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LAPLACE TRANSFORMS
Integral transform is a particular kind of mathematical operator which arises in the analysis of
some boundary value and initial value problems of classical Physics. A function g defined by a
relation of the form g(y) = b
a
K x,y f x dx is called the integral transform of f(x). The function
K(x, y) is called the kernel of the transform. The various types of integral transforms are Laplace
transform, Fourier transform, Mellin transform, Hankel transform etc.
Laplace transform is a widely used integral transform in Mathematics with many applications
in physics and engineering. It is named after Pierre-Simon Laplace, who introduced the
transform in his work on probability theory. The Laplace transform is used for solving
differential and integral equations. In physics and engineering it is used for analysis of linear
time-invariant systems such as electrical circuits, harmonic oscillators, optical devices, and
mechanical systems.
Definition
The Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s),
defined by:
L {f (t)} = -st
0
F s e f t dt
. The parameter s is a complex number.
L is called the Laplace transformation operator.
Transforms of elementary functions
1. L {1} =1
s, s > 0.
Proof: -st
0
1L {1} e .1dt = if s>0
s
2.
n+1n
n+1
n!when n = 0,1,2,...
sL t =
Γ n+1otherwise
s
Proof:
-stn -st n n n-1
0 0
n-2
e nL t = e t dt = t + L t
-s s
n-1 n-1n n 1= L t =...= ... L 1
s s s s s
n+1
n+1
n!if n is a positive integer
n-1n 1 1 s= ... =
n+1s s s sotherwise
s
3. 1atL e = , s > as-a
Proof:
- s-a tat -st atL e = e e dt = e dt0 0
- s-a te 1
= = , s>a- s-a s-a0
4. 2 2
aL sinat = , 0
s +as
Proof:
0
2 2
0
2 2
sin cosL sinat sin
, 0
st
ste s at a at
e atdts a
as
s a
5. 2 2
sL cosat = , 0
s +as
Proof:
-st
-st 0
2 2
0
2 2
e -scosat+asinatL cosat e cosatdt
s +a
s, 0
s +as
6. 2 2
aL sinha t = ,
s -as a
Proof:
2 2
1 1 1L sinhat
2 2
,
at ate eL
s a s a
as a
s a
7. 2 2
sL coshat = ,
s -as a
Proof:
1 1 1
L coshat2 2
,2 2
a t a te e
Ls a s a
ss a
s a
Properties of Laplace transforms
1. Linearity property.
If a, b, c be any constants and f, g, h be any functions of t then
L af t +bg t -ch t =aL f t +bL g t -cL h t
Because of the above property of L, it is called linear operator.
2. First shifting property.
If L{f (t)} = F(s) then atL e f t = F s-a .
Proof:
- s-a tat -st at
0 0
-rt
0
L e f t = e e f t dt = e f t dt
= e f t dt where r =s-a.
= F(r) = F s-a .
Application of this property leads to the following results.
1.
n+1
at n
n+1
n!when n = 0,1,2,...
s-aL e t =
Γ n+1otherwise
s-a
2.
at
2 2
bL e sinbt =
s-a +b
3.
at
2 2
s-aL e cosbt =
s-a +b
4.
at
2 2
bL e sinhbt =
s-a - b
5.
at
2 2
s-aL e coshbt =
s-a -b
where in each case s > a.
Examples:
Find the Laplace transform of the following.
1. Sin32t
Solution:
3
3
3
2 2 2 2
sin 6 3sin 2 4sin 2
3 1sin 2 sin 2 sin 6
4 4
3 1sin 2 sin 2 sin 6
4 4
3 2 1 6 48. .
4 4 4 36 4 36
t t t
t t t
L t L t L t
s s s s
2. 4te sin2tcost
Solution:
4t 4t
2 2
1L e sin2tcost L e sin3t +sint
2
1 3 1
2 4 9 4 1s s
3. 2 2coste t
Solution:
2 2 2
2
1cos 1 cos 2
2
21 1
2 2 2 4
t tL e t L e t
s
s s
4.
