Chapter 1

The pure tone

Everything in this course — phase noise, power spectral densities, Allan deviations, LIGO's sensitivity curve — is a story about one equation with three knobs. This chapter is about the equation itself, so that when we start adding noise to the knobs, nothing else will be new.

A signal with three knobs

Take the most boring signal in physics: a voltage (or a laser field, a trapped ion's drive, the output of a quartz crystal, anything that oscillates) doing exactly the same thing forever,

$$ V(t) = A \cos\!\big(2\pi \nu_0 t + \varphi\big). $$

There are only three things you can adjust:

Drag each slider below and watch which aspect of the curve it controls. The faint curve is a frozen reference ($A=1$, $\nu_0=2\,$Hz, $\varphi=0$) so you can see what changed.

The three knobs of a pure tone

Amplitude scales the curve vertically, frequency squeezes it horizontally, phase slides it sideways without changing its shape. Hover to read off values.

The eternal 2π trap
Physicists use two frequencies: the ordinary frequency $\nu$ (cycles per second, Hz) and the angular frequency $\omega = 2\pi\nu$ (radians per second). Spectrum analyzers, data sheets and this course all use $\nu$ (and later $f$ for "offset" frequencies). Theory papers often use $\omega$. Roughly half of all wrong noise calculations in the wild are a factor of $2\pi$ (or $\sqrt{2\pi}$) inserted or dropped at this step. When in doubt, write the cosine out and check that its argument is dimensionless.

The phasor: one picture instead of trigonometry

Cosines are awkward to manipulate. The clean way to think about an oscillation — and the picture that will carry us through modulation and noise — is a rotating arrow. Euler's formula says

$$ V(t) = A\cos(2\pi\nu_0 t + \varphi) = \mathrm{Re}\Big[\, A\, e^{\,i(2\pi\nu_0 t + \varphi)} \Big], $$

so a pure tone is the shadow (real part) of an arrow of fixed length $A$ rotating counter-clockwise in the complex plane at $\nu_0$ turns per second, having started at angle $\varphi$. Press play:

The rotating-arrow (phasor) picture

Left: the arrow $A e^{i\Phi(t)}$ in the complex plane. Right: its horizontal shadow traced over time — that is your oscilloscope trace. Amplitude sets the arrow's length; frequency its rotation rate; phase its starting angle.

The angle of the arrow at time $t$,

$$ \Phi(t) = 2\pi\nu_0 t + \varphi, $$

is called the instantaneous phase (also "total phase"). It is the single most important quantity in this course. Notice what the two other knobs are in terms of it: the phase $\varphi$ is just $\Phi(0)$, and the frequency is how fast the phase advances:

$$ \nu(t) \;=\; \frac{1}{2\pi}\,\frac{d\Phi}{dt}. $$

For a pure tone this derivative is the constant $\nu_0$. But keep the formula: frequency is the slope of phase. When the phase starts wandering randomly in chapter 6, this one line is what will connect "phase noise" and "frequency noise" — they are the same wandering, looked at either as position or as velocity.

The master equation of this course
A real oscillator is never pure. Its amplitude and phase both wander, so the honest description is $$ V(t) = \big[A_0 + \delta\!A(t)\big]\, \cos\!\big(2\pi\nu_0 t + \varphi(t)\big). $$ Every kind of "noise" we will meet lives in one of the two wiggle terms: $\delta\!A(t)$ is amplitude noise, $\varphi(t)$ is phase noise, and the derivative $\delta\nu(t)=\dot\varphi/2\pi$ is frequency noise. The whole course is the study of this equation.

What a pure tone looks like on a spectrum analyzer

The second home of every signal is the frequency domain. For a truly eternal cosine, all the power sits at exactly one frequency: the spectrum is an infinitely narrow spike at $\nu_0$. But you never observe a signal forever — you record for some time $T$ and ask your FFT (or your spectrum analyzer) what it saw. The result is a peak of finite width

$$ \Delta\nu \sim \frac{1}{T}, $$

the Fourier uncertainty relation: to know a frequency to 1 Hz you must watch for about a second; to know it to 1 mHz, about $10^3\,$s. This width has nothing to do with noise — it is the resolution limit of a finite observation. Try it:

A finite observation smears the spike

The spectrum of a perfect 8 Hz tone observed for a time $T$. The dashed line marks 8 Hz. Doubling $T$ halves the width of the peak (and the plot's frequency gridding gets finer, because $1/T$ is also the FFT's frequency resolution).

Lab note
On a swept spectrum analyzer the same physics appears as the resolution bandwidth (RBW): an RBW of 1 kHz means the instrument effectively averages for $\sim$1 ms per point, and no feature narrower than 1 kHz can be resolved. A "carrier" on the screen always has the shape of the RBW filter, not the shape of the signal — until the signal's own noise is wider than the RBW. That crossover is exactly where chapter 6 begins.

