Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 13.6-2
The resistor value in Fig. 13.6-1 under Example 13.6-1 is changed to 2
W
. (i) Find the step response of 
v
o
(
t
) (ii) Determine the zero-state response of 
v
o
(
t
) if 
v
S
(
t
) 
= 
e
-
t
u
(
t
) V.
Solution
The differential equation governing the output voltage with 2
W
resistor instead of 3
W
is 
d v t
dt
dv t
dt
v t
v t
o
o
o
s
2
2
2
( )
( )
( )
( )
+
+
=
.
The System Function 
H s
V s
V s
s
s
s
o
s
( )
( )
( )
(
)
=
=
+
+
=
+
1
2
1
1
1
2
2
.
The roots of denominator polynomial are –1 and –1. The factors of the denominator polynomial are 
(
s
+ 
1) and (s
+ 
1). Therefore, the root at –1 has a multiplicity of 2.
(i) With 
v
S
(
t
) 
= 
u
(
t
) the Laplace transform of output is 
V
o
(
s
) 
= 
1
1
2
s s
(
)
+
. Expressing this in partial 
fractions, 
V s
s s
A
s
A
s
A
s
o
( )
(
)
(
)
(
)
=
+
=
+
+
+
+
1
1
1
1
2
1
2
2
3
. We can find out the residues by applying the 
expressions developed earlier in this section. Or we may proceed as below.
1
1
1
1
1
1
1
2
1
2
2
3
1
2
2
3
2
s s
A
s
A
s
A
s
A s
A s A s s
s s
(
)
(
)
(
)
(
)
(
)
(
)
+
=
+
+
+
+
=
+
+
+
+
+
==
+
+
+
+
+
+
s A
A
s A
A
A
A
s s
2
1
3
1
2
3
1
2
2
1
(
)
(
)
(
)
Now comparing the coefficients of various powers of 
s
in the numerator, we get
A
A
A
A
A
A
1
1
2
3
1
3
1 2
0
0
=
+
+
=
+
=
;
and
Solving these equations, we get, 
A
1
= 
1, 
A
2
= -
1 and 
A
3
= -
1.
∴
+
= +
−
+
+
−
+
1
1
1
1
1
1
1
2
2
s s
s
s
s
(
)
(
)
(
)
\
The step response 
v
o
(
t
) 
= 
(
) ( )
1
-
-
-
-
te
e
u t
t
t
V
(ii) With 
v
S
(
t
) 
= 
e
-
t
u
(
t
) the Laplace transform of input is 
1
1
(
)
s
+
the Laplace transform of output is
 
V s
s
o
( )
(
)
=
+
1
1
3
. There is no need for partial fractions in this case. 
v
o
(
t
) 
= 
t
e u t
t
2
2
-
( )
V.
example: 13.6-3
Find the impulse response of the circuit in Fig. 13.5-1 under Example 13.5-1 in this chapter.
Solution
The System Function was shown to be 
V s
V s
s
s
s
o
s
( )
( )
=
+ +
+
1
2
1
3
2
in Example 13.5-1. 

---

Some Theorems on Laplace Transforms 
13.19
The characteristic equation is 
s
3
+ 
s
2
+ 
2
s
+ 
1 
= 
0 and its roots are 
s
1
= -
0.215
+ 
j
1.307, 
s
2
= -
0.215
- 
j
1.307 and 
s
3
= -
0.57. There is a pair of 
complex conjugate
roots.
Impulse response is obtained by inverting the System Function.
H s
s
s
s
s
j
s
j
s
( )
(
.
.
)(
.
.
)(
. )
=
+ +
+
=
+
−
+
+
+
1
2
1
1
0 215
1 307
0 215
1 307
0 57
3
2
∴
=
+
−
+
+
+
+
+
H s
A
s
j
A
s
j
A
s
( )
(
.
.
) (
.
.
) (
. )
1
2
3
0 215
1 307
0 215
1 307
0 57
A
s
j
s
j
s
j
1
0 215
1 307
1
0 215
1 307
0 57
1
2 614
0 355
=
+
+
+
=
×
=−
+
(
.
.
)(
. )
.
( .
.
.
++
=
∠ −
°
j
1 307
0 283
164 8
.
)
.
.
A
2
will be conjugate of 
A
1
and therefore 
A
2
= 
0.283
∠
164.8
°
A
s
j
s
j
j
s
3
0 57
1
0 215
1 307
0 215
1 307
1
0 355
1 307
=
+
−
+
+
=
−
=−
(
.
.
)(
.
.
)
( .
.
.
))( .
.
)
.
0 355
1 307
0 545
+
=
j
∴
=
+
−
−
+
−
h t
e
e
e
e
j
j
t
j
( )
.
.
.
(
.
.
)
.
(
.
0 283
0 283
164 8
0 215
1 307
164 8
0 21
0
0
55
1 307
0 57
0 215
1 307 164 8
0 545
0 283
0
−
−
−
−
+
=
+
j
t
t
t
j
t
e
e
e
e
.
)
.
.
( .
. )
.
.
[
−−
−
−
−
+
=
−
( .
. )
.
.
]
.
.
cos( .
j
t
t
t
e
e
t
1 307 164 8
0 57
0 215
0
0 545
0 566
1 307
1664 8
0 545
0
0 57
. )
.
.
+
−
e
t
13.7 
some theorems on laplace transforms
The property of linearity of Laplace transforms was already noted and made use of in earlier sections. 
We look at other interesting properties of Laplace transform in this section.
13.7.1 
time-shifting theorem
If 
v 
(
t
) 
= 
f 
(
t
) 
u 
(
t
) has a Laplace transform 
V 
(
s
) then 
v
d 
(
t
) 
= 
v 
(
t 
- 
t
d 
) 
= 
f 
(
t 
- 
t
d 
) 
u 
(
t 
- 
t
d 
) has a 
Laplace transform 
V s
V s e
d
st
d
( )
( )
.
=
−
The shifting operation implied in this theorem is illustrated in Fig. 13.7-1. Note that there is a 
difference between 
f 
(
t 
- 
t
d
) 
u
(
t
) and 
f 
(
t 
- 
t
d
) 
u
(
t 
- 
t
d
).
0.5
Time
f
(
t
), 
u
(
t
)
–1
–2
1
4
3
2
1
0.5
Time
v
(
t
) = 
f
(
t
) 
u
(
t
)
–1
–2
1
4
3
2
1
0.5
Time
f
(
t –
1), 
u
(
t –
1)
–1
–2
1
4
3
2
1
v
d
(
t
) = 
v
(
t –
1) = 
f
(
t –
1)
u
(
t –
1)
0.5
Time
–1
–2
1
4
3
2
1
Fig. 13.7-1 
Illustrating the time-shift operation envisaged in shifting theorem on laplace 
transforms