1, 0 1
( ) , 1 2
0, 2
t
f t t t
t
Solution:
1 2
-st -st -st
0 1 2
2 2
2 2
L f(t) e .1 e .tdt e .0dt
1 2 s s s
dt
e e e
s s s s
Exercises:
Find the Laplace transform of
1. 5sin t 2. 3
1 tte 3. cosh sinat at
4. sin , 0
0,
t tf t
t
5.
3
1t
t
Transforms of integrals
t
0
1If L f(t) = F s then L f(u)du = F s
s
Proof:
t
0
t
0
Let φ t = f u du, then φ t = f t and φ 0 =0.
\ L φ t =sL φ t -φ 0
1Hence L φ t = L φ t
s
1i.e. L f u du = F s .
s
Multiplication by tn
n
nn
n
If L f t = F s
dthen L t f t = -1 F s , n =1,2,3,...
ds
Proof:
We have -st
0
e f t dt = F s
-st
0
-st
0
-st
0
d dConsider e f t dt = F s
ds ds
ByLeibnitz's rule for differentiation under the integral sign
de f t dt = F s
s ds
de tf t dt = - F s
ds
which proves the theorem for n = 1.
Now assume the theorem to true for n = m(say), so that
m-st m
m
0
m+1-st m
m+1
0
m+11-st m+1
m+1
0
de t f t dt = -1 F s
ds
dThen e t f t dt = -1 F s
ds
By Leibnitz's rule,
de t f t dt = -1 F s
ds
m
m
m
d
ds
This shows that if the theorem is true for n = m, it is also true for n = m+1. But it is true for n
= 1. Hence it is true for n = 1+1 = 2, n = 2+1 = 3 and so on.
Thus the theorem is true for all positive integral values of n.
Division by t
s
1If L f t = F s then L f t = F s ds
t
Proof:
-s t
0
-st
0
We have F s = e f(t)dt
Consider F s ds e f(t)dt dss s
-st
0
-st
0
e f(t)dsdt changing theorder of integration
f(t) e ds dt t is independent of s
1L f(t)
t
s
s
Examples:
Find the Laplace transform of
1. tcosat
Solution:
2 2
2 2
22 2 2 2
sL cosat =
s +a
s s -aL tcosat
s +a s +a
d
ds
2. te-tsin3t
Solution:
2
22 2
3sin 3
9
3 6sin 3
9 9
L ts
d sL t t
ds s s
2 2
2 2
Now using first shifting property,we get
6 1 6 1sin 3
2 101 9
ts s
L e t ts ss
3. cosat-cosbt
t
Solution:
2 2 2 2
2 2 2 2
s
1 22 2
2 2
s sL cosat-cosbt = -
s +a s +b
cosat-cosbt s sL = - ds
t s +a s +b
s +b= log
s +a
Exercises:
Find the Laplace transforms of the following functions.
1. tsin2t 2. te
2tsin3t 3.
-te sint
t
4. ate -cosbt
t 5.
1- cos3t
t
Periodic functions:
Suppose that the function ( )F t is periodic with period . Then its Laplace transform is given by
( 1)
00
( ) ( ) ( ) .
n
st st
n n
L F t e F t dt e F t dt
Let us put .t n Then 0 0
( ) exp( ) ( ) .n
L F t sn s F n d
But ( ) ( )F n F , we get
0 00 0
0
( ) exp( ) exp( ) ( ) exp( ) exp( ) ( )
1exp( ) ( ) .
1
n
n n
sw
L F t sn s F d s s F d
s F de
Theorem
If ( )F t has a Laplace Transform and if 0
( )
( ) ( ), ( ) .1
s
sw
e F d
F t F t L F te
Example (a)
Find the transform of the function 1, 0
( , ) ; ( 2 , ) ( , ).0, 2
t ct c t c c t c
c t c
0
2
1( , ) .
1 (1 )
c
s
cs cs
e d
L t ce s e
Example (b)
Find the transform of the square-wave function
1, 0( , ) ; Q( 2 , ) ( , ).
1, 2
t cQ t c t c c Q t c
c t c
2
/ 2
0
2 / 2
(1 ) (1 ) 1( , ) tanh .
1 (1 ) (1 ) 2
c c
s s
cs cs cs
c
cs cs cs cs
e d e de e e cs
L Q t ce s e s e e s
Exercise:
1. Define a triangular-wave function ( , )T t c by
, 0( , ) ; T( 2 , ) ( , ).
