5 Laplace Transform

THE LAPLACE TRANSFORM

Chapter 5

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Why?

••ThereThere areare DifferentialDifferential EquationsEquations thatthat willwill bebedifficultdifficult forfor usus toto solvesolve justjust knowingknowing whateverwhatevermethodsmethods wewe havehave learntlearnt soso farfar..

••TheThe LaplaceLaplace transformtransform isis usedused toto produceproduce ananeasilyeasily solvablesolvable algebraicalgebraic equationequation fromfrom ananordinaryordinary differentialdifferential equationequation..

••ItIt hashas importantimportant applicationsapplications inin Mathematics,Mathematics,PhysicsPhysics andand EngineeringEngineering..


Plan

I - Definition and basic properties

II - Inverse Laplace transform and solutions of DE

III - Operational Properties


I – Definitions and basic properties

At the end of the lesson you should be

able to :

• Define Laplace Transform.

• Find the Laplace Transform of different type of functions using the definition.

Learning objective


Integral Transform

If f ( t ) is defined for and

is a function of two variables, then the

improper integral

is an integral transform.

0t

dttftsKdttftsKb

b)(),(lim)(),(

00

),( tsK


If the limit exist, then the integral exists or

is convergent. (In general, the limit will exist

for only certain values of the variable s).

If the limit does not exist, the integral does

not exist and is divergent.

Integral Transform


Definition: Laplace Transform

Let f be a function defined for

Then the integral

is said to be the Laplace transform of f,

provided that the integral converges.

.0t

dttfetfL st )()(0


Notations

)()(

;)()(;)()(

sYtyLand

sGtgLsFtfL


Use the definition to find the values of the following:

}{.2

}1{.1

tL

L

Example 1


Use the definition to find the values of the following:

}{.3

}{.2

}1{.1

3teL

tL

L

Example 1


?}1{L

Solution

dteL st 1}1{0

dteL st

0

}1{

Nst

NdteL

0lim}1{

Nst

N s

eL

0

lim}1{

s

eL

sN

N

1lim}1{

sL

1}1{


?}{tL

Solution

s

evedv

dtdutust

st

dt

22

0

2

110}{

sss

e

s

tetL

stst

dttetL st

0}{

dtess

tetL st

st

00

1}{


?}{ 3teL Solution

dteeeL tstt 3

0

3 )(

dte tst

0

3

dte ts

0

)3(

0

)3(

)3(s

e ts

0

)3()3(

)3()3(

s

e

s

e tsts

)3(

10

s

)3(

1

s


Theorem: Transforms of some Basic Functions

22

2222

221

}{sinh.4

}{cosh.7}{sin.3

}{cos.6!

}{.2

1}{.5

1}1{.1

ks

kktL

ks

sktL

ks

kktL

ks

sktL

s

ntL

aseL

sL

n

n

at


is a Linear TransformL

)}({)}({)}()({ 2121 tgLctfLctgctfcL


Example2

15)( ttf

Find the Laplace transform of the function


Example2

}51{ tL

}5{}1{}51{ tLLtL

}{5}1{ tLL

2

15

1

ss

22

551

s

s

ss


Transform of a Piecewise function

Given

dttfetfL st )()(0

11

101)(

t

ttf

Find

Example 3


1

1

0

)}({ dtedtetfL stst

1

1

0

11 stst es

es

)0(1

)1(1

ss es

es

Solution

)12(1

ses


Laplace Transform of a Derivative

Let

0

'' )()( dttfetfL stFind

)()( tfLsF



0

'' )()( dttfetfL st

dttfestfe stst )()(00

)}({)0( tfLsf

)0()( fssF

f(t)v(t)dt fdv

dtsedueu'

stst



dttfetfL st )()(0

dttfestfe stst )()(00

)}({)0(0 ' tfsLf

)]0()([)0( fssFsf

)0()()0( 2 sfsFsf

)0()0()(2 fsfsFs



Theorem

)0(....)0()0()()}({ )1(21)( nnnnn ffsfssFstfL

)()( tfLsF where



Example

tyyy 423 '''

