Foundations Chapter Exercises (with Full Solutions)

β: This English translation is in beta — the Traditional-Chinese original is the authoritative version.

Prerequisites: psd_phase_noise_jitter · lti_vs_ltv | Next: 03 core-theory chapter exercises

This page is the exercise set for Chapter 02 (Foundations). Problem types cover derivation problems (derive the expressions by hand), numerical problems (plug in numbers, carry the units, do the dimension check), and reverse-design problems (given a target spec, back-solve the required parameters).

How to use this page: work each problem yourself first, then expand the "Solution" to compare. Every solution follows the same format: step-by-step substitution (with units) → result → dimension check → one-line Python verification. All Python verifications call this site's real library simulations/common/ (no invented APIs), so you can paste them straight into a REPL.

Core formulas involved (all from the spec and this chapter's pages, using the same notation):

• phase→time: $\Delta t=\dfrac{\Delta\phi}{2\pi f_0}$ (spec Eq. 17)
• phase variance: $\sigma_\phi^2=\displaystyle\int_{f_1}^{f_2}S_\phi(f)\,df$ (spec Eq. 18)
• rms jitter: $\sigma_t=\dfrac{1}{2\pi f_0}\sqrt{\displaystyle\int_{f_1}^{f_2}S_\phi(f)\,df}$ (spec Eq. 19)
• SSB↔phase PSD (small-angle): $\mathcal{L}(\Delta f)\approx\tfrac12 S_\phi(\Delta f)$ (spec Eq. 16)
• Parseval: $\displaystyle\sum_{n=0}^{\infty}c_n^2=\frac{1}{\pi}\int_0^{2\pi}\lvert\Gamma(x)\rvert^2dx=2\,\Gamma_{rms}^2$ ([P1] Eq.(20), p.185)
• Lorentzian linewidth (FWHM): $\Delta f_{3\mathrm{dB}}=\dfrac{D}{\pi}$, phase diffusion $\operatorname{Var}[\Delta\phi(t)]=2D\lvert t\rvert$ (spec §11.2; linked to [E2] Demir 2000, not among the 5 source PDFs)
• Allan slope reference: white FM $\to\tau^{-1/2}$, flicker FM $\to\tau^{0}$, random-walk FM $\to\tau^{+1/2}$ (spec §11.2; Allan is external literature)

---

Problems

Exercise 1 (numerical) — phase ↔ time conversion

An oscillator at $f_0=5$ GHz has an instantaneous excess-phase offset $\Delta\phi=5\times10^{-4}$ rad. Find the corresponding timing error $\Delta t$ (fs), and convert $\Delta\phi$ to degrees.

Quick check (work it out yourself, then check)

Try it yourself first: Δt = ? (Δφ = 5×10⁻⁴ rad, f₀ = 5 GHz; answer in fs)

fs

Graded correct within ±5% relative error; scientific notation is accepted.

Exercise 2 (numerical) — rms jitter from phase variance

A clock has rms phase $\sigma_\phi=14.07$ mrad (over the 1→100 MHz integration band), with $f_0=5$ GHz. Find the rms timing jitter $\sigma_t$ (fs).

Exercise 3 (derivation + numerical) — Parseval: $\Gamma_{rms}$ and $\sum c_n^2$ from the ISF

The ISF of an ideal LC oscillator is $\Gamma(\theta)=-\sin\theta$.

(a) Using the definition $\Gamma_{rms}^2=\dfrac{1}{2\pi}\displaystyle\int_0^{2\pi}\Gamma^2(\theta)\,d\theta$, compute $\Gamma_{rms}$ by hand. (b) Using Parseval (spec Eq. 11), find $\sum_{n=0}^{\infty}c_n^2$, and state which $c_n$ is the only nonzero one.

Exercise 4 (numerical) — phase variance from PSD integration

An oscillator's one-sided phase PSD at offset $f$, in the $1/f^2$ region, can be written $S_\phi(f)=\dfrac{K}{f^2}$, with $K=10^{-4}\ \text{rad}^2\cdot\text{Hz}$ (i.e., $S_\phi(1\,\text{Hz})=10^{-4}$). Find the phase variance $\sigma_\phi^2$ and $\sigma_\phi$ (mrad) over the integration band $f_1=10^3$ Hz to $f_2=10^6$ Hz.

