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The Laplace Transform

CHAPTER 4

Copyright © Jones and Bartlett；滄海書局 Ch4_2

Chapter Contents

4.1 Definition of the Laplace Transform4.2 The Inverse Transform and Transforms of Derivat

ives4.3 Translation Theorems4.4 Additional Operational Properties4.5 The Dirac Delta Function4.6 Systems of Linear Differential Equations


4.1 Definition of Laplace Transform

Basic DefinitionIf f(t) is defined for t 0, then

(1)

b

bdttftsKdttftsK

00)(),(lim)(),(

If f(t) is defined for t 0, then(2)

is said to be the Laplace Transform of f.

Definition 4.1.1 Laplace Transform

0)()}({ dttfetf stL


Evaluate L {1}

Solution: Here we keep that the bounds of integral are 0 and in mind.From the definition

Since e-st 0 as t , for s > 0.

Example 1 Using Definition 4.1.1

sse

se

dtedte

sb

b

bst

b

b st

b

st

11limlim

lim)1()1(

0

00

L


Evaluate L {t}

Solution:


2

00

111}1{

1

1}{

ssss

dtess

tet st

st

L

L


Evaluate L {e-3t}

Solution:


3,3

1

3

}{

0

)3(

0

)3(

0

33

ss

se

dtetdeee

ts

tststtL


Evaluate L {sin 2t}

Solution:


00

0

2cos22sin

2sin}{sin2

dttess

te

dttet

stst

stL

00,2cos

2sdtte

sst


Example 4 (2)

}2{sin4222 t

ssL

0,4

2}2{sin 2

s

stL

00

2sin22cos2

dttess

tes

stst

0,02coslim ste st

t Laplace transform of sin 2t↓


L is a Linear Tramsform

We can easily verify that

(3) )()(

)}({)}({

)}()({

sGsF

tgtf

tgtf

LL

L


(a)

(b) (c)

(d) (e)

(f) (g)

Theorem 4.1.1 Transform of Some Basic Functions

s1

}1{ L

,3,2,1,!

}{ 1 nsn

t nnL

ase ta

1

}{L

22}{sinks

ktk

L 22}{cos

kss

tk

L

22){sinks

ktk

L

22}{coshks

stk

L


Fig 4.1.1 Piecewise-continuous function


A function f(t) is said to be of exponential order,

if there exists constants c, M > 0, and T > 0, such that

|f(t)| Mect for all t > T. See Fig 4.1, 4.2.

Definition 4.1.2 Exponential Order


Fig 4.1.2 Function f is of exponential order


Fig 4.1.3 Functions with blue graphs are of exponential order

tet ||tet 2cos2 tt ee ||


If f(t) is piecewise continuous on [0, ) and of

exponential order c, then L {f(t)} exists for s > c.

Theorem 4.1.2 Sufficient Conditions for Existence


Example 5

Find L {f(t)} for

Solution:

0 ,22

20)}({

3

3

3

3

0

s

se

se

dtedtetf

sst

ststL

3,2

30,0)(

t

ttf


Fig 4.1.5 Piecewise-continuous function in Ex 5


4.2 The Inverse Transform and Transform of Derivatives

(a)

(b) (c)

(d) (e)

(f) (g)

Theorem 4.2.1 Some Inverse Transform

s1

1 1L

,3,2,1,!1

1

nsn

t nn L

ase ta 11L

221sin

ksk

tk L

221cos

kss

tk L

221sinh

ksk

tk L

221cosh

kss

tk L


Example 1 Applying Theorem 4.2.1

Find the inverse transform of

(a) (b)

Solution:(a)

(b)

51 1

sL

71

21

sL

45

15

1

241!4

!411

tss

LL

tss

7sin7

17

77

17

12

12

1

LL


L -1 is also linear

We can easily verify that

(1))}({)}({

)}()({11

1

sGsF

sGsF

LL

L


Find

Solution:

(2)

Example 2 Termwise Division and Linearity

462

21

ss

L

462

21

ss

L

42

26

42

46

42

21

21

221

sss

sss

LL

L

tt 2sin32cos2


Example 3 Partial Fractions and Linearity

Find

Solution:Using partial fractions

Then

(3)

