In mathematics, the Poisson summation formula is an equation that relates the Fourier series coefficients of the periodic summation of a function to values of the function's continuous Fourier transform. Consequently, the periodic summation of a function is completely defined by discrete samples of the original function's Fourier transform. And conversely, the periodic summation of a function's Fourier transform is completely defined by discrete samples of the original function. The Poisson summation formula was discovered by Siméon Denis Poisson and is sometimes called Poisson resummation. For a smooth, complex valued function s ( x ) {\displaystyle s(x)} on R {\displaystyle \mathbb {R} } which decays at infinity with all derivatives (Schwartz function), the simplest version of the Poisson summation formula states that
where S {\displaystyle S} is the Fourier transform of s {\displaystyle s} , i.e., S ( ξ ) ≜ ∫ − ∞ ∞ s ( x ) e − i 2 π ξ x d x . {\textstyle S(\xi )\triangleq \int _{-\infty }^{\infty }s(x)\ e^{-i2\pi \xi x}\,dx.} The summation formula can be restated in many equivalent ways, but a simple one is the following. Suppose that f ∈ L 1 ( R n ) {\displaystyle f\in L^{1}(\mathbb {R} ^{n})} (L1 for L1 space) and Λ {\displaystyle \Lambda } is a unimodular lattice in R n {\displaystyle \mathbb {R} ^{n}} . Then the periodization of f {\displaystyle f} , which is defined as the sum f Λ ( x ) = ∑ λ ∈ Λ f ( x + λ ) , {\textstyle f_{\Lambda }(x)=\sum _{\lambda \in \Lambda }f(x+\lambda ),} converges in the L 1 {\displaystyle L^{1}} norm of R n / Λ {\displaystyle \mathbb {R} ^{n}/\Lambda } to an L 1 ( R n / Λ ) {\displaystyle L^{1}(\mathbb {R} ^{n}/\Lambda )} function having Fourier series f Λ ( x ) ∼ ∑ λ ′ ∈ Λ ′ f ^ ( λ ′ ) e 2 π i λ ′ x {\displaystyle f_{\Lambda }(x)\sim \sum _{\lambda '\in \Lambda '}{\hat {f}}(\lambda ')e^{2\pi i\lambda 'x}} where Λ ′ {\displaystyle \Lambda '} is the dual lattice to Λ {\displaystyle \Lambda } . (Note that the Fourier series on the right-hand side need not converge in L 1 {\displaystyle L^{1}} or otherwise.)
Periodization of a function Let s ( x ) {\textstyle s\left(x\right)} be a smooth, complex valued function on R {\displaystyle \mathbb {R} } which decays at infinity with all derivatives (Schwartz function), and its Fourier transform S ( f ) {\displaystyle S\left(f\right)} , defined as
S ( f ) = ∫ − ∞ ∞ s ( x ) e − 2 π i x f d x . {\displaystyle S(f)=\int _{-\infty }^{\infty }s(x)e^{-2\pi ixf}dx.}
Then S ( f ) {\displaystyle S(f)} is also a Schwartz function, and we have the reciprocal relationship that
s ( x ) = ∫ − ∞ ∞ S ( f ) e 2 π i x f d f . {\displaystyle s(x)=\int _{-\infty }^{\infty }S(f)e^{2\pi ixf}df.}
The periodization of s ( x ) {\displaystyle s(x)} with period P > 0 {\displaystyle P>0} is given by
s P ( x ) ≜ ∑ n = − ∞ ∞ s ( x + n P ) . {\displaystyle s_{_{P}}(x)\triangleq \sum _{n=-\infty }^{\infty }s(x+nP).}
Likewise, the periodization of S ( f ) {\displaystyle S(f)} with period 1 / T {\displaystyle 1/T} , where T > 0 {\displaystyle T>0} , is
S 1 / T ( f ) ≜ ∑ k = − ∞ ∞ S ( f + k / T ) . {\displaystyle S_{1/T}(f)\triangleq \sum _{k=-\infty }^{\infty }S(f+k/T).}
Then Eq.1, ∑ n = − ∞ ∞ s ( n ) = ∑ k = − ∞ ∞ S ( k ) , {\displaystyle \sum _{n=-\infty }^{\infty }s(n)=\sum _{k=-\infty }^{\infty }S(k),} is a special case (P=1, x=0) of this generalization:
which is a Fourier series expansion with coefficients that are samples of the function S ( f ) . {\displaystyle S(f).} Conversely, Eq.2 follows from Eq.1 by applying the known behavior of the Fourier transform under translations (see the Fourier transform properties time scaling and shifting). Similarly:
also known as the important Discrete-time Fourier transform.
