Chapter 13 Laplace Transform Analysis -...
Transcript of Chapter 13 Laplace Transform Analysis -...
-
Chapter 13 Laplace Transform Analysis
-
Chapter 13: Outline
-
Laplace transforms
-
Laplace Transform
l s-domain phasor analysis:
l Laplace transform:)cos()( xt
temXtx φ+ωσ= xmXX φ∠=
+≤≤−∞
-
Laplace Transform (Three conditions)
l Unilateral (one-sided Laplace transform): Laplacetransform holds for t=0. Previous effects are included in the initial conditions at t=0-.
l Existence condition: l The resulting transform is a function of s.
. ,0lim )( ctetft σσ
σ >=−∞→
-
One-Sided Waveform
[ ] [ ] +− ≤≤∞<
=⇒
><
=
00,)( if ,
0),(0,0
)()(0,10,0
)( :stepunit
ttfLL
ttft
tutftt
tu
f(t)f(t)u(t)
-
Sufficient Existence Condition
econvergenc of abscissa :order lexponentia of is )(
,0)( if finite remains integral The
)()()()(
finite is )(
c
0t
000 0
0
lim
σ
σσσ
σ
⇒⇒
>=⇒
==≤
∫∫∫∫ ∫
∫
∞ −
∞→
∞ −∞ −∞ ∞ −−
∞ −
−
−−− −
−
tf
dtetf
dtetfdtetfdtetfdtetf
dtetf
ct
tststst
st
-
Inverse Laplace Transform
l Inverse Laplace transform for t=0:
l Inverse Laplace transform is usually done by partial fraction expansion (will be covered in section 13.2).
ccjc dsstesF
jsFtf jc σ
ωπ ω
>+≡−= ∫ − ,)(21)]([1L)(
-
Example 13.1
[ ]
[ ]
[ ] [ ][ ] )0Re complex, becan :(Note
1)(1 ,0 if
when ,1
0 if ,0lim)(lim
0
)(
0
)(
>+
===
−>+
===
>+==
∫∫∞ +−∞ −−−
+−
∞→
−
∞→
−
−−
σ
σ
σσ
aas
tuLLa
aas
dtedteeeL
aeetf
eL
ctsastatat
ta
t
at
t
at
-
Example 13.1
-
Example 13.2
[ ] ( )
0
112
1
21
sin
22
=
+=
+
−−
=
−= −
c
tjtj
sjsjsj
eej
LtL
σ
ββ
ββ
β ββ
-
Transform Properties
– Linear combination:
[ ] [ ][ ][ ] )()(
)()()()()()( ),()(
sAFtAfLsBGsAFtBgtAfL
sGtgLsFtfL
=+=+
==
-
Transform Properties
– Multiplication by e-at:
[ ][ ] )()(
)()(
asFtfeL
sFtfLat +=
=−
-
Transform Properties
– Multiplication by t:
[ ])()(
)()(
0
0
ttfLdtettf
dtetfdsd
sFdsd
st
st
−=−=
=
∫
∫∞ −
∞ −
−
−
-
Transform Properties
– Time delay:
[ ] )()(),(
,0)()()(
0
00
000
sFetgL
ttttf
ttttuttftg
st−=
>−<
=−−=
-
Transform Properties
– Differentiation:
[ ]
[ ] )0()0()()(
)0()()(
)(
2 −−
−
′−−=′′
−=
=′
fsfsFstfL
fssFdt
tdfLtfL
-
Transform Properties
– Integration:
[ ] )(1)()(
,0)0( ,)()(
)()(
0
0
sFs
dfLtgL
se
ddtegdttftdg
dftg
t
stst
t
=
=
−
===
=
∫
∫
−
−
−−−
λλ
λλ
-
Example 13.3
se
se
ssF
DtutuDt
tf
sDsD
−− −=−+=
−−=
-
Example 13.4
[ ]
[ ] [ ]
[ ]
[ ] [ ] 22
22
2222
22
sincossinsincoscos)cos(
cos
)0sin(cos)(
)(sin)(
βφβφ
βφβφφβ
ββ
ββ
ββ
ββ
ββ
β
+−
=−=+
+=
+=−
+==′
+=⇒=
−
ss
ttLtL
ss
tL
ss
ss
tLtfL
stfLttf
-
Table 13.1
-
Table 13.2
-
Solving Differential Equations– The transformation automatically incorporates
the initial conditions (zero-input response).– Transformation converts linear differential
equations to s-domain algebraic equations.– Transformation is similar to the s-domain
phasor analysis. Denominator of the s-domain function includes the characteristic polynomial.
