se 207: modeling and simulation introduction to laplace transform dr. samir al-amer term 072
TRANSCRIPT
SE 207: Modeling and Simulation Introduction to Laplace Transform
Dr. Samir Al-Amer
Term 072
Why do we use them We use transforms to transform the problem
into a one that is easier to solve then use the inverse transform to obtain the solution to the original problem
4.274110A
0.63080.38560.1553 0.089943.2log43.1log23.1loglog
43.2*43.1*23.1:
0.6308
A
AcomputeproblemOriginal
Laplace Transform
t is a real variable
f(t) is a real function
Time Domain
s is complex variable
F(s) is a complex valued function
Frequency Domain
2
1)()( 2
ssFetf t
LLaplace Transform
L-1
InverseLaplace Transform
Use of Laplace Transform in solving ODE
Differential Equation Laplace Transform
Algebraic Equation
Solution of theAlgebraic Equation
Inverse Laplace
transform
Solution of the Differential Equation
2
1)()(
0)(21)(1)0(,0)(2)(
2
ssXetx
sXssXxtxtx
t
Definition of Laplace Transform
0
)()}({)( dtetftfLsF st
Sufficient conditions for existence of the Laplace transform
•
•
0
0
)(
thatsuch,,exist There
continuous piecewise is )(
ttforeMtf
tM
tf
t
Examples of functions of exponential order
0,1,12)(
0,0,1)sin()(
0,0,100
01)(
0
0
0
tMTaketf
tMTakettf
tMTaket
ttf
x
Exampleunit step
sss
edtedtf(t)eF(s)
ssF
t
ttf
st
stst 1101
:Proof
_______________________________________
1)(
00
01)(
0
00
ExampleShifted Step
s
eA
s
eA
s
eAsF
dtAedtedtf(t)eF(s)
s
eAsF
t
tAtf
ssst
ststst
s
222
2
0 20
2
0)(
0
:Proof
_______________________________________
)(20
2)(
Integration by parts
02
00000
0
000
1100
1
1,
1,,
?
ss
te
sdte
ss
tevduuvdtte
dtes
vdus
teuv
es
vdtdudtedvtuLet
dtte
vduuvudv
stst
stst
stst
stst
st
Example
ExampleRamp
stst
ststststst
es
vdtdudtedvtuLet
vduuvudv
ss
te
sdte
stedttedtf(t)eF(s)
ssF
t
tttf
1,,
partby n integratio UsingDone
1100
1
:Proof
_______________________________________
1)(
00
0)(
000
02
000
0
2
ExampleExponential Function
asdtedteedtf(t)eF(s)
assFetf
tasstatst
at
1
:Proof
_______________________________________
1)()(
0
)(
00
Examplesine Function
shortlygiven be willproof theof Details
)sin(
:Proof
_______________________________________
)()sin()(
2200
22
sdtetdtf(t)eF(s)
ssFttf
stst
Examplecosine Function
2200
22
)cos(
:Proof
_______________________________________
)()cos()(
s
sdtetdtf(t)eF(s)
s
ssFttf
stst
ExampleRectangle Pulse
)1(0
:Proof
_______________________________________
)1()(],(0
],0[)(
00
sLL
L
ststst
sL
es
AdtedtAedtf(t)eF(s)
es
AsF
Lt
LtAtf
Properties of Laplace TransformAddition
000
)()()()()(
:Proof
)()()()(
__________________________________________________
)()()()(
)()(
)()(
dtetgdtetfdtetgf(t)tgtfL
tgLtfLtgtfL
sGsFtgtf
sGtg
sFtf
ststst
Properties of Laplace TransformMultiplication by a constant
)()()(
:Proof
)()(
__________________________________________________
)()(
)()(
00
tfLadtetfadtf(t)eatfaL
tfLatfaL
sFatfa
sFtf
stst
Properties of Laplace TransformMultiplication by exponential
__________________________________________________
)()(
)()(
asFetf
sFtfat
)()(
:Proof
)()(
0
)(
0
