Derivation of the Poisson Summation Formula
📂Distribution Theory Derivation of the Poisson Summation Formula Let f : R → C f : \mathbb{R} \to \mathbb{C} f : R → C ∑ n ∈ Z f ( n ) = ∑ k ∈ Z f ^ ( k )
\sum_{n \in \mathbb{Z}} f(n) = \sum_{k \in \mathbb{Z}} \widehat{f}(k)
n ∈ Z ∑ f ( n ) = k ∈ Z ∑ f ( k )
Schwartz functions f ∈ C ∞ ( R ) f \in C^{\infty}(\mathbb{R}) f ∈ C ∞ ( R ) ∣ f ( x ) ∣ \left| f (x) \right| ∣ f ( x ) ∣ 0 0 0 x → ± ∞ x \to \pm \infty x → ± ∞ For f f f γ ∈ R \gamma \in \mathbb{R} γ ∈ R f ^ ( γ ) \widehat{f}(\gamma) f ( γ ) Fourier transform :
f ^ ( γ ) = ∫ R f ( x ) e − 2 π i γ x d x
\widehat{f} ( \gamma ) = \int_{\mathbb{R}} f(x) e^{-2 \pi i \gamma x} dx
f ( γ ) = ∫ R f ( x ) e − 2 πiγ x d x Proof Let
F ( x ) : = ∑ n ∈ Z f ( x + n )
F(x) := \sum_{n \in \mathbb{Z}} f ( x + n )
F ( x ) := n ∈ Z ∑ f ( x + n ) F F F 1 1 1 F ^ k \widehat{F}_{k} F k F ^ k = ∫ 0 1 ∑ n ∈ Z f ( x + n ) e − 2 π i k z d x = ∑ n ∈ Z ∫ 0 1 f ( x + n ) e − 2 π i k z d x = ∑ n ∈ Z ∫ n n + 1 f ( x ) e − 2 π i k z d x = ∫ R f ( x ) e − 2 π i k z d x = f ^ ( k )
\begin{align*}
\widehat{F}_{k} &= \int_{0}^{1} \sum_{n \in \mathbb{Z}} f(x+n) e^{-2\pi i k z } dx
\\ =& \sum_{n \in \mathbb{Z}} \int_{0}^{1} f(x+n) e^{-2\pi i k z } dx
\\ =& \sum_{n \in \mathbb{Z}} \int_{n}^{n+1} f(x) e^{-2\pi i k z } dx
\\ =& \int_{\mathbb{R}} f(x) e^{-2\pi i k z } dx
\\ =& \widehat{f} (k)
\end{align*}
F k = = = = = ∫ 0 1 n ∈ Z ∑ f ( x + n ) e − 2 πik z d x n ∈ Z ∑ ∫ 0 1 f ( x + n ) e − 2 πik z d x n ∈ Z ∑ ∫ n n + 1 f ( x ) e − 2 πik z d x ∫ R f ( x ) e − 2 πik z d x f ( k ) F F F ∑ n ∈ Z f ( x + n ) = F ( x ) = ∑ k ∈ Z F ^ k e i k x = ∑ k ∈ Z f ^ ( k ) e i k x
\sum_{n \in \mathbb{Z}} f(x+n) = F(x) = \sum_{k \in \mathbb{Z}} \widehat{F}_{k} e^{i k x } = \sum_{k \in \mathbb{Z}} \widehat{f} (k) e^{i k x}
n ∈ Z ∑ f ( x + n ) = F ( x ) = k ∈ Z ∑ F k e ik x = k ∈ Z ∑ f ( k ) e ik x x = 0 x = 0 x = 0 ∑ n ∈ Z f ( n ) = ∑ k ∈ Z f ^ ( k )
\sum_{n \in \mathbb{Z}} f(n) = \sum_{k \in \mathbb{Z}} \widehat{f}(k)
n ∈ Z ∑ f ( n ) = k ∈ Z ∑ f ( k )
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