Quick check (work it out yourself, then check)

Try it yourself first: σ_φ = ? (K = 10⁻⁴ rad²·Hz, f₁ = 10³ Hz, f₂ = 10⁶ Hz; answer in mrad)

mrad

Graded correct within ±6% relative error; scientific notation is accepted.

Exercise 5 (concept + derivation) — LTI vs LTV: same impulse, different phase

The ideal-LC ISF is $\Gamma(\theta)=-\sin\theta$. The same charge impulse $\Delta q$ is injected (a) at the waveform zero-crossing ($\theta=\pi/2$, where the $\cos$ waveform has its maximum slope and $-\sin$ is at its extremum) and (b) at the waveform peak ($\theta=0$). Using $\Delta\phi=\Gamma(\theta)\,\Delta q/q_{max}$, show how much the two phase effects differ, and explain in one sentence why this is "LTV, not LTI". Take $\Delta q=1$ fC, $q_{max}=1$ pC.

Quick check (work it out yourself, then check)

Try (a) yourself first: at the zero-crossing θ=π/2, Δφ_a = ? (Δq=1 fC, q_max=1 pC; answer in mrad, include the sign)

mrad

Graded correct within ±2% relative error; scientific notation is accepted.

Exercise 6 (reverse design) — phase diffusion from the Lorentzian linewidth

A free-running oscillator's measured carrier 3-dB linewidth (FWHM) is $\Delta f_{3\mathrm{dB}}=1$ kHz.

(a) Back-solve the phase-diffusion coefficient $D$ (rad²/s). (b) Estimate how long $t$ it takes for the phase to accumulate a variance $\operatorname{Var}[\Delta\phi]=1\ \text{rad}^2$ (the phase has "wandered by about 1 rad" and coherence has largely collapsed).

Quick check (work it out yourself, then check)

Try (a) yourself first: back-solve D = ? (Δf₃dB = 1 kHz; answer in rad²/s)

rad²/s

Graded correct within ±2% relative error; scientific notation is accepted.

Exercise 7 (concept + slope reading) — Allan deviation slopes

On a log–log Allan-deviation plot $\sigma_y(\tau)$ you measure three segments with different slopes: $-1/2$, $0$, $+1/2$. Which FM noise type does each correspond to? Also explain why "flicker FM" forms a flat plateau (floor) in the ADEV.

Exercise 8 (numerical) — from a single $\mathcal{L}(\Delta f)$ point to $S_\phi$ to jitter

A spur-free oscillator measures $\mathcal{L}(1\,\text{MHz})=-120$ dBc/Hz at $\Delta f=1$ MHz. (a) Find $S_\phi$ at that offset (rad²/Hz). (b) If over the 1-decade span 1→10 MHz the PSD follows $1/f^2$ (i.e., $S_\phi=K/f^2$, with $K$ set by (a)), and $f_0=5$ GHz, find the rms jitter $\sigma_t$ (fs) over this band.

Quick check (work it out yourself, then check)

Try (b) yourself first: σ_t = ? (L(1 MHz) = −120 dBc/Hz, 1/f² integrated 1→10 MHz, f₀ = 5 GHz; answer in fs)

fs

Graded correct within ±5% relative error; scientific notation is accepted.

---

Solutions

Exercise 1 solution (phase ↔ time conversion)

Step-by-step substitution (with units). Using phase→time (spec Eq. 17):

$$\Delta t=\frac{\Delta\phi}{2\pi f_0}=\frac{5\times10^{-4}\ \text{rad}}{2\pi\times5\times10^{9}\ \text{Hz}}=\frac{5\times10^{-4}}{3.1416\times10^{10}}\ \text{s}.$$$$\Delta t\approx1.59\times10^{-14}\ \text{s}=15.9\ \text{fs}.$$

Converting to degrees:

$$\Delta\phi=5\times10^{-4}\times\frac{180}{\pi}\approx0.0286^\circ.$$

Result: $\Delta t\approx15.9$ fs, $\Delta\phi\approx0.0286^\circ$.

