Chapter 12 Laplace transform Masatsugu Sei Suzuki ...

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Chapter 12 Laplace transform

Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton

(Date October 27, 2010)

Pierre-Simon, marquis de Laplace (23 March 1749 – 5 March 1827) was a French

mathematician and astronomer whose work was pivotal to the development of

mathematical astronomy and statistics. He summarized and extended the work of his

predecessors in his five volume Mécanique Céleste (Celestial Mechanics) (1799–1825).

This work translated the geometric study of classical mechanics to one based on calculus,

opening up a broader range of problems. In statistics, the so-called Bayesian

interpretation of probability was mainly developed by Laplace. He formulated Laplace's

equation, and pioneered the Laplace transform which appears in many branches of

mathematical physics, a field that he took a leading role in forming. The Laplacian

differential operator, widely used in applied mathematics, is also named after him.

http://en.wikipedia.org/wiki/Pierre-Simon_Laplace ________________________________________________________________________ Oliver Heaviside (18 May 1850 – 3 February 1925) was a self-taught English electrical engineer, mathematician, and physicist who adapted complex numbers to the study of electrical circuits, invented mathematical techniques to the solution of differential equations (later found to be equivalent to Laplace transforms), reformulated Maxwell's field equations in terms of electric and magnetic forces and energy flux, and

2

independently co-formulated vector analysis. Although at odds with the scientific establishment for most of his life, Heaviside changed the face of mathematics and science for years to come

http://en.wikipedia.org/wiki/Oliver_Heaviside ________________________________________________________________________ Thomas John I'Anson Bromwich (1875 – 1929) was an English mathematician, and a Fellow of the Royal Society. http://en.wikipedia.org/wiki/Thomas_John_l%27Anson_Bromwich ________________________________________________________________________ 12.1. Definition

The Laplace transform is an integral transform perhaps second only to the Fourier transform in its utility in solving physical problems. The Laplace transform is particularly useful in solving linear ordinary differential equations such as those arising in the analysis of electronic circuits.

The (unilateral) Laplace transform is defined by

0

)()]([)( tfdtetfLsF st ,

with

f(t): original function F(s): image function

3

((Link)) http://www.intmath.com/Laplace-transformation/Intro.php Laplace transform of elementary functions. (1) atetf )(

asdteedteeL tasatstat

1

][0

)(

0

for Re(s)>Re(a).

(2) nttf )(

0

10

0

|][ nststn

nstn tdtes

n

s

ettdtetL .

For Re(s)>0,

0limlim )(

tsR

t

nstn

tetet .

Then we have

121 !

....][1

][][

n

nnn

s

ntL

s

n

s

ntL

s

ntL .

(3) f(t) = 1

ss

edtetfL

stst 1

|]1)([ 0

0

.

12.2 Laplace transform of derivative

0

0

0

)()(|)()(')]('[ tfdtestfetfdtetfL ststst .

)(lim)(lim )Re( tfetfeI ts

t

st

t

.

If atMetf )( , then we have

0lim )]Re([

tsa

tMeI for Re(s)>0.

Then

4

)0()()0()]([)]('[ fssFftfsLtfL .

Similarly

)0(')0()(

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

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

2

)2(

fsfsFs

fftfsLs

ftfsLtfL

,

)0(")]0(')]('[[)0(")]("[)]([ )3( fftfsLsftfsLtfL ,

or

)0(")]0(')]0()]([[)]([ 2)3( fsfftfsLstfL , or

)0(")]0(')0()]([)]([ 23)3( fsffstfLstfL . 12.3. Laplace transform of the integral

0

0

00 00

)(1

|)(1

[)(])([ tfdtes

pdpfes

pdpfdtetdtfL stt

stt

stt

.

We consider the first term If atMetf )( , then we have

)1

()(00 a

eMdteMpdpf

attat

t ,

0lim)(1

lim)(Re)(Re

0

as

eeMpdpfe

s

tstas

t

tst

t.

Thus for Re(s)>a

)(1

])([0

sFs

tdtfLt

.

Next we consider the function

5

(1) Substitution

atnettf )( .

1

!][

nn

s

ntL , 1

0 )(

!][

n

atnstatn

as

netdteetL .

More generally,

)]([)()(0

)( tfeLtfdteasF attas

.

(2) Derivative

)]([)()(

0

ttfLttfdteds

sdF st

.

(3) Faltung (Convolution)

0

11 )()]([ ufduetfL su ,

0

22 )()]([ vfdvetfL sv ,

0

21)(

0

21 )()(][][ vfufdvedufLfL vus ,

tvu , u = and v = t - .

[ 0u and 0v ]→[ t0 and t0 ]

Fig.1 Jacobian:

6

dtddtddtd

t

v

t

u

vu

dudv

10

11.

Then we have

]*[)()()()(][][ 21

0

21

00

21)(

0

21 ffLtffddtevfufdvedufLfLt

stvus

,

with

t

tffdff0

2121 )()(*

((Note)) There is a relation,

1221 ** ffff , or

)*)*()*(* 321321 ffffff .

(4) Laplace transform of t

xdxftg0

)()(

)()(' tftg ,

)()0()( sFgssG and g(0) = 0,

or

s

sFsG

)()( .

