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!


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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