The Laplace Transform Objective To study the Laplace transform and inverse Laplace transform
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The Laplace Transform Objective To study the Laplace transform and inverse Laplace transform. Modules Module I- Laplace transform of some elementary functions Module II- Properties of Laplace transform Module III- Existence conditions Module IV- Inverse Laplace transform Introduction A Transformation is a mathematical device which converts one function into another. For example, when the differential operator D operates on f(x) = sin x, it gives a new function g(x) = D f(x) = cos x.
Transcript of The Laplace Transform Objective To study the Laplace transform and inverse Laplace transform
The Laplace Transform
Objective
To study the Laplace
transform and inverse
Laplace transform.
Modules
Module I- Laplace
transform of some
elementary functions
Module II- Properties
of Laplace transform
Module III- Existence
conditions
Module IV- Inverse
Laplace transform
Introduction
A Transformation is a
mathematical device which
converts one function into
another. For example, when
the differential operator D
operates on f(x) = sin x, it
gives a new function g(x) = D
f(x) = cos x.
Laplace transform or Laplace
transformation is widely used
by scientists and engineers. It
is particularly effective in
solving linear differential
equations. It is very useful in
system analysis where initial
conditions can be easily
included to give system
response. We begin this
session by the definition of
Laplace transform.
Definitions
There are two types of
Laplace transforms. The
transform defined by
dtetfs st)()( ,
where s is a parameter which
may be real or complex is
known as two sided or
bilateral Laplace transform of
the function f(t), provided the
integral exists.
For some functions the above
transform cause problem of
convergence. This can be
almost avoided by restricting
the range of integration to
between 0 and and
considering f(t) = 0 for t < 0.
Thus the transformation
defined by
0
)(.)( dttfes st
where t > 0 and s is a
parameter which may be real
or complex is known as
unilateral or simply, Laplace
transform of the function f(t),
provided the integral exists. It
is also denoted as L{f(t)} or
)(sf .
In this session we will be
used this second definition.
Module I- Laplace
transform of some
elementary functions
I. f(t) = k
Therefore
0
.)( kdtes st
0,
0
ss
k
s
ek
st
.
II. f(t) = eat
Therefore
0
.)( dtees atst
0
)( dte tas
asasas
e tas
,1
)(0
)(
.
III. f(t) = tn, (n > -1)
Therefore
0
.)( dttes nst
0
..s
dk
s
ke
n
k , by putting
s
dkdtkst ,
0
1..
1dkke
s
nk
n
0
1)1(
1..
1dkke
s
nk
n
1
)1(
ns
n, if n > -1 and s >
0.
Note. If n is a positive
integer, we have !)1( nn
Therefore, 1
!
n
n
s
ntL
IV. f(t) = cos at
Therefore
0
cos.)( atdtes st
0
22)sincos( ataats
as
e at
22 as
s
.
V. f(t) = sin at
Therefore
0
sin.)( atdtes st
0
22)cossin( ataata
as
e at
22
1
as .
Module II- Properties of
Laplace transform
I. Linearity property
If c1, c2 are two constants and
f1, f2 are two functions of t,
then
)()()()( 22112211 tfLctfLctfctfcL
.
Proof.
By definition, we have
0
22112211 )()()()( dttfctfcetfctfcL st
0
22
0
11 )()( dttfecdttfec stst
)()( 2211 tfLctfLc .
This result may be
generalized as
)(...)()(
)(...)()(
2211
2211
tfLctfLctfLc
tfctfctfcL
nn
nn
,
for n constants c1, c2, …, cn
and n functions f1, f2, …, fn.
Examples
We have 2
coshatat ee
at
.
Therefore
2}{cosh
atat eeLatL
atat eLeL 2
1
asas
11
2
1
22 as
s
.
Similarly,
2}{sinh
atat eeLatL
atat eLeL 2
1
asas
11
2
1
22 as
a
.
II. First shifting property
If )()}({ stfL , then
)()}(.{ astfeL at .
