MAE40 Linear Circuits 168
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?
1. Easier than ODEs2. Easier to perform engineering design
3. Frequency response ideas - filtering
MAE40 Linear Circuits 169
Element Transformations
Voltage sourceTime domain
v(t) =vS(t)i(t) = depends on cct
Transform domainV(s) =VS(s)=L(vS(t))I(s) = L(i(t)) depends on cct
Current sourceI(s) =L(iS(t))V(s) = L(v(t)) depends on cct
+_
i(t)
vS
iS
v(t)
MAE40 Linear Circuits 170
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 and linearity
)()()()()()()()()()()()()()()()(
2121
2121
2121
2121
sgVsItgvtisrIsVtritvsIsItitisVsVtvtv
=⇔=
=⇔=
=⇔=
=⇔=
ββ
µµ
)()()()(0)(0)(0)(0)(
sVsVtvtvsItisVtv
PNPN
OCOC
SCSC
=⇔=
=⇔=
=⇔=
MAE40 Linear Circuits 171
Element Transformations contd
Resistors
Capacitors
Inductors
)()()()()()()()(sGVsIsRIsVtGvtitRitv
RRRR
RRRR==
==
iR
vR
sisV
sLsILissLIsV
idvL
tidttdiLtv
LLLLLL
tLLL
LL
)0()(1)()0()()(
)0()(1)()()(0
+=−=
+== ∫ ττ
RsYRsZ RR
1)()( ==
sCsYsC
sZ CC == )(1)(
sLsYsLsZ LL
1)()( ==
vC
iC
vL
iL+ + +
svsI
sCsVCvssCVsI
vdiC
tvdttdvCti
CCCCCC
CtCC
CC
)0()(1)()0()()(
)0()(1)()()(0
+=−−=
+== ∫ ττ
MAE40 Linear Circuits 172
Element Transformations contdResistor
Capacitor
Note the source transformation rules apply!
IR(s)
VR(s)
+
-
vR(t)
+
-
vC(t)
+
-
iR(t)
iC(t)
R R
CsC1
)0(CCvVC(s)
+
-
IC(s)
VC(s)
+
-
IC(s)
+_s
vC )0(
sC1
)()( sRIsV RR =
svsI
sCsVCvssCVsI C
CCCCC)0()(1)()0()()( +=--=
MAE40 Linear Circuits 173
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)+
-
sL siL )0(
MAE40 Linear Circuits 174
Example 10-1, T&R, 5th ed, 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
sC1
)0(CCv
I2(s)
I1(s)
vC+
-
)(1)()()()( 21 ssCV
sC
sVsIRsVsI C
CC ===
AC Vv =)0(
)()(1)( tueVtv
RCs
VsV RCt
AcA
C−
=+
=
MAE40 Linear Circuits 175
Example 10-2 T&R, 5th ed, p 457
Solve for i(t)
KVL around loop
Solve
Invert
+_
R
VAu(t) Li(t) +_
+_
R
sLI(s)sVA
LiL(0)
)(sVL
+
-
0)0()()( =++− LA LisIsLRsV
€
i(t) =VAR−VARe−RtL + iL (0)e
−Rt L#
$ %
&
' ( u(t) Amps
€
I(s) =VA
Ls s+ RL( )
+iL (0)s+ RL
=VA
Rs
+iL (0) −
VAR
# $ % &
' (
s+ RL
MAE40 Linear Circuits 176
Impedance and Admittance
Impedance (Z) is the s-domain proportionality factor relating the transform of the voltage across a two-terminal element to the transform of the current through the element with all initial conditions zero
Admittance (Y) is the s-domain proportionality factor relating the transform of the current through a two-terminal element to the transform of the voltage across the element with initial conditions zero
Impedance is like resistanceAdmittance is like conductance
€
V (s) = Z(s)I(s)
€
I(s) = Y (s)V (s)
MAE40 Linear Circuits 177
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=+=
MAE40 Linear Circuits 178
Example 10-3 T&R, 5th ed, 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)+
-sC1
W+
++=
++=+=
1111)(
2
RCsRLsRLCs
sCR
sLsC
RsLsZeq
)()()()()( 121
12 sV
RsLRLCsRsV
sZsZsV
eq!"