12.4. Inverse Laplace transformation

The inverse Laplace transformation is defined as

)()]([)( 1 tfsFLtf .

7

If

dttf )( converges, then we have

iti efdedtf )(2

1)( . (Fourier Integral)

Since

)(2

1 )(

ted ti ,

then we have

)()()( tfdtf .

Now we consider

)()()( ttfetg t . Then we have

iti egdedtg )(2

1)( ,

or

0

)(2

1)(

itit eefdeedtf ,

with the choice of variable

is ,

)(2

1)( )(

iFedtf ti

.

Since ds = id

)(2

1)( sFdse

itf

i

i

st

.

8

(Bromwich integral , or Bromwich -Wagner integral)

Re s

Im s

g-Â¶

g+Â¶t<0

Re s

Im s

g-Â¶

g+Â¶t>0

Contours for Inverse Laplace Transform

Fig.2 ((Bromwich integral)) from Wikipedia

An integral formula for the inverse Laplace transform, called the Bromwich integral, the Fourier-Mellin integral, and Mellin's inverse formula, is given by the line integral:

dssFei

tfsFLi

i

st )(2

1)()]([1

where the integration is done along the vertical line x = γ in the complex plane such that γ is greater than the real part of all singularities of F(s). This ensures that the contour path is in the region of convergence. If all singularities are in the left half-plane, then γ can be set to zero and the above inverse integral formula above becomes identical to the inverse Fourier transform.

In practice, computing the complex integral can be done by using the Cauchy residue theorem. It is named after Hjalmar Mellin (Finland 1854 – 1933), Joseph Fourier, and Thomas John I'Anson Bromwich (1875-1929). 12.5 Example

We now consider the example of the Bromwich integral of as

sF

1

)( .

9

Fig.3

dsas

ei

tfsFLi

i

st

1

2

1)()]([1

Using the Jordan’s lemma, this integral can be rewritten as

dz

aze

itf zt 1

2

1)(

along the contour C1 (counter-clock wise) for t>0 and the contour C2 (clock wise) for t<0. There is a single pole at z = -a (<0). Using the residue the torem, we get

ateazstf )(Re)( for t>0 and

0)( tf for t<0. 12.6 LaplaceTransform and InverseLaplaceTransform (Mathematica)

Using the mathematica, one can calculate the Laplace transfrom and inverse Laplace transform.

10

(a) LaplaceTransform[f(t), t, s] To represent the Laplace transform of the function f(t) with respect to the variable t and the kernel.

(b) InverseLaplaceTransform[f(s), s, t]

to represents the inverse Laplace transform of the function f(s) with respect to the variable s and the kernel

12.6.1 Example-1: Laplace transformation

Clear"Global`"LaplaceTransformf't, t, sf0 s LaplaceTransformft, t, s

LaplaceTransformf''t, t, ss f0 s2 LaplaceTransformft, t, s f0

LaplaceTransformf'''t, t, ss2 f0 s3 LaplaceTransformft, t, s s f0 f0

LaplaceTransform0

tgt1 t1, t, s

LaplaceTransformgt, t, ss

12.6.2 Example-2: Laplace transformation

12

12.6.3 Example-3: inverse Laplace transformation

Clear"Global`"f1 LaplaceTransformExpa t2, t, s

s24 a Erfc s

2 a

2 a

s24 a Erfc s

2 a

2 a

SimplifyInverseLaplaceTransformf1, s, t, a 0a t2

((Note))

Erfcx 1 Erfx,

Erfx 1

0

Expu2 u

13

Erf x

Erfcx

1 2 3 4x

0.2

0.4

0.6

0.8

1.0

Fig.4 Plot of erf(x) and erfc(x) as a function of x. 12.6.4 Example-4: inverse Laplace transformation

14

Clear"Global`"InverseLaplaceTransforms12 Exp1s, s, tCosh2 t

t

InverseLaplaceTransformExpass

, s, t

BesselI0, 2 a t

InverseLaplaceTransformExpass

, s, t

BesselJ0, 2 a t

InverseLaplaceTransform 1

s s, s, t

2 t


s 1 s , s, t

1 t Erfc t


1 s , s, t

1

t t Erfc t

15


s 1 s , s, t

t 1

tt Erfc 1

1t

InverseLaplaceTransformLogs a

s b, s, t

a t b t

t

InverseLaplaceTransform s

s 1, s, t

1

t t Erf t


s b s , s, t

1 b2 t Erfcb t b


s, s, t

1

t

InverseLaplaceTransforms32 Exp1ss

, s, t

t BesselI1, 2 t

16

12.7 Laplace transformation of the Bessel function The Bessel function of the first kind can be expressed by

datniatJn )sin(exp(2

1)( .

The Laplace transform of Jn(at) is

sin

1)exp(

2

1

])sin(exp[)exp(2

1

)sin(exp()exp(2

1)]([

0

0

iasdin

dttiasdin

datnidtstatJL n

,

for Re[s]>0. When iaez , izdidaedz i , and

)(2

1sin

2

z

azia .

we have

C

n

n

n

n

n

aszz

zdz

ai

z

azs

dza

z

iatJL

22

2

1

2

21

2

1

)(2

11

2

1)]([

.

