Wavelength and Frequency — Question 8

Question 8

Spatial sampling can hide a shorter wavelength

A sinusoidal snapshot is sampled only at positions $x_j=j\Delta x$, where $\Delta x=.10\,\mathrm{m}$. Compare wave numbers $k_1=4\pi\,\mathrm{rad/m}$ and $k_2=24\pi\,\mathrm{rad/m}$ with equal amplitudes and zero phase.

1. Show that every sampled displacement is identical for the two profiles.

2. Find their wavelengths and propose a way to remove this ambiguity.

Question 8 - Solution

Spatial sampling can hide a shorter wavelength

Method and mathematical working

Evaluate both phases on the sampling lattice. An integer multiple of $2\pi$ at every sample produces identical cosine values even if the curves differ between samples.$$\begin{aligned}(k_2-k_1)x_j&=(k_2-k_1)j\Delta x=2\pi j.\\[4pt](k_2-k_1)j\Delta x&=(24\pi-4\pi)j(0.10)=2\pi j,\\\lambda_1&=0.50\,\mathrm m,\qquad\lambda_2=1/12\,\mathrm m.\end{aligned}$$

(a) Calculation and conclusion

The phase difference at sample $j$ is $(k_2-k_1)x_j=20\pi(.10j)=2\pi j$. Thus the two cosines agree at all sampled positions.

(b) Interpretation and limits

The wavelengths are $.50\,\mathrm{m}$ and $1/12=.08333\,\mathrm{m}$. Finer sampling with a justified shortest-wavelength bound, or additional nonuniform sample positions, can distinguish them. Agreement on a coarse regular grid does not identify the continuous spatial pattern.

Physical check

This is a spatial alias, not evidence that the underlying functions are equal. Extra measurements must break the repeating sample-phase relation or establish a valid shortest-wavelength bound.