It is said that the Fourier transform $\hat{f}(\omega)$ of a function $f(t)$ and the Fourier transform $\hat{b}(\omega)$ of its samples $b(k)=f(t)|_{t=k}$ are related by Poisson's summation formula and it's given by
$$\hat{b}(\omega)=\displaystyle\sum_{k\in{\mathbb{Z}}}\hat{f}(\omega+2\pi{k})$$
where
$$\hat{b}(\omega)=\displaystyle\sum_{k\in\mathbb{Z}}b(k)e^{-i\omega k}, \omega\in{\mathbb{R}}.$$
I just fail to see why, is this equation even right?
Best Answer
Here is a formal approach, taken from Rudin's book on real and complex analysis. Let $$ \varphi(t)=\frac{1}{2\pi} \int f(x)e^{-itx}\, dx $$ and $$ F(x) = \sum_{k=-\infty}^{+\infty} f(x+2k\pi). $$ Then $F$ is $2\pi$-periodic and its $n$th Fourier coefficient is $\varphi(n)$, hence $F(x)=\sum \varphi(n) e^{inx}$. In particular, $$ \sum_{k=-\infty}^{+\infty} f(2k\pi) = \sum_{n=-\infty}^{+\infty} \varphi(n). $$ More generally, $$ \sum_{k=-\infty}^{+\infty} f(k \beta) = \sum_{n=-\infty}^{+\infty} \varphi(n\alpha) $$ whenever $\alpha>0$, $\beta>0$, $\alpha \beta = 2\pi$.