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

Core-Theory Chapter Exercises (with Full Solutions)

Prerequisite reading: this chapter's theory pages isf_definition, fourier_series_of_isf, rms_isf, white_noise_to_phase_noise, flicker_noise_upconversion, effective_isf (finish them before attempting the problems).

This page is the complete exercise set for Chapter 03, ISF Core Theory. The problems span derivations, numerical problems, and design back-calculations, all built around the [P1] Hajimiri–Lee ISF framework, using the site-wide notation.

Format: every solution = step-by-step substitution (with units) → result → dimension check → one-line Python verification. Python always imports from simulations/common/ (real functions, nothing fabricated).

Authoritative formulas involved (verbatim from spec Section 3, with citations):

• impulse→phase (operational ISF): $\Delta\phi=\dfrac{\Gamma(\omega_0\tau)}{q_{max}}\,\Delta q$ (spec formula 5)
• ISF Fourier series: $\Gamma(\omega_0\tau)=\dfrac{c_0}{2}+\displaystyle\sum_{n=1}^{\infty}c_n\cos(n\omega_0\tau+\theta_n)$ ([P1] Eq.(12), p.183)
• Parseval / rms ISF: $\displaystyle\sum_{n=0}^{\infty}c_n^2=\frac{1}{\pi}\int_0^{2\pi}\lvert\Gamma(x)\rvert^2dx=2\,\Gamma_{rms}^2$ ([P1] Eq.(20), p.185)
• The signature white-noise 1/f² result: $\mathcal{L}\{\Delta\omega\}=10\log_{10}\!\left(\dfrac{\Gamma_{rms}^2}{q_{max}^2}\cdot\dfrac{\overline{i_n^2}/\Delta f}{4\,\Delta\omega^2}\right)$ ([P1] Eq.(21), p.185)
• 1/f³ corner: $\Delta\omega_{1/f^3}=\omega_{1/f}\cdot\dfrac{c_0^2}{2\,\Gamma_{rms}^2}\approx\omega_{1/f}\left(\dfrac{c_0}{c_1}\right)^2$ ([P1] Eq.(24), p.185)
• effective ISF (cyclostationary): $\Gamma_{eff}=\Gamma\cdot\alpha$ ([P1] Eqs.(25)–(27), p.186)

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Problems

Exercise 1 (numerical) — impulse → phase step

Ideal LC ($\Gamma(\theta)=-\sin\theta$), $q_{max}=1$ pC, $f_0=5$ GHz. A single charge impulse of $\Delta q=1$ fC is injected. Find: (a) the phase step $\Delta\phi$ (rad) and the timing error $\Delta t$ (fs) for injection at $\theta=3\pi/2$ (where $\Gamma$ takes its maximum $+1$). (b) $\Delta\phi$ for injection at the peak, $\theta=0$.

Quick check (work it out yourself, then check)

Work out (a) yourself first: injection at θ = 3π/2 gives timing error Δt = ? (Δq = 1 fC, q_max = 1 pC, Γ = +1, f₀ = 5 GHz; answer in fs)

fs

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

Exercise 2 (derivation + numerical) — $\Gamma_{rms}$ from the ISF

A toy ISF is the two-harmonic waveform $\Gamma(\theta)=\cos\theta+\tfrac12\cos(2\theta)$. (a) Write down the Fourier coefficients $c_0,c_1,c_2$ directly. (b) Use Parseval to find $\sum c_n^2$ and $\Gamma_{rms}$.

Exercise 3 (numerical) — white noise → $\mathcal{L}$ (applying Eq.(21))

$f_0=5$ GHz, $\Delta f=1$ MHz, $q_{max}=1$ pC, $\Gamma_{rms}=0.5$, $S_i=\overline{i_n^2}/\Delta f=10^{-24}\ \text{A}^2/\text{Hz}$. Use [P1] Eq.(21) to find $\mathcal{L}(1\,\text{MHz})$ (dBc/Hz).

Quick check (work it out yourself, then check)

Work it out yourself first: L(1 MHz) = ? (Γ_rms = 0.5, q_max = 1 pC, S_i = 10⁻²⁴ A²/Hz; answer in dBc/Hz, remember the minus sign)

dBc/Hz

Graded correct within ±1% relative error; scientific notation is accepted.