If we set s = 1, 2, −4, then

)4)(2)(1(962

1

sssss

L

)4)(2)(1(962

sssss

)2)(1()4)(1()4)(2(

962

ssCssBssA

ss

421

sC

sB

sA


(4) Thus

(5)

30/1,6/25,5/16 cBA

Example 3 (2)

)4)(2)(1(962

1

sssss

L

ttt eee 42

301

625

516

41

301

21

625

11

516 111

sssLLL


Transform of Derivatives

(6)

(7)(8)

)}({ tf L

000)()()( dttfestfedttfe ststst

)}({)0( tfsf L

)0()()}({ fssFtf L

)}({ tf L

000)()()( dttfestfedttfe ststst

)}({)0( tfsf L)0()]0()([ ffssFs

)0()0()()}({ 2 fsfsFstf L)0()0()0()()}({ 23 ffsfssFstf L


If are continuous on [0, ) and are of

exponential order and if f(n)(t) is piecewise-continuous

On [0, ), then

where

Theorem 4.2.2 Transform of a Derivative

)1(,,, nfff

.)}({)( tfsF L

)0()0()0()(

)}({)1(21

)(

nnnn

n

ffsfssFs

tf

L


Solving Linear ODEs

Then

(9)

(10)

)(01

1

1 tgyadt

yda

dtyd

a n

n

nn

n

n

1)1(

10 )0(,)0(,)0( n

n yyyyyy

)}({}{01

1

1 tgyadt

yda

dtyd

a n

n

nn

n

n LLLL

)]0()0()([ )1(1 nnnn yyssYsa

)(

)(

)]0()0()([

0

)2(211

sG

sYa

yyssYsa nnnn


We have

(11)

where

)()()()( sGsQsYsP

)()(

)()(

)(sPsG

sPsQ

sY

01

1)( asasasP nn

nn


Find unknown that satisfies a DE and Initial condition

)(tyTransformed DE becomes an algebraic equation In )(sY

Apply Laplace transform L

Solve transformed equation for )(sY

Solution of original IVP

)(ty Apply Inverse transform 1L


Solve

Solution:

(12)

(13)

Example 4 Solving a First-Order IVP

6)0(,2sin133 ytydtdy

}2{sin13}{3 tydtdy

LLL

426

)(36)( 2

ssYssY

426

6)()3( 2

ssYs

)4)(3(506

)4)(3(26

36

)( 2

2

2

sss

ssssY


We can find A = 8, B = −2, C = 6Thus

43)4)(3(506

22

2

sCBs

sA

sss

)3)(()4(506 22 sCBssAs

462

38

)4)(3(506

)( 22

2

ss

ssss

sY

42

34

23

18)( 2

12

11

sss

sty LLL

ttety t 2sin32cos28)( 3

Example 4 (2)


Example 5 Solving a Second-Order IVP

Solve

Solution:

(14)

Thus

5)0(',1)0(,2'3" 4 yyeyyy t

}{}{23 42

2tey

dtdy

dtyd

LLLL

41

)(2)]0()([3)0()0()(2

ssYyssYysysYs

41

2)()23( 2

sssYss

)4)(2)(1(96

)4)(23(1

232

)(2

22

sssss

ssssss

sY

ttt eeesYty 421

301

625

516

)}({)( L


4.3 Translation Theorems

Proof:L {eatf(t)} = e-steatf(t)dt = e-(s-a)tf(t)dt = F(s – a)

If L {f} = F(s) and a is any real number, then

L {eatf(t)} = F(s – a)

Theorem 4.3.1 First Translation Theorem

assat tftfe )}({)}({ LL


Fig 4.3.1 Shift on s-axis


Evaluate (a) (b)

Solution:

(a) (b)

