s-Domain Circuit Analysis
Operate directly in the s-domain with capacitors,
inductors and resistors
Key feature – linearity – is preserved
Ccts described by ODEs and their ICs
Order equals number of C plus number of L
Element-by-element and source transformation
Nodal or mesh analysis for s-domain cct variables
Solution via Inverse Laplace Transform
Why?
Easier than ODEs
Easier to perform engineering design
Frequency response ideas - filtering
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132
Element Transformations
Voltage source
Time domain
v(t) =vS(t)
i(t) = depends on cct
i(t)
vS +
_
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
MAE140 Linear Circuits
iS
v(t)
133
Element Transformations contd
Controlled sources
v1(t ) = µv2 (t ) ! V1( s ) = µV2 ( s )
i1(t ) = "i2 (t ) !
I1( s ) = "I 2 ( s )
v1(t ) = ri2 (t ) ! V1( s ) = rI 2 ( s )
i1(t ) = gv2 (t ) !
I1( s ) = gV2 ( s )
Short cct, open cct, OpAmp relations
vSC (t ) = 0
! VSC ( s ) = 0
iOC (t ) = 0
!
I OC ( s ) = 0
v N (t ) = vP (t ) ! VN ( s ) = VP ( s )
Sources and active devices behave identically
Constraints expressed between transformed variables
This all hinges on uniqueness of Laplace Transforms and
linearity
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134
Element Transformations contd
iR
+
vR
Resistors
vR (t ) = RiR (t )
iR (t ) = GvR (t )
VR ( s ) = RI R ( s )
I R ( s ) = GVR ( s )
Capacitors
Z R ( s) = R
iC
YR ( s ) =
diL (t )
vL (t ) = L
dt
v ( 0)
1
IC (s) + C
sC
s
1
Z C (s) =
sC
1t
iL (t ) = " vL (# )d# + iL (0)
L0
VL ( s ) = sLI L ( s ) ! LiL (0)
I L (s) =
I C ( s ) = sCVC ( s ) ! CvC (0!)
Inductors
MAE140 Linear Circuits
+
vL
+
vC
1t
vC (t ) = " iC (# )d# + vC (0)
C0
dv (t )
iC (t ) = C C
dt
iL
1
R
VC ( s ) =
1
i ( 0)
VL ( s ) + L
sL
s
Z L ( s ) = sL
YC ( s ) = sC
YL ( s ) =
1
sL
135
Element Transformations contd
Resistor
IR(s)
iR(t)
+
+
vR(t)
R
VR(s)
-
Capacitor
+
-
-
IC(s)
IC(s)
iC(t)
vC(t)
R
+
C
VR ( s ) = RI R ( s )
VC(s)
-
+
1
sC
CvC (0)
1
sC
VC(s)
+
_
I C ( s ) = sCVC ( s ) − CvC (0−)
1
v (0)
VC ( s ) =
IC (s) + C
sC
s
-
vC (0)
s
Note the source transformation rules apply!
MAE140 Linear Circuits
136
Element Transformations contd
Inductors
VL ( s ) = sLI L ( s ) − LiL (0)
IL(s)
iL(t)
sL
VL(s)
-
+
VL(s)
sL
i L ( 0)
s
+
_
vL(t)
1
i (0)
VL ( s ) + L
sL
s
IL(s)
+
+
I L (s) =
LiL(0)
-
MAE140 Linear Circuits
137
Example 10-1 T&R p 456
RC cct behavior
Switch in place since t=-∞, closed at t=0. Solve for vC(t).
t=0
VA
R I1(s)
I2(s)
R
+
vC
-
1
sC
C
CvC (0)
vC (0) = V A
Initial conditions
s-domain solution using nodal analysis
V (s)
V (s)
I1( s ) = C
I 2 (s) = C
= sCVC ( s )
1
R
sC
t-domain solution via inverse Laplace transform
VA
VC ( s ) =
1
s+
RC
MAE140 Linear Circuits
vc (t ) = V Ae
!t
RC u (t )
138
Example 10-2 T&R p 457
R
Solve for i(t)
VA
s
R
i(t)
Invert
!
L + iL (0) =
s s+ RL s+ RL
(
LiL(0)
-
)
#
&
%iL (0) " V A (
R'
R +$
s
s+ RL
VA
#V A V A " Rt
" Rt L &
L
i(t) = %
"
e
+ iL (0)e
(u(t) Amps
R
R
$
'
MAE140 Linear Circuits
!
VL (s )
VA
! ( R + sL) I ( s ) + LiL (0) = 0
s
VA
I(s) =
sL
I(s)
L
KVL around loop
Solve
+
_
+
_
VAu(t) +
_
+
139
Impedance and Admittance
Impedance is the s-domain proportionality factor
relating the transform of the voltage across a twoterminal element to the transform of the current
through the element with all initial conditions zero
Admittance 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 resistance
Admittance is like conductance
MAE140 Linear Circuits
140
Circuit Analysis in s-Domain
Basic rules
The equivalent impedance Zeq(s) of two impedances Z1(s)
and Z2(s) in series is Z eq ( s ) = Z1( s ) + Z 2 ( s )
Same current flows
V ( s ) = Z1( s ) I (s )+ Z 2 ( s ) I ( s ) = Z eq ( s ) I (s )
I(s)
+
Z1
V(s)
Z2
-
The equivalent admittance Yeq(s) of two admittances Y1(s)
and Y2(s) in parallel is Yeq ( s ) = Y1( s ) + Y2 ( s )
Same voltage
I ( s ) = Y1( s )V ( s ) + Y2 ( s )V ( s ) = Yeq ( s )V ( s )
I(s)
+
V(s)
Y1
Y2
MAE140 Linear Circuits
141
Example 10-3 T&R p 461
Find ZAB(s) and then find V2(s) by voltage division
A
L
+
v1(t) +
_
C
R
v2(t)
A
V1(s) +
_
-
B
sL
1
sC
B
+
R
V2(s)
-
2+
+R
1
1
RLCs
Ls
= sL +
=
Ω
Z eq ( s ) = sL + R
+
1+
sC
RCs 1
sC
R
& Z1( s ) #
&
#
R
V2 ( s ) = $
V
(
s
)
=
V1( s )
! 1
$
!