Proof.
By definition, we have
0
)(..)}(.{ dttfeetfeL atstat
0
)( )(. dttfe tas
0
)(. dttfe rt, where r = s –
a
)()( asr .
Examples.
1. If n is positive integer, we
know that
1
!
n
n
s
ntL .
Therefore by first shifting
property,
1)(
!
n
nat
as
nteL .
2. We know that
22 3
3cos
s
stL .
By shifting property,
22
2
3)2(
)2(3cos
s
steL t .
III. Change of scale
property
If )()}({ stfL , then
a
s
aatfL
1)}({ .
Proof.
By definition, we have
0
)(.)}({ dtatfeatfL st
0
)(.a
drrfe a
sr
, where at = r
and a
dudt
0
)(.1
drrfea
pr , where
a
sp
a
s
ap
a
1)(
1.
Examples.
1. We know that
1
1}{sinh
2
stL .
Therefore by change of scale
property, we have
4
2
12
1.
2
1}2{sinh
22
sstL
.
2. Let 2
1)(
2
s
stfL , then
18
3
23
13.
3
1)3(
22
s
s
s
s
tfL
.
IV. Change of scale shifting
property
If )()}({ stfL , then
a
bs
aatfeL bt
1)}(.{ .
Proof is similar to the proof
of above property.
Example. Let
1
}{cos2
s
stL , then
1
3
2
3
2
.3
13cos2
s
s
teL t
22 3)2(
2
s
s.
Module III- Existence
conditions
The Laplace transform does
not exists for all functions. If
it exists, it is uniquely
determined. The following
conditions are to be satisfied:
Let f(t) be the given
function. If
1. f(t) is piecewise
continuous on every finite
interval
and 2. f(t) satisfies the
inequality atebtf .)( for all
0t and for some constants
a and b,
then L{f(t)} exists.
The function which satisfies
the condition 2 is known as
exponential order.
For example, cos ht < et for
all t > 0,
tn < n! e
t (n =
0, 1, 2, …) for all t > 0.
But att bee 2
, whatever may
be a and b. So 2teL does not
exist. Similarly, t
1 does not
have Laplace transform.
Example. Find the Laplace
transforms of
(i) sin 2t cos 3t
(ii) sin t
(i) Here
)2sin3cos2(2
13cos2sin tttt
)sin5(sin2
1tt
.
Therefore
)sin5(sin2
13cos2sin ttLttL
tLtL sin5sin2
1
2222 1
1
5
5
2
1
ss
)1)(25(
)5(222
2
ss
s.
(ii) We have
...
!5!3sin
53
tt
tt
Therefore
...!5!3
sin
53
tttLtL
...!5
1
!3
1 2/52/32/1 tLtLtL
...)2/7(
!5
1)2/5(
!3
1)2/3(2/72/52/3
sss
...
.2
1.
2
3.
2
5
120
1.
2
1.
2
3
.6
1.
2
1
2/72/52/3
sss
...
4
1
!3
1
4
1
!2
1
4
11
2
32
2/3 ssss
2/3
4/1
2
.
s
e s
.
Example. Find the Laplace
transform of t e-4t
sin3t.
Solution.
We know that 2
1
ttL .
Therefore by first shifting
theorem, we have
2
3
)3(
1
isteL it
22
)3)(3(
)3(
isis
is
or
22
2
)9(
6)9()3sin3(cos
s
isstittL
Equating the imaginary parts
on both sides, we get
22 )9(
6)3sin
s
sttL
.
Again applying the first
shifting theorem, we have
22
4
9)4(
)4(6)3sin.
s
stteL t
22 258
)4(6
ss
s.
Module IV- Inverse
Laplace transform
If )()( stfL , then f(t) is
called the inverse Laplace
transform of )(s and is
denoted by
)()(1 tfsL .
Here 1L denotes the inverse
Laplace transform.
For example, since
2
12
seL t , we have
tes
L 21
2
1
.