2 , 2
t t cT t c t c c T t c
c t c t c
Sketch ( , )T t c and find its
Laplace Transform.
2. Define the function ( )G t by ( ) ,0 ; G( ) ( ), 0.tG t e t c t c G t t Sketch the
graph of ( )G t and find its Laplace Transform.
3. Define the function ( )S t by ( ) 1 ,0 1; S( 1) ( ), 0.S t t t t S t t Sketch the
graph of ( )S t and find its Laplace Transform.
4. Sketch a half-wave rectification of the function sin t , as described below, and find its
transform.
sin , 02
( ) ; F ( ).2
0,
t t
F t t F tw
tw
5. Find ( )L F t where ( ) for 0 and ( ) ( ).F t t t F t F t
A Step Function:
Applications frequently deal with situations that change abruptly at specified times. We
need a notation for a function that will suppress a given term up to a certain value of t and
insert the term for all larger t. The function we are about to introduce leads us to a
powerful tool for constructing inverse transforms.
Let us define function ( )t by 0, 0
( ) .1, 0
tt
t
The definition says that ( )t is zero when the argument is negative and ( )t is unity
when the argument is positive or is zero. It follows that
0,( ) .
1,
t ct c
t c
The Laplace Transform of ( ) ( )t c F t c is
( ) ( ) ( ) .st
c
L t c F t c e F t c dt
Now put t c v in the integral to obtain
( )
0
( ) ( ) ( ) ( ) .s c v csL t c F t c e F v dv e L F t
Theorem
If 1 ( ) ( ),L f s F t if 0,c and if ( )F t be assigned values (no matter what ones) for
10, ( ) ( ) ( ).csc t L e f s F t c t c
Example (a):
Find ( )L y t where
2 , 0 2( ) .
6, 2
t ty t
t
2 2 2
2 2 2 2
3 2 3
( ) ( 0) ( 2) 6 ( 2) ( ) (6 ) ( 2)
2 2 4 2and ( ) 6 ( 2) .s s
y t t t t t t t t t
L y t L t e L t es s s s
Example (b) :
Find and sketch a function ( )g t for which
31
2 2
3 4 4( ) .
s se eg t L
s s s
( ) 3 4( 1) ( 1) 4( 3) ( 3)
3, 0 1
7 4 , 1 3
5, 3
g t t t t t
t
t t
t
Exercises:
1. Sketch the graph of the given function for 0.t
(i) ( 1) 2 ( 2) 3 ( 4).t t t
(ii) 2 2 ( 2).t t t
2. Express ( )F t in terms of the function and find ( )L F t .
(i) 4, 0 2
( ) .2 1, 2
tF t
t t
(ii)
2 , 0 2
( ) 1, 2 3 .
7, 3
t t
F t t t
t
(iii) sin3 , 0
( ) .0,
t tF t
t
t
g(t)
-5
3
1 3
3. Find and sketch an inverse Laplace transform of
35.
s se e
s s
4. Evaluate
41
3.
( 2)
seL
s
5. If ( )F t is to be continuous for 0t and
2 21
2
(1 )(1 3 )( ) ,
s se eF t L
s
evaluate
(1), (3), (5).F F F
INVERSE LAPLACE TRANSFORMS
Introduction:
Let L{f(t)} = F(s). Then f(t) is defined as the inverse Laplace transform of F(s) and is denoted
by L-1
{F(s)}.
Thus L-1
F(s) = f(t) ……………..(1)
L-1
is known as the inverse laplace transform operator and is such that
111 LLLL
In the inverse problem (1), F(s) is given (known) and f(t) is to be determined.
Properties of Inverse Laplace transform
For each property on Laplace transform, there is a corresponding property on inverse Laplace
transform, which readily follow from the definitions.