There is a simple pole at 22 assz inside the closed path (|z|<a). From the Residue theorem,

n

n

n a

sas

asatJL

)(1)]([

22

22

.

For n = 1,

17

220

1)]([

asatJL

.

12.8 Laplace transform of the Bessel differential equation (a)

)]([1

)]([ 01 atJdt

dL

aatJL ,

with J0(0)=1.

a

sas

as

ass

a

JatJsLa

atJL

22

22

22

001

1

)11

(1

)]0()]([[1

)]([

.

(b)

n

nn

nnn

ssn

sssssn

tJLtJLnt

tJL

)1(1

})1()1{(12

1

)]}([)]([{2

1]

)([

2

1212

2

11

.

(c) Laplace transform of the modified Bessel function of the second kind

22

22

22

22

)(

)(1

)]([{)]([

asa

ass

ia

sas

asi

iatJLiatIL

n

n

nn

nn

nn

n

.

12.9 Laplace transform of Bessel functions (Mathematica)

18

Laplace transformation of the Bessel function

Clear"Global`"LaplaceTransformBesselJ0, t, t, s

1

1 s2

LaplaceTransformBesselJ1, t, t, s1

1 s2 s 1 s2

LaplaceTransformBesselJ2, t, t, s1

1 s2 s 1 s2 2

LaplaceTransformBesselJn, t, t, ss 1 s2 n

1 s2

19

LaplaceTransformBesselI0, t, t, s Simplify, s 1 &

1

1 s2


1 s

1 s2


1 2 s2 2 s 1 s2

1 s2

LaplaceTransformBesselIn, t, t, ss 1 s2 n

1 s2

12.10 Faltung (convolution): Example-1

The Faltung (convolution) can be used for solving the integral equations. (a) Example-1

nt

tdt

g

0

)(

. (1)

The Laplace transforms:

)]([)( tgLsG , st

LsK

]1

[][ ,

1

!][

nn

s

ntL .

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From the property of the Faltung (convolution), the Laplace transform of Eq.(1) can be obtained as

1

!)(

ns

n

ssG

,

or

2/1

!)(

ns

nsG

.

Then we have

2

11

)2

1(

!)]([)(

nt

n

nsGLtg

.

((Mathematica))

Ks_ LaplaceTransform 1

t, t, s

s

Gs_ n

sn12;

InverseLaplaceTransformGs, s, tt

12 n n

Gamma 12 n

________________________________________________________________________ 12.11 Faltung (convolution): Example-2

We solve

1)()(0

t

dgtg ,

by using the Laplace transform. Laplace transform of this equation is obtained as

21

ssG

ssG

1)(

1][ ,

or

1

1)(

ssG .

Then we have

tesGLtg )]([)( 1 ____________________________________________________________________ 12.12 Faltung (convolution): Example-3

)sin(cos2)()()(0

ttdgttgt

Laplace transform of this equation:

)1

(2)(1

)( 22 s

ssG

ssG

or

)1)(1(

2)(

2

2

ss

ssG

The inverse Laplace transform (Bromwich iontegral):

tte

ississssdsesGi

dsesGi

tg

t

C

st

i

i

st

cossin

)(Re)(Re)1(Re)(2

1

)(2

1)(

22

Fig.5 ((Mathematica))

ResidueFs, s, 1t

ResidueFs, s, ResidueFs, s, FullSimplify

Cost Sint ________________________________________________________________________ 12.13 Faltung (convolution): Example-4

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

1 atdgtaJatgt

The Laplace transform of this equation:

2222

22

)](1)[(as

s

as

sassG

or

22

1)(

assG

.

23

Then we have

)()( 0 atJtg

12.14 Arfken: Example 15-10-3, p.983

We consider the Laplace transform of the Bessel equation with n = 0

xyyxy '" We put x = t, y = f(t) Initial condition; f(0) = 1 and f’(0) = 0

0 tffft Noting that

])([)]0(')0()([][][ 22 ssFsds

dfsfsFs

ds

dfL

ds

dftL

)(][][ sFds

dfL

ds

dtfL

we have

0)(1)(])([ 2 sFds

dssFssFs

ds

d

or

0)('1)(]1)(2)('[ 2 sFssFssFsFs or

0)()(')1( 2 ssFsFs or

1)(

2

s

CsF

When C = 1, f(t) = J0(t);

24

1

1)]([

20

s

tJL

where J0(t) is the Bessel function of the first kind. ((Mathematica))


s2 1

, s, t

BesselJ0, t 12.15 Bromwich integral Arfken 15-12-4

][22

1

ks

sL

(a) Partial fraction expansion

)11

(2

1

))(()(

22 ksksksks

s

ks

ssF

)(2

1)]([)( 1 ktkt eesFLtf

(b) Bromwich integral

)(2

1

)(Re)(Re

2

1

2

1)(

2

1)(

22

22

ktkt

C

st

i

i

sti

i

st

ee

ksskss

dseks

s

i

dseks

s

idsesF

itf

where we use the Jordan's lemma.

