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MAE140 Linear Circuits132
s-Domain Circuit Analysis
Operate directly in the s-domain with capacitors,inductors and resistorsKey feature – linearity – is preservedCcts described by ODEs and their ICs
Order equals number of C plus number of L
Element-by-element and source transformationNodal or mesh analysis for s-domain cct variables
Solution via Inverse Laplace TransformWhy?
Easier than ODEsEasier to perform engineering designFrequency response ideas - filtering
MAE140 Linear Circuits133
Element Transformations
Voltage sourceTime domain
v(t) =vS(t)i(t) = depends on cct
Transform domain
V(s) =VS(s)=L(vS(t))I(s) = L(i(t)) depends on cct
Current source
I(s) =L(iS(t))V(s) = L(v(t)) depends on cct
+_
i(t)
vS
iS
v(t)
MAE140 Linear Circuits134
Element Transformations contd
Controlled sources
Short cct, open cct, OpAmp relations
Sources and active devices behave identicallyConstraints expressed between transformed variables
This all hinges on uniqueness of Laplace Transforms andlinearity
)()()()(
)()()()(
)()()()(
)()()()(
2121
2121
2121
2121
sgVsItgvti
srIsVtritv
sIsItiti
sVsVtvtv
=!=
=!=
=!=
=!=
""
µµ
)()()()(
0)(0)(
0)(0)(
sVsVtvtv
sIti
sVtv
PNPN
OCOC
SCSC
=!=
=!=
=!=
MAE140 Linear Circuits135
Element Transformations contd
Resistors
Capacitors
Inductors
)()()()(
)()()()(
sGVsIsRIsV
tGvtitRitv
RRRR
RRRR
==
==
iR
vR
s
isV
sLsILissLIsV
idvL
tidt
tdiLtv
LLLLLL
t
LLLL
L
)0()(
1)()0()()(
)0()(1
)()(
)(
0
+=!=
+== " ##
RsYRsZ RR
1)()( ==
sCsYsC
sZ CC == )(1)(
sLsYsLsZ LL
1)()( ==
vC
iC
vL
iL+ + +
s
vsI
sCsVCvssCVsI
vdiC
tvdt
tdvCti
CCCCCC
C
t
CCC
C
)0()(
1)()0()()(
)0()(1
)()(
)(
0
+=!!=
+== " ##
MAE140 Linear Circuits136
Element Transformations contdResistor
Capacitor
Note the source transformation rules apply!
IR(s)
VR(s)
+
-
vR(t)
+
-
vC(t)
+
-
iR(t)
iC(t)
R R
CsC
1
)0(CCvVC(s)
+
-
IC(s)
VC(s)
+
-
IC(s)
+_s
vC )0(
sC
1
)()( sRIsV RR =
svsI
sCsVCvssCVsI C
CCCCC)0()(1)()0()()( +=−−=
MAE140 Linear Circuits137
Element Transformations contd
Inductorss
isVsL
sILissLIsV LLLLLL
)0()(1)()0()()( +=−=
iL(t)+_
vL(t)
-
+sL
LiL(0)
IL(s)+
VL(s)
-
IL(s)
VL(s)+
-
sLs
iL )0(
MAE140 Linear Circuits138
Example 10-1 T&R p 456
RC cct behaviorSwitch in place since t=-∞, closed at t=0. Solve for vC(t).
Initial conditionss-domain solution using nodal analysis
t-domain solution via inverse Laplace transform
R
VA
t=0
C
R
sC
1)0(CCv
I2(s)
I1(s)
vC+
-
)(1
)()(
)()( 21 ssCV
sC
sVsI
R
sVsI C
CC ===
AC Vv =)0(
)()(1
)( tueVtv
RCs
VsV RC
t
AcA
C
!=
+
=
MAE140 Linear Circuits139
Example 10-2 T&R p 457
Solve for i(t)
KVL around loop
Solve
Invert
+_
R
VAu(t) Li(t) +_
+_
R
sLI(s)s
VA
LiL(0)
)(sVL
+
-
0)0()()( =++! LA
LisIsLRs
V
!
i(t) =VA
R"VA
Re"Rt
L + iL (0)e"Rt
L#
$ %
&
' ( u(t) Amps
!
I(s) =
VAL
s s+ RL( )
+iL (0)
s+ RL
=
VAR
s+
iL (0) "VA
R
# $ %
& ' (
s+ RL
MAE140 Linear Circuits140
Impedance and Admittance
Impedance is the s-domain proportionality factorrelating the transform of the voltage across a two-terminal element to the transform of the currentthrough the element with all initial conditions zero
Admittance is the s-domain proportionality factorrelating the transform of the current through atwo-terminal element to the transform of thevoltage across the element with initial conditionszero
Impedance is like resistanceAdmittance is like conductance
MAE140 Linear Circuits141
Circuit Analysis in s-Domain
Basic rulesThe equivalent impedance Zeq(s) of two impedances Z1(s)
and Z2(s) in series is
Same current flows
The equivalent admittance Yeq(s) of two admittances Y1(s)and Y2(s) in parallel is
Same voltage
)()()( 21 sZsZsZeq +=
I(s)
V(s) Z2
Z1+
-
( ) ( )sIsZsIsZsIsZsV eq )()()()()( 21 =+=
)()()( 21 sYsYsYeq +=
I(s)
V(s) Y1
+
-Y2)()()()()()()( 21 sVsYsVsYsVsYsI eq=+=
MAE140 Linear Circuits142
Example 10-3 T&R p 461
Find ZAB(s) and then find V2(s) by voltage division
+_
L
v1(t) RC
A
B
v2(t)+
-
+_
sL
V1(s) R
A
B
V2(s)+
-sC
1
Ω+
++=
++=+=
1111)(
2
RCsRLsRLCs
sCR
sLsC
RsLsZeq
)()()(
)()( 121
12 sV
RsLRLCs
RsV
sZ
sZsV
eq!"
