Fundemental period of $x(t) = \sum_{k=-\infty}^{\infty}\operatorname{rect}(\frac{t+k}{2c})$

Question

My effort for this signal: $$x(t) = x(t+T) = \sum_{k=-\infty}^{\infty}\operatorname{rect}\left(\frac{t+k+T}{2c}\right)$$ How can I go on?

Answer 1

Perfect time to use the Poisson sumation formula! Recall that for suitable functions $\phi:\Bbb R\to \Bbb C$,

$$\sum_{n\in\Bbb Z}\phi(x+n)=\sum_{k\in\Bbb Z}\hat\phi(k)~\mathrm e^{ 2\pi \mathrm ikx}$$ Where $$\hat{\phi}(\xi)=\mathcal F(\phi)(\xi)=\int_{\Bbb R} \phi(x)\mathrm e^{-2\pi\mathrm i\xi x}\mathrm dx$$ Is the Fourier transform. With that in mind, define $\phi(x)=\operatorname{rect}(x/(2c))$. You can easily calculate $$\mathcal F(\phi)(\xi)=2|c|\operatorname{sinc}(2\pi c~\xi)$$ Where of course $\operatorname{sinc}(x)=(\sin x)/x$. Hence, $$\sum_{n\in\Bbb Z}\operatorname{rect}\left(\frac{x+n}{2c}\right)=\sum_{n\in\Bbb Z}\phi(x+n) \\ =\sum_{k\in\Bbb Z}\hat\phi(k)~\mathrm e^{2\pi\mathrm ikx}=2|c|\sum_{k\in \Bbb Z}\operatorname{sinc}(2\pi c~k)~\mathrm e^{2\pi \mathrm ikx} $$

Perhaps you can finish up from here?