Example 1 Using the First Translation Theorem

}{ 35 te tL }4cos{ 2 te tL

45

45335

)5(6!3

}{}{

sstte

ssss

t LL

16)2(2

16

}4{cos}4cos{

22

2

)2(2

ss

ss

tte

ss

sst LL


Inverse Form of Theorem 4.3.1

(1)

where

)(})({)}({ 11 tfesFasF atass

LL

.)}({)( 1 sFtf L


Evaluate (a) (b)

Solution:(a)

we have A = 2, B = 11

(2)

Example 2 Partial Fractions and Completing the square

21

)3(52

ss

L

643/52/

21

sss

L

22 )3(3)3(52

sB

sA

ss

BsAs )3(52

22 )3(11

32

)3(52

ssss


Example 2 (2)

And

(3)

From (3), we have

(4)

211

21

)3(1

113

12

)3(52

ssss

LLL

teess tt 33

21 112

)3(52

L

tess

t

ss

3

32

12

1 1)3(

1

LL


Example 2 (3)

(b) (5)

(6)

(7)

2)2(3/52/

643/52/

22

ss

sss

2)2(1

32

2)2(2

21

643/52/

21

21

21

sss

sss

LL

L

22

1

22

1

22

232

221

ssss sss

LL

tete tt 2sin32

2cos21 22


Solve

Solution:

Example 3 IVP

17)0(',2)0(,9'6 32 yyetyyy t

)(9)]0()([6)0()0()(2 sYyssYysysYs 3)3(2s

)()96( 2 sYss3)3(

252

ss

)()3( 2 sYs3)3(

252

ss

)(sY52 )3(

2)3(52

sss


Example 3 (2)

(8)

52 )3(2

)3(11

32

)(

sss

sY

51

211

)3(!4

!42

)3(1

113

12

)(

sss

ty

LLL

,1 3

32

1 t

ss

tes

L t

ss

ets

34

35

1 !4

L

ttt etteety 3433

121

112)(


Example 4 An IVP

Solve

Solution:

0)0(',0)0(,164 yyeyyy t

)(6)]0()([4)0()0()(2 sYyssYysysYs1

11

ss

)()64( 2 sYss)1(

12

sss

)(sY)64)(1(

122

sssss

643/52/

13/16/1

)( 2

sss

sssY


Example 4 (2)

tetee ttt 2sin32

2cos21

31

61 22

2)2(2

232

2)2(2

21

11

311

61

)(

21

21

11

sss

sssY

LL

LL


The Unit Step Function U (t – a) is

Definition 4.3.1 Unit Step Function

at

atat

,1

0,0)(U

See Fig 4.3.2.


Also a function of the type

(9)is the same as

(10)Similarly, a function of the type

(11)

can be written as (12)

atth

attgtf

),(

0),()(

)()()()()()( atthattgtgtf UU

bt

btatg

at

tf

,0

),(

0,0

)(

)]()()[()( btattgtf UU


Express

in terms of U (t).Solution:From (9) and (10), with a = 5, g(t) = 20t, h(t) = 0

f(t) = 20t – 20tU (t – 5)

Consider the function

(13) See Fig 4.3.5.

Example 5 A Piecewise-Defined Function

5,0

50,20)(

t

tttf

atatf

atatatf

),(

0,0)()( U


Fig 4.3.6 Shift on t-axis


Proof:

If F(s) = L {f}, and a > 0, then L {f(t – a)U (t – a)} = e-asF(s)

Theorem 4.3.2 Second Translation Theorem

dtatatfedtatatfea

sta st )()()()(

0UU

)}()({ atatf UL

0)( dtatfe st


Theorem 4.3.2 proof

Let v = t – a, dv = dt, then

If f(t) = 1, then f(t – a) = 1, F(s) = L {1} = 1/s,

(14)eg: The L.T. of Fig 4.3.4 is

0

)( )( dvvfe avs )}({)(0

tfedvvfee assvas L

se

atas

)}({UL

se

se

s

tttfss 32

31

2

)}3({)}2({3}1{2)}({

ULULLL

)}()({ atatf UL


Inverse Form of Theorem 4.7

(15))()()}({1 atatfsFe as UL


Example 6 Using Formula (15)

Evaluate (a) (b)

