laplace transform example solution

7/23/2019 Laplace Transform Example Solution

1/105

Laplace Transform


2/105

Laplace Transform

Example 1:

Find the Laplace transforms of

a. f(t) = a where: a = constantb. f(t) = eat

c. f(t) = cos(at)

d. f(t) = sinh(at)


3/105

Solution to Example 1

(a) f(t) = a where: a = constant

s

a

s

ae

s

aeaL

s

aeaL

dteaaL

dtaeaL

ss

st

st

st

)0()(

0

0

0

}{

}{

}{

)(}{


4/105


(b) f(t) = eat

asas

e

as

eeL

as

e

eL

dteeL

dteeeL

asasat

tasat

tasat

atstat

1}{

}{

}{

)(}{

)0)(())((

0

)(

0

)(

0


5/105


(c) f(t) = cos(at)

0 22}{cos

2cos

sincos

sincos

dtee

eee

LatL

ee

at

atjate

atjate

IdentityEulers

jatjatst

jatjat

jatjat

jat

jat


6/105


(c) f(t) = cos(at)

2222

)0)(())(()0)(())((

0

)(

0

)(

0

)()(

)()(

2

111

2

1}{cos

2

1}{cos

2

1}{cos

2

1}{cos

as

s

as

jasjas

jasjasatL

jas

e

jas

e

jas

e

jas

eatL

jas

e

jas

eatL

dteeatL

jasjasjasjas

tjastjas

tjastjas


7/105


(d) f(t) = sinh(at)

0 22}{sinh

2

sinh

sinhcosh

sinhcosh

dtee

eee

LatL

eeat

atate

atate

IdentityEulers

atatst

atat

atat

at

at


8/105


(d) f(t) = sinh(at)

2222

)0)(())(()0)(())((

0

)(

0

)(

0

)()(

)()(

2

111

2

1}{sinh

21}{sinh

2

1}{sinh

2

1}{sinh

as

a

as

asas

asasatL

ase

ase

ase

aseatL

as

e

as

eatL

dteeatL

asasasas

tastas

tastas


9/105

Laplace Transform

Example 2:

Find the Laplace transforms of

1. x(t) = -1 0


10/105


(1)

se

se

sse

se

se

setx

s

e

s

etx

dtedtedtetx

ssssss

stst

ststst

42)4()()0()2(

4

2

0

4

4

2

2

0

1)}({

)}({

)1()0()1()}({


11/105


(2)

3

3

2

3

2

2

3

2

2

0

3

3

2

3

2

3

2

2

0

3

3

2

3

2

2

0

3

3

2

2

0

21042)}({

21042)}({

)2()10()4()2()}({

)2()104()2()}({

s

e

s

e

s

e

s

et

s

etL

se

sedt

se

set

setL

s

evdtdu

dtedvtu

dtedtedttedtetL

dtedttedtetL

ststststst

ststststst

st

st

stststst

ststst


12/105


(2)

2

2

2

3

323

2

2

2

3232

)3()()2()3(

2

)2(

2

)3()2()3()0()2(

442)}({

210104481222)}({

210

)2()3(42)}({

s

e

s

e

stL

s

e

s

e

s

e

s

e

s

e

s

e

s

e

ss

etL

s

e

s

e

s

e

s

e

s

e

s

e

s

e

s

e

s

e

s

etL

ss

ssssssss

ssss

ssssss


13/105

Seatwork 1

Derive the following f(t):

1. sin(at)

2. cosh(at)3. tn

4. tcost


14/105

Laplace Transform

Example 4:

Use Table 1 to determine the Laplace transform of eachof the following functions:

2cos:)(

:)(

4sin:)(

:)(

:)(

2

7

3

te

ed

tc

tb

ta

t

tej

tei

tth

tg

tf

t

t

cos:)(

2sin:)(

4sin:)(

5cosh:)(

3sinh:)(

3


15/105


(a) (c)

(b) (d)

413

3

1

3

6!3

!

ss

t

s

ntt

n

n

817

7

1

7

5040!7

!