Dimension check: $\dfrac{[\text{rad}]}{[\text{rad/s}]}=[\text{s}]$ ✓ (the unit of $2\pi f_0$ is rad/s; rad is dimensionless, so the ratio gives seconds). This is exactly the timing error of canonical Example A in spec §8.

from simulations.common.noise_utils import phase_to_time_error
dt = phase_to_time_error(5e-4, f0=5e9)
print(dt*1e15, "fs")   # -> 15.92 fs

Exercise 2 solution (rms jitter from phase variance)

Step-by-step substitution (with units). rms jitter (spec Eq. 19). Here $\sigma_\phi$ is given directly, so $\sqrt{\int S_\phi df}=\sigma_\phi$ and the formula reduces to the rms version of phase→time:

$$\sigma_t=\frac{\sigma_\phi}{2\pi f_0}=\frac{14.07\times10^{-3}\ \text{rad}}{2\pi\times5\times10^{9}\ \text{Hz}}=\frac{1.407\times10^{-2}}{3.1416\times10^{10}}\ \text{s}.$$$$\sigma_t\approx4.479\times10^{-13}\ \text{s}=447.9\ \text{fs}.$$

Result: $\sigma_t\approx447.9$ fs. This corresponds to canonical Example C in spec §8 (the lab_08 result for $f_0=5$ GHz, $\mathcal{L}(1\text{MHz})=-100$ dBc/Hz, 1/f² slope, integrated 1→100 MHz).

Dimension check: $\dfrac{[\text{rad}]}{[\text{rad/s}]}=[\text{s}]$ ✓.

import numpy as np
sigma_phi = 14.07e-3            # rad
f0 = 5e9
sigma_t = sigma_phi/(2*np.pi*f0)
print(sigma_t*1e15, "fs")      # -> 447.9 fs

(For the full integral version see numerical_feeling and integrate_rms_jitter in simulations/common/noise_utils.py.)

Exercise 3 solution (Parseval: $\Gamma_{rms}$ and $\sum c_n^2$ from the ISF)

(a) Compute $\Gamma_{rms}$ by direct integration. Using the fact that $\sin^2\theta$ averages to $\tfrac12$ over one period:

$$\Gamma_{rms}^2=\frac{1}{2\pi}\int_0^{2\pi}(-\sin\theta)^2\,d\theta=\frac{1}{2\pi}\int_0^{2\pi}\sin^2\theta\,d\theta=\frac{1}{2\pi}\cdot\pi=\frac12.$$$$\Gamma_{rms}=\frac{1}{\sqrt2}\approx0.707.$$

(b) Find $\sum c_n^2$ with Parseval. Spec Eq. 11 ([P1] Eq.(20), p.185):

$$\sum_{n=0}^{\infty}c_n^2=2\,\Gamma_{rms}^2=2\times\frac12=1.$$

Which $c_n$ is nonzero: $\Gamma(\theta)=-\sin\theta=\cos(\theta+\tfrac{\pi}{2})$ is a pure first harmonic, so only $c_1=1$ (with $\theta_1=\pi/2$); all others $c_0=c_2=\dots=0$. Check: $\sum c_n^2=c_1^2=1$ ✓.

Result: $\Gamma_{rms}=1/\sqrt2\approx0.707$, $\sum c_n^2=1$, the only nonzero coefficient is $c_1=1$.

Dimension check: $\Gamma$ is dimensionless (spec notation table), so $\Gamma_{rms}^2$ and $\sum c_n^2$ are both dimensionless ✓.

import numpy as np
from simulations.common.isf_utils import gamma_lc_ideal, gamma_rms, compute_fourier_coefficients
theta = np.linspace(0, 2*np.pi, 4096, endpoint=False)
g = gamma_lc_ideal(theta)                     # = -sin(theta)
print(gamma_rms(theta, g))                    # -> 0.7071
a0, a, b, c, ph = compute_fourier_coefficients(theta, g, n_harmonics=5)
print(np.sum(c**2))                           # -> ~1.0  (== 2*Gamma_rms^2)

Exercise 4 solution (phase variance from PSD integration)

Step-by-step substitution (with units). Phase variance (spec Eq. 18), substituting $S_\phi(f)=K/f^2$:

$$\sigma_\phi^2=\int_{f_1}^{f_2}\frac{K}{f^2}\,df=K\left[-\frac1f\right]_{f_1}^{f_2}=K\left(\frac{1}{f_1}-\frac{1}{f_2}\right).$$

Substituting $K=10^{-4}\ \text{rad}^2\cdot\text{Hz}$, $f_1=10^3$ Hz, $f_2=10^6$ Hz:

$$\sigma_\phi^2=10^{-4}\left(\frac{1}{10^3}-\frac{1}{10^6}\right)=10^{-4}\times(10^{-3}-10^{-6})=10^{-4}\times9.99\times10^{-4}=9.99\times10^{-8}\ \text{rad}^2.$$$$\sigma_\phi=\sqrt{9.99\times10^{-8}}\approx3.16\times10^{-4}\ \text{rad}=0.316\ \text{mrad}.$$

Result: $\sigma_\phi^2\approx9.99\times10^{-8}\ \text{rad}^2$, $\sigma_\phi\approx0.316$ mrad.

