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I have a goal of arriving to the fourier transform of dirac-comb $\omega_T(t)$ so i calculated the fourier series $\omega_T(t) = \frac{1}{T} + \frac{2}{T} \sum_{n=1}^\infty \cos(n \frac{2 \pi}{T} t)$ i think the next step to obtaining the fourier transform is evaluating the series, isnt it?

so $F(\omega_T(t))(z) = F(\frac{1}{T}(1 + 2 \sum_{n=1}^\infty \cos(n \frac{2 \pi}{T} t)))(z) = \frac{1}{T} F(1)(z) + 2 F(\sum_{n=1}^\infty \cos(n \frac{2 \pi}{T} t))(z) = \frac{1}{T} \omega(z) + \frac{1}{T}\sum_{n=1}^\infty F(2\cos(n \frac{2 \pi}{T} t))(z)$ do i have the right to switch the transform and the infinite series?

HellBoy
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  • Here is a nice answer to a similar question. https://math.stackexchange.com/questions/1898432/the-fourier-transform-of-a-comb-function-is-a-comb-function – mark leeds Nov 11 '23 at 12:36
  • i did see that solution before but im not allowed to use the exponential coefficient , i need to use the expression i had above @markleeds – HellBoy Nov 11 '23 at 12:44
  • I posted an answer because you specifically asked about the switching between series and Fourier transform. In any case, you can simply rewrite $\cos(\omega t)=(e^{i\omega t}+e^{-i\omega t})/2$, with $\omega=2\pi n/T$, and then follow what has been done in the linked post. – Lorenzo Pompili Nov 11 '23 at 14:02

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You can switch $F$ and the infinite series because the series converges in the space of tempered distributions $\mathscr S’(\mathbb R)$ and $F$ is continuous and linear from $\mathscr S’(\mathbb R)$ to itself.

Why is the series converging in $\mathscr S’(\mathbb R)$? Well, it has to be checked of course. Take $f\in \mathscr S(\mathbb R)$, consider the quantities $$ \int_{\mathbb R}\cos(2\pi n t/T)f(t)dt. $$ Since the Fourier transform of cosine functions are sums of two Dirac deltas, the above quantity is the same as $$ \frac{F(f)(2\pi n t/T)+ F(f)(-2\pi n t/T)}{2} $$ and since the Fourier transform preserves $\mathscr S(\mathbb R)$, $F(f)$ decays fast at infinity, so that in particular the series

$$ \sum_n\int_{\mathbb R}\cos(2\pi n t/T)f(t)dt. $$

is well-defined and converges. This shows that the Fourier series you wrote is well defined and converges in $f\in \mathscr S(\mathbb R)$, so exchanging series and Fourier transform is justified from what I said above.

With this argument of testing against a Schwartz function $f\in \mathscr S(\mathbb R)$, you might as well just compute the Fourier transform of the Dirac comb directly from the definition of Fourier transform (it’s basically the same argument computationally speaking).

Lorenzo Pompili
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  • very good answer, last question is this step correct : $\frac{1}{T} \int_{0}^T \omega_T (t) exp(-i \frac{2 \pi n}{T} t) dt = \frac{1}{T} \int_{0}^T \sum_{k=-\infty}^\infty \delta(t - kT) exp(-i \frac{2 \pi n}{T} t) dt = \frac{1}{T} \sum_{k=-\infty}^\infty \int_{0}^T \delta(t - kT) exp(-i \frac{2 \pi n}{T} t) dt = \frac{1}{T} \sum_{k=-\infty}^\infty exp(- i \frac{2 \pi n}{T} * kT) $ ? will i get this expression to evaluate to 1/T in the end? – HellBoy Nov 11 '23 at 15:04