Title: Default for 20.5 The Convolution Theorem
Course: HELM Workbooks 2023-2024

Convolution

Let \(f(t)\) and \(g(t)\) be two functions of \(t\). The convolution of \(f(t)\) and \(g(t)\) is also a function of \(t\), denoted by \((f*g)(t)\) and is defined by the relation \[(f*g)(t)=\int_{-\infty}^{\infty}f(t-x)g(x)\,dx\] However if \(f\) and \(g\) are both causal functions then (strictly) \(f(t), g(t)\) are written \(f(t)u(t)\) and \(g(t)u(t)\) respectively, so that \[(f*g)(t)= \int_{-\infty}^{\infty}f(t-x)u(t-x)g(x)u(x)\,dx =\int_0^tf(t-x)g(x)\,dx\] because of the properties of the step functions: \(u(t-x)=0\) if \(x>t\) and \(u(x)=0\) if \(x<0\).

Key Point

Convolution

If \(f(t)\) and \(g(t)\) are causal functions then their convolution is defined by: \[(f*g)(t)=\int_0^tf(t-x)g(x)\,dx\]

This is an odd looking definition but it turns out to have considerable use both in Laplace transform theory and in the modelling of linear engineering systems. The reader should note that the variable of integration is \(x\). As far as the integration process is concerned the \(t\)-variable is (temporarily) regarded as a constant.

Example

Find the convolution of \(f\) and \(g\) if \(f(t)=tu(t)\) and \(g(t)=t^2u(t)\).

Solution

\[f(t-x)=(t-x)u(t-x)\quad \text{ and}\quad g(x)=x^2u(x)\] Therefore \[\begin{eqnarray*} (f*g)(t)=\int_0^t(t-x)x^2\,dx &=& \left[\tfrac{1}{3}x^3 t- \tfrac{1}{4}x^4\ \right]_0^t\\ &=& \tfrac{1}{3}t^4 - \tfrac{1}{4} t^4= \tfrac{1}{12}t^4\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{eqnarray*}\]

Example

Find the convolution of \(f(t)=t.u(t)\) and \(g(t)=\sin t.u(t)\).

Solution

Here \(f(t-x)=(t-x)u(t-x)\) and \(g(x)=\sin x.u(x)\) and so \[(f*g)(t)=\int_0^t(t-x)\sin x \,dx\] We need to integrate by parts. We find, remembering again that \(t\) is a constant in the integration process, \[\begin{eqnarray*} \int_0^t(t-x)\sin x\,dx&=&\biggl [-(t-x)\cos x\biggr ]_0^t-\int_0^t(-1)(-\cos x)\,dx\qquad\qquad\qquad\qquad\\ &=&\left[0+t\right]-\int_0^t\cos x\,dx\\&=&t-\biggl[\sin x\biggr ]_0^t=t-\sin t \end{eqnarray*}\] so that
\(\qquad\displaystyle (f*g)(t)=t-\sin t\quad\) or, equivalently, in this case \(\quad (t*\sin t)(t)=t-\sin t\).

Exercise

In the previous Example we found the convolution of \(f(t)=t.u(t)\) and \(g(t)=\sin t.u(t)\).
In this Exercise you are asked to find the convolution \((g*f)(t)\) that is, to reverse the order of \(f\) and \(g\).

1. Begin by writing \((g*f)(t)\) as an appropriate integral:

\[(g*f)(t) = \int\] |  |  | \(\mathrm{d}x\)

2. Now evaluate the convolution integral:
\((g*f)(t) = \)
Feedback: 1. We have that, \(g(t-x)=\sin(t-x).u(t-x)\) and \(f(x)=xu(x)\), so \[ (g*f)(t)=\int_0^t\sin(t-x).x\,dx\] 2. Evaluating the integral,we get: \[\begin{eqnarray*} (g*f)(t)&=&\int_0^t\sin (t-x).x\,dx\\&=&\biggl [x\cos (t-x)\biggr]_0^t-\int_0^t\cos (t-x)\,dx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\ &=&\left[t-0\right]+\left[\phantom{\dfrac{1}{1}}\!\!\!\sin(t-x)\right]_0^t=t-\sin t \end{eqnarray*}\]

This Task illustrates the general result in the following Key Point:

Key Point

Commutativity Property of Convolution

\[(f*g)(t)=(g*f)(t)\] In words: the convolution of \(f(t)\) with \(g(t)\) is the same as the convolution of \(g(t)\) with \(f(t)\).