Intuition: the $1/f^2$ integral is dominated by the low-frequency end $f_1$ ($1/f_1\gg1/f_2$) — which is exactly why close-in phase noise contributes most of the total jitter.

Dimension check: $[K]\cdot[1/f]=(\text{rad}^2\cdot\text{Hz})\cdot(1/\text{Hz})=\text{rad}^2$ ✓ ($\sigma_\phi^2$ is in rad²).

import numpy as np
K, f1, f2 = 1e-4, 1e3, 1e6
var = K*(1/f1 - 1/f2)
print(var, "rad^2 ->", np.sqrt(var)*1e3, "mrad")   # -> 9.99e-8 rad^2 -> 0.316 mrad

Exercise 5 solution (LTI vs LTV: same impulse, different phase)

Step-by-step substitution (with units). Using the operational ISF (spec Eq. 5) $\Delta\phi=\Gamma(\theta)\,\Delta q/q_{max}$. $\Delta q/q_{max}=10^{-15}/10^{-12}=10^{-3}$.

(a) Zero-crossing $\theta=\pi/2$ ($\Gamma=-\sin(\pi/2)=-1$, maximum sensitivity):

$$\Delta\phi_a=(-1)\times10^{-3}=-1\times10^{-3}\ \text{rad}=-1\ \text{mrad}.$$

(b) Peak $\theta=0$ ($\Gamma=-\sin 0=0$, zero sensitivity):

$$\Delta\phi_b=0\times10^{-3}=0\ \text{rad}.$$

Result: injection at the zero-crossing produces a phase step of $\lvert\Delta\phi\rvert=1$ mrad; injection at the peak produces 0 phase step (there the charge only changes the amplitude, which is subsequently pulled back by amplitude restoring).

Why this is LTV, not LTI: the impulse response of an LTI (linear time-invariant) system depends only on the elapsed time $t-\tau$, not on the injection instant $\tau$. Here, however, the same $\Delta q$ at $\theta=\pi/2$ versus $\theta=0$ produces completely different $\Delta\phi$ — the response explicitly depends on the absolute injection phase $\Gamma(\omega_0\tau)$, which is precisely the defining signature of an LTV (linear time-varying) system (see lti_vs_ltv).

Dimension check: $\Gamma$ dimensionless $\times\ \Delta q/q_{max}$ (C/C, dimensionless) $=$ rad ✓.

from simulations.common.isf_utils import gamma_lc_ideal, impulse_to_phase_step
import numpy as np
dq, qmax = 1e-15, 1e-12
for name, theta in [("zero-crossing", np.pi/2), ("peak", 0.0)]:
    g = gamma_lc_ideal(theta)                       # -sin(theta)
    dphi = impulse_to_phase_step(dq, g, qmax)
    print(name, dphi*1e3, "mrad")                   # -> -1.0 mrad ; 0.0 mrad

Exercise 6 solution (phase diffusion from the Lorentzian linewidth)

Note: Lorentzian linewidth and phase diffusion are external literature ([E2] Demir 2000), not among the 5 source PDFs; the formulas are taken verbatim from spec §11.2.

(a) Back-solve $D$. FWHM linewidth (spec §11.2):

$$\Delta f_{3\mathrm{dB}}=\frac{D}{\pi}\quad\Longrightarrow\quad D=\pi\,\Delta f_{3\mathrm{dB}}=\pi\times10^{3}\ \text{Hz}=3.142\times10^{3}\ \text{rad}^2/\text{s}.$$

(b) Estimate the time to accumulate 1 rad². Phase diffusion (spec §11.2) $\operatorname{Var}[\Delta\phi(t)]=2D\lvert t\rvert$; setting it $=1\ \text{rad}^2$:

$$t=\frac{1}{2D}=\frac{1}{2\times3.142\times10^{3}}\approx1.59\times10^{-4}\ \text{s}=159\ \mu\text{s}.$$

Result: $D\approx3.14\times10^{3}\ \text{rad}^2/\text{s}$; accumulating $\approx1\ \text{rad}^2$ of phase variance takes about 159 µs (i.e., a coherence time of order $\tau_c\sim1/(2D)$). Note that $1/\Delta f_{3\mathrm{dB}}=1$ ms is of the same order — the narrower the linewidth, the longer the coherence time; the two are reciprocals.