asFdtf(t)edtef(t)eetfL
asFetfL
tsastatat
at
Properties of Laplace TransformExamples Multiplication by exponential
__________________________________________________
)()(
)()(
asFetf
sFtfat
22
2222
2222
)(
1,
1
)(
)()cos(,)cos(
)()sin(,)sin(
asetL
stL
as
asetL
s
stL
asetL
stL
at
at
at
Useful Identities
jj
jj
j
j
eej
ee
je
je
2
1)sin(
2
1)cos(
)sin()cos(
)sin()cos(
Examplesin Function
22
0 00
00
22
11
2
1
2
1
2
1
)sin(
_______________________________________
)()sin()(
sjsjsj
dtedtej
dteeej
dtetdtf(t)eF(s)
ssFttf
sttjsttjsttjtj
stst
Examplecosine Function
22
0 00
00
22
11
2
1
2
1
2
1
)cos(
______________________________________________
)()cos()(
s
s
jsjs
dteedteedteee
dtetdtf(t)eF(s)
s
ssFttf
sttjsttjsttjtj
stst
Laplace Transform
Inverse Laplace Transform
Properties of Laplace TransformMultiplication by time
43
3
1322
2
62)(
!)(
21)(
11)(
1)(
_________________________________
)()(
ssds
dtutL
s
ntutL
ssds
dtutL
ssds
dtutL
stuL
sFds
dtftL
nn
Properties of Laplace Transform
s
sXxssXxsxsXs
tutxtxtxL
ffsfssFsdt
tfdL
fsfsFsdt
tfdL
fssFdt
tdfL
1)(2)]0()([3)0()0()(
)()(2)(3)(
__________________________________________________________
)0()0()0()()(
)0()0()()(
)0()()(
2
233
3
22
2
Properties of Laplace TransformIntegration
)()()(.6
______________________________
)(1
)( 5.0
sFeatuatfL
sFs
dfL
sa
t
Properties of Laplace TransformDelay
)()()(
)()()(
)()(
sFeatuatfL
sFeatuatf
sFtf
sa
sa
Properties of Laplace Transform
LsLs es
ALes
As
AsF
LtALuLtuLtAtuAttf
Lt
LtAt
t
tf
11)(
)()()()()(
0
0
00
)(
22
Slope =A
L
Properties of Laplace Transform4
Slope =A
L L
Slope =A_ _A L
Slope =A
L
=
Summary
SE 207: Modeling and SimulationLesson 3: Inverse Laplace Transform
Dr. Samir Al-Amer
Term 072
Properties of Laplace Transform
)0()0(...)0()()(
)0()0()0()()(
)0()0()()(
)0()()(
)1()2(1
233
3
22
2
nnnnn
n
ffsfssFsdt
tfdL
ffsfssFsdt
tfdL
fsfsFsdt
tfdL
fssFdt
tdfL
Solving Linear ODE using Laplace Transform
???)(
23
1)(
)(
1)(2)(3)(
1)(2)]0()([3)0()0()(
0)0()0(),()(2)(3)(
2
2
2
tx
nsformLaplaceTrainverseuse
ssssX
sXforsolves
sXssXsXs
ssXxssXxsxsXs
nsformLaplaceTrause
xxtutxtxtx
Inverse Laplace Transform
them.sum and each term of inverse obtain thethen
sorder term second andfirst of sum theasF(s) Expand
ExpansionFraction Partial
)()(
Transform Laplace Inverse1 sFLtf
Notation
F(s) of thecalled are D(s) of roots
F(s) of thecalled are N(s) of roots
spolynomial are D(s) andN(s) whereD(s)
N(s)F(s)
as expressed becan it , sin function rational isF(s)
poles
zeros
Notation
3:,5.0:12
3s
2:,4,3:3)4)(s(s
2s
127s
2s2
zeropoles
zeropoless
Notation
F(s) of thecalled are D(s) of roots
F(s) of thecalled are N(s) of roots
spolynomial are D(s) andN(s) whereD(s)
N(s)F(s)
as expressed becan it , sin function rational isF(s)
poles
zeros
Examples
properaress
s
s
properstrictlyaress
12
2s,
127s
23s,
127s
1
127s
2s,
127s
1
2
2
2
22
Partial Fraction Expansion
polescomplex *
poles repeated *
polesdistict *
cases hreeconsider t willWe
Partial Fraction Expansion
2j-2j,-11-at polescomplex wo t
2)-at poles (double -2sat pole repeatedone
0,-1at poles realdistict twohas)(
