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2012/12/24
1
Introduction to Laplace Transforms•Introduction•Definition of the Laplace Transform•Properties of the Laplace Transform•The Inverse Laplace Transform•The Convolution Integral•Application to Integrodifferential Equations•Summary
Introduction•A very powerful tool for analyzing circuits•Differential equations are converted into
algebraic equations•Laplace transform method
–Transformation from time domain to frequencydomain
–Obtain the solution–Transformation from frequency domain to time
domain by applying inverse Laplace transform
•It provides the total response (natural/forced)in one single operation
2012/12/24
2
Definition of the Laplace Transform
dtetfsF
js
tutftfdtetfsFtfL
tf
st
st
)()(
bygivenistransformLaplacesided-twoThe
)sunit(where
)()()(,)()()(
bygivenissided)-(onetransformLaplaceits,)(functiongivenaFor
1
0
Existence of Laplace Transform
Convergenceregion(>c)
ct
t
ct
t
c
t
t
tjttjt
stst
tfe
tfe
tfe
tfttf
edttfedttfee
jsdttfedtetfsF
if)(lim
if0)(lim
e.convergencofabscissathecalledis
0)(lim
such thatexistsconstantpositivereal,aif
orderlexponentiaofbetosaidis)(.asorderlexponentiaofis)(ifconvergesintegralThe
1)()(
)()()()(
converges.integraltheifexiststransformLaplaceThe
00
00
2012/12/24
3
Cont’d
exists.transformLaplacethe
,0for0)(0for)(For*
exist.tdon'stransformLaplacethe,)(For*
sin,,)(For*
0sin,sin,)(For*
2
22
tTttfTtetf
teetf
cteteetf
tttttf
t
tt
cctctct
c
Determine L[u(t)]
sss
es
dte
dtetutuL
stst
st
1)1(
1)0(
1
11
)()(
00
0
2012/12/24
4
Determine L[e-atu(t)]
as
asas
eas
dtee
dtetuetueL
tas
stat
statat
1
)1(1
)0(1
1
)()(
0
)(
0
0
Determine L[(t)]
1)()0()()0(
)()0(
)()(But
)()(
0
0
0
0
0
0
0
0
etL
fdttf
dttf
dtttf
dtettL st
2012/12/24
5
Properties of the Laplace Transform
22
2211
2021012211
1121
21
21
21
)(cos
:Example
)()(
)()()()(
:PropertyLinearity
ss
jsjs
eLeLeeLtutL
sFasFa
dtetfadtetfatfatfaL
tjtjtjtj
stst
Scaling Property
2222
22
)(
0
)(
00
42
)2(21
)(2sin
)(sin
:Example
1
)(1
)(Let)()()(
sstutL
stutL
as
Fa
dxexfa
atxadx
exfdteatfatfL
asx
asxst
2012/12/24
6
Time Shift Property
22
22
0
)(
0
0
)()(cos
)(cos
:Example
)(
)(Let)()(
)()()()()(
:ShiftTime
ss
eatuatL
ss
tutL
sFe
atxdxexfedxexf
dteatfdteatuatfatuatfL
as
as
sxasaxs
st
a
st
Frequency Shift Property
22
22
2222
)(
00
)(sin
)(cos
)(sin,)(cos
:Example
)()()()(
:ShiftFrequency
ωasu(t)ωteL
ωasas
u(t)ωteL
ωstuωt
ωss
tuωt
asF
dtetfdtetfetutfeL
at
at
tasstatat
2012/12/24
7
Time Differentiation Property
)0()()()(
)()(
)(
)(
)(
'
0
000
0
|
fssFtfLtudtdf
L
ssFtudtdf
Lfe
dtfesdtedtdf
dtdtfed
fesedtdf
dtde
fedtdf
dtfed
dtdx
ydtdy
xdtxyd
dtedtdf
tudtdf
L
st
ststst
ststst
stst
st
(Cont’d)
)0()0()0()(
)0()0()(
)0()0()()0()(
)1(0'21
'2
'''2
2
nnnn
n
n
