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94 CHAPTER 3 The Laplace Transform

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0
1 2 3 4 5
t
FIGURE 3.19 E out (t) in Example 3.13.

Let q j (s) be the polynomial of degree n − 1 formed by omitting the factor s − a j from q(s),
for j = 1,2,··· ,n. For example,

q 1 (s) = c(s − a 2 )···(s − a n ).
Then
n
p(a j )
−1 a j t
L [F(s)](t) = e .
q j (a j )
j=1
This is called Heaviside’s formula. In applying the formula, start with a 1 , evaluate p(a 1 ), then
substitute a 1 into the denominator with the term (s − a 1 ) removed. This gives the coefficient of
a 1 t
−1
e . Continue this with the other a j ’s and sum to obtain L [F].
Before showing why Heaviside’s formula is true, here is a simple example with
s s
F(s) = = .
2
(s + 4)(s − 1) (s − 2i)(s + 2i)(s − 1)
Here p(s) = s, and q(s) = (s − 2i)(s + 2i)(s − 1). Write a 1 = 2i, a 2 =−2i, and a 3 = 1. Then
2i −2i 1
2it
L[F(s)](t) = e + e −2it + e t
4i(2i − 1) −4i(−2i − 1) (1 − 2i)(1 + 2i)
−1 − 2i 2it −1 + 2i −2it 1 t
= e + e + e
10 10 5
1 2i 1
2it
2it
=− (e + e −2it ) − (e − e −2it ) + e t
10 10 5
1 2 1
t
=− cos(2t) + sin(2t) + e .
5 5 5
We have used the fact that
1 1
iθ
iθ
cos(θ) = (e + e −iθ ) and sin(θ) = (e − e −iθ ).
2 2i
These can be obtained by solving for cos(θ) and sin(θ) in Euler’s formulas
iθ
e = cos(θ) + i sin(θ) and e −iθ = cos(θ) − i sin(θ).

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October 14, 2010 14:14 THM/NEIL Page-94 27410_03_ch03_p77-120