Poisson Summation Formula

As shown in §B.14 above, the Fourier transform of an impulse train is an impulse train with inversely proportional spacing:

$$\psi_R(t)\;\longleftrightarrow\;{\frac{1}{R}}\cdot\psi_{\frac{1}{R}}(f) = \,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(Rf)$$ (B.56)

where

$$\psi_R(t) \isdefs \sum_{m=-\infty}^\infty \delta(t-mR) = \frac{1}{R}\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}\left(\frac{t}{R}\right).$$ (B.57)

Using this Fourier theorem, we can derive the continuous-time PSF using the convolution theorem for Fourier transforms:^B.1

$$\sum_m w(t-mR) = \psi_R \ast w \;\longleftrightarrow\; \Psi_R \cdot W = \frac{1}{R}\cdot\psi_{\frac{1}{R}}\cdot W$$ (B.58)

Using linearity and the shift theorem for inverse Fourier transforms, the above relation yields

$$\begin{eqnarray*} \sum_m w(t-mR) &=& \frac{1}{R} \hbox{\sc IFT}_t \left[W(f)\sum_k\delta\left(f-k\frac{1}{R}\right) \right] \quad\left(\mbox{define $f_k\isdef \frac{k}{R}$}\right) \\ [5pt] &=& \frac{1}{R} \hbox{\sc IFT}_t \left[\sum_k W(f_k)\cdot\delta\left(f-f_k\right) \right]\\ [5pt] &=& \frac{1}{R} \sum_k W(f_k)\cdot\hbox{\sc IFT}_t \left[\delta\left(f-f_k\right) \right]\\ [5pt] &=& \frac{1}{R} \sum_k W(f_k)e^{j 2\pi f_k t}. \end{eqnarray*}$$

We have therefore shown

$$\zbox {\sum_{m=-\infty}^{\infty} w(t-mR) = \frac{1}{R} \sum_{k=-\infty}^{\infty} W(f_k)e^{j 2\pi f_k t}, \quad f_k\isdef \frac{k}{R}.} \protect$$ (B.59)

Compare this result to Eq.(8.30). The left-hand side of (B.60) can be interpreted $s_R(t)=\hbox{\sc Alias}_R(w)$ , i.e., the time-alias of $w$ on a block of length $R$ . The function $s_R(t)$ is periodic with period $R$ seconds. The right-hand side of (B.60) can be interpreted as the inverse Fourier series of $W(f)$ sampled at intervals of $1/R$ Hz. This sampling of $W(\omega)$ in the frequency domain corresponds to the aliasing of $w(t)$ in the time domain.