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11C.T. PanC.T. Pan 11
LAPLACE LAPLACE TRANSFORM AND TRANSFORM AND ITS APPLICATION ITS APPLICATION IN CIRCUIT IN CIRCUIT ANALYSISANALYSIS
22C.T. PanC.T. Pan
12.1 Definition of the Laplace Transform12.1 Definition of the Laplace Transform
12.2 Useful Laplace Transform Pairs12.2 Useful Laplace Transform Pairs
12.3 Circuit Analysis in S Domain12.3 Circuit Analysis in S Domain
12.4 The Transfer Function and the Convolution 12.4 The Transfer Function and the Convolution IntegralIntegral
22
33C.T. PanC.T. Pan
12.5 The Transfer Function and the Steady state 12.5 The Transfer Function and the Steady state Sinusoidal ResponseSinusoidal Response
12.6 The Impulse Function in Circuit Analysis12.6 The Impulse Function in Circuit Analysis
33
44C.T. PanC.T. Pan 44
12.1 Definition of the Laplace Transform12.1 Definition of the Laplace Transform
Pierre Simon Laplace (1749-1827) :
A French astronomer and mathematician First presented the Laplace transform and its applications to differential equations in 1979.
55C.T. PanC.T. Pan 55
12.1 Definition of the Laplace Transform12.1 Definition of the Laplace Transform
Definition:
[ ]0
( ) ( ) ( )
a complex variable
stL f t F s f t e dt
s jσ ω
−
∞ −= =
= +∫
The Laplace transform is an integral transformation of a function f(t) from the time domain into the complex frequency domain, F(s).
66C.T. PanC.T. Pan 66
12.1 Definition of the Laplace Transform12.1 Definition of the Laplace Transform
[ ] 1
1
1 1( ) ( ) ( )2
Look-up table , an easier way for circuit application ( ) ( )
j stj
L F s f t F s e dsj
f t F s
+−−
= =
⇔
∫σ ω
σ ωπ
One-sided (unilateral) Laplace transformTwo-sided (bilateral) Laplace transform
77C.T. PanC.T. Pan 77
12.1 Definition of the Laplace Transform12.1 Definition of the Laplace Transform
Similar to the application of phasor transform to
solve the steady state AC circuits , Laplace
transform can be used to transform the time domain
circuits into S domain circuits to simplify the
solution of integral differential equations to the
manipulation of a set of algebraic equations.
88C.T. PanC.T. Pan 88
Functions f(t) , t> F(s)
impulse 1
step
ramp t
exponential
sine
0−
( )tδ
( )u t
2
1S
ate− 1S a+
1S
sin tω 2 2Sω
ω+
12.2 Useful Laplace Transform Pairs12.2 Useful Laplace Transform Pairs
99C.T. PanC.T. Pan 99
Functions f(t) , t> F(s)
cosine
damped ramp
damped sine
damped cosine
0−
cos tω 2 2
SS ω+
atte−
cosate tω−
( )21
S a+
( )2 2S aω
ω+ +sinate tω−
( )2 2
S aS a ω
+
+ +
12.2 Useful Laplace Transform Pairs12.2 Useful Laplace Transform Pairs
1010C.T. PanC.T. Pan 1010
12.2 Useful Laplace Transform Pairs12.2 Useful Laplace Transform Pairs
[ ]
[ ]
[ ]
[ ]
-0
0
0
( ) ( ) ( 0 )
( ) ( )
( ) ( ) ( ) , 0
( ) ( )
1 ( ) ( ) , 0
( ) ( )
( ) ( )
l i m l i ml i m l i m
t
a s
a t
st
t s
dL f t s F s fd t
F sL f dS
L f t a u t a e F s a
L e f t F s a
sL f a t F aa a
f t s F s
f t s F s+
−
−
−
→ ∞→
→ ∞ →
∆ = −
∆ = ∆ − − = >
∆ = +
∆ = >
∆ =
∆ =
∫
τ τ
1111C.T. PanC.T. Pan 1111
12.2 Useful Laplace Transform Pairs12.2 Useful Laplace Transform PairsExample Use the Laplace transform to
solve the differential equation.
