Chapter 13
The Laplace Transform in Circuit
Analysis
13.1
13.2-3
13.4-5
13.6
13.7
13.8
Circuit Elements in the s Domain
Circuit Analysis in the s Domain
The Transfer Function and Natural Response
The Transfer Function and the Convolution
Integral
The Transfer Function and the SteadyState Sinusoidal Response
The Impulse Function in Circuit Analysis
1
Key points

How to represent the initial energy of L, C in the
s-domain?

Why the functional forms of natural and steadystate responses are determined by the poles of
transfer function H(s) and excitation source X(s),
respectively?

Why the output of an LTI circuit is the
convolution of the input and impulse response?
How to interpret the memory of a circuit by
convolution?
2
Section 13.1
Circuit Elements in the s
Domain
1.
Equivalent elements of R, L, C
3
A resistor in the s domain

iv-relation in the time domain:
v (t )  R  i (t ).

By operational Laplace transform:
Lv(t )  LR  i (t )  R  Li (t ) ,
 V ( s )  R  I ( s ).

Physical units: V(s) in volt-seconds, I(s) in
ampere-seconds.
4
An inductor in the s domain

iv-relation in the time domain:
d
v (t )  L  i (t ).
dt

By operational Laplace transform:
Lv (t )  LL  i(t )  L  Li(t ) ,
 V ( s )  L  sI ( s )  I 0   sL  I ( s )  LI 0 .
initial current
5
Equivalent circuit of an inductor

Series equivalent:

Parallel equivalent:
Thévenin 
Norton
6
A capacitor in the s domain

iv-relation in the time domain:
d
i (t )  C  v (t ).
dt

By operational Laplace transform:
Li (t )  LC  v(t )  C  Lv(t ) ,
 I ( s )  C  sV ( s )  V0   sC  V ( s )  CV0 .
initial voltage
7
Equivalent circuit of a capacitor

Parallel equivalent:

Series equivalent:
Norton 
Thévenin
8
Section 13.2, 13.3
Circuit Analysis in the s
Domain
1.
2.
3.
4.
5.
6.
Procedures
Nature response of RC circuit
Step response of RLC circuit
Sinusoidal source
MCM
Superposition
9
How to analyze a circuit in the s-domain?
1.
Replacing each circuit element with its s-domain
equivalent. The initial energy in L or C is taken
into account by adding independent source in
series or parallel with the element impedance.
2.
Writing & solving algebraic equations by the
same circuit analysis techniques developed for
resistive networks.
3.
Obtaining the t-domain solutions by inverse
Laplace transform.
10
Why to operate in the s-domain?

It is convenient in solving transient responses of
linear, lumped parameter circuits, for the initial
conditions have been incorporated into the
equivalent circuit.

It is also useful for circuits with multiple essential
nodes and meshes, for the simultaneous ODEs
have been reduced to simultaneous algebraic
equations.

It can correctly predict the impulsive response,
which is more difficult in the t-domain (Sec. 13.8).
11
Nature response of an RC circuit (1)

Q: i(t), v(t) = ?

Replacing the charged capacitor by a Thévenin
equivalent circuit in the s-domain.

KVL,  algebraic equation & solution of I(s):
V0
I
CV0
V0 R

 IR,  I ( s ) 

.
1
s sC
1  RCs s  ( RC )
12
Nature response of an RC circuit (2)

The t-domain solution is obtained by inverse
Laplace transform:
 V0 R  V0 t ( RC ) 1  1 
i (t )  L 
L  
 e
1 
s
 s  ( RC )  R
V0 t ( RC )
 e
u(t ).
R
1

i(0+) = V0/R, which is true for vC(0+) = vC(0-) = V0.

i() = 0, which is true for capacitor becomes
open (no loop current) in steady state.
13
Nature response of an RC circuit (3)

To directly solve v(t), replacing the charged
capacitor by a Norton equivalent in the s-domain.

