Chapter 6
Noisy oscillators: amplitude, phase & frequency noise
This is the chapter the whole course has been building toward. Back to the master equation: $V(t) = [A_0 + \delta\!A(t)]\cos\!\big(2\pi\nu_0 t + \varphi(t)\big)$. In chapter 2 we wiggled $\delta\!A$ and $\varphi$ deliberately, with clean sinusoidal modulation. A real oscillator wiggles them randomly — and everything we need is already on the table: just apply chapter 4's power spectral densities to the two wiggle terms.
The cast of characters
Four random processes describe one noisy oscillator, and each has its own PSD with its own units. Fix the conventions once (one-sided PSDs, ordinary frequency in Hz throughout, $f$ always the offset frequency of the noise, distinct from the carrier $\nu_0$):
| process | definition | PSD | units of PSD |
|---|---|---|---|
| fractional amplitude noise | $a(t) = \delta\!A(t)/A_0$ | $S_a(f)$ | $1/\text{Hz}$ |
| phase noise | $\varphi(t)$ | $S_\varphi(f)$ | $\text{rad}^2/\text{Hz}$ |
| frequency noise | $\delta\nu(t) = \dot\varphi(t)/2\pi$ | $S_{\delta\nu}(f)$ | $\text{Hz}^2/\text{Hz}$ |
| fractional frequency noise | $y(t) = \delta\nu(t)/\nu_0$ | $S_y(f) = S_{\delta\nu}(f)/\nu_0^2$ | $1/\text{Hz}$ |
Amplitude noise mostly minds its own business (it returns in the spectrum analyzer section below). The real action is between $\varphi$ and $\delta\nu$, and it is governed by one identity — the identity of oscillator physics:
$$ \boxed{\; S_{\delta\nu}(f) \;=\; f^{2}\, S_\varphi(f) \;} $$
The derivation takes three lines. Recall from chapter 1 that frequency is the slope of phase, and from chapter 4 that differentiation multiplies a Fourier amplitude by $i\,2\pi f$:
$$ \begin{aligned} \delta\nu(t) &= \frac{\dot\varphi(t)}{2\pi} &&\text{(slope of phase)}\\[2pt] \widetilde{\delta\nu}(f) &= \frac{(i\,2\pi f)\,\tilde\varphi(f)}{2\pi} = i f\,\tilde\varphi(f) &&\text{(derivative theorem)}\\[2pt] S_{\delta\nu}(f) &= |if|^2\, S_\varphi(f) = f^2\, S_\varphi(f). &&\text{(square the magnitude)} \end{aligned} $$
Note how tidily the $2\pi$'s cancel: differentiation brings a factor $2\pi f$ to the amplitude spectrum, and the $1/2\pi$ in $\delta\nu = \dot\varphi/2\pi$ removes it, leaving a bare $f^2$ in power. Phase noise and frequency noise are the same wandering with two bookkeepings — position versus velocity of the phasor's angle. A single physical process therefore has two names, one per bookkeeping. Multiplying by $f^2$ shifts a power law by two rungs, which generates the famous ladder of oscillator noise types:
| phase PSD shape | frequency PSD shape | name |
|---|---|---|
| $\propto f^{0}$ (flat) | $\propto f^{2}$ | white phase noise (white PM) |
| $\propto f^{-1}$ | $\propto f^{1}$ | flicker phase noise (flicker PM) |
| $\propto f^{-2}$ (random-walk phase) | $\propto f^{0}$ (flat) | white frequency noise (white FM) |
| $\propto f^{-3}$ | $\propto f^{-1}$ | flicker frequency noise (flicker FM) |
| $\propto f^{-4}$ | $\propto f^{-2}$ | random-walk frequency noise (random-walk FM) |
Real oscillators show several rungs at once: typically white PM at high offsets, then white FM, then flicker FM and random-walk FM as you move toward the carrier. Every color of chapter 5 appears — just attached to the phase of an oscillator instead of to a voltage.