2
% RLCs + sL + R "
$% Z eq ( s ) !"
MAE140 Linear Circuits
142
Superposition in s-domain ccts
The s-domain response of a cct can be found as the
sum of two responses
1. The zero-input response caused by initial condition
sources with all external inputs turned off
2. The zero-state response caused by the external sources
with initial condition sources set to zero
Linearity and superposition
Another subdivision of responses
1. Natural response – the general solution
Response representing the natural modes (poles) of the
cct
2. Forced response – the particular solution
Response containing modes due to the input
MAE140 Linear Circuits
143
Example 10-6 T&R p 466
t=0
The switch has been open for a long time and is closed
at 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
L
R
+
+
v(t)
V(s)
C
-
IA
s
sL
R
1
sC
RCIA
-
IA
IA
C
V
(
s
)
=
Z
(
s
)
=
1
RLs
zs
eq
=
Z eq ( s) =
s s2 + 1 s + 1
2
1 +1+
sC RLCs + Ls + R
RC
LC
sL R
RI A s
Vzi ( s ) = Z eq ( s ) RCI A =
1
1
s2 +
s+
RC
LC144
MAE140 Linear Circuits
Example 10-6 contd
IA
I
C
Vzs ( s ) = Z eq ( s ) A =
s s2 + 1 s + 1
IA
RC
LC
s
RI A s
Vzi ( s ) = Z eq ( s ) RCI A =
1
1
s2 +
s+
RC
LC
Substitute values
+
sL
1
sC
R
V(s)
RCIA
-
6000
3
"3
=
+
(s + 1000)(s + 3000) s + 1000 s + 3000
v zs (t) = 3e"1000t " 3e"3000t u(t)
Vzs (s) =
[
]
1.5s
"0.75
2.25
=
+
(s + 1000)(s + 3000) s + 1000 s + 3000
v zi (t) = "0.75e"1000t + 2.25e"3000t u(t)
Vzi (s) =
!
MAE140 Linear Circuits
!
[
]
145
Example 10-11 T&R (not in 4th ed)
Formulate node voltage equations in s-domain
v1(t) +
_
A
V1(s) +
_
R2
C2
R1
C1
R1
R3
R2
B
1
sC1
C1vC1(0)
+
vx(t)
-
R3
1
sC2
MAE140 Linear Circuits
C
+
Vx(s)
-
+
+ µvx(t) v (t)
2
-
D
+
+ µVx(s) V (s)
2
-
C2vC2(0)
146
Example 10-11 contd
A
V1(s) +
_
R1
B
1
sC1
C1vC1(0)
Node A: V A ( s) = V1( s)
Node B:
Node C:
R2
R3
1
sC2
C
+
Vx(s)
-
D
+
+ µVx(s) V (s)
2
-
C2vC2(0)
Node D: VD ( s) = µVx ( s) = µVC ( s)
VB (s) " V A (s) VB (s) " VD (s) VB (s)
+
+
1
R1
R2
sC1
V (s) " VC (s)
+ B
" C1vC1 (0) " C2vC 2 (0) = 0
1
sC2
"sC2VB (s) + [ sC2 + G3 ]VC (s) = "C2vC 2 (0)
!
MAE140 Linear Circuits
!
147
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
R2
C
+
-
vS(t) +
_
R1
D
C
Convert to s-domain
VA(s)
VS(s) +
_
MAE140 Linear Circuits
R1
+
vO(t)
1
VB(s) sC VC(s) R2 VD(s)
CvC(0)
+
-
B
A
+
VO(s)
148
Nodal Analysis
VA(s)
R1
VS(s) +
_
+
-
Example 10-16 contd
1
VB(s) sC VC(s) R2 VD(s)
+
VO(s)
CvC(0)
Node A: V A ( s ) = VS ( s )
Node D: VD ( s ) = VO ( s )
Node C: VC ( s ) = 0
Node B: (G1 + sC )VB ( s ) ! G1VS ( s ) = CvC (0)
Node C KCL: ! sCVB ( s ) ! G2VO ( s ) = !CvC (0)
( sG1C %
Solve for VO(s)
(
%
R
s
G
&
2 #V ( s ) = " & 1 !
#V ( s )
VO ( s ) = " &
S
& R2 s + 1
# S
G1 + sC #
R1C $
&'
#$
'
(
%
R
s
1
# 1 = " R1 !
= "& 1 !
& R2 s + 1
# s R2 s + 1
R1C $
R1C
'
Invert LT
MAE140 Linear Circuits
− t
− R1 R C
vO (t ) =
e 1 u (t )
R2
149
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
MAE140 Linear Circuits
150
Features of s-domain cct analysis
A cct is stable if all of its poles are located in the
open left half of the complex s-plane
A key property of a system
Stability: 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 stable
With 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
MAE140 Linear Circuits
151