Page 132 - Circuit Analysis II with MATLAB Applications

Chapter 4 The Laplace Transformation

Solution:

We start with the definition of the Laplace transform, that is,

f
L ft ` = Fs = ³ ft e – st dt
^
0
For this example,

f e – st f 1
L u t ^ 0 ` = ³ 1e – st dt = --------- = 0 – § © – 1 · -- - ¹ = -- s -
0 s 0 s

Thus, we have obtained the transform pair

1
u t œ --- (4.38)
0
s
for Re s^` = V ! . 0 *

Example 4.2

Find L u t `
^
1
Solution:
We apply the definition
f
^
L ft ` = Fs = ³ ft e – st dt
0
and for this example,
f
L u t ^ 1 ` = L t^` = ³ te – st dt
0
We will perform integration by parts recalling that

d
d
³ uv = uv – ³ v u (4.39)
We let
u = t and dv = e – st
then,
–
st
e –
du = 1 and v = -----------
s
By substitution into (4.39),

* This condition was established in (4.9).

4-14 Circuit Analysis II with MATLAB Applications

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