1) Linearity Property
Let L-1
{F(s)} = f(t) and L-1
{G(s)} = g(t) and a and b be any two constants. Then
L-1
[a F(s) b G(s)] = a L-1
{F(s)} b L-1
{ G(s)}
2) Shifting Property
If L-1
{F(s)}=f(t) then L-1
[F(s-a)]= ate L
-1{F(s)}
3) Inverse transform of derivative
If L-1
{F(s)}=f(t) then L-1
{Fn(s)}= )}({)1( 1 sFLt nn
4) Division by s
If L-1
{F(s)}=f(t) then 1
0
F(s)( )
s
t
L f t dt
Table of Inverse Laplace Transforms of some standard functions
F(s) )()( 1 sFLtf
0,1
ss
1
asas
,1
ate
0,22
sas
s
Cos at
0,1
22
s
as a
atSin
asas
,1
22
a
ath Sin
asas
s
,
22
ath Cos
0,1
1
s
sn
n = 0, 1, 2, 3, . . .
!n
t n
0,1
1
s
sn
n > -1
1 n
tn
Examples
1. Find the inverse Laplace transforms of the following:
22
)(as
bsii
22 9
94
254
52)(
s
s
s
siii
52
1)(
si
Solution:
2
5
11
2
1
25
1
2
1
52
1)(
t
es
Ls
Li
ata
bat
asLb
as
sL
as
bsLii sincos
1)(
22
1
22
1
22
1
9
29
4
4
252
5
4
2
9
84
254
52)(
2
1
2
1
22
1
s
sL
s
sL
s
s
s
sLiii
thth
tt3sin
2
33cos4
2
5sin
2
5cos
2
1
Exercise:
Find the inverse Laplace transform of the following
25
14
36
2)(
22
s
s
s
si (ii)
6
3)2(
s
s (iii)
8
2532
s
s (iv)
sssss
8312
Evaluation of L-1
F(s – a)
We have, if L{ f(t)} = F(s), then L[eat f(t)] = F(s – a), and so
L-1
F(s – a) = eat f(t) = e
at L
-1 F(s)
Examples
41
1
13: Evaluate 1.
s
sL
41
3
1
4
1-
1
12
1
13
1
11-1s3 L Given
sL
sL
s
4
1
3
1 12
13
sLe
sLe tt
Using the formula
getweandntakingandn
t
sL
n
n,32
!
11
1
32
3 32 teteGiven
tt
52s-s
2sL : Evaluate 2.
2
1-
41
13
41
1
41
31
41
2 L
2
1
2
1
2
1
2
1-
sL
s
sL
s
sL
s
s
4
13
4 2
1
2
1
sLe
s
s Le t-t
tet e tt 2sin
2
32cos
(3)Evaluate13
12:
2
1
ss
sL
45
1
45
2
45
12L
2
2
3
1
2
2
3
2
3
1
2
2
3
2
3
1-
sL
s
sL
s
s
45
1
45
22
12
3
2
12
3
sLe
s
sLe
tt
ththe
t
2
5sin
5
2
2
5cos2 2
3
Exercise
Find the inverse Laplace transforms of the following
(i) 136
52
ss
s (ii)
944
472
ss
s (iii)
2
4
)2(
s
e s
(iv) 4)1(
)2(
s
es s
INVERSE LAPLACE TRANSFORM OF e-as
F(s)
Evaluation of L-1
[e-as
F(s)]
We have, if L{f(t)} = F(s), then L[f(t-a) H(t-a) = e-as
F(s), and so
L-1
[e-as
F(s)] = f(t-a) H(t-a)
Examples
4
51
2:)1(
s
eLEvaluate
s
Here
4
2 31 1 2 1
4 4
32 551
4
15, ( )
2
1 1( ) ( )
62
5( ) ( ) 5
62
tt
ts
a F ss
e tTherefore f t L F s L e L
ss
Thus
e teL f t a H t a H t
s
41:)2(
2
2
2
1
s
se
s
eLEvaluate
ss
22coscos
222cossin
)1(
2cos4
)(
sin1
1)(
)1(22
2
12
2
11
21
tHttHt
tHttHtGiven
asreadsrelationNow
ts
sLtf
ts
LtfHere
tHtftHtfGiven
Exercise:
Find the inverse Laplace transform of the following
(i)
s
e
s
e
s
ss 2
32
323 (ii) 23
2cosh
se
ss
(iii) 2
31
s
e s (iv) 22
2
s
ese ss
INVERSE LAPLACE TRANSFORM BY PARTIAL FRACTION AND
LOGARITHMIC FUNCTION
Examples
(1) )2)(1(
1
ss
Let F(s)= )2)(1(
1
ss
By applying partial fraction we get
)2)(1(
1
ss=
)2)(1(
)1()2(
)2()1(
ss
sBsA
s
B
s
A
1= )1()2( sBsA
put s=2 3
131 BB
put s=-13
131 AA
Therefore
F(s)= )2(
3
1
)1(
3
1
ss
F(s)= )1(
1
3
1
)2(
1
3
1
ss
Taking inverse laplace transform, we get
)}({1 sFL)1(
1
3
1
)2(
1
3
1 11
sL
sL = tt ee
3
1
3
1 2
sss
ss
2
452 : Evaluate 2.