25

Fig.6 12.16 Bromwich integral Arfken 15-12-5c Bromwich integral

])1(

1[

21

ssL ,

)1(

1)(

2

sssF ,

t

ii

e

ii

ee

ssississ

dsessi

dsessi

dsesFi

tf

ititt

C

st

i

i

sti

i

st

cos1

)2)(()2(

)0(Re)(Re)(Re

)1(

1

2

1

)1(

1

2

1)(

2

1)(

0

2

2

.

where we use the Jordan's lemma.

26

Fig.7 ________________________________________________________________________ 12.17 Bromwich integral Arfken 15-12-6

Fig.8

27

ssF

1)( .

s = 0, s = ∞ branch point.

]})()()(

)()([)({2

1

)(2

1)]([

654

32

1

C

st

C

st

C

st

C

st

C

stst

i

i

st

dssFedssFedssFe

dssFedssFedssFei

dssFei

sFL

.

Since there is no pole, we get

0)( dssFest ,

from the Cauchy's theorem.

])()()(

)()([2

1)]([

654

32

1

C

st

C

st

C

st

C

st

C

st

dssFedssFedssFe

dssFedssFei

sFL

.

The integral along the line C5 (path 5 → 4)

a1

23

4 56

s

s

Fig.9

iqes 0 (q>0) on the line between the points 5 and 4.

28

ttdq

qe

dqeq

ei

dqeqe

ei

sFdsei

I

qt

iqt

i

i

qt

C

st

2

1)2

1(

2

11

2

1

1

2

1

1

2

1

)(2

1

0

0

2/

0

5

5

.

The integral along the line C3 (path 3 → 2):

iqes 0 (q>0).

ttdq

qe

dqeq

ei

dqeqe

ei

sFdsei

I

qt

iqt

i

i

qt

C

st

2

1)2

1(

2

11

2

1

1

2

1

1

2

1

)(2

1

0

0

2/

0

3

3

'

From the Jordan's lemma,

0)(lim2

C

st

RdssFe , and 0)(lim

6

C

st

RdssFe ,

when

0)(lim

sFR

,

where R (∞) is the radius of the path C2 and the path C6. The integral around the small circle (C4) centered at s = 0:

ies 0 , dieds i ,

29

02

2

1

1

2

1

)(2

1)(

2

1

2/

2/

4

de

dieei

diee

ei

dssFei

dssFei

i

ite

i

i

te

st

C

st

i

i

,

as 0 . Then we have

tIIsFL

1

)]([ 531 .

((Mathematica))


s, s, t

1

t ________________________________________________________________________ 12.18 Bromwich integral Arfken 15-12-7 Show that

).(]1

[)]([ 022

11 atJas

LsFL

by evaluation of the Bromwich integral, where

22

1)(

assF

We note that the line connecting between s = ia, s = -ia, form a cut line.

30

]})()()()(

)()()(

)()([)({2

1

)(2

1)]([

10987

654

32

1

C

st

C

st

C

st

C

st

C

st

C

st

C

st

C

st

C

stst

i

i

st

dssFedssFedssFedssFe

dssFedssFedssFe

dssFedssFedssFei

dssFei

sFL

Fig.10 Since there is no pole inside the closed path, we have

0)( dssFest . (Cauchy's theorem).

From the Jordan's lemma,

0)(2

C

st dssFe , and 10

)(C

st dssFe for t>0.

We note that F(s) is continuous when crossing between the lines C3 and C9. In other words, the integrals along the C3 line and the line C9 cancel out each other.

31

0)()(93

C

st

C

st dssFedssFe .

On the circle C5 with the radius (→0),

ieias ( = 3/2 → -/2).

0)2(

1

)2(

)2(

))(()(

2/

2/3

2/

2/

2/3

2/

2/

2/3

55

deia

i

deeia

ei

ideeeia

e

dsiasias

edssFe

i

i

i

te

i

ii

te

C

st

C

st

i

i

,

On the circle C7 with the radius (→0),

ieias ( = /2 → -3/2)

0)2(

1

)2(

)2(

))(()(

2/

2/3

2/

2/

2/3

2/

2/3

2/

77

deia

i

deeia

ei

ideeeia

e

dsiasias

edssFe

i

i

i

te

i

ii

te

C

st

C

st

i

i

as → 0.

0)(7

C

st dssFe .

Then we have

32

])()()([2

1)]([

864

1

C

st

C

st

C

st dssFedssFedssFei

sFL

.

On the line C4 and C8:

ia

-ia

1

2

3

4 5

6

7

8

s

Fig.11

2/iqeias , (0≤q<2a)

2/3)2( ieqaias

a tqai

a

ii

itqeia

CC

st

CC

st

qaq

dqei

eqaqe

dqee

dsiasias

eds

as

e

i

2

0

)(

2

02/32/

2/)(

848422

)2(

)2(

))((2/

.

On the line C6:

33

ia

-ia

1

2

3

4 5

6

7

8

s

Fig.12

2/5iqeias , (0≤q<2a)

2/3)2( ieqaias . Then we have

a tqai

aii

itqeia

C

st

C

st

qaq

dqei

eqaqe

dqee

dsiasias

eds

as

e

i

2

0

)(

0

22/32/5

2/5)(

6622

)2(

)2(

))((2/5

.

Finally we have

34

a taqi

a tqaia tqai

qaq

dqe

qaq

dqei

qaq

dqei

isFL

2

0

)(

2

0

)(2

0

)(1

)2(

1

])2()2(

[2

1)]([

.

