# Calculate the Fourier Transform

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

• Well..first of all, $e^{i w_0 n t - i w_0 t} \neq e^n$ Commented Jun 5, 2019 at 15:57
• @DaveNine Of course not, corrected it Commented Jun 5, 2019 at 16:02
• 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. Commented Jun 5, 2019 at 16:02
• You have a little bit of an issue in your series when $n=0$, by the way. Commented Jun 5, 2019 at 16:04