Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

Partial inversion integral for step function for 
s
 
= 
1 and (a) 
w
0
= 
10 (b) 
w
0
= 
20 
(c) 
w
0
= 
50 
13.3 
laplace transforms of some common rIght-sIded functIons
Integral of sum of two functions is sum of integrals of each function. Thus, Laplace transformation is 
a linear operation. If 
v
1
(
t
) and 
v
2
(
t
) are two right-sided functions and 
a
1
and 
a
2 
are two real numbers, 
then, 
a v t
a v t
a V s
a V s
1 1
2 2
1 1
2 2
( )
( )
( )
( )
+
⇔
+
is a Laplace transform pair. This is called 
Property of 
Linearity of Laplace transforms. 
Now we work out the
 
Laplace transforms for many commonly 
used right-sided functions using the defining integral and property of linearity.
Let 
v t
e u t
s t
o
( )
( )
=
be a right-sided complex exponential function with a complex frequency of 
s
o
. 
Then,
V s
e e dt
e
dt
e
s s
s
s t
st
s s t
s s t
o
o
o
o
( )
(
)
(
(
)
(
)
=
=
=
−
=
−
−
−
∞
−
− −
∞
− −
∞
∫
∫
0
0
0
1
−−
>
s
s
o
o
)
Re( )
with ROC
s
Therefore, 
e u t
s s
s t
o
o
( )
/ (
)
⇔
−
1
is a Laplace transform pair with ROC Re(
s
) > Re(
s
o
).
The special case of 
v
(
t
) 
= 
u
(
t
) is covered by this transform pair with 
s
o
= 
0.
Therefore, 
u t
s
( )
/
⇔
1 is a Laplace transform pair with ROC Re(
s
) > 0.
The special case of 
v
(
t
) 
= 
cos
w
o
t
u
(
t
) is covered by expressing 
v
(
t
) as 
= 
(
) /
e
e
j
t
j
t
o
o
w
w
+
−
2 by 
employing Euler’s formula and then applying property of linearity of Laplace transforms.
∴
=
−
+
+
=
+
V s
s
j
s
j
s
s
o
o
o
( )
.
.
0 5
0 5
2
2
w
w
w
Therefore, cos
( )
w
w
o
o
t u t
s
s
⇔
+
2
2
is a Laplace transform pair with ROC Re(
s
) > 0.

---

13.8
Analysis of Dynamic Circuits by Laplace Transforms
Similarly, sin
( )
w
w
w
o
o
o
t u t
s
⇔
+
2
2
is a Laplace transform pair with ROC Re(
s
) > 0.
Consider 
v
(
t
) 
= 
e
-
a
t
cos
b
t
u
(
t
). This can be expressed as 
= 
[
] /
(
)
(
)
e
e
j
t
j
t
− +
− −
+
a
b
a
b
2 by Euler’s 
formula. Then,
∴
=
+ −
+
+ +
=
+
+
+
V s
s
j
s
j
s
s
( )
.
.
(
)
(
)
0 5
0 5
2
2
a
b
a
b
a
a
b
Therefore, 
e
t u t
s
s
t
−
⇔
+
+
+
a
b
a
a
b
cos
( )
(
)
(
)
2
2
is a Laplace transform pair with ROC of Re(
s
) > 
-
a
. 
Similarly, 
e
t u t
s
t
−
⇔
+
+
a
b
b
a
b
sin
( )
(
)
2
2
is a Laplace transform pair with ROC of Re(
s
) > 
-
a
.
Now consider 
v t
e
e
s
u t
s
s t
s t
o
o
( )
( )
(
)
=
−
+∆
∆
The Laplace transform of this function can be found from the defining integral as 
V s
s s s
s
s s
s s
s s s
o
o
o
o
( )
(
)(
)
=
− −
+
−





 =
− −
−
1
1
1
1
∆
∆
∆
Now we send 
v
(
t
) to a limit as 
D
s
→
0.
lim ( )
lim
( )
lim
(
)
(
)
∆
∆
∆
∆
∆
∆
s
s
s
s t
s t
s
s
s t
v t
e
e
s
u t
e
e
o
o
o
→
→
+
→
+
=
−
=
−
0
0
0
ss t
s t
o
o
s
u t
te u t
∆