One noise, two spectra
Seeing is believing. Below, one random phase record $\varphi(t)$ is generated (sampled at $f_s = 256$ Hz for 128 s), its numerical derivative gives $\delta\nu(t)$, and both PSDs are estimated with Welch's method from chapter 4. On the $S_{\delta\nu}$ panel two curves are drawn: the direct estimate (Welch of the differenced record) and $f^2 \times$ the measured $S_\varphi$ from the panel to its left. If the boxed identity is right, they must lie on top of each other — for every noise type, at every level.
Pick a noise type and watch all four panels. On the bottom-right panel the dashed curve is $f^2 S_\varphi(f)$ computed from the bottom-left panel — it lands on the directly estimated $S_{\delta\nu}(f)$ for every noise type, over three decades of $f$. (Two estimation artifacts, neither of them physics: the direct estimate droops slightly in the last half-decade before Nyquist — the finite-difference derivative — and the few leftmost bins wobble with chapter-4 leakage.)
Stare at white FM for a moment, because it is the canonical case: the frequency record $\delta\nu(t)$ is featureless white noise, yet the phase — its running integral — wanders off like a drunkard, and its spectrum $S_\varphi \propto 1/f^2$ diverges at low $f$ exactly as chapter 5's random walk did. Nothing "went wrong" at low frequencies; integration is just division by $f$ in amplitude, $f^2$ in power.
What a spectrum analyzer shows: $\mathcal{L}(f)$
Now connect $S_\varphi$ to what you actually see when you feed the oscillator into a spectrum analyzer: the RF spectrum of $V(t)$ itself, around the carrier at $\nu_0$. Chapter 2 already did the hard work. For small phase excursions ($\varphi_{\mathrm{rms}} \ll 1$ rad — narrowband PM), a sinusoidal phase wiggle of amplitude $m$ at rate $f_m$ puts a pair of sidebands at $\nu_0 \pm f_m$, each with power $m^2/4$ relative to the carrier. Random phase noise is just chapter 4's sum of independent sinusoids, one per Fourier bin, so each slice of $S_\varphi$ makes its own little sideband pair. Now chain the two facts together. The 1-Hz slice of phase noise at offset $f$ carries variance $S_\varphi(f) \times 1\,\mathrm{Hz}$ (chapter 4: area under a PSD is variance). Treat that slice as a single sinusoidal wiggle of amplitude $m$: a sinusoid's mean square is $m^2/2$, so $m^2/2 = S_\varphi(f) \times 1\,\mathrm{Hz}$, i.e. $m^2 = 2 S_\varphi(f)\,\mathrm{Hz}$. That wiggle puts power $m^2/4$ (relative to the carrier) into each of its two sidebands. So the sideband power in 1 Hz at offset $f$, on one side of the carrier, divided by the carrier power, is
$$ \mathcal{L}(f) \;=\; \frac{m^2}{4}\ \text{per Hz} \;=\; \frac{2\,S_\varphi(f)}{4} \;=\; \tfrac{1}{2}\, S_\varphi(f) . $$
The $\tfrac12$ is because the noise power splits evenly between the upper and lower sidebands and $\mathcal{L}$ counts only one of them ("single-sideband phase noise"). Datasheets quote it logarithmically, in dBc/Hz — decibels relative to the carrier, per hertz:
$$ \mathcal{L}_{\mathrm{dB}}(f) = 10 \log_{10}\!\big[\tfrac{1}{2} S_\varphi(f)\big]\ \text{dBc/Hz}. $$
Worked conversion, since this trips everyone up once: a synthesizer spec of $\mathcal{L} = -100$ dBc/Hz means $\mathcal{L}(f) = 10^{-10}$ per Hz, so $S_\varphi(f) = 2\mathcal{L}(f) = 2\times 10^{-10}\ \text{rad}^2/\text{Hz}$. Factor of two, no $2\pi$ anywhere.