fsfsfssFs
dtfd
L
fsfsFs
ffssFsftfsLdt
fdL
2012/12/24
8
Time Integration Property
)(1
)0(1
)(11
)(1
)(
)()(
)(
)(,)(Let
)()(
|
|
00
0
000
0
000
sFs
us
sFs
ues
sFs
dttfL
dttfsLsF
dtuesdtfeue
dtuesdueudedueued
dttfdudxxfu
dtedxxfdttfL
stt
t
ststst
ststststst
t
sttt
(Cont’d)
13232
2
0
20
01
1
!,
2
1112
111)(
1)(,)()(Let:Examples
)()0(where
)0(1
)(1
)(
nn
t
t
t
sn
tLs
tL
ssst
LtdtL
ssstLdttfL
ssFtutf
dttff
fs
sFs
dttfL
2012/12/24
9
Frequency Differentiation Property
2
00
0
)(11
)(
:Example
)()1()(
)()(
)(
))(())(()(
)()(
asasdsd
tuteL
dssFd
tftL
dssdF
ttfL
ttfL
dtettfdttetfds
sdF
dtetfsF
at
n
nnn
stst
st
Time Periodicity Property
)2()2()()()(
)()()()(
1
11
321
TtuTtfTtuTtftf
tftftftf
+
+
2012/12/24
10
(Cont’d)
Ts
Ts
TsTs
TsTs
as
esF
sF
se
eesF
esFesFsFsF
sFeatuatfL
1)(
)(
.0)Re(is,that,1If
1)(
)()()()(
)()()(
givespropertyshifttimeThe
1
21
2111
Initial-Value Theorem
)(lim)0(
0)0()(lim
0,Let
)0()(
givespropertyationdifferentiThe
0
0
ssFf
fssF
dtedtdf
s
dtedtdf
dtdf
LfssF
s
s
st
st
2012/12/24
11
Final-Value Theorem
)(01
lim)(lim)(
11
)()(sin)(:2Example
)(1)(lim)(1
)()()(:1Example
0.)ifonlyandifvalidis(It)(lim)(
)0()()0()(lim
,0Let
)0()(
givespropertyationdifferentiThe
200
2
0
c0
0
0
00
0
ss
ssFf
ssFtuttf
ssFfs
sFtutf
ssFf
ffdfdtedtdf
fssF
s
dtedtdf
dtdf
LfssF
ss
s
s
s
st
Summary
)0(
)0(
)0()(
)0()0()(
)0()(ationdifferentiTime
)()(shiftFrequency)()()(shiftTime
1)(caling
)()()()(Linearity)()(Property
)1(
'2
1
'22
2
22112211
n
n
nn
n
n
at
as
f
fs
fssFs
dtfd
fsfsFsdt
fd
fssFdtdf
asFtfesFeatuatf
as
Fa
atfS
sFasFatfatfasFtf
2012/12/24
12
Summary
)()()()(nConvolutio)(lim)(valueFinal
)(lim)0(valueInitial1
)()()(yperiodicitTime
)()(
nintegratioFrequency
)(ationdifferentiFrequency
)(1
)(nintegratioTime
)()(Property
2121
0
1
0
sFsFtftfssFf
ssFfesF
nTtftf
dssFttf
sFdsd
ttf
sFs
dttf
sFtf
s
s
sT
s
Summary
1
2
1
2
)(!
)(1
!
1
1
1)(
1)()()(
natn
at
nn
at
asn
et
aste
sn
t
st
ase
stu
tsFtf
22
22
22
22
22
22
)(cos
)(sin
sincos)cos(
cossin)sin(
cos
sin
)()(
asas
te
aste
ss
t
ss
t
ss
t
st
sFtf
at
at
2012/12/24
13
The Inverse Laplace Transform
•Zeros: the roots of N(s) = 0•Poles: the roots of D(s) = 0•Steps to find the inverse Laplace transform.
–Decompose F(s) into simple terms using partialfraction expansion.
–Find the inverse of each term.
)()(
)(sDsN
sF
Case 1: Simple Poles
n
n
n
ji
n
n
kkk
psk
psk
psk
sF
sDsN
jipp
ppps
pspspssN
sF
...,,,:Residues
)(
,)(thanlessis)(ofdegreetheAssuming
forand
...,,,:Poles
)())(()(
)(
21
2
2
1
1
21
21
2012/12/24
14
Simple Poles (Cont’d)
)()(
)()(
|)()(s
|)()(ssSet
)()()()(s
:findingformethodResidue
)(
21
1
21
1
111
1
2
2111
1
2
2
1
1
tuekekektf
tukeaskL
sFpk
ksFpp
pskps
pskps
ksFp
k
psk
psk
psk
sF
tpn
tptp
at
psii
ps
n
n
n
n
n
i
Case 2: Repeated Poles
||
||
)()(!