)52(s2)1)(ss(s
1)(
22
sF
ssF
Partial Fraction Expansion
n
i
tsi
ssii
n
i i
i
i
i
eAf(t)
sFs-sA
where
s-s
AF(s)
1
1
Transform Laplace inverse
)(
as expressed becan F(s)
distict are of poles all andproper strictly is F(s) If
Example
032Transform Laplace inverse
3)1(
5)(4)(
2)4(
5)(1)(
)4()1(
distinct41at are poles proper,strictly
)4)(1(
5
45
5
4
1
4122
1111
21
1
2
2
1
tforeeeAf(t)
s
ssFssFs-sA
s
ssFssFs-sA
where
s
A
s
A
s-s
AF(s)
,
ss
s
ss
sF(s)
ttn
i
tsi
ssss
ssss
n
i i
i
i
Example
032
Transform Laplace inverse
)4(
3
)1(
2
)4)(1(
5
distinct41at are poles proper,strictly
)4)(1(
5
45
5
4
2
tforeef(t)
ssss
s
,
ss
s
ss
sF(s)
tt
Alternative Way of Obtaining Ai
032Transform Laplace inverse
3,2;54,1
)4)(1(
5
)4)(1(
)4()()(
)4)(1(
)1()4(
)4()1(
distinct41at are poles proper,strictly
)4)(1(
5
45
5
4
1
212121
2121
2121
2
tforeeeAf(t)
AAsolveAAAA
ss
s
ss
AAAAssF
ss
sAsA
s
A
s
AF(s)
,
ss
s
ss
sF(s)
ttn
i
tsi
i
Repeated poles
1)2(
165)()5(
1)5(
)165()5(5
)5(
165)()2(
2)5(
165)()2(
)5()2()2(
2at are poles reperated and 5at poledistict proper,strictly
)5()2(
165
5252
2
2
22
212
22
211
2122
11
2
ss
sss
ss
s
ssFsA
s
ss
s
s
ds
dsFs
ds
dA
s
ssFsA
s
A
s
A
s
AF(s)
ss
sF(s)
Repeated poles
1)5(
)165()5(5
)5(
165)()2(
sother term oft coefficien for the formula New *
2)5(
165)()2(
case poledistict in as obtained is order termhigest oft coefficien The *
presentare)2()2(
bothNote*
1)2(
165)()5(before as obtained is poledistict oft coefficien The *
2at are poles reperated and 5at poledistict )5()2()2(
2
2
22
212
22
211
122
11
5252
2122
11
sss
ss
ss
s
ss
s
s
ds
dsFs
ds
dA
s
ssFsA
s
Aand
s
A
s
ssFsA
s
A
s
A
s
AF(s)
Repeated poles
0112Transform Laplaceinverse
)(1)2(
165)()5(
1)5(
)165()5(5
)5(
165)()2(
2)5(
165)()2(
)5()2()2(
52,2at are poles proper,strictly )5()2(
165
modifiedis obtain tousedformula thepoles repeated has If
522
5252
2
2
22
212
22
211
2122
11
2
1
tforeetef(t)
poledisticts
ssFsA
s
ss
s
s
ds
dsFs
ds
dA
s
ssFsA
s
A
s
A
s
AF(s)
,ss
sF(s)
AF(s)
ttt
ss
sss
ss
Repeated poles
05.0Transform Laplaceinverse
)()()4(
)()1(
)()1(
)()1(
)4()1()1()1(
4,11,1at are poles proper,strictly )4()1(
3
modifiedis obtain tousedformula thepoles repeated has If
421312
211
42
1
32
2
13
1
312
1
311
2132
123
11
3
1
tforeAeAteAetAf(t)
poledistictsFsA
sFsds
dA
sFsds
dA
sFsA
s
A
s
A
s
A
s
AF(s)
,ss
sF(s)
AF(s)
tttt
s
s
s
s
2!1
Common Error
)4()1()1()1(
as expanded be shouldIt general.invalidnotisThis
)4()1(
)4()1(
3 expandmaySome
2132
123
11
23
11
3
s
A
s
A
s
A
s
AF(s)
s
A
s
AF(s)
asss
sF(s)
Complex Poles
?)(
)()21(
)()21(
)21()21(
j2-1- andj21-at polescomplex twohas52
84
)21(2
)21(1
1212
211
21
2
tjtj
js
js
ekektf
ksFjsk
sFjsk
js
k
js
kF(s)
ss
sF(s)
Complex Poles
)sin()cos()(
2,1,8,4
2)1(41252
)( as expressed becan
52
84
WayeAlternativ
2222
222
teaBC
tBetf
aCB
sssss
as
CBs
ss
sF(s)
atat
What do we do if F(s) is not strictly proper
023)()(
)2(
2
)1(
31
)1)(2(
41
23
41
23
64
proper.strictly isG(s) andnumber real a isk where
G(s)kF(s) asit express odivision t long use
proper strictly not but proper isF(s)If
2
22
2
tforeettf
