lecture 25 laplace transform hung-yi lee. reference textbook 13.1, 13.2
TRANSCRIPT
Lecture 25Laplace Transform
Hung-yi Lee
Reference
• Textbook 13.1, 13.2
Laplace TransformMotivation and Introduction
Laplace Transform
dtetf st
0)(sF
tf sF
dsesFj
tf stj
j
)(
2
1
c
c
se
R
Laplace Transform ( L[f(t)] )
Inverse Laplace Transform (L-1[F(s)])
Time domain s-domain
Note
dtetftf st
0)(L
dtetfdtetf stst
0
0
0)()(
Always 0?When f(0)=∞, it may not be zero
Note
tf
sF
Laplace Transform (L)
Inverse Laplace Transform (L-1)
Time domain
s-domain
tf̂=
When t≥0
1
,,1
tutu
tu
Domain
台北市羅斯福路四段一號博理館
25°1'9"N 121°32'31"E
Different domains means view the same thing in different perspectives
Position:
Domain
x
y 1,1 xy 0,2
Different domains means view the same thing in different perspectives
Linear Algebra:
Domain
1tfTime domain
ss /1F s-domain
SignalMuggle
Engineer LaplaceTransform
Inverse LaplaceTransform
Transform: switch between different domains
Fourier Series
Fourier Transform
TimeDomain
TimeDomain
FrequencyDomain
FrequencyDomain
https://www.youtube.com/watch?v=r4c9ojz6hJg
Fourier Transform
dtetf tj
)(F
deFt tj
)(
2
1f
Laplace Transform
dtetfs st
0)(F
dsesFj
tf stj
j
)(
2
1
Laplace Transform
v.s. Fourier
Transform
dtetfs st
0)(F
dtetf tj
)(F
Fourier Transform:
Laplace Transform:
dteetf tjt
0)(
dtetuetf tjt
)(
Laplace Transform of f(t)
Fourier Transform of f(t)e-σtu(t)
=
Multiply e-σt
Do Fourier Transform
Laplace Transform
v.s. Fourier
TransformMultiply u(t)
dtetfs st
0)(F
dtetuetf tjt
)(
Transformable Function
Fourier Series
Fourier Transform
Laplace Transform
All Functions2te
te
1
0 D
Periodic Functions
Why we do Laplace transform?
Transfer Function
H(s)(circuit, filter …)
)(tx
)(ty
Laplace transform can help us find y(t) easily
Transfer Function ttx 0
t cosAe)( 0
00t
0 cosAe||)( 0 sHtsHty
H(s)(circuit, filter …)
The signal with complex frequency s0 = σ0 + ω0 t0
t sinjAe 0
00t
0 sinAe|| 0 sHtsHj
tj 0eAe s
tjHj0
00 eAee|H| sss
Transfer Function
H(s) H(s)
(z is complex)
tj 0eAe stx
tjHj
000 eAee|H| sss
ty
(H(s0) is complex)
t0Zestx
t0
0ZessHty
Transfer Function
H(s)
dsesFj
tx stj
j
)(
2
1
dsesFsHj
ty stj
j
)(
2
1
)(L sFtx
)(HL sFsty We do not know y(t), but we know its Laplace transform
Transfer Function
H(s)
ty
tx )(sF
)(H sFs
L
-1L
Laplace TransformLaplace Transform Pairs
Laplace Transform Pairs (1/4)
1)( tf
01)(L dtetf st
101
s s
1
t
t
stes 0
1
If Re[s]=σ>0
Time domain: 1 s-domain: 1/s
ROC: σ > 0
ROC
Time domain: u(t)(Only consider t>0)
Laplace Transform Pairs (2/4)
00
)(L dtedteetf tsastat
atetf )(
t
t
tsaesa 0
1 10
1
sa as
1
If Re[a+s]=a+σ>0
Time domain: e-at s-domain: 1/(s+a)ROC: σ > -a
ROC-a
Laplace Transform Pairs (3/4)
0 2dte
j
ee sttjtj
sincos
sincos
je
jej
j
ttf sin)(
0
sin)(L dtettf st
002
1 dtedtej
tjstjs
222
sj
jsjs
jsjsj
11
2
122
s
j
ee
ee
jj
jj
2sin
2cos
If Re[s]=σ>0
Laplace Transform Pairs (4/4)
0 2dte
ee sttjtj
ttf cos)(
0
cos)(L dtettf st
002
1 dtedte tjstjs
222
s
jsjs
jsjs
11
2
122
s
sIf Re[s]=σ>0
sincos
sincos
je
jej
j
j
ee
ee
jj
jj
2sin
2cos
Summary of Transform Pairs
• More complete Transform Pairs: http://www.vibrationdata.com/math/Laplace_Transforms.pdf
The 4 transform pairs are sufficient to imply all transform pairs in Table 13.2.
