2 The convolution theorem

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

Example 5

Find the inverse transform of $\frac{6}{s\left(s^2+9\right)}$ .

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

Solution

1. $\frac{6}{s\left(s^2+9\right)}=\frac{\left(2∕3\right)}{s}-\frac{\left(2∕3\right)s}{s^2+9}$ and so $$\begin{array}{rcll}{\mathcal{L}}^{-1}\left\{\frac{6}{s\left(s^2+9\right)}\right\}=\frac{2}{3}{\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}-\frac{2}{3}{\mathcal{L}}^{-1}\left\{\frac{s}{s^2+9}\right\}=\frac{2}{3}u\left(t\right)-\frac{2}{3}cos3t.u\left(t\right)& & & \end{array}$$
2. Let us choose $F\left(s\right)=\frac{2}{s}\text{ and}G\left(s\right)=\frac{3}{s^2+9}$ then

   $f\left(t\right)={\mathcal{L}}^{-1}\left\{F\left(s\right)\right\}=2u\left(t\right)\text{ and}g\left(t\right)={\mathcal{L}}^{-1}\left\{G\left(s\right)\right\}=sin3t.u\left(t\right)$

   So

   $$\begin{array}{rcll}{\mathcal{L}}^{-1}\left\{F\left(s\right)G\left(s\right)\right\}& =& \left(f\ast g\right)\left(t\right)\text{(by the convolution theorem)}& \\ & =& {\int \; <!--nolimits\; -->}_0^{t}2u\left(t-x\right)sin3x.u\left(x\right)dx& \end{array}$$

   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 $u\left(x\right)=0$ . We conclude that

   $\left(f\ast g\right)\left(t\right)=0\text{ if}t<0$

   that is, $\left(f\ast g\right)\left(t\right)$ 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

   $u\left(t-x\right)=1u\left(x\right)=1$

   since both $t-x$ and $x$ are non-negative. Therefore

   $$\begin{array}{rcll}{\mathcal{L}}^{-1}\left\{F\left(s\right)G\left(s\right)\right\}& =& {\int \; <!--nolimits\; -->}_0^{t}2sin3xdx& \\ & =& {\left[-\frac{2}{3}cos3x\right]}_0^t=-\frac{2}{3}\left(cos3t-1\right)t\ge 0& \end{array}$$

   Hence

   ${\mathcal{L}}^{-1}\left\{\frac{6}{s\left(s^2+9\right)}\right\}=-\frac{2}{3}\left(cos3t-1\right)u\left(t\right)$

   which agrees with the value obtained above using the partial fraction approach.

Task!

Use the convolution theorem to find the inverse transform of $H\left(s\right)=\frac{s}{\left(s-1\right)\left(s^2+1\right)}$ .

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

Answer

Now construct the convolution integral:

Answer

Answer

Exercises

1. Find the convolution of
   1. $2tu\left(t\right)$ and $t^{3}u\left(t\right)$
   2. ${\text{e}}^{t}u\left(t\right)$ and $tu\left(t\right)$
   3. ${\text{e}}^{-2t}u\left(t\right)$ and ${\text{e}}^{-t}u\left(t\right)$ .

       In each case reverse the order to check that $\left(f\ast g\right)\left(t\right)=\left(g\ast f\right)\left(t\right)$ .

2. Use the convolution theorem to determine the inverse Laplace transforms of
   1. $\frac{1}{s^2\left(s+1\right)}$
   2. $\frac{1}{\left(s-1\right)\left(s-2\right)}$
   3. $\frac{1}{{\left(s^2+1\right)}^2}$

Answer