Example

Obtain the Laplace transforms of \(f(t)=t.u(t)\) and \(g(t)=\sin t.u(t)\) and \((f*g)(t)\).

Solution

Begin by finding \({\cal L}\{f(t)\}\), \({\cal L}\{g(t)\}\):

\[{\cal L}\{f(t)\} ={1\over {s^2}}\qquad {\cal L}\{g(t)\}={1\over {s^2+1}}\]

Now find \({\cal L}\{(f*g)(t)\}\):

From the previous Example \((f*g)(t)=t-\sin t\) and so \[ {\cal L}\{(f*g)(t)\}={\cal L}\{t-\sin t\}={1\over {s^2}}-{1\over {s^2+1}}\]

Now compare \({\cal L}\{f(t)\}\times {\cal L}\{g(t)\}\) with \({\cal L}\{f*g(t)\}\). What do you observe?

\[{\cal L}\{(f*g)(t)\}={1\over {s^2}}-{1\over {s^2+1}}={1\over {s^2}}\left({1\over {s^2+1}}\right) ={\cal L}\{f(t)\}{\cal L}\{g(t)\} =F(s)G(s)\] We see that the Laplace transform of the convolution of \(f(t)\) and \(g(t)\) is the product of their separate Laplace transforms. This, in fact, is a general result which is expressed in the statement of the convolution theorem which we discuss in the next subsection.

The convolution theorem

Let \(f(t)\) and \(g(t)\) be causal functions with Laplace transforms \(F(s)\) and \(G(s)\) respectively, i.e. \({\cal L}\{f(t)\}=F(s)\) and \({\cal L}\{g(t)\}=G(s)\). Then it can be shown that

Key Point

The Convolution Theorem

\[{\cal L}^{-1}\{F(s)G(s)\}=(f*g)(t)\qquad\text{ or equivalently}\qquad {\cal L}\{(f*g)(t)\}=F(s)G(s)\]

Example

Find the inverse transform of \(\displaystyle{{6\over {s(s^2+9)}}}\).

1. Using partial fractions
2. Using the convolution theorem.

Solution

1. \(\quad\displaystyle {6\over {s(s^2+9)}}={(2/3)\over s}-{(2/3)s\over{s^2+9}}\) and so \[\begin{eqnarray*} {\cal L}^{-1}\left\{{6\over {s(s^2+9)}}\right\}= \tfrac{2}{3}{\cal L}^{-1}\left\{{1\over s}\right\}- \tfrac{2}{3}{\cal L}^{-1}\left\{{s\over {s^2+9}}\right\}= \tfrac{2}{3}u(t)-\tfrac{2}{3}\cos 3t.u(t) \end{eqnarray*}\]
2. Let us choose \(\quad\displaystyle F(s)={2\over s}\qquad\text{ and}\qquad G(s)={3\over {s^2+9}}\) then \[f(t)={\cal L}^{-1}\{F(s)\}=2u(t)\quad \text{ and}\quad g(t)={\cal L}^{-1}\{G(s)\}=\sin 3t.u(t)\] So \[\begin{eqnarray*} {\cal L}^{-1}\{F(s)G(s)\}&=&(f*g)(t)\qquad \text{(by the convolution theorem)}\qquad\qquad\qquad\qquad\qquad\\&=&\int_0^t2u(t-x)\sin 3x. u(x)\,dx \end{eqnarray*}\] Now the variable \(t\) can take any value from \(-\infty\) to \(+\infty\). If \(t<0\) then the variable of integration, \(x\), is negative and so \(u(x)=0\). We conclude that \[(f*g)(t)=0\quad \text{ if}\quad t<0\] that is, \((f*g)(t)\) is a causal function. Let us now consider the other possibility for \(t\), that is the range \(t\ge 0\). Now, in the range of integration \(0\le x\le t\) and so \[u(t-x)=1\qquad u(x)=1\] since both \(t-x\) and \(x\) are non-negative. Therefore \[\begin{eqnarray*} {\cal L}^{-1}\{F(s)G(s)\}&=&\int_0^t2\sin 3x\,dx\\&=&\left[\phantom{\dfrac{1}{1}}\!\!\!-\dfrac{2}{3}\cos 3x \right]_0^t=-\dfrac{2}{3}(\cos 3t -1)\qquad t\ge 0\qquad\qquad\qquad\qquad\qquad \end{eqnarray*}\] Hence \[{\cal L}^{-1}\{{6\over {s(s^2+9)}}\}=-\dfrac{2}{3}(\cos 3t -1)u(t)\] which agrees with the value obtained above using the partial fraction approach.

