Sampling & aliasing

I sample a 7 kHz tone 8000 times a second. What do the samples sound like? Watch the dots, then the wave through them.

The same dots, two waves

First the words. The real tone, which goes on in continuous time, I will write $x_c(t)$. Sampling keeps its value every $T_s$ seconds, so the samples are

where $n$ is the sample number and $f_s$ is the sample rate, here 8000 samples a second, or 8 kHz. You met this in Sinusoids (3.2). The dots come every $T_s=1/8000$ s, which is 0.125 ms.

Press “Hear the samples” in the picture at the top of the page: one second of the samples sounds at 1 kHz.

Look at the slow wave and find a dot it misses. There is none. It passes through all seventeen. It is a clean, smooth cosine, only the wrong one: the samples are exactly those of a 1 kHz cosine, so nothing in them says “7 kHz”. That is an alias: a clean, slower, wrong tone, not noise.

A ceiling fan filmed by a phone can seem to turn slowly, for the same reason. The camera sees the blades only once per frame.

Where a tone lands

Why 1 kHz and not some other number? The rule is the one from Sinusoids (3.2), written in hertz instead of radians per sample.

The digital frequency is $\Omega=2\pi f/f_s$ (3.2 §4). Adding $f_s$ to $f$ adds $2\pi$ to $\Omega$, and an extra $2\pi$ gives the same samples (3.2 §5). So $f$ and $f+f_s$ give the same samples. A cosine is even, so $f$ and $f_s-f$ give the same samples too. Every tone therefore has a twin between 0 and half the sample rate. That half is the Nyquist frequency,

For the 7 kHz tone, 3.2’s rule goes like this: $\Omega=2\pi\cdot7/8=1.75\pi$ rad/sample. Subtract $2\pi$ to get $-0.25\pi$. Its mirror twin is $0.25\pi=2\pi\cdot1/8$, which is 1 kHz at 8 kHz.

Watch the dot as the input frequency climbs. Its landing point rises with it until the tone reaches $f_N=4$ kHz. Then it turns back down, reaching 0 at $f_s=8$ kHz, and climbs again.

Notice the rings stamped along the path: six different tones, 1, 7, 9, 15, 17 and 23 kHz, all land on 1 kHz. When the animation has finished, the instrument is yours. Drag across the plot to choose the input frequency, pick 7 and check the dot sits at 1.

The rule behind the zig-zag has two steps. First take away whole sample rates: the remainder $f_r=f\bmod f_s$. For example $25\bmod 8=1$, because $25=3\cdot8+1$. Then fold the remainder back below $f_N$:

The fold is the cosine’s two arrows again. A cosine is two arrows spinning opposite ways (see Complex exponentials & phasors (3.4)). Sampling moves each arrow’s frequency up or down by whole multiples of $f_s$. The arrow at $-f$ moved up by $f_s$ sits at $f_s-f$. The page The sampling theorem (10.2) draws the whole picture.

Now run it backwards. The dots trace a 1 kHz wave, and you know the tone was 7 kHz. The sample rate was between 5 and 10 kHz. What was it? Try $f_s=8$: the tone lands at $\min(7,1)=1$ kHz. Try $f_s=6$: $7\bmod6=1$ and $\min(1,5)=1$ kHz. So the sample rate was 8 kHz or 6 kHz.

That is the catch. Once the dots are all you have, the tone and its alias cannot be told apart, and the aliasing cannot be undone. Converters filter before they sample, which is the subject of Anti-aliasing and practical converters (10.4).

A square wave aliases at any note

So far there was one tone. Real sounds hold many. In Signals as sums of sinusoids (7.1) you built a square wave from spinning arrows at the odd harmonics $3f,5f,7f,\dots$, with lengths $4/(\pi k)$. They never stop.

That is the problem. However low the note, some harmonic lies above $f_N$, and it lands somewhere else. Here is a 1.5 kHz square wave sampled at 8 kHz. Watch its first four harmonics move, one at a time, into the sampled panel.

The first stays where it is; the other three land at 3.5, 0.5 and 2.5 kHz. Only 1.5 kHz belongs to the note, and the others are not multiples of it, so it sounds out of tune. Press “Hear the samples” to hear the sampled square wave. A cheap synthesiser whose high notes sound out of tune does exactly this.

The four lines shown carry 0.9496 of the square wave’s power (Fourier series coefficients (7.2)). The rest of the harmonics fold too. A caution: this works out of tune only when the note does not divide $f_s$ evenly. A 1 kHz square wave at 8 kHz folds its harmonics onto multiples of 1 kHz. At 1.5 kHz and 8 kHz, as here, they do not.

Worked example

1. A CD chain. A 25 kHz interferer meets $f_s=44.1$ kHz, so $f_N=22.05$ kHz. The remainder is $f_r=25$ kHz, less than $f_s$, and above $f_N$. So $f_a=44.1-25=19.1$ kHz, inside the audio band.
2. 9 kHz at 10 kHz. $f_r=9$ kHz, and $f_a=\min(9,1)=1$ kHz.
3. Through 3.2. 7 kHz at 8 kHz is $\Omega=2\pi\cdot7/8=1.75\pi$ rad/sample. Minus $2\pi$ gives $-0.25\pi$, whose twin is $0.25\pi=2\pi\cdot1/8$. That is 1 kHz.
4. The 1.5 kHz square wave at 8 kHz. The harmonics at 1.5, 4.5, 7.5, 10.5, 13.5, 16.5, 19.5 and 22.5 kHz land at 1.5, 3.5, 0.5, 2.5, 2.5, 0.5, 3.5 and 1.5 kHz.
5. Reverse. A 7 kHz tone that shows as 1 kHz, with $5<f_s<10$ kHz, means $f_s=8$ or $6$ kHz.
6. Audio. To keep a 20 kHz tone below $f_N$ you need $f_s>40$ kHz. The page The sampling theorem (10.2) explains why that is enough.

Where you’ll meet this

Every audio interface, camera and sensor samples. The next pages build on this one. The page The sampling theorem (10.2) shows the copies of the spectrum behind the fold. The page Reconstruction (10.3) rebuilds the wave from its dots. The page Anti-aliasing and practical converters (10.4) filters first, so the alias never forms.

Reference card