fail to apply if the equation
involve ( )tf in the form of piecewise continuous functions, periodic functions, step
function and delta function. By using this method, it provides us another way for
solving differential equations. In general, the steps in solving differential equations
shown as follows algorithm.
Theorem 3.10 (Transform of Derivatives)
If ( ) ( )sYty =L , then
( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )
( )( ) ( ) ( ) ( ) ( )( )000
000
00
0
121
23
2
−−− −−′−−=
′′−′−−=′′′
′−−=′′
−=′
=
nnnnn yysyssYsty
yysyssYsty
ysysYsty
yssYty
sYty
L
MMMMMMM
L
L
L
L
L
98
Method of Solution
1. Transformed differential equation becomes an algebraic equation into ( )sY .
2. Substitute initial conditions ( ) bay = and ( ) cay =′ .
3. Solve for ( )sY .
4. Apply inverse Laplace transform, ( ) sY1−L .
5. Obtained ( )ty which is original solution for given differential equation.
3.7.1.1 FIRST ORDER LINEAR DIFFERENTIAL EQUATION
Consider an initial value problem of first order linear differential equation
( )tfbyya =+′ , ( ) 00 yy =
where a, b and 0y are constants. By taking Laplace transform , we get
( ) .tfybya LLL =+′ (3.2)
From Theorem 3.10, we have
( ) ( )sYty =L
( ) ( ) ( ).0yssYty −=′L
Let ( ) ( )sFtf =L , then equation (3.2) becomes
( ) ( )[ ] ( ) ( )sFsbYyssYa =+− 0
( )[ ] ( ) ( )sFsbYyssYa =+− 0
( ) ( ) ( ).0 sFaysYbas =−+
Therefore, we obtain the solution of the transform as
( ) ( ).0
bas
aysFsY
+
+=
Finally by taking the inverse of the Laplace transform, we obtain the solution
( ) ( ).01
+
+= −
bas
aysFty L
Example 3.18
Use the method of Laplace transform to find the solution of the initial value problem.
( ) 00,12 ==+′ yyy
Solution
Let ( ) ( )sYty =L . Taking Laplace transform both side equation, then
12 LLL =+′ yy
( ) ( )[ ] ( )[ ]s
sYyssY1
20 =+− .
99
Substitute initial condition gives
( ) ( )
( ) ( )s
sYs
ssYssY
12
12
=+
=+
( )( )21
+=
sssY .
By using partial fraction, given function can be written as a sum of two functions.
( )( ) ( )22
1
++=
+=
s
B
s
A
sssY
where 2
1and
2
1−== BA . Then
( )( )
+
−
=+
=2
1
2
11
2
1
2
1
sssssY .
Taking the inverse Laplace transform gives the value of ( )ty .
( )( )
tessss
ty 2111
2
1
2
1
2
1
2
11
2
1
2
1 −−−− −=
+
−
=
+= LLL .
Exercises
Solve the initial value problem using Laplace transform. ( ) 10,1 ==−′ yyy
Answer
( ) tety 21+−=
3.7.1.2 SECOND ORDER LINEAR DIFFERENTIAL EQUATION
Consider an initial value problem of second order linear differential equation
( )tfcyybya =+′+′′ , ( ) ( ) 10 00 yyandyy =′=
where a, b, c, 0y and 1y are constants. By taking Laplace transform ,we get
( ) .tfyyy cba LLLL =++ ′′′ (3.3)
From Theorem 3.10, we have
( ) ( )sYty =L
( ) ( ) ( )0yssYty −=′L
( ) ( ) ( ) ( ).00 2 ysysYsty ′−−=′′L
Let ( ) ( )sFtf =L , then equation (3.3) becomes
100
( ) ( ) ( )[ ] ( ) ( )[ ] ( ) ( )sFscYyssYbysysYsa =+−+′−− 0002
( ) ( ) ( ) ( )sFscYbysbsYayasysYas =+−+−− 010
2
( )[ ] [ ] ( ).10
2 sFaybasycbsassY =−+−++
Therefore, we obtain the solution of the transform as
( ) ( ) [ ].
2
10
cbsas
aybasysFsY
+++++
=
Finally by taking the inverse of the Laplace transform, we obtain the solution
( ) ( ) [ ].
2
101
++
+++= −
cbsas
aybasysFty L
Example 3.19
Solve the initial value problem using Laplace transform.
( ) ( ) .10,00,2 =′==+′′ yyyy
Solution
Let ( ) ( )sYty =L . Taking Laplace transform both side equation, then
2LLL =+′′ yy
( ) ( ) ( )[ ] ( )[ ]s
sYysysYs2
002 =+′−− .
