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:
- the amplitude $A$ — how big the swing is (volts, tesla, watts1/2… whatever your signal is measured in);
- the frequency $\nu_0$ — how many full cycles happen per second, in hertz;
- the phase $\varphi$ — where in its cycle the oscillation is at $t=0$, in radians.
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.
Amplitude scales the curve vertically, frequency squeezes it horizontally, phase slides it sideways without changing its shape. Hover to read off values.
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:
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.
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:
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).
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.
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
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.
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.
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.
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.