23
2
havewe
1212
452
2
452
2
452 2
2
2
23
2
s
C
s
B
s
A
sss
ss
sss
ss
sss
ss
Then 2s2+5s-4 = A(s+2) (s-1) + Bs (s-1) + Cs (s+2)
For s = 0, we get A = 2, for s = 1, we get C = 1 and for s = -2, we get B = -1. Using these values
in (1), we get 1
1
2
12
2
45223
2
ssssss
ss
tt eess
ssL
2
22
21 2
25
452
21
54: Evaluate 3.
2
1
ss
sL
Consider
21121
5422
s
C
s
B
s
A
ss
s
Then
4s + 5 = A(s + 2) + B(s + 1) (s + 2) + C (s + 1)2
For s = -1, we get A = 1, for s = -2, we get C = -3
Comparing the coefficients of s2, we get B + C = 0, so that B = 3. Using these values in (1), we
get
2
3
1
3
1
1
21
5422
sssss
s
Hence
sLe
sLe
sLe
ss
sL ttt 1
31
31
21
54 121
2
1
2
1
ttt eete 233
44
31: Evaluate 4.
as
sL
Let
)1(2244
3
as
DCs
as
B
as
A
as
s
Hence
s3 = A(s + a) (s
2 + a
2) + B (s-a)(s
2+a
2)+(Cs + D) (s
2 – a
2)
For s = a, we get A = ¼; for s = -a, we get B = ¼; comparing the constant terms, we get
D = a(A-B) = 0; comparing the coefficients of s3, we get
1 = A + B + C and so C = ½. Using these values in (1), we get
2244
3
2
111
4
1
as
s
asasas
s
Taking inverse transforms, we get
ateeas
sL atat cos
2
1
4
144
31
athat coscos2
1
1: Evaluate 5.
24
1
ss
sL
Consider
11
2
2
1
111 222224 ssss
s
ssss
s
ss
s
1
1
1
1
2
1
11
11
2
12222
22
ssssssss
ssss
4
3
1
4
3
1
2
1
1
4
3
1
4
3
1
2
1
2
12
1
2
12
1
24
1
2
2
12
2
1
s
Le
s
Less
sL
ss
tt
2
3
2
3sin
2
3
2
3sin
2
12
1
2
1 t
e
t
ett
2sin
2
3sin
3
2 tht
Transform of logarithmic and inverse functions
.ttfL then F(s), f(t)} L{ if have, We HencesFds
d )(1 ttfsF
ds
dL
Examples
bs
asLEvaluate log:)1( 1
b
eetfThus
eetftor
eesFds
dLthatSo
bsassF
ds
dThen
bsasbs
assFLet
atbt
atbt
btat
1
11
logloglog)(
(2) Evaluate 1L
s
a1tan
oftransformInverse
s
sF
Since
t
s
sFdttfL
0
we have
t
dttfs
sFL
0
1
a
attf
attftor
thatsoatsFds
dLor
as
asF
ds
dThen
s
asFLet
sin
sin
sin
tan)(
1
22
1
Examples:
22
1 1:)1(
assLEvaluate
2
0
1
22
1
1
22
cos1
sin1
sin)(
1
a
at
dta
at
s
sFL
assLThen
a
atsFLtf
thatsoas
sFdenoteusLet
t
1211
1111
get we this,Using
parts.by n integratioon,111
1
1
1:)2(
3
0
222
1
2
0
2
1
2
1
22
1
atat
t
at-
at
t
at
at
eeata
dtateaass
L
atea
dtteass
LHence
teas
Lhavewe
assLEvaluate
CONVOLUTION THEOREM AND L.T. OF CONVOLUTION
INTEGRAL
The convolution of two functions f(t) and g(t) denoted by f(t) g(t) is defined as
f(t) g(t) =
t
duugutf0
)()(
Property:
f(t) g(t) = g(t) f(t)
Proof :- By definition, we have
f(t) g(t) =
t
duugutf0
)()(
Setting t-u = x, we get
f(t) g(t) =
0
))(()(t
dxxtgxf
=
t
tftgdxxfxtg0
)()()()(
This is the desired property. Note that the operation is commutative.
