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.