Derivations We prove that, if s ∈ L 1 ( R ) {\displaystyle s\in L^{1}(\mathbb {R} )} , then the (possibly divergent) Fourier series of s P ( x ) {\displaystyle s_{P}(x)} is
s P ( x ) ∼ ∑ k = − ∞ ∞ 1 P S ( k P ) e 2 π i k P x . {\displaystyle s_{_{P}}(x)\sim \sum _{k=-\infty }^{\infty }{\frac {1}{P}}S\left({\frac {k}{P}}\right)e^{2\pi i{\frac {k}{P}}x}.}
When s ( x ) {\displaystyle s(x)} is a Schwartz function, this establishes equality in Eq.2 of the previous section. First, the periodization s P ( x ) {\displaystyle s_{P}(x)} converges in L 1 {\displaystyle L^{1}} norm to an L 1 ( [ 0 , P ] ) {\displaystyle L^{1}([0,P])} function which is periodic on R {\displaystyle \mathbb {R} } , and therefore integrable on any interval of length P . {\displaystyle P.} We must therefore show that the Fourier series coefficients of s P ( x ) {\displaystyle s_{_{P}}(x)} are 1 P S ( k P ) {\textstyle {\frac {1}{P}}S\left({\frac {k}{P}}\right)} where S ( f ) {\textstyle S\left(f\right)} is the Fourier transform of s ( x ) {\textstyle s\left(x\right)} . (Not S [ k ] {\textstyle S\left[k\right]} , which is the Fourier coefficient of s P ( x ) {\displaystyle s_{_{P}}(x)} .) Proceeding from the definition of the Fourier coefficients we have:
S [ k ] ≜ 1 P ∫ 0 P s P ( x ) ⋅ e − i 2 π k P x d x = 1 P ∫ 0 P ( ∑ n = − ∞ ∞ s ( x + n P ) ) ⋅ e − i 2 π k P x d x = 1 P ∑ n = − ∞ ∞ ∫ 0 P s ( x + n P ) ⋅ e − i 2 π k P x d x , {\displaystyle {\begin{aligned}S[k]\ &\triangleq \ {\frac {1}{P}}\int _{0}^{P}s_{_{P}}(x)\cdot e^{-i2\pi {\frac {k}{P}}x}\,dx\\&=\ {\frac {1}{P}}\int _{0}^{P}\left(\sum _{n=-\infty }^{\infty }s(x+nP)\right)\cdot e^{-i2\pi {\frac {k}{P}}x}\,dx\\&=\ {\frac {1}{P}}\sum _{n=-\infty }^{\infty }\int _{0}^{P}s(x+nP)\cdot e^{-i2\pi {\frac {k}{P}}x}\,dx,\end{aligned}}}
where the interchange of summation with integration is justified by dominated convergence. With a change of variables ( τ = x + n P {\displaystyle \tau =x+nP} ), this becomes the following, completing the proof of Eq.2:
S [ k ] = 1 P ∑ n = − ∞ ∞ ∫ n P ( n + 1 ) P s ( τ ) e − i 2 π k P τ e i 2 π k n ⏟ 1 d τ = 1 P ∫ − ∞ ∞ s ( τ ) e − i 2 π k P τ d τ ≜ 1 P ⋅ S ( k P ) . {\displaystyle {\begin{aligned}S[k]={\frac {1}{P}}\sum _{n=-\infty }^{\infty }\int _{nP}^{(n+1)P}s(\tau )\ e^{-i2\pi {\frac {k}{P}}\tau }\ \underbrace {e^{i2\pi kn}} _{1}\,d\tau \ =\ {\frac {1}{P}}\int _{-\infty }^{\infty }s(\tau )\ e^{-i2\pi {\frac {k}{P}}\tau }d\tau \triangleq {\frac {1}{P}}\cdot S\left({\frac {k}{P}}\right)\end{aligned}}.}
This proves Eq.2 for L 1 {\displaystyle L^{1}} functions, in the sense that the right-hand side is the (possibly divergent) Fourier series of the left-hand side. Similarly, if S ( f ) {\displaystyle S(f)} is in L 1 ( R ) {\displaystyle L^{1}(\mathbb {R} )} , a similar proof shows the corresponding version of Eq.3. Finally, if s P ( x ) {\displaystyle s_{_{P}}(x)} has an absolutely convergent Fourier series, then Eq.2 holds as an equality almost everywhere. This is the case, in particular, when s ( x ) {\displaystyle s(x)} is a Schwartz function. Similarly, Eq.3 holds when S ( f ) {\displaystyle S(f)} is a Schwartz function.