Inverse Laplace transform
follows all the properties of
Laplace transform.
From the results of Laplace
transforms, we have
(1) ks
kL
1 , k being
constant.
(2) ateas
L
11
(3) )!1(
1 11
n
t
sL
n
n if n is
positive integer. Otherwise
)(
1
n
t n
.
(4)
)!1()(
1 11
n
te
asL
nat
n
(5) ataas
L sin11
22
1
(6)
ateaabs
L bt sin1
)(
122
1
(7) atas
sL cos
22
1
(8)
ateabs
bsL bt cos
)( 22
1
(9) ataas
L sinh11
22
1
(10) atas
sL cosh
22
1
Example. Find the inverse
Laplace transform of
(i)
5
22
2
13
s
s (ii)
2516
1542
s
s
Solution.
(i) We have
5
24
5
22
2
363
2
13
s
ss
s
s
53
1.
2
31.3
1.
2
3
sss
Therefore
5
1
3
11
5
221 1
.2
31.3
1.
2
3
2
13
sL
sL
sL
s
sL
!42
3
!23)1(
2
3 42 tt
42
16
1
2
3
2
3tt .
(ii) We have
16
2516
154
2516
154
22
s
s
s
s
2
2
2
2
4
5
1.
16
15
4
5.
4
1
ss
s
.
Therefore
2
2
1
2
2
1
2
1
4
5
1.
16
15
4
5.
4
1
2516
154
s
L
s
sL
s
sL
tt
4
5sinh
4/5
1.
16
15
4
5cosh
4
1
tt
4
5sinh
4
3
4
5cosh
4
1.
Inverse Laplace transforms
using method of partial
fractions
If )(s is rational algebraic
function, then we have to
express )(s in terms of
partial fractions in order to
find the inverse Laplace
transform.
Example. Find the inverse
Laplace transform of
sss
s
23
123
2
Solution.
Here the denominator can be
written as
)23(23 223 ssssss
)2)(1( sss
Let
21)2)(1(
12
s
C
s
B
s
A
sss
s
.
Then
)1()2()2)(1(12 sCssBsssAs
.
Putting s = 0,
we get 1 = 2A
A = ½
Putting s = -1,
we get 2 = -B
B = -2
Putting s = -2,
we get 5 = 2C
C = 5/2.
Therefore
2
1.
2
5
1
1.2
1.
2
1
)2)(1(
12
ssssss
s
and
2
1.
2
5
1
1.2
1.
2
1
)2)(1(
1 1112
1
sL
sL
sL
sss
sL
tt ee 2
2
52
2
1 .
Summary
In the session we have
discussed the Laplace
transform of various
functions and properties of
Transform. Also we
discussed the inverse Laplace
transform and the method of
finding inverse transform by
the partial fraction method.
Assignment questions
1. Define Laplace transform.
2. Find the Laplace
transforms of the following
functions:
(i) f(t) = sin at sin bt
(ii) f(t) = cos3 2t
(iii) f(t) = e-2t
sin 4t
(iv) f(t) = e-3t
(sin 2t –
2t cos 2t)
(v)
2,0
20,cos)(
t
tttf
3. Find the inverse Laplace
transforms of:
(i) 4
622
s
s
(ii) 168
1242
ss
s
(iii) )2)(3(
322
sss
ss
(iv)
)22)(1(
22
sss
s
(v))4)(1( 22 ss
s.
Reference
1. The Laplace
Transform, Shaum
Outline Series, Shaum
Publishing Company,
New York.
2. Advanced
Engineering
Mathematics by E.
Kreyszig, John Wylie
& Sons, New York
(1999).
Quiz
1. The Laplace transform of
sin at is
a. 22
1
as b.
22 as
s
c.
22
1
as
2. If 3
2 !2
stL , then the
value of 23 teL t is
a. 3)3(
!2
s b.
3)3(
!2
s c.
3)3(
!2s
.
