Chapter 2

Wiggling the knobs: modulation

Chapter 1 gave us a signal with three knobs, $V(t) = A\cos(2\pi\nu_0 t + \varphi)$. Noise, when it arrives in later chapters, will be nothing but these knobs being wiggled randomly. So before touching randomness we wiggle each knob with the simplest possible motion — a single sinusoid — and learn everything there is to know about sidebands in a setting where every line can be checked by hand.

Amplitude modulation

Wiggle the amplitude knob sinusoidally at a modulation frequency $f_m$, with a fractional modulation depth $m$:

$$ V(t) = A_0\big[1 + m\cos(2\pi f_m t)\big]\cos(2\pi\nu_0 t). $$

In the time domain this is a carrier whose envelope breathes. To see what it is in the frequency domain, just multiply out and use the product-to-sum identity $\cos a \cos b = \tfrac12\cos(a+b) + \tfrac12\cos(a-b)$:

$$ \begin{aligned} V(t) = A_0\cos(2\pi\nu_0 t) &\;+\; \frac{mA_0}{2}\cos\!\big(2\pi(\nu_0 + f_m)t\big) \\[2pt] &\;+\; \frac{mA_0}{2}\cos\!\big(2\pi(\nu_0 - f_m)t\big). \end{aligned} $$

No approximation was made: an amplitude-modulated tone is exactly three pure tones. The original carrier at $\nu_0$ is untouched, and two new lines — sidebands — appear symmetrically at $\nu_0 \pm f_m$, each with amplitude $mA_0/2$. Since power goes as amplitude squared, each sideband carries a fraction

$$ \frac{P_{\text{sideband}}}{P_{\text{carrier}}} = \frac{m^2}{4}, \qquad\text{i.e.}\qquad 20\log_{10}\!\frac{m}{2}\ \text{dBc} $$

of the carrier power ("dBc" = decibels relative to the carrier). A 20% modulation depth puts each sideband 20 dB below the carrier; 2% puts it 40 dB down. Check the prediction against the live spectrum:

Note what the two sidebands are doing, because it echoes the beats of chapter 1 in reverse: there, two tones made a breathing envelope; here, a breathing envelope is two tones flanking the carrier. Sidebands are not a metaphor — a spectrum analyzer really shows three resolvable lines, and you can filter one out with a sharp enough filter.

Phase modulation

Now wiggle the phase knob instead, by $\beta$ radians at frequency $f_m$:

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

The number $\beta$ is called the modulation index and it is measured in radians — it is the peak phase excursion. For small excursions, expand with $\cos(x + \epsilon) \approx \cos x - \epsilon\sin x$:

$$ V(t) \approx A\cos(2\pi\nu_0 t) \;+\; \frac{A\beta}{2}\cos\!\big(2\pi(\nu_0 + f_m)t\big) \;-\; \frac{A\beta}{2}\cos\!\big(2\pi(\nu_0 - f_m)t\big), $$

where the last step used $\sin a \sin b = \tfrac12\cos(a-b) - \tfrac12\cos(a+b)$. Again a carrier and two sidebands at $\nu_0 \pm f_m$, each of amplitude $A\beta/2$ — the same sideband amplitude as an AM signal with $m = \beta$. The only difference is that minus sign on the lower sideband.

What happens when $\beta$ is not small? The expansion doesn't stop at first order. The exact result is one of the prettiest identities in applied mathematics — the Jacobi–Anger expansion in terms of Bessel functions $J_n$:

$$ e^{\,i\beta\sin\theta} \;=\; \sum_{n=-\infty}^{\infty} J_n(\beta)\, e^{\,in\theta}, $$

which turns the phase-modulated tone into an exact sum of pure tones,

$$ V(t) = A \sum_{n=-\infty}^{\infty} J_n(\beta)\, \cos\!\big(2\pi(\nu_0 + n f_m)t\big). $$

Phase modulation generates a ladder of sidebands at every harmonic $\nu_0 \pm n f_m$, with amplitudes $J_n(\beta)$ (and $J_{-n} = (-1)^n J_n$, so the ladder is symmetric in magnitude). Three facts to take away:

• For $\beta \ll 1$, $J_0 \approx 1$, $J_1 \approx \beta/2$, and higher orders scale as $\beta^n$ — recovering the small-$\beta$ result.
• Total power is conserved: $\sum_n J_n^2(\beta) = 1$. The envelope never changes (the arrow keeps length $A$), so power taken by sidebands is stolen from the carrier.
• The carrier amplitude is $J_0(\beta)$, which passes through zero at $\beta \approx 2.405$: drive the phase hard enough and the carrier vanishes entirely, with all the power in sidebands.

The phasor picture of sidebands

Here is the deepest intuition in this chapter — the picture that will let you see the difference between amplitude and phase noise in chapter 6. Climb into the frame rotating with the carrier at $\nu_0$ (chapter 1's rotating arrow, viewed from a co-rotating camera). In this frame the carrier is a frozen arrow of length $A$. The upper sideband, at $\nu_0 + f_m$, rotates faster than the frame, so it appears as a small arrow turning counter-clockwise at $f_m$; the lower sideband turns clockwise at $f_m$. Two small counter-rotating arrows riding on the tip of a big one — that is what modulation looks like.

