APPLICATION OF THE LAPLACE TRANSFORM
TO CIRCUIT ANALYSIS
LEARNING GOALS
Laplace circuit solutions
Showing the usefulness of the Laplace transform
Circuit Element Models
Transforming circuits into the Laplace domain
Analysis Techniques
All standard analysis techniques, KVL, KCL, node,
loop analysis, Thevenin’s theorem are applicable
Prof. Julio Bolacell
LAPLACE CIRCUIT SOLUTIONS
We compare a conventional approach to solve differential equations with a
technique using the Laplace transform
)()()( tdt
diLtRitvS :KVL
tCC
CC eKtit
dt
diLtRi
)(0)()(
equationary Complement
pC iii
L
ReLKeRK
tC
tC 0)(
pp Kti )(
case thisfor solution Particular
pS RKv 1
tL
R
CeKR
ti
1
)(0000 ) i( for t(t)vS
conditionsboundary Use
0;11
)(
teR
tit
L
R
“Take Laplace transform” of the equation
dt
diLsRIsVS L)()(
)()0()( ssIissIdt
di
L
)()(1
sLsIsRIs
)(
1)(
LsRssI
LRs
K
s
K
sLRs
LsI
/)/(
/1)( 21
RsILRsK
RssIK
LRs
s
1|)()/(
1|)(
/2
01
0;11
)(
teR
tit
L
R
)()()( tdt
diLtRitvS
LInitial conditions
are automatically
included
No need to
search for
particular
or comple-
mentary
solutions
Only algebra
is needed
P
a
r
t
i
c
u
l
a
r
Comple
mentary
LEARNING BY DOING 0),( ttv Find
KCL using Model
dt
dvC
R
vv S
Sv
0
R
vv
dt
dvC S
Svvdt
dvRC
L)()( sVsV
dt
dvRC S
L
)()0()( ssVvssVdt
dv
L
ssVtuv SS
1)()(
0)0(0,0)( vttvSInitial condition
given in implicit
form
In the Laplace domain the differential
equation is now an algebraic equation
ssVsRCsV
1)()(
)/1(
/1
)1(
1)(
RCss
RC
RCsssV
Use partial fractions to determine inverse
RCs
K
s
K
RCss
RCsV
/1)/1(
/1)( 21
1|)()/1(
1|)(
/12
01
RCs
s
sVRCsK
ssVK
0,1)(
tetv RC
t
CIRCUIT ELEMENT MODELS
The method used so far follows the steps:
1. Write the differential equation model
2. Use Laplace transform to convert the model to an algebraic form
For a more efficient approach:
1. Develop s-domain models for circuit elements
2. Draw the “Laplace equivalent circuit” keeping the interconnections and replacing
the elements by their s-domain models
3. Analyze the Laplace equivalent circuit. All usual circuit tools are applicable and all
equations are algebraic.
)()()()( sRIsVtRitv
)()(
)()(
sIti
sVtv
SS
SS
sourcest Independen
...
)()()()(
)()()()(
sBVsItBvti
sAIsVtAitv
CDCD
CDCD
sources Dependent
Resistor
Capacitor: Model 1
)0()(1
)(0
vdxxiC
tvt
)0()()( CvsCsVsI
Source transformation
)0(1
)0(
Cv
Cs
s
v
Ieq
s
vsI
CssV
)0()(
1)(
Impedance in series
with voltage source
Capacitor: Model 2
Impedance in parallel
with current source
s
sIdxxi
t)(
)(0
L
LEARNING BY DOING
Ai 1)0(
Determine the model in the s-domain and the expression for
the voltage across the inductor
Inductor with
initial current
Equivalent circuit in
s-domain
1
1)()(1)(
ssVsIsV
Law sOhm'
)()1(1 sIs :KVL
Steady state for t<0
ANALYSIS TECHNIQUES
All the analysis techniques are applicable in the s-domain
LEARNING EXAMPLE Draw the s-domain equivalent and find the voltage in both
s-domain and time domain
0)0(0,0)( oS vtti
One needs to determine the initial voltage
across the capacitor
1
3)(
ssIS
)(1
||)( sICs
RsV So
1
103
/1
/1)(
1)(
3
sRCs
CsI
CsR
Cs
R
sV So
25.0)1025)(1010( 63 RC
14)1)(4(
120)( 21
s
K
s
K
sssVo
40|)()1(
40|)()4(
12
41
so
so
sVsK
sVsK
)(40)( 4tueetv
tto
LEARNING EXAMPLE Write the loop equations in the s-domain
)0()0()0(
)( 1121 iLs
v
s
vsVA
1 Loop
))()(())()((1
)(1
)( 21121
2
1
1
11 sIsIsLsIsIsC
sIsC
sIR
)()0()0(