Two oscillators at $\nu_0 = 64$ Hz: one with only white phase noise, one with only white amplitude noise (both band-limited to 48 Hz, so the pedestal spans $\nu_0 \pm 48$ Hz and no sideband folds past the band edges). The dashed guide is the predicted pedestal height $S_\varphi/4$ — half the power in each sideband, so $\mathcal{L} = S_\varphi/2$ relative to the carrier's power of $\tfrac12$. Slide $S_a$ up to equal $S_\varphi$ and watch the two spectra become pixel-identical: a spectrum analyzer cannot tell AM noise from PM noise.
The punchline of the demo deserves italics: with equal $S_a$ and $S_\varphi$ the pedestals are identical. Both AM and PM noise produce symmetric sidebands of the same power; the difference is only in the sidebands' phase relationship to the carrier (chapter 2: AM sidebands add along the carrier phasor, PM sidebands add at right angles to it), and a power spectrum throws phase information away. To separate them you must be phase-sensitive: demodulate against a reference.
The universal floor: −174 dBm/Hz
How quiet can $\mathcal{L}(f)$ possibly be? Thermodynamics sets a hard floor, and chapter 5 already handed us the number: any resistor delivers into a matched load the available noise power density $N_0 = k_B T$, independent of the resistance, which at the RF reference temperature of 290 K is $4.0\times 10^{-21}$ W/Hz $=$ $-174$ dBm/Hz. Every input of every amplifier, mixer and cable in the warm world sits in this bath. What is new in this chapter is what that bath does to a carrier's phase.
Now superpose that noise on a carrier. Take a clean carrier of power $C$ riding on white noise of density $N_0$, and look at the 1-Hz slice of noise at $\nu_0 + f$. Over short times that slice behaves like a little sinusoid offset by $f$ from the carrier, so it can be drawn as a phasor — a beautiful echo of chapter 2's rotating-frame demo. In the frame rotating with the carrier, the noise phasor turns at the offset rate $f$: when it points along the carrier it stretches or shrinks it (amplitude modulation); a quarter turn later it tips the carrier sideways (phase modulation). Averaged over the rotation, half the superposed noise power reads as amplitude noise and half as phase noise — the same even AM/PM split that the spectrum analyzer demo above could not resolve, now seen from the phasor side.
Press play. In the frame rotating with the carrier (gray), the 1-Hz noise slice is a phasor of amplitude $b$ (orange) turning at the offset rate $f$. Its projection along the carrier (blue) stretches and shrinks the amplitude; its projection across the carrier (green) tips the phase. Right: the two projections in time — the same sinusoid a quarter turn apart, each with mean square $b^2/2$ (dashed RMS guides). Half the noise power reads as AM, half as PM, whatever $f$ and $b$ you choose.
That factor of one half, plus the definition of $\mathcal{L}$, gives the additive phase-noise floor. Send a carrier of power $C$ through any chain of amplifiers, cables and mixers with overall noise figure $F$ ($F$ is the amplifier-chain noise figure — the factor by which the chain inflates $k_B T$ — quantified properly in chapter 8): the chain adds noise density $Fk_B T$ in each 1-Hz slice near the carrier, half of it is phase noise, and $\mathcal{L}$ counts one sideband: $$ \mathcal{L}_{\text{floor}}(f) \;=\; \frac{F\,k_B T}{2C}, \qquad \text{flat in } f. $$ In decibels, with $-174$ for $k_B T$ and $-3$ for the half: $$ \mathcal{L}_{\text{floor}} \;=\; -177 + F_{\mathrm{dB}} - C_{\mathrm{dBm}}\ \ \text{dBc/Hz}. $$ Worked number: a $+10$ dBm carrier through a chain with $F = 6$ dB floors out at $-177 + 6 - 10 = -181$ dBc/Hz, and no amount of engineering skill can put the far-from-carrier noise below that.
Spurs: the lines that aren't noise
Put a real source on a spectrum analyzer and you will see, besides the smooth noise pedestal, discrete lines — "spurs", in the trade. The usual suspects: mains harmonics (a ladder at multiples of 50 or 60 Hz), switching-supply ripple (tens to hundreds of kHz), vibration and microphonics (fans, vacuum pumps, slammed doors — each mechanical tone phase-modulates the source through its cavity or crystal), and in synthesizers, mixer products and reference breakthrough. Two rules keep the bookkeeping honest.
