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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
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.We start out with the convolution integral,
$y(t)=\int_{()} \,d \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 .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:
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 .
Thus, we have the following results for our four regions:
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.
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