Poisson Summation Formula - Derivation

Derivation

We can also prove that Eq.3 holds in the sense that if s(t) ∈ L1(R), then the right-hand side is the (possibly divergent) Fourier series of the left-hand side. This proof may be found in either (Pinsky 2002) or (Zygmund 1968). It follows from the dominated convergence theorem that sP(t) exists and is finite for almost every t. And furthermore it follows that sP is integrable on the interval . The right-hand side of Eq.3 has the form of a Fourier series. So it is sufficient to show that the Fourier series coefficients of sP(t) are . Proceeding from the definition of the Fourier coefficients we have:

\begin{align} S\ &\stackrel{\text{def}}{=}\ \frac{1}{P}\int_0^{P} s_P(t)\cdot e^{-i 2\pi \frac{k}{P} t}\, dt\\ &=\ \frac{1}{P}\int_0^{P} \left(\sum_{n=-\infty}^{\infty} s(t + nP)\right) \cdot e^{-i 2\pi\frac{k}{P} t}\, dt\\ &=\ \frac{1}{P} \sum_{n=-\infty}^{\infty} \int_0^{P} s(t + nP)\cdot e^{-i 2\pi\frac{k}{P} t}\, dt, \end{align}
where the interchange of summation with integration is once again justified by dominated convergence. With a change of variables (τ = t + nP) this becomes:
\begin{align} S = \frac{1}{P} \sum_{n=-\infty}^{\infty} \int_{nP}^{nP + P} s(\tau) \ e^{-i 2\pi \frac{k}{P} \tau} \ \underbrace{e^{i 2\pi k n}}_{1}\,d\tau \ =\ \frac{1}{P} \int_{-\infty}^{\infty} s(\tau) \ e^{-i 2\pi \frac{k}{P} \tau} d\tau = \frac{1}{P}\cdot \hat s\left(\frac{k}{P}\right) \end{align} QED.

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