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8/18/2019 S domain analysis
1/20
MAE140 Linear Circuits132
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 transformationNodal or mesh analysis for s-domain cct variables
Solution via Inverse Laplace Transform
Why?
Easier than ODEsEasier to perform engineering design
Frequency response ideas - filtering
8/18/2019 S domain analysis
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MAE140 Linear Circuits133
Element Transformations
Voltage sourceTime domain
v(t) =vS (t)
i(t) = depends on cct
Transform domain
V(s) =V S (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)
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MAE140 Linear Circuits134
Element Transformations contd
Controlled sources
Short cct, open cct, OpAmp relations
Sources and active devices behave identically
Constraints expressed between transformed variables
This all hinges on uniqueness of Laplace Transforms andlinearity
)()()()(
)()()()(
)()()()(
)()()()(
2121
2121
2121
2121
s gV s I t gvt i
srI sV t rit v
s I s I t it i
sV sV t vt v
=!=
=!=
=!=
=!=
" "
µ µ
)()()()(
0)(0)(
0)(0)(
sV sV t vt v
s I t i
sV t v
P N P N
OC OC
SC SC
=!=
=!=
=!=
8/18/2019 S domain analysis
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MAE140 Linear Circuits135
Element Transformations contd
Resistors
Capacitors
Inductors
)()()()(
)()()()(
sGV s I s RI sV
t Gvt it Rit v
R R R R
R R R R
==
==
i R
v R
s
i sV
sL s I Li s sLI sV
id v L
t idt
t di Lt v
L L L L L L
t
L L L
L
L
)0()(
1)()0()()(
)0()(
1
)(
)(
)(0
+=!=
+==
" # #
R sY R s Z R R
1)()( ==
sC sY sC
s Z C C == )(
1)(
sL sY sL s Z L L
1)()( ==
vC
iC
v L
i L+ + +
s
v s I
sC sV Cv s sCV s I
vd iC
t vdt
t dvC t i
C C C C C C
C
t
C C
C
C
)0()(
1)()0()()(
)0()(
1
)(
)(
)(0
+=!!=
+==
" # #
8/18/2019 S domain analysis
5/20
MAE140 Linear Circuits136
Element Transformations contd
Resistor
Capacitor
Note the source transformation rules apply!
I R(s)
V R(s)
+
-
v R(t)
+
-
vC (t)
+
-
i R(t)
iC (t)
R R
C sC
1
)0(C CvV C (s)
+
-
I C (s)
V C (s)
+
-
I C (s)
+ _ s
vC )0(
sC
1
)()( s RI sV R R
=
s
v
s I sC sV Cv s sCV s I C
C C C C C
)0(
)(
1
)()0()()( +=!!
=
8/18/2019 S domain analysis
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MAE140 Linear Circuits137
Element Transformations contd
Inductors si sV sL s I Li s sLI sV L L L L L L )0()(1)()0()()( +=!=
i L(t)
+ _
v L(t)
-
+
sL
Li L(0)
I L(s)
+
V L(s)
-
I L(s)
V L(s)
+
-
sL s
i L )0(
8/18/2019 S domain analysis
7/20
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 conditions
s-domain solution using nodal analysis
t-domain solution via inverse Laplace transform
R
V A
t=0
C
R
sC
1)0(C Cv
I 2(s)
I 1(s)
vC +
-
)(1
)()(
)()( 21 s sCV
sC
sV s I
R
sV s I C
C C ===
AC V v =)0(
)()(1
)( t ueV t v
RC s
V sV RC
t
Ac A
C
!
=
+
=
8/18/2019 S domain analysis
8/20
MAE140 Linear Circuits139
Example 10-2 T&R p 457
Solve for i(t)
KVL around loop
Solve
Invert
+ _
R
V Au(t) Li(t) + _
+ _
R
sL I(s) s
V A
Li L(0)
)( sV L
+
-
0)0()()( =++! L A
Li s I sL R s
V
i(t ) =V A
R" V A
Re" Rt
L+ i L (0)e
" Rt L
#
$ %
&
' ( u(t ) Amps
I (s) =
V A L
s s + R
L( )
+
i L (0)
s+ R L
=
V A R
s+
i L (0) "V A
R
# $ %
& ' (
s+ R L
8/18/2019 S domain analysis
9/20
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 resistance
Admittance is like conductance
8/18/2019 S domain analysis
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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 s Z s Z s Z eq +=
I(s)
V(s) Z2
Z1+
-( ) ( ) s I s Z s I s Z s I s Z sV eq )()()()()( 21 =+=
)()()( 21 sY sY sY eq +=
I(s)
V(s) Y1
+
-
Y2)()()()()()()( 21 sV sY sV sY sV sY s I eq=+=
8/18/2019 S domain analysis
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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
V 1(s) R
A
B
V 2(s)
+
-
sC
1
#+
++=
+
+=+=
11
11)(
2
RCs
R Ls RLCs
sC R
sL sC
R sL s Z eq
)()()(
)()( 121
12 sV
R sL RLCs
R sV
s Z
s Z sV
eq!"
