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Properties of Laplace Transform
.

2
1 { d } d [ s + 3s + 1 ]
2
A 11 = s F (s) = =
p
1! ds ds s + 1
s=0 s=0
2
s + 3s + 1 2s + 3
= + = 4.
(s + 1) 2 s + 1 s=0
Therefore,
4 1 1
F(s) = 1 + + − .
s s 2 s + 1
From Table of Laplace transforms the inverse transform is

−t
f(t) = δ(t) + 4 + t − e , for t > 0.
. . . .
If the function F(s) exists in proper fractional form as the quotient of two

polynomials, we can employ the Heaviside expansion theorem in the
determination of f(t) from F(s). This theorem is an efficient method for finding
the residues of F(s). Let

P(s) A 1 A 2 A k
F(s) = = + + · · · +
Q(s) s − s 1 s − s 2 s − s k

where P(s) and Q(s) are polynomials with no common factors and with the

degree of P(s) less than the degree of Q(s).
Suppose that the factors of Q(s) are distinct constants. Then, as in (2.15) we
find
[ ]
s − s k
A = lim P(s)
k
s→s k A(s)
Also, the limit P(s) is P(s ). Now, because
k
s − s k 1 1
lim = lim = ,
(1)
s→s k Q(s) s→s k Q (s) Q (s )
(1)
k
then
P(s )
k
A = .
k
(1)
Q (s )
k
Thus,
k
P(s) ∑ P(s ) 1
n
F(s) = = · (2.20)
(1)
Q(s) Q (s ) s − s n
n=1 n
From this, the inverse transformation becomes
{ } k
P(s) ∑ P(s )
−1 n s n t
f(t) = L = e
(1)
Q(s) Q (s )
n=1 n
This is the Heaviside expansion theorem. It can be written in formal form.

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