Convolution Example 1: Smoothing a Rectangular Pulse

Figure 7.3: Illustration of the convolution of a rectangular pulse $x=[0 ,0 ,0 ,0,1,1,1,1,1,1,0,0,0,0]$ and the impulse response of an ``averaging filter'' $h=[1/3,1/3,1/3,0,0,0,0,0,0,0,0,0,0,0]$ ($N=14$ ).

Filter input signal $x(n)$ .

Filter impulse response $h(n)$ .

Filter output signal $y(n)$ .

Figure 7.3 illustrates convolution of

$$x = [ 0, 0, 0,0,1,1,1,1,1,1,0,0,0,0]$$

with

$$h = \left[\frac{1}{3},\frac{1}{3},\frac{1}{3},0,0,0,0,0,0,0,0,0,0,0\right]$$

to get

$$y = x\circledast h = \left[0,0,0,0,\frac{1}{3},\frac{2}{3},1,1,1,1,\frac{2}{3},\frac{1}{3},0,0\right] \protect$$ (7.2)

as graphed in Fig.7.3(c). In this case, $h$ can be viewed as a ``moving three-point average'' filter. Note how the corners of the rectangular pulse are ``smoothed'' by the three-point filter. Also note that the pulse is smeared to the ``right'' (forward in time) because the filter impulse response starts at time zero. Such a filter is said to be causal (see [71] for details). By shifting the impulse response left one sample to get

$$h=\left[\frac{1}{3},\frac{1}{3},0,0,0,0,0,0,0,0,0,0,\frac{1}{3}\right]$$

(in which case $\hbox{\sc Flip}(h)=h$ ), we obtain a noncausal filter $h$ which is symmetric about time zero so that the input signal is smoothed ``in place'' with no added delay (imagine Fig.7.3(c) shifted left one sample, in which case the input pulse edges align with the midpoint of the rise and fall in the output signal).