S-Domain Analysis
s-Domain Circuit Analysis
Time domain
(t domain)
Linear
Circuit
Differential
equation
Complex frequency
domain (s domain)
Laplace Transform
L
Laplace Transform
L
Classical
techniques
Response
waveform
Transformed
Circuit
Algebraic
equation
Algebraic
techniques
Inverse Transform
L-1
Response
transform
Kirchhoff’s Laws in s-Domain
t domain
s domain
i2 (t )
Kirchhoff’s current law (KCL)
i1 (t )
i3 (t )
i4 (t )
i1 (t ) + i2 (t ) − i3 (t ) + i4 (t ) = 0
I1 ( s ) + I 2 ( s ) − I 3 ( s ) + I 4 ( s ) = 0
+ v2 (t ) −
Kirchhoff’s voltage law (KVL)
− v1 (t ) + v2 (t ) + v3 (t ) = 0
+ v4 (t ) −
+
+
+
v1 (t )
v3 (t )
v5 (t )
−
−
−
− V1 ( s ) + V2 ( s ) + V3 ( s ) = 0
Signal Sources in s Domain
t domain
s domain
i (t )
+
v(t )
v(t ) = vS (t )
i (t ) = depends _
on circuit
vS (t )
L
V (s )
V ( s ) = VS ( s )
I ( s ) = depends
on circuit
I (s )
_
+
_
i (t ) = iS (t )
v(t )
v(t ) = depends +
on circuit
VS (s )
_
i (t )
Current Source:
Voltage Source:
+
_
+
Voltage Source:
I (s )
_
iS (t )
L
V (s )
+
Current Source:
I S (s)
I ( s ) = VS ( s )
V ( s ) = depends
on circuit
Time and s-Domain Element Models
Impedance and Voltage Source for Initial Conditions
Resistor:
vR (t ) = RiR (t )
Inductor:
diL (t )
vL (t ) = L
dt
Capacitor:
iR (t )
s-Domain
I R (s )
+
+
vR (t )
R
L
_
VR (s )
R
_
iL (t )
I L (s )
+
vL (t )
_
L
L
iC (t )
+
1 t
vC (t ) = ∫ iC (τ )dτ
vC (t )
C 0
_
+ vC (0)
Ls
VL (s )
_
_
+
I C (s )
L
VC (s )
_
VR ( s ) = RI R ( s )
LiL (0)
VL ( s ) = LsI L ( s ) −
LiL (0)
Capacitor:
+
C
Resistor:
Inductor:
+
1
_
+
Time Domain
1 Cs VC ( s ) =
I C ( s) +
Cs
vC (0)
vC( 0 )
s
s
Impedance and Voltage Source for Initial
Conditions
• Impedance Z(s)
Z ( s) =
voltage transform
current transform
with all initial conditions set to zero
• Impedance of the three passive elements
Z R ( s) =
VR ( s )
=R
I R (s)
VL ( s )
= Ls
I L (s)
with iL (0) = 0
VC ( s ) 1
=
I C ( s ) Cs
with vC ( 0 ) = 0
Z L ( s) =
Z C (s) =
Time and s-Domain Element Models
Admittance and Current Source for Initial Conditions
Time Domain
Resistor:
1
iR (t ) = vR (t )
R
Inductor:
iR (t )
+
+
vR (t )
iC (t ) = C
dvC (t )
dt
R
L
_
VR (s )
Resistor:
R
iL (t )
I L (s )
+
L
VL (s )
Ls
_
iC (t )
I C (s )
+
vC (t )
_
Inductor:
+
L
+
C
L
VC (s )
_ 1 Cs
1
VR ( s )
R
I R (s) =
_
1 t
iL (t ) = ∫ vL (τ ) dτ
vL (t )
L 0
_
+ iL (0)
Capacitor:
s-Domain
I R (s )
I L (s) =
i L ( 0)
s
1
VL ( s ) +
Ls
i L( 0 )
s
Capacitor:
I C ( s ) = CsVC ( s ) −
CvC (0)
CvC (0)
Admittance and Current Source for Initial
Conditions
• Admittance Y(s)
Y ( s) =
current transform
1
=
voltage transform Z ( s)
with all initial conditions set to zero
• Admittance of the three passive elements
YR ( s ) =
I R ( s) 1
=
VR ( s ) R
YL ( s ) =
I L ( s) 1
=
VL ( s ) Ls
with iL (0) = 0
YC ( s ) =
I C ( s)
= Cs
VC ( s )
with vC( 0 ) = 0
Example: Solve for Current Waveform i(t)
R
L
i (t )
L
VA
s
+ VR (s ) −
_
+
_
+
V Au (t )
R
I (s )
Ls
_
+
+
VL (s )
LiL (0) _
VA
By KVL: −
+ VR ( s ) + VL ( s ) = 0
s
Resistor: VR ( s ) = RI ( s )
Inductor: VL ( s ) = LsI ( s ) − LiL (0)
V
− A + RI ( s) + LsI ( s) − LiL (0) = 0
s
VA L
iL (0)
+
I (s) =
s ( s + R L) s + R L
VA R VA R
i L ( 0)
=
−
+
s
s+R L s+R L
R
− t
V A V A − RL t
Inverse Transform: i (t ) =  − e + iL (0)e L u (t )