CONVOLUTION THEOREM:
L[f(t) g(t)] = L{f(t)}.L{g(t)}
Proof: Let us denote
f(t) g(t) = (t) =
t
duugutf0
)()(
Consider
0 0
)]()([)]([
t
st dtugutfetL
=
0 0
)()(
t
st duugutfe (1)
We note that the region for this double integral is the entire area lying between the lines u =0 and
u = t. On changing the order of integration, we find that t varies from u to and u varies from 0
to .
u
u=t
t=u t =
0 u=0 t
Hence (1) becomes
L[(t)] =
0
)()(u ut
st dtduugutfe
=
0
)( )()( dudtutfeugeu
utssu
=
0 0
)()( dudvvfeuge svsu , where v = t-u
=
0 0
)()( dvvfeduuge svsu
= L g(t) . L f(t)
Thus
L f(t) . L g(t) = L[f(t) g(t)]
This is desired property.
Examples:
1. Verify Convolution theorem for the functions f(t) and g(t) in the following cases :
(i) f(t) = t, g(t) = sint (ii) f(t) =t, g(t) = et
(i) Here,
f g =
t
duutguf0
)()( =
t
duutu0
)sin(
Employing integration by parts, we get
f g = t – sint
so that
L [f g] = )1(
1
1
112222
ssss
(1)
Next consider
L f(t) . L g(t) = )1(
1
1
112222
ssss
(2)
From (1) and (2), we find that
L [f g] = L f(t) . L g(t)
Thus convolution theorem is verified.
(ii) Here f g =
t
ut duue0
Employing integration by parts, we get
f g = et – t – 1
so that
L[f g] = )1(
111
1
122
sssss
(3)
Next
L f(t) . L g(t) = )1(
1
1
1122
ssss
(4)
From (3) and (4) we find that
L[f g] = L f(t) . L g(t)
Thus convolution theorem is verified.
2. By using the Convolution theorem, prove that
t
tLfs
dttfL0
)(1
)(
Let us define g(t) = 1, so that g(t-u) = 1
Then
t t
gfLdtutgtfLdttfL0 0
][)()()(
= L f(t) . L g(t) = L f(t) . s
1
Thus
t
tLfs
dttfL0
)(1
)(
This is the result as desired.
3. Using Convolution theorem, prove that
t
u
ssduuteL
0
2 )1)(1(
1)sin(
Let us denote, f(t) = e-t g(t) = sin t, then
t t
u duutgufLduuteL0 0
)()()sin( = L f(t) . L g(t)
= )1(
1
)1(
12
ss
= )1)(1(
12 ss
This is the result as desired.
