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partial integral of the form 
v t
e
t
t
d
t
( )
[
cos
sin
]
≈
+
+
∫
1
2
2
2
2
2
0
0
p
s
w
w
w
s
w
w
s
w
for various values of 
s
and 
w
0
. Figure 13.2-2 shows 
the resulting waveforms for 
s
= 
0.1 and 
w
0
= 
10, 20 and 50.
1
0.5
1
–1
2
(a)
Time(s)
3
1
0.5
1
–1
2
(b)
Time(s)
3
1
0.5
1
–1
2
(c)
Time(s)
3
Fig. 13.2-2 
Partial inversion integral for unit step function for 
s
 
= 
0.1 and (a) 
w
0
= 
10 (b) 
w
0
= 
20 (c) 
w
0
= 
50
Even a short range of 10 rad/s shows the tendency of the integral to approach step waveform. With 
w
0
= 
50 rad/s the integral has more or less yielded step waveform 
- 
at least in the range –1 s to 4 s. 
We also observe the familiar Gibb’s oscillations at discontinuities. Figure 13.2-3 shows the results of 
partial evaluation of inversion integral for 
s
= 
1 and 
w
0
= 
10, 20 and 50.
This set of simulation result shows that we have to include more and more components in the 
partial integral to converge to unit step waveshape in a given time-interval as we let the components 

---

Laplace Transforms of Some Common Right-Sided Functions 
13.7
grow at a faster rate, 
i.e.,
for higher values of 
s
. And, keeping 
s
at a fixed value, we would need to 
include more and more frequency components when we increase the time-range over which we want 
convergence. However, we have infinite components at our disposal and it will be possible to include 
enough of them to recover the 
u
(
t
) shape up to any finite 
t
however large it may be.
Therefore, Laplace transform expands a transient right-sided time-function in terms of infinitely 
many complex exponential functions of infinitesimal amplitudes. The ROC of such a Laplace transform 
will include portions of right-half of 
s
-plane and hence the time-domain waveform gets constructed by 
growing complex exponential functions though that appears counter-intuitive.
1
0.5
1
–1
2
(a)
Time(s)
3
1
0.5
1
–1
2
(b)
Time(s)
3
1
0.5
1
–1
2
(c)
Time(s)
3
Fig. 13.2-3