Frequency in discrete time

An arrow that turns by $\Omega$ each sample, marked once per sample. Watch the samples slow down again once $\Omega$ passes $\pi$.

An arrow, looked at once per sample

Sinusoids (3.2) gave you the digital frequency $\Omega=2\pi f/f_s$ in rad/sample, and showed that $\Omega$ and $\Omega+2\pi$ give the same samples. Sampling & aliasing (10.1) used that to find where a tone lands in hertz. This page makes $\Omega$ an axis, and the axis turns out to be a circle.

Start from the arrow of Complex exponentials & phasors (3.4). Look at it once per sample, at $n=0,1,2,\dots$:

Each sample number, the arrow has turned $\Omega$ further. Its horizontal position is the sample, $x[n]=\cos(\Omega n)$, which is the reading you used in Fourier series coefficients (7.2). So $\Omega$ is not mysterious: it is the turn per sample, in radians.

One turn per sample, on a wheel

The picture at the top of the page draws this arrow on a wheel, once per sample. At $\Omega=0$ the arrow does not turn: every mark sits at 1, and every sample is 1. At $0.25\pi$ rad/sample it turns an eighth of a turn per sample. Eight samples make one full turn, and the stems trace one cycle every 8 samples.

At $\pi$ it turns half a turn per sample. The marks alternate between 1 and −1, and so do the samples: nothing sampled can change faster. Watch the stems here. They flip between the top and the bottom of their range at every sample, which is the fastest a sequence can move.

Then $\Omega$ keeps growing, and the stems slow down again. At $1.5\pi$, three quarters of a turn forward lands where a quarter turn back would. The samples are those of $-0.5\pi$, and cos makes them the same as $0.5\pi$. At $2\pi$ a whole turn per sample looks like no turn. Marks and samples are back where $\Omega=0$ put them: the frequency axis is a circle.

A clock’s hand photographed once an hour makes the same point. Eleven hours forward looks like one hour back.

When the clip ends, mark 1 becomes a handle. Drag it round the wheel, or use the arrow keys, to set $\Omega$ from 0 to $2\pi$. The angle keeps counting past $\pi$, so you can follow the arrow all the way round. Try $\Omega=0.75\pi$ and then $1.25\pi$. Notice that the stems are identical, even though the marks are at mirrored places on the wheel.

Why the axis is a circle

Put the observation into symbols. Adding $2\pi$ to $\Omega$ adds a whole turn per sample, and a whole turn changes nothing:

because $e^{j2\pi n}=1$ for every whole $n$. So $\Omega$ and $\Omega+2\pi$ are one frequency. Move forward by $2\pi$ and you are where you began, which is what a circle does. I call the stretch from $-\pi$ to $\pi$ the principal interval. It holds every distinct frequency once, half-open at the bottom: $\Omega\in(-\pi,\pi]$, the same convention as the angle of a complex number.

Why is $\pi$ the fastest? A forward turn of $\Omega$ between $\pi$ and $2\pi$ lands where a backward turn of $2\pi-\Omega$, shorter than $\pi$, would land. The largest turn that is not better described as a shorter one the other way is half a turn, $\Omega=\pi$, where $e^{j\pi n}=(-1)^n$. At the other end, $\Omega=0$ is no turn, so $x[n]=1$ for every $n$. Low frequencies sit near 0 and high ones near $\pm\pi$.

For the real signal $\cos(\Omega n)$ there is one more step. Cosine is even, so $\Omega$ and $-\Omega$ give the same samples, which is why the 1.5π clip frame matches 0.5π.

Every function of $\Omega$ that is built from samples repeats every $2\pi$, so a page draws one turn, $(-\pi,\pi]$, and that shows all of it. Here is the same fact in the terms of The sampling theorem (10.2). A shift by $f_s$ hertz is a shift by $2\pi$ in $\Omega$, since $\Omega=2\pi f/f_s$. The copies of a sampled spectrum, which sit every $f_s$ in hertz, therefore sit every $2\pi$ in rad/sample.