Dimension check: (a) $[\,\Delta f\,]=\text{Hz}=1/\text{s}$, multiplied by $\pi$ (dimensionless), gives $D$ in rad²/s (rad² being the square of the dimensionless "rad" of phase variance, per second) ✓. (b) $\dfrac{\text{rad}^2}{\text{rad}^2/\text{s}}=\text{s}$ ✓.

import numpy as np
df_3db = 1e3                       # Hz (FWHM)
D = np.pi*df_3db                   # rad^2/s
t = 1/(2*D)                        # Var = 2 D t = 1 rad^2
print(D, "rad^2/s ;", t*1e6, "us")   # -> 3141.6 rad^2/s ; 159.2 us

(Background and full derivation: lorentzian_linewidth.)

Exercise 7 solution (reading Allan-deviation slopes)

Note: The Allan variance is external literature (Allan 1966), not among the 5 source PDFs; the slope table is taken verbatim from spec §11.2.

Slope reference (spec §11.2). On a log–log $\sigma_y(\tau)$ plot:

Measured slope $\sigma_y(\tau)\propto\tau^{?}$	Noise type	Physics
$\tau^{-1/2}$	white FM (white frequency modulation)	frequency is white noise → phase is a random walk; the longer you average, the more stable
$\tau^{0}$ (plateau/floor)	flicker FM (1/f frequency)	$\sigma_y$ is independent of the gate time $\tau$ → ADEV forms a flat floor
$\tau^{+1/2}$	random-walk FM (random walk of frequency)	the frequency itself drifts → the longer you average, the worse it gets; slope turns upward

Why flicker FM forms a plateau: substitute $S_y(f)\propto1/f$ (flicker FM) into the spec §11.2 expression $\sigma_y^2(\tau)=2\displaystyle\int_0^\infty S_y(f)\dfrac{\sin^4(\pi f\tau)}{(\pi f\tau)^2}\,df$ and change variables to $u=\pi f\tau$: $\tau$ cancels completely out of the integrand — the integral becomes a pure numerical constant independent of $\tau$, so $\sigma_y^2(\tau)=$ constant $\Rightarrow\sigma_y(\tau)\propto\tau^0$, a horizontal line (floor) on the log–log plot. This floor is the signature long-term-stability feature of quartz/atomic clocks, marking the limit that no amount of extra averaging time can push down.

Result: $\tau^{-1/2}\to$ white FM, $\tau^{0}\to$ flicker FM (floor), $\tau^{+1/2}\to$ random-walk FM.

Dimension check: $\sigma_y(\tau)$ is the dimensionless fractional-frequency stability ($y=\Delta f/f_0$); its power-law relationship with $\tau$ only compares slopes and carries no dimension ✓.

import numpy as np
# Read off by the slope definition: fit the log-log slope of a sigma_y(tau) segment
tau = np.logspace(-3, 1, 200)
for label, slope in [("white FM", -0.5), ("flicker FM", 0.0), ("random-walk FM", 0.5)]:
    sy = tau**slope
    fit = np.polyfit(np.log10(tau), np.log10(sy), 1)[0]
    print(label, "slope =", round(fit, 2))   # -> -0.5 ; 0.0 ; 0.5

(Full derivation and plots: allan_variance.)

Exercise 8 solution (from a single $\mathcal{L}$ point to $S_\phi$ to jitter)

(a) $\mathcal{L}\to S_\phi$. First convert dBc/Hz back to linear:

$$\mathcal{L}_{\text{lin}}(1\,\text{MHz})=10^{-120/10}=10^{-12}\ \text{(per Hz)}.$$

The small-angle relation (spec Eq. 16) $\mathcal{L}\approx\tfrac12 S_\phi\Rightarrow S_\phi=2\mathcal{L}_{\text{lin}}$:

$$S_\phi(1\,\text{MHz})=2\times10^{-12}=2\times10^{-12}\ \text{rad}^2/\text{Hz}.$$

(b) Set $K$, then integrate for $\sigma_t$. With the $1/f^2$ model $S_\phi(f)=K/f^2$, use (a) to set $K$ at $f=10^6$ Hz:

$$K=S_\phi(10^6)\cdot(10^6)^2=2\times10^{-12}\times10^{12}=2\ \text{rad}^2\cdot\text{Hz}.$$