[ ]
2
2
2
6 8 2 ( )
(0) 1'(0) 2
Take Laplace transfrom2( ) (0) '(0) 6 ( ) (0) 8 ( )
d v dv v u tdtdt
vv
s V s sv v sV s v V ss
+ + =
== −
− − + − + =
1212C.T. PanC.T. Pan 1212
12.2 Useful Laplace Transform Pairs12.2 Useful Laplace Transform Pairs
[ ]2
22
2 2
2
2 4
2( ) (0) '(0) 6 ( ) (0) 8 ( )
4 2( 6 8) ( )
4 2 4 2( )( 2)( 4)( 6 8)
1( ) (1 2 ) ( )4
t t
s V s sv v sV s v V ss
s ss s V ss
s s s sV ss s ss s s
v t e e u t− −
− − + − + =
+ ++ + =
+ + + +∴ = =
+ ++ +
= + +
1313C.T. PanC.T. Pan 1313
12.3 Circuit Analysis in S Domain12.3 Circuit Analysis in S Domain(1) KCL , ( ) 0 , for any node.
Take Laplace Transform ( ) 0 , for any node.
(2) KVL , ( ) 0 , for any loop.
Take Laplace Transform ( ) 0 ,
kn
kn
km
km
i t
I s
v t
V s
=
=
=
=
∑
∑
∑
∑ for any loop.
1414C.T. PanC.T. Pan 1414
12.3 Circuit Analysis in S Domain12.3 Circuit Analysis in S Domain
(3) Circuit Component Modelsresistor ( ) ( ) ( ) ( ) ( ) ( )
R R
R R
R R
v t Ri tV s RI sI s GV s
===
1515C.T. PanC.T. Pan 1515
12.3 Circuit Analysis in S Domain12.3 Circuit Analysis in S Domain
0
inductor
( )
1 ( ) (0 ) ( )
( ) ( ) (0 )
( ) (0 ) ( )
LL
tL L L
L L L
L L
div t Ldt
i t i v dL
V s sLI s Li
V s iI ssL s
−
−
−
−
=
= +
= −
= +
∫ τ τ (0 )LLi −
(0 )Lis
−
1616C.T. PanC.T. Pan 1616
12.3 Circuit Analysis in S Domain12.3 Circuit Analysis in S Domain
0
capacitor
1 ( ) (0 ) ( )
( ) ( ) (0 )
(0 ) 1 ( ) ( )
CC
tC C C
C C c
CC C
dvi Cdt
v t v i dC
I s sCV s Cv
vV s I ss sC
−
−
−
−
=
= +
= −
= +
∫ τ τ
1sC
(0 )Cvs
−
1sC
(0 )CCv −
1717C.T. PanC.T. Pan 1717
12.3 Circuit Analysis in S Domain12.3 Circuit Analysis in S Domain
M i2i1
L1 L2v1 v2
+
- -
+
Coupling inductors
V1(s)=L1SI1(s)-L1i1(0)+MSI2(s)-Mi2(0)V2(s)= MSI1(s)- Mi1(0)+L2SI2(s)-L2i2(0)
dtdL
dtdM
dtdM
dtdL
22
12
2111
iiv
iiv
+=
+=
1818C.T. PanC.T. Pan 1818
12.3 Circuit Analysis in S Domain12.3 Circuit Analysis in S DomainFor zero initial conditions
( )impedance ( )( )( ) 1admittance ( )( ) ( )
( ) ( ) ( ) '
V s Z sI SI S Y sV s Z s
V s Z s I s ohm s law in s domain
=
= =
= −
@
@
1919C.T. PanC.T. Pan 1919
12.3 Circuit Analysis in S Domain12.3 Circuit Analysis in S Domain
The elegance of using the Laplace transform in
circuit analysis lies in the automatic inclusion of
the initial conditions in the transformation process,
thus providing a complete (transient and steady
state) solution.