R R
forced response
natural response
Series Equivalence and Voltage Division
I1 ( s )
I (s )
Rest
of
Circuit
+ V1 ( s ) −
Z1
+
V (s)
−
V1 ( s ) = Z1 ( s ) I1 ( s ) = Z1 ( s ) I ( s )
+
Z2 V2 ( s) I 2 ( s )
−
V2 ( s ) = Z 2 ( s ) I 2 ( s ) = Z 2 ( s ) I ( s )
KVL: V ( s ) = V1 ( s ) + V2 ( s )
= ( Z1 ( s ) + Z 2 ( s )) I ( s )
I (s )
Rest
of
Circuit
+
V (s) Z
EQ
−
Z EQ = Z1 + Z 2
Z EQ ( s ) = Z1 ( s ) + Z 2 ( s )
V1 ( s) =
Z1 ( s )
V (s)
Z EQ ( s )
V2 ( s ) =
Z 2 (s)
V (s)
Z EQ ( s)
Parallel Equivalence and Current Division
I1 ( s ) = Y1 ( s )V ( s )
I (s )
Rest
of
Circuit
+
I1 ( s )
I 2 (s)
V (s) Y
Y2
1
−
I 2 ( s ) = Y2 ( s )V ( s )
KCL: I ( s ) = I1 ( s ) + I 2 ( s )
= (Y1 ( s ) + Y2 ( s ))V ( s )
I (s )
Rest
of
Circuit
+
V (s) Y
EQ
−
YEQ = Y1 + Y2
YEQ ( s ) = Y1 ( s ) + Y2 ( s )
I1 ( s ) =
Y1 ( s )
I ( s)
YEQ ( s )
I 2 (s) =
Y2 ( s)
I (s)
YEQ ( s )
Example:
Equivalence Impedance and Admittance
A
L
v1 (t )
R
_
+
B
L
A
V1 ( s )
B
Z EQ ( s ) = Ls + Z EQ1 ( s) = Ls +
Ls
Z EQ ZR
EQ1
Z EQ
Inductor current = 0
at t = 0
+
capacitor voltage = 0
C v2 (t ) Find equivalent impedance at A and B
_
Solve for v2(t)
RCs + 1
1
1
= + Cs =
YEQ1 ( s ) =
Z EQ1 ( s ) R
R
Z EQ1
RLCs 2 + Ls + R
+
=
RCs + 1
1
V2 ( s )
Z EQ1 ( s )
Cs
_
V2 ( s ) =
V1 ( s )
Z EQ
R
=
V1 ( s)
2
RCLs + Ls + R
R
RCs + 1
_
+
General Techniques for s-Domain Circuit
Analysis
• Node Voltage Analysis (in s-domain)
–
–
–
–
Use Kirchhoff’s Current Law (KCL)
Get equations of node voltages
Use current sources for initial conditions
Voltage source
current source
• Mesh Current Analysis (in s-domain)
–
–
–
–
Use Kirchhoff’s Voltage Law (KVL)
Get equations of currents in the mesh
Use voltage sources for initial conditions
Current source
voltage source
(Works only for “Planar” circuits)
Formulating Node-Voltage Equations
Step 0: Transform the circuit into the s domain using current
sources to represent capacitor and inductor initial conditions
Step 1: Select a reference node. Identify a node voltage at each
of the non-reference nodes and a current with every element
in the circuit
Step 2: Write KCL connection constraints in terms of the
element currents at the non-reference nodes
Step 3: Use the element admittances and the fundamental
property of node voltages to express the element currents in
terms of the node voltages
Step 4: Substitute the device constraints from Step 3 into the
KCL connection constraints from Step 2 and arrange the
resulting equations in a standard form
Example: Formulating Node-Voltage Equations
L
iS (t )
R
C
Step 0: Transform the circuit into the s domain using
current sources to represent capacitor and
inductor initial conditions
Step 1: Identify N-1=2 node voltages and a current
with each element
t domain
VA (s )
I 2 ( s)
I S (s)
Reference
node
Step 2: Apply KCL at nodes A and B:
iL (0)
−
− I1 ( s ) − I 2 ( s ) = 0
I
s
Node
A
:
(
)
L
S
s
iL (0)
i (0)
+ I1 ( s ) − I 3 ( s ) = 0
Node B : CvC (0) + L
s
s
Step 3: Express element equations in terms of node
Ls
VB (s )
voltages
1
I1 ( s ) I 3 ( s )
I1 ( s ) = YL ( s )[VA ( s ) − VB ( s )] = [VA ( s ) − VB ( s )]
Ls
1
R
Cs CvC (0) I 2 ( s ) = YR ( s)VA ( s ) = GVA ( s) where G = 1 R
s domain
I 3 ( s ) = YC ( s )VB ( s) = CsVB ( s )
Formulating Node-Voltage Equations (Cont’d)
Step 2: Apply KCL at nodes A and B:
iL (0)
− I1 ( s ) − I 2 ( s ) = 0
Node A : I S ( s ) −
s
i (0)
+ I1 ( s ) − I 3 ( s ) = 0
Node B : CvC (0) + L
s
Step 3: Express element equations in terms of node voltages
1
I1 ( s ) = YL ( s)[VA ( s) − VB ( s)] = [VA ( s ) − VB ( s )]
Ls
I 2 ( s ) = YR ( s )VA ( s) = GVA ( s ) where G = 1 R
I 3 ( s ) = YC ( s )VB ( s ) = CsVB ( s)
Step 4: Substitute eqns. in Step 3 into eqns. in Step 2 and collect
common terms to yield node-voltage eqns.
i (0)
1 
 1 