– Inverse transformation is required to obtain the resultant time domain function.
-
First-Order Example
Vv 6)0( =−
[ ]
)5(75
56
)5(756
)(
15)()0()(2.0
15)()(2.0
++
+=
++
=
=+−
=+′
−
ssssss
sV
ssVvssV
tvtv
[ ]
( ) 0,9151)2.0(756)5(
755
6)()(
555
111
≥−=−−+=
+
+
+==
−−−
−−−
teee
ssL
sLsVLtv
ttt
-
Second-Order Example
11
2
1
)0( ,)0(
0 t,0 t,
RIvIi
II
iin
==
><
=
−−
-
Second Order Example (Cont.)
[ ][ ]
[ ]
0for ,)()(
switching) (no if/1)/(
/)/()(
/)()(
)()()()()(
)()()(
0
11
12
221
21
21
1
2
≥=⇒=
=++++
=
=+−=+−
⇒
=+′=+′
>
tItisI
sI
IILCsLRss
LCIsILRsIsI
sIsIRIssVC
sVsRIIssILItItvC
tvtRitiL
t
Characteristic polynomial
-
Transform Inversion
-
Partial-Fraction Expansion
l Partial-fraction expansion of a strictly proper rational function:
1,
011
1
011
1)()()( −≤
+++−−+
+++−−+== nmasansna
ns
bsbmsmbmsmb
sDsNsF
L
L
l Three cases will be considered: distinct real poles, complex poles and repeated poles.
-
Case 1: Distinct Real Poles
l (Heaviside's theorem, cover-up rule.)
K
L
L
,2,1 ,)()(
0 ,)(
)()(
)(
2121
2
2
1
1
=−=
≥+++=
−++
−+
−==
= isFssA
teAeAeAtf
ssA
ssA
ssA
sDsN
sF
i
n
ssii
tsn
tsts
n
n
-
Example 13.5: Inversion of a Third-Order Function (Heaviside’s theorem)
=−=
==Ω=
AIAI
FCHLR
2,2201
,1,12
21
++
++=
LCsLRss
LCIsILRsIsI
/1)/(2
/21)/(2
1)(
0,52)(10
12
52)(
1)()10(
5)()2(
2)(
102)10)(2(40242
)(
102
103
22
01
3212
≥+−=⇒+
++
−+=
=+=
−=+=
==
++
++=
+++−−
=
−−
−=
−=
=
teetisss
sI
sIsA
sIsA
ssIA
sA
sA
sA
sssss
sI
tt
s
s
s
-
Example 13.5: (Method of Undetermined Coefficients)
133
311113
52
3314
)(
10252
)10)(2(40242
)(
3
331
32
=⇒
+=+−==
++
+−=
+++−−
=
=
A
AAsI
sA
sssssss
sI
s
-
Case 2: Complex Poles
)()3)(202
2()( nsssssssD −−ω+α+= L
)cos(21
21
)(
)()(2
21
21
)(
)()(
, , if
)()(
3
22021
20
2
φβ
βα
βαβα
αωαβαωα
αβαβα
φβα
+=+=
=−+=
+++
−+=
−+=
−±−=±−=<
−+−∗−−
+−=
∗
=∑
teKeKKetg
eKsFjsK
js
K
js
KsG
ssA
sGsF
jjss
tm
tjtj
jmjs
n
i i
i
-
Case 2: Complex Poles (Cont.)
[ ]
KjKKCBjB
KCB
sCBBs
ssCBs
sG
sKKs
tgL
teKtg
mm
mm
tm
=+=−+⇒⇒
−++−−+
=++
+=
++−+
=
+= −
φφβαωβα
αωααα
ωα
βαφβφα
φβα
sincos/)(
. determine then ts,coefficien compare , and , , , Find
)()()(
2)(
)(sincos)(
)(
)cos()(
Or,...