Find the Laplace transform of the following IVP

1000 )( y)y( '



)4()23( ''' tLyyyL

)(4)(2)(3)( ''' tLyLyLyL

Solution tyyy 423 '''

2

'2 4)(2)0()(3)0()0()(

ssYyssYysysYs

2

2 4)(2)(31)(

ssYssYsYs


2

2 4)(2)(31)(

ssYssYsYs


Solution

2

2 41)()23(

ssYss

2

22 4

)()23(s

ssYss

)1(

2

)1)(2(

4)(

22

2

ss

s

sss

ssY


II – Inverse Laplace Transform and solutions of DEs


able to :

• Define Inverse Laplace Transform.

• Solve ODEs using the Laplace Transform.

Learning objective


Inverse Transforms

If F (s) represents the Laplace transform of a

function f (t), i.e.,

L {f (t)}=F (s)

then f (t) is the inverse Laplace transform of F (s)

and, )}({)( 1 sFLtf


Theorem : Some Inverse Transforms

22

1

22

1

22

1

22

1

1

1

11

sinh.4

cosh.7sin.3

cos.6...,3,2,1,!

.2

1.5

11.1

ks

kLkt

ks

sLkt

ks

kLkt

ks

sLktn

s

nLt

asLe

sL

n

n

at


Application

ts

Ls

L

ts

Ls

Ls

L

6sin6

1

6

6

6

1

36

1.2

60

1!5

!5

212

2.1

22

1

2

1

5

15

1

15

1

6

1


is a Linear Transform1L

)}({)}({)}()({ 111 sGLsFLsGsFL

Where F and G are the transforms of some functions f and g . This can also extend to any finite combination of Laplace transform


Division and Linearity

Find

4

432

1

s

sL


Division and Linearity

22

1

22

1

2

1

2

4

2

3

4

43

sL

s

sL

s

sL

22

1

22

1

2

2)2(

2)3(

sL

s

sL

tt 2sin)2(2cos)3(


Partial Fractions in Inverse Laplace

Find

)2)(1(

11

ssL


Partial Fractions in Inverse Laplace

1

1

2

1

)2)(1(

1

ssss

)2)(1(

11

ssL

1

1

2

11

ssL

1

1

2

1 11

sL

sL

tt ee 2

,


Method to solve DE

DEApply Laplace

transform Lequation in )(sY

Apply Inverse Laplace transform

1L

Solution )(sYSolutionof the initial DE

)(ty


Example 1

Solve the partial given IVP by

Laplace transform.

3)0(,02 yydt

dy


Solution 1

3)0(,02 yydt

dy

02

y

dt

dyL

02

yL

dt

dyL

0)()0()(2 sYyssY


Solution 1

0)()0()(2 sYyssY

06)()12( sYs

12

6)(

ssY

12

6)( 1

sLty

))2

1((2

61

s

L


Solution 1

)2

1(

13)( 1

s

Lty

23)(t

ety


Example 2

Solve the given IVP DE by Laplace transform

6)0(,2sin133 ytydt

dy


Solution 2

tLyLdt

dyL 2sin133

4

213)(3)0()(

2ssYysYs

4

26)(36)(

2

ssYsYs

4

)4(6

4

26)(3

2

2

2

s

s

ssYs


Solution 2

)4)(3(

506

4

24626

3

1)(

2

2

2

2

ss

s

s

s

ssY

)4)(3(

506)(

2

21

ss

sLty

)4)(3(

)3)(()4(

43)4)(3(

5062

2

22

2

ss

sCBssA

s

CBs

s

A

ss

s

8A 2B 6C……………………


Solution 2

4

62

3

8

)4)(3(

506)(

2

11

2

21

s

sL

sL

ss

sLty

4

16

42

3

18

2

1

2

11

sL

s

sL

sL

4

2

2

62cos28

2

13

sLte t

tte t 2sin32cos28 3


III – Operational Properties


able to use translation theorems.