Phase variance (spec Eq. 18), integrated $f_1=10^6\to f_2=10^7$ Hz:

$$\sigma_\phi^2=\int_{f_1}^{f_2}\frac{K}{f^2}df=K\!\left(\frac{1}{f_1}-\frac{1}{f_2}\right)=2\left(\frac{1}{10^6}-\frac{1}{10^7}\right)=2\times9\times10^{-7}=1.8\times10^{-6}\ \text{rad}^2.$$$$\sigma_\phi=\sqrt{1.8\times10^{-6}}\approx1.342\times10^{-3}\ \text{rad}=1.342\ \text{mrad}.$$

rms jitter (spec Eq. 19), with $f_0=5$ GHz:

$$\sigma_t=\frac{\sigma_\phi}{2\pi f_0}=\frac{1.342\times10^{-3}}{2\pi\times5\times10^{9}}\approx4.27\times10^{-14}\ \text{s}=42.7\ \text{fs}.$$

Result: $S_\phi(1\text{MHz})=2\times10^{-12}\ \text{rad}^2/\text{Hz}$, $\sigma_\phi\approx1.34$ mrad, $\sigma_t\approx42.7$ fs (1→10 MHz band).

Dimension check: (a) $\mathcal{L}$ and $S_\phi$ are both per-Hz (rad²/Hz) ✓; (b) $[K]\cdot[1/f]=\text{rad}^2\cdot\text{Hz}\cdot\text{Hz}^{-1}=\text{rad}^2$ ✓; $\dfrac{\text{rad}}{\text{rad/s}}=\text{s}$ ✓.

import numpy as np
from simulations.common.noise_utils import integrate_rms_jitter
# Analytic path (direct substitution)
L_dbc = -120.0
S_phi_1M = 2*10**(L_dbc/10)              # = 2e-12 rad^2/Hz
K = S_phi_1M*(1e6)**2                    # = 2 rad^2*Hz
var = K*(1/1e6 - 1/1e7)
sigma_phi = np.sqrt(var)
sigma_t = sigma_phi/(2*np.pi*5e9)
print(sigma_phi*1e3, "mrad ;", sigma_t*1e15, "fs")   # -> 1.342 mrad ; 42.7 fs

# Numeric path (fold L(f)=K/f^2 back to dBc/Hz, then integrate); matches the analytic result
f = np.logspace(6, 7, 2000)
L_curve = 10*np.log10(0.5*K/f**2)        # L = S_phi/2 = (K/f^2)/2
st, sp = integrate_rms_jitter(f, L_curve, f0=5e9, fmin=1e6, fmax=1e7)
print(st*1e15, "fs (numeric)")           # -> ~42.7 fs

---

Key takeaways

• phase↔time: $\Delta t=\Delta\phi/(2\pi f_0)$; rms version $\sigma_t=\sigma_\phi/(2\pi f_0)$ (Exercises 1, 2).
• Parseval: $\sum c_n^2=2\Gamma_{rms}^2$; for $-\sin$, $\Gamma_{rms}=1/\sqrt2$ and $\sum c_n^2=1$ (Exercise 3).
• PSD integration: in the $1/f^2$ region $\sigma_\phi^2$ is dominated by the low-frequency end; $\sigma_\phi^2=K(1/f_1-1/f_2)$ (Exercises 4, 8).
• The essence of LTV: the response depends on the absolute injection phase; the same impulse at different phases gives different $\Delta\phi$ (Exercise 5).
• Lorentzian: $D=\pi\Delta f_{3\mathrm{dB}}$, coherence time $\sim1/(2D)$ (Exercise 6, external literature).
• Allan slopes: $\tau^{-1/2}/\tau^0/\tau^{+1/2}\to$ white/flicker/random-walk FM; flicker FM forms a floor (Exercise 7, external literature).
• All Python verifications call simulations/common/ (noise_utils, isf_utils) and can be reproduced directly.

Further reading

• Phase/amplitude and jitter definitions: oscillator_phase, phase_vs_amplitude_noise
• PSD / phase noise / jitter: psd_phase_noise_jitter
• LTI vs LTV: lti_vs_ltv
• Allan variance: allan_variance
• Lorentzian linewidth: lorentzian_linewidth
• More exercises: 03 core-theory chapter exercises · 06 design chapter exercises · 04 simulation chapter worked examples