sst

s

ntt

n

n

16

4

4

44sin

sin4sin

222

22

ss

t

stt

2

1

1

2

2

se

as

ee

t

att


16/105


(e) (g)

(f) (h)

41)21(2cos

cos2

cos

222

22

s

s

s

st

s

st

t

9

3

3

33sinh

sinh3sinh

222

22

sst

stt

2555cos

cosh5cosh

222

22

s

s

s

st

s

stt

22222

222

16

8

4

)4(24sin

2sin4sin

s

s

s

stt

s

stttt


17/105


(i) (j)

4)1(

2

2)1(

22sin

)(sin2sin

222

22

sste

astete

t

att

1)3(

)3(

1)3(

)3(cos

)(

)(coscos

222

3

22

3

s

s

s

ste

as

astete

t

att


18/105

Laplace Transform Property

Example 5:

Find the Laplace transforms of the

following functions:

tetb

ta

25:)(

23:)(

2


19/105


(a)

(b)

211

23)!1(23)(

sssssF

1210

1)1(2)!2(5)( 312 ssss

sF


20/105


Example 6:

Find the Laplace transforms of the following:

)cos(sin:)(65sin23cos5:)(

21

3

ttebttta

t


21/105


(a)

(b)

422

132222

36

25

10

9

5

)(

)!3(6

5

)5(2

3

5)(

sss

s

sF

sss

ssF

1

)(21

1

21

1

1)(

1

)(21

1

)1(21

1

1)(

22

2222

s

s

sssF

s

s

ss

sF


22/105


Example 7:

a.) Use Table1 to find the Laplace transform

b.) Use the first shift theorem to write down

tttf 5sin)(

tteL t 5sin3


23/105


(a)

(b)

222222 510

5

)5(2)(

s

s

s

ssF

222222

5)3(

3010

5)3(

)3(10)(

s

s

s

ssF


24/105


Example 8:

The Laplace transform of a function, f(t) is

given by:

State the Laplace transform of

)1(12)(

ss

ssF

)(:)(

)(:)(3

2

tfeb

tfeat

t


25/105


(a)

(b)

65

52

)3)(2(

142

]1)2)[(2(

1)2(2)(

2

ss

s

ss

s

ss

ssF

6552

)2)(3(162

]1)3)[(3(1)3(2)(

2

sss

sss

ssssF


26/105


Example 9:

Given:

Use the second shift theorem with d=2.

)2()2(:,92

)( tftLfinds

s

tfL


27/105


9

)2(

)(

2

s

se

sF

s


28/105


Example 10:

The Laplace transform of a function is .

Find the function.

2

3

s

e s


29/105


ttf

tftL

d

s

esF

s

)(

33

3,)1(

)(2

3


30/105


Example 11:

Verify the final value theorem fortetf 2)(


31/105


00

20

1)0(

lim2

1

lim

)(lim)(lim

)(2

2

0

0

e

ess

tfssF

t

ts

ts


32/105


Example 12:

Verify the initial value theorem for

25)2()2(4)( 2

s

ssF


33/105


294

)2(4lim5cos4lim

25)2(

)2(4lim5cos4lim

)(lim)(lim

5cos4)(;25)2(

)2(4)(

2

22

0

2

2

0

0

2

2

ss

sste

s

sste

ssFtf

tetfs

ssF

s

t

t

s

t

t

st

t


34/105


44

)(25

)()(41

)(

)(21

4)]0(5cos[4

2541

21

4lim5cos4lim

22

2)0(2

22

22

0

e

ss

s

s

s

tes

t

t


35/105


Example 13:

Find the laplace transform of

using differentiation of laplace transform.

tttf 2sin)(


36/105


22

22

12

2

2

4

42sin

4

)02)(1(2)4(02sin

4

2

2sin

4

22sin

s

stt

s

sstt

sds

d

tt

st


37/105


Example 14:

Given x(0) = 2 and x(0) = -1

Write expressions for the Laplacetransforms of:

(a) 2x 3x + x

(b) -x + 2x + x


38/105


(a)

132

84)(

84)(132

624)()(3)(2

0)(6)(324)(2

0)(2)(312)(2

)(

2)()0()('

12)()0(')0()("

2

2

2

2

2

22

ss

ssX

ssXss

ssXssXsXs

sXssXssXs

sXssXssXs

sXx

ssXXssXx

ssXsXsXsXsx


39/105


(b)

12

52)(

52)(12

52)()(2)(

412)()(2)(

0)(4)(212)(

0)(2)(212)(

)(

2)()0()('

12)()0(')0()("

2

2

2

2

2

2

22

ss

ssX

ssXss

ssXssXsXs

ssXssXsXs

sXssXssXs

sXssXssXs

sXx

ssXXssXx

ssXsXsXsXsx


40/105

Seatwork 2

Find the laplace transform of the following

function

tt

yandyywheretyyy

tttt

tt

tt

tt

t

t

t

t

2cosh

2

1.5

0)0(,2)0(',0)0(":0cosh2"4'".4

2sinh2cosh.33,

32,62

21,22

10,1

)(.2

1

1

1.1

2

2

2


41/105

Seatwork 2

6. Perform the following operations:

a. Find the final value of

b. Find the initial value of

2

35)(

ss

sX

s

e

s

ee

ssX

sss 2

2

22)(


42/105

Inverse Laplace Transform


43/105


Example 1:Find the inverse Laplace transforms of thefollowing:

1

1:)(

1

1:)(

1:)(

16:)(

2

:)(

2

2

2

3

3

s

se

sd

s

sc

sb

sa


44/105


tttf

ss

s

s

ssFe

ttf

s

sFd

ttfs

ssFc

ttfs

sFb

ttfs

sFa

sincos)(

1

1

11

1)()(

sin)(;

1

1)()(

cos)(;1

)()(

8)(;16

)()(

)(;2

)()(

222

2

2

2

3

2

3


45/105


Example 2:

Find the inverse Laplace transforms of the


9)1(

15:)(

4)1(

)1(:)(

)2(10:)(

2

2

4

sc

s

sb

sa


46/105


tetf

ss

sFc

tetfs

ss

ssFb

tetfss

sFa

t

t

t

3sin5)(;

3)1(

)3(5

9)1(

15)()(

2cos)(;2)1(

)1(4)1(

)1()()(

6

10)(;

)2(

!3

!3

10

)2(

10)()(

222

222

32

44


47/105


Example 3:

Find the inverse Laplace transforms of the


1182

1:)(

136

32:)(

1363:)(

2

2

2

ss

sc

ss

sb

sssa


48/105


tetf

s

s

ss

ssFa

t 2cos)(23

3

496

3)()(

3

222


49/105


tetetf

ss

s

s

ssF

s

s

ss

ssFb

tt 2sin32cos2)(

23

)2(3

23

)3(2

23

6)3(2)(

23

692

496

32)()(

33

222222

222


50/105


tetetf

ss

ssF

s

s

s

ssF

ss

s

ss

ssFc

tt 5.1sin5.1

35.1cos)(

5.12

3

5.1

5.1

5.12

2)(

5.12

32

5.12

1)(

5.144

1

5.54

1)()(

22

2222

2222

22

I L l T f i


51/105

Inverse Laplace Transform using

Partial Fraction Expansion

Example 4:

Find the inverse Laplace transform of:

23

86:)(

14:)(

2

2

ss

sb

ss

sa


52/105


t

s

s

etfss

sF

ss

ssB

ss

ssA

s

B

s

A

ss

s

ss

sa

31)(1

31)(

31

3

)1(

14)1(

11

1

)1(

14

1)1(

1414)(

1

0

2

tt

s

s

eetf

sssF

ss

ssB

ss

ssA

s

B

s

A

ss

s

ss

sb

24)(

1

2

2

4)(

21

2

)1)(2(

86)1(

41

4

)1)(2(

86)2(

12)1)(2(

86

23

86)(

2

1

2

2

I L l T f i


53/105



Example 5:


sss

ss

23

26323

2


54/105


tt

s

s

s

eetf

ssssF

sss

sssC

sss

sssB

sss

sssA

s

C

s

B

s

A

sss

ss

sss

ss

2

2

2

1

2

0

2

2

2

2

1)(

2

1

1

11)(

12

2

)1)(2(

263)2(

11

1

)1)(2(

263)1(

12

2

)1)(2(

263

21)1)(2(

263

)23(

263

I L l T f i


55/105



Example 6:


234 45

2

)( ssssF


56/105


tt

s

ss

s

s

eettf

sssssF

ssssA

ss

s

ssss

ds

dB

ssssC

ssssD

s

D

s

C

s

B

s

A

sssssssF

3

2

24

1

8

5

2

1)(

1

3

2

4

24

1

8

5

2

1

)(

2

1

4

2

)1)(4(

2)(

8

5

16

10

45

)52(2

)1)(4(

2)(

)!12(

1

24

1

48

2

)1)(4(

2

)4(

3

2

)1)(4(

2)1(

14)1)(4(

2

45

2)(

4

2

02

2

0

220

2

2

)12(

4

2

1

2

22234



57/105


Complex Numbers

Example 7:

Find the inverse Laplace transform of :

13632

sss


58/105


23

2

462

166

2

)13(4366

js

js

s

s

23

:)()(3

))((

3

136

32

jas

letasBbsAs

bs

B

as

A

bsas

s

ss

s


59/105


bsasss

s

B

BA

A

jAbjAj

11

2

1

136

3

1

)4()23(2

2

21

21

te

tjttjte

eee

ee

ee

bsasL

ss

sL

t

t

tjtjt

tjtj

btat

2cos...

2sin2cos2sin2(cos2

1...

)(2

1...

2

1...

)(2

1....

11

2

1

136

3

3

3

223

)23()23(

1

2

1


60/105

Convolution Theorem

Example 8:

Find the convolution of 2t and t3.


61/105


3

3

)(

)(2)(

)(

2)(

vvg

vtvtf

ttg

ttf

10..........

20452

542..........

5

0

54

)0(

4

)(2

542..........

)(2...........

)(2*2

)(2)(*)(

5

5555

5544

0

54

0

43

0

33

0

3

t

tttt

ttttvtv

dvvtv

dvvvttt

dvvvttgtf

t

t

t

t


62/105

Convolution Theorem

Example 9:

Show that f*g = g*f

where f(t)=2t and g(t) = t3


63/105


vvf

andvtvtg

sottg

dvvfvtgtfg

ttttfg

t

2)(

,)()(

,)(

)()())(*(

10*2))(*(

3

3

0

55

10.........

54

3

22.........

)33(2.........

2)33(.........

2)(.........

2**

5

0

5432

23

0

43223

0

3423

0

3

3

t

vtv

vt

vt

dvvtvvtvt

vdvvtvvtt

vdvvt

ttfg

t

t

t

t



64/105


Convolution Theorem

Example 10

Use the convolution theorem to find the inverse

Laplace transform of the following functions:

)4(

3

)(

)3)(2(

1)(

2

ssb

ssa


65/105


t

t

esGLtgesFLtf

then

ssG

ssFa

31

21

)}({)()}({)(

,

3

1)(,

2

1)(:)(

tt

tt

tvt

tvt

tvvt

tvvt

ee

ee

ee

dvee

dveee

dvee

tgf

sGsFLss

L

32

2

0

2

0

2

0

322

0

3)(2

11

..............................

)1(..............................

..............................

..............................

..............................

..............................

))(*(..............................

)}()({)3)(2(

1


66/105


ttgtf

then

ssG

ssFb

2sin)(,.......3)(.....

,

4

1)(,.....3)()..(

21

2

)2cos1(4

3.........................

2

2cos

2

3

.........................

2sin2

3.........................

2

2sin

3.........................

))(*(.........................