Solution:(a) then

(b) then

se

s21

41

L

2/

21

9se

ss L

tesFssFa 41 )}({),4/(1)(,2 L

)2(4

1 )2(421

tee

sts UL

tsFsssFa 3cos)}({),9/()(,2/ 12 L

223cos

92/

21 tte

ss s UL


Alternative Form of Theorem 4.3.2

Since , then

The above can be solved. However, we try another approach.Let u = t – a,

That is,

(16)

4)2(4)2( 22 ttt

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

)}2({2

2

ttttt

tt

UUUL

UL

0

)( )()()}()({ duaugedttgeattg aus

a

stUL

)}({)}()({ atgeattg as LUL


Find

Solution:With g(t) = cos t, a = , then

g(t + ) = cos(t + )= −cos tBy (16),

Example 7 Second Translation Theorem – Alternative Form

)}({cos ttUL

ss es

stett

1}{cos)}({cos 2LUL


Solve

Solution:We find f(t) = 3 cos tU (t −), then

(17)

Example 8 An IVP

tt

ttf

,sin3

0,0)(

5)0(,)(' ytfyy

)()0()( sYyssY ses

s

13 2

)()1( sYs ses

s

13

5 2

sss e

ss

es

ess

sY

111

11

23

15

)( 22


Example 8 (2)

It follows from (15) with a = , then

Thus

(18)

See Fig 4.3.7.

)()sin(1

1,)(

11

21)(1

tte

stee

ssts ULUL

)()cos(12

1

tte

ss s UL

)()cos(23

)()sin(23

)(23

5)( )( ttttteety tt UUU

)(]cossin[23

5 )( tttee tt U

tttee

tett

t

,cos2/3sin2/32/35

0,5)(


Fig 4.3.7 Graph of function (18) in Ex 8


Remember that the DE of a beam is

(19)

Beams

)(4

4

xwdx

ydEI


A beam of length L is embedded at both ends as Fig 4.3.8. Find the deflection of the beam when the load is given by

Solution:We have the boundary conditions: y(0) = y(L) = 0, y’(0) = y’(L) = 0. By (10),

Example 9

LxL

LxxL

wxw

2/,0

2/0,2

1)( 0

22

12

1)( 00

Lxx

Lwx

Lwxw U

222

2 0 Lx

Lxx

LL

wU


Fig 4.3.8 Embedded beam with a variable load in Ex 9


Example 9 (2)

Transforming (19) into

where c1 = y”(0), c3 = y(3)(0)

2

220 1122 Lse

sssL

L

w

2665

043

31

222

0)3(4

1122)(

1122)0()0(")(

Ls

Ls

esss

LEIL

w

s

c

sc

sY

esss

LEIL

wysysYs

)0()0()0()0()( 234 yysysyssYsEI


Thus

412

311 !3

!3!2

!2

)(

sc

sc

xy

LL

2/

61

61

510 !5

!51!5

!51!4

!42/2 Lse

sssL

EILw

LLL

2225

6062

55403221 L

xL

xxxL

EIL

wx

cx

cU

Example 9 (3)


Example 9 (4)

Applying y(L) = y’(L) = 0, then

Thus

01920

49

62

40

3

2

2

1 EI

LwLc

Lc

0960

85

2

302

21 EI

LwLcLc

EILwcEILwc 40/9,960/23 022

01

2225

60

803

192023

)(

5540

3022

0

Lx

Lxxx

LEIL

w

xEILw

xEILw

xy

U


4.4 Additional Operational Properties

Multiplying a Function by tn

that is, Similarly,

)}({)()]([

)(

00

0

ttfdtttfedttfes

dttfedsd

dsdF

stst

st

L

)}({)}({ tfdsd

ttf LL

)}({)}({

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

2

2

2

tfdsd

tfdsd

dsd

ttfdsd

ttfttft

LL

LLL


If F(s) =L {f(t)} and n = 1, 2, 3, …, then

Theorem 4.4.1 Derivatives of Transform

)()1()}({ sFdsd

tft n

nnn L


Example 1 Using Theorem 4.4.1

Find L {t sin kt}

Solution:With f(t) = sin kt, F(s) = k/(s2 + k2), then

22222 )(2

}{sin}sin{

ksks

ksk

dsd

ktdsd

ktt

LL


Different approaches

Theorem 4.3.1:

Theorem 4.4.1:

23

233

)3(11

}{}{

sstte

ssss

t LL

2233

)3(1

)3(3

1}{}{

ss

sdsd

edsd

te tt LL


Solve

Solution:

or

From example 1,

Thus

Example 2 An IVP

1)0(,0)0(,4cos16 xxtxx

222

22

)16(161

)(

161)()16(

ss

ssX

ss

sXs

kttks

kssin

)(2

2221

L

ttst

ss

stx

4sin81

sin41

)16(8

81

164

41

)( 221

21

LL


Convolution

A special product of f * g is defined by

(2)and is called the convolution of f and g. The convolution is a function of t, eg:

(3)

Note: f * g = g * f

t

dtgfgf0

)()(*

)cossin(21

)sin(sin0

ttt ettdtete


Proof:

If f(t) and g(t) are piecewise continuous on [0, ) and

of exponential order, then

Theorem 4.4.2 Convolution Theorem

)()()}({)}({}{ sGsFtgtfgf LLL

0

)(

0

0 0

)(

00

)()(

})()(

)()()()(

dgedf

ddgfe

dgedfesGsF

s

ss


Theorem 4.4.2 proof

Holding fixed, let t = + , dt = d

The integrating area is the shaded region in Fig 4.4.1.Changing the order of integration:

dttgedfsGsF st )()()()(0

}{

)()(

)()()()(

00

00

gf

dtdtgfe

dtgfdtesGsF

tst

tst

L


Fig 4.4.1 Changing order of integration from t first to first


Example 3 Transform of a Convolution

Find

Solution:Original statement = L {et * sin t}

t

dte0

)sin( L

)1)(1(1

11

11

22

ssss


Inverse Transform of Theorem 4.9

L -1{F(s)G(s)} = f * g(4)

Look at the table in Appendix III,

(5)

222

3

)(2

}cos{sinks

kktktkt

L


Example 4 Inverse Transform as a Convolution

Find

Solution:Let

then

(6)

22

21

)(1ks

L

22

1)()(

kssGsF

ktkks

kk

tgtf sin11

)()( 221

L

t

dtkkkks 02222

1 )(sinsin1

)(1 L


Example 4 (2)

Now recall thatsin A sin B = (1/2) [cos (A – B) – cos (A+B)]

If we set A = k, B = k(t − ), then

3

02

022221

2cossin

cos)2(sin21

21

]cos)2([cos21

)(1

kktktkt

kttkkk

dkttkkks

t

t

L


Transform of an Integral

When g(t) = 1, G(s) = 1/s, then

(7)

(8)

ssF

dft )(

)(0

L

s

sFdf

t )()( 1

0L


Examples:

ttdss

ttdss

tdss

t

t

t

cos121

)sin()1(

1

sin)cos1()1(

1

cos1sin)1(

1

2

0231

0221

021

L

L

L


Volterra Integral Equation

t

dthftgtf0

)()()()( (9)


Example 5 An Integral Equation

Solve

Solution:First, h(t-) = e(t-), h(t) = et. From (9)

Solving for F(s) and using partial fractions

)(for )(3)(0

2 tfdefettft tt

11

)(1

123)( 3

s

sFss

sF

12166

)( 43

sssssF

tett

sssstf

213

11

21!3!2

3)(

32

114

13

1 LLLL


Series Circuits

From Fig 4.4.2, we have

(10)

which is called the integrodifferential equation.