#$%
&
++=
!!"
#
$$%
&=
MAE40 Linear Circuits 179
Example
Formulate node voltage equations in the 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
1sC
C1vC1(0)2
1sC
A B C D
MAE40 Linear Circuits 180
Example contd
Node A: Node D:
Node B:
Node C:
+_
R1
V1(s) +-
R2
C2vC2(0)
R3 Vx(s)+
-µVx(s) V2(s)
+
-1
1sC
C1vC1(0)2
1sC
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)
MAE40 Linear Circuits 181
Example
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)+
sC1
CvC(0)
VA(s) VB(s) VC(s) VD(s)
MAE40 Linear Circuits 182
Example
Nodal AnalysisNode A:Node D:Node C: Node B:Node C KCL:Solve for VO(s)
Invert LT
+_
R1
VS(s) +-
R2
VO(s)+
sC1
CvC(0)
VA(s) VB(s) VC(s) VD(s)
)()( sVsV SA =
)()( sVsV OD =
)0()()()( 11 CSB CvsVGsVsCG =−+
0)( =sVC
)0()()( 2 COB CvsVGssCV −=−−
€
VO(s) = −
sG1CG2
G1 + sC
#
$
% % %
&
'
( ( ( VS (s) = −
R2R1×
ss+ 1
R1C
#
$
% %
&
'
( ( VS (s)
= −R2R1×
ss+ 1
R1C
#
$
% %
&
'
( ( 1s
=−R2R1
×1
s+ 1R1C
)()( 112 tue
RRtv CR
t
O
--=
MAE40 Linear Circuits 183
Superposition in s-domain ccts
The s-domain response of a cct can be found as the sum 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 cct2. Forced response – the particular solution
Response containing modes due to the input
MAE40 Linear Circuits 184
Example 10-6, T&R, 5th ed, p 466
The switch has been open for a long time and is closed at t=0.
Find the zero-state and zero-input components of V(s)Find v(t) for IA=1mA, L=2H, R=1.5KW, C=1/6 µF
IA
v(t)
t=0
R CL
+
-
V(s)RsL
+
-sI A
sC1
RCIA
RLsRLCsRLs
sCRsL
sZeq++
=++
=211
1)(
LCs
RCs
sRIRCIsZsV
LCs
RCs
CI
sIsZsV
AAeqzi
AA
eqzs
11)()(
11)()(
2
2
++==
++==
MAE40 Linear Circuits 185
Example 10-6 contd
Substitute values
LCs
RCs
sRIRCIsZsV
LCs
RCs
CI
sIsZsV
AAeqzi
AA
eqzs
11)()(
11)()(
2
2
++==
++==
€
Vzs(s) =6000
(s+1000)(s+ 3000)=
3s+1000
+−3
s+ 3000vzs(t) = 3e−1000t − 3e−3000t[ ]u(t)
€
Vzi (s) =1.5s
(s+1000)(s+ 3000)=
−0.75s+1000
+2.25
s+ 3000vzi (t) = −0.75e−1000t + 2.25e−3000t[ ]u(t)
V(s)RsL
+
-sI A
sC1
RCIA
What are the natural and forced responses?
MAE40 Linear Circuits 186
Features of s-domain cct analysis
The response transform of a finite-dimensional, lumped-parameter linear cct with input being a sum of exponentials is a rational function and its inverse Laplace Transform is a sum of exponentials
The exponential modes are given by the poles of the response transform
Because the response is real, the poles are either real or occur in complex conjugate pairs
The natural modes are the zeros of the cct determinant and lead to the natural response
The forced poles are the poles of the input transform and lead to the forced response
MAE40 Linear Circuits 187
Features of s-domain cct analysis
A cct is if all of its poles are located in the open 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 worst marginally 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 “Positive Real” or energy dissipating