Page 138 - Circuit Analysis II with MATLAB Applications

Chapter 4 The Laplace Transformation

n – at n!
t e u t œ ------------------------- (4.58)
0
s + a n + 1

n
where is a positive integer, and V ! – a Thus, for n = 1 , we get the transform pair
1
te – at u t œ ------------------ (4.59)
0
s + a 2

for V ! – . a

For n = 2 , we get the transform

2 – at 2!
t e u t œ ------------------ (4.60)
0
s + a 3

and in general,

n – at n!
t e u t œ ------------------------- (4.61)
0
s + a n + 1

for V ! – a
Example 4.8

Find L sin^ Ztu t ` 0
Solution:

f a
L sin^ Ztu t ` 0 ³ sin Z = t e – st dt = ³ lim sin Zt e – st dt

0 a o f 0
and from tables of integrals *

ax asin bx bcos bx
–
e
ax
³ e sin bxdx = ------------------------------------------------------
2
2
a + b
Then,
1
* This can also be derived from sin Zt = ----- e 1 jZt – e – jZt , and the use of (4.55) where e – at u t œ ----------- . By the lin-
j2 0 s + a
earity property, the sum of these terms corresponds to the sum of their Laplace transforms. Therefore,
1
Z
1 ·
1 §
L sin Ztu t = ----- -------------- – -------------- = ----------------- 2
@
>
0
©
jZ ¹
j2 sjZ–
s +
2
s +
Z
4-20 Circuit Analysis II with MATLAB Applications
Orchard Publications