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I will unashamedly say that this was at least spurred by homework. However I have gone far beyond the syllabus of the course and still can't find an authoritative answer. And it seems an interesting question to me that I doubt the professor will answer (if I wasn't ashamed to ask).

I am asked to calculate the Fourier transform of the convolution of two signals, for generality:

$\mathrm{F}\left\lbrace\sin^3\left(at+b\right)*\cos^3\left(ct+d\right)\right\rbrace$.

I have tried two approaches.

First, take the product of the Fourier transforms of the sinusoids. This leads to an expression that contains terms of the form $\delta(\omega-a)\delta(\omega-b)$. According to [1] and unless I missed it, the product of two distributions, unlike other operations, is not defined.

Secondly calculate the convolution directly. This leads me to an integral of the form:

$\int_{-\infty}^{\infty}\sin^3\left(a\tau+b\right)\cos^3\left(c(\tau-t)+d\right)\mathrm{d}\tau$

This also I think is non-convergent.

So am I right to think that this convolution and its Fourier transform are not defined?

[1] Zemanian: Distribution Theory and Transform Analysis

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    $\begingroup$ What is it you want to calculate? The Fourier transform of the convolution or the convolution itself? In the integral you wrote down things seem to get mixed up. $\endgroup$ Commented Jun 1, 2012 at 14:53
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    $\begingroup$ Warn: in your first equation you use $*$ for convolution, in the second (inside the integral) it's a product. $\endgroup$ Commented Jun 1, 2012 at 15:04
  • $\begingroup$ Thanks and sorry. I have corrected it. I want to calculate the Fourier transform, but if the convolution does not exist, neither does its Fourier transform, does it? $\endgroup$ Commented Jun 1, 2012 at 18:15
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    $\begingroup$ In general, sin * cos is not a tempered distribution, so you cannot expect the existence of its Fourier Transform. The same is true for the product of its separate Fourier Transforms. There are specific situations in which you can try to extend either the convolution or the multiplication, but for your case I don't expect anything useful. $\endgroup$ Commented Jun 1, 2012 at 21:35
  • $\begingroup$ Thanks. Why don't you post that as an answer? Also I am curious: is sin*cos a distribution at all? $\endgroup$ Commented Jun 2, 2012 at 6:33

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It is not difficult to find a test function $\varphi$ so that $$ (\sin \ast \cos) (\varphi) = \int \sin x\int \cos t \varphi(t-x) dt dx $$ does not exist. Hence $sin \ast cos$ is not a (tempered) distribution, so you cannot expect the existence of its Fourier Transform. The same is true for the product of its separate Fourier Transforms.

There are specific situations in which you can try to extend either the convolution or the multiplication, but for your case I don't expect anything useful.

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  • $\begingroup$ Thanks. Shouldn't that be $\int\sin x\int\cos(t-x)\phi(t)dtdx$ or is it the same thing? $\endgroup$ Commented Jun 2, 2012 at 19:50
  • $\begingroup$ I haven't really checked, but upto a sign it is the same thing. $\endgroup$ Commented Jun 2, 2012 at 22:42

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