equation. integral thesolve tomethod Transform LaplaceEmploy 4)
t
0
sinlf(t) duutuf
equation. integralgiven theofsolution theis This
21
1)(
1)(
1
)(1sin)(
1
get wehere, n theorem,convolutio usingBy
sin1
get weequation,given theof transformLaplace Taking
2
3
21
3
2
2
t
0
t
s
sLtfor
s
stfL
Thus
s
tfL
stLtLf
sL f(t)
duutufLs
L f(t)
Exercise:
Solve the following problems
1. Verify convolution theorem for the following pair of functions:
(i) f(t) = cosat, g(t) = cosbt
(ii) f(t) = t, g(t) = t e-t
(iii) f(t) = et g(t) = sint
2. Using the convolution theorem, prove the following:
(i)
t
u
ss
sudueutL
0
22
1
)1()1(cos)(
(ii) 22
0)(
1)(
assduueutL
t
au
t
t
u
fduuutfttf
dueutftf
0
0
2
4)0(,cos)(')4
21)()3
t
duugutftgtfsGsFL
soandsGsFtLgtLf
0
1 )()()()(
)()()()(g(t) f(t)L
thenG(s), Lg(t) and F(s) L(t) if have, We
theorem nconvolutio using by F(s) of transform Inverse
transformLaplace inversefor n theoremconvolutio thecalled is expression This
Examples
Employ convolution theorem to evaluate the following :
bsasL
1)1( 1
ba
ee
ba
ee
dueedueebsas
L
e, g(t) ef(t)
bssG
as
atbttbaat
t
ubaat
t
buuta-
-bt-at
1
1
n theorem,convolutioby Therefore,
get weinverse, theTaking
1)(,
1 F(s) denote usLet
00
1
222
1)2(as
sL
a
att
a
auatatu
duauatat
a
duauutaaas
sL
at , g(t) at
f(t)
as
ssG
asF(s)
t
t
t-
2
sin
2
2cossin
2a
1
formula angle compound usingby ,2
2sinsin1
cossin1
n theorem,convolutioby Hence
cosa
sin
Then)(,1
denote usLet
0
0
0
222
1
2222
11)3(
2
1
ss
sL
Here
1)(,
1
1
2
s
ssG
sF(s)
Therefore
ttettee
uue
eduuess
L
t , g(t) ef(t)
ttt
tu
tut-
t
cossin2
11cossin
2
cossin2
sin11
1
have wen theorem,convolutioBy
sin
0
2
1
By employing convolution theorem, evaluate the following:
21
54)6(
1)3(
1
1)5(
11)2(
,)4(11
1)1(
2
1
222
1
22
1
22
1
2222
21
2
1
ss
sL
asL
ssL
ss
sL
babsas
sL
ssL
Applications of Laplace transform
Laplace transform is very useful for solving linear differential equations with constant
coefficients and with given initial conditions. We take the Laplace transforms of the differential
equations and then make use of the initial conditions, which transforms the differential equations
to an algebraic equation. Solve for Laplace transform from this algebraic equation and the
required solution is obtained by taking the inverse of this transform.
Laplace transform of derivatives:
1. ' (0)L f s L f f
2. '' 2 0 '(0)L f s L f s f f
3. ''' 3 2 ' 0 0 ''(0)L f s L f s f s f f and so on.
Problems:
1. Solve'' 2 cosy a y at given
'0, 0y y when 0t .
Solution: Taking Laplace transforms,
2 ' 2
2 2 0 0 Y
ss Y s y y a
s a
On using the initial condition,
2
2 2
sY
s a
Taking inverse,
1
sin2
y t ata
2. Solve '' ' 25 6 ty y y e , given '0 0 1y y
Solution: Taking Laplace transforms,
2 ' 1 0 0 5 0 6
2s Y sy y sY y Y
s
On using the initial condition,
2 2
22
8 13 8 13
2 5 6 3 2
s s s sY
s s s s s
Resolving into partial fractions,
2
2 3 1
3 2 2Y
s s s
Taking the inverse Laplace transforms,
3 2 2 1 2 3 2
2
12 3 ( ) 3 2t t t t t ty e e e L y e e te
s
Exercises :-
Solve the following:
1. '' ' 2 '3 2 1 , 0 0 1ty y y e y y
2. ''' '' ' ' ''4 4 1 , 0 0 0 0y y y y t y y y
Electrical circuits:
Consider a simple circuit comprised of an inductance of magnitude L (henrys ), a resistance of
magnitude R (ohms), and capacitance of magnitude C (farads) connected in series.
If E is the emf (volts) applied to an LRC circuit, then the current i (amperes) in the circuit at
time t is governed by the differential equation,
di q
L Ri Edt C
Here q is the charge(coulomb) is related to i through the relation dq
idt
. If (0) 0q then the
above equation can be rewritten as,
0
1t
diL Ri i dt E
dx C
Examples:
1. A LR circuit carries an emf of voltage 0 sinE E t , where 0E and are constants. Find
the current i in the circuit if initially there is no current in the circuit.