Rule one: spurs are quoted in dBc, not dBc/Hz. A spur is a sinusoid — a power ratio to the carrier — while the pedestal is a power density. This is exactly chapter 4's line-versus-floor trap on the analyzer screen: narrow the resolution bandwidth and the displayed noise floor drops dB for dB while the spur's height does not move. A number in dBc/Hz for a spur (or dBc for noise) is a category error, and datasheets that mix them up are telling you something about their authors.
Rule two: a small PM spur is chapter 2's narrowband PM in disguise. A pair of phase-modulation sidebands at $\nu_0 \pm f_m$, each at $-X$ dBc, is narrowband PM with modulation index $\beta = 2\times 10^{-X/20}$ (each sideband carries $\beta^2/4$ of the carrier power), i.e. an equivalent RMS phase wobble $$ \varphi_{\mathrm{rms}} \;=\; \frac{\beta}{\sqrt{2}} \;=\; \sqrt{2}\times 10^{-X/20}\ \text{rad} $$ Worked numbers: a $-60$ dBc sideband pair is $\varphi_{\mathrm{rms}} = \sqrt2\times10^{-3} \approx 1.4$ mrad of phase wobble. A fan line at $-40$ dBc (pair) on a 10 GHz source is 14 mrad — divided by $2\pi\nu_0$ that is $2.3\times10^{-13}$ s $\approx 0.23$ ps of timing wobble, often the single largest entry in the jitter budget of an otherwise excellent synthesizer.
When the carrier dissolves: linewidth
The sideband picture assumed $\varphi_{\mathrm{rms}} \ll 1$ rad. What if the phase wanders by many radians — as it always eventually does for any FM-type noise, since random-walk phase grows without bound? Then the "carrier plus small sidebands" decomposition fails: the sidebands merge with the carrier and the line acquires a nonzero width of its own, a linewidth, independent of how long you measure.
The classic solvable case is pure white frequency noise, $S_{\delta\nu}(f) = h_0$ (one-sided, in $\text{Hz}^2/\text{Hz}$). Sketch of the argument: $\varphi$ is then a random walk, so the phase accumulated over a lag $\tau$ is Gaussian with variance $\langle\Delta\varphi^2\rangle = 2\pi^2 h_0 \tau$ (integrate the white driving noise). The field's coherence is $\langle e^{i\Delta\varphi}\rangle = e^{-\langle\Delta\varphi^2\rangle/2} = e^{-\pi^2 h_0 \tau}$ — exponential decay of correlation, and the Fourier transform of an exponential is a Lorentzian. The result:
$$ \text{the line is Lorentzian with FWHM}\quad \Delta\nu = \pi\, h_0 . $$
A curiously clean formula: the linewidth doesn't depend on any bandwidth or measurement time, just on the white FM level, with a bare $\pi$ in front. Watch it happen:
A 64 Hz oscillator with pure white FM at level $h_0$. At small $h_0$ the line is a resolution-limited spike (width set by $1/T_{\mathrm{seg}}$, chapter 1); raise $h_0$ and it fattens into the predicted Lorentzian (dashed), FWHM $=\pi h_0$. Note the wings always agree — they are the small-angle sidebands $S_\varphi/4 = h_0/4f^2$, valid far from the carrier even when the center is not.
For noise spectra other than white FM, which parts of $S_\varphi(f)$ end up in the linewidth and which in the wings is answered by the β-separation line $S_\varphi(f) = 8\ln 2\, f/\pi^2$: roughly, spectral regions where $S_\varphi$ lies above this line contribute to the (Gaussian-ish) core and its width, while regions below it only decorate the wings — a very useful rule of thumb when you have a measured $S_\varphi$ and want a number to quote.