Two tones: the simplest interesting signal

Before we modulate anything, add just one more cosine. Two equal tones at nearby frequencies $\nu_1$ and $\nu_2$ give, by a trigonometric identity,

$$ \cos(2\pi\nu_1 t) + \cos(2\pi\nu_2 t) = 2\, \underbrace{\cos\!\big(2\pi \tfrac{\nu_1-\nu_2}{2}\, t\big)}_{\text{slow envelope}} \;\underbrace{\cos\!\big(2\pi \tfrac{\nu_1+\nu_2}{2}\, t\big)}_{\text{fast carrier}} . $$

In the time domain you see beats: a fast oscillation at the average frequency whose amplitude swells and shrinks at the difference frequency. In the frequency domain you see the truth with no trigonometry at all: just two spikes.

Beats: two spikes, one envelope

Left: time trace with its slow envelope. Right: the spectrum — two clean peaks at $\nu_1$ and $\nu_2$. The beat you see in time is the spacing you read in frequency.

This is worth internalizing now, because it is the pattern of everything to come: a signal that looks complicated in one domain is often trivially simple in the other. A wobbling envelope in time = two spectral lines. And when we modulate a carrier in the next chapter, the modulation will show up as sidebands — extra spikes placed symmetrically around the carrier, exactly like these two.

The same thing in code

Every plot on this page is generated live by a few lines of JavaScript. Here is the essential part (the full source is in this page — view source), in case reading code is how the mathematics clicks for you:

// a pure tone, sampled at fs for T seconds
const fs = 200, T = 4, N = Math.round(fs * T);
const t = [], V = [];
for (let i = 0; i < N; i++) {
  t.push(i / fs);
  V.push(A * Math.cos(2 * Math.PI * nu0 * t[i] + phi));
}

// its spectrum: |FFT|^2, scaled to a power spectral density
const { f, psd } = FFT.periodogram(V, fs);   // chapter 4 explains the scaling

Exercises

Exercise 1.1 — phase as a time shift

A phase offset $\varphi$ slides the cosine in time: $\cos(2\pi\nu_0 t + \varphi) = \cos\!\big(2\pi\nu_0 (t + \varphi/2\pi\nu_0)\big)$. How many milliseconds does $\varphi = 90^\circ$ shift a 2 Hz tone? Check your answer with the first demo.

Solution

$90^\circ = \pi/2$ rad shifts by $\Delta t = \varphi/(2\pi\nu_0) = (\pi/2)/(2\pi\cdot 2\,\text{Hz}) = 1/8\,\text{s} = 125$ ms — a quarter of the 500 ms period, earlier in time. On the phasor picture, the arrow simply starts a quarter-turn ahead.

Exercise 1.2 — Fourier resolution

You beat two lasers on a photodiode and record the beat note for 0.1 s. Roughly how well can you determine the beat frequency? What recording length do you need to resolve two lines 1 Hz apart? Verify with the third demo.

Solution

The linewidth of the measured peak is $\sim 1/T = 10$ Hz, so the frequency is uncertain at roughly that level (you can split the line somewhat better if the signal-to-noise is high). To resolve two lines 1 Hz apart you need the peaks narrower than their spacing: $T \gtrsim 1$ s.

Exercise 1.3 — reading a beat

In the beats demo, set $\nu_1 = 40$ Hz and $\nu_2 = 43$ Hz. What do you expect the envelope period to be? Why does the intensity envelope repeat at $\nu_1-\nu_2 = 3$ Hz and not at the $(\nu_1-\nu_2)/2 = 1.5$ Hz that appears in the identity?

Solution

The field envelope $2\cos(2\pi\cdot 1.5\,t)$ goes through zero twice per cycle, and the eye (or a power detector) sees the envelope's magnitude — which peaks every $1/3$ s. Detected power goes as the square, $|2\cos(\pi \Delta\nu\, t)|^2 = 2 + 2\cos(2\pi \Delta\nu\, t)$: a DC term plus a clean beat at the difference frequency $\Delta\nu = 3$ Hz. This is why heterodyne detection gives you the difference frequency directly.

Exercise 1.4 — a phasor warm-up for chapter 2

Two phasors of equal length rotate at slightly different rates $\nu_1$ and $\nu_2$. Describe geometrically what their sum does. What is the length of the sum arrow as a function of time?

Solution

In a frame rotating at the average frequency, the two arrows rotate slowly in opposite directions at $\pm(\nu_1-\nu_2)/2$. Their sum always lies along the bisector, with length $2A|\cos(\pi(\nu_1-\nu_2)t)|$ — it breathes between $2A$ and $0$. The sum arrow is the beat envelope; the beating you saw in the last demo is two phasors alternately aligning and cancelling.