Solve V(s), perform inverse Laplace transform:
V0
V
.
CV0  sCV  ,  V ( s ) 
1
s  ( RC )
R


 v (t )  L1V0 s  ( RC ) 1   V0e t ( RC )u(t )  Ri (t ).
14
Step response of a parallel RLC (1)

Q: iL(t) = ?
iL(0-) = 0
vC (0-) = 0
15
Step response of a parallel RLC (2)

KCL,  algebraic equation & solution of V(s):
I dc
V V
I dc C
 sCV   ,  V ( s )  2
.
1
1
s
R sL
s  ( RC ) s  ( LC )

Solve IL(s):
V ( s)
I dc ( LC ) 1
 2
I L ( s) 
sL
s s  ( RC ) 1 s  ( LC ) 1



3.84  107
s s  (6.4  10 ) s  (1.6  10 )
2
4
9


.
16
Step response of a parallel RLC (3)

Perform partial fraction expansion and inverse
Laplace transform:
24
20127
20  127
I L ( s) 


(mA  s).
s s  ( 32k  j 24k ) s  ( 32k  j 24k )

iL (t )  24u(t )  20e
j127 ( 32 k) t
e


e j ( 24 k ) t u(t )  c.c.

 24  40e ( 32 k)t cos ( 24k)t  127  u(t ) (mA)
 24  e ( 32 k)t 24cos(24k )t  32sin(24k )t  u(t ) (mA).
17
Transient response due to a sinusoidal source (1)

For a parallel RLC circuit, replace the current
source by a sinusoidal one: ig (t )  I mcost  u(t ).
The algebraic equation changes:
V V
sI m
sCV  
 Ig  2
,
2
R sL
s 

I m C s 2
 V ( s)  2
,
2
2
1
1
s    s  ( RC ) s  ( LC )


V
I m ( LC ) 1 s
 I L ( s) 
 2
.
2
2
1
1


sL s    s  ( RC ) s  ( LC )


18
Transient response due to a sinusoidal source (2)

Perform partial fraction expansion and inverse
Laplace transform:
K1
K1*
K2
K 2*
.
I L ( s) 



s  j s  j s  (   j ) s  (   j )
Driving
frequency
Neper
frequency
Damped
frequency
iL (t )  2 K1 cost  K1   2 K 2 e t cost  K 2  u(t ).
Steady-state
response (source)
Natural response (RLC
parameters)
19
Step response of a 2-mesh circuit (1)

Q: i1(t), i2(t) = ?
i1 (0-) = 0
i2(0-)
=0
20
Step response of a 2-mesh circuit (2)

MCM,  2 algebraic equations & solutions:
336

 (1)
8.4 sI1  42( I1  I 2 ) 
s

42( I 2  I1 )  (10s  48) I 2  0 ( 2)
 42   I1  336
42  8.4 s

   

90  10s   I 2   0
  42
s
.

1 
15 14
 42 
 I1  42  8.4 s
336 s   s  s  2  s  12 

.


 I     42


90  10s 
 0   7  8.4  1.4 
 2 
 s s  2 s  12 
1
21
Step response of a 2-mesh circuit (3)

Perform inverse Laplace transform:
i1 (t )  15  14e
i2 (t )  7  8.4e
2 t
2 t
e
 1.4e
12 t
12 t
336
u(t )  (42 // 48)  15 A.
42
u(t )  15 42  48  7 A.
22
Use of superposition (1)

Given 2 independent sources vg, ig and initially
charged C, L,  v2(t) = ?
23
Use of superposition: Vg acts alone (2)
1
2
 1
Vg

1
 V1  Vg V1 V1  V2


 R  sL  ( sC ) 1  0,  R  sL  sC V1  sCV2  R ,
1
 1


1



 sCV    1  sC V   0.
V2  V1  V2  0.
1
 2
R
 ( sC ) 1 R2


 2

24
Use of superposition (3)

For convenience, define admittance matrix:
1
1



sC

sC
 R sL
 V1
 1
 
1

V
2



 sC
 sC
R2


Y11 Y12  V1 Vg R1 

   
.


Y12 Y22  V2  0 
 Y12 R1
 V2 
V.
2 g
Y11Y22  Y12
25
Use of superposition: Ig acts alone (4)
1
2
Y11 Y12  V1  0 
Y11
Y Y   V    I ,  V2  Y Y  Y 2 I g .
 12 22   2   g 
11 22
12
Same matrix
Same denominator
26
Use of superposition: Energized L acts alone (5)
1
Y11 Y12  V1  
Y Y   V    0
 12 22   2  
Same matrix
2
s
Y12 s
,  V2 Y Y  Y 2  .