Exercise 4 (design back-calculation) — solving for the required $q_{max}$

Keep the numbers from Exercise 3, but the target spec is now $\mathcal{L}(1\,\text{MHz})=-160$ dBc/Hz (cleaner than Exercise 3). With all other parameters unchanged ($\Gamma_{rms}=0.5$, $S_i=10^{-24}$, $\Delta f=1$ MHz), by how much must $q_{max}$ be scaled up?

Quick check (work it out yourself, then check)

Work it out yourself first: q_max needed to reach −160 dBc/Hz = ? (the original q_max = 1 pC gives −148 dBc/Hz; answer in pC)

pC

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

Exercise 5 (derivation + numerical) — $c_0\to1/f^3$ corner

An oscillator's measured ISF has $c_0=0.2$, $c_1=1.0$ (i.e., an appreciable DC offset — the waveform is up/down asymmetric), and the device 1/f corner is $f_{1/f}=1$ MHz (i.e., $\omega_{1/f}=2\pi\times10^6$ rad/s). Use [P1] Eq.(24) to estimate the $1/f^3$ corner frequency $\Delta f_{1/f^3}$ (with the $c_0/c_1$ approximation). If the circuit is made symmetric ($c_0\to0.02$), what does the corner become?

Quick check (work it out yourself, then check)

Try it yourself first: for the asymmetric case (c₀=0.2, c₁=1.0), the 1/f³ corner Δf ≈ ? (f_1/f = 1 MHz; answer in kHz)

kHz

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

Exercise 6 (derivation) — the frequency-translation meaning of the Fourier coefficients

For a single tone injected near $n\omega_0$, $i(\tau)=I_0\cos((n\omega_0+\Delta\omega)\tau)$, use the product-to-sum identity to prove by hand that after weighting by the $n$-th ISF harmonic $c_n\cos(n\omega_0\tau+\theta_n)$ and integrating, the surviving slow term gives $\phi_n(t)\approx\dfrac{I_0 c_n}{2q_{max}}\cdot\dfrac{\sin(\Delta\omega t-\theta_n)}{\Delta\omega}$, and explain why this is exactly "the oscillator acting as a mixer, downconverting noise near $n\omega_0$ to $\Delta\omega$".

Exercise 7 (numerical) — effective ISF (cyclostationary)

A certain noise source conducts only during one half-cycle of the waveform. Approximate its noise modulating function (NMF) $\alpha(\theta)$ as square-wave gating: $\alpha(\theta)=1$ for $\theta\in[0,\pi)$, $\alpha(\theta)=0$ for $\theta\in[\pi,2\pi)$. The base ISF is still $\Gamma(\theta)=-\sin\theta$. Find $\Gamma_{eff,rms}$ of the effective ISF $\Gamma_{eff}=\Gamma\cdot\alpha$, and compare with the always-conducting case $\Gamma_{rms}=1/\sqrt2$.

Quick check (work it out yourself, then check)

Try it yourself first: after half-cycle gating, Γ_eff,rms = ? (Γ(θ)=−sinθ, α=1 on [0,π), 0 on [π,2π))

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

Exercise 8 (design back-calculation) — $\Gamma_{rms}/q_{max}$ from $\mathcal{L}$

A 5 GHz LC oscillator measures $\mathcal{L}(1\,\text{MHz})=-130$ dBc/Hz, and its white-noise source is known to be $S_i=2\times10^{-23}\ \text{A}^2/\text{Hz}$ (multi-source equivalent). Assuming the $1/f^2$ region is white-noise dominated, apply Eq.(21) to back-solve the effective $\Gamma_{rms}/q_{max}$ (units $1/\text{C}$). If $q_{max}=1.5$ pC, roughly what is $\Gamma_{rms}$?

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Solutions

Exercise 1 solution (impulse → phase step)

(a) $\theta=3\pi/2$. $\Gamma=-\sin(3\pi/2)=-(-1)=+1$. Using spec formula 5:

$$\Delta\phi=\frac{\Gamma}{q_{max}}\Delta q=\frac{1\times(1\times10^{-15}\ \text{C})}{1\times10^{-12}\ \text{C}}=1\times10^{-3}\ \text{rad}=1\ \text{mrad}.$$

Timing error (spec formula 17):

$$\Delta t=\frac{\Delta\phi}{2\pi f_0}=\frac{10^{-3}}{2\pi\times5\times10^{9}}\approx3.18\times10^{-14}\ \text{s}=31.8\ \text{fs}.$$

(b) $\theta=0$. $\Gamma=-\sin0=0\Rightarrow\Delta\phi=0$ (injection at the peak only changes the amplitude, which the restoring force pulls back).