=
( )
( )
Therefore, Laplace transform of 
te u t
s t
o
( )
= 
lim
(
)(
) (
)
∆
∆
s
o
o
o
s s
s s s
s s
→
− −
−
=
−
0
2
1
1
Therefore, 
te u t
s s
s t
o
o
( )
/ (
)
⇔
−
1
2
is a Laplace transform pair with ROC Re(
s
) > Re(
s
o
).
The special case of 
v
(
t
) 
= 
t
u
(
t
) is covered by this transform pair with 
s
o
= 
0.
Therefore, 
tu t
s
( )
/
⇔
1
2
is a Laplace transform pair with ROC Re(
s
) > 0.
And finally, we consider 
v
(
t
) 
= 
d
(
t
). 
V s
t e dt
t e dt
st
( )
( )
( )
=
=
=
−
∞
−
−
+
∫
∫
d
d
0
0
0
0
1 . Thus, 
d
( )
t
⇔
1 is a 
Laplace transform pair with ROC of entire 
s
-plane. It requires all complex exponential functions with 
equal intensity to synthesise an impulse function in time-domain.
These commonly used Laplace transform pairs are listed in Table 13.3-1. Some of them have been 
derived in this section. Others will be taken up later.

---

The 
s
-Domain System Function 
H 
(
s
) 
13.9
table 13.3-1
Basic Laplace Transform Pairs
Time-Function
Laplace Transform
Region of Convergence
d
(
t
) 
1
Entire 
s-
plane
u
(
t
) 
1
s
Re(
s
) > 0
e u t
s t
o
( )
1
s s
o
-
Re(
s
) > Re(
s
o
)
e
u t
j
t
o
w
( )
1
s
j
o
-
w
Re(
s
) > 0
e u t
t
-
a
( )
1
s
+
a
Re(
s
) > 
-
a
cos
( )
w
o
t u t
s
s
o
2
2
+
w
Re(
s
) > 0
sin
( )
w
o
t u t
w
w
o
o
s
2
2
+
Re(
s
) > 0
e
t u t
t
-
a
b
cos
( )
(
)
(
)
s
s
+
+
+
a
a
b
2
2
Re(
s
) > 
-
a
e
t u t
t
-
a
b
sin
( )
b
a
b
(
)
s
+
+
2
2
Re(
s
) > 
-
a
tu t
( )
1
2
s
Re(
s
) > 0
t u t n
n
( ),
, ,...
=
1 2
n
s
n
!
+
1
Re(
s
) > 0
te u t
s t
o
( )
1
2
(
)
s s
o
-
Re(
s
) > Re(
s
o
)
t e u t n
n s t
o
( ),
, ,....
=
1 2
n
s s
o
n
!
(
)
−
+
1
Re(
s
) > Re(
s
o
)
13.4 
the 
s
-domaIn system functIon 
H
(
s
)
We saw in Section 13.1 that when an input 
e
st
is applied to a linear time-invariant circuit described by 
an 
n
th
-order differential equation 
d y
dt
a
d
y
dt
a
dy
dt
a y
b
d x
dt
b
d
x
dt
n
n
n
n
n
m
m
m
m
m
m
+
+ +
+
=
+
−
−
−
−
−
−
1
1
1
1
0
1
1
1
++ +
+
b
dx
dt
b x
1
0