We put

cosaaq , where 0 . Since dadq )sin( , we have

)(1

sin

)sin(1)]([ 0

0

cos0 cos

1 atJdeda

aesFL tia

tia

.

((Mathematica))


s2 a2

, s, t

BesselJ0, a t ____________________________________________________________ 12.18 Bromwich integral Arfken 15-12-10

Evaluate the Bromwich integral for

i

i

stdsesFi

tf

)(

2

1)(

with

222 )()(

as

ssF

Bromwich integral:

C

sti

i

st dsesFi

dsesFi

tf )(2

1)(

2

1)(

from the Jordan's lemma (t>0). There are two poles at s = ia and -ia. From the Residue thorem, we have

35

a

attiassiassdsesF

itf

C

st

2

)sin()(Re)(Re)(

2

1)(

((Mathematica))

ResidueGs, s, a ResidueGs, s, a ExpToTrig

t Sina t2 a

Fig.13 Here note that

32 )(

2))((]

)([

ias

sesteeias

ias

se

ds

d stststst

,

iatiatiatiat

eia

t

ia

iaeiateeiaiass

4)2(

2)(2)(Re 3

,

36

iatiatiatiat

eia

t

ia

iaeiateeiaiass

4)2(

2))(2()(Re 3 .

Therefore

)sin(2

)sin(24

)(4

)( ata

tati

ia

tee

ia

ttf iatiat .

________________________________________________________________ 12.19 Bromwich integral; Arfken 15-12-13 You have a Laplace transform;

))((

1)(

bsassF

. ( )( ba

Invert this transform by each of three methods. (a) Partial fractions and use of tables.

)11

(1

))((

1)(

bsasabbsassF

)(1

)]([)( 1 btat eeab

sFLtf

for ba

(2) Faltung (convolution).

)(1

|][1

)]([

0)(

0

)(

0

)(1

btat

tbaat

tbaat

tbta

eeab

eba

e

deedeesFL

.

(3) Bromwich integral.

37

Fig.14

)(1

)(Re)(Re

)(2

1)(

2

1)(

btat

C

sti

i

st

eeab

bssass

dsesFi

dsesFi

tf

,

from the Jordan's lemma and the residue theorem. ((Mathematica))

Fs_ 1

s a s b;

ResidueFs s t, s, a ResidueFs s t, s, b Simplify

a t b t

a b 12.20 Cut lines

38

The choice of variables in the path integrals when the multi-valued function is included

(a) s

Fig.15

Path (1→2): iqes 0

Path (3→4):

iqes 0

(b) )1)(1( ss

Check on the possibility of taking the line segment joining s+1 and s-1 as a cut line

ires 1 ies 1

where 20 and 20 , depending on the position of s in the complex plane.

2/)()( iersf

39

Fig.16 points (+)/2 1 0 0 0 2 0 3 0 4 5 6 2 /2 7 2 /2 8 2 The phase at points 6 and 7 is not the same as the points 2 and 3. This behavior can be expected at a branch point cut line.

(b) ))(( isis

Check on the possibility of taking the line segment joining s+i and s-i as a cut line

ireis ieis

where 2

22

and

22

2

, and the values of and are discrete

but not continuous.

2/)()( iersf

40

ia

-ia

1

2

3

4 5

6

7

8

Fig.17 Check on the possibility of taking the line segment joining s+i and s-i as a cut line points (+)/2 1 /2 /2 /2 2 /2 3/2 3 /2 3/2 4 3/2 3/2 /2 5 3/2 3/2 6 5/2 3/2 7 5/2 3/2 8 5/2 5/2 The phase at points 6 and 7 is not the same as the points 2 and 3. This behavior can be expected at a branch point cut line. ((Note)) Path (3→2):

41

2

i

qeis , 2

3

)2(

ieqis

Path (7→6):

2

3i

qeis , 2

5

)2(

ieqis

______________________________________________________________________ 12.21 Diffusion with constant boundary condition

Suppose that ),( tx satisfies a diffusion equation

tx

1

2

2

for x>0, subject to the boundary condition

0 for x>0 and t<0

)(),0( 0 ttx for x = 0 and t>0,

where 0 is constant.

Laplace transformation:

),(1

)]0,(),([1),(

2

2

sxsxsxsx

sx

with

ssx 0),0(

)exp(),0()exp(),(

),(1),(

0

2

2

sx

ssx

sxsx

sxsx

sx

The inverse Laplace transformation (Bromwich integral)

i

i

st

i

i

st

dss

xs

ei

dssxei

tx

)exp(1

2

),(2

1),(

0

42

Fig.18 Using the Jordan’s lemma, we have

0)exp(1

C

st dss

xs

e

since there is no pole inside the path C. Then we have

543)()exp(1

543

IIIdss

xs

eCCC

i

i

st

Calculation of I4

iididdss

xs

eIC

st

2)exp(

1

4

4

43

Calculation of I3

duu

eedq

q

eeds

sx

seI

ixutuq

ixqt

C

st

00

3

2

3

2)exp(1

where

iqes 0 , 2uq Calculation of I5

duu

eedq

q

eeds

sx

seI

ixutuq

ixqt

C

st

00

5

2

5

2)exp(1

,

where

iqes 0 , 2uq , So we have

u

duxueii

duu

eedu

u

eeiIII

tu

ixutuixutu

0

00

543

)sin(42

)(22

2

22

.