From phase noise to jitter
Managers and datasheets want the whole $\mathcal{L}(f)$ curve boiled down to one number. The honest way to do it is to integrate: the total RMS phase excursion contributed by a band $[f_1, f_2]$ of the spectrum is $$ \varphi_{\mathrm{rms}} \;=\; \sqrt{\int_{f_1}^{f_2} S_\varphi(f)\,df} \;=\; \sqrt{2\!\int_{f_1}^{f_2} \mathcal{L}(f)\,df} $$ (with $\mathcal{L}$ in linear units, not dB), and dividing radians by $2\pi\nu_0$ converts phase to seconds of timing jitter — the time-error bookkeeping of the map below. One warning, straight from chapter 5: this integral is almost never band-independent. Under a flat (white-PM) region it grows linearly with the upper cutoff $f_2$; under a $1/f$ region, logarithmically with both cutoffs; under anything steeper it blows up as $f_1$ pushes toward the carrier. A real $\mathcal{L}(f)$ is steep close in and flat far out — divergent at both ends — so a quoted jitter number is meaningless without its integration band. Hence the datasheet conventions — telecom quotes 12 kHz–20 MHz, and a jitter number without a band deserves the same suspicion as a linewidth without a measurement time.
Worked example, matching the default preset of the demo below: an oscillator whose $\mathcal{L}$ is flat at $-170$ dBc/Hz across 12 kHz–20 MHz has $\varphi_{\mathrm{rms}} = \sqrt{2\times 10^{-17} \times 2.0\times 10^{7}} = 2.0\times 10^{-5}$ rad $= 20\ \mu$rad; at $\nu_0 = 100$ MHz that is $\varphi_{\mathrm{rms}}/2\pi\nu_0 = 3.2\times10^{-14}$ s — about 32 fs of timing jitter.
A model oscillator: Leeson's formula
Everything so far describes oscillator noise; nothing yet predicts it. The classic predictive model (D. B. Leeson, 1966) is disarmingly simple and earns its keep in every RF design review. An oscillator is just two parts in a ring: an amplifier (noise figure $F$, carrier power $C$) whose output is filtered by a resonator and fed back to its own input —
The resonator is summarized by one number, its quality factor $Q = \nu_0/\Delta\nu$: the resonance frequency divided by the resonance's full width — equivalently, roughly the number of oscillation cycles before the stored energy rings down. In a circuit some energy leaks out through the couplings, so the working number is the lower loaded quality factor $Q_L$, and the resonator only responds within $\nu_0/2Q_L$ of resonance.
Now feed the amplifier's additive phase noise — the flat floor $Fk_B T/2C$ of the previous section — into this ring, and ask what comes out. Three regions, from far away moving in toward the carrier:
- Beyond the resonator bandwidth ($f > \nu_0/2Q_L$): the resonator cannot respond this fast, the loop does nothing, and the floor passes straight to the output. Flat — white PM.
- Inside the resonator bandwidth ($f < \nu_0/2Q_L$): the loop converts the amplifier's phase fluctuations into frequency fluctuations of the whole oscillator. White phase in becomes white frequency out, and by the boxed identity white FM means $S_\varphi \propto 1/f^2$: the noise rises 20 dB per decade toward the carrier.
- Below the flicker corner ($f < f_c$): the transistor's own $1/f$ flicker noise phase-modulates the carrier directly, multiplying whatever is there by a further $1/f$.
Stack the three effects and you have Leeson's formula:
$$ \mathcal{L}(f) \;=\; \frac{F\,k_B T}{2C}\, \left[\,1 + \Big(\frac{\nu_0}{2 Q_L f}\Big)^{\!2}\right] \left(1 + \frac{f_c}{f}\right). $$
Read from the carrier outward: an $f^{-3}$ region (flicker FM), then $f^{-2}$ (white FM), then the flat additive floor (white PM), with the breakpoints at $f_c$ and at $\nu_0/2Q_L$. When the two breakpoints swap order — as in high-$Q$ crystal oscillators, where $\nu_0/2Q_L$ can be tens of hertz — a $1/f$ flicker-PM shelf appears between them instead; the demo shows it.
