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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$ ).

\includegraphics{eps/smoother-x}
Filter input signal $ x(n)$ .


\includegraphics{eps/smoother-h}
Filter impulse response $ h(n)$ .


\includegraphics{eps/smoother-y}
Filter output signal $ y(n)$ .


Figure 7.3 illustrates convolution of

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

with

$\displaystyle 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

$\displaystyle y = x\circledast h = \left[0,0,0,0,\frac{1}{3},\frac{2}{3},1,1,1,1,\frac{2}{3},\frac{1}{3},0,0\right] \protect$ (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 [70] for details). By shifting the impulse response left one sample to get

$\displaystyle 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).


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``Mathematics of the Discrete Fourier Transform (DFT), with Audio Applications --- Second Edition'', by Julius O. Smith III, W3K Publishing, 2007, ISBN 978-0-9745607-4-8.
Copyright © 2014-04-21 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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