Page 117 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

INTEGRAL TRANSFORMS: THE LAPLACE TRANSFORM 107
6.5.2 Repeated Roots
We now consider the case when q(s ) has a repeated root (s + a) n+1 .Then

p(s ) φ(s )
F(s ) = = n = 1, 2, 3,...
q(s ) (s − a) n+1

A a A 1 A n
= + + ··· + + H(s ) (6.57)
(s − a) (s − a) 2 (s − a) n+1
It follows that

n
φ(s ) = A 0 (s − a) + ··· + A m (s − a) n−m + ··· + A n + (s − a) n+1 H(s ) (6.58)

By letting s →a we see that A n = φ(a). To find the remaining A’s, differentiate φ (n – r)times
and take the limit as s → a.

φ (n−r) (a) = (n − r)!A r (6.59)

Thus

n (n−r)
φ (a) 1

F(s ) = + H(s ) (6.60)
(n − r)! (s − a) r+1
r=0
If the inverse transform of H(s ) (the part containing no repeated roots) is h(t) it follows from
the shifting theorem and the inverse transform of 1/s m that
n (n−r)
φ (a) r at
f (t) = t e + h(t) (6.61)
(n − r)!r!
r=0
Example 6.8. Inverse transform with repeated roots

s A 0 A 1 A 2 C
F(s ) = = + + +
(s + 2) (s + 1) (s + 2) (s + 2) 2 (s + 2) 3 (s + 1)
3
3
Multiply by (s + 2) .
s 2 C(s + 2) 3
= A 0 (s + 2) + A 1 (s + 2) + A 2 + = φ(s )
(s + 1) (s + 1)
Take the limit as s →−2,

A 2 = 2