Convolution.
We have seen that they Laplace transform has the linearity property:
L(af (t) + bg(t)) = aL(f (t)) + bL(g(t)).
I am sure that all of you at this stage has wondered if it is true that:
L(f (t)g(t)) = L(f (t))L(g(t)).
I am afraid that it is not true!!
Counter Example: It is an example to to show that the property that you
consider is not verified:
Suppose f (t) = et and g(t) = 1. Then f (t)g(t) = et that has Laplace transform
L(f (t)g(t)) = L(et ) =
1
s−1
(1)
However:
L(f (t))L(g(t)) = L(et )L(1) =
1
s(s − 1)
(2)
Clearly equation (1) is different to equation (2)! What a tragedy! :o|
We cannot leave it like that! We must find a function that is constructed from
f and g, h say, that verifies:
L(h(t)) = L(f (t))L(g(t)).
And it will be called convolution of f with g.
We will define a new operation between functions denoted by ∗ (convolution).
(We know already some operations between functions: sum, product, quotient
for example).
Z
h(t) = f ∗ g(t) =
t
f (τ )g(t − τ )dτ
0
1
Be careful!!! In the integral above t acts as a constant and we have to integrate
with respect τ. It can be proven (not for us though, but you can check it in the
book Advanced Engineering Mathematics (Erwin Kreyszig)) that:
L(f ∗ g (t)) = L(f (t))L(g(t))
I am sure you are thinking ” but what for all this? ” well, it can simplify some
calculations with inverses of Laplace transforms whose partial fraction decomposition is not easy to do... Let us compute some convolutions to have an idea
of how they work, and after we will apply to inverses of Laplace transform and
ODE’s:
Example 1 Compute t ∗ t:
t∗t =
Rt
=
Rt
0
0
τ (t − τ )dτ
τ t − τ 2 dτ
(recall that t is like a constant and τ is the variable of integration)
h 2
3 iτ =t
= t τ2 − τ3
τ =0
3
3
3
= t2 − t3 = t6
Example 2 Compute eat ∗ ebt if a 6= b:
eat ∗ ebt =
Rt
=
Rt
=
=
=
=
=
0
eaτ eb(t−τ ) dτ
eaτ ebt e−bτ dτ, using ex+y = ex ey
Rt
ebt 0 eaτ e−bτ dτ, because t behaves like a constant⇒ebt behaves as a constant too
Rt
ebt 0 e(a−b)τ dτ
h
iτ =t
ebt 1 e(a−b)τ
a−b
τ =0
bt 1
(a−b)t
e
e
− ebt 1
a−b
a−b
1 (eat − ebt )
a−b
0
Example 3 What is L−1 (
1
)?
(s − 3)(s + 5)
2
We have:
1
1
1
= s−
3s+5
(s − 3)(s + 5)
= L(e3t )L(e−5t )
= L(e3t ∗ e−5t )
So to find the inverse transform of
1
we need to compute the con(s − 3)(s + 5)
volution e3t ∗ e−5t :
L−1 (
1
) = L−1 (L(e3t ∗ e−5t )) = e3t ∗ e−5t
(s − 3)(s + 5)
But using Example 2 we conclude:
L−1 (
1
) = 81 (e3t − e−5t )
(s − 3)(s + 5)
Note: You could have computed the partial fraction decomposition of
1
(s − 3)(s + 5)
and obtain the same result.
Example 4 Compute the inverse transform of
1
:
s(s + a)
1
1 1 = L(e−at )L(1) = L(e−at ∗ 1)
= s+
as
s(s + a)
Therefore:
L−1 (
h
i
Rt
1
1 e−aτ τ =t = 1 (1 − e−at )
) = e−at ∗ 1 = 0 e−aτ 1dτ = −a
a
s(s + a)
τ =0
Example 5 In this example we will find the f such that
L(f (t)) =
s
2
s +1
2
2
s
Thus, we want to compute the inverse transform of 2
s +1
Using the Convolution Theorem. In fact, applying the theorem is the easy bit
L(f (t)) = [L(cos t)]2 = L(cos t ∗ cos t)
3
(3)
and so f (t) = cos t ∗ cos t. Now we need to work out the convolution:
Z
t
cos τ cos (t − τ )dτ
cos t ∗ cos t =
0
We can then use the formula
2 cos A cos B = cos (A + B) + cos (A − B)
which can be found in the beginning of the notes. Then:
1
cos t ∗ cos t =
2
t
Z
1
cos tdτ +
2
0
Z
t
cos (2τ − t)dτ
0
The first integral is easy to compute, since t is constant with respect τ then cos t
is constant as well. Therefore:
1
2
Z
0
t
Z
1
cos tdτ = cos t
2
t
dτ =
0
1
1
[τ cos t]ττ =t
=0 = t cos t
2
2
To solve the second integral we use the following change ot variables
2τ − t = x ⇒
dx
= dτ
2
the limits of integration change as well:
τ = t ⇒ x = 2t − t = t,
τ = 0 ⇒ x = −t
Then:
Z
0
τ
1
cos(2τ − t)dτ =
2
Z
t
1
cos xdx =
sin x
2
−t
x=t
= sin t,
x=−t
because sin t = − sin(−t). Therefore:
Rt
Rt
cos t ∗ cos t = 21 0 cos tdτ + 12 0 cos (2τ − t)dτ
= 21 t cos t + 21 sin t
1
, ω constant? give it a try! (it is
(s2 + ω 2 )2
rather similar to Example 28). The solution is 1 3 (sin ωt − ωt cos ωt).
2ω
What is the inverse transform of
4