#$%
&
++=
!!"
#
$$%
&=
MAE140 Linear Circuits143
Superposition in s-domain ccts
The s-domain response of a cct can be found as thesum of two responses1. The zero-input response caused by initial condition
sources with all external inputs turned off2. The zero-state response caused by the external sources
with initial condition sources set to zeroLinearity and superposition
Another subdivision of responses1. Natural response – the general solution
Response representing the natural modes (poles) of thecct
2. Forced response – the particular solutionResponse containing modes due to the input
MAE140 Linear Circuits144
Example 10-6 T&R p 466
The switch has been open for a long time and is closedat t=0.
Find the zero-state and zero-input components of V2(s)Find v(t) for IA=1mA, L=2H, R=1.5KΩ, C=1/6 µF
IA
v(t)
t=0
R CL
+
-
V(s)RsL
+
-s
I A
sC
1
RCIA
RLsRLCsRLs
sCRsL
sZeq ++=
++=
2111)(
LCs
RCs
sRIRCIsZsV
LCs
RCs
CI
s
IsZsV
AAeqzi
AA
eqzs
11)()(
11)()(
2
2
++
==
++
==
MAE140 Linear Circuits145
Example 10-6 contd
Substitute values
LCs
RCs
sRIRCIsZsV
LCs
RCs
CI
s
IsZsV
AAeqzi
AA
eqzs
11)()(
11)()(
2
2
++
==
++
==
!
Vzs(s) =6000
(s+1000)(s+ 3000)=
3
s+1000+
"3
s+ 3000
vzs(t) = 3e"1000t
" 3e"3000t[ ]u(t)
!
Vzi (s) =1.5s
(s+1000)(s+ 3000)=
"0.75
s+1000+
2.25
s+ 3000
vzi (t) = "0.75e"1000t + 2.25e"3000t[ ]u(t)
V(s)RsL
+
-s
I A
sC
1
RCIA
MAE140 Linear Circuits146
Example 10-11 T&R (not in 4th ed)
Formulate node voltage equations in s-domain
+_
R1
v1(t) +-
R2C2
C1 R3 vx(t)+
-µvx(t) v2(t)
+
-
+_
R1
V1(s) +-
R2
C2vC2(0)
R3 Vx(s)+
-µVx(s) V2(s)
+
-1
1
sC
C1vC1(0)2
1
sC
A B C D
MAE140 Linear Circuits147
Example 10-11 contd
Node A: Node D:
Node B:
Node C:
+_
R1
V1(s) +-
R2
C2vC2(0)
R3 Vx(s)+
-µVx(s) V2(s)
+
-1
1
sC
C1vC1(0)2
1
sC
A B C D
!
VB (s) "VA (s)
R1
+VB (s) "VD (s)
R2
+VB (s)
1sC1
+VB (s) "VC (s)
1sC2
"C1vC1(0) "C2vC2(0) = 0
)()( 1 sVsVA = )()()( sVsVsV CxD µµ ==
!
"sC2VB (s) + sC2 +G3[ ]VC (s) = "C2vC2(0)
MAE140 Linear Circuits148
Example 10-16 T&R (not in 4th ed)
Find vO(t) when vS(t) is a unit step u(t) and vC(0)=0
Convert to s-domain
+_R1
vS(t) +-
C R2 vO(t)+
A B C D
+_
R1
VS(s) +-
R2
VO(s)+
sC
1
CvC(0)
VA(s) VB(s) VC(s) VD(s)
MAE140 Linear Circuits149
Example 10-16 contd
Nodal AnalysisNode A:Node D:Node C:Node B:Node C KCL:Solve for VO(s)
Invert LT
+_
R1
VS(s) +-
R2
VO(s)+
sC
1
CvC(0)
VA(s) VB(s) VC(s) VD(s)
)()( sVsV SA =
)()( sVsV OD =
)0()()()( 11 CSB CvsVGsVsCG =!+
0)( =sVC
)0()()( 2 COB CvsVGssCV !=!!
CRsR
R
sCR
s
s
R
R
sV
CRs
s
R
RsV
sCG
G
CsG
sV SSO
12
1
12
1
12
1
1
21
1
11
1
)(1
)()(
+!
"=
##
$
%
&&
'
(
+!"=
##
$
%
&&
'
(
+!"=
###
$
%
&&&
'
(
+"=
)()( 121 tue
RRtv CR
t
O
−−=
MAE140 Linear Circuits150
Features of s-domain cct analysis
The response transform of a finite-dimensional,lumped-parameter linear cct with input being asum of exponentials is a rational function and itsinverse Laplace Transform is a sum of exponentials
The exponential modes are given by the poles of theresponse transform
Because the response is real, the poles are eitherreal or occur in complex conjugate pairs
The natural modes are the zeros of the cctdeterminant and lead to the natural response
The forced poles are the poles of the input transformand lead to the forced response
MAE140 Linear Circuits151
Features of s-domain cct analysis
A cct is stable if all of its poles are located in theopen left half of the complex s-planeA key property of a systemStability: the natural response dies away as t→∞
Bounded inputs yield bounded outputs
A cct composed of Rs, Cs and Ls will be at worstmarginally stableWith Rs in the right place it will be stable
Z(s) and Y(s) both have no poles in Re(s)>0
Impedances/admittances of RLC ccts are “PositiveReal” or energy dissipating
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