2020C.T. PanC.T. Pan 2020
12.3 Circuit Analysis in S Domain12.3 Circuit Analysis in S Domain
Circuit analysis in s domain
nStep 1 : Transform the time domain circuit into s-domain circuit.
nStep 2 : Solve the s-domain circuit.e.g. Nodal analysis or mesh analysis.
nStep 3 : Transform the solution back into time domain.
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12.3 Circuit Analysis in S Domain12.3 Circuit Analysis in S DomainExample Find vo(t) given vo(0)=5V
S-domain equivalent circuit
101s +
10s
2222C.T. PanC.T. Pan 2222
12.3 Circuit Analysis in S Domain12.3 Circuit Analysis in S Domain
2
Nodal analysis10 ( ) ( ) ( )1 2 0.5
10 10 10 /25 35 10 15( )
( 1)( 2) 1 2
( ) (10 15 ) ( ) V
oo o
o
t to
V s V s V sss
sV ss s s s
v t e e u t− −
−+ + + = +
+∴ = = +
+ + + +
∴ = +
101s +
10s
2323C.T. PanC.T. Pan 2323
12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
Given a linear circuit N in s domain as shown below
Transfer function H(s) is defined as
)s(X)s(Y)s(H =
2424C.T. PanC.T. Pan 2424
12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
( ) ( ) , ( ) ( ) ; ( )
( ) ( ) , ( ) ( ) ; ( )
( ) ( ) , ( ) ( ) ; ( )
( ) ( ) , ( ) ( ) ; ( )
o i
o i
If Y s V s X s V s then H s voltage gain
If Y s I s X s I s then H s current gain
If Y s V s X s I s then H s impedance
If Y s I s X s V s then H s admittance
= = =
= = =
= = =
= = =
2525C.T. PanC.T. Pan 2525
12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
Given the transfer funtion H(s) and input X(s) , then Y(s)=H(s)X(s)
If the input is δ(t) , then X(s)=1 and Y(s)=H(s)
Hence , the physical meaning of H(s) is in fact the Laplace transform of the impulse response of the corresponding circuit.
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12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
in time domain
Y(s)=H(s)X(s) , in s-domain
( ) ( ) ( ) ( ) ( ) y t h t x d h t x tτ τ τ∞
−∞
= − ∗∫ @
Geometrical interpretation of finding the
convolution integral value at t=tk is based on :
2727C.T. PanC.T. Pan 2727
12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
(1)Approximating the input function by using a series of impulse functions.
(2)Shifting property of linear systems input x(t)→output y(t)x(t-τ)→output y(t-τ)
(3)Superposition theorem for linear systems
(4)Definition of integral : finding the area
2828C.T. PanC.T. Pan 2828
12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
(1)Input x(τ) is approximated using impulsefunctions , x(τ)=0 , for τ<0
0
1 1 1
2 2 2
3 3 3
, ( ) , 0 , (0) , (0)
, ( ) , ( )2 , ( ) , ( )3 , ( ) , ( )
k kinstant x value areax xx xx xx x
τ ττ ττ τ τ τ ττ τ τ τ ττ τ τ τ τ
====
VV V
V VV V
0 1τ 2τ 3τ 4τ 5τ
(0)x( )x τ
0 1τ
( )x τ
2τ
0 1 1 2 2
0
( ) ( ) ( ) ( )
( ) ( )k
x f f f
x k k∞
=
≅ + − + − +
−∑
L
@ V V V
τ δ τ δ τ τ δ τ τ
τ τδ τ τ
2929C.T. PanC.T. Pan 2929
12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
(2) Use the linearity property
1 1 1 1
2 2 2 2
( )(0) ( ) (0) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )
input output responsex x hx x hx x h
up to
τδ τ τ τ
τ τδ τ τ τ τ τ τ
τ τδ τ τ τ τ τ τ
τ
→
→
− → −
− → −
V VV VV V
M M
kt=
3030C.T. PanC.T. Pan 3030
12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
(3) Use superposition theorem to find the total approximate response
$0
( ) ( ) ( )
[ ]
n
k kk
k
y t x k h t k
tn integer
=
= −
=
∑ V V V
V
τ τ τ
τ
3131C.T. PanC.T. Pan 3131
12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
(4) Take the limit , Δτ→dτ,
( ) ( ) ( )k ky t x h t d∞
−∞
= −∫ τ τ τ
Due to causality principle , h(t-τ)=0 for t≥tkand x(τ)=0 for τ<0
0
( ) ( )kt
kh t x d= −∫ τ τ τ
$( ) ( )k ky t y t→
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12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
( ) ( ) , ( ) ( ) , (4) ,
( ) ( ) ( )
(4) (4 ) ( )
tGiven x t u t h t e u t find y
y t h t x d
y h x d
−
∞
−∞
∞
−∞
= =
= −
∴ = −
∫
∫
τ τ τ
τ τ τ
Example:
3333C.T. PanC.T. Pan 3333
12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
Step1.