11 22
12
Same denominator
27
Use of superposition: Energized C acts alone (6)
1

2
""
Y
Y
 11 12  V1   C 
Y Y    ""    C ,

 12 22  V2  
 (Y11  Y12 )C
""
.
 V2 
2
Y11Y22  Y12
The total voltage is: V2  V2  V2 V2 V2"" .
28
Section 13.4, 13.5
The Transfer Function and
Natural Response
29
What is the transfer function of a circuit?

The ratio of a circuit’s output to its input in the
s-domain:
Y ( s)
H ( s) 
X (s)

A single circuit may have many transfer
functions, each corresponds to some specific
choices of input and output.
30
Poles and zeros of transfer function

For linear and lumped-parameter circuits, H(s)
is always a rational function of s.

Poles and zeros always appear in complex
conjugate pairs.

The poles must lie in the left half of the s-plane
if bounded input leads to bounded output.


Im
Re
31
Example: Series RLC circuit
input

If the output is the loop current I:
I
sC
1

 2
H ( s) 
.
1
Vg R  sL  ( sC )
s LC  sRC  1

If the output is the capacitor voltage V:
( sC ) 1
1
V
.
H ( s) 

 2
1
Vg R  sL  ( sC )
s LC  sRC  1
32
How do poles, zeros influence the solution?

Since Y(s) = H(s) X(s),  the partial fraction
expansion of the output Y(s) yields a term K/(s-a)
for each pole s = a of H(s) or X(s).

The functional forms of the transient (natural)
and steady-state responses ytr(t) and yss(t) are
determined by the poles of H(s) and X(s),
respectively.

The partial fraction coefficients of Ytr(s) and Yss(s)
are determined by both H(s) and X(s).
33
Example 13.2: Linear ramp excitation (1)

Q: vo(t) = ?
50tu(t)
50/s2
34
Example 13.2 (2)

Only one essential node,  use NVM:
Vo  Vg
Vo
Vo

 6  0,
1000 250  0.05s 10 s
Vo
1000( s  5000)
 H ( s) 
 2
.
7
Vg s  6000s  2.5  10 

H(s) has 2 complex conjugate poles:
s  3000  j 4000.

Vg(s) = 50/s2 has 1 repeated real pole: s = 0(2).
35
Example 13.2 (3)

The total response in the s-domain is:
Vo ( s )  H ( s )Vg ( s ) 
5  104 ( s  5000)
s 2 s 2  6000s  2.5  107 
 Ytr  Yss
expansion coefficients depend on H(s) & Vg(s)
5 5  104 80 5 5  104   80 10 4  104
.


 2
s  3000  j 4000 s  3000  j 4000 s
s
poles of H(s): -3k  j4k

pole of Vg(s): 0(2)
The total response in the t-domain:
vo (t )  ytr  yss 
 5 10

3 3, 000 t
e


cos(4,000t  80 ) u (t )
 10t  4 10  4 u (t ).
36
Example 13.2 (4)
Steady state
component yss(t)
Total response
 = 0.33 ms, impact of ytr(t)
37
Section 13.6
The Transfer Function and
the Convolution Integral
1.
2.
3.
4.
Impulse response
Time invariant
Convolution integral
Memory of circuit
38
Impulse response

If the input to a linear, lumped-parameter circuit
is an impulse (t), the output function h(t) is
called impulse response, which happens to be
the natural response of the circuit:
X ( s )  L (t )  1, Y ( s )  H ( s )  1  H ( s ),
y (t )  L1Y ( s )  L1H ( s )  h(t ).

The application of an impulse source is
equivalent to suddenly storing energy in the
circuit. The subsequent release of this energy
gives rise to the natural response.
39
Time invariant

For a linear, lumped-parameter circuit, delaying
the input x(t) by  simply delays the response y(t)
by  as well (time invariant):
X ( s, )  Lx (t   )u(t   )  e s X ( s ),
Y ( s, )  H ( s ) X ( s, )  e s H ( s ) X ( s )  e sY ( s ),
y (t , )  L1Y ( s, )  L1Y ( s )
t t 
 y (t   )u(t   ).
40
Motivation of working in the time domain

The properties of impulse response and timeinvariance allow one to calculate the output
function y(t) of a “linear and time invariant (LTI)”
circuit in the t-domain only.