Learning objective


First translation theorem

)()}({ sFtfL

a

)()}({ asFtfeL at

If

and is any real number, then

.



.

}{ 35 teL t

45

45335

)5(

6!3}{}{

ss

tLteLss

sst

Example 1:



.

Example 2:

}4cos{ 2 teL t

16)2(

2

4}4{cos}4cos{

22

2222

s

s

s

stLteL

ss

sst


.

)()}({ asFtfeL at

)}({)}({ 11 asFLtfeLL at

)}({)( 1 asFLtfeat

Inverse form of First translation theorem


.

Example 1:

2

1

)3(

52

s

sL

22 )3(

11

3

2

)3(

52

sss

s

2

11

2

1

)3(

111

3

12

)3(

52

sL

sL

s

sL

3

2

13

)(

111)1(2

ss

t

sLe

tt tee 33 112

Inverse form of First translation theorem


Exercise

.0)0(,0)0(,164 yyeyyy t

Solve


}{}1{}{6}{4}{ teLLyLyLyL

1

11)(6)]0()([4)0()0()(2

sssYyssYysysYs

1

11)(6)(4)(2

sssYssYsYs

Solution


1

11)()64( 2

sssYss

)1(

12)()64( 2

ss

ssYss

)64)(1(

12)(

2

ssss

ssY

Solution


)64(1)64)(1(

1222

ss

DCs

s

B

s

A

ssss

s

)64(3

5

21

3

1

6

1

)64)(1(

1222

ss

s

ssssss

s

Solution


)64(3

5

21

3

1

6

1

)64)(1(

12)(

22

ss

s

ssssss

ssY

})64(

3

5

2{}1

3

1

{}6

1

{)}({2

1111

ss

s

Ls

Ls

LsYL

Solution


})64(

3

5

2{}1

3

1

{}6

1

{)}({2

1111

ss

s

Ls

Ls

LsYL

}2)2(

2{

23

2}

2)2(

2{

2

1}

1

1{

3

1}

1{

6

1)}({

2

1

2

1111

sL

s

sL

sL

sLsYL

.2sin23

22cos

2

1

3

1

6

1)( 22 teteety ttt

Solution


Unit Step Function or Heaviside Function

The unit step function is defined as

)( atU

.,1

0,0)(

at

atatU

U

t

1


When a function, say for ismultiplied by the unit stepfunction “ turn off ” a portion of the graphof that function.

)(tf 0t)( atU

Remark


What happen when is

multiplied by the Heaviside function

32)( ttf

Example

?)2()( tUtf


32)( ttf

Example

)2()( tUtf

-3

t

f

t

-3

2

f

0 0


The Second Translation Theorem

If and then )}({)( tfLsF ,0a

).()}()({ sFeatUatfL as


Example 1

Let

sLsF

atftf

1}1{)(

1)(1)(

.)}({s

eatUL

as

then


Example 2

Find

3,2

30,)(

t

t ttg

)3(2))3(1()( tUtUttg

)}({ tgL where

)}.3({2)}3({}{)}({ tULttULtLtgL

)3(2)3( tUttUt


Example 2

)}.3({2)}3({}{)}({ tULttULtLtgL

.2)}3({1 3

2 s

ettUL

s

s

For We would like to apply the previous theorem )}3({ ttUL

So, we write ttf )3(

3)( ttf


Example 2

Then

where

)()}3({ 3 sFettUL s

Finally

sstLtfLsF

31}3{)}({)(

2

.1231

)}({3

2

333

2

3

2 s

e

s

e

s

e

s

e

s

e

stgL

sssss


The Inverse Second Translation Theorem

If then )}({)( 1 sFLtf

).()()}({1 atUatfsFeL as


Example

Evaluate

}4

1{ 21 se

sL


)2(}4

1{ )2(421

tUee

sL ts

.}4

1{ ,

4

1)( ,2

}4

1{

41

21

t

s

es

Ls

sFa

es

L

therefore,

Example


5 Laplace Transform

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