Two morals. First, inside the resonator bandwidth $\mathcal{L} \propto F/Q_L^2$: the figure of merit is $F/Q^2$, so doubling $Q_L$ buys 6 dB where halving $F$ buys only 3 — a factor of two in $Q$ is worth four times more than a factor of two in noise figure. That is why the quietest sources are built around resonators of extreme quality factor — quartz crystals ($Q_L \sim 10^4$–$10^6$), then microwave cavities, and ultimately atomic resonances. Second, at fixed $F$, $Q_L$ and $C$ the model scales as $(\nu_0/f)^2$, so $\mathcal{L} \propto \nu_0^2$. But an ideal frequency multiplier $\times N$ multiplies phase — hence $S_\varphi$ and $\mathcal{L}$ — by exactly $N^2$, a cost of $+20\log_{10}N$ dB. The two scalings cancel: frequency multiplication is noise-neutral in this model. A 100 MHz oscillator multiplied $\times 100$ to 10 GHz lands on the same curve as a hypothetical 10 GHz oscillator with the same $Q_L$, $F$, $C$. The real reason to build low and multiply is that the best resonators are low-frequency objects: crystals with $Q_L \sim 10^6$ exist at 5–120 MHz and nowhere else. Chapter 10 leans on this constantly.
Pick a preset, then pull the sliders. The dashed vertical guides mark the two breakpoints $f_c$ and $\nu_0/2Q_L$; the dashed horizontal guide is the additive floor $Fk_B T/2C$; the short gray segments label the local slope. The readouts integrate the model over the telecom band 12 kHz–20 MHz. With the "×100" preset the entire OCXO curve rides up rigidly by $+40$ dB — that is what frequency multiplication does to phase noise.
Race the presets against each other. Close to the carrier the multiplied crystal wins by a mile: at 1 kHz offset the OCXO$\,\times 100$ sits some 30 dB below the 10 GHz dielectric-resonator oscillator, because no microwave resonator comes near a crystal's $Q_L$. Far from the carrier the DRO wins instead — multiplication dragged the OCXO's noise floor up by the same 40 dB, and the readouts show the DRO with the smaller 12 kHz–20 MHz jitter. (Note also that the OCXO's timing jitter is unchanged by multiplication: $\varphi_{\mathrm{rms}}$ grows $\times 100$ but so does $\nu_0$.) The best 10 GHz source is therefore neither: it is the DRO phase-locked to the multiplied crystal, quiet crystal close in, quiet DRO far out — precisely the architecture chapter 10 builds.
The bookkeeping map
Four equivalent descriptions of the same phase wandering, and the conversion factors between their PSDs — this half-page is worth pinning above your desk:
| process | units | PSD, from the phase PSD | who uses it |
|---|---|---|---|
| phase error $\varphi(t)$ | rad | $S_\varphi(f)$ | RF engineers, via $\mathcal{L} = S_\varphi/2$ |
| frequency error $\delta\nu(t) = \dot\varphi/2\pi$ | Hz | $S_{\delta\nu}(f) = f^2\, S_\varphi(f)$ | laser people (linewidth $=\pi h_0$) |
| fractional frequency $y(t) = \delta\nu/\nu_0$ | — | $S_y(f) = \dfrac{f^2}{\nu_0^2}\, S_\varphi(f)$ | clock people (input to $\sigma_y(\tau)$) |
| time error $x(t) = \varphi/2\pi\nu_0$ | s | $S_x(f) = \dfrac{S_\varphi(f)}{(2\pi\nu_0)^2}$ | timekeeping & navigation |
Two sanity checks on the factors. First, $y = \dot x$ (fractional frequency is the slope of the time error), so the derivative theorem demands $S_y = (2\pi f)^2 S_x$ — and indeed $(2\pi f)^2 \cdot S_\varphi/(2\pi\nu_0)^2 = (f^2/\nu_0^2) S_\varphi$. Second, dividing by $\nu_0$ or $2\pi\nu_0$ never changes the shape in $f$; only the $\varphi \leftrightarrow \delta\nu$ hop, a time derivative, tilts the spectrum by $f^2$. In chapter 7, the whole function $S_y(f)$ gets compressed into a single curve, the Allan deviation $\sigma_y(\tau)$ — the natural language for "how good is this clock at averaging time $\tau$".
