Poisson Summation Sampling and Lattice Duality — Examples

Each off-diagonal entry of $A$ vanishes, so for every permutation $\sigma \mathrm{\ne }\mathrm{id}$ some factor $A_{\sigma (i),i}$ with $\sigma (i)\mathrm{\ne }i$ is zero and only $\sigma =\mathrm{id}$ contributes: $\det A={\prod }_{i=1}^n{a}_i\mathrm{\ne }0$ by [F1]. Hence $A$ is invertible with inverse $\mathrm{diag}(a_1^{-1},\dots ,a_n^{-1})$, as the product computation $(A{A}^{-1}{)}_{ij}={\delta }_{ij}$ shows [F1]. Therefore $\mathrm{covol}(\mathrm{\Lambda })=\mathrm{\mid }\det A\mathrm{\mid }={\prod }_{i=1}^n\mathrm{\mid }{a}_i\mathrm{\mid }$, and since a diagonal matrix equals its transpose, $A^{-T}=A^{-1}=\mathrm{diag}(a_i^{-1})$, so ${\mathrm{\Lambda }}^{\ast }=\mathrm{diag}(a_1^{-1},\dots ,a_n^{-1}){\mathbb{Z}}^n$, which is the displayed set of tuples $(m_1\mathrm{/}{a}_1,\dots ,m_n\mathrm{/}{a}_n)$ with $m\in {\mathbb{Z}}^n$ by [F2] and the definition of $A{\mathbb{Z}}^n$.

For $n=1$ the matrix is $(a_1)$ with $a_1=a\mathrm{\ne }0$, giving $\mathrm{covol}(a\mathbb{Z})=\mathrm{\mid }a\mathrm{\mid }$ and $(a\mathbb{Z}{)}^{\ast }=a^{-1}\mathbb{Z}$; at the sampling spacing $a=h>0$ this is exactly the pair $h\mathbb{Z}$, $h^{-1}\mathbb{Z}$ used by the sampling results, whose dual lattice is again full-rank and satisfies $\mathrm{covol}({\mathrm{\Lambda }}^{\ast })=\mathrm{covol}(\mathrm{\Lambda }{)}^{-1}=h^{-1}$

By [F1] the $L^1$ transform of $\mathbf{1}$ is $\mathrm{sinc}$, so the $L^2$ transform of $\mathbf{1}$ is represented by $\mathrm{sinc}$ [F2]. Since $\mathrm{sinc}$ is bounded by the two bounds of The normalised sinc function, it lies in $L^2$, and ${\mathcal{F}}_2(\mathrm{sinc})={\mathcal{F}}_2({\mathcal{F}}_2\mathbf{1})=R\mathbf{1}=\mathbf{1}$, the last equality because $\mathbf{1}$ is even and $\mathbf{1}\circ (-1)=\mathbf{1}$.

Thus $\mathrm{sinc}$ is band-limited with $h=1$ and band $[-1\mathrm{/}2,1\mathrm{/}2]$, and its samples are $\mathrm{sinc}(0)=1$ and $\mathrm{sinc}(k)=0$ for nonzero integers $k$ (The normalised sinc function). Its Shannon series therefore has the single nonzero term at $k=0$: ${\sum }_k\mathrm{sinc}(k)\mathrm{sinc}(x-k)=\mathrm{sinc}(x)$ for every $x$, and ${\sum }_k\mathrm{\mid }\mathrm{sinc}(k)\mathrm{\mid }=1<\mathrm{\infty }$, so both clauses of [F3] hold and the reconstruction is pointwise everywhere. The normalisation check [F4] also matches: here $\widehat{f}=\mathbf{1}$, so $G_1(\theta )=\mathbf{1}(-\theta )=1$ on the fundamental interval and $\widehat{G_1}(k)=1$ for $k=0$ and $0$ otherwise, which equals $1^{1\mathrm{/}2}\mathrm{sinc}(k)$; the Fourier pair is the interval indicator and $\mathrm{sinc}$, with ${\mathcal{F}}_2\mathrm{sinc}={\mathbf{1}}_{[-1\mathrm{/}2,1\mathrm{/}2]}$. Values at the band endpoints do not affect these $L^2$ classes.

For every $k\in \mathbb{Z}$ the addition law [F1] gives $f_2(hk)=e^{2\pi i\xi hk}{e}^{2\pi i(m\mathrm{/}h)hk}=f_1(hk)e^{2\pi imk}$, and $e^{2\pi imk}=1$ because $mk\in \mathbb{Z}$ and $2\pi imk\in 2\pi i\mathbb{Z}$ [F2]. Hence $f_2$ and $f_1$ have identical samples on $h\mathbb{Z}$.

The two functions are distinct: at $x=h\mathrm{/}(2m)$, which is a real number because $m\mathrm{\ne }0$, one has $f_2(x)f_1(x{)}^{-1}=e^{2\pi i(m\mathrm{/}h)(h\mathrm{/}(2m))}=e^{i\pi }=-1$ by [F1] and [F2], and $-1\mathrm{\ne }1$, so $f_2\mathrm{\ne }{f}_1$. Together with step 1.1 this refutes the displayed statement: the frequencies $\xi$ and $\xi +m\mathrm{/}h$ are distinct, yet every procedure reading only the samples on $h\mathbb{Z}$ sees the same data. The witness consists of pure frequencies of constant modulus one, which are bounded but not square-integrable on $\mathbb{R}$; it therefore does not contradict the $L^2$ reconstruction theorem of the A page.