as 1ate
s
1122
s
tsin
22 ss tcos
s-domainTime domain
Note: Impulse function
• What is L-1[1]? L-1[1]=δ(t) (impulse function, Dirac delta function)
0""
00
t
tt
0""
00
0
00 t
ttt
1dtt
dt/1
dt
tdut 0
1
0t
1
Laplace TransformProperties
The six properties in Table 13.1 (P585)
Property 1: Linear Combination• Let L[f(t)]=F(s) and L[g(t)]=G(s)
)()()()(0
dtetgtftgtfL st
00
)()( dtetgdtetf stst
)()( sGsF
)()(0
dtetAftAfL st )(sAF
Property 2: Multiplication by e-at
• Let L[f(t)]=F(s)
0
)()(L dtetfetfe statat
0
)( dtetf tas )( asF
as 1ate
s
11
ROC
ROC-a
Multiplication by e-at
Property 3: Multiplication by t• Let L[f(t)]=F(s)
)()(L sFds
dttf
0
)( dtetfds
dsF
ds
d st
0dt
ds
detf
st
0
dtettf st
0
dtettf st
Property 4: Time Delay
00
0
00
)(
0
)( )()(
ttttf
tt
ttuttftg
0tt
Delay by t0 and zero-padding up to t0
t
tt
stdtettftg0
)()(L 0
0tt 0tt
0
0)( def ts
ddt
0
)(0 defe sst sFe st0
Property 5: Differentiation
• Let L[f(t)]=F(s) b
a
b
a
b
adtuvuvudtv
Integration by parts:
00
)()()( dtestfetf stst
0
)()(L dte
dt
tdftf st
0
)()0( dtetfsf st )0()( fssF
v’
v v
u
u u’
Property 5: Differentiation
• Let L[f(t)]=F(s)
0)(L fssFtf
1
1LsF
s
tf tetf
te tf 1
1L
s
tf
Example
0)(L fssFtf 11
1
ss
1
1
s
Property 5: Differentiation
0)(L fssFtf
sFtf )(L
00)(L ffssFstf
002 fsfsFs
……
000)(L )1(21 nnnnn ffsfssFstf
Property 6: Integration
tdftg
0)()(
• Let L[f(t)]=F(s)
0 0)()(L dtedftg stt
0 0)( dtdef
t st
0)(
ddtef st
(You can use integration by parts as in P584)
t
t
dtdef st)(
tdftg
0)()(
• Let L[f(t)]=F(s)
Property 6: Integration
0)(L
ddteftg st
0)(
ddtef st
0
1)( de
sf st
0
1)( des
f s
0
)(1 defs
s
ss
F1
Laplace Transform Properties
)(1
)( n Integratio
)0()( )( ation Differenti
)( )()( delay Time
)( )( by tion Multiplica
)( )( by tion Multiplica
)()( )()( n CombinatioLinear
-
0
0
00
sFs
df
fssFtf
sFettuttf
dssdFttft
asFtfee
sBGsAFtBgtAf
t
st
atat
Table 13.1 Laplace Transform Properties (P585)Operation Time Function Laplace Transform
More properties (in Homework)
Time-scaling property
Periodic function
Integral of F(s)
sFtf /L
tftdssFs
1L
tf tgf(t)=0 outside 0<t<T
TtuTtfTtuTtftftg 22
…..
sTe
sFtg
1
L
22
22
22
1
2
1
2
sincos )cos(
sincos )cos(
sin
!
1
1
!
1
1 )()(
as
aste
s
st
st
as
ret
aste
ase
s
rt
st
s
eDtutu
s
AA
at
ratr
at
at
rr
sD
Table 13.2 Laplace Transform Pairs (P585) f(t) F(s)
1
0 D
D-ttuL u
D-tLtuL u sD-e11
ss
22
22
22
1
2
1
2
sincos )cos(
sincos )cos(
sin
!
1
1
!