]2

[

]]2

[1[

])sin(2

1[

])sin(42[2

)(2

),(

0

0

0

0

0

0

5430

2

2

t

xErfc

t

xErf

u

duxue

u

duxueii

i

IIIi

tx

tu

tu

.

where Erfc is the complementary error function.

44

________________________________________________________________________ 12.22 Ladder circuit Frequency domain The impedance of inductance is Ls. The impedance of capacitance is 1/(Cs). The impedance of resistance is R. The source is the Laplace transformation of the time-dependent voltage source.

Fig.19 We assume that

00

1 1V

VI

0101

11

1V

sI

sVV

01

2

11

1V

s

VI

00213

12)

11(1 V

sV

sIII

00312

122)

12(

11 V

ssV

ss

ssIVV

02

022

4 122)1

22(1

VssVs

sssV

s

VI

02

02

435 )1

223(1221

2 Vs

ssVsss

III

02

02

253 )2

245()1

221

223(1 Vs

ssVs

ss

ssVIV

02

3 )2

245( Vs

ssVEs

45

sss

EV s

2245 2

0

Then we can get the following expressions

sE

sss

sss

Vs

ssI

2245

1223

)1

223(2

2

02

5

sE

sss

ssVssI

2245

122122

2

2

02

4

sE

sss

sVs

I

2245

121

22

03

sE

sss

sVs

I

2245

111

12

02

sE

sss

VI

2245

1

201

sEV 3

sE

sss

ss

Vs

sV

2245

1221

222

02

46

sE

sss

sVs

V

2245

111

12

01

sss

EV s

2245 2

0

((Mathematica))

Clear"Global`"

V0 E1s

5 4 s 2 s2 2

s Simplify

s E1s2 5 s 4 s2 2 s3

V1 1 1

s E1s

5 4 s 2 s2 2

s Simplify

1 s E1s2 5 s 4 s2 2 s3

V2 2 2 s 1

s E1s

5 4 s 2 s2 2

s

Simplify

1 2 s 2 s2 E1s2 5 s 4 s2 2 s3

V3 E1sE1s

47

I1 E1s

5 4 s 2 s2 2

s Simplify

s E1s2 5 s 4 s2 2 s3

I2 1 1

s E1s

5 4 s 2 s2 2

s Simplify

1 s E1s2 5 s 4 s2 2 s3

I3 2 1

s E1s

5 4 s 2 s2 2

s Simplify

1 2 s E1s2 5 s 4 s2 2 s3

I4 2 s 2 s2 1 E1s5 4 s 2 s2

2

s

Simplify

s 1 2 s 2 s2 E1s2 5 s 4 s2 2 s3

48

I5 3 2 s 2 s2 1

s E1s

5 4 s 2 s2 2

s

Simplify

1 3 s 2 s2 2 s3 E1s2 5 s 4 s2 2 s3

rule1 E1 1 &E1 1

1&

v0 V0 . rule1

1

2 5 s 4 s2 2 s3

v0 InverseLaplaceTransformV0 . rule1, s, t;




i1 InverseLaplaceTransformI1 . rule1, s, t;





49

pl1 PlotEvaluatev3, v2, v1, v0, t, 0.1, 10,

PlotRange 0.1, 10, 0.2, 1,

PlotStyle TableHue0.2 i, Thick, i, 0, 4,

Background LightGray

4 6 8 10

-0.2

0.2

0.4

0.6

0.8

1.0

Fig.20

50

pl2 PlotEvaluatei5, i4, i3, i2, i1, t, 0.1, 20,

PlotRange 0.1, 20, 0.2, 1,

PlotStyle TableHue0.2 i, Thick, i, 0, 5,

Background GrayLevel0.7

10 15 20

-0.2

0.2

0.4

0.6

0.8

1.0

Fig.21 12.23 Ladder circuit: Example-2

51

Fig.22 We now consider the infinite ladder circuit consisting of L = 1 H and C = 1 F. Recurrence relation

1

1

1

1

11

1

n

n

n

n

n sZ

Zs

Zs

ZssZ

In the limit of n→∞

sZ

ZsZ

1

or

2

4)(

2sssZ

The current I(s) is given by

52

4

)(2

)(

)()(

2

ss

sE

sZ

sEsI

)]([)( 1 sILti

((Mathematikca))

Ladder circuit second exampl We consider a infinite ladder consisting of L = 1 H and C = 1 F. What happens to the current when the unitstep is applied at t = 0.

K1 Solvex s x

1 s x, x

x 1

2s 4 s2 , x

1

2s 4 s2

K2 x . K12;

I1 1s

K2

2

s s 4 s2

eq1 InverseLaplaceTransformI1, s, tt HypergeometricPFQ1

2, 3

2, 2, t2

Ploteq1, t, 0, 20, PlotStyle Thick, Red, Background GrayLevel0.7,

PlotRange 0, 20, 0.2, 1.2, AxesLabel "t", "it"

5 10 15 20t

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2it

Fig.23

53

When the Dirac Delta function is applied at t = 0, what happens?