4 2 0 2 4
0.2
0.4
0.6
0.8
1.0x
4 2 0 2 4
0.2
0.4
0.6
0.8
1.0h
( )x τ
τ
τ
( )h τ−
2−
2−
4−
4−
3434C.T. PanC.T. Pan 3434
12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
Step2. Shift to tk=4
4 2 0 2 4
0.2
0.4
0.6
0.8
1.0h 4 (4 )h − τ
4− 2−τ
3535C.T. PanC.T. Pan 3535
12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
Step3. Find the product h(4-τ)x(τ)
3636C.T. PanC.T. Pan 3636
12.4 The Transfer Function and the 12.4 The Transfer Function and the Convolution IntegralConvolution Integral
Step4. Find the integral (area)4 4 4(4 ) 4 4
00 04 4 4
(4) =
= ( 1) (1 )
y e d e e d e e
e e e
τ τ ττ τ− − − −
− −
= =
− = −
∫ ∫
Step5. Check
4
1 11 1( ) ( ) ( )
1( ) (1 ) , 1
(4) (1 )
t
a aY s H s X ss a s s a s
y t e aa
y e
−
−
−
= = × = ++ +
∴ = − =
∴ = −
3737C.T. PanC.T. Pan 3737
12.5 The Transfer Function and the 12.5 The Transfer Function and the Steady State Sinusoidal ResponseSteady State Sinusoidal Response
From definition of transfer function
( )( )( )
( ) ( ) ( )
Y sH sX s
Y s H s X s
=
⇒ =
Assume input X(t)=Acos(wt+Ф) and H(s) is given, then one can get the steady state solution without needing a separate phasor analysis.
3838C.T. PanC.T. Pan 3838
12.5 The Transfer Function and the 12.5 The Transfer Function and the Steady State Sinusoidal ResponseSteady State Sinusoidal Response
2 2
*1 1
*1 1
1
: ( ) cos cos sin sin(cos sin )( )
( ) ( ) ( )
:
( )
1 ( )2
( ) ( cos sin )
ss
j
proof X t A t A tA sX s
sY s H s X s
K Ks j s j
other terms due to polesunder steady state
K KY ss j s j
K
H j Ae
s j
H s A s
s j
φ
φ φφ φ
φ φ
= ω − ω− ω
∴ =+ ω
∴ =
= +− ω + ω
+∑
∴ = +− ω + ω
=
= ω
= ω
−ω
+ ω
3939C.T. PanC.T. Pan 3939
12.5 The Transfer Function and the 12.5 The Transfer Function and the Steady State Sinusoidal ResponseSteady State Sinusoidal Response
( )( ) | ( ) |
( ) | ( ) | cos[ ( )]. . ( ( ))
( ( )) | ( ) | ( )
j
ss
ss
Let H j H j eand take inverse Laplace transform
Then y t A H j ti e P X t Athen P y t A H j
θ
φ θ
φφ θ
ωω = ω
= ω ω + + ω
= ∠= ω ∠ + ω
4040C.T. PanC.T. Pan 4040
Example : The transfer function H(s) of the circuit given below is known.Find the steady state solution of Vo(t) for the given Vg(t).
