The Convolution Theorem
Chapter Sixty-Three
Syllabus topic Module 1, "1.2 The Laplace Transform"
Pages 139 to 141 of 303
In one line
The transform of a convolution is the product of the transforms, and a convolution is not a product.
The problem it solves
The linearity chapter warned that the transform of a product is not the product of the transforms. So the natural question is: if you multiply two transforms together, what have you got the transform of?
The answer is not f times g. It is a new operation on f and g called their convolution, and the theorem that says so is MU's "Convolution Theorem".
What a convolution is
The convolution of f and g, written f star g, is defined by the integral below.
(f star g)(t) = integrate(f(u) g(t - u), (u, 0, t))
Read the integrand carefully. As u runs from 0 to t, f is evaluated forwards from 0 to t and g is evaluated backwards from t to 0. The two functions slide past each other in opposite directions, and the integral adds up all the products.
That is why it is not a product: at a given time t, the convolution depends on the whole history of both functions up to t, not just on their values at t.
The theorem
L{(f star g)(t)} = F(s) G(s)
So the correct statement is: a product of transforms corresponds to a convolution of functions.
What it means physically, which is why it exists
A linear system, given an impulse at time zero, produces some response. Call it g. Now feed the system an arbitrary input f.
Think of f as a great many impulses, one at each instant u, of size f(u). Each one produces a copy of the response g, starting at time u, scaled by f(u). At time t, the impulse that arrived at time u has been running for t minus u, so its contribution is f(u) times g(t minus u). Adding up all of them gives exactly the convolution integral.
So: the output of a linear system is the convolution of the input with the impulse response. That single sentence is the foundation of signal processing, and the convolution theorem is what turns it into a multiplication.
It is also why a digital filter is implemented as a convolution, and why the Fast Fourier Transform is used to do it: transform both, multiply, transform back, because multiplying is cheap and convolving is not.
Worked convolutions, from the definition
integrate(1 * 1, (u, 0, t)) = t
integrate(u, (u, 0, t)) = t^2/2
integrate(e^u, (u, 0, t)) = e^t - 1
The first says the convolution of 1 with 1 is t. Check it with the theorem: the transform of 1 is one over s, so the product is one over s squared, which is the transform of t.
The Convolution Theorem
L{t} = 1/s^2
The second says the convolution of t with 1 is t squared over 2, and the theorem agrees: one over s squared times one over s is one over s cubed, the transform of t squared over 2.
L{t^2/2} = 1/s^3
A more interesting one: the convolution of e to the t with 1.
integrate(e^u, (u, 0, t)) = e^t - 1
And by the theorem: one over (s minus 1) times one over s. Splitting that into partial fractions gives one over (s minus 1) minus one over s, which inverts to e to the t minus 1. The two agree.
L⁻¹{1/(s (s - 1))} = e^t - 1
The properties of convolution
These are set as short questions and each is proved by a substitution in the integral.
| Property | Statement |
|---|---|
| Commutative | f star g equals g star f |
| Associative | (f star g) star h equals f star (g star h) |
| Distributive | f star (g plus h) equals f star g plus f star h |
| Identity | f star delta equals f |
The last row is worth pausing on. The convolution identity is the impulse, not the constant function 1. That is the algebraic statement of what an impulse response is: feed a system an impulse and the output is the impulse response itself.
Commutativity is the surprising one, because the definition treats f and g so differently. The proof is the substitution v = t minus u, which swaps their roles.
The commonest misuse
L{f(t) g(t)} = F(s) G(s)
That is still false, and the convolution theorem does not rescue it. What is true is the theorem with a star, not a product, on the left.
Worked counterexample, on the simplest functions. Take f = g = t.
L{t t} = 2/s^3
The product of the transforms is one over s to the fourth, which is the transform of t cubed over 6. And t cubed over 6 is indeed the convolution of t with t.
integrate(u (t - u), (u, 0, t)) = t^3/6
L{t^3/6} = 1/s^4
So the product of the transforms corresponds to the convolution, t cubed over 6, and not to the product, t squared. Both statements are now in front of you and they are visibly different.
Check yourself
integrate(1 * 1, (u, 0, t)) = t
integrate(u^2, (u, 0, t)) = t^3/3
integrate(sin(u), (u, 0, t)) = 1 - cos(t)
L{1 - cos(t)} = 1/s - s/(s^2 + 1)
The third line of the first block is the convolution of sin t with 1, and the theorem says its transform should be one over (s squared plus 1) times one over s. Check that against the last line.
The Convolution Theorem
1/s - s/(s^2 + 1) = 1/(s(s^2 + 1))
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.