Result: (a) $\Delta\phi=1$ mrad, $\Delta t\approx31.8$ fs; (b) $\Delta\phi=0$.

Intuition: this is the full-scale version of canonical Example A ($\Gamma=0.5$ gives 15.9 fs) — $\Gamma=1$ gives twice that, i.e. 31.8 fs.

Dimension check: $\Gamma$ dimensionless $\times$ (C/C) $=$ rad ✓; $\dfrac{\text{rad}}{\text{rad/s}}=\text{s}$ ✓.

import numpy as np
from simulations.common.isf_utils import gamma_lc_ideal, impulse_to_phase_step
from simulations.common.noise_utils import phase_to_time_error
for th in (3*np.pi/2, 0.0):
    dphi = impulse_to_phase_step(1e-15, gamma_lc_ideal(th), qmax=1e-12)
    print(round(dphi*1e3,3), "mrad ;", round(phase_to_time_error(dphi,5e9)*1e15,1), "fs")
# -> 1.0 mrad ; 31.8 fs    and    0.0 mrad ; 0.0 fs

Exercise 2 solution ($\Gamma_{rms}$ from the ISF)

(a) Read off the coefficients. Match $\Gamma(\theta)=\cos\theta+\tfrac12\cos(2\theta)$ against the Fourier series $\Gamma=\tfrac{c_0}{2}+\sum c_n\cos(n\theta+\theta_n)$: no constant term $\Rightarrow c_0=0$; first-harmonic amplitude $c_1=1$ ($\theta_1=0$); second-harmonic amplitude $c_2=\tfrac12$ ($\theta_2=0$); $c_{n\ge3}=0$.

(b) Parseval. Spec formula 11:

$$\sum_{n=0}^{\infty}c_n^2=c_0^2+c_1^2+c_2^2=0+1^2+\left(\tfrac12\right)^2=1.25.$$$$\Gamma_{rms}^2=\frac{\sum c_n^2}{2}=\frac{1.25}{2}=0.625\quad\Longrightarrow\quad\Gamma_{rms}=\sqrt{0.625}\approx0.791.$$

Result: $c_0=0,\ c_1=1,\ c_2=0.5$; $\sum c_n^2=1.25$; $\Gamma_{rms}\approx0.791$.

Cross-check (direct integration): $\Gamma_{rms}^2=\tfrac{1}{2\pi}\int_0^{2\pi}(\cos\theta+\tfrac12\cos2\theta)^2d\theta$; the cross term $\int\cos\theta\cos2\theta=0$ (orthogonality), leaving $\tfrac12+\tfrac12\cdot\tfrac14=0.5+0.125=0.625$ ✓.

Dimension check: $c_n$ and $\Gamma_{rms}$ are all dimensionless ✓.

import numpy as np
from simulations.common.isf_utils import gamma_rms, compute_fourier_coefficients
theta = np.linspace(0, 2*np.pi, 8192, endpoint=False)
g = np.cos(theta) + 0.5*np.cos(2*theta)
print(gamma_rms(theta, g))                          # -> 0.7906
a0, a, b, c, ph = compute_fourier_coefficients(theta, g, n_harmonics=3)
print(c[:3], np.sum(c**2))                          # -> ~[1, 0.5] ... ; 1.25

Exercise 3 solution (white noise → $\mathcal{L}$, applying Eq.(21))

Step-by-step substitution (with units). This is canonical Example B.

1. $\Delta\omega=2\pi\Delta f=2\pi\times10^6=6.283\times10^6\ \text{rad/s}$, $\Delta\omega^2=3.948\times10^{13}$.
2. $\dfrac{\Gamma_{rms}^2}{q_{max}^2}=\dfrac{0.25}{(10^{-12})^2}=2.5\times10^{23}\ \text{C}^{-2}$.
3. $\dfrac{S_i}{4\Delta\omega^2}=\dfrac{10^{-24}}{4\times3.948\times10^{13}}=6.332\times10^{-39}$.
4. Multiply: $2.5\times10^{23}\times6.332\times10^{-39}=1.583\times10^{-15}$.
5. $\mathcal{L}=10\log_{10}(1.583\times10^{-15})=-148.0\ \text{dBc/Hz}$.

