Page 105 - Advanced engineering mathematics

3.3 Shifting and the Heaviside Function 85

THEOREM 3.3 First Shifting Theorem

For any number a,
at
L[e f (t)](s) = F(s − a). (3.4)

This conclusion is also called shifting in the s variable. The proof is a straightforward appeal
to the definition:
∞

at −st at
L[e f (t)](s) = e e f (t)dt
0
∞

= e −(s−a) f (t)dt = F(s − a).
0
EXAMPLE 3.4
2
at
2
We know from the table that L[cos(bt)]= s/(s + b ) = F(s). For the transform of e cos(bt),
replace s with s − a to get
s − a
at .
L[e cos(bt)](s) =
2
(s − a) + b 2
EXAMPLE 3.5
3
4
Since L[t ]= 6/s , then
6
3 7t .
L[t e ](s) =
(s − 7) 4
Every formula for the Laplace transform of a function is also a formula for the inverse
Laplace transform of a function. The inverse version of the first shifting theorem is
at
−1
L [F(s − a)]= e f (t). (3.5)
EXAMPLE 3.6
Compute

4
−1
L .
2
s + 4s + 20
The idea is to manipulate the given function of s to the form F(s −a) for some F and a. Then we
can apply the inverse form of the shifting theorem, which is equation (3.5). Complete the square
in the denominator to write
4 4
= = F(s + 2)
2
s + 4s + 20 (s + 2) + 16
2
if
4
F(s) = .
2
s + 16
From the table, F(s) has inverse f (t) = sin(4t). By equation (3.5),

4
−1
L
s + 4s + 20
2
−1
= L [F(s + 2)]
= e −2t f (t) = e −2t sin(4t).
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October 14, 2010 14:14 THM/NEIL Page-85 27410_03_ch03_p77-120