Signals and Systems with MATLAB Applications, Second Edition 4-1 Orchard Publications Chapter 4 Circuit Analysis with Laplace Transforms his chapter presents applications of the Laplace transform. Several examples are given to illus- trate how the Laplace transformation is applied to circuit analysis. Complex impedance, com- plex admittance, and transfer functions are also defined. 4.1 Circuit Transformation from Time to Complex Frequency In this section we will derive the voltage-current relationships for the three basic passive circuit devices, i.e., resistors, inductors, and capacitors in the complex frequency domain. a. Resistor The time and complex frequency domains for purely resistive circuits are shown in Figure 4.1. Figure 4.1. Resistive circuit in time domain and complex frequency domain b. Inductor The time and complex frequency domains for purely inductive circuits is shown in Figure 4.2. Figure 4.2. Inductive circuit in time domain and complex frequency domain c. Capacitor The time and complex frequency domains for purely capacitive circuits is shown in Figure 4.3. T v R t Ri R t = i R t v R t R ----------- = R + Time Domain v R t i R t R + Complex Frequency Domain V R s I R s V R s RI R s = I R s V R s R ------------- = + Time Domain L v L t i L t v L t L di L dt ------- = i L t 1 L -- v L t d – t = + Complex Frequency Domain + sL Li L 0 V L s I L s V L s sLI L s Li L 0 – = I L s V L s Ls ------------- i L 0 s -------------- + =
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Signals and Systems with MATLAB Applications, Second Edition 4-1Orchard Publications
Chapter 4
Circuit Analysis with Laplace Transforms
his chapter presents applications of the Laplace transform. Several examples are given to illus-
trate how the Laplace transformation is applied to circuit analysis. Complex impedance, com-
plex admittance, and transfer functions are also defined.
4.1 Circuit Transformation from Time to Complex Frequency
In this section we will derive the voltage-current relationships for the three basic passive circuit
devices, i.e., resistors, inductors, and capacitors in the complex frequency domain.
a. Resistor
The time and complex frequency domains for purely resistive circuits are shown in Figure 4.1.
Figure 4.1. Resistive circuit in time domain and complex frequency domain
b. Inductor
The time and complex frequency domains for purely inductive circuits is shown in Figure 4.2.
Figure 4.2. Inductive circuit in time domain and complex frequency domain
c. Capacitor
The time and complex frequency domains for purely capacitive circuits is shown in Figure 4.3.
T
vR t ! RiR t !=
iR t !vR t !
R------------=
"
R
+
Time Domain
vR t !iR t !
R
+
"
Complex Frequency Domain
VR s ! IR s !
VR s ! RIR s !=
IR s !VR s !
R--------------=
"
+
Time Domain
L vL t ! iL t !
vL t ! LdiL
dt-------=
iL t !1
L--- vL td
#–
t
$=
+
"
Complex Frequency Domain
+"
sL
LiL 0
" !
VL s !
IL s !
VL s ! sLIL s ! LiL 0" !–=
IL s !VL s !
Ls-------------
iL 0" !
s--------------+=
Chapter 4 Circuit Analysis with Laplace Transforms
4-2 Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
Figure 4.3. Capacitive circuit in time domain and complex frequency domain
Note:
In the complex frequency domain, the terms and are called complex inductive impedance,
and complex capacitive impedance respectively. Likewise, the terms and and are called com-
plex capacitive admittance and complex inductive admittance respectively.
Example 4.1
Use the Laplace transform method to find the voltage across the capacitor for the circuit of
Figure 4.4, given that .
Figure 4.4. Circuit for Example 4.1
Solution:
We apply KCL at node as shown in Figure 4.5.
Figure 4.5. Application of KCL for the circuit of Example 4.1
Then,
or
+
Time Domain
+C
vC t! "iC t! "
iC t! " CdvC
dt---------=
vC t! "1
C---- iC td
#–
t
$=
+
Complex Frequency Domain
+
+
VC s! "
vC 0 ! "
s----------------
IC s! "
1
sC------ IC s! " sCVC s! " CvC 0
! "–=
VC s! "IC s! "
sC------------
vC 0 ! "
s----------------+=
sL 1 sC%
sC 1 sL%
vC t! "
vC 0 ! " 6 V=
R
C
+
+
V 1 F
&'(
12u0 t! "
vS
vC t! "
A
R
C
+
+
V 1 F
&'(
12u0 t! "
vS
vC t! "
AiR
iC
iR iC+ 0=
Signals and Systems with MATLAB Applications, Second Edition 4-3Orchard Publications
Circuit Transformation from Time to Complex Frequency
or
(4.1)
The Laplace transform of (4.1) is
or
or
By partial fraction expansion,
Therefore,
Example 4.2
Use the Laplace transform method to find the current through the capacitor for the circuit of
Figure 4.6, given that .