20
220
220
2
22
Undetermined coefficients
-
Example 13.6: Inversion with Complex Poles
432
2
21
2
1
3212
2
)256)(2(71615
2
5)()2(
2564321
4321
)(
)(2
)(
43, ,2)256)(2(
71615)(
js
s
sssss
K
sFsA
ssCBs
js
K
js
KsG
sGs
AsF
jjssssss
sssF
+−=
−=
∗
+++−−
=
=+=
+++
=++
+−+
=
++
=
±−=±−=−=⇒+++
−−= βα
-
Example 13.6: (Cont., method of undetermined coefficients)
0 ),4.674cos(265)(5)(
4.67262410
66252
52527
)0(
25610
25
)256)(2(71615
)(
10515 ,2562
515)(
25625
)256)(2(71615
)(
0322
0
22
2
2
2
3
3
22
2
≥++=+=
∠=+=−
+=
−=⇒+=⋅−
=
+++
++
=+++
−−=
=⇒+=∞→++
++
+=
++
=⇒
+++
++
=+++
−−=
−−− tteetgetf
jCB
jBK
CC
F
ssCs
ssssss
sF
BBsss
CsBss
sss
ssF
ssCBs
ssssss
sF
ttt
βα
LL
)cos()( φβα += − teKtg tm
-
Case 3: Repeated Poles
[ ]i
i
ssii
ssii
n
j j
jiiiii
ii
ii
i
i
i
i
n
n
sFssdsdA
sFssA
ssA
sssssFss AA
tsteAtseAtg
ssA
ssAsG
ssA
ssAsGsF
=
=
=
−=
−=⇒
−−−=− ∑++
+=
−+
−=
−++
−+=
)()(
)()(
)()()()(
21
22
3
221
2
21
221
3
3
)(
poles) (double )(
)(
)()( L
-
Case 3: Repeated Poles (Cont.)
[ ][ ]
[ ]i
i
i
ssii
ssii
ssii
ii
ii
ii
i
i
i
i
i
i
sFssdsd
A
sFssdsd
A
sFssA
tsetAtsteAtseAtg
ssA
ssA
ssAsG
=
=
=
−=
−=
−=⇒
++=
−+
−+
−=
)()(!2
1
)()(
)()(21
32
2
1
32
33
2321
33
221
)(
)()()(
-
Example 13.7: Inversion with a Triple Pole
0 ,26
)(
)4(6
)4(0
41
051
646
1641
32014
)0(
10)(lim
)4(6
)4(4)5()4(
142
6)()4( ,1)()5(
)4()4(4)5()4(
142
5424
32
1212
11
321211
3
2
43
1354
43
132
12113
2
51)(
51)(
5)(
≥++−=
+++
=⇒+++−==
−=⇒=
+++++
+−−
=+==+=
+++++
+−−
−−−
∞→
−=−=
++++−=
++++==
++++==
teetetf
sss
AA
F
AssF
ssA
sA
ss
ss
sFsAsFsA
As
As
AsA
ss
ss
ttt
s
ss
ssF
ssF
ssF
-
Time DelayInitial-Value TheoremFinal-Value Theorem
-
Time Delay
[ ] [ ]
0 ),()()()(
)()()(
)()()(
)()()(
0021
2121
00
00
0
≥−−+=
+=+
=
=−−
−−
−
tttuttftftf
esFsFsD
esNsNsF
etgLttuttgL
stst
st
-
Example 13.8: Inversion with Time Delay
[ ] 0 ),3(8844)3()3(2)()(44)(
544
)5(20
)(
)(2)()5(
4020)(
4020)(5)(
0)0( ),3(40)(20)()(5)(
)3(40)(20)( :excitation
)3(5511
51
1
311
3
3
≥−−−−=−−−=
−=
−−+=
−−=
−=−
+−=
+−=−
=−+−=−=−′
−−=
−
−−
−
−
ttueetutftfty
etf
sssssF
esFsFss
esY
se
sYssY
ytututxtyty
tututx
tt
t
st
s
-
Initial-Value Theorem
( )properstrictly is )()(lim)0(
sF
ssFfs ∞→
+ =
-
Initial-Value Theorem (Cont.)