12.5 The Transfer Function and the 12.5 The Transfer Function and the Steady State Sinusoidal ResponseSteady State Sinusoidal Response
4141C.T. PanC.T. Pan 4141
12.5 The Transfer Function and the 12.5 The Transfer Function and the Steady State Sinusoidal ResponseSteady State Sinusoidal Response
2 6
1000( 5000)( )6000 25 10
( ) 120cos(5000 30 )g
sH ss s
V t t V°
+=
+ + ⋅= +
4242C.T. PanC.T. Pan 4242
12.5 The Transfer Function and the 12.5 The Transfer Function and the Steady State Sinusoidal ResponseSteady State Sinusoidal Response
6 6
: 50001000( 5000 5000)( 5000)
25 10 5000(6000) 25 10
2 456
2( ) 120 cos(5000 30 45 )6
20 2cos(5000 15 )
sso
Solution Let s j jjEvaluate H j
j
Then V t t
t V
°
° °
°
= ω=+
=− × + + ×
= ∠−
= × + −
= −
4343C.T. PanC.T. Pan 4343
nIn theory , the relationship between H(s) and
H(jw) provides a link between the time domain
and the frequency domain.
nIn some cases , we can determine H(jw)
experimentally and then construct H(s) from the
data.
12.5 The Transfer Function and the 12.5 The Transfer Function and the Steady State Sinusoidal ResponseSteady State Sinusoidal Response
4444C.T. PanC.T. Pan 4444
12.5 The Transfer Function and the 12.5 The Transfer Function and the Steady State Sinusoidal ResponseSteady State Sinusoidal Response
Example: Find the impulse response of the Example: Find the impulse response of the following circuit.following circuit.
inv
R+
_ov
+
_C
4545C.T. PanC.T. Pan 4545
12.5 The Transfer Function and the 12.5 The Transfer Function and the Steady State Sinusoidal ResponseSteady State Sinusoidal Response
(a) Time domain solution(a) Time domain solution
0
( )
0 , (0 ) 01 ( ) 0 , (0 )
1
oo
o
t
o
dvRC v tdt
At t vtAt t v dt
C R
VRC
δ
δ
− −
+ +
+ =
= =
= =
=
∫( )tδ ( )ov t
4646C.T. PanC.T. Pan 4646
12.5 The Transfer Function and the 12.5 The Transfer Function and the Steady State Sinusoidal ResponseSteady State Sinusoidal Response
0 , ( ) 0For t tδ+> =
R
1( ) , (0 )ov t vRC
+ =+
_C
0
1 ( ) ( )
oo
tRC
o
dvRC vdt
v t e u tRC
−
+ =
∴ =
4747C.T. PanC.T. Pan 4747
12.5 The Transfer Function and the 12.5 The Transfer Function and the Steady State Sinusoidal ResponseSteady State Sinusoidal Response
(b) s(b) s--domain solutiondomain solution
. .
( )( )( )
o
in zero I C
V sH sV s
=
Find the transfer functionFind the transfer function
Transform into sTransform into s--domain circuitdomain circuit
( )inV s
R+
_( )oV s1
Cs
4848C.T. PanC.T. Pan 4848
12.5 The Transfer Function and the 12.5 The Transfer Function and the Steady State Sinusoidal ResponseSteady State Sinusoidal Response
-1
1( ) 1 ( ) 1( ) 1
( ) [ ( )]
1 ( )
o
in
tRC
V s CsH sV s RCsR
Csh t H s
e u tRC
−
∴ = = =++
∴ =
=
L
( )inV s
R+
_( )oV s1
Cs
Same answer.Same answer.
4949C.T. PanC.T. Pan 4949
12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
Example 1: Impulse voltage source excitationExample 1: Impulse voltage source excitation
( )OV tδ
( )i t
(0 ) 0i − =
5050C.T. PanC.T. Pan 5050
12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
(a) Time domain solution(a) Time domain solution
0
0, (0 ) 01 (0 ) ( )
( )
t
O
O
At t i
i V x dxLV AL
δ−
−
+
= =
∴ =
=
∫ ( )OV tδ
(0 )i +
The impulse voltage source has stored energy,The impulse voltage source has stored energy,, in the inductor as an initial current, in the inductor as an initial current
in an infinitesimal moment.in an infinitesimal moment.