Figure 4.6. Circuit for Example 4.2
vC t ! 12u0 t !–
1------------------------------------- 1
dvC
dt---------"+ 0=
dvC
dt--------- vC t !+ 12u0 t !=
sVC s ! vC 0#
! VC s !+–12
s------=
s 1+ !VC s !12
s------ 6+=
VC s !6s 12+
s s 1+ !-------------------=
VC s !6s 12+
s s 1+ !-------------------
r1
s----
r2
s 1+ !----------------+= =
r16s 12+
s 1+ !------------------
s 0=
12= =
r26s 12+
s------------------
s 1–=
6–= =
VC s !12
s------
6
s 1+-----------–= 12 6e
t––$ 12 6e
t–– !u0 t ! vC t != =
iC t !
vC 0#
! 6 V=
C
#
++#
V 1 F
%&'
12u0 t !
vS
vC t !
iC t !
Chapter 4 Circuit Analysis with Laplace Transforms
4-4 Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
Solution:
This is the same circuit as in Example 4.1. We apply KVL for the loop shown in Figure 4.7.
Figure 4.7. Application of KVL for the circuit of Example 4.2
and with and , we get
(4.2)
Next, taking the Laplace transform of both sides of (4.2), we get
or
Check: From Example 4.1,
Then,
(4.3)
The presence of the delta function in (4.3) is a result of the unit step that is applied at .
R
C
+
+
V 1 F
!"#
12u0 t$ %
vS
vC t$ %iC t$ %
RiC t$ %1
C---- iC t$ % td
&–
t
'+ 12u0 t$ %=
R 1= C 1=
iC t$ % iC t$ % td&–
t
'+ 12u0 t$ %=
IC s$ %IC s$ %
s------------
vC 0 $ %
s----------------+ +
12
s------=
11
s---+( )
* + IC s$ % 12
s------
6
s---–
6
s---= =
s 1+
s-----------
( )* + IC s$ % 6
s---=
IC s$ % 6
s 1+-----------= iC t$ % 6e
t–u0 t$ %=,
vC t$ % 12 6et–
–$ %u0 t$ %=
iC t$ % CdvC
dt---------
dvC
dt---------
td
d12 6e
t––$ %u0 t$ % 6e
t–u0 t$ % 6- t$ %+= = = =
t 0=
Signals and Systems with MATLAB Applications, Second Edition 4-5Orchard Publications
Circuit Transformation from Time to Complex Frequency
Example 4.3
In the circuit of Figure 4.8, switch closes at , while at the same time, switch opens. Use
the Laplace transform method to find for .
Figure 4.8. Circuit for Example 4.3Solution:
Since the circuit contains a capacitor and an inductor, we must consider two initial conditions One
is given as . The other initial condition is obtained by observing that there is an initial
current of in inductor ; this is provided by the current source just before switch
opens. Therefore, our second initial condition is .
For , we transform the circuit of Figure 4.8 into its s-domain* equivalent shown in Figure 4.9.
Figure 4.9. Transformed circuit of Example 4.3
In Figure 4.9 the current in has been replaced by a voltage source of . This is found from the
relation
(4.4)
The polarity of this voltage source is as shown in Figure 4.9 so that it is consistent with the direction
of the current in the circuit of Figure 4.8 just before switch opens.
* Henceforth, for convenience, we will refer the time domain as t-domain and the complex frequency domain as s-domain
S1 t 0= S2
vout t ! t 0"
#
+
C
#
+
1 $
2 A
1 F2 $
0.5 H
0.5 H
vC 0#
! 3 V=
t 0=
t 0=
is t !
vout t !
iL1 t !
L2
R1
R2
L1
S1
S2
vC 0#
! 3 V=
2 A L1 2 A S2
iL1 0#
! 2 A=
t 0"
#
++#
1
2 0.5s
0.5s+#
1/s
3/s
1 V
1
Vout s !
L1 1 V
L1iL1 0#
!1
2--- 2% 1 V= =
iL1 t ! S2
Chapter 4 Circuit Analysis with Laplace Transforms
4-6 Signals and Systems with MATLAB Applications, Second Edition
Orchard Publications
The initial capacitor voltage is replaced by a voltage source equal to .
Applying KCL at node , we get
(4.5)
and after simplification
(4.6)
We will use MATLAB to factor the denominator of (4.6) into a linear and a quadratic factor.