Proving the Convolution Theorem, and Using It Forwards
Chapter Sixty-Four
Syllabus topic Module 1, "1.2 The Laplace Transform"
Pages 142 to 144 of 303
In one line
The proof reverses the order of a double integral, and the forward direction of the theorem computes transforms that no other rule reaches.
The proof
Write the product of the two transforms as a double integral.
F(s) is the integral over one variable, say v, of e to the minus sv times f(v). G(s) is the integral over another, say w, of e to the minus sw times g(w). Multiplying them gives a double integral over the whole quarter plane where v and w are both non-negative, of e to the minus s(v plus w) times f(v) times g(w).
Now change variables. Put t = v plus w and keep v, so w is t minus v. The Jacobian of that change is 1, so the area element is unchanged.
The region matters and is where the work is. In the (v, w) quarter plane, v and w are both at least zero. In the (v, t) variables that becomes: v at least zero, and t at least v, since w = t minus v must be non-negative. So for a given t, v runs from 0 to t, and t runs from 0 to infinity.
Reverse the order so that t is the outer variable.
The inner integral over v, from 0 to t, of f(v) times g(t minus v), is exactly the convolution. The outer integral over t of e to the minus st times that convolution is the defining integral of its transform.
So the product of the transforms is the transform of the convolution, which is the theorem.
The one step worth practising is drawing the region and reading off the new limits. Getting v from 0 to t rather than 0 to infinity is the whole of it, and a sketch of the wedge between the two lines settles it.
Using it forwards: a transform no other rule gives
The forward direction is the half students never practise, and it is set.
Find the transform of the integral below.
integrate(sin(u) cos(t - u), (u, 0, t))
That is the convolution of sin t with cos t. By the theorem, its transform is the product of their transforms.
L{sin(t)} = 1/(s^2 + 1)
L{cos(t)} = s/(s^2 + 1)
So the answer is s over (s squared plus 1) squared, with no integration at all.
(1/(s^2 + 1))(s/(s^2 + 1)) = s/(s^2 + 1)^2
And the convolution itself, if you want it, can be done by a trigonometric identity and gives t sin t over 2.
integrate(sin(u) cos(t - u), (u, 0, t)) = t sin(t)/2
Check that against the table: the transform of t sin at is 2as over (s squared plus a squared) squared, so with a = 1 the transform of t sin t over 2 is s over (s squared plus 1) squared. The two routes agree.
Proving the Convolution Theorem, and Using It Forwards
L{t sin(t)/2} = s/(s^2 + 1)^2
More worked convolutions
integrate(e^u (t - u), (u, 0, t)) = e^t - t - 1
integrate(e^(2u) e^(3(t - u)), (u, 0, t)) = e^(3t) - e^(2t)
integrate(cos(u), (u, 0, t)) = sin(t)
The second is the convolution of two exponentials, and it is worth knowing as a pattern: convolving e to the at with e to the bt gives (e to the bt minus e to the at) over (b minus a) when a and b differ. Here b minus a is 1, so there is no divisor.
Check it with the theorem.
L⁻¹{1/((s - 2)(s - 3))} = e^(3t) - e^(2t)
The transforms multiply to one over (s minus 2)(s minus 3), and inverting that by partial fractions gives the same answer. Two routes, one answer, again.
Using it backwards: inverting
The reverse reading is the subject of its own chapter in the inverse section, but the idea in one line: if F(s) factorises into two pieces you recognise, the inverse is the convolution of their two inverses.
L⁻¹{1/(s(s^2 + 1))} = 1 - cos(t)
The factors are one over s, inverting to 1, and one over (s squared plus 1), inverting to sin t. Convolving 1 with sin t is the integral of sin u from 0 to t, which is 1 minus cos t.
integrate(sin(u), (u, 0, t)) = 1 - cos(t)
When to reach for it
Honestly, not often in this paper. Partial fractions is usually faster. Reach for convolution in three situations.
The factors do not split into partial fractions usefully. One over (s squared plus 1) squared is the classic case: partial fractions does nothing to it, and convolution gives the answer.
One factor is an exponential in s, meaning a delayed input, and you want the answer as an integral rather than in pieces.
The question asks for it. MU's label is "Use of Convolution Theorem", so a question may require the method by name even where another would be quicker.
Worked, the classic case.
L⁻¹{1/(s^2 + 1)^2} = (sin(t) - t cos(t))/2
By convolution: both factors invert to sin t, so the answer is the convolution of sin t with itself.
integrate(sin(u) sin(t - u), (u, 0, t)) = (sin(t) - t cos(t))/2
The integral is done by turning the product of sines into a difference of cosines, and the result is the printed answer. No partial fraction would have got there.
Check yourself
integrate(1 * (t - u), (u, 0, t)) = t^2/2
integrate(e^(-u) e^(-(t - u)), (u, 0, t)) = t e^(-t)
Proving the Convolution Theorem, and Using It Forwards
L{t e^(-t)} = 1/(s + 1)^2
The second convolution is e to the minus t with itself, and the theorem says the transform should be one over (s plus 1) squared, which the last line confirms. Notice that convolving a function with itself does not square it: it produced a t out of nowhere.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.