---

13.10
Analysis of Dynamic Circuits by Laplace Transforms
the response is given by 
H
(
s
)
e
st
where
H s
Y s
X s
b s
b
s
b s b
s
a s
a s a
m
m
m
m
o
n
n
n
o
( )
( )
( )
=
=
+
+ +
+
+
+ +
+
−
−
−
−
1
1
1
1
1
1
H
(
s
) in this context is the ratio of complex amplitude of forced response component in output to 
the complex amplitude of input complex exponential function with a complex frequency of 
s
. There is 
only forced response in this context and forced response itself is the total response.
In Section 13.2, we observed that a right-sided function 
x
(
t
) can be expressed as a sum of infinitely 
many complex exponential functions of frequency between 
s
- 
j
∞
to 
s
+ 
j
∞
with the line Re(
s
) 
= 
s
falling within the ROC of Laplace transform of 
x
(
t
). We combine these two facts along with 
superposition principle to arrive at the zero-state response of a linear time-invariant circuit to a right-
sided input function.
Consider a particular value of complex frequency 
s
and a small band of complex frequency 
D
s
centered on it. This band contributes complex exponential functions of frequencies between 
(
s 
- 
0.5
D
s
) and (
s
+ 
0.5
D
s
). For sufficiently small 
D
s
, we may take all these complex exponential 
functions to be evolving approximately at the centre frequency of the band, 
i.e.,
at 
s
itself. In that case, 
all the infinitesimal contributions coming from this band may be consolidated into a signal 
≈
X
(
s
) 
D
s
e
st
.
This single complex frequency component with complex amplitude of 
X
(
s
) 
D
s
will produce a total 
response component of 
H
(
s
) 
X
(
s
) 
D
s
e
st
in the output. We get the zero-state response of the circuit by 
adding all such contributions over the line Re(
s
) 
= 
s
falling within the ROC of 
X
(
s
) and sending the 
sum to a limit by making 
D
s
→
0. The result will be the following integral.
∴
=
− ∞
+ ∞
∫
y t
H s X s e ds
j
j
st
( )
( ) ( )
s
s
(13.4-1) 
Compare Eqn. 13.4-1 with the synthesis equation of Laplace transform given by Eqn. 13.2-2. It 
is evident that Eqn. 13.4-1 is the synthesis equation of the Laplace transform 
H
(
s
)
X
(
s
). But then, a 
synthesis equation which returns 
y
(
t
) must be synthesising it from the Laplace transform 
Y
(
s
) of the 
time-function 
y
(
t
). Therefore, 
Y
(
s
) 
= 
H
(
s
)
X
(
s
). This important result requires restatement.
The Laplace transform of zero-state response 
= 
Laplace transform of input source 
function 
×
Ratio of complex amplitude of forced response to the complex amplitude of 
input complex exponential function at a complex frequency of 
s
.
Now comes a definition. 
The ratio of Laplace transform of zero-state response to Laplace transform 
of input source function is defined as the s-domain System Function.
And these two are seen to be the 
same.
\
The 
s
-domain System Function, 
H s
Y
s
X s
b s
b
s
b s b
s
a s
a s a
m
m
m
m
o
n
n
n
( )
( )
( )
=
=
+
+ +
+
+
+ +
+
−
−
−
−
zsr
1
1
1
1
1
1
oo
(13.4-2) 
Note carefully that System Function is independent of initial conditions in the circuit since it is the 
zero-state response to a right-sided input that is employed in its definition. This function is also called 
a 
Transfer Function
when both 
x
and 
y
are similar quantities, 
i.e.,
when 
x
and 
y
are voltages or 
x
and 
y
are currents and is denoted by 
T
(
s
). It is called an 
Input Impedance Function
and is denoted by 
Z
i
(
s
) 

---

Poles and Zeros of System Function and Excitation Function 
13.11
if 
y 
is the voltage across a terminal pair and 
x
is the current entering the positive terminal. It is called 
an 
Input Admittance Function
and is denoted by 
Y
i
(
s
) if 
y 
is the current into a terminal pair and 
x
is 
the voltage across the terminal pair. These two, 
i.e.,
Z
i
(
s
) and 
Y
i
(
s
), together are at times referred to as 
immittance functions
.
If the quantities 
x
and 
y
are voltage–current or current–voltage pair and they refer to different 
terminal pairs in the circuit, we call the 
s
-domain System Function a 
Transfer Impedance Function
or 
Transfer Admittance Function
, as the case may be. They are represented by 
Z
m
(
s
) and 
Y
m
(
s
), respectively.
We have an expression for 
H
(
s
) as a ratio of rational polynomials in 
s
in Eqn. 13.4-2. Rational 
polynomials are polynomials containing only integer powers of the independent variable. However, 
there is another more interesting interpretation possible for 
H
(
s
).
Let us try to find the impulse response of the circuit by this transform technique. We remember that 
‘impulse response’ means ‘zero-state response to unit impulse input’ by definition. Hence, we can use 
the System Function to arrive at the Laplace transform of impulse response as 
H
(
s
)
X
(
s
). But 
x
(
t
) 
= 
d
(
t
) 
and therefore 
X
(
s
) 
= 
1. Hence, for a linear time-invariant circuit, the following statement holds.
Laplace transform of impulse response 
= 
s-
domain System Function, and, Impulse 
response 
= 
inverse Laplace transform of 
s
-domain system function This result was 
anticipated in naming the System Function as 
H 
(
s
).
Once the System Function and Laplace transform of input source function are known, one can 
obtain the Laplace transform of zero-state response by inverting the product of input transform and 
System Function. We will take up the task of inverting Laplace transforms in later sections.