The Convolution Theorem/Derivatives & Integrals of Transforms
If we take the Laplace transform of 𝑥 ′′ + 𝑥 = cos 𝑡 , 𝑥 (0) = 𝑥 ′ (0) = 0 we
get:
𝑠
(𝑠 2 𝑋(𝑠) − 𝑠𝑥 (0) − 𝑥 ′ (0)) + 𝑋(𝑠) =
𝑠 2 +1
𝑠
(𝑠 2 + 1)𝑋(𝑠) =
𝑠2 +1
𝑠 1
𝑋(𝑠) = ( ) ( ) = ℒ(cos 𝑡) ℒ(sin 𝑡)
𝑠2 +1 𝑠2 +1
Unfortunately, ℒ(cos 𝑡) ℒ(sin 𝑡) ≠ ℒ(cos 𝑡 ∙ sin 𝑡)
1 1
= ℒ( sin 2𝑡) = .
2 𝑠2 +4
However, given 𝐻 (𝑠) = 𝐹(𝑠)𝐺(𝑠) there is a function ℎ(𝑡) such that,
ℒ(ℎ(𝑡)) = 𝐻 (𝑠) = 𝐹 (𝑠)𝐺 (𝑠).
This function is:
𝑤=𝑡
ℎ(𝑡) = ∫𝑤=0 𝑓(𝑤)𝑔(𝑡 − 𝑤) 𝑑𝑤
where ℒ(𝑓 (𝑡)) = 𝐹(𝑠) and ℒ(𝑔(𝑡)) = 𝐺 (𝑠).
We call ℎ(𝑡) the convolution of 𝑓 (𝑡) and 𝑔(𝑡) and write it as:
𝑤=𝑡
ℎ(𝑡) = 𝑓 ∗ 𝑔(𝑡) = ∫𝑤=0 𝑓(𝑤)𝑔(𝑡 − 𝑤)𝑑𝑤
and ℒ(𝑓(𝑡) ∗ 𝑔(𝑡)) = [ℒ(𝑓(𝑡))] ∙ [ℒ(𝑔(𝑡))].
, 2
Ex. When we took the Laplace transform of 𝑥 ′′ + 𝑥 = cos 𝑡, where
𝑥 (0) = 𝑥 ′ (0) = 0 we got:
𝑠 1
𝑋(𝑠) = ( )(𝑠2+1) = ℒ(cos 𝑡) ∙ ℒ(sin 𝑡).
𝑠2 +1
Thus, ℒ (cos 𝑡 ∗ sin 𝑡) = ℒ(cos 𝑡) ∙ ℒ(sin 𝑡) and 𝑥 (𝑡) = cos 𝑡 ∗ sin 𝑡
(we will calculate this shortly).
Ex. Let’s calculate (𝑓 ∗ 𝑔)(𝑡) when:
𝑓(𝑡) = 5 for 0 ≤ 𝑡 ≤ 2 and 0 otherwise
𝑔(𝑡) = 10 for 4 ≤ 𝑡 ≤ 7 and 0 otherwise.
𝑤=𝑡
(𝑓 ∗ 𝑔)(𝑡) = ∫𝑤=0 𝑓 (𝑤)𝑔(𝑡 − 𝑤)𝑑𝑤 .
Let’s graph 𝑓 (𝑤 ), 𝑔(𝑤 ), 𝑔(−𝑤 ), and 𝑔(𝑡 − 𝑤 )
𝑔(−𝑤) 𝑔(𝑤)
𝑓(𝑤)