Aliasing & Moiré

Sampling a sinusoid

Sampling a continuous signal $x(t)$ at rate $f_s$ keeps only the values $x[n] = x(n / f_s)$. For a sinusoid of frequency $f$ and any integer $k$, the sinusoids at $f + k f_s$ and at $k f_s - f$ pass through exactly the same samples, since

and the cosine is even, so $-f$ with phase $-\varphi$ gives the same values as $f$ with $\varphi$. The samples cannot tell these frequencies apart. Exactly one of them lies in $[0, f_s / 2]$: the frequency that the samples appear to have, the alias

In the spectrum, sampling turns the two lines at $\pm f$ into copies repeated at every multiple of $f_s$. Whatever copy lands inside the band $[-f_s/2, f_s/2]$ is what a reconstruction from the samples produces.

The Nyquist rate

If $f < f_s / 2$, the alias is the signal itself and nothing is lost. This is the sampling theorem: a signal that contains no frequencies above $f_\text{max}$ is determined by its samples if

$2 f_\text{max}$ is the Nyquist rate, and $f_s / 2$ the Nyquist frequency of a given sampling rate: more than two samples per period. Above the Nyquist frequency, the alias folds back: raising $f$ from $f_s / 2$ to $f_s$ lowers $f_a$ from $f_s / 2$ to 0. This is why wheels in films seem to turn slowly or backwards: the frame rate samples the rotation too sparsely.

Aliasing in images

An image is a 2D signal, sampled once per pixel, so its Nyquist frequency is $\tfrac{1}{2}$ cycle per pixel: a stripe pattern with a period of 2 pixels, black and white. Keeping only every $k$-th pixel lowers the sampling rate by $k$, and the new Nyquist frequency is $\tfrac{1}{2k}$ cycles per pixel. Every finer detail folds back into coarser patterns that are not in the scene: moiré. The zone plate shows it best, since its frequency grows steadily from the center outwards; after subsampling, new rings appear around new centers. The same happens at the fence in the landscape, and the aliased stripes of a fine pattern can even run in a different direction than the original ones.

Anti-aliasing: filter first, then subsample

Once aliasing has happened, it cannot be undone: the samples of the fine pattern and of its alias are the same. The only remedy is to remove the frequencies above the new Nyquist frequency before sampling, with a low-pass filter. The Gaussian is a good choice; its frequency response is again a Gaussian,

with $f$ in cycles per pixel: a wider Gaussian in space is narrower in frequency. A rule of thumb for subsampling by $k$ is $\sigma \approx k / 2$. A smaller $\sigma$ leaves more aliasing, a larger one blurs away detail that the smaller image could still show. Image pyramids use this “blur, then subsample” step at every level. A camera does it optically: the lens and the sensor pixel area blur the image before it is sampled (compare cell center and cell average in sampling & quantization), and many cameras add an optical low-pass filter in front of the sensor. The Gaussian itself is covered in Gaussian & separability.

Try this

• Start with $f = 8$ Hz at $f_s = 10$ Hz: the samples trace a 2 Hz sinusoid. Lower $f$ slowly to 5 Hz and watch the alias rise to meet it.
• Set $f = 10$ Hz: one sample per period, and the samples are constant. The sinusoid has aliased to 0 Hz.
• Set $f = 5$ Hz exactly at $f_s = 10$ Hz and change the phase: at 90° all samples are 0 and the sinusoid disappears. The theorem needs $f_s > 2f$, strictly.
• With $f = 8$ Hz, change the phase: the alias moves in the opposite direction, because it comes from the mirrored copy $f_s - f$.
• Pick the zone plate, subsample by 4 and set $\sigma = 0$: outside the center, the rings turn into new ring patterns around new centers. Increase $\sigma$ until they fade to gray.
• Look at the line groups of the test chart subsampled by 3 without a pre-filter: the finest groups show stripes with the wrong period, or none at all.
• Set $\sigma = 4$ with $k = 2$: no aliasing, but the result is blurrier than necessary. Compare the Gaussian's response at $\tfrac{1}{4k}$ in the values panel.