Page 114 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

104 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
But

∞

e −s τ G(s ) = e −st g(t − τ)dt (6.48)
t=0
so that

∞ t

F(s )G(s ) = e −st f (τ)g(t − τ)dτ dt (6.49)
t=0 τ=0
where we have used the fact that g(t − τ) = 0 when τ> t. The inverse transform of F(s )G(s )
is

t

−1
L [F(s )G(s )] = f (τ)g(t − τ)dτ (6.50)
τ=0

6.5 PARTIAL FRACTIONS
In the example differential equation above we determined two roots of the polynomial in the
denominator, then separated the two roots so that the two expressions could be inverted in
forms that we already knew. The method of separating out the expressions 1/(s + 1) and
1/(s + 3) is known as the method of partial fractions. We now develop the method into a more
user friendly form.

6.5.1 Nonrepeating Roots
Suppose we wish to invert the transform F(s ) = p(s )/q(s ), where p(s )and q(s ) are polynomi-
als. We first note that the inverse exists if the degree of p(s ) is lower than that of q(s ). Suppose
q(s ) can be factored and a nonrepeated root is a.
φ(s )
F(s ) = (6.51)
s − a
According to the theory of partial fractions there exists a constant C such that

φ(s ) C
= + H(s ) (6.52)
s − a s − a
Multiply both sides by (s − a) and take the limit as s → a and the result is

C = φ(a) (6.53)