Result: $\mathcal{L}(1\,\text{MHz})\approx-148.0$ dBc/Hz (the theoretical floor for a single ideal white-noise source).

Dimension check: inside the bracket, $\text{C}^{-2}\cdot\dfrac{\text{A}^2/\text{Hz}}{(\text{rad/s})^2}$; with $\text{C}=\text{A·s}$ this reduces to $\text{s}$ (per-Hz), and taking $10\log_{10}$ reads as dBc/Hz ✓. See white_noise_to_phase_noise.

import numpy as np
gamma_rms, qmax, Si = 0.5, 1e-12, 1e-24
dw = 2*np.pi*1e6
L = 10*np.log10((gamma_rms**2/qmax**2)*(Si/(4*dw**2)))
print(round(L,1), "dBc/Hz")   # -> -148.0 dBc/Hz

Exercise 4 solution (solving for the required $q_{max}$)

Back-calculation strategy. $\mathcal{L}\propto1/q_{max}^2$ (denominator of Eq.(21)). The target is $\Delta L=-160-(-148)=-12$ dB below Exercise 3. Writing $\mathcal{L}$ in linear form with everything else fixed, $\mathcal{L}_{\text{lin}}\propto1/q_{max}^2$:

$$\frac{q_{max,\text{new}}^2}{q_{max,\text{old}}^2}=\frac{\mathcal{L}_{\text{lin,old}}}{\mathcal{L}_{\text{lin,new}}}=10^{(-148-(-160))/10}=10^{12/10}=10^{1.2}=15.85.$$$$\frac{q_{max,\text{new}}}{q_{max,\text{old}}}=\sqrt{15.85}=3.98\quad\Longrightarrow\quad q_{max,\text{new}}\approx3.98\times1\ \text{pC}=3.98\ \text{pC}.$$

Direct check (solving Eq.(21) for $q_{max}$ outright): $q_{max}=\sqrt{\dfrac{\Gamma_{rms}^2}{\mathcal{L}_{\text{lin}}}\cdot\dfrac{S_i}{4\Delta\omega^2}}$, with $\mathcal{L}_{\text{lin}}=10^{-16}$:

$$q_{max}=\sqrt{\frac{0.25}{10^{-16}}\times6.332\times10^{-39}}=\sqrt{1.583\times10^{-23}}=3.98\times10^{-12}\ \text{C}=3.98\ \text{pC}.$$

Result: $q_{max}$ must be scaled up by about 4× to $\approx3.98$ pC (i.e., every 6 dB of phase-noise reduction costs $q_{max}$ $\times2$).

Intuition: this quantifies "increasing the signal swing is the most direct knob for lowering $1/f^2$ phase noise" (claim C3), but 12 dB demands 4× the charge swing, paid for in power/area — exactly the trade-off in tank_swing.

Dimension check: $\sqrt{\text{C}^{-2}\text{·s}^{-1}\cdots}$ inverts back to $q_{max}$ in C ✓.

import numpy as np
gamma_rms, Si, dw = 0.5, 1e-24, 2*np.pi*1e6
L_lin = 10**(-160/10)
qmax = np.sqrt((gamma_rms**2/L_lin)*(Si/(4*dw**2)))
print(qmax*1e12, "pC")        # -> 3.98 pC

Exercise 5 solution ($c_0\to1/f^3$ corner)

Step-by-step substitution (with units). Use the $c_0/c_1$ approximation of [P1] Eq.(24), $\Delta\omega_{1/f^3}\approx\omega_{1/f}\left(\dfrac{c_0}{c_1}\right)^2$, then convert via $\Delta f_{1/f^3}=\Delta\omega_{1/f^3}/(2\pi)$; since $\omega_{1/f}=2\pi f_{1/f}$, the $2\pi$ cancels: $\Delta f_{1/f^3}\approx f_{1/f}\left(\dfrac{c_0}{c_1}\right)^2$.

Asymmetric case ($c_0=0.2$, $c_1=1.0$):

$$\Delta f_{1/f^3}\approx10^6\times\left(\frac{0.2}{1.0}\right)^2=10^6\times0.04=4\times10^{4}\ \text{Hz}=40\ \text{kHz}.$$

After symmetrization ($c_0=0.02$, $c_1=1.0$):

$$\Delta f_{1/f^3}\approx10^6\times\left(\frac{0.02}{1.0}\right)^2=10^6\times4\times10^{-4}=400\ \text{Hz}.$$

Result: asymmetric, the $1/f^3$ corner is $\approx40$ kHz; after symmetrization ($c_0$ down 10×) the corner drops $100$× to $\approx400$ Hz.

