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I'm stuck calculating the Fourier Transform of the following (periodic) signal and grateful for any help:

First, I calculated the complex Fourier Series $x_p(t) = \sum_{n=-\infty}^{+\infty} {\frac{U_{0}\sin(n\pi T_{1})}{n\pi} e^{iw_{0}nt}}$ to simplify things.

Now I start to calculate the Fourier Transform $$X_p(w)=\int_{-\infty}^{+ \infty} {x_p(t)\cdot e^{-iw_0t}dt} =$$

$$ \int_{-\infty}^{+ \infty} {\sum_{n=-\infty}^{+\infty} {\frac{U_{0}\sin(n\pi T_{1})}{n\pi} e^{iw_{0}nt}}\cdot e^{-iw_0t}dt} =$$

$$ \int_{-\infty}^{+ \infty} {\sum_{n=-\infty}^{+\infty} {\frac{U_{0}\sin(n\pi T_{1})}{n\pi} e^{iw_{0}nt-iw_0t}}dt} = $$

$$ U_{0} \cdot\int_{-\infty}^{+ \infty} {\sum_{n=-\infty}^{+\infty} {\frac{\sin(n\pi T_{1})}{n\pi}}e^{iw_0t(n-1)}dt} = ?$$

Now I think we could interchange summation and integration, but that doesn't help me much further

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    $\begingroup$ Well..first of all, $e^{i w_0 n t - i w_0 t} \neq e^n$ $\endgroup$
    – DaveNine
    Commented Jun 5, 2019 at 15:57
  • $\begingroup$ @DaveNine Of course not, corrected it $\endgroup$
    – WhatAMesh
    Commented Jun 5, 2019 at 16:02
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    $\begingroup$ After that, explain why you $can$ exchange the integral, do it, and integrate the only thing in there that is w.r.t $t$. Then write the infinite sum going from 1 to $\infty$ using a famous identity. $\endgroup$
    – DaveNine
    Commented Jun 5, 2019 at 16:02
  • $\begingroup$ You have a little bit of an issue in your series when $n=0$, by the way. $\endgroup$
    – DaveNine
    Commented Jun 5, 2019 at 16:04

1 Answer 1

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Hint: Start with computing the Fourier transform of the Rectangle function and then revisit the Sinc() function.

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