Page 108 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

98 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
This is a function that is shifted in the s -domain and hence Eq. (6.6) is applicable. Noting that
3
−1
2
2
L (1/s ) = t / (3) = t /2 from Eq. (6.16)
t 2
f (t) = e t
2
Or we could use Eq. (6.17) directly.
Example 6.2. Find the inverse transform of the function

3
F(s ) = e −s
2
s + 4
The inverse transform of

2
F(s ) =
2
s + 4
is, according to Eq. (6.11)

3
f (t) = sin(2t)
2
The exponential term implies shifting in the time domain by 1. Thus

f (t) = 0, t < 1
3
= sin[2(t − 1)], t > 1
2
Example 6.3. Find the inverse transform of
s
F(s ) =
2
(s − 2) + 1
The denominator is shifted in the s -domain. Thus we shift the numerator term and write F(s )
as two terms
s − 2 2
F(s ) = +
2
(s − 2) + 1 (s − 2) + 1
2
Equations (6.6), (6.12), and (6.13) are applicable. The inverse transform of the first of these is
a shifted cosine and the second is a shifted sine. Therefore each must be multiplied by exp(2t).
The inverse transform is

2t
2t
f (t) = e cos(t) + 2e sin(t)