Example

Use the convolution theorem to find the inverse transform of \(\displaystyle{H(s)={s\over {(s-1)(s^2+1)}}}\).

Solution

Begin by choosing two functions of \(s\), that is, \(F(s)\) and \(G(s)\):

Although there are many possibilities it would seem sensible to choose \[ F(s)={{1\over {s-1}}}\qquad \text{and} \qquad G(s)={{s\over {s^2+1}}}\] since, by inspection, we can write down their inverse Laplace transforms: \[\qquad\displaystyle f(t)={\cal L}^{-1}\{F(s)\}=\text{e}^tu(t)\qquad\text{ and}\qquad g(t)={\cal L}^{-1}\{G(s)\}=\cos t.u(t)\]

Now construct the convolution integral:

\[\begin{eqnarray*} h(t)&=&{\cal L}^{-1}\{H(s)\}\\&=&{\cal L}^{-1}\{F(s)G(s)\}\\&=&\int_0^tf(t-x)g(x)\,dx=\int_0^t\text{e}^{t-x}u(t-x)\cos x.u(x)\,dx\qquad\qquad\qquad\qquad\qquad\qquad \end{eqnarray*}\]

Now complete the evaluation of the integral, treating the cases \(t<0\) and \(t\ge 0\) separately:

You should find \(h(t)={1\over 2}(\sin t-\cos t +\text{e}^t)u(t)\) since \(h(t)=0\) if \(t<0\) and \[\begin{eqnarray*} h(t)&=&\int_0^t\text{e}^{t-x}\cos x\,dx\qquad \text{ if}\quad t\ge 0\\&=&\biggl[\text{e}^{t-x}\sin x\biggr]_0^t-\int_0^t(-1)\text{e}^{t-x}\sin x\,dx\qquad \text{(integrating by parts)}\qquad\qquad\qquad\qquad \\ &=&\sin t+\biggl[- \text{e}^{t-x}\cos x \biggr]_0^t-\int_0^t(-\text{e}^{t-x})(-\cos x)\,dx\\&=&\sin t-\cos t +\text{e}^t-h(t)\\\text{ or}\qquad 2h(t)&=&\sin t-\cos t+\text{e}^t\qquad t\ge 0 \end{eqnarray*}\] Finally \(h(t)={1\over 2}(\sin t-\cos t+\text{e}^t)u(t)\).

Exercise

Find the convolution of

1. \(2tu(t)\) and \(t^3u(t)\):
2. \(\text{e}^tu(t)\) and \(tu(t)\):
3. \(\text{e}^{-2t}u(t)\) and \(\text{e}^{-t}u(t)\):

In each case reverse the order to check that \((f*g)(t)=(g*f)(t)\).

Exercise

Use the convolution theorem to determine the inverse Laplace transforms of

1. \(\displaystyle{{1\over {s^2(s+1)}}}\):
2. \(\displaystyle{{1\over {(s-1)(s-2)}}}\):
3. \(\displaystyle{{1\over {(s^2+1)^2}}}\):