Validity (approximate vs exact form): the $(c_0/c_1)^2$ approximation assumes the ISF is fundamental-dominated ($\Gamma_{rms}^2\approx c_1^2/2$). Here $c_0=0.2$ is not negligible; using the exact form $\Delta f_{1/f^3}=f_{1/f}\cdot\dfrac{c_0^2}{2\Gamma_{rms}^2}$, with $\Gamma_{rms}^2=(c_0^2+c_1^2)/2=(0.04+1)/2=0.52$, gives $\Delta f_{1/f^3}=10^6\times\dfrac{0.04}{2\times0.52}\approx38.5$ kHz — about 4% off the approximate 40 kHz. After symmetrization ($c_0=0.02$, $\Gamma_{rms}^2\approx0.5$) the two forms nearly coincide.

Design message: $1/f^3$ corner $\propto c_0^2$. Making the waveform up/down symmetric (suppressing $c_0$) is the most effective way to push flicker-upconverted close-in $1/f^3$ noise away from the carrier (see symmetry, flicker_noise_upconversion).

Dimension check: $(c_0/c_1)^2$ dimensionless $\times\ f_{1/f}$ (Hz) $=$ Hz ✓.

f_1f = 1e6
for c0 in (0.2, 0.02):
    f_corner = f_1f*(c0/1.0)**2
    print(c0, "->", f_corner, "Hz")   # -> 0.2 -> 40000.0 Hz ; 0.02 -> 400.0 Hz

Exercise 6 solution (frequency-translation meaning of the Fourier coefficients — derivation)

Target expression. The phase contribution of the $n$-th harmonic term ([P1] Eq.(13)):

$$\phi_n(t)=\frac{1}{q_{max}}\int^{t}\!\!c_n\cos(n\omega_0\tau+\theta_n)\,I_0\cos\big((n\omega_0+\Delta\omega)\tau\big)\,d\tau.$$

Step (i): product-to-sum. Let $A=n\omega_0\tau+\theta_n$, $B=(n\omega_0+\Delta\omega)\tau$, and use $\cos A\cos B=\tfrac12[\cos(A-B)+\cos(A+B)]$:

$$A-B=\theta_n-\Delta\omega\tau,\qquad A+B=(2n\omega_0+\Delta\omega)\tau+\theta_n.$$

Integrand $=\dfrac{I_0c_n}{2}\Big[\underbrace{\cos(\Delta\omega\tau-\theta_n)}_{\text{slow term, }\approx\Delta\omega}+\underbrace{\cos((2n\omega_0+\Delta\omega)\tau+\theta_n)}_{\text{fast term, }\approx2n\omega_0}\Big]$.

Step (ii): the integrator is a low-pass filter — the fast term averages out.

• Slow term: $\int\cos(\Delta\omega\tau-\theta_n)d\tau=\dfrac{\sin(\Delta\omega\tau-\theta_n)}{\Delta\omega}$ — the denominator is only the tiny $\Delta\omega$, so it gets amplified and survives.
• Fast term: $\dfrac{\sin(\cdots)}{2n\omega_0+\Delta\omega}$ — the denominator is the huge $2n\omega_0$; the amplitude is crushed by a factor $\sim\Delta\omega/(2n\omega_0)$, hence negligible.

Step (iii): keep only the slow term.

$$\phi_n(t)\approx\frac{1}{q_{max}}\cdot\frac{I_0c_n}{2}\cdot\frac{\sin(\Delta\omega t-\theta_n)}{\Delta\omega}=\frac{I_0 c_n}{2q_{max}}\cdot\frac{\sin(\Delta\omega t-\theta_n)}{\Delta\omega}.\qquad\blacksquare$$

Physical meaning (mixer view): the $n$-th ISF harmonic $c_n$ acts like one tooth of an LO (local-oscillator) comb, downconverting noise at the injection frequency $n\omega_0+\Delta\omega$ to $\Delta\omega$ in baseband; the fast term (sum frequency $\approx2n\omega_0$) is filtered out by the integrator's low-pass action. The oscillator is itself a mixer sampling its own harmonics — this is why $c_n$ is "the conversion coefficient from each harmonic to the phase output" (see fourier_series_of_isf).

