CONVOLUTION EXAMPLE(DISCUSSED IN SS CLASS BY KAUNDINYA)

Convolve together two square pulses, xtxt and htht, as shown in Figure


Two basic signals that we will convolve together.

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 the origin, the point of the function that was originally on the origin, is labeled as point t. This step is shown in Figure,
h(t-τ).
Reflected square pulse.
Reflected and shifted square pulse.
h(-τ)and h(t-τ)
Note that in the above Figure ττ is the 1st axis variable while tt is a constant (in this figure). Since convolution is commutative it will never matter which function is reflected and shifted; however, as the 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 The value of the function y at time t is given by the amount of overlap(to be precise the integral of the overlapping region) between h(t-τ) and x(τ) .
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-τ) over x(τ) see Figure 3.




In this case we will have the following four regions. Compare these limits of integration to the four illustrations of h(t-τ) and x(τ)in Figure 3.



Four Limits of Integration

Using the Convolution Integral
Finally we are ready for a little math. Using the convolution integral, let us integrate the product of x h(t-τ) x(τ) . For our first and fourth region this will be trivial as it will always be 0. The second region, 0≤t<1>

Convolution Results



common sense aproach
By looking at Figure we can obtain the system output, ytyt, by "common" sense. For
t<0 there is no overlap, so y(t) is 0. As tt 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 tt 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 square pulses the result will always be a triangular pulse. Its origin, peak value and strech will, of course, vary.