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For signal $x(t)$ with bandwidth $B_x$ and signal $y(t)$ with bandwidth $B_y$, what will the bandwidth of the signal $z(t)=x(t)y(t)$ be?

j.e.rogers
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1 Answers1

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The bandwidths simply add.

You can break down both signal into their sinusoidal components and pairwise multiply them. For every pair of sines you get the sum and difference frequencies. This is a simple consequence of the multiplication theorems for sine waves. The highest frequency of the product will be the sum of the highest frequencies in the individual signals. This is equivalent to a convolution in the frequency domain.

Hilmar
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    I think that the statement "The bandwidths simply add." needs some qualification since it does not necessarily hold for all definitions of bandwidth that engineers might use. If $B_x$ and $B_y$ are the supports of $X(f)$ and $Y(f)$ respectively, then it is true that the support of $Z(f)$ will be $B_x+B_y$, but (i) support is not the only measure of bandwidth (do $99%$ containment bandwidths add?), and (ii) in many signal models, the support of $X(f)$ and $Y(f)$ is $(-\infty,\infty)$ and so adding supports is meaningless. – Dilip Sarwate Oct 31 '12 at 12:39
  • Isn't multiplication in the time domain means convolution in the frequency domain? So how is it simply adding? – JobHunter69 Mar 20 '17 at 04:34