---

13.20
Analysis of Dynamic Circuits by Laplace Transforms
This theorem follows from the defining equation for Laplace transforms.
V s
v t e dt
v t t e dt
d
d
st
d
st
( )
( )
(
)
=
=
−
−
∞
−
∞
−
−
∫
∫
0
0
Use variable substittution 
t
t
t
t
t
t
= −
=
=
−
+
−
∞
−
−
−
∞
∫
t t
V s
v
e
d
e
v
e
d
d
s
t
t
st
s
t
d
d
d
d
( )
( )
( )
(
)
∫∫
d
t
But 
is zero for all 
0
v
V s
e
v
e d
e
d
st
s
d
( )
.
( )
( )
t
t
t
t
t
≤
∴
=
=
−
−
−
∞
−
−
∫
0
sst
d
V s
( )
example: 13.7-1
Find the zero-state response of a series 
RC
circuit with a time constant of 2 s excited by a rectangular 
pulse voltage of 10 V height and 2 s duration starting from 
t
= 
0. The voltage across the capacitor is 
the output variable.
Solution
The differential equation governing the voltage across capacitor in a series 
RC
circuit excited by a voltage 
source is 
dv
dt
RC
v
RC
v
s
+
=
1
1
, where 
v
is the voltage across the capacitor and 
v
S
is the source voltage. 
In this case, the equation is 
dv
dt
v
v
s
+
=
0 5
0 5
.
.
. Therefore, the System Function is 
H s
s
( )
. / (
. )
=
+
0 5
0 5
.
The rectangular pulse voltage can be expressed as the sum of 10
u
(
t
) and 
-
10
u
(
t 
- 
2). 
i.e.,
v
S
= 
10[
u
(
t
) 
- 
u
(
t 
- 
2)]. Therefore, its Laplace transform is 
10 10
10 1
2
2
s
e
s
e
s
s
s
−
=
−
−
−
[
]
.
Therefore, 
V
(
s
) 
= 
0 5
0 5
.
.
s
+
× 
10 1
2
[
]
-
-
e
s
s
= 
5
0 5
1
2
s s
e
s
(
. )
[
]
+
−
−
We express 
5
0 5
s s
(
. )
+
in partial fractions as 
A
s
A
s
1
2
0 5
(
. )
+
+
and determine 
A
1 
and 
A
2 
as 
A
s
s s
A
s
s s
s
s
1
0 5
2
0
0 5
5
0 5
10
5
0 5
10
= +
×
+
= −
=
×
+
=
=−
=
(
. )
(
. )
( )
(
. )
.
and
.
∴
=
+
−
=
−
−
−
+
−
−
−
V s
s s
e
e
s
e
s
s
s
s
( )
(
. )
[
]
[
]
[
]
(
. )
5
0 5
1
10 1
10 1
0 5
2
2
2
. Inverse transform of 
-
-
10
2
e
s
s
/
is 
-
10
u
(
t 
- 
2) by Time-shifting Theorem. Similarly, inverse transform of 
−
+
−
10
0 5
2
e
s
s
/ (
. )
is –10 
e
-
0.5(
t
-
2)
u
(
t 
- 
2). Therefore, the output voltage is given by
v t
u t
u t
e
u t
e
u t
t
t
( )
[ ( )
(
)]
[
( )
(
)]
(
.
. (
)
=
−
−
−
−
−
=
−
−
−
−
10
2
10
2
10 1
0 5
0 5
2
ee
u t
e
u t
t
t
−
−
−
−
−
−
0 5
0 5
2
10 1
2
.
. (
)
) ( )
(
) (
) V

---

Some Theorems on Laplace Transforms 
13.21
Figure 13.7-2 shows the two components of 
response in dotted curves.