The same thing in code
Synthesizing an oscillator with a prescribed white FM level $h_0$ is four lines — the recipe behind the last demo, and the discrete-time version of "phase integrates frequency" (recall from chapter 4 that discrete white noise of standard deviation $\sigma$ at rate $f_s$ has one-sided PSD $2\sigma^2/f_s$, hence $\sigma = \sqrt{h_0 f_s/2}$):
// oscillator with one-sided white FM level S_dnu(f) = h0 [Hz^2/Hz]
const sigma = Math.sqrt(h0 * fs / 2); // std of delta-nu per sample, in Hz
let phi = 0;
for (let i = 0; i < N; i++) {
const dnu = sigma * Noise.randn(rand); // instantaneous frequency error
phi += 2 * Math.PI * dnu / fs; // phase integrates frequency
V[i] = Math.cos(2 * Math.PI * nu0 * i / fs + phi);
}
Exercises
A 10 MHz quartz oscillator datasheet quotes $\mathcal{L}(10\,\text{kHz}) = -110$ dBc/Hz. Convert this to (i) $S_\varphi$ at that offset, (ii) $S_{\delta\nu}$ at that offset, and (iii) $S_y$ at that offset.
Solution
(i) $\mathcal{L} = 10^{-11}$ per Hz, so $S_\varphi = 2\mathcal{L} = 2\times 10^{-11}\ \text{rad}^2/\text{Hz}$. (ii) $S_{\delta\nu} = f^2 S_\varphi = (10^4)^2 \cdot 2\times 10^{-11} = 2\times 10^{-3}\ \text{Hz}^2/\text{Hz}$. (iii) $S_y = S_{\delta\nu}/\nu_0^2 = 2\times 10^{-3}/(10^7)^2 = 2\times 10^{-17}\ /\text{Hz}$. Note how unremarkable-looking numbers in one bookkeeping become extreme in another — always write the units.
An oscillator has flat $S_\varphi = 10^{-10}\ \text{rad}^2/\text{Hz}$ from 1 Hz to 100 kHz (and negligible noise outside). What is $\varphi_{\mathrm{rms}}$, and is the narrowband-PM formula $\mathcal{L} = S_\varphi/2$ trustworthy here?
Solution
$\varphi_{\mathrm{rms}}^2 = \int S_\varphi\, df = 10^{-10} \times (10^5 - 1) \approx 10^{-5}\ \text{rad}^2$, so $\varphi_{\mathrm{rms}} \approx 3.2$ mrad. That is spectacularly small compared to 1 rad, so yes — the carrier is intact and $\mathcal{L} = S_\varphi/2$ describes the screen perfectly.
A diode laser's frequency noise is measured to be white, $S_{\delta\nu}(f) = h_0 = 1\ \text{Hz}^2/\text{Hz}$, over the relevant band. What is its linewidth?
Solution
$\Delta\nu = \pi h_0 = \pi \times 1 \approx 3.1$ Hz (FWHM, Lorentzian). A pleasingly literal appearance of $\pi$: 1 unit of white frequency noise makes $\pi$ units of linewidth.
An oscillator drifts linearly: $\nu(t) = \nu_0 + Dt$ with $D$ in Hz/s (a warming quartz crystal, an aging laser cavity). Find $\varphi(t)$ and explain why no stationary $S_\varphi(f)$ or $S_{\delta\nu}(f)$ describes this signal.
Solution
$\varphi(t) = 2\pi\!\int_0^t (\nu(t')-\nu_0)\,dt' = 2\pi\cdot \tfrac{1}{2}Dt^2 = \pi D t^2$. This grows deterministically and without bound — its statistics depend on absolute time, so the process is not stationary and the PSD machinery of chapter 4 does not apply (an FFT of a finite record shows a smeared chirp whose shape depends on the record length, not a well-defined spectrum). Drift needs a different tool: it appears in chapter 7 as the $\sigma_y \propto \tau$ ramp at long averaging times in the Allan deviation.