This is beneficial when x(t), h(t) are known only
through experimental data.
41
Decompose the input source x(t)

We can approximate x(t) by a series of
rectangular pulses rec(t-i) of uniform width :
sq  (t )

sq  (t  i )
By having ,  rec(t-i)/ (t-i), x(t)
converges to a train of impulses:

x(t )   x(i )  lim rec  (t  i )
i 0
 0

  x(i )     (t  i ).
i 0
42
Synthesize the output y(t) (1)

Since the circuit is LTI:
 (t )  h(t ),

 (t  i )  h(t  i ),
 a x (t )  a y (t );
i i
 i i

  x (i )   (t  i )
i 0

  x (i )  h(t  i ).
i 0
43
Synthesize the output y(t) (2)

As , summation  integration:
if x(t) extends (-, )


y (t )   x ( )h(t   )d   x ( )h(t   )d .

0

By change of variable u = t-, 

y (t )   x (t  u )h(u )du.


The output of an LTI circuit is the convolution of
input and the impulse response of the circuit:
y (t )  x (t )  h (t )




  x ( )h(t   )d   x (t   )h( )d .
44
Convolution of a causal circuit

For physically realizable
circuit, no response can
occur prior to the input
excitation (causal),  {h(t)
= 0 for t < 0}.

Excitation is turned on at t
= 0,  {x(t) = 0 for t < 0}. 
y (t )  x (t )  h (t )
t
  x (t   )h( )d .
0
45
Effect of x(t) is weighted by h(t)

The convolution integral
t
y (t )   x (t   )h( )d
0
shows that the value of y(t)
is the weighted average of
x(t) from t = 0 to t = t [from 
= t to  = 0 for x(t-)].

0
t
If h(t) is monotonically decreasing, the highest
weight is given to the present x(t).
46
Memory of the circuit

If h(t) only lasts from t
= 0 to t = T, the
convolution integral
T
t
y (t )   x (t   )h( )d .
0
implies that the circuit
has a memory over a
finite interval t = [t-T,t].

If h(t) = (t), no memory, output at t only depends
on x(t),  y(t) = x(t)*(t) = x(t), no distortion.
47
Example 13.3: RL driven by a trapezoidal source (1)

Q: vo(t) = ?
1
Vo
1
Vo 
Vi ,  H ( s )  
.
s 1
Vi s  1
 1  t
 h (t )  L 
  e u(t ).
 s  1
1
48
Example 13.3 (2)
t
vo (t )   vi (t   )h( )d .

Separate into 3 intervals:
0
49
Example 13.3 (3)

Since the circuit has certain memory, vo(t) has
some distortion with respect to vi(t).
50
Section 13.7
The Transfer Function and
the Steady-State Sinusoidal
Response
51
How to get sinusoidal steady-state response by H(s)?

In Chapters 9-11, we used phasor analysis to
get steady-state response yss(t) due to a
sinusoidal input x (t )  A cost   .

If we know H(s), yss(t) must be:
yss (t )  H ( j ) A cost     ( ),
where H ( j )  H ( s ) s  j  H ( j ) e j ( ) .

The changes of amplitude and phase depend
on the sampling of H(s) along the imaginary axis.
52
Proof
x (t )  A cost     A cos  cos t  A sin  sin t.
X ( s )  A cos 

s cos    sin 
s

A
sin


A
.
2
2
2
2
2
2
s 
s 
s 
s cos    sin 
Y ( s)  H ( s) X ( s)  H ( s)  A
 Ytr ( s )  Yss ( s ),
2
2
s 
K1
K1*

where Yss ( s) 
, K1  Y ( s )( s  j ) s  j
s  j s  j
s cos    sin 
 H ( s) A
s  j
s  j
j cos    sin  H ( j ) Ae j
 H ( j ) A

.
2 j
2
j (  )
j


H
(
j

)
e
Ae
1
yss (t )  L 
  c.c.  A H ( j ) cost     ( ).
2( s  j )


53
Obtain H(s) from H( j)

We can reverse the process: determine H( j)
experimentally, then construct H(s) from the
data (not always possible).