)()(1

)(0

tEdiC

tRidtdi

Lt


Example 6 An Integrodifferential Equation

Determine i(t) in Fig 4.4.2, when L = 0.1 h, R = 2 , C = 0.1 f, i(0) = 0, and

E(t) = 120t – 120tU (t – 1)

Solution:Using the data, (10) becomes

And then

)1(120120)(10)(21.00

ttditidtdi t

U

ss e

se

ssssI

sIssI111

120)(

10)(2)(0.1 22


Example 6 (2)

sss

s

ss

es

es

es

essss

es

essss

sI

22

2

222

)10(1

)10(10/1

10100/1

100/1)10(

10/110

100/11/1001200

)10(1

)10(1

)10(1

1200)(

)1()1(1080120

)]1([12)]1(1[12)()1(1010

)1(1010

tette

teettitt

tt

U

UU


Example 6 (3)

Written as a piecewise- defined function

(11)

1 ,)1(10801201212

10 ,1201212)(

)1(1010)1(1010

1010

tetteee

tteeti

tttt

tt


Fig 4.4.3


Post Script – Green’s Function Redux

By applying the Laplace transform to the initial-value problem

where a and b are constants, we find that the transform of y(t) is

where F(s) = L {f(t)}. By rewriting the foregoing transform as the product

,0)0( ,0)0( ),( yytfbyyay

basssF

sY

2

)()(

)(1

)( 2 sFbass

sY


We can use the inverse form of the convolution theorem (4) to write the solution of the IVP as

(12)

where and On the other hand, we know from (9) of Section 3.10

that the solution of the IVP is also given by

(13)where G(t, ) is the Green’s function for the differential equation.

t

dftgty0

)()()(

)(1

21 tg

bass

L ).()}({1 tfsF L

t

dftGty0

)() ,()(


By comparing (12) and (13) we see that the Green’s function for the differential equation is related to by

(14)For example, for the initial-value problem y(0) = 0, y’(0) = 0 we find

Thus from (14) we see that the Green’s function for the DE is G(t, ) = g(t - ) = 1/2 sin 2(t - ). See Example 4 in Section 3.10.

)(1

21 tg

bass

L

)() ,( tgtG),(4 tfyy

)(4 tfyy

)(2sin21

41

21 tgt

s

L


Periodic Function

f(t + T) = f(t)

If f(t) is piecewise continuous on [0, ), of exponential order, and periodic with period T, then

Theorem 4.4.3 Transform of a Periodic Function

T st

sT dttfee

tf0

)(1

1)}({L


Use the same transform method

T

stT st dttfedttfetf )()()}({0

L

Theorem 4.4.3 proof

T stsT

sTT st

sT

T

st

dttfee

tf

tfedttfetf

tfedttfe

0

0

)(1

1)}({

)}({)()}({

)}({)(

L

LL

L


Find the L. T. of the function in Fig 4.4.4.

Solution:We find T = 2 and

From Theorem 4.4.3,

(15)

Example 7

21 ,0

10 ,1)(

t

tTE

)1(

111

101

11

)}({ 2

1

02 s

s

sst

s esse

edte

etE

L


Fig 4.4.4


Example 8 A Periodic Impressed Voltage

The DE

(16)Find i(t) where i(0) = 0, E(t) is as shown in Fig 4.4.4.

Solution:

or

(17)

Because and

)(tERidtdi

L

s

s

eLRssL

sI

essRIsLsI

11

)/(/1

)(

)1(1

)()(

sss-s eee

e321

11

LRsRL

sRL

LRss ///

)/(1


Then i(t) is described as follows and see Fig 4.4.5:

(15)

...1111

)( 2

ss eeLRssR

sI

43 ,

32 ,1

21 ,

10 ,1

)(

)3()2()1(

)2()1(

)1(

teeee

teee

tee

te

ti

tttt

ttt

tt

t

Example 8 (2)


Fig 4.4.5


4.5 The Dirac Delta Function

Unit ImpulseSee Fig 4.5.1(a). Its function is defined by

(1)

where a > 0, t0 > 0.

For a small value of a, a(t – t0) is a constant function of large magnitude. The behavior of a(t – t0) as a 0, is called unit impulse, since it has the property . See Fig 4.5.1(b).

1)(0 0 dttt

att

attata

att

tta

0

00

0

0

,0

,21

0 ,0

)(


Fig 4.5.1 Unit impulse


The Dirac Delta Function

This function is defined by (t – t0) = lima0 a(t – t0)(2)The two important properties:

(i)

(ii) , x > t0

The unit impulse (t – t0) is called the Dirac delta function.