LaplaceTransformDiracDeltat, t, s1

I2 1

K2

2

s 4 s2

eq4 InverseLaplaceTransformI2, s, tBesselJ1, 2 t

t

Ploteq4, t, 0, 20, PlotStyle Red, Thick, Prolog AbsoluteThickness1.5,

Background GrayLevel0.7, PlotRange 0, 20, 0.2, 1,

AxesLabel "t", "it"

5 10 15 20t

-0.2

0.0

0.2

0.4

0.6

0.8

1.0it

Fig.24 12.24 Laplace transform of periodic functions

Suppose that f(t) is a periodic function with a period T. The Laplace transform of f(t) is given by

......)()()()()(3

2

2

00

T

T

stT

T

stT

stst dtetfdtetfdtetfdtetfsF

or

Tst

sT

n

TstTns

n

nT

Tn

st

dtetfe

dtetfedtetfsF

0

1 0

)1(

1 )1(

)(1

1

)()()(

54

_______________________________________________________________________ 12.24.1 Laplace transform of periodic function: Example-1 Clear"Global`"ft_ : 1 ; 0 t 1; ft_ : 0 ; 1 t 2;

extenst_ : ft ; 0 b t 2; extenst_ : extenst 2 ; t 2;

extenst_ : extenst 2 ; t 0;

Plotextenst, t, 5, 5, PlotStyle Red, Thick,

AspectRatio Automatic, AxesLabel "t", "ft",

Ticks 0, 4, 1, 0, 1, 1

0 41t

11f t

T = 2; f[t] := 1 /; 0 § t § 1;

eq1 Integrate 1 Exps t, t, 0, 11 s

s

Fs_ eq1

1 Exp2 s Simplify

s

s s s ______________________________________________________________________ 12.24.2 Laplace transform of periodic function: Example-2

55

Clear"Global`"ft_ : 1 ; 0 t 1; ft_ : 0 ; 1 t 3;





Ticks 0, 4, 1, 0, 1, 1

0 41t

11f t

T = 3; f[t] := 1 /; 0 § t § 1;

eq1 Integrate 1 Exps t, t, 0, 11 s

s

Fs_ eq1

1 Exp3 s Simplify

2 s

s s s 2 s s ________________________________________________________________________ 12.24.3 Laplace transform of periodic function: Example-3

56

Clear"Global`"ft_ : 0 . 1 t 0; ft_ : t ; 0 t 1;

ft_ : t 2 ; 1 t 2;





Ticks 0, 4, 1, 0, 1, 1

0 41t

11f t

T = 3; f[t] := 1 /; 0 § t § 1; f[t] := 2 - t /; 1 § t §2;

eq1

Integrate 1 Exps t, t, 0, 1

Integrate2 t Exps t, t, 1, 2 Simplify

2 s s s

s2

Fs_ eq1

1 Exp3 s Simplify

s 1 s 2 s s1 3 s s2

________________________________________________________________________ 12.24.4 Laplace transform of periodic function: Example-4

57

Clear"Global`"ft_ : t ; 1 t 0; ft_ : t ; 0 t 1;





Ticks 0, 4, 1, 0, 1, 1

0 41t

11f t

T = 2; f[t] := t /; 0 § t § 1; f[t] := 2 - t /; 1 § t §2;

eq1

Integrate t Exps t, t, 0, 1

Integrate2 t Exps t, t, 1, 2 Simplify

2 s 1 s2

s2

Fs_ eq1

1 Exp2 s Simplify

1 s

1 s s2

______________________________________________________________________ 12.24.5 Laplace transform of periodic function: Example-5

58

Clear"Global`"ft_ : t ; 0 t 1;





Ticks 0, 4, 1, 0, 1, 1

0 41t

11f t

T = 1; f[t] := t /; 0 § t § 1;

eq1 Integrate t Exps t, t, 0, 1 Simplify

1 s 1 ss2

Fs_ eq1

1 Exp s Simplify

1 s s

1 s s2

_______________________________________________________________________ 12.24.5 Laplace transform of periodic function: Example-5

59

Clear"Global`"ft_ : Sin2 t ; 0 t 12;

ft_ : Sin2 t ; 12 t 1





Ticks 3, 2, 1, 0, 1, 2, 3, 0, 1, 1

-3 -2 -1 0 1 2 3t

11f t

T = 1/2; f[t] := Sin[2 p t] /; 0 § t § 1/2; f[t]

eq1 Integrate Sin2 t Exps t, t, 0, 12 Simplify

2 1 s2 4 2 s2

Fs_ eq1

1 Exp s2 Simplify

2 1 s2 1 s2 4 2 s2

________________________________________________________________________ 12.25 Solving of differential equations using Laplace transform (Mathematica) ((Example-1))