1
1 )()(
as
aste
s
st
st
as
ret
aste
ase
s
rt
st
s
eDtutu
s
AA
at
ratr
at
at
rr
sD
Table 13.2 Laplace Transform Pairs (P585) f(t) F(s)
1
0 1
1-ttuL us
-se1
1
0 1
1-ttu u
1-ttut u
t
1-ttutL udss
d-se1
2
-s-s e1e
s
s
2
-s-s e-e-1
s
s
Example for Periodic function
1
0 1
2
-s-s e-e-1
s
sL
1
0 1 2 3 4 5
L
TtuTtfTtuTtftftg 22
sTe
sFtg
1
L
2
-s-s e-e-1
s
sse 21
1
22
22
22
1
2
1
2
sincos )cos(
sincos )cos(
sin
!
1
1
!
1
1 )()(
as
aste
s
st
st
as
ret
aste
ase
s
rt
st
s
eDtutu
s
AA
at
ratr
at
at
rr
sD
Table 13.2 Laplace Transform Pairs (P585) f(t) F(s)
1 1s
tds
ds 1
2s
2tds
ds 2
32 s
3tds
sd 32
432 s
rt 132 rrs 1! rsr
L
L
L
L……
L
22
22
22
1
2
1
2
sincos )cos(
sincos )cos(
sin
!
1
1
!
1
1 )()(
as
aste
s
st
st
as
ret
aste
ase
s
rt
st
s
eDtutu
s
AA
at
ratr
at
at
rr
sD
Table 13.2 Laplace Transform Pairs (P585) f(t) F(s)
tcos
sinsincoscos tt
tcosL
2222
sincos
ss
s
Note: Euler’s formula
tjte tj00 sincos0
0
1
js 20
2 ss
20
20
s
j
0
0
0
1
js
js
js
20
20
s
js
Note: Multiplication
)()( sGsFtgtfL
tsinL22
s
t2sinL
2
22
s
2
2cos1sin 2 t
t
t2sinL 22 22
1
2
1
s
s
s
Laplace TransformInverse Laplace Transform
Partial-Fraction Expansions
• Rational Function
)(
)()(
011
1
011
1
sD
sN
asasas
bsbsbsbsF
nn
n
mm
mm
n
n
ss
A
ss
A
ss
A
2
2
1
1
nssssss
sN
21
)( s1, s2, ……, sn are the roots of D(s) ( poles of F(s) )
We only consider the case that m < n.(strictly proper rational function)
One pole, one term
Partial-Fraction Expansions
• Rational Function
)(
)()(
011
1
011
1
sD
sN
asasas
bsbsbsbsF
nn
n
mm
mm
sDsR
bs m
)(F
If m = n
δ(t)
If m = n + 1
sDsR
csbs m
)(F
1
0fssFtf
sdδ(t)/dt ……
multiply sdifferentiate
Partial-Fraction Expansions
• Rational Function
n
n
ss
A
ss
A
ss
AsF
2
2
1
1)(
ateas
L
11
tsn
tsts neAeAeAtfsFL 2121
1 )(
tsi
i
i ieAss
AL
1
0t
There are three tips you should know.
Tip 1: How to find A1,A2,∙∙∙∙∙∙∙,An
•Example 13.5
102102
40242)(F 321
2
s
A
s
A
s
A
sss
sss
Panacea: reduce to a common denominator, and then compare the coefficients
21010240242 3212 ssAssAssAss
ssAssAssA 2102012 23
22
21
2321 AAA
2421012 321 AAA
4020 1 A
Tip 1: How to find A1,A2,∙∙∙∙∙∙∙,An
•cover-up rule
212
22
)(then , setting A
ss
sNss
2
2
1
1
21
)()(
ss
A
ss
A
ssss
sNsF
2
211
2
)(
ss
AssA
ss
sN
1
21
11
)(then , setting A
ss
sNss
2
1
21
1
)(A
ss
ssA
ss
sN
Tip 1: How to find A1,A2,∙∙∙∙∙∙∙,An
• Example 13.5: cover-up rule
102102
40242)(F 321
2
s
A
s
A
s
A
sss
sss
Take A2 as example. Multiplying s+2 at both sides
Set s=-2:
102
2102
402422 321
2
s
A
s
A
s
As
sss
sss
10
22
10
40242 32
12
s
AsA
s
As
ss
ss
2
2
1022
4022422A
52 A
Tip 2: For Complex Poles
• Example 13.6: 2562
71615)(
2
2
sss
sssF Find f(t) = L-1[F(s)]
43 ,2 js
43432)( 22211
js
A
js
A
s
AsF
It is not easy to
find A21 and A22.