)}()({)4(

3

0

0

0

1

2

1

t

v

vdv

dv

v

tgf

sGsFLss

L

t

t

t

S l i Li C t t C ffi i t Diff ti l


67/105

Solving Linear Constant Coefficient Differential

Equations using Laplace Transforms

Example 11

Solve the differential equation using Laplace

transform

3)0(,.....0 xxdtdx


68/105


0)(3)(

0)]([)]0()([

0)(

sXssX

sXxssX

xLdt

dxL

1

3)(

3)()1(

3)()(

ssX

sXs

sXssX

tetx 3)(

S l i Li C t t C ffi i t Diff ti l


69/105

Solving Linear Constant Coefficient Differential

Equations using Laplace Transforms

Example 12

Solve the equation using Laplace transform

3)0(,........9 2

xexdt

dx t


70/105


23

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

2

)1(3)()1(

233

26393

29)()1(

2

9)(3)(

29)()0()(

sssssX

s

ssXs

ss

ss

ssXs

ssXssX

ssXxssX

tetx 23)(


71/105

Seatwork 3

1. Find the inverse laplace of the following:

44

1.

136.

42

2.

1021.

12.

2

2

2

2

4

sse

sssd

s

sc

ssb

sa


72/105

Seatwork 3

2. Use complex operation in finding the inverselaplace:

3. Use partial fraction expansion to find the inverse

laplace transform

84

32.

136

13.

22

ss

sb

ss

sa

22

3 )1(12.

432.

sssb

ssa


73/105

Seatwork 3

4. Find the convolution of the followingfunctions

a.

b.

ttgttf 3sin)()(

ttgetf t

2cos)()(


74/105

Seatwork 3

5. Find the inverse laplace of the given

laplace equations using convolution theorem

a. b.

6. Find the solution to

)9(322

1

ssL

)1)(1(1

2

1

ssL

4)0(';1)0(:

4)(4)('4)(" 2

xxwhere

etxtxtx t


75/105

Electrical Circuits


76/105

Example

Find i(t) of the circuit


77/105

Solution

S


78/105

Solution

VC

qV

s

E

s

V

Cs

sIsRI

s

E

s

CV

s

sI

CsRI

iiiEdtiC

Ri

EVV

VVE

CC

C

CRcR

CR

CR

5.370004.0

015.0)0(;

)0()()(

)0()(1)(

;1

0

Apply Kirchoffs Voltage Law

S l i


79/105

Solution

teti

ssI

sIs

sIssI

s

ssss

sIsI

ssssIsI

25025.1)(

)250(

25.1)(

25.1)()250(

25.1)(250)(

10

5.125.3750)(2500)(10

505.370004.0

)()(10

E l


80/105

Example

The network shows that the switch isclosed when t = 0,assume that i(0) = 0;

q(0) = 0. Find i(t)

S l ti


81/105

Solution

S l ti


82/105

Solution

VC

qV

s

E

s

V

Cs

sIsLsIsRI

Ii

s

E

s

CV

s

sI

C

IssILsRI

iiiiEdtiCdt

diLRi

EVVV

VVVE

CC

C

CLRcL

R

CLR

CLR

000005.0

0)0()0(;

)0()()()(

0)0()0(;)0()(1

)0()()(

;1

0


S l ti


83/105

Solution

teti

ssI

sIs

sIssIsIs

s

ss

sIssIsI

ss

sIssIsI

t1000

2

2

2

3000)(

)1000(3000)(

3000)()1000(

3000)(1000000)(2000)(02.0

60)(20000)(02.0)(40

60

00005.0

)()(02.0)(40

D i ti


84/105

Derivation

s

V

Cs

sIidt

CL

VCqCs

q

Cs

sI

s

q

s

sI

Cidt

CL

idtqs

idt

s

sI

CidtCL

s

dttf

s

sFdttfL

C

C

)0()(1

/;)0()()0()(11

;)(11

)()(

)(

0

0

E l


85/105

Example

Find the currents in the network shownwhen the switch is closed at t = 0 and the

capacitor carries a charge of 0.03

coulomb.

