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Properties of Laplace Transform

denominator polynomial; the division is carried forward until the numerator

polynomial of the remainder is one degree less than the denominator. This
results in a polynomial in s plus a proper fraction. The proper fraction can be
expanded into a partial fraction expansion. The result of such an expansion is an
expression of the form

A 1 A 2 A p1 A p2 A pr
′
F (s) = B +B s+· · ·+ + +· · ·+ + +· · ·+ .
1
0
s − s 1 s − s 2 s − s p (s − s ) 2 (s − s ) r
p
p
(2.13)
This expression has been written in a form to show three types of terms;
polynomial, simple partial fraction including all terms with distinct roots, and
partial fraction appropriate to multiple roots.
To find the constants A , A , . . . the polynomial terms are removed, leaving the
1
2
proper fraction
′
F (s) − (B + B s + · · · ) = F(s) (2.14)
1
0
where
A A A A A A
F(s) = 1 + 2 + · · · k + p1 + p2 + · · · + pr .
s − s 1 s − s 2 s − s k s − s p (s − s ) 2 (s − s ) r
p
p
To find the constants A that are the residues of the function F(s) at the simple
k
poles s , it is only necessary to note that as s → s the term A (s−s ) will become
k
k
k
k
large compared with all other terms. In the limit
A = lim (s − s )F(s). (2.15)
k
k
s→s k
Upon taking the inverse transform for each simple pole, the result will be a simple
exponential of the form

( )
a
−1 k s k t
L = A e . (2.16)
k
s − s k
Note also that because F(s) contains only real coefficients, if s is a complex pole
k
with residue A , there will also be a conjugate pole s with residue A . For such
∗
∗
k
k
k
complex poles
( )
A A ∗
∗
−1 k k s k t ∗ s t
k
L + = A e + A e .
k
k
s − s k s − s ∗
k
These can be combined in the following way:
response = (a + ib )e (σ k +iω k )t + (a − ib )e (σ k −ib k )t =
k
k
k
k
e σ k t ((a + ib )(cos ω t + i sin ω t) + (a − ib )(cos ω t + i sin ω t)) =
k
k
k
k
k
k
k
k
= 2e σ k t (a cos ω t − b sin ω t) = 2A e σ k t cos(ω t + θ ) (2.17)
k
k
k
k
k
k
k
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