You beat two nominally identical lasers on a photodiode and the beat note is Lorentzian with a 2 kHz FWHM. Assuming white FM noise and equal, independent lasers, what is each laser's $h_0$?
Solution
For independent Lorentzian lines the beat's linewidth is the sum of the two linewidths (the frequency noises add: $S_{\delta\nu,\mathrm{beat}} = 2h_0$), so each laser has $\Delta\nu = 1$ kHz $= \pi h_0$, giving $h_0 = 1000/\pi \approx 318\ \text{Hz}^2/\text{Hz}$. This is the standard way to measure laser linewidths — a single laser has nothing to beat against, so you build two.
A $-20$ dBm signal passes through a receiver chain with noise figure $F = 10$ dB. What is the best possible phase-noise floor of the signal at the chain's output? Why doesn't the chain's gain appear in the answer?
Solution
$\mathcal{L}_{\text{floor}} = -177 + F_{\mathrm{dB}} - C_{\mathrm{dBm}} = -177 + 10 - (-20) = -147$ dBc/Hz. Gain is absent because it amplifies the carrier and the added noise alike, leaving the ratio untouched — only the input-referred noise ($Fk_B T$) and the carrier power matter. Compare the $+10$ dBm, $F = 6$ dB example in the text ($-181$ dBc/Hz): 34 dB of floor separates a strong signal in a good chain from a weak signal in a mediocre one.
A 10 GHz oscillator is evaluated at 1 kHz offset. (a) Its loaded quality factor is raised from $Q_L = 10^3$ to $10^4$ — by how much does $\mathcal{L}(1\,\text{kHz})$ improve? (b) What if, instead, the amplifier's noise figure were halved? (Check first that 1 kHz lies inside the resonator half-bandwidth in both cases.)
Solution
The half-bandwidth $\nu_0/2Q_L$ is 5 MHz at $Q_L = 10^3$ and 500 kHz at $10^4$ — both $\gg$ 1 kHz, so the offset is deep inside the $1/f^2$ region where $\mathcal{L} \propto F/Q_L^2$. (a) Ten times the $Q_L$ means $100\times$ less noise power: a 20 dB improvement. (b) Halving $F$ buys only 3 dB. That asymmetry is the $F/Q^2$ moral: engineer the resonator first, the transistor second.
A 10 GHz synthesizer shows a PM sideband pair at $\pm f_m$, each sideband at $-50$ dBc. What equivalent RMS phase wobble does the pair represent, and how much timing wobble is that?
Solution
$\varphi_{\mathrm{rms}} = \sqrt{2}\times 10^{-50/20} = \sqrt{2}\times 3.16\times10^{-3} \approx 4.5$ mrad. In time: $\varphi_{\mathrm{rms}}/2\pi\nu_0 = 4.47\times10^{-3}/(2\pi\times 10^{10}) = 7.1\times 10^{-14}$ s $\approx 71$ fs — from one pair of lines 50 dB down. Spurs punch far above their apparent weight on a log plot.
A 100 MHz OCXO has $\mathcal{L}(f) = -160$ dBc/Hz, flat from 12 kHz to 20 MHz, with negligible contribution outside that band. Compute the RMS phase jitter over the telecom band and the corresponding timing jitter.
Solution
$\varphi_{\mathrm{rms}}^2 = 2\!\int \mathcal{L}\,df = 2\times 10^{-16} \times (2.0\times 10^{7} - 1.2\times 10^{4}) \approx 4.0\times 10^{-9}\ \text{rad}^2$, so $\varphi_{\mathrm{rms}} = 6.3\times 10^{-5}$ rad. Timing: $6.3\times10^{-5}/(2\pi\times 10^{8}) = 1.0\times10^{-13}$ s $= 0.10$ ps. Note the lower band edge barely matters here — a flat spectrum is dominated by its upper edge. Move the same $-160$ dBc/Hz into a $1/f$ region and the edges would rule instead, which is why the band must always be quoted.