1)(0 0 x

dttt

0

00 ,0

,)(

tt

tttt


Proof:

The Laplace Transform is

(4)

For t0 > 0, (3)

Theorem 4.5.1 Transform of the Dirac Delta Function

0)}({ 0stett L

))](()(([21

)( 000 attatta

tta UU

saee

e

se

se

att

sasast

atsats

a

2

21

)}({

0

00 )()(

0L


When a 0, (4) is 0/0. Use the L’Hopital’s rule, then (4) becomes 1 as a 0.Thus ,

Now when t0 = 0, we have

00

2lim)(lim)(

00

00

stsasa

a

sta

ae

saee

etttt

LL

1)( tL


Example 1 Two Initial-Value Problems

Solve subject to (a) y(0) = 1, y’(0) = 0(b) y(0) = 0, y’(0) = 0Solution:(a) s2Y – s + Y = 4e-2s

Thusy(t) = cos t + 4 sin(t – 2)U (t – 2)

Since sin(t – 2) = sin t, then

(5)See Fig 4.5.2.

),2(4" tyy

14

1)( 2

2

2

se

ss

sYs

2 ,sin4cos

20,cos)(

ttt

ttty


Fig 4.5.2


Example 1 (2)

(b)

Thus y(t) = 4 sin(t – 2)U (t – 2)and

(6)

14

)( 2

2

se

sYs

2 ,sin4

20,0

)2()2sin(4)(

tt

t

ttty U


Fig 4.5.3


4.6 Systems of Linear DEs

Coupled StringsIn example 1, we will deal with

(1))(

)(

12222

1221111

xxkxm

xxkxkxm


Example 1 Example 4 of Section 3.12 Revised

Use L.T. to solve

(2)where x1(0) = 0, x1’(0) = 1, x2(0) = 0, x2’(0) = −1.

Solution: s2X1(s)– sx1(0) – x1’(0) + 10X1(s) – 4X2(s) = 0 −4X1(s) + s2X2(s) – sx2(0) – x2’(0) + 4X2(s) = 0

Rearrange:(s2 + 10)X1(s) – 4 X2(s) = 1 −4X1(s) + (s2 + 4)X2(s) = −1 (3)

044

04 10

221

211

xxx

xxx


Example 1 (2)

Solving (3) for X1:

Use X1(s) to get X2(s)

tttx

sssss

sX

32sin53

2sin10

2)(

125/6

25/1

)12)(2()(

1

2222

2

1

tttx

sssss

sX

32sin10

32sin

52

)(

125/3

25/2

)12)(2(6

)(

2

2222

2

2


Example 1 (3)

Then

(4)tttx

tttx

32sin10

32sin

52

)(

32sin53

2sin10

2)(

2

1


Networks


(5) 0

)(

122

2

uidtdi

RC

tERidtdi

L


Solve (5) where E(t) = 60 V, L = 1 h, R = 50 ohm, C = 10-4 f, i1(0) = i2(0) = 0.

Solution:We have

Then sI1(s) + 50I2(s) = 60/s−200I1(s) + (s + 200)I2(s) = 0

Example 2 An Electric Network

0)10(50

6050

1224

21

iidtdi

idtdi


Example 2 (2)

Solving the above

Thus

222

221

)100(120

1005/65/6

)100(12000

)(

)100(60

1005/65/6

)100(1200060

)(

ssssssI

ssssss

sI

tt

tt

teeti

teeti

1001002

1001001

12056

56

)(

6056

56

)(


Double Pendulum


(6)

0)()( 1121221212

121 glmmllmlmm

022212122222 glmllmlm


Please verify that when

the solution of (6) is

(7)

See Fig 4.6.3.

Example 3 Double Pendulum

,1)0(,1)0(,16,1,3 212121 llmm

0)0(',0)0(' 21

ttt

ttt

2cos23

32

cos21

)(

2cos43

32

cos41

)(

2

1


Fig 4.6.3

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