Once we know H(s), we can find the response to
other excitation sources.
54
Section 13.8
The Impulse Function
in Circuit Analysis
55
E.g. Impulsive inductor voltage (1)

Q: vo(t) = ?
i1(0-)=10 A

i2(0-)=0
The opening of the switch forces the two
inductor currents i1, i2 change immediately by
inducing an impulsive inductor voltage [v = Li'(t)].
56
E.g. Equivalent circuit & solution in the s-domain (2)
initial current
V0  (100 s  30)
V0

 0,
10  3s
15  2s
improper rational
2(6s 2  65s  150)
60 10
 V0 ( s ) 
.
 12  
s ( s  5)
s s5
57
E.g. Solutions in the t-domain (3)
60 10 

5 t
v0 (t )  L 12 


12

(
t
)

(
60

10
e
)u(t ).

s s  5

1

To verify whether this solution vo(t) is correct, we
need to solve i(t) as well.
100 s  30
4
2
I (s) 
 
,  i (t )  (4  2e 5t )u (t ).
10  3s  15  2s s s  5
jump
jump
58
Impulsive inductor voltage (4)

The jump of i2(t) from 0 to 6 A causes i2 (t )  6 (t ) ,
contributing to a voltage impulse L2i2 (t )  12 (t ).

After t > 0+,
vo (t )  (15)i2 (t )  ( 2 H )i2 (t )
 15( 4  2e 5t )  2( 10e 5t )  60  10e 5t ,
consistent with that solved by Laplace transform.
59
Key points

How to represent the initial energy of L, C in the
s-domain?

Why the functional forms of natural and steadystate responses are determined by the poles of
transfer function H(s) and excitation source X(s),
respectively?

Why the output of an LTI circuit is the
convolution of the input and impulse response?
How to interpret the memory of a circuit by
convolution?
60
Practical Perspective
Voltage Surges
61
Why can a voltage surge occur?

Q: Why a voltage surge is created when a load
is switched off?

Model: A sinusoidal voltage source drives three
loads, where Rb is switched off at t = 0.

Since i2(t) cannot change abruptly, i1(t) will jump
by the amount of i3(0-),  voltage surge occurs.
62
Example

Let Vo = 1200 (rms), f = 60 Hz, Ra = 12 , Rb = 8
, Xa = 41.1  (i.e. La = Xa/ = 109 mH), Xl = 1  (i.e.
Ll = 2.65 mH). Solve vo(t) for t > 0-.

To draw the s-domain circuit, we need to
calculate the initial inductor currents i2(0-), i0(0-).
63
Steady-state before the switching

The three branch currents (rms phasors) are:
I1 = Vo/Ra = (1200)/(12 ) = 100 A,
I2 = Vo/(jXa) = (1200)/(j41.1 ) = 2.92-90 A,
I3 = Vo/Rb = (1200)/(8 ) = 150 A,

The line current is: I0 = I1 + I2 + I3 = 25.2-6.65 A.

Source voltage: Vg = Vo + I0(jXl) = 125-11.5 V.

The two initial inductor currents at t = 0- are:
 i2(t) = 2.92(2)cos(120t-90),
 i2(0-) = 0;
 i0(t) = 25.2(2)cos(120t-6.65),
 i0(0-) = 35.4 A.
64
S-domain analysis

The s-domain circuit is:
(I0 = 35.4 A)
Vg =
12511.5
V (rms)

By NVM:
(Ll = 2.65 mH)
(12 )
Vo  Ll I 0  Vg
sLl
(La = 109
mH)
Vo Vo


 0,
Ra sLa
Ra Ll Vg ( s)  I 0 Ra
253
Vo 

s  Ra ( La  Ll ) La Ll  s  1475
866.85 86  6.85


s  j120
s  j120
65
Inverse Laplace transform
253
866.85 86  6.85


,
 Given Vo ( s ) 
s  1475 s  j120
s  j120


 vo (t )  253e 1475t  173 cos(120t  6.85 )  u (t ).
66