Dimension check: the units of $\phi_n=\dfrac{I_0 c_n}{2q_{max}}\cdot\dfrac{\sin(\cdots)}{\Delta\omega}$ are $\dfrac{[\text{A}]\cdot(\text{dimensionless})}{[\text{C}]\cdot[\text{rad/s}]}$; substituting $\text{C}=\text{A·s}$ and treating rad as dimensionless, $=\dfrac{\text{A}}{(\text{A·s})\cdot(1/\text{s})}=\dfrac{\text{A}}{\text{A}}=1$ (dimensionless), so $\phi$ is dimensionless (phase in rad) ✓.

import numpy as np
# Numerical check: do the integral directly; the slow term survives, the fast term vanishes
n, w0, dw, c_n, I0, qmax, th_n = 1, 1.0, 1e-3, 1.0, 1.0, 1.0, 0.4
tau = np.linspace(0, 2000*np.pi, 4_000_000)         # spans many slow periods
integrand = c_n*np.cos(n*w0*tau+th_n)*I0*np.cos((n*w0+dw)*tau)
phi = np.cumsum(integrand)*(tau[1]-tau[0])/qmax
analytic = I0*c_n/(2*qmax)*np.sin(dw*tau-th_n)/dw
print(np.max(np.abs(phi-analytic-np.mean(phi-analytic)))/np.max(np.abs(analytic)))
# -> slow-term envelope matches (relative error ~5e-4; the residual is the averaged-out fast term)

Exercise 7 solution (effective ISF, cyclostationary)

Step-by-step substitution. effective ISF ([P1] Eqs.(25)–(27)): $\Gamma_{eff}(\theta)=\Gamma(\theta)\,\alpha(\theta)$. Here $\alpha$ is square-wave gating (conducting for half the cycle), so

$$\Gamma_{eff}(\theta)=\begin{cases}-\sin\theta,&\theta\in[0,\pi)\\[2pt]0,&\theta\in[\pi,2\pi).\end{cases}$$

Take the rms:

$$\Gamma_{eff,rms}^2=\frac{1}{2\pi}\int_0^{2\pi}\Gamma_{eff}^2\,d\theta=\frac{1}{2\pi}\int_0^{\pi}\sin^2\theta\,d\theta=\frac{1}{2\pi}\cdot\frac{\pi}{2}=\frac14.$$$$\Gamma_{eff,rms}=\frac12=0.5.$$

Comparison: always conducting gives $\Gamma_{rms}=1/\sqrt2\approx0.707$; with half-cycle gating, $\Gamma_{eff,rms}=0.5$. The ratio is $0.5/0.707=1/\sqrt2$ — gating off half the phase drops the rms by $\sqrt2$ (halves the power).

Result: $\Gamma_{eff,rms}=0.5$ ($\approx3$ dB lower in power than the always-on $0.707$).

Design message: letting the noise conduct only at phases where the ISF is small greatly reduces its effective contribution — this is the design intuition of steering noise current away from the high-sensitivity region (the zero crossings) (see effective_isf). The square-wave gating here is an illustrative toy; the real NMF $\alpha(\theta)$ is set by the device's bias-dependent thermal noise.

Dimension check: $\Gamma$, $\alpha$, $\Gamma_{eff}$ are all dimensionless ✓.

import numpy as np
from simulations.common.isf_utils import gamma_lc_ideal, gamma_rms, effective_isf
theta = np.linspace(0, 2*np.pi, 8192, endpoint=False)
g = gamma_lc_ideal(theta)
alpha = (theta < np.pi).astype(float)          # half-cycle square-wave gating
g_eff = effective_isf(g, alpha)                # = g*alpha
print(gamma_rms(theta, g_eff))                 # -> 0.5

Exercise 8 solution ($\Gamma_{rms}/q_{max}$ from $\mathcal{L}$)

Back-calculation strategy. Solve Eq.(21) for $\dfrac{\Gamma_{rms}^2}{q_{max}^2}$:

$$\frac{\Gamma_{rms}^2}{q_{max}^2}=\frac{\mathcal{L}_{\text{lin}}}{S_i/(4\Delta\omega^2)}=\mathcal{L}_{\text{lin}}\cdot\frac{4\Delta\omega^2}{S_i}.$$

Step-by-step substitution (with units).

