# 2.3 Convolution - complete example

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Shows a full example of convolution including math and figures.

## Basic example

Let us look at a basic continuous-time convolution example to help express some of the important ideas. We will convolve together two square pulses, $x(t)$ and $h(t)$ , as shown in

## Reflect and shift

Now we will take one of the functions and reflect it around the y-axis. Then we must shift the function, such that theorigin, the point of the function that was originally on the origin, is labeled as point $t$ . This step is shown in , $h(t-\tau )$ .

Note that in $\tau$ is the 1st axis variable while $t$ is a constant (in this figure).Since convolution is commutative it will never matter which function is reflected and shifted; however, asthe functions become more complicated reflecting and shifting the "right one" will often make the problem much easier.

## Regions of integration

We start out with the convolution integral, $y(t)=\int_{()} \,d \tau$ x τ h t τ . The value of the function $y$ at time $t$ is given by the amount of overlap(to be precise the integral of theoverlapping region) between $h(t-\tau )$ and $x(\tau )$ .

Next, we want to look at the functions and divide the span of the functions into different limits of integration.These different regions can be understood by thinking about how we slide $h(t-\tau )$ over $x(\tau )$ , see .

In this case we will have the following four regions. Compare these limits of integration to thefour illustrations of $h(t-\tau )$ and $x(\tau )$ in .

## Four limits of integration

• $t< 0$
• $0\le t< 1$
• $1\le t< 2$
• $t\ge 2$

## Using the convolution integral

Finally we are ready for a little math. Using the convolution integral, let us integrate the product of $x(\tau )h(t-\tau )$ . For our first and fourth region this will be trivial as it will always be $0$ . The second region, $0\le t< 1$ , will require the following math:

$y(t)=\int_{0}^{t} 1()\,d \tau =t$
The third region, $1\le t< 2$ , is solved in much the same manner. Take note of the changes in our integration though. As we move $h(t-\tau )$ across our other function, the left-hand edge of the function, $t-1$ , becomes our lowlimit for the integral. This is shown through our convolution integral as
$y(t)=\int_{t-1}^{1} 1()\,d \tau =1-t-1=2-t$
The above formulas show the method for calculating convolution; however, do not let the simplicity of thisexample confuse you when you work on other problems. The method will be the same, you will just have to deal withmore math in more complicated integrals.

Note that the value of $y(t)$ at all time is given by the integral of the overlapping functions. In this example $y$ for a given $t$ equals the gray area in the plots in .

## Convolution results

Thus, we have the following results for our four regions:

$y(t)=\begin{cases}0 & \text{if t< 0}\\ t & \text{if 0\le t< 1}\\ 2-t & \text{if 1\le t< 2}\\ 0 & \text{if t\ge 2}\end{cases}()$
Now that we have found the resulting function for each of the four regions, we can combine them together and graph theconvolution of $(x(t), h(t))$ .

## Common sense approach

By looking at we can obtain the system output, $y(t)$ , by "common" sense.For $t< 0$ there is no overlap, so $y(t)$ is 0. As $t$ goes from 0 to 1 the overlap will linearly increase with a maximum for $t=1$ , the maximum corresponds to the peak value in the triangular pulse.As $t$ goes from 1 to 2 the overlap will linearly decrease. For $t> 2$ there will be no overlap and hence no output.

We see readily from the "common" sense approach that the output function $y(t)$ is the same as obtained above with calculations. When convolving to squarepulses the result will always be a triangular pulse. Its origin, peak value and strech will, of course, vary.

• Introduction
• Convolution - Discrete time
• Convolution - Analog
• Properties of convolution

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