Everything is in how the two small arrows are phased. Write the complex envelope (the signal with the carrier rotation divided out), with $\theta = 2\pi f_m t$:

$$ \text{AM:}\quad 1 + \tfrac{m}{2}e^{\,i\theta} + \tfrac{m}{2}e^{-i\theta} = 1 + m\cos\theta \qquad\text{(real: } \parallel \text{ carrier)}, $$

$$ \text{PM:}\quad 1 + \tfrac{\beta}{2}e^{\,i\theta} - \tfrac{\beta}{2}e^{-i\theta} = 1 + i\beta\sin\theta \qquad\text{(imaginary: } \perp \text{ carrier)}. $$

For AM the two arrows are mirror images about the carrier: their sum always lies along the carrier, so the tip slides in and out — the length breathes, the angle never moves. For PM the lower arrow is flipped by 180° (that minus sign): the pair's sum is always perpendicular to the carrier, so the tip swings side to side — the angle wobbles, the length (to first order) never changes. Same two arrows, one relative sign, completely different physics.

One more payoff hides in the PM picture. The first-order tip path is a straight line tangent to the circle of radius $A$ — but the true PM signal has constant length, so its tip really moves on the faint circular arc. The gap between the straight line and the arc is second order in $\beta$, and it is precisely what the higher-order Bessel sidebands are for: adding $J_2, J_3, \dots$ terms bends the straight line onto the arc. Small $\beta$: line ≈ arc, two sidebands suffice. Large $\beta$: you need the whole ladder.

Frequency modulation

The third knob. Wiggle the frequency sinusoidally with peak frequency deviation $\Delta\nu$:

$$ \nu(t) = \nu_0 + \Delta\nu \cos(2\pi f_m t). $$

Chapter 1's one-liner — frequency is the slope of phase, $\nu(t) = \dot\Phi/2\pi$ — tells us exactly what signal this is. To get the phase, integrate:

$$ \Phi(t) = 2\pi\!\int_0^t \nu(t')\,dt' = 2\pi\nu_0 t + \frac{\Delta\nu}{f_m}\,\sin(2\pi f_m t). $$

Compare with the PM waveform above: FM is PM with modulation index

$$ \beta = \frac{\Delta\nu}{f_m}. $$

There is no third kind of modulation — only the amplitude knob and the phase knob, with FM being the phase knob driven through an integrator. But the formula $\beta = \Delta\nu/f_m$ has a consequence that trips people up constantly: the same frequency deviation produces a very different spectrum depending on how fast you wiggle.

• Narrowband FM ($f_m \gg \Delta\nu$, so $\beta \ll 1$): two puny sidebands at $\nu_0 \pm f_m$, carrier barely touched. Fast, gentle wiggling hardly disturbs the phase.
• Wideband FM ($f_m \ll \Delta\nu$, so $\beta \gg 1$): a dense forest of Bessel sidebands filling the band the instantaneous frequency actually sweeps, $\nu_0 \pm \Delta\nu$.

A serviceable estimate covering both limits is Carson's rule for the bandwidth containing ~98% of the power:

$$ B \approx 2\,(\Delta\nu + f_m). $$

Noise is modulation at every frequency at once

Now the point of the whole exercise. A noisy oscillator is $V(t) = A\cos(2\pi\nu_0 t + \varphi(t))$ with $\varphi(t)$ a random wiggle. But any $\varphi(t)$ can be decomposed into Fourier components — a sum of sinusoidal phase modulations, one at every frequency $f$, each with its own tiny index $\beta(f) \ll 1$. Each component does exactly what this chapter taught: it plants one pair of sidebands at $\nu_0 \pm f$, of power $\beta(f)^2/4$ relative to the carrier, with the PM phase signature. Sum over all components and the discrete sidebands merge into a continuous noise pedestal hugging the carrier — the "skirt" you see around every real oscillator's line on a spectrum analyzer. A phase-noise pedestal is nothing but a continuum of PM sidebands. Making this quantitative — how the pedestal's shape encodes the spectral density of $\varphi(t)$ — needs the language of random processes (chapter 3) and spectral densities (chapter 4), and pays off in chapter 6, where the picture becomes the working definition of phase noise $\mathcal{L}(f)$.

The same thing in code

The honest way to synthesize a phase- or frequency-modulated signal — and the way the FM demo above actually does it — is to accumulate the phase sample by sample, advancing it at the instantaneous frequency. This works for any $\nu(t)$, including the random ones of later chapters:

// synthesize V(t) = A cos Φ(t)  with  ν(t) = ν0 + Δν cos(2π f_m t)
let phi = 0;                       // Φ(0)
for (let i = 0; i < N; i++) {
  const t  = i / fs;
  const nu = nu0 + dnu * Math.cos(2 * Math.PI * fm * t);
  V[i] = A * Math.cos(phi);
  phi += 2 * Math.PI * nu / fs;    // dΦ = 2π ν(t) dt
}

The single line phi += 2 * Math.PI * nu / fs is chapter 1's "frequency is the slope of phase", run forward as a numerical integral. Replace the cosine by a random sequence and you have synthesized frequency noise; that is literally how the demos of chapter 6 work.

Exercises