[ ]( )0 ),0()()0()( i.e.,
)()0()0()()(
>−=−
−−=+−
−+
tftfftf
tufftftf
c
c
-
Initial-Value Theorem (Cont.)
[ ][ ]
( )[ ]
0)0()(lim
0)0()0()0()(
00over finite is )(
)0()(lim0)(lim
)()0()()(
/)0()0()()(
0
0
=−
=−−−⇒
≤≤′
−==′
′=−=′
−−=
+
∞→
−−+
+−
−
∞→
∞ −
∞→
∞ −−
−+
∫∫
−
−
fssF
fffssF
ttf
fssFdtetf
dtetffssFtfL
sffsFsF
s
c
ccs
stcs
stcccc
c
-
Initial Slope
[ ]
)()()0()(
lim)0(
)0()()(
)0()()(
known is)0( assuming )0( slope initial Find
sDsDfssN
sf
fsDssN
fssFtfL
ff
s
−
∞→
+
−−
−+
−=′
−=−=′
′
-
Final-Value Theorem
=∞→
axisimaginary on the poles no origin, at the poles multiple no
RHP, in the poles no
)(lim)(0
ssFfs
[ ]
)0()()()0()(lim
)()0()()(
00
0
−∞−
→
∞ −−
−∞=′=−
′=−=′
∫∫
−
−
ffdttffssF
dtetffssFtfL
s
st
-
Example 13.9: Calculating Initial and Final Values
5)0( =−f
proper)(strictly )8009018(
16005)()(
)( 233
+++−
==ssss
ssDsN
sF
9090
lim)0(
proper)(strictly )8009018(
560045090)(
)()0()( :)0(
)0(5)(lim)0(
4
3
23
23
−=+
−−=′
+++−−−
=−′
===
∞→
+
−+
−
∞→
+
LL
ss
sf
sssssss
sDsDfssN
f
fssFf
s
s
-
Example 13.9: (Cont.)
5)0( =−f
proper)(strictly )8009018(
16005)()(
)( 233
+++−
==ssss
ssDsN
sF
2)(lim)(locations polecheck values,finalFor
0−==∞
→ssFf
s
-
Transform circuit analysis
-
Transform Circuit Analysis
l Given a circuit with some initial state at t=0-and an excitation x(t) starting at t=0, find the resulting behavior of any voltage or current y(t) for t=0.
l Zero-state response, natural response, forced response, zero-input response and complete response.
-
Zero-State Responsel Zero-state response: a circuit with no stored energy at t=0-.
( )functionnetwork domain - theas same
)(
)state" initial zero(" )(
)]([)()],([)(
network,order th -nan For
011
1
011
1
0101
s
asasasabsbsbsb
XY
sH
sYsdt
ydL
tyLsYtyLsY
xbdtdx
bdt
xdbya
dtdy
adt
yda
nn
nn
mm
mm
kk
k
m
m
mn
n
n
++++++++
=≡
⇓
=
==⇓
+++=+++
−−
−−
LL
LL
-
Zero-State Response
[ ]
polynomial sticcharacteri :)()()(
)( ,)()(
)(
)()( (3).
)()()( ),()( (2).)()(
)(obtain diagram,domain - Draw (1).
:Steps
1
sPsDsN
sXsPsN
sH
sYLty
sXsHsYsXtxsXsY
sHs
X
XH ==
=
=→
=
−
-
Step Response (zero initial state, by definition)
0,1)()( >== ttutx
ssX /1)( =
)()(1
)()(ssPsN
ssHsY H=×=
A pole at the origin
-
Example 13.10: Step Response
Z(s)
ssVsssZ
sH
sHsVsI
s
s
1)( ,)16/2()(
)(1
)()()(
=+==
=
-
Example 13.10: (Cont.)
