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So I've been struggling to understand how can I solve the following problem. Here's how it goes:

Given $m(t) = \cos(2\cdot \pi\cdot 100\cdot t)$, sketch the frequency spectrum of $s(t) = m(t)\cdot \cos(2\cdot \pi\cdot 1000\cdot t)$.

So, basically, here's what I've done so far:

The Fourier Transform of $m(t)$ is $$\mathcal{F}(m(t)) = M(\omega) = \pi \cdot (\delta(\omega - 2\cdot \pi\cdot 100) + \delta(\omega + 2\cdot \pi\cdot 100))$$

But, as the property of the product in time domain implies a convolution in the frequency one, then I would just need to convolute the two given cosines, which returns me: $$M(\omega)*\mathcal{F}(\cos(2\cdot \pi\cdot 1000\cdot t)) = \mathcal{F}(\cos(2\cdot \pi\cdot 1000\cdot t))(\omega - (2\cdot \pi\cdot 100)) + \mathcal{F}(\cos(2\cdot \pi\cdot 1000\cdot t))(\omega + (2\cdot \pi\cdot 100))$$

Or, verbosely: The convolution between the two Fourier Transforms should be equal its values when the impulses occur (because of sampling property). But, then, should the result be of the convolution be $0$? Does it make sense at all? I mean, I cannot find the error (it it even exists) by myself, so a little help would be much appreciated.

P.S.: and also, because convolution is commutative, the result should be the same if I convoluted the other way around. But it would be still zero, by my line of thought, right?

Thank you very much!

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It seems to me that it is sufficient to give a characterization of $\delta_a * \delta_b$. From the definition of convolution of two distributions one of which has compact support, this is $\phi \mapsto \langle \delta_a,\psi \rangle=\psi(a)$ where $\psi(x)=\langle \delta_b,\tau_{-x} \phi \rangle = \phi(b+x)$, thus $(\delta_a * \delta_b)(\phi)=\phi(a+b)$, i.e. $\delta_a * \delta_b = \delta_{a+b}$.

Note that this makes sense if you recall that $f \mapsto \delta_0 * f$ is the identity operator on functions and thus you should expect that $T \mapsto \delta_0 * T$ is the identity operator on distributions.

A more direct way to proceed is to look at $4\cos(ax)\cos(bx)=\left ( e^{iax} + e^{-iax} \right ) \left ( e^{ibx} + e^{-ibx} \right ) = e^{i(a+b)x} + e^{-i(a+b)x} + e^{i(a-b)x} + e^{i(b-a)x}$ which immediately reveals the four Dirac deltas that will appear in the Fourier transform of the product.

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