Why does the amplitude of a discrete convolution depend on the time step?

Question

I've just finished covering convolutions in my signals class, and I've been playing around with the conv function in MATLAB, but there's something I don't quite understand. Say I have two discrete signals, $a$ and $b$ sampled at $dt$. As $dt \rightarrow 0$, the amplitude of the result increases. Since the convolution integral is just $$ y(t)=\int_{-\infty}^\infty f(\tau)h(t-\tau)d\tau, $$

why does the resulting convolution seem to depend on what my value is for $dt$? By decreasing the sampling rate, shouldn't I just have a better approximation?

Answer 1

Very informally, you can think of the integrand as the product of $f(\tau)h(t-\tau)$ and the differential $dt$. In Matlab, you can approximate this by multiplying the discrete convolution by the sampling interval, as in this example:

% sample rate is 0.01 seconds
dt = 0.01;
t = 0:dt:1;
f = exp(-t/2);
h = ones(1,length(t));
c = dt*conv(f,h);
% sample rate is now 0.001 seconds
dt2 = 0.001;
t2 = 0:dt2:1;
f2 = exp(-t2/2);
h2 = ones(1,length(t2));
c2 = dt2*conv(f2,h2);
% confirm that the two convolutions are the same
plot(t,c(1:length(t)),t2,c2(1:length(t2)))

Note that what you're trying to do is approximate a result from continuous signal processing using discrete signal processing. Usually, and as this example shows, you have to be careful. Similar situations arise when trying to estimate a signal's energy or its Fourier transform.

As a final note, as CMDoolittle mentions, the correct discrete convolution is calculated by conv(f,h), without including dt.

Answer 2

Your computer doesn't compute the continuous integral, it does discrete convolution, which is just a sum of products at each time step. When you increase dt, you get more points in each signal vector, which increases the sum at each time step. You must normalize the result of conv() according to the length of the vectors involved.