Poisson Summation Formula Derivation

First consider the summation of N complex exponentials:

$$\begin{eqnarray*} x(n) &\mathrel{\stackrel{\Delta}{=}}& \frac{1}{N} \sum_{k=0}^{N-1}e^{j\omega_kn} \eqsp \left\{ \begin{array}{ll} 1 & n=0 \quad (\hbox{\sc mod}\ N) \\ 0 & \mbox{elsewhere} \\ \end{array} \right. \\ [10pt] &=& \hbox{\sc IDFT}_{N,n}(1 \cdots 1) \;=\; \delta(n) + \delta(n-N) + \delta(n+N) + \cdots \end{eqnarray*}$$

where $\omega_k \mathrel{\stackrel{\Delta}{=}}2\pi k / N$ .

Setting $N=R$ (the FFT hop size) gives

$$\zbox{\sum_m \delta(n-mR) = \frac{1}{R} \sum_{k=0}^{R-1}e^{j\omega_kn}}$$

where $\omega_k \mathrel{\stackrel{\Delta}{=}}2\pi k/R$ (harmonics of the frame rate).

Let us now consider these equivalent signals as inputs to an LTI system, with an impulse response given by $w(n)$ , and frequency response equal to $W(\omega)$ .

Looking across the top of the above figure, for the case of input signal $x(n)=\sum_m \delta(n-mR)$ we have:

$$y(n) = \sum_m w(n-mR)$$

and looking across the bottom of the above figure, for the case of input signal $\frac{1}{R} \sum_{k=0}^{R-1}e^{j\omega_kn}$ , we have:

$$y(n) = \frac{1}{R} \sum_{k=0}^{R-1} W(\omega_k)e^{j\omega_kn}$$

Since the inputs were equal, the corresponding outputs must be equal too. This derives the Poisson Summation Formula: