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Properties of Laplace Transform
If we define g(t) = 0 for t < 0, then g(t − τ) = 0 for t < τ and we can write (2.21)
as
∫ ∫
∞ ∞
L(f(t)) · L(g(t)) = e −st f(τ)g(t − τ)dtdτ.
0 0
Due to the hypotheses on f and g, the Laplace integrals of f and g converge absolutely
and hence, in view of the preceding calculation,
∫ ∫
∞ ∞
|e −st f(τ)g(t − τ)|dtdτ
0 0
converges. This fact allows us to reverse the order of integration, ∗ so that
∫ ∫
∞ ∞
L(f(t)) · L(g(t)) = e −st f(τ)g(t − τ)dτdt =
0 0
∫ (∫ )
∞ t
= e −st f(τ)g(t − τ)dτ dt =
0 0
∫ (∫ )
∞ t
= e −st f(τ)g(t − τ)dτ dt = L[(f ∗ g)(t)]. 2
0 0
. . . .
.
Example 2.38,
1
at
bt
L(e ∗ e ) = .
(s − a)(s − b)
Moreover,
( ) ∫ t
at
1 e − e bt
−1 at bt aτ b(t−τ)
L (s − a)(s − b) = e ∗ e = e e dτ = a − b , a ̸= b.
. 0
. . . . .
Example 2.39, Find
( )
1
−1
L .
2
s (s − 1)
.
. . . . .
Solution. Previously we applied a partial fraction decomposition. But we
can also write
1 1 1
= · ,
2
s (s − 1) s 2 s − 1
2
t
where L(t) = 1/s , L(e ) = 1/(s − 1). By the convolution theorem,
1 1 t
· = L(t ∗ e ),
s 2 s − 1
and thus
( )
1
−1 t t
L = t ∗ e = e − t − 1.
. . . . s (s − 1)
2
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