Substitute initial condition gives
( ) ( )s
sYsYs2
12 =+−
( ) ( )s
s
ssYs
+=+=+2
12
12
( ) ( )12
2 +
+=
ss
ssY .
By using partial fraction, given function can be written as a sum of two functions.
( ) ( ) ( )11
222 +
++=
+
+=
s
CBs
s
A
ss
ssY
where 1and2,2 =−== CBA . Then
( ) ( ) ( ) ( ) ( )11
1
22
1
122
1
22222 +
++
−=++−
+=++
=ss
s
ss
s
sss
ssY .
Taking the inverse Laplace transform gives the value of ( )ty .
( ) ( ) ttss
s
sss
sty sincos22
1
1
12
12
1
222
1
22
11
2
1 +−=
+
+
+
−
=
+
+= −−−− LLLL .
101
Exercises
Solve the initial value problem using Laplace transform ,0256 =+′+′′ yyy
( ) ( ) .10,00 =′= yy
Answer
( ) tety t 4sin4
1 3−=
3.7.2 ELECTRIC CIRCUITS ANALYSIS
Electric circuits are conducted by three basic element; resistor having resistance, R,
capacitor having capacitance, C, and Inductor having inductance, L. Resistor is
measured in ohm (Ω ) and capacitor and inductor are measured in farad (F) and Henry
(H).
Figure 1
There are currents, i(t) and charges, q(t) moving around the circuit when the voltage
source is on. In order to obtain the currents and charges through the circuit,
Kirchhoff’s current law and Kirchhoff’s voltage law are introduced. According to
Kirchhoff’s current law, the algebraic sum of the current flowing into any junction
point must be zero. However, Kirchhoff’s voltage law said that the algebraic sum of
instantaneous charges in potential (voltages drops) around any closed loop must be
zero. To apply this law, we have to know the voltage drops across each element of
the circuit where
(i) voltage drops across a resistor
Ri=
(ii) voltage drops across an inductor
dt
diL=
(i) voltage drops across a capacitor
qC
1=
Hence, we have
)(1
tEqC
Ridt
diL =++ …………………….(i)
R
L
C
E(t)
102
as Kirchhoff’s voltage law. Well known that through electric circuit E(t) the voltage
supplied to the circuit at time t and the current is the instantaneous rate of charge,
.dtdqi =
Example 1
According to figure 1, an RLC series circuit has a voltage source given by
ttE 100sin)( = volts, a resistor of 0.02 ohms, an inductor of 0.001 henrys, and a
capacitor of 2 farads. If the initial current and initial charge on the capacitor is both
zero, determine the current in the circuit for .0>t
Solution
Given ,100sin)( ttE = ,02.0 Ω=R ,001.0 HL = FC 2= and 0)0()0( == qi
Substitute these parameters into Kirchhoff’s voltage law, gives
tqidt
di100sin
2
102.0001.0 =++
Differentiate every term with respect to t, we’ll get
tdt
dq
dt
di
dt
id100cos100
2
102.0001.0
2
2
=++
tidt
di
dt
id100cos000,10050020
2
2
=++
Taking Laplace transform, we have
tidt
di
dt
id100cos10000050020
2
2
LLLL =+
+
( )22
2
100
100000)(500)0(20)(20)0(')0(
+=+−+−−s
ssIissIisisIs
( )22
2
100
100000)(500)(20
+=++s
ssIssIsIs
[ ]22
2
100
10000050020)(
+=++s
ssssI
[ ]50020100
100000)(
222 +++=
sss
ssI
[ ]( )[ ]2222 2010100
100000)(
+++=
ss
ssI
By using partial fraction decomposition and inverse Laplace transform, the current for
this circuit is
ttttetit
100sin425.9
20100cos
425.9
9520sin
85.18
10520cos
425.9
95)(
10 +−
−= −
103
Example 2
Figure 2
At time, ,0=t the charge on the capacitor in the electric network shown in Figure (2)
is 2 coulombs, while the current through the capacitor is zero. Determine the current
in the various branches of the network at any time .0>t
Solution
Kirchhoff current law gives
321 iii +=
Figure 2 shows that this network has two loop, so we’ll apply Kirchhoff’s voltage law
to both loop. Hence, we have
Loop 1: )(12 tERi
dt
diL =++
5)(20 322 =++ iidt
di
52020 322 =++ ii
dt
di………………….(i)
Loop 1: )(1
32 tEq
Cdt
diL =+−
0160 32 =+− q
dt
di
0160 3
2
2
2
=+−dt
dq
dt
id
0160 32
2
2
=+− idt
id………………….(ii)
By using Laplace transform, we’ll get
s
sIsIissI5
)(20)(20)0()( 3222 =++−
s
sIsIssI5
)(20)(20)( 322 =++
ssIsIs
5)(20)()20( 32 =++ ……………………(iii)
and
0)(160))0(')0()(( 3222
2 =+−−− sIisisIs
0)(160)( 32
2 =+− sIsIs …………………………..(iv)
E(t)