1. $\mathcal{L}_{\text{lin}}=10^{-130/10}=10^{-13}$.
2. $\Delta\omega=2\pi\times10^6=6.283\times10^6$, $4\Delta\omega^2=4\times3.948\times10^{13}=1.579\times10^{14}$.
3. $\dfrac{4\Delta\omega^2}{S_i}=\dfrac{1.579\times10^{14}}{2\times10^{-23}}=7.896\times10^{36}$.
4. $\dfrac{\Gamma_{rms}^2}{q_{max}^2}=10^{-13}\times7.896\times10^{36}=7.896\times10^{23}\ \text{C}^{-2}$.
5. $\dfrac{\Gamma_{rms}}{q_{max}}=\sqrt{7.896\times10^{23}}=8.886\times10^{11}\ \text{C}^{-1}$.

If $q_{max}=1.5$ pC:

$$\Gamma_{rms}=\frac{\Gamma_{rms}}{q_{max}}\times q_{max}=8.886\times10^{11}\times1.5\times10^{-12}=1.33.$$

Result: $\Gamma_{rms}/q_{max}\approx8.89\times10^{11}\ \text{C}^{-1}$; if $q_{max}=1.5$ pC, then $\Gamma_{rms}\approx1.33$.

Intuition check: $\Gamma_{rms}\approx1.33$ is somewhat above the ideal $-\sin$ value $0.707$ — reasonable, because this part's measured phase noise ($-130$ dBc/Hz) sits about 18 dB above the canonical single ideal white-noise source ($-148$), reflecting the reality of multiple sources, cyclostationarity, and a larger ISF. The back-calculation serves as a health check: does the measured PN imply an effective $\Gamma_{rms}$ that is too large?

Dimension check: $\mathcal{L}_{\text{lin}}$ (per-Hz $=$ s) $\times\dfrac{(\text{rad/s})^2}{\text{A}^2/\text{Hz}}=\text{s}\cdot\dfrac{\text{s}^{-2}}{\text{A}^2\text{s}}=\text{A}^{-2}\text{s}^{-2}=\text{C}^{-2}$ ✓.

import numpy as np
L_lin, Si, dw = 10**(-130/10), 2e-23, 2*np.pi*1e6
ratio2 = L_lin*(4*dw**2/Si)              # (Gamma_rms/qmax)^2
ratio = np.sqrt(ratio2)
print(ratio, "1/C ;", ratio*1.5e-12, "= Gamma_rms")   # -> 8.89e11 1/C ; 1.33

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Key takeaways

• impulse→phase: $\Delta\phi=\Gamma\,\Delta q/q_{max}$; $\Gamma=1$ at 5 GHz gives 31.8 fs (Exercise 1).
• $\Gamma_{rms}$ from the ISF: Parseval $\sum c_n^2=2\Gamma_{rms}^2$; two-harmonic example $\Gamma_{rms}=0.791$ (Exercise 2).
• White noise→$\mathcal{L}$: canonical Example B $\approx-148$ dBc/Hz (Exercise 3); back-solving $q_{max}$: every 6 dB lower costs $\times2$ (Exercise 4).
• $c_0\to1/f^3$ corner: corner $\propto c_0^2$; symmetrizing by 10× → corner drops 100× (Exercise 5).
• Fourier coefficients = mixer conversion: $c_n$ downconverts noise near $n\omega_0$ to $\Delta\omega$; fast terms are filtered out by the integrator (Exercise 6).
• effective ISF: half-cycle gating gives $\Gamma_{eff,rms}=0.5 < 0.707$; steering noise away from high-sensitivity regions lowers noise (Exercise 7).
• Back-solving $\Gamma_{rms}/q_{max}$: measured PN can health-check whether the effective ISF is too large (Exercise 8).
• All Python verifications import from simulations/common/ (isf_utils, noise_utils).

Further reading

• impulse→phase: impulse_to_phase_shift
• white noise→$\mathcal{L}$: white_noise_to_phase_noise
• Fourier series: fourier_series_of_isf
• $\Gamma_{rms}$ and Parseval: rms_isf
• flicker upconversion and symmetry: flicker_noise_upconversion
• effective ISF: effective_isf

Other exercises

• Foundations chapter exercises (PSD / jitter dialects / random processes): 02 Foundations exercises
• Design chapter exercises (swing / topology / PLL budget / SerDes back-calculation): 06 Design chapter exercises
• Graded worked examples (basic conversions / ISF→PN / jitter integration / design back-calculation; each with a step-by-step solution + Python verification): worked_examples