Poisson Summation Formula

Consider the summation of N complex sinusoids having frequencies uniformly spaced around the unit circle [264]:

$$\begin{eqnarray*} x(n) &\mathrel{\stackrel{\Delta}{=}}& \frac{1}{N} \sum_{k=0}^{N-1}e^{j\omega_kn} = \left\{ \begin{array}{ll} 1 & n=0 \quad (\hbox{\sc mod}\ N) \\ 0 & \mbox{elsewhere} \\ \end{array} \right. \\ [5pt] &=& \hbox{\sc IDFT}_n(1 \cdots 1) \end{eqnarray*}$$

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

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

$$\zbox {\sum_m \delta(n-mR) = \frac{1}{R} \sum_{k=0}^{R-1}e^{j\omega_kn}}$$ (9.26)

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 Fig.8.16, for the case of input signal $\sum_m \delta(n-mR)$ we have

$$y(n) = \sum_m w(n-mR).$$ (9.27)

Looking across the bottom of the figure, for the case of input signal

$$x(n) = \frac{1}{R} \sum_{k=0}^{R-1}e^{j\omega_kn},$$ (9.28)

we have the output signal

$$y(n) = \frac{1}{R} \sum_{k=0}^{R-1} W(\omega_k)e^{j\omega_kn}.$$ (9.29)

This second form follows from the fact that complex sinusoids $e^{j\omega_kn}$ are eigenfunctions of linear systems--a basic result from linear systems theory [264,263].

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

$$\zbox {\underbrace{\sum_m w(n-mR)}_{\hbox{\sc Alias}_R(w)} = \underbrace{\frac{1}{R}\sum_{k=0}^{R-1} W(\omega_k)e^{j\omega_k n}}_{\hbox{\sc DFT}_R^{-1} \left[\hbox{\sc Sample}_{\frac{2\pi}{R}}(W)\right]}} \quad \omega_k \isdef \frac{2\pi k}{R} \protect$$ (9.30)

Note that the PSF is the Fourier dual of the sampling theorem [270], [264, Appendix G].

The continuous-time PSF is derived in §B.15.