5.0)(lim)(
0)(lim)0(
0 ,5.05.0)(
)4(1
45.05.0
)4(81
)()(
0
44
22
==∞
==
≥−−=
+−
+−=
++
=×=
→
∞→
+
−−
ssIi
ssIi
tteeti
ssssss
ssHsI
s
s
tt
Steady state response natural response
-
Zero-state AC response
)cos()( xm tXtx φβ +=
22)( β+= ssDX
ymymF
FN
FNFN
XH
YXjHYtYtytytyty
ssN
sPsN
sYsYssP
sNsNsY
φβφβ
ββ
∠==+=+=
++=+≡
+=
)( where)cos()()()()(
)()()(
)()())(()()(
)( 2222
Natural responsefrom Y(s)
Forced responseFrom phasor analysis
-
Example 13.11: Zero-State AC Response (from Fig. 13.9)
ttN
FNF
s
ss
teeti
sIsIs
sNss
ssss
sVsHsI
ss
sVtttv
44
22
22
2
2
10)(
)()(64
)()4(
104
1)64()4(
40050)()()(
6450
)(0 ,8cos50)(
−− −−=
+≡+
++
−+−
=
+++
==
+=⇒≥=
Phasor analysis
-
Example 13.11: (Cont.)
0)9.81cos(07.71)0(
0 ),9.818cos(07.7)101()()()(
)9.818cos(07.7)(
9.8107.7)8(
9.81141.0)8( ,050
0
04
0
0
00
=−+−=
≥−++−=+=
−=
−∠=⋅=
−∠=∠=
+
−
i
ttettititi
tti
VjHI
jHV
tFN
F
s
s
-
Natural Response and Forced Response
response) forced thefind toanalysisphasor use(or 0 ,5105)(
35
210
15
)3)(2)(1(10
)()()(
)2,1( 3 :(1) case10)(
)( ,)2)(1(
1)(Let
32
000
221
0
≥+−=
++
+−
++
=+++
==
−≠−≠−==
+=++
=
−−−
−
−−
teeety
sssssssXsHsY
sssetx
eAeAtyss
sH
ttt
ts
ttN
-
Natural Response and Forced Response(Cont.)
0 ,101010)(
)2(10
210
110
)2)(1(10
)(
analysis transformuse
frequency natural a as same theisfrequency excitation when applicablenot is analysisphasor
)2(
2 :(2) case
22
22
0
≥−−=
+−
++
−+
+=
++=
⇒
⇒∞→−
−=
−−− tteeety
ssssssY
H
s
ttt
natural mixed forced
-
Zero-Input Response
l The excitation equals zero for t=0 but the circuit contains stored energy at t=0-.
l Thévenin/Norton equivalent circuits can be established.
-
Zero-Input Response (Cont.)
[ ]
−=
+=⇒
−=
′=
−
−
−
)0()()( :Norton
)(1)0(
)( :Thevenin
)0()()(
)()(
CCC
CC
C
CCC
CC
CvssCVsI
sIsCs
vsV
vssVCsI
tvCti
-
Zero-Input Response (Cont.)
[ ]
+=
−=⇒
−=
′=
−
−
−
)(1)0(
)( :Norton
)0()()( :Thevenin
)0()()(
)()(
sVsLs
isI
LissLIsV
issILsV
tiLtv
LL
L
LLL
LLL
LL
-
Example 13.12: Calculating a Zero-Input Response
VvAi
tAti
CL
s
120206)0( ,6)0(
:analysis statesteady DC0 ,6)(
=⋅==
-
Example 13.12: (Cont.)
0 ,)455cos(26
)cos()(
452666 ,5 ,5
25010606
)(
12012)(
100202
:responseinput -zero
05
0
20
22
≥−=
+=
−∠=−===⇒
+++
=++
+=
+=
++
−
−
tAte
teKti
jK
ssCBs
sss
sI
ssI
ss
t
tmL
L
L
φβ
βα
ωα
α
-
Complete Response
l Complete response: complete response=zero-input response + zero-state response (with fictitious sources)
-
Example 13.13: Calculating a Complete Response
ssVVvAi
t
tvtV
tVtv
SCL
Cs
20)( ,20)0( ,4)0(
0for analysis statesteady DC
)( find 0 ,20
0 ,20)(
−===
<
≥−<
=
−−
-
Example 13.13: (Cont.)
0 ,)6.268cos(52020)(808
)(
5.05.0
/202)(
5.01
51
40
04
21
≥−+−=++
++=
+−
=
++
− tVtetvss
CBssA
sV
ss
sVs
s
tC
C
C
Overshoot due to underdamped
-
Chapter 13: Problem Set
l 6, 10, 16, 21, 25, 28, 31, 35, 38, 42, 46, 51, 57, 58, 63, 65