212 (0 )( )iL +
5151C.T. PanC.T. Pan 5151
12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
0 , ( ) 0For t tδ+> =
( )i t
/
0 ,
(0 )
( ) ( ) ,
O
tO
diL Ri natural responsedt
Vi ALV Li t e u tL R
τ τ
+
−
+ =
=
∴ = =
Note that the impulse source just builds up an Note that the impulse source just builds up an initial inductor current but does not contribute initial inductor current but does not contribute to any forced response.to any forced response.
5252C.T. PanC.T. Pan 5252
12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
(b) s(b) s--domain solutiondomain solution
/
/ ( )
( ) ( )
O O
tO
V V LI s RR sL sL
Vi t e u tL
τ−
∴ = =+ +
∴ =OV
R
sL
( )I s
Same answer but much easier.Same answer but much easier.
5353C.T. PanC.T. Pan 5353
12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
Example 2: Impulse current source excitationExample 2: Impulse current source excitation
R C+
_( ) , (0 ) 0v t v − =( )oI tδ
5454C.T. PanC.T. Pan 5454
12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
(a) Time domain solution(a) Time domain solution 0, (0 ) 0 (0) 0 ,
At t vv short circuit
−= =∴ =
0
1(0 ) ( )
t
o
o
v I x dxCIC
δ+ =
=
∫
The impulse current source has stored energy,The impulse current source has stored energy,, in the capacitor as an initial voltage, in the capacitor as an initial voltage
in an infinitesimal moment.in an infinitesimal moment.
212 (0 )( )vC +
( )oI tδ R
5555C.T. PanC.T. Pan 5555
12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
0 ,
(0 )
( ) ( ) ,
o
to
dv vC natural responsedt R
IvCIv t e u t RCC
τ τ
+
−
+ =
=
∴ = =
0 , ( ) 0, For t t open circuitδ+> =
( )v t
Note that the impulse current just builds up an Note that the impulse current just builds up an initial capacitor voltage but does not contribute initial capacitor voltage but does not contribute to any forced response.to any forced response.
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12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
(b) s(b) s--domain solutiondomain solutionTransform into sTransform into s--domain circuitdomain circuit
R C+
_( )V soI
1/ V( ) 1 1
( ) ( )
oo
to
R I CCss IR s
Cs RCIv t e u tC
τ−
×∴ = =
+ +
∴ =
Same answer but much easier.Same answer but much easier.
5757C.T. PanC.T. Pan 5757
12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
Example 3: Impulse caused by switching operationExample 3: Impulse caused by switching operation
1 2(0 ) (0 )Note that v v− −≠
The switch is closed at t=0 in the following circuit.The switch is closed at t=0 in the following circuit.
5858C.T. PanC.T. Pan 5858
12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
Transform into sTransform into s--domaindomain
1 2
1 2 12
1 2 2 1 2
( ) 1 1
, ( )
o
o e
o e oe
VsI s V C
sC sCC C V C V CC V sC C sC s C C
= ⋅+
⋅ ⋅= = = ⋅
+ +
@
12
1 2
( ) ( )
( )
o e
o
i t V C tCv t V
C C
δ∴ =
=+
5959C.T. PanC.T. Pan 5959
12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
At t=0 , a finite charge of CAt t=0 , a finite charge of C11 is transferred to Cis transferred to C22instantaneously.instantaneously.
Note that , as the switch is closed , the voltage Note that , as the switch is closed , the voltage across Cacross C22 does not jump to Vdoes not jump to Voo of Cof C11 but to its final but to its final value of the two paralleled capacitors. value of the two paralleled capacitors.