S l ti


86/105

Solution

At t>0, using Kirchoffs Law

S l ti


87/105

Solution

At t>0, using Kirchoffs Law

S l ti


88/105

Solution

Using Kirchoffs Current Law

)1(

0

321

321

eqnIII

III

S l ti


89/105

Solution

Using Kirchoffs Voltage Law in loop 1

)2(1

0

211

1

1

eqndtIC

IRE

VVE

VVE

CR

CR

S l ti


90/105

Solution

Using Kirchoffs Voltage Law in loop 2

)3(

0

3211

21

21

eqnIRIREVVE

VVE

RR

RR

Sol tion


91/105

Solution

Solving eqn. 2

)4(0)(

200)(

10)(2000)(100

60

0005.0

)()(10

60

600005.003.0)0(;)0()()(

1

21

21

21

211

211

eqns

sIsI

ssIsI

ss

sIsI

s

VCqV

sV

CssIsIR

sE

dtIC

IRE

CC

Solution


92/105

Solution

Solving eqn. 3

)(2)(

6

10)(20)(1060

)()(

31

31

3211

3211

sIsIs

sIsIs

sIRsIR

s

E

IRIRE

but

213 III

substitute

)5()(2)(36

)(2)(2)(6

)()(2)(6

21

211

211

eqnsIsIs

sIsIsIs

sIsIsIs

Solution


93/105

Solution

Combining eqn (4) and eqn (5)

ssIsI

s

sIsI

6)(2)(3

30)(

200)(

21

21

26)(

600)(2

6)(2)(3

0)(

600)(3

22

21

21

ss

sIsI

ssIsI

s

sIsI

)(3

300

3)(

3300)(

33001)(

3)(300)(

300

2

2

2

2

22

answereI

ssI

ss

ssI

sssI

ss

sIsI

t

Solution


94/105

Solution

Substituting I2(s) in eqn (4)

300)300(

600)(

0)300(

600)(

0)300(

3200)(

0300

3

200)(

1

1

1

1

s

B

s

A

sssI

sssI

sssI

s

ssI

)(22

300

22)(

2300

600

)300(

600)300(

2300

600

)300(

600

30 01

1

30 0

0

answereI

sssI

sssB

sssA

t

s

s

Solution


95/105

Solution

Substituting the value of I1(s) and I2(s) ineqn (1)

)(2

322

300

3

300300

3

213

321

answereI

eeI

III

III

t

tt

Example


96/105

Example

The switch is closed at t = 0 with thenetwork is in the steady state prior to t = 0.

Find i(t).

Solution


97/105

Solution

For t


98/105

Solution

AV

RR

Ei

iiiERiRi

EVV

VVE

2)48(

24)0(

)0(;

0

48

4848

48

48

Apply Kirchoffs Voltage Law (t=0)

Solution


99/105

Solution

For t > 0

Solution


100/105

Solution

For t > 0

Solution


101/105

Solution

s

ELsLsIsRI

AIis

EIssILsRI

tiiiEdt

diLRi

EVV

VVE

LRL

R

LR

LR

2)()(

2)0()0(;)0()()(

)(;

0


Solution


102/105

Solution

4)4(

122)(

122)()4(

122)(4)(

2

244)(2)(8

244)(2)(8

2

s

B

s

A

ss

ssI

ssIss

sssIsIs

s

s

sssIsI

sssIsI

Solution


103/105

Solution

t

s

s

eti

sssI

ss

ssB

ss

ssA

4

4

0

3)(

4

13)(

1

4

4

)4(

122)4(

34

12

)4(

122

Seatwork 4


104/105

Seatwork 4

1. The circuit is in dc steady state prior to t= 0. Find v(t) when the switch is opened

at time t=0.

Answer: v(t) = 25 e-2t V

Seatwork 4


105/105

Seatwork 4

2. In the network shown, the switch is closedat t = 0. Find i(t).

Answer: i(t) = 6 - 6e-2t

laplace transform example solution

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