Four rulers for one frequency

A frequency has four names in common use, and only one of them needs the sample rate. You have two already: $\Omega$ in rad/sample, and $f$ in hertz. The other two are rescalings of $\Omega$.

The first is cycles per sample, $\nu=\Omega/2\pi=f/f_s$. The fastest sequence turns half a cycle per sample, so $\nu$ runs from 0 to 0.5. The second is the normalised frequency of SciPy and MATLAB, $\Omega/\pi=2f/f_s$, which runs from 0 to 1 and puts 1 at the Nyquist frequency of 10.1. It is the same quantity as $\Omega$ written in units of $\pi$ rad/sample. Hertz needs $f_s$: $f=\Omega f_s/2\pi$.

The normalised scale is the one that trips people up. In SciPy, butter(4, 0.25) asks for a cutoff of $0.25\pi$ rad/sample, not 0.25 cycles per sample.

The next picture stacks the four scales, with one marker across them all. Watch which ruler changes when the sample rate does.

The clip begins with the marker at $0.25\pi$ and $f_s=8$ kHz: 0.125 cycles per sample, 0.25 on SciPy’s scale, and 1000 Hz. The marker then moves to $0.5\pi$, a quarter cycle per sample, 0.5 on SciPy’s scale and 2000 Hz at 8 kHz. Then it moves to $\pi$, the fastest: half a cycle per sample, 1 on SciPy’s scale, and $f_s/2=4000$ Hz, the Nyquist frequency of 10.1.

Then the marker returns to $0.25\pi$ and $f_s$ eases from 8 kHz to 48 kHz. Watch the rulers. The hertz labels and the hertz ruler’s title change in place, and nothing else moves: the same $0.25\pi$ rad/sample is now 6000 Hz.

When the clip ends, drag the marker, or use the arrow keys, to choose a frequency from 0 to $\pi$ at $f_s=48$ kHz.

In SciPy, freqz(b, a) returns $\Omega$ from 0 to $\pi$, while freqz(b, a, fs=8000) returns hertz from 0 to 4000. Some analysers draw 0 to $f_s$ instead, which is one full turn with the top half holding the negative frequencies. Check which one a plot uses before you read a number off it.

Worked example

1. Concert A. 440 Hz at $f_s=8$ kHz: $\Omega=2\pi\cdot440/8000=0.3456$ rad/sample $=0.110\pi$. Then $\nu=0.0550$ and the normalised value is 0.110. Since $\nu=11/200$, the samples repeat every 200 samples, as 3.2 found.
2. 1 kHz. At 8 kHz: $\Omega=0.25\pi=0.7854$ rad/sample, $\nu=0.125$, normalised 0.25. At 48 kHz: $\Omega=0.1309$ rad/sample $=0.0417\pi$, $\nu=0.02083$, normalised 0.04167. And $0.25\pi$ at 48 kHz is 6000 Hz.
3. Past $\pi$. 7 kHz at 8 kHz is $\Omega=1.75\pi$, the same samples as $-0.25\pi$ and, since cosine is even, as $0.25\pi$: 1 kHz, the fold of 10.1.
4. Samples. $\cos(\pi n)=1,-1,1,-1,\dots$, and $\cos(1.5\pi n)=\cos(0.5\pi n)=1,0,-1,0,\dots$
5. SciPy. butter(4, 0.25) and butter(4, 1000, fs=8000) return identical coefficients.
6. Copies. The signal of 10.2, reaching 4 kHz, sampled at $f_s=10$ kHz reaches $2\pi\cdot4/10=0.8\pi$, with copies centred at $\pm2\pi$.

Where you’ll meet this

Every spectrum in the next chapters is drawn against $\Omega$ on $(-\pi,\pi]$. The DTFT (12.2) builds the spectrum of a sequence on this axis. The DFT (13.2) samples it at $2\pi k/N$. Filter specifications (18.1) give cutoffs in the normalised scale, so the $0.25\pi$ trap above comes back there.

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