ssss
sF(s)
ss
s
ss
ssF(s)
tt
Solving for the Response
23
5)(
0)(2]1)([32)(
0)(2)]0()([3)0()0()(
2)0(,1)0(,0)(2)(3)(
__________________________________________________________
)0()0()()(
)0()()(
2
2
2
22
2
ss
ssX
sXssXssXs
sXxssXxsxsXs
xxtxtxtxSolve
fsfsFsdt
tfdL
fssFdt
tdfL
Final value theorem
0)40)(10(
0
)4)(1(
2lim)(
03
2
3
2)(
03
2
3
2
)4(32
)1(32
)4)(1(
2
0
4
4
ss
sf
eef
tforeef(t)
ssssF(s)
s
tt
Final value theorem
parts real positive with poles no
hasF if theorem valuefinalapply can eRemember w
0)40)(10(
0
)4)(1(
2lim)(
5
2
5
2)(
05
2
5
2
)4(52
)1(52
)4)(1(
2
0
4
4
validNotss
sf
eef
tforeef(t)
ssssF(s)
s
tt
Step function
s
AU(s)
t
tAu(t)
00
0
A
impulse function
0,1
1)(00
dt(t)
sFtfor(t)
impulse function
propertysampling
otherwise
bacifcfdttfc)(t
dt(t)
(t)Ltfor(t)
b
a
0
],[)()(
0,1
100
Initial Value& Final Value Theorems
Transfrom Laplace inverse
obtain toneed out thewith
F(s) fromdirectly obtained becan
)( ValueFinal)0( Value Initial fandf
Initial Value Theorem
)(limlimit theby taking obtained is
timeinitial at thefunction theof valuethe
)(lim)0(
sFs
sFsf
s
s
Final Value Theorems
).()(
3)2)(ss(s
4s)(,
3)-2)(s(s
5sG(s)
:Examples
origin. at the pole single of
exception possible a with and planecomplex theof
halfright in the on the poles no has F(s) provided
)(lim)(0
gnotbutfobtaincanWe
sF
sFsfs
SE 207: Modeling and SimulationLesson 4: Additional properties of Laplace transform and solution of ODE
Dr. Samir Al-Amer
Term 072
Outlines What to do if we have proper function? Time delay Inversion of some irrational functions Examples
Step function
s
AU(s)
t
tAu(t)
00
0
A
impulse function
0,1
1)(00
dt(t)
sFtfor(t)
impulse function
1
Area=1
You can consider the unit impulse as the limiting case for a rectangle pulse with unit area as the width of the pulse approaches zero
impulse function
otherwise
bacifcfdttfc)(t
propertysampling
dt(t)
(t)Ltfor(t)
b
a 0
],[)()(
0,1
100
Sample property of impulse function
03
3
)6cos()3cos(2
2
5
35
1
5
1
dte)(t
edte)(t
dtt)(t
t
t
Time delay
f(t) F(s)g(t) G(s)
sesFsG
tftg2)()(
)2()(
What do we do if F(s) is not strictly proper
rationalNotess
ssF
cannotnotss
ss
s22
2
2
45
6)(
earlier. discussed s techniquethe
apply Weproper,strictly 45
65H(s)
earlier. discussed techniques
apply thecan Weproper,strictly 4)3)(s(s
1G(s)
What do we do if F(s) is not strictly proper
022)()(
)2(
2
)1(
21
)1)(2(
641
23
641
23
87
proper.strictly isG(s) andnumber real a isk where
G(s)kF(s) asit express odivision t long use
proper strictly not but proper isF(s)If
2
22
2
tforeettf
ssss
sF(s)
ss
s
ss
ssF(s)
tt
Example
57
882
3244
2
44
32
2
22
2
2
s
ss
ssss
ss
ssF(s)
44
572
2
ss
s
− − −
Example
tt
ss
s
etetss
Ltf
sds
d
s
ss
ds
dB
s
ssA
s
B
s
A
s
s
ss
s
ss
ssF(s)
222
1
22
2
2
2
2
2
2
222
2
79)(22
7
2
92)(
7572
572
92
572
222
2
572
44
572
44
32
Solving for the Response
23
5)(
0)(2]1)([32)(
0)(2)]0()([3)0()0()(
2)0(,1)0(,0)(2)(3)(
__________________________________________________________
)0()0()()(
)0()()(
2
2
2
22
2
ss
ssX
sXssXssXs
sXxssXxsxsXs
xxtxtxtxSolve
fsfsFsdt
tfdL
fssFdt
tdfL