)0( 222
11 sViLs
viL B
2 Loop
)()())()((1
))()(( 22212
2
121 sIRsLsIsIsC
sIsIsL
Do not increase
number of loops
LEARNING EXAMPLE Write the node equations in the s-domain
s
ivC
s
isI A
)0()0(
)0()( 2
111
1 VNode
)(1
)(11
21
2
11
21
1 sVsCsL
sVsCsLsL
G
)(1
)(1
1
2
12
2
122 sVsL
sCsVsL
sCsCG
s
ivCvCsIB
)0()0()0()( 2
1122
2 VNode
Do not increase number
of nodes
LEARNING EXAMPLE
theorem. sNorton' and sThevenin' tion,transforma source
ion,superposit analysis, loop analysis, node using Find )(tvo
Assume all initial conditions are zero
Node Analysis
01
)()(12
)(4 1
1
s
sVsV
s
ssV
s
o
1 V@KCL
01
)()(
2
)( 1
s
sVsVsV oo
oKCL@V
s
2
0)()21()(2
124)()()1(
1
21
2
sVsssV
s
ssVssVs
o
o
)1( 2s
s2
2)1(
)3(8)(
s
ssVo
Could have
used voltage
divider here
s
sV
12
)(1
Loop Analysis
ssI
4)(1
1 Loop
ssIsI
ssIsIs
12)(2)(
1))()(( 2212
2 Loop
22)1(
)3(4)(
s
ssI
22)1(
)3(8)(2)(
s
ssIsVo
Source Superposition Applying current source
Current divider
'2I
s
s
ssVo
4
21
2
2)('
Voltage divider
ss
s
sVo
12
12
2)("
2
"'
)1(
)3(8)()()(
s
ssVsVsV
ooo
Applying voltage source
Source Transformation
The resistance is redundant
Combine the sources and use current
divider
2
124
21
2)(ss
ss
ssVo
2)1(
)3(8)(
s
ssVo
Using Thevenin’s Theorem
Reduce this part
s
s
ss
ssVOC
124412)(
Only independent sources
s
ss
sZTh
11 2
s
s
s
ssVo
124
12
2)(
2
Voltage
divider
2)1(
)3(8)(
s
ssVo
Using Norton’s Theorem
2
124/124)(
s
s
s
s
ssISC
Reduce this part
sZTh
2
124
21
2)(s
s
ss
ssVo
2)1(
)3(8)(
s
ssVo
Current
division
LEARNING EXAMPLE zero be to conditions initial all Assume voltagethe Determine ).(tvo
. Three loops, three non-reference nodes
. One voltage source between non-reference
nodes - supernode
. One current source. One loop current known
or supermesh
. If v_2 is known, v_o can be obtained with a
voltage divider
Selecting the analysis technique:
Transforming the circuit to s-domain
ssVsV
12)()( 12 :constraint Supernode
01
)()(2
/2
)(
2
)( 211
s
sVsI
s
sVsV :supernode KCL@
2
)()( 1 sV
sI : variablegControllin
)(1
1)( 20 sV
ssV
:divider Voltage
ssVsV /12)()( 21 :algebra the Doing
ssVsI /62/)()( 2
0)1/()(
)/62/)((2/12)()1)(2/1(
2
22
ssV
ssVssVs
)54(
)3)(1(12)(
22
sss
sssV
)54(
)3(12)(
2
sss
ssVo
Continued ... theorem sThevenin' using Compute )(sVo
-keep dependent source and controlling
variable in the same sub-circuit
-Make sub-circuit to be reduced as simple
as possible
-Try to leave a simple voltage divider after
reduction to Thevenin equivalent
sVOC /12
2
/12'
0'2/2
/12
2
/12
sVI
Is
sVsV
OC
OCOC
ssVOC
12)(
0'0'2)/2/()'2(' IIsII
0)/2/("2""2 sIIIISC
s/12
sI /6"
s
sISC
)3(6 3
2
)(
)(
ssI
sVZ
SC
OCTH
s
ss
sVo
12
3
21
1)(
Continued … Computing the inverse Laplace transform
Analysis in the s-domain has established that the Laplace transform of the
output voltage is
)54(
)3(12)(
2
sss
ssVo
)12)(12(542jsjsss
)12()12()12)(12(
)3(12)(
*11
js
K
js
K
s
K
jsjss
ssV o
o
)()cos(||2)()(
11
*11 tuKteK
js
K
js
K t
5
36|)( 0 soo ssVK
57.16179.343.19879.3
)902(43.1535
45212
)2)(12(
)11(1212
|)()12(1 jj
j
jss
oVjsK
)(57.161cos(59.75
36)( 2
tutetvt
o
1)2( 2 s
1)2(1)2(
)2(
12
)3(12)(
22
21
2
s
C
s
sC
s
C
ss
ssV o
o
)(]sincos[)()(
)(2122
222
1 tutCtCes
C
s
sC t
5/36|)( 0 soo ssVC
])2([)1)2(()3(12 212
CsCssCs o 5/12610/362122 22 CCCs o
5/360: 11 CCCo2s of tscoefficien Equating
)(sin5
12)cos1(
5
36)( 22
tutetetvtt
o
One can also use
quadratic factors...
