# Can the (point-wise) multiplication of two discrete band-limited functions be inherently aliased?

I'm not fully used to the language of DSP, so forgive me if I confuse some of the notation, or if the below is trivial.

I'm interested in the case where we want to do the point-wise multiplication of two discrete signals, $$f[x_n]=g[x_n]h[x_n]$$ for $$n=0,\dots,M-1$$. This is easy enough to do in the computer. However, I'll take a detour and formulate the problem in the (Discrete) Fourier domain, where the multiplication becomes a convolution: $$\widehat{f}[k_n] = \sum_{j=-M}^M \widehat{g}[k_j]\widehat{h}[k_n-k_j].$$ Interesting is that (just like the MATLAB conv command) we get $$2M-1$$ outputs; i.e., the convolved signal is inherently longer than the original signal. If we take the result back to the sample domain, we have to throw away $$M$$ samples to get back to the original domain (or get back to sample domain and pick every second entry).

I ran the following test in MATLAB that seems to confirm my thought,

M=100;
x=[0:M-1];
g=sin(x/M*6*pi);
h=cos(x/M*6*pi);
f=g.*h;

ff = fft(f);gf = fft(g);hf = fft(h);
v  = ifftshift( (-1/2:1/M:1/2-1/M) + mod(M,2)/(2*M) ); % Normalized frequencies

figure(1); plot( x, g, x, h, x, f )
title('Discrete multiplication')
legend('g','h','f')

figure(2); plot( v, abs(gf), v, abs(hf), v, abs(ff) )
title('Spectrum')


This generates the following two figures: and that appears to confirm my thoughts. The spectrum of $$f[x_n]$$ (in yellow) is broader than the spectrum of either input signals. Then to get to the question in the title: (1) is it possible that the point-wise multiplication of two discrete, band-limited functions is aliased itself? And (2) although I was able to recreate the $$f$$ also using the convolution theorem using

ghf = conv( fftshift(gf), fftshift(hf)./M, 'same' );
gh  = ifft(ifftshift(ghf),'symmetric'); % = identical to the vector 'f'


would it be possible to circumvent the aliasing by outputing the 'full' convolution, thereby getting $$2M-1$$ samples back, rather than $$M$$ samples?

• In order to be band-limited, the signal needs to be infinitely long. Hence it can't have "$M$" samples. – Hilmar Feb 8 '20 at 9:44

Yes. Or at least it's possible that the result won't fit in a tidy manner into your sampling rate. Because multiplying two signals together creates energy at different frequencies than the frequencies of the original (this is easiest to do by a single-frequency example: $$2 \cos \omega_1 t \cdot \cos \omega_2 t = \cos (\omega_1 - \omega_2)t + \cos(\omega_1 + \omega_2)t$$. If that $$\omega_1 + \omega_2$$ is to great to fit into your sampling rate, then it'll alias.
(2) would it be possible to circumvent the aliasing by outputing the 'full' convolution, thereby getting $$2𝑀−1$$ samples back?
Yes and no. Yes, you could, but to be consistent the result would have to be at a sample rate $$\frac{2M - 1}{M}$$ times the original.