Distributional formulation These equations can be interpreted in the language of distributions for a function s {\displaystyle s} whose derivatives are all rapidly decreasing (see Schwartz function). The Poisson summation formula arises as a particular case of the Convolution Theorem on tempered distributions, using the Dirac comb distribution and its Fourier series:
∑ n = − ∞ ∞ δ ( x − n T ) ≡ ∑ k = − ∞ ∞ 1 T ⋅ e − i 2 π k T x ⟺ F 1 T ⋅ ∑ k = − ∞ ∞ δ ( f − k / T ) . {\displaystyle \sum _{n=-\infty }^{\infty }\delta (x-nT)\equiv \sum _{k=-\infty }^{\infty }{\frac {1}{T}}\cdot e^{-i2\pi {\frac {k}{T}}x}\quad {\stackrel {\mathcal {F}}{\Longleftrightarrow }}\quad {\frac {1}{T}}\cdot \sum _{k=-\infty }^{\infty }\delta (f-k/T).}
In other words, the periodization of a Dirac delta δ , {\displaystyle \delta ,} resulting in a Dirac comb, corresponds to the discretization of its spectrum which is constantly one. Hence, this again is a Dirac comb but with reciprocal increments. For the case T = 1 , {\displaystyle T=1,} Eq.1 readily follows:
∑ k = − ∞ ∞ S ( k ) = ∑ k = − ∞ ∞ ( ∫ − ∞ ∞ s ( x ) e − i 2 π k x d x ) = ∫ − ∞ ∞ s ( x ) ( ∑ k = − ∞ ∞ e − i 2 π k x ) ⏟ ∑ n = − ∞ ∞ δ ( x − n ) d x = ∑ n = − ∞ ∞ ( ∫ − ∞ ∞ s ( x ) δ ( x − n ) d x ) = ∑ n = − ∞ ∞ s ( n ) . {\displaystyle {\begin{aligned}\sum _{k=-\infty }^{\infty }S(k)&=\sum _{k=-\infty }^{\infty }\left(\int _{-\infty }^{\infty }s(x)\ e^{-i2\pi kx}dx\right)=\int _{-\infty }^{\infty }s(x)\underbrace {\left(\sum _{k=-\infty }^{\infty }e^{-i2\pi kx}\right)} _{\sum _{n=-\infty }^{\infty }\delta (x-n)}dx\\&=\sum _{n=-\infty }^{\infty }\left(\int _{-\infty }^{\infty }s(x)\ \delta (x-n)\ dx\right)=\sum _{n=-\infty }^{\infty }s(n).\end{aligned}}}
Similarly:
∑ k = − ∞ ∞ S ( f − k / T ) = ∑ k = − ∞ ∞ F { s ( x ) ⋅ e i 2 π k T x } = F { s ( x ) ∑ k = − ∞ ∞ e i 2 π k T x ⏟ T ∑ n = − ∞ ∞ δ ( x − n T ) } = F { ∑ n = − ∞ ∞ T ⋅ s ( n T ) ⋅ δ ( x − n T ) } = ∑ n = − ∞ ∞ T ⋅ s ( n T ) ⋅ F { δ ( x − n T ) } = ∑ n = − ∞ ∞ T ⋅ s ( n T ) ⋅ e − i 2 π n T f . {\displaystyle {\begin{aligned}\sum _{k=-\infty }^{\infty }S(f-k/T)&=\sum _{k=-\infty }^{\infty }{\mathcal {F}}\left\{s(x)\cdot e^{i2\pi {\frac {k}{T}}x}\right\}\\&={\mathcal {F}}{\bigg \{}s(x)\underbrace {\sum _{k=-\infty }^{\infty }e^{i2\pi {\frac {k}{T}}x}} _{T\sum _{n=-\infty }^{\infty }\delta (x-nT)}{\bigg \}}={\mathcal {F}}\left\{\sum _{n=-\infty }^{\infty }T\cdot s(nT)\cdot \delta (x-nT)\right\}\\&=\sum _{n=-\infty }^{\infty }T\cdot s(nT)\cdot {\mathcal {F}}\left\{\delta (x-nT)\right\}=\sum _{n=-\infty }^{\infty }T\cdot s(nT)\cdot e^{-i2\pi nTf}.\end{aligned}}}
Or:
∑ k = − ∞ ∞ S ( f − k / T ) = S ( f ) ∗ ∑ k = − ∞ ∞ δ ( f − k / T ) = S ( f ) ∗ F { T ∑ n = − ∞ ∞ δ ( x − n T ) } = F { s ( x ) ⋅ T ∑ n = − ∞ ∞ δ ( x − n T ) } = F { ∑ n = − ∞ ∞ T ⋅ s ( n T ) ⋅ δ ( x − n T ) } as above . {\displaystyle {\begin{aligned}\sum _{k=-\infty }^{\infty }S(f-k/T)&=S(f)*\sum _{k=-\infty }^{\infty }\delta (f-k/T)\\&=S(f)*{\mathcal {F}}\left\{T\sum _{n=-\infty }^{\infty }\delta (x-nT)\right\}\\&={\mathcal {F}}\left\{s(x)\cdot T\sum _{n=-\infty }^{\infty }\delta (x-nT)\right\}={\mathcal {F}}\left\{\sum _{n=-\infty }^{\infty }T\cdot s(nT)\cdot \delta (x-nT)\right\}\quad {\text{as above}}.\end{aligned}}}
The Poisson summation formula can also be proved q