43
43
2
43)(43 22
211
js
AjsA
s
AjssFjs
432
71615 2
jss
ss
Set s=-3+j4 to find A21…….
Tip 2: For Complex Poles
• Example 13.6: 2562
71615)(
2
2
sss
sssF Find f(t) = L-1[F(s)]
2562
5)(
2
ss
CBs
ssF Do not split the complex poles
Find B and C
2562
71615
2562
221253052
2
2
22
sss
ss
sss
CCsBsBsss
66 72125
10 155
CC
BB
256
6610
2
5)(
2
ss
s
ssF
Tip 2: For Complex Poles
• Example 13.6: 2562
71615)(
2
2
sss
sssF Find f(t) = L-1[F(s)]
2562
5)(
2
ss
CBs
ssF Do not split the complex poles
Find B and C (another approach)
2562
515)(
2
2
3
3
ss
CsBs
s
s
s
sssF
BssFs
515lim
252
5
252
7)0(
CF
10B
66C
ssFs lim
)0(F
Tip 2: For Complex Poles
• Example 13.6: 2562
71615)(
2
2
sss
sssF Find f(t) = L-1[F(s)]
256
6610L
21
ss
sFind
221
21
43
96310L
256
6610L
s
s
ss
s
(Refer to P593)
te
s
s t
cosL22
1
te
st
sinL
221
22
1
221
43
4L24
43
3L10
ss
s
256
6610
2
5)(
2
ss
s
ssF
Tip 3: Repeated Poles
• Exercise 13.31
22
34
10124)(
ss
sssF
34321
s
AA
s
A
221
2
2
3434
10124)(
s
A
s
A
ss
sssF
order=2order=3
212
2
43
10124
AsAs
ss
212
2
496
10124
AsAss
ss
41 A
126 21 AA
1049 21 AA
?
334321
s
A
s
A
s
A
Tip 3: Repeated Poles
• Exercise 13.31
2321
2
2
33434
10124)(
s
A
s
A
s
A
ss
sssF
(Refer to P596 – 597)
232
12
2
3
4
3
4
3
10124
s
As
s
AsA
s
ss61 A
32
122
34
3
4
10124AAs
s
As
s
ss
103 A
We cannot find A2 by multiplying (s+3)
34
3
34
10124 32
12
s
AA
s
As
ss
ss
Tip 3: Repeated Poles
• Exercise 13.31
22
2
2
3
10
34
6
34
10124)(
ss
A
sss
sssF
22
2
2
3
10
34
6
34
10124)(
s
s
s
sA
s
s
ss
sssssF
s 4 6 02A 2A2
Exercise 13.10
2
8
3
4235F
s
esss
s
ses
s
s8
23
42
3
5
te 35
L-1
)(0 sFe st)()( 00 ttuttf Time delay
Delay by 8 and zero-padding up to 8
?
L-1
23
42F
s
ss
221
3
A
3
A
ss
21 AA342 ss
10A2 3s
Exercise 13.10
2
8
3
4235F
s
esss
s
ses
s
s8
23
42
3
5
te 35
L-1
)(0 sFe st)()( 00 ttuttf Time delay
Delay by 8 and zero-padding up to 8
?
L-1
23
42F
s
ss
21
3
10
3
A
ss
23
42Fs
s
sss
21
3
10
3
A
s
sss
s 1A2
Exercise 13.10
2
8
3
4235F
s
esss
s
ses
s
s8
23
42
3
5
te 35
L-1
)(0 sFe st)()( 00 ttuttf Time delay
Delay by 8 and zero-padding up to 8
?
L-1
23
42F
s
ss
23
10
3
2
ss
te 32
L-1
tte 310 L-1
88
5tf 3
tutf
e t
tt teetf 33 102
Initial and Final Values
• We can find the value f(0+) and f(∞) from F(s)
Initial-value Theorem
sFtf LWe know
)(lim)0( ssFfs
Because F(s) is strictly proper, )(lim ssF
s is defined.
