# 3.9 Evaluation of convolution integrals

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The key to evaluating a convolution integral such as

$x\left(t\right)*h\left(t\right)={\int }_{-\infty }^{\infty }x\left(\tau \right)h\left(t-\tau \right)d\tau$

is to realize that as far as the integral is concerned, the variable $t$ is a constant and the integral is over the variable $\tau$ . Therefore, for each $t$ , we are finding the area of the product $x\left(\tau \right)h\left(t-\tau \right)$ . Let's look at an example that illustrates how this works.

Example 3.1 Find the convolution of $x\left(t\right)=u\left(t\right)$ and $h\left(t\right)={e}^{-t}u\left(t\right)$ . The convolution integral is given by

$h\left(t\right)*x\left(t\right)={\int }_{-\infty }^{\infty }{e}^{-\tau }u\left(\tau \right)u\left(t-\tau \right)d\tau$

[link] shows the graph of ${e}^{-\tau }u\left(\tau \right)$ , ${e}^{-t}u\left(t\right)$ , and their product. From the graph of the product, it is easy to see the the convolution integral becomes

${\int }_{0}^{t}{e}^{-\tau }d\tau =\left\{\begin{array}{cc}1-{e}^{-t},& t\ge 0\\ 0,& t<0\end{array}\right)$

Signals which can be expressed in functional form should be convolved as in the above example. Other signals may not have an easy functional representation but rather may be piece-wise linear. In order to convolve such signals, one must evaluate the convolution integral over different intervals on the $t$ -axis so that each distinct interval corresponds to a different expression for $x\left(t\right)*h\left(t\right)$ . The following example illustrates this:

Example 3.2 Suppose we attempt to convolve the unit step function $x\left(t\right)=u\left(t\right)$ with the trapezoidal function

$h\left(t\right)=\left\{\begin{array}{cc}t,& 0\le t<1\\ 1,& 1\le t<2\\ 0,& \text{elsewhere}\end{array}\right)$

From [link] , it can be seen that on the interval $0\le t<1$ , the product $x\left(t-\tau \right)h\left(\tau \right)$ is an equilateral triangle with area ${t}^{2}/2$ . On the interval $1\le t<2$ , the area of $x\left(t-\tau \right)h\left(\tau \right)$ is $t-1/2$ . This latter area results by adding the area of an equilateral triangle having a base of 1, and the area of a rectangle having a base of $t-1$ and a height of 1. For all values of $t$ greater than 2, the convolution is 1.5 since $x\left(t-\tau \right)h\left(\tau \right)=h\left(\tau \right)$ and $h\left(\tau \right)$ is a trapezoid having an area of 1.5. Finally, for $t<0$ , the convolution is zero since $x\left(t-\tau \right)h\left(\tau \right)=0$ .

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