#$%
&
++
=
!!"
#
$$%
&=
8/18/2019 S domain analysis
12/20
MAE140 Linear Circuits143
Superposition in s-domain ccts
The s-domain response of a cct can be found as thesum of two responses
1. The zero-input response caused by initial conditionsources with all external inputs turned off
2. The zero-state response caused by the external sources
with initial condition sources set to zeroLinearity and superposition
Another subdivision of responses
1. Natural response – the general solution
Response representing the natural modes (poles) of the
cct2. Forced response – the particular solution
Response containing modes due to the input
8/18/2019 S domain analysis
13/20
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=1.5K#, C=1/6 µF
I A
v(t)
t = 0
R C L
+
-
V(s)
R sL
+
- s
I A
sC
1
RCI A
R Ls RLCs
RLs
sC R sL
s Z eq++
=
++
=211
1)(
LC s
RC s
s RI RCI s Z sV
LC s
RC s
C
I
s I s Z sV
A Aeq zi
A
Aeq zs
11)()(
11)()(
2
2
++
==
++
==
8/18/2019 S domain analysis
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MAE140 Linear Circuits145
Example 10-6 contd
Substitute values
LC s
RC s
s RI RCI s Z sV
LC s
RC s
C
I
s
I s Z sV
A Aeq zi
A
Aeq zs
11)()(
11)()(
2
2
++
==
++
==
V zs(s) =6000
(s+1000)(s+ 3000)=
3
s+1000+
"3
s+ 3000
v zs(t ) = 3e"1000t
" 3e"3000t [ ]u(t )
V zi (s) =1.5s
(s+1000)(s+ 3000)=
"0.75s+1000
+2.25
s+ 3000
v zi (t ) = "0.75e"1000t
+ 2.25e"3000t [ ]u(t )
V(s)
R sL
+
- s
I A
sC
1
RCI A
8/18/2019 S domain analysis
15/20
MAE140 Linear Circuits146
Example 10-11 T&R (not in 4th ed)
Formulate node voltage equations in s-domain
+ _
R1
v1(t) +
-
R2C 2
C 1 R3 v x(t)
+
-
µ v x(t) v2(t)
+
-
+ _
R1
V 1(s) +
-
R2
C 2vC2(0)
R3 V x(s)
+
-
µ V x(s) V 2(s)
+
-1
1
sC
C 1vC1(0)
2
1
sC
A B C D
8/18/2019 S domain analysis
16/20
MAE140 Linear Circuits147
Example 10-11 contd
Node A: Node D:
Node B:
Node C:
+ _
R1
V 1(s) +
-
R2
C 2vC2(0)
R3 V x(s)
+
-
µ V x(s) V 2(s)
+
-1
1
sC
C 1vC1(0)
2
1
sC
A B C D
V B (s)"V A (s)
R1
+
V B (s)"V D(s)
R2
+
V B (s)
1sC 1
+
V B (s) "V C (s)
1sC 2
"C 1vC 1(0)"C 2vC 2(0) = 0
)()( 1 sV sV A = )()()( sV sV sV C x D µ µ ==
"sC 2V B (s)+ sC 2 +G3[ ]V C (s) = "C 2vC 2(0)
8/18/2019 S domain analysis
17/20
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
+ _
R1vS (t) +
-C
R2 vO(t)+
A B C D
+ _
R1
V S (s) + -
R2
V O(s)
+
sC
1
CvC (0)
V A(s) V B(s) V C (s) V D(s)
8/18/2019 S domain analysis
18/20
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
V S (s) + -
R2
V O
(s)+
sC
1
CvC (0)
V A(s) V B(s) V C (s) V D(s)
)()( sV sV S A =
)()( sV sV O D =
)0()()()( 11 C S B Cv sV G sV sC G =!+
0)( = sV C
)0()()( 2 C O B Cv sV G s sCV !
=
!!
C R s R
R
sC R
s
s
R
R
sV
C R s
s
R
R sV
sC G
G
C sG
sV S S O
12
1
12
1
12
1
1
21
111
1
)(1
)()(
+
!"=##
$
%
&&
'
(
+
!"=
##
$
%
&&
'
(
+
!"=
###
$
%
&&&
'
(
+
"=
)()( 1
2
1 t ue R
Rt v
C R
t
O
!!
=
8/18/2019 S domain analysis
19/20
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 the
response transformBecause 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 responseThe forced poles are the poles of the input transform
and lead to the forced response
8/18/2019 S domain analysis
20/20
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-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 worstmarginally 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 “PositiveReal” or energy dissipating