Example 13.9
• Find the initial value f(0+)
8009018
16005)(F
23
3
ssss
ss 50 f
)(lim)0( ssFfs
8009018
16005lim
23
3
sss
ss
5
Example 13.9
• Find the initial slope f’(0+)
8009018
16005)(F
23
3
ssss
ss 50 f
)(lim)0( sFsfs
)(FsfL s
0)(FsfL fss 58009018
1600523
3
sss
s
s
sss
ssf
s5
8009018
16005lim)0(
23
3
8009018
8009018516005lim
23
233
sss
sssssss
90
∞ - ∞ ?
Initial and Final Values
• We can find the value f(0+) and f(∞) from F(s)
)(lim)(0
ssFfs
Initial-value Theorem
Final-value Theorem
sFtf LWe know
)(lim)0( ssFfs
If the final value exists(Can be known from the poles)
Because F(s) is strictly proper, )(lim ssF
s is defined.
Final Values
4 regions
Region A
Region D
Region C
Region B
Final Values
Region A
as
As
)(F
rr
as
A
as
A
as
As
2
21)(F
cbss
Css
2
B)(F
a>0
a>0
b<0
atet )(f
No final value
No final value
attet )(f
22
s
CsB
α<0 teC
teBtt
t
sin
cos)(f
No final value
Final Values
Region B
as
As
)(F
rr
as
A
as
A
as
As
2
21)(F
cbss
Css
2
B)(F
a<0
a<0
b>0
atet )(fFinal value = 0
Final value = 0
attet )(f
22
s
CsB
α>0 teC
teBtt
t
sin
cos)(f
Final value = 0
Region C
Final Values
22
B)(F
s
Css
tCtBt sincos)(f
No final value
2222CB
ss
s
Final Values
Region D
s
As )(F A)(f t
final value = constant
rr
s
A
s
A
s
As
221)(F tt 2A)(f No final value
Summary for Final Values
Final value exists (1) Poles on the left plane, or (2) single pole at the origin
(P601)
Non zero final valuesingle pole at the origin
Final Values )(lim)(0
ssFfs
Final-value Theorem
22
F
s
ss
The final value exists iff the poles are in this region
Only one pole 0)(lim0
ssFs
ts cosf The final value not exists
The final-value theorem gives the wrong answer when the final value does not exist.
as
s
1
F
0a0)(lim
0
ssF
s
ates f The final value not exists
Final Values )(lim)(0
ssFfs
Final-value Theorem
The final value exists iff the poles are in this region
Only one pole
The final value is not zero iff there is only one pole at the origin
2
2
1
1)(Fas
A
as
A
s
As
The final value is clearly A
The final value is 0
The final value is A
)(lim)(A0
ssFfs
Example 13.9
• Find the final value
8009018
16005)(F
23
3
ssss
ss 50 f
Four poles: 0, -10, -4+8j, -4-8j The final value exists.
The final value is not zero.
)(lim)(0
ssFfs
8009018
16005lim
23
3
0
sss
ss
2
Laplace TransformApplication
Differential Equation
Find v(t)
6VV15812
8)0(
v
BVtvtRi )()( )()( tvCti
BVtvtvRC )()(
15)()(60
112 tvtv
15)()(2.0 tvtv
15L)(L)(L2.0 tvtv
)(L)( tvsV
6)(
)0()(L)(L
ssV
vtvstv
s
1515L
s
sVssV15
)(6)(2.0
Differential Equation
0t
s
sVssV15
)(6)(2.0
55
756
12.0
152.1
)(
s
B
s
A
ss
s
sssV
0s55
756)(
s
sBA
s
sssV 15A
B5756
)(5
s
As
s
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Homework
• 13.6, 13.9, 13.10, 13.16, 13.25, 13.28, 13.35, 13.38. 13.46
Thank you!
Answer
• 13.6: derive by yourself• 13.9: proof by yourself• 13.10: proof by yourself• 13.16: F(s)=2(1-3se-2s-e-3s)/s2
• 13.25: f(t)=-2+5e-2t-3e-4t-e-6t
• 13.28: f(t)=5-5e-4t+10e-3tcos(t-36.9 。 )• 13.35: f(t)=2te-tcos(2t-180 。 )• 13.46: f(0+)=2, f’(0+)=-5, f(∞) not exist
Appendix
Fourier Series
• Periodic Function: f(t) = f(t+nT)
Fourier Series:
Laplace Transform Pairs (1/4)
s
11L ste11 0s
σ=0
0 ω
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