LEARNING EXTENSION equations node using Find )(tio
supernode
oVSo VV
Assume zero initial conditions
Implicit circuit transformation to s-domain SV
KCL at supernode
02
)()(2))()((
sV
Ls
sV
ssVsVCs oo
So
2
)()(,
12)(
sVsI
ssV o
oS
16
15
4
1
61
15.0
61)(
22
s
s
ss
ssIo
algebra the Doing
4
15
4
1
4
15
4
1
4
15
4
1
4
15
4
1
61)(
*
1 1
js
K
js
K
jsjs
ssIo
)()cos(||2)()(
11
*11 tuKteK
js
K
js
K t
4
152
4
15
4
161
|)(4
15
4
1
4
15
4
11
j
j
sIjsKjs
o
72.15653.6
9097.0
72.6633.61K
72.1564
15cos06.13)( 4 teti
t
o
LEARNING EXTENSION equations loop using Find )(tvo
supermesh
12
2II
s
source to due constraint
0212
21
2211 Is
sIIIs
supermesh onKVL
Solve for I2
)73.3)(27.0(
216
)14(
216)(
22
sss
s
sss
ssI
73.327.0)( 210
2
s
K
s
K
s
KsI
2|)( 020 sssIK
48.2)73.327.0)(27.0(
2)27.0(16|)()27.0( 27.021
ssIsK
47.4)27.073.3)(73.3(
2)73.3(16|)()73.3( 73.322
ssIsK
)(2)(
)(47.448.22)(
2
73.327.02
titv
tueeti
o
tt
Determine inverse transform
TRANSIENT CIRCUIT ANALYSIS USING LAPLACE TRANSFORM
For the study of transients, especially transients due to switching, it is important
to determine initial conditions. For this determination, one relies on the properties:
1. Voltage across capacitors cannot change discontinuously
2. Current through inductors cannot change discontinuously
LEARNING EXAMPLE 0),( ttvo Determine
AiVv LC 1)0(,1)0(
itshortcircu are inductors
circuit open are capacitors case DCFor
Assume steady state for t<0 and determine
voltage across capacitors and currents
through inductors
)0( Li
)0(Cv
Circuit for t>0
Use mesh analysis
Laplace
14
)1( 21 s
sIIs
11
)2
1( 21 s
Is
ssI
Solve for I2 s
)1( s
232
12)(
22
ss
ssI
ssI
ssVo
1)(
2)( 2
232
72)(
2
ss
ssVo
Now determine the inverse transform
roots conjugatecomplex 042acb
4
7
4
3
4
7
4
3)(
*
1 1
js
K
js
KsVo
4
7
4
31 )(
4
7
4
3
js
o sVjsK
5.7614.2
)()cos(||2)()(
11
*11 tuKteK
js
K
js
K t
)5.764
7cos(28.4)( ttvo
Circuit for t>0
LEARNING EXTENSION 0),(1 tti Determine
Initial current through inductor
Aii LL 1)0()0(
12
6 s2s
1)(1 sI
ss
ssI
1
182
2)(1
Current
divider
)()(9
)( 911 tueti
s
ssI
t
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