60

üx'[t]+y[t]==t, 4 x[t]+ y'[t]==0

with the initial condition x[0]==1, y[0]==-1

Clear"Global`"eq1 x't yt t;

eq2 4 xt y't 0;

eq3 LaplaceTransformeq1, t, s . x0 1

1 s LaplaceTransformxt, t, s LaplaceTransformyt, t, s

1

s2

eq4 LaplaceTransformeq2, t, s . y0 1

1 4 LaplaceTransformxt, t, s s LaplaceTransformyt, t, s 0

eq5 Solveeq3, eq4,

LaplaceTransformxt, t, s, LaplaceTransformyt, t, s

LaplaceTransformxt, t, s 1 s s2

s 4 s2,

LaplaceTransformyt, t, s 4 4 s2 s3

s2 4 s2

61

Xs_ LaplaceTransformxt, t, s . eq51

1 s s2

s 4 s2

Ys_ LaplaceTransformyt, t, s . eq51

4 4 s2 s3

s2 4 s2

x1t_ InverseLaplaceTransformXs, s, t

1

4

3 2 t

8

7 2 t

8

y1t_ InverseLaplaceTransformYs, s, t3 2 t

4

7 2 t

4 t

Plotx1t, y1t, t, 0, 2,

PlotStyle Red, Thick, Blue, Thick,


0.5 1.0 1.5 2.0

-80

-60

-40

-20

20

40

62

((Exampl-2)) _______________________________________________________________________ ü

Solve the differential equation

y''[t]+3 y'[t]+2 y[t]==f[t]

withf[t]=periodic function: y[0]==0,y'[0]==0

Clear"Global`"ft_

k0

15

UnitStept 2 k UnitStept 1 2 k UnitStept 1 2 k

UnitStept 2 2 k;

Plotft, t, 0, 25, PlotStyle Red, Thick,


5 10 15 20 25

-1.0

-0.5

0.5

1.0

eq1 y''t 3 y't 2 yt ft;

eq2 LaplaceTransformeq1, t, s . y0 0, y'0 0

63

2 LaplaceTransformyt, t, s 3 s LaplaceTransformyt, t, s s2 LaplaceTransformyt, t, s

1

s32 s

s

2 31 s

s

2 30 s

s

2 29 s

s

2 28 s

s

2 27 s

s

2 26 s

s

2 25 s

s

2 24 s

s

2 23 s

s

2 22 s

s

2 21 s

s

2 20 s

s

2 19 s

s

2 18 s

s

2 17 s

s

2 16 s

s

2 15 s

s

2 14 s

s

2 13 s

s

2 12 s

s

2 11 s

s

2 10 s

s

2 9 s

s

2 8 s

s

2 7 s

s

2 6 s

s

2 5 s

s

2 4 s

s

2 3 s

s

2 2 s

s

2 s

s

eq3 Solveeq2, LaplaceTransformyt, t, s Simplify

LaplaceTransformyt, t, s

1

s 2 3 s s2 32 s 1 s2 1 2 s 4 s 6 s 8 s 10 s 12 s

14 s 16 s 18 s 20 s 22 s 24 s 26 s 28 s 30 s

eq4 InverseLaplaceTransformeq3, s, t Simplify;

yt_ yt . eq41;

Plotyt, t, 0, 50, PlotStyle Red, Thick,


10 20 30 40 50

-0.10

-0.05

0.05

0.10

0.15

0.20

________________________________________________________________________ 12.26 Damped simple harmonics (Mathematica)

64

We consider the motion of the damped oscillator using the Laplace transformation.

Clear"Global`"eq11 x''t 2 x't 02 xt ft02 xt 2 xt xt ft

eq12

LaplaceTransformeq11, t, s .

LaplaceTransformxt, t, s Xs,

LaplaceTransformft, t, s Fs Simplify

Fs s x0 2 x0 x0 s2 2 s 02 Xs

eq13 Solveeq12, XsXs

Fs s x0 2 x0 x0s2 2 s 02

Xs_ Xs . eq131Fs s x0 2 x0 x0

s2 2 s 02

65

f[t] = f0 UnitStep[t]; step pulseInitial condition x[0] = 0; x'[0]=0

eq21 Fs_ LaplaceTransformf0 UnitStept, t, sf0

s

Initial x0 0, x'0 0x0 0, x0 0

X1s_ Xs . Initial Simplify

f0

s s2 2 s 02

x1t_

InverseLaplaceTransformX1s, s, t .

2 02 ,1

2 02

ExpToTrig Simplify

1

02f0 1 Cosht Cost Sint Cost Sint Sinht

Limitx1t, t 0, Direction 1f0 f0 2

02

66

rule1 1, 0 4, f0 1, 15 1, 0 4, f0 1, 15

Plotx1t . rule1, t, 0, 10, PlotRange All,

PlotStyle Thick, Red, Background LightGray,

AxesLabel "t", "xt"

2 4 6 8 10t

-0.8

-0.6

-0.4

-0.2

0.2

0.4

xt

_____________________________________________________________________ APPENDIX Useful inverse Laplace transform

67


sExpa s , s, t

Simplify, a 0 &

a2

4 t

t

InverseLaplaceTransform1

sExpa s , s, t

Simplify, a 0 &

Erfc a

2 t

InverseLaplaceTransform Expa s , s, t Simplify, a 0 &

a a2

4 t

2 t32

InverseLaplaceTransforms32 1s, s, t Simplify

Sinh2 t

InverseLaplaceTransforms12 1s, s, t Simplify

Cosh2 t t

Chapter 12 Laplace transform Masatsugu Sei Suzuki ...

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Transcript of Chapter 12 Laplace transform Masatsugu Sei Suzuki ...