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12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
2
1 22 2 2
1 2
11 1 2
1 2
1 2 1
1 1 2
1 2 1
, 0
, 0
+ , 0, 0 , , 0
.
o
o
o
o
o
C CN ote Q C V V tC C
CQ C V V tC C
Q Q C V tA lso a t t Q C V Q
Q Q C VC onserva tion o f charge
+
+
+
−
⋅= ⋅ = >
+
= ⋅ = >+
= >
= = =∴ + =
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12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
If we consider charged capacitors as voltage sources , If we consider charged capacitors as voltage sources , then we should not connect two capacitors with then we should not connect two capacitors with unequal voltages in parallel.unequal voltages in parallel.
Due to violation of KVL , an impulse will occur which Due to violation of KVL , an impulse will occur which may damage the components. may damage the components.
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12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
Example 4: Impulse caused by switching operationExample 4: Impulse caused by switching operationThe switch is opened at The switch is opened at t t = 0 in the following circuit.= 0 in the following circuit.
100V
10Ω 3H
1i 1L
2i
2L 2H
15Ω
0t = ov
+
_
1
2
0 , 100 (0 ) 10 10
(0 ) 0
At t steady state solutionVi A
i A
−
−
−
=
= =Ω
=
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12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
For t > 0, the SFor t > 0, the S--domain circuit isdomain circuit is10Ω 3s
( )I s 1L
2L 2s
15Ω
( )oV s100
s
30
5
5
(100 / ) 30 4 2( )5 25 5
60 10( ) ( ) (2 15) 125
( ) (4 2 ) ( ) .....................( ) ( ) 12 ( ) (60 10 ) ( ) .....( )
o
t
t
sI ss s s
V s I s ss s
i t e u t Av t t e u t Bδ
−
−
+= = +
+ +
= + = + ++
∴ = +
= + +
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12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
1 2
1 2
0 , (0 ) 10 , (0 ) 0
0 , ( ), (0 ) 6 , (0 ) 6 , ( ), 12 ( ) ( ).
L L
L L
o
Note t i A i At from A i A i A
Also from B there exists t at v tδ
− −
+ + +
= = =
= = =
10Ω 3s
( )I s 1L
2L 2s
15Ω
( )oV s100
s
30
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12.6 The Impulse Function in Circuit 12.6 The Impulse Function in Circuit AnalysisAnalysis
Thus, if we consider an inductor current as a Thus, if we consider an inductor current as a current source, then two inductors with unequal current source, then two inductors with unequal currents should not be connected in series.currents should not be connected in series.
Due to violation of KCL, it will result in impulse Due to violation of KCL, it will result in impulse voltage which may damage the components.voltage which may damage the components.
10Ω 3s
( )I s 1L
2L 2s
15Ω
( )oV s100
s
30
SUMMARYSUMMARY
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Objective 1 : Know the component models in sObjective 1 : Know the component models in s--domain.domain.
Objective 2 : Be able to transform a time domain circuit Objective 2 : Be able to transform a time domain circuit into the sinto the s--domain circuit.domain circuit.
Objective 3 : Know how to analyze the sObjective 3 : Know how to analyze the s--domain circuitdomain circuitand transform the solution back to time and transform the solution back to time domain.domain.
SUMMARYSUMMARYObjective 4 : Understand the significance of transfer Objective 4 : Understand the significance of transfer
function and be able to calculate the function and be able to calculate the transfer function from the stransfer function from the s--domain circuit.domain circuit.
Objective 5 : Know the geometrical interpretation of Objective 5 : Know the geometrical interpretation of convolution integral and be able to convolution integral and be able to calculate the integral.calculate the integral.
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SUMMARYSUMMARYObjective 6 : Know the relation between the Objective 6 : Know the relation between the phasorphasor
solution technique for finding sinusoidal solution technique for finding sinusoidal steady state solution and the ssteady state solution and the s--domain domain solution technique .solution technique .
Objective 7 : Know how to use sObjective 7 : Know how to use s--domain solution domain solution technique to solve a circuit containing technique to solve a circuit containing impulse sources or a switching circuit impulse sources or a switching circuit which may result in impulse functions.which may result in impulse functions.
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SUMMARYSUMMARY
Chapter problems : Chapter problems : 13.1313.1313.2013.2013.2713.2713.3613.3613.5713.5713.8513.8513.8813.88
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