Inverting by Convolution
Chapter Seventy-Three
Syllabus topic Module 1, "1.3 Inverse Laplace Transform"
Pages 162 to 163 of 303
In one line
Split F into two factors you recognise, invert each, and convolve the two answers.
MU's label
"Use of Convolution Theorem". So the method is asked for by name, and a question may require it even where partial fractions would be faster.
The rule
L⁻¹{F(s) G(s)} = integrate(f(u) g(t - u), (u, 0, t))
Three steps: factorise F(s) into two recognisable pieces, invert each, and do the convolution integral.
Worked, where partial fractions would also work
F(s) = 1/(s(s - 2))
Factorise as one over s times one over (s minus 2). Those invert to 1 and to e to the 2t.
The convolution of 1 with e to the 2t is the integral of e to the 2u from 0 to t.
integrate(e^(2u), (u, 0, t)) = e^(2t)/2 - 1/2
L⁻¹{1/(s(s - 2))} = e^(2t)/2 - 1/2
Partial fractions gives the same answer in about the same time. So this is not where convolution earns its keep.
Worked, where partial fractions cannot help
F(s) = 1/(s^2 + 4)^2
Partial fractions does nothing to this: the denominator is a repeated irreducible quadratic and the fraction is already in its simplest form.
Factorise as one over (s squared plus 4) times itself. Each factor inverts to sin 2t over 2.
So the answer is the convolution of sin 2t over 2 with itself.
integrate(sin(2u) sin(2(t - u))/4, (u, 0, t)) = (sin(2t) - 2 t cos(2t))/16
L⁻¹{1/(s^2 + 4)^2} = (sin(2t) - 2 t cos(2t))/16
The integral is done by turning the product of sines into a difference of cosines, which is the identity from the trigonometric-transform chapter. The standard result is that the convolution of sin(at) with itself is (sin at minus at cos at) over 2a, so with a = 2 and the two factors of one half in front, the divisor is 16.
That is the case convolution is for, and it is examined.
Worked, a product with an exponential factor
F(s) = 1/(s(s^2 + 1))
Factorise as one over s times one over (s squared plus 1), inverting to 1 and sin t.
integrate(sin(u), (u, 0, t)) = 1 - cos(t)
L⁻¹{1/(s(s^2 + 1))} = 1 - cos(t)
Notice that convolving with 1 is the same as integrating from 0 to t, which is the integral theorem of the forward section. The two statements are the same statement.
The convolution integrals you will actually meet
Four shapes cover nearly everything, and each is worth having done once.
Two exponentials.
integrate(e^(2u) e^(3(t - u)), (u, 0, t)) = e^(3t) - e^(2t)
An exponential with 1.
integrate(e^(-u), (u, 0, t)) = 1 - e^(-t)
A sine with a cosine. Use the product-to-sum identity first.
integrate(sin(u) cos(t - u), (u, 0, t)) = t sin(t)/2
Inverting by Convolution
A sine with itself.
integrate(sin(u) sin(t - u), (u, 0, t)) = (sin(t) - t cos(t))/2
The last two produce a t multiplying a trigonometric function, which is the signature of convolving two things of the same frequency. In a physical system that is resonance, and the growing t is the amplitude building up.
Choosing between convolution and partial fractions
| Situation | Better method |
|---|---|
| the denominator factorises into distinct linear factors | partial fractions |
| a repeated irreducible quadratic | convolution |
| the question says to use the convolution theorem | convolution |
| one factor is an exponential in s | the second shifting theorem |
| you want the answer as an integral, not in closed form | convolution |
The last row is worth a sentence. Sometimes the convolution integral cannot be done in closed form at all, and then the convolution is the answer. That is perfectly respectable: an answer written as an integral is an answer, and in signal processing it is the answer that gets implemented.
The check
Transform your answer and see whether you get F(s) back. That is a complete check and it uses only the forward table.
For the repeated-quadratic example: the answer was (sin 2t minus 2t cos 2t) over 32, and its transform should be one over (s squared plus 4) squared.
L{(sin(2t) - 2 t cos(2t))/16} = 1/(s^2 + 4)^2
It is.
Check yourself
integrate(e^(-2u), (u, 0, t)) = 1/2 - e^(-2t)/2
integrate(u e^(t - u), (u, 0, t)) = e^t - t - 1
L⁻¹{1/(s(s + 2))} = 1/2 - e^(-2t)/2
L⁻¹{1/(s^2 (s - 1))} = e^t - t - 1
L⁻¹{1/(s^2 + 1)^2} = (sin(t) - t cos(t))/2
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.