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Let $X(f)$ be the Fourier transform of $x(t)$:

$$ X(f) \triangleq \mathscr{F}\Big\{ x(t) \Big\} = \int\limits_{-\infty}^{\infty} x(t)\,e^{-j 2 \pi f t} \ \mathrm{d}t $$

$$ x(t) \triangleq \mathscr{F}^{-1}\Big\{ X(f) \Big\} = \int\limits_{-\infty}^{\infty} X(f)\,e^{j 2 \pi f t} \ \mathrm{d}f $$

The question here it's how can I mathematically relate the Fourier transform of a function with its Fourier coefficients $c_n$ (complex form)

$$ x(t) = \sum\limits_{n=-\infty}^{\infty} c_n e^{j 2 \pi (n/T)t} $$ where $$ c_n = \frac{1}{T} \int\limits_{t_0}^{t_0+T} x(t)\,e^{-j 2 \pi (n/T)t} \ \mathrm{d}t \qquad \qquad -\infty < t_0 < +\infty $$

I mean, it's quite obvious that for a non periodic function $ T\cdot c_n = \mathscr{F}\Big\{ x(t) \Big\} $

However, if I have $x(t)$ periodic and take $x_m(t)=x(t) \cdot \operatorname{rect}\left( \tfrac{t}{T} \right)$

(that means we take a period of $x(t)$)

$$\mathscr{F}\Big\{ x_m(t) \Big\} = X(\nu) * T \operatorname{sinc}(T \nu)$$

($\nu$ is the frequency variable in Hz)

On the other side we may consider that

$$x_m(t) = \sum\limits_{n=-\infty}^{\infty} c_n e^{j 2 \pi (n/T)t} \longleftrightarrow \mathscr{F}\Big\{ x_m(t) \Big\} = \sum\limits_{n=-\infty}^{\infty} c_n \delta(\nu-\tfrac{n}{T}) $$

So what I've seen people doing was saying that we could replace $c_n$ for $\frac{1}{T}\mathscr{F}\Big\{ x_m(t) \Big\}$ and then if I have a repeated spectrum in time I can only calculate it's transform and I'll have the amplitudes on the frequency spectrum for each dirac.

I cannot see it mathematically nor can say if it's true.

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  • $\begingroup$ you need to fix your mathematical notation. you are using "$f$" for two different mathematical objects. $\endgroup$ – robert bristow-johnson Mar 8 at 22:33
  • $\begingroup$ also, do you know how to do $\LaTeX$ markup here? would you like me to edit your question and show you? $\endgroup$ – robert bristow-johnson Mar 8 at 22:34
  • $\begingroup$ did i express your question correctly (as you intended), Lucas? you just can't have two $f$ symbols meaning different things, so i left $f$ as frequency and changed your function to $x(t)$. $\endgroup$ – robert bristow-johnson Mar 8 at 23:41
  • $\begingroup$ Oh, come on, Robert, what's the harm if Lucas uses $f$ to mean two different things? After all, even you (and Lucas too, I think) use $x_m(t)$ to denote both one period of $x(t)$ and all of $x(t)$ since you equate $x_m(t)$ with the Fourier series expression for $x(t)$. $\endgroup$ – Dilip Sarwate Mar 9 at 3:56
  • $\begingroup$ @DilipSarwate i saw that too. but i was trying to translate Lucas's statement of the problem faithfully, but with eliminating one of the two uses of $f$. $\endgroup$ – robert bristow-johnson Mar 9 at 4:43

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