Node A :  G + VA ( s ) −  VB ( s ) = I S ( s ) − L
s
Ls 
 Ls 

i (0)

 1
 1 
Node B : −  VA ( s ) +  + Cs VB ( s) = CvC (0) + L
s

 Ls
 Ls 
Solving s-Domain Circuit Equations
G + 1 Ls
− 1 Ls
• Circuit Determinant: ∆( s ) =
− 1 Ls Cs + 1 Ls
= (G + 1 Ls )(Cs + 1 Ls ) − (1 Ls ) 2
GLCs 2 + Cs + G
=
Ls
Depends on circuit element parameters: L, C, G=1/R,
not on driving force and initial conditions
• Solve for node A using Cramer’s rule:
− 1 Ls
I S ( s ) + i L ( 0) s
∆ A ( s) iL (0) s + CvC (0) Cs + 1 Ls
=
VA ( s ) =
∆(s)
∆(s)
( LCs 2 + 1) I S ( s) − LCsiL (0) + CvC (0)
=
+
2
GLCs + Cs + G
GLCs 2 + Cs + G
Zero State
when initial condition
sources are turned off
Zero input
when input sources
are turned off
Solving s-Domain Circuit Eqns. (Cont’d)
• Solve for node B using Cramer’s rule:
G + 1 Ls
I S ( s ) − iL (0) s
− 1 Ls iL (0) s + CvC (0)
∆ B (s)
=
∆( s)
∆( s)
I S ( s)
GLiL (0) + (GLs + 1)CvC (0)
=
+
2
GLCs + Cs + G
GLCs 2 + Cs + G
VB ( s ) =
Zero State
Zero input
Network Functions
Network function =
Zero - state Response Transform
Input Signal Transform
• Driving-point function relates the voltage and
current at a given pair of terminals called a port
V (s)
1
=
Z ( s) =
I ( s) Y ( s)
• Transfer function relates an input and response at
different ports in the circuit
V (s)
TV ( s ) = Voltage Transfer Function = 2
V1 ( s )
I 2 (s)
TI ( s ) = Current Transfer Function =
V1
I1 ( s )
I ( s)
In
TY ( s) = Transfer Admittance = 2
V1 ( s)
V (s)
V1
TZ ( s) = Transfer Impedance = 2
I1 ( s )
_
+
_
+
In
I (s )
+
Circuit
in the
zero-state
V (s )
−
Circuit Output
in the
V1 or I1
V or I
zero-state 2 2
Input
I2
+
V2
−
TV (s )
Out
I1 TI (s )
In
Out
I2
TY (s )
+
V2
−
I1 TZ (s )
Out
In
Out
Calculating Network Functions
Z1
V1 ( s )
+
Z2 V2 ( s)
−
I 2 (s)
I1 ( s )
Y1
Y2
• Driving-point impedance
Z EQ ( s ) = Z1 ( s) + Z 2 ( s )
• Voltage transfer function:


Z 2 (s)
V2 ( s ) = 
V1 ( s )
 Z1 ( s ) + Z 2 ( s ) 
V (s)
Z 2 ( s)
TV ( s) = 2
=
V1 ( s ) Z1 ( s ) + Z 2 ( s )
• Driving-point admittance
YEQ ( s ) = Y1 ( s ) + Y2 ( s)
• Voltage transfer function:
 Y2 ( s )

I 2 (s) = 
Y1 ( s)
 Y1 ( s) + Y2 ( s ) 
I (s)
Y2 ( s )
TI ( s ) = 2
=
I1 ( s ) Y1 ( s ) + Y2 ( s)
_
+
Impulse Response and Step Response
• Input-output relationship in s-domain
Input
Y ( s) = T ( s) X ( s)
X (s )
• When input signal is an impulse x(t ) = δ (t )
Y ( s) = T ( s) ×1 = T ( s)
T(s)
Circuit
Output
Y (s )
– Impulse response equals network function
– H(s) = impulse response transform
– h(t) = impulse response waveform
• When input signal is a step x(t ) = u (t )
– G(s) = step response transform
– g(t) = step response waveform
T (s) H (s)
=
G (s) =
s
s
g ( s) = ∫ h(τ )dτ ,
t
0
(=) means equal almost everywhere,
dg (t )
excludes those points at which g(t)
h(t )(=)
dt
has a discontinuity