3. If )()}({ stfL , then
a
s
aatfL
1)}({ is known
as
a. Shifting property
b. Change of scale
shifting property
c. Change of scale
property
4. The inverse Laplace
transform of 22)(
1
bas is
a. bteb
at sin1
b. btea
bt sin1
c. ateb
at sin1
5. The inverse Laplace
transform of 9)1(
32
s
s is
a.
tte t 3sin
3
23cos b.
ttet 3sin
2
33cos c.
ttet 3sin
3
23cos
Answers
1.a 2.b 3.c
4.a 5.c
Glossary
Function: It is an assignment
f from a set A into another set
B; the set A is called domain
of f and the set of all function
values is called the range of f.
Partial fraction: Suppose that
xg
xf is a proper rational
function and xg is a
product of polynomials. Then
xg
xfcan be expressed as
sum of simpler rational
functions, each of which is
called a partial fraction. This
process is called
decomposition of xg
xfinto
partial fractions.
The decomposition depends
on the nature of the factors of
xg .
If
nxxxxg ...21
. i.e., a product of non-
repeating linear functions,
then
n
n
x
A
x
A
x
A
xg
xf
...
2
2
1
1
, where iA are constants.
If
r
kxxxxg ...1
i.e., some factors repeating,
then
r
r
k
k
x
B
x
B
x
A
x
A
x
A
xg
xf
......
1
1
2
21
, where ii BA , are
constants.
If some factors are quadratic,
but non-repeating, then
corresponding to these
factors, the partial
fraction is in the
formfactor quadratic
offunction linear a x.
FAQs
1. Define Laplace transform.
Answer.
The transformation defined
by
0
)(.)( dttfes st
where t > 0 and s is a
parameter which may be real
or complex is known as the
Laplace transform of the
function f(t), provided the
integral exists.
2. State and prove the first
shifting property of Laplace
transform
Answer.
Statement
If )()}({ stfL , then
)()}(.{ astfeL at .
Proof.
By definition, we have
0
)(..)}(.{ dttfeetfeL atstat
0
)( )(. dttfe tas
0
)(. dttfe rt , where r = s –
a
)()( asr .
3. Find the Laplace transform
of the function f(t) defined by
Tt
TtT
t
tf
,1
0,)( .
Answer.
By definition, we have
0
)()}({ dttfetfL st
T
st
T
st dtedtT
te 1..
0
T
stT stT
st
s
edt
s
e
s
et
T00
.1.1
2
1
Ts
e sT .
4. Find the inverse Laplace
transform of 1
12
ss
s.
Answer.
We have
4
3
2
1
2
1
2
1
1
122
s
s
ss
s
2222
2
3
2
1
1.
2
1
2
3
2
1
2
1
ss
s
.
Therefore
2222
1
2
1
2
3
2
1
1.
2
1
2
3
2
1
2
1
1
1
ss
s
Lss
sL
tetett
2
3sin
3
2.
2
1
2
3cos 2
1
2
1
ttet
2
3sin
2
3cos3
3
12
1
.
5. Find the inverse Laplace
transform of
)52)(1(
352
sss
s.
Answer.
Let
)52()1()52)(1(
3522
ss
CBs
s
A
sss
s
i.e.,
)1)(()52(35 2 sCBsssAs
.
Putting s = 1, we get 8 = 8A
or A = 1.
Equating the coefficients of
s2,
0 = A +
B i.e., B = -A = -1
Putting s = 0, we get 3 = 5A
– C i.e., C = 5A – 3 = 2
Thus
)52(
2
)1(
1
)52)(1(
3522
ss
s
ssss
s
4)1(
3)1(
)1(
12
s
s
s
2222 2)1(
1.3
2)1(
)1(
)1(
1
ss
s
s
Therefore
2222
1
2
1
2)1(
1.3
2)1(
)1(
)1(
1
)52)(1(
35
ss
s
sL
sss
sL
22
1
22
11
2)1(
1.3
2)1(
)1(
)1(
1
sL
s
sL
sL
tetee ttt 2sin2
32cos .
