Post on 30-Jan-2022
1
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-¶
g+¶t<0
Re s
Im s
g-¶
g+¶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
11
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
InverseLaplaceTransform 1
s 1 s , s, t
1 t Erfc t
InverseLaplaceTransform 1
1 s , s, t
1
t t Erfc t
15
InverseLaplaceTransform 1
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
InverseLaplaceTransform 1
s b s , s, t
1 b2 t Erfcb t b
InverseLaplaceTransform 1
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)
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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
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LaplaceTransformBesselI0, t, t, s Simplify, s 1 &
1
1 s2
LaplaceTransformBesselI1, t, t, s Simplify, s 1 &
1 s
1 s2
LaplaceTransformBesselI2, t, t, s Simplify, s 1 &
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))
InverseLaplaceTransform 1
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))
InverseLaplaceTransform 1
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))
InverseLaplaceTransform 1
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;
v1 InverseLaplaceTransformV1 . rule1, s, t;
v2 InverseLaplaceTransformV2 . rule1, s, t;
v3 InverseLaplaceTransformV3 . rule1, s, t;
i1 InverseLaplaceTransformI1 . rule1, s, t;
i2 InverseLaplaceTransformI2 . rule1, s, t;
i3 InverseLaplaceTransformI3 . rule1, s, t;
i4 InverseLaplaceTransformI4 . rule1, s, t;
i5 InverseLaplaceTransformI5 . 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;
extenst_ : ft ; 0 b t 3; extenst_ : extenst 3 ; t 3;
extenst_ : extenst 3 ; 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 = 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;
extenst_ : ft ; 1 b t 2; extenst_ : extenst 3 ; t 2;
extenst_ : extenst 3 ; t 1;
Plotextenst, t, 5, 5, PlotStyle Red, Thick,
AspectRatio Automatic, AxesLabel "t", "ft",
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;
extenst_ : ft ; 1 b t 1; extenst_ : extenst 2 ; t 1;
extenst_ : extenst 2 ; t 1;
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] := 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;
extenst_ : ft ; 0 b t 1; extenst_ : extenst 1 ; t 1;
extenst_ : extenst 1 ; 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 = 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
extenst_ : ft ; 0 b t 1; extenst_ : extenst 1 ; t 1;
extenst_ : extenst 1 ; t 0;
Plotextenst, t, 3, 3, PlotStyle Red, Thick,
AspectRatio Automatic, AxesLabel "t", "ft",
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,
Background LightGray
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,
Background LightGray
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,
Background LightGray
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
InverseLaplaceTransform 1
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