Solution: The differential equation governing the current i is,
0 sindi
L Ri E tdt
or, 0 sinEdi R
i tdt L L
Taking Laplace transforms on both sides,
02 2
(0)E
sI i aIL s
, where
Ra
L
Applying the initial condition, (0) 0i , we get,
0
2 2
1EI
L s a s
By resolving into partial fractions,
02 2 2 2 2 2
1E sI a
L s aa s s
Taking inverse Laplace transforms,
02 2
1sin cosatE
i e a t tL a
2. A resistance R in series with an inductance L is connected with emf ( )E t . The current i
is given by, ( )di
L Ri E tdt . The switch is connected at time 0t and disconnected at
time 0t a Find the current i in terms of t , given that the emf is constant when the
switch is on.
Solution: 0
( )0
E t aE t
t a
Here consider E as a constant.
( ) ( ) ( ) , 0E t E H t H t a t
The governing equation becomes,
( ) ( )di R E
i H t H t adt L L
Taking Laplace transforms and applying the initial condition, (0) 0i ,
1 asE esI I
L s s
, where R
L
1 1 11
( )
asasE e E
I eL s s R s s
Taking inverse Laplace transforms,
( )1 1 ( )t t aEi e e H t a
R
1 0
1
t
t a
Ee t a
Ri
Ee e t a
R
Exercises:
1. A voltage atEe is applied at 0t to a circuit of inductance L and resistance R . Show
that current at time t isRt
at LE
e eR aL
.
2. A simple electrical circuit consists of resistance R and inductance L in a series with
constant emf E . If the switch is closed when 0t , find the current at any time t .
Mass spring systems
Consider a spring of length x , tied at one end to a support and the other end is tied to a fixed
mass m which is free. If F is the force acting on the object, then from Newton’s second law of
motion,
2
2
d xm kx
dt
Suppose the medium through the setup is worked is resisting with a velocity of '( )x t we have,
2
2'
d xm kx cx
dt
''( ) "( ) ( ) 0mx t cx t kx t
The auxiliary equation is given by, 2 0mD cD k
The roots of the quadratic equation are, 2 4
2
c c mkD
m
The value of 2 4c mk determines the sensitivity of the medium.
1. 2 4 0c mk implies motion is under-damped.
2. 2 4 0c mk implies the motion is critically damped.
3. 2 4 0c mk implies the motion is over-damped.
Here k is the stiffness of the spring and can be given by the weight of the object per unit total
length of the spring, w
kb
, where b is the length of the spring entirely and w mg
1. A spring can extend 20cm when 0.5kg of mass is attached to it. It is suspended vertically
from a support and set into vibration by pulling it down 10cm and imparting a velocity of
5 /cm s vertically upwards. Find the displacement from its equilibrium.
Solution: Let ( )x t be the displacement from its equilibrium.
Here 20 , 500 , (0) 10 , '(0) 5 /b cm m gm x cm x cm s
We can thus calculate the value of k , i.e. 24500 /k dy cm
The equation of motion is , 500 ''( ) 24500 ( ) 0x t x t
''( ) 49 ( ) 0x t x t
Taking Laplace transforms,
2 (0) '(0) 49 0s X sx x X
2 2
10 5
49 49
sX
s s
Taking the inverse Laplace transform,
10 5( ) cos7 sin 7
7 7x t t t
2. A spring of stiffness k has a mass m attached to one end. It is acted upon by external
force sinA t . Discuss its motion in general.
Solution: We know from the Newton’s law of motion, 2
2sin
d xm kx a t
dt ,
0 0(0) , '(0)x x x v
''( ) ( ) sink
x t x t A tm
Taking Laplace transforms,
2
2 2(0) '(0)
k As X s x x X
m m s
By applying the initial conditions,
0 02 22 2 2
1
k kkm mm
A sX x v
m s ss s
Taking the inverse Laplace transforms,
0 02 2
sinsin
( ) cos
kt
A t k Amx t x t v
m kk m k mm
Exercise:
1. A spring is stretched 6 inches by a 12pound weight. Let the weight be attached to the
spring and pulled down 4 inches below the equilibrium point. If the weight id started
with an upward velocity of 2 feet per second, describe the motion. No damping or
impressed force is present.
2. A spring is such that 4 lb weight stretches it 6 inches. An impressed force cos8
2
tis acting
on the spring. If the 4 pound weight is started from the equilibrium point with an upward
velocity of 4 feet per second, describe the motion.
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