1)()(
!21
)()()()(
bydeterminedis
.atpoleahavetdoesn'that)(ofpartremainingtheiss)(where
)()()(
)(
,atpolesrepeatednhas)(Suppose
2
2
2
1
1
11
11
ps
nm
m
mnps
nn
ps
nnps
nn
i
nn
nn
sFpsdsd
mksFps
dsd
k
sFpsdsd
ksFpsk
k
pssFF
sFps
kps
kps
ksF
pssF
2012/12/24
15
Repeated Poles (Cont’d)
)()(
)!1(
!2)(
)()!1()(
1
11
2321
11
tftuet
nk
etk
tekektf
tun
etas
L
ptnn
ptptpt
atn
n
Case 3: Complex Poles
)()(sincos)()(
)()()(
)(
)(and)(2Let
.polescomplexofpairthishavetdoesn'that)(ofpartremainingtheiss)(where
)()(
111
1221
221
1121
222222
1
1221
tftuteBteAtf
sFs
Bs
sAsF
BsAAsAsssbass
sFF
sFbass
AsAsF
tt
2012/12/24
16
Example 1
)(782)(7
)1()3(129
)2(12
)()3(
8)1)(2(
124)3(
12)()2(
2)3)(2(
12)3)(2(
12)(
givesmethodresidueApplying32
)(Let:Sol
)3)(2(12
)(if)(Find
32
3
2
3
2
2
2
0
2
0
2
||
||
||
tueetf
sss
sFsC
sss
sFsB
sss
ssFA
sC
sB
sA
sF
ssss
sFtf
tt
ss
ss
ss
Example 2
)(2213141)(
134
52)1(410
)()2(
22)1()2(
44)1(410
)()2(
14)1)(1(
14)2(410
)()1(
1)2)(1(
4)2)(1(
410)(
givesmethodresiduetheApplying2)2(1
)(Let:Sol
)2)(1(410
)(if)(Find
22
2
2
2
2
2
2
2
2
212
2
1
202
2
0
2
2
2
||
||
||
||
tuteeetf
sss
dsd
sVsdsd
D
sss
sVsC
sss
sVsB
sss
ssVA
sD
sC
sB
sA
sV
ssss
sVtv
ttt
ss
ss
ss
ss
2012/12/24
17
Example 3
102
344)34)(4(20
2537520
10Set
2258
20)()3(
givesmethodresiduetheApplying2583
)(Let:Sol
)258)(3(20
)(if)(Find
||323
2
2
CB
CBA
CA
,sss
sHsA
ssCBs
sA
sH
ssssHth
ss)(
3sin32
3cos22)(
9)4(3
32
9)4()4(2
32
258102
32
)(
4
43
22
2
tute
teetf
sss
s
sss
ssH
t
tt
The Convolution Integral•Useful in finding the response y(t) of a system to an
excitation x(t) when the system impulse response h(t)is defined.
)()()(or
)()()(
thtxty
dthxty
dthxdtHx
dtxHtxHty
thtHdtxtx
)()()()(
)()()]([)(
)()(,)()()(
system,linearaofpropertylinearityApply the
2012/12/24
18
dtxtx
txtx
txtx
dtttdutdt
tdu
tutuxtx
txtxtxtx
ii
iiii
i
)()()(or
)()()()(
)()()(
)()(),()(
)()()()(
)()()()(
0
11
00
10
(Cont’d)
(Cont’d)
)()(
)()()(
)()(
)()(lim
)()()()(
)()()(
)()()()(
)()()(
0
0
00
11
00
11
00
thtx
dthxty
dthx
thx
thxthx
thxty
txtx
txtx
iii
ii
ii
2012/12/24
19
(Cont’d)
dthx
dthx
thtxtyttth
tth
dthxdthxty
ttx
t)()(
)()(
)()()(.or0for0)(then
,0for0)(i.e.causal,isresponseimpulestheifSecond,
)()()()()(
,0for0)(ifFirst,
0
0
0
Properties of Convolution Integral
dfdtuftutf
tfdtfttf
ttftttf
tfdtfttf
tytxtftytxtf
tytftxtftytxtf
txththtx
t)()()()()(.7
)()()()()(.6
)()()(.5
)()()()()(.4
)()()()()()(:eAssociativ.3
)()()()()()()(:veDistributi.2
)()()()(:eCommutativ.1
'''
00
2012/12/24
20
)()()()(
)()()(
)()()()()(
)()()()()(
givespropertyshifttimeThe
)()()()()()(
)()(,)()(
)0for0)()(()()()()(
21
2100
2100
201021
2022
21010221
202101
2121021
tftfL
dtedtff
dtdetutff
ddtetutffsFsF
dtetutftutfLesF
desFfdefsFsFsF
defsFdefsF
ttftfdtfftftf
stt
st
st
sts
ss
ss
t
)()()]()([ 2121 sFsFtftfL
0,202
201
201
42
5
)()()()(
5)(,4)(If
2
11
1
2
tee
ssL
ssL
sXsHLtxth
ethetx
tt
tt
Example
2012/12/24
21
• Folding: Take the mirror image of h() about theordinate axis to obtain h().
• Displacement: Shift or delay h() by t to obtainh(t).
• Multiplication: Find the product of h(t) and x().• Integration: For a given time t, calculate the area
under the product h(t) x() for 0
2012/12/24
22
Example 1 (Cont’d)
0 < t < 1 1 < t < 2 2 < t < 3
3 < t < 4 t > 4
Step 3
)()( 21 txtx
Solving Linear, Time-Invariant,Differential Equations
•Steps–By taking the Laplace transform, the time-
domain differential equation is converted into ans-domain algebraic equation.
–Obtain the solution of the s-domain algebraicequation.
–The time-domain solution is obtained byapplying inverse Laplace transform.
)0()0()0()( )1(0'21
nnnnn
n
fsfsfssFsdt
fdL
2012/12/24
23
Example
42)4)(2(24
)(
2)0(,1)0(
2)(8)0()0(6
)0()0()(
)(2)(8)(
6)(
:Sol2)0(,1)0(
)(2)(8)(
6)(
equationaldifferentitheSolve:Q
2
'
'2
2
2
'
2
2
sC
sB
sA
sssss
sV
vvs
sVvsv
vsvsVs
tuLtvdt
tdvdt
tvdL
vv
tutvdt
tdvdt
tvd
)(2141
)(
41
,21
,41
method,residuetheUsing
42 tueetv
CBA
tt