= Spatial sampling of a wave
{title2=$|k|<\pi/p$}
In <spatial sampling of a wave>, a detector records a <wave> at positions $x_n=np$ separated by pitch $p>0$. Angular <wavenumbers> differing by $2\pi m/p$, with <integers> $m$, give identical samples of a complex <Fourier mode>:
$$
e^{i(k+2\pi m/p)np}=e^{iknp}.
$$
This is spatial <aliasing>. To distinguish arbitrary <sinusoids> within a prescribed band, require $|k|<\pi/p$, equivalently $|\nu_x|<1/(2p)$, the <Nyquist spatial frequency>. Thus $p<\lambda_{\min}/2$ for the shortest allowed <wavelength>. At equality a <sine> can vanish at every sample, so phase information can be lost. The <Nyquist–Shannon sampling theorem> states the precise reconstruction conditions for a <bandlimited function>; a finite scan introduces further limitations.
The synthetic two-tone snapshot is
$$
\frac{u(x)}A=\cos(2\pi\nu_1x)+\frac12\cos(2\pi\nu_2x),\qquad \nu_1=2\,\mathrm{m^{-1}},\quad \nu_2=8\,\mathrm{m^{-1}},
$$
where both $\nu_j$ count cycles per metre. The component <wavelengths> are $0.50\,\mathrm m$ and $0.125\,\mathrm m$. With $p=0.05\,\mathrm m$, both lie below the <Nyquist spatial frequency> of $10$ cycles per metre. With $p=0.10\,\mathrm m$, the $8$-cycle component aliases to the $2$-cycle component: at every $x_n=n/10$, $\cos(2\pi\cdot8x_n)=\cos(2\pi\cdot2x_n)$. The sampled sum is therefore indistinguishable from $1.5\cos(4\pi x)$, although the continuous curves differ between samples. Drawing lines between sample points cannot resolve this ambiguity.
\Image[/two-tone-spatial-sampling.png]
{title=Synthetic two-tone spatial waveform: fine samples distinguish the components, coarse samples hide the shorter wavelength}
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