R
L C
104
By using partial fraction and inverse Laplace transform, we’ll get
tetesi tt 12sin3
6412cos16
4
1)( 44
2
−− +−=
tesi t 12sin3
80)( 4
3
−−=
Hence, from kirchhoff’s current law
tetesi tt 12sin3
1612cos16
4
1)( 44
1
−− −−=
105
PRACTICE 3
1. Find the Laplace transform of the following functions.
a) ( ) tttf 3sinh64 3 −= b) ( ) 422 tttf −= c) ( ) 63 −= tetf
d) ( ) ttetf 55= e) ( ) tettf 22 −= f) ( ) ( )31 ttf +=
g) ( ) tetft 2sinh−= h) ( ) tttf
4
3cosh= i) ( ) ( )231 tetf +=
j) ( ) tetft cosh6 2−= k) ( ) ( ) t
ettf−−= 21 l) ( ) t
ettttf422sin −+=
m) ( ) tttf 2sinh2= n) ( ) ( )2tt eetf −−= o) ( ) tetttf −+= 3cossin4
p) ( ) ( )23tettf −=
2. Find the inverse Laplace transform of:
a) ( )65
1
+−=
ssF b) ( )
( )33
1
−=s
sF c) ( )9
372 +−
=s
ssF
d) ( )( )42+
=s
ssF e) ( )
25
532 ++−
=s
ssF f) ( )
( )352
8
+=
ssF
g) ( ) 2
4
8 20F s
s s=
+ + h) ( )
34
12 +−
−=
ss
ssF i) ( )
346
102 ++
=ss
sF
j) ( )19
632 ++
=s
ssF k) ( )
2
22 −
−=s
ssF l) ( )
12
12
+
+
=
s
ssF
3. Find the inverse Laplace transform of:
a) ( )( )12
−=
sssF b) ( )
( )22
2 −+
=ss
ssF c) ( )
( )( )53
1
−−=
sssF
d) ( ) ( )28
2 ++
=ss
ssF e) ( )
9
422 −+
=s
ssF f) ( )
( )( )163 2 ++=
ss
ssF
4. Use the method of Laplace transform to find the solution of the following initial value problems.
a) ( ) 00,sin3 ==+′ ytyy b) ( ) 00,122 4 ==+′ yeyyt
c) ( ) ( ) 00,00,84 2 =′==′−′′ yyeyy t d) ( ) ( ) 10,20,32 =′==−′+′′ yytyyy
e) ( ) ( ) 50,00,44 2 =′==−′′ yyeyy t f) ( ) ( ) 000,34 =′==−′′ yytyy
g) ( ) ( ) 00,00,8634 =′=−=+′−′′ yytyyy
Answer
1. a) 9
182424 −
−ss
b) 53
244
ss− c)
6
3
e
s
−
−
d) ( )25
5
−s e)
( )32
2
+s f)
ssss
1366234+++
106
g) ( ) 41
22 −+s
h) ( )2
2
916
144256
−
+
s
s i)
6
1
3
21
−+
−+
sss
j) ( )
( ) 12
262 −+
+
s
s k)
( ) ( )3 5
1 4 24
1 1 1s s s− +
+ + +
l) ( ) ( )322 4
2
1
4
++
+ ss
s m)
( )32
2
4
1612
−
+
s
s n)
2
12
2
1
++−
− sss
o) 1
3
4
42 +
++ ss
p) ( ) 6
1
3
2223 −+
−−
sss
2. a) t
e 5
6
5
1 −− b) 2
2
1tet c) tt 3sin3cos7 −
d)
−−tet
t
3
1
2
122 e) tt 5sin5cos3 +− f) 5
221
2
t
e t−
g) te t 2sin2 4− h) ( )tte t sinhcosh2 + i) te t 5sin2 3−
j)
+ tt3
1sin18
3
1cos3
9
1 k) t2cosh2− l)
−−
ttet
sin2
1cos2
1
3. a) ( )te−12 b) tet 21 +−− c) ( )tt ee 53
2
1+−
d) tt 2sin2
12cos22 +− e) ( )tt ee 335
3
1 −+ f) tte t 4sin25
44cos
25
3
25
3 3 ++− −
4. a) ( )tett 3sin3cos10
1 −++− b) ( )tt ee 242 −− c) tt ee 4221 +−
d) ( ) tt eet 221 −−+ e) ( ) tt eett −+++2
1
5
2sincos
10
1 2
f) ( )tt 2sinh368
1+− g) tt eet 32 −+
