Frequency and phase modulation

Here a message moves a carrier’s angle, not its size, and a slow tone turns the spectrum into a comb of lines. Watch the carrier’s line shrink and vanish as the swing grows, its power spreading into more and more lines.

The index decides the sidebands

In “Lift the message, and the envelope carries it” of Amplitude modulation (27.1), the message rode on the carrier’s size. The carrier’s peaks traced $1+\mu x[n]$, and an envelope detector read them back. Anything that changes the level on the way changes that message too.

This page keeps the size fixed and moves the angle instead. The signal is

with the carrier at $\Omega_1=2\pi f_1/f_s$. In “Where the wave starts” of Sinusoids (3.2), the phase $\phi$ was one fixed number. Here it is a signal, $\phi[n]$, that changes from sample to sample.

Phase or frequency

There are two ways to put a message $x[n]$ into that angle. In phase modulation (PM), the phase is the message times a constant, so the wave runs ahead where $x$ is positive and behind where it is negative. In frequency modulation (FM), the message sets how fast the phase turns.

In “Reading a chirp” of Spectrograms & the STFT (15.5), a tone’s frequency was the slope of its phase, divided by $2\pi$. In FM that slope is set so that the frequency at sample $n$ is

I scale the message so that its largest size is 1. Then $f_\text{dev}$, the peak frequency deviation, is the furthest the frequency moves away from the carrier.

A tone as the message

Let the message be one slow tone, $\cos(2\pi f_\text{mod}t)$, at the modulating frequency $f_\text{mod}$. The frequency then swings as $f_1+f_\text{dev}\cos(2\pi f_\text{mod}t)$, so the phase must be $\beta\sin(2\pi f_\text{mod}t)$. Its slope is $2\pi f_\text{mod}\beta\cos(2\pi f_\text{mod}t)$, and divided by $2\pi$ that is $f_\text{dev}\cos(2\pi f_\text{mod}t)$ when

This $\beta$ is the modulation index: the largest swing of the phase, in radians. On this page $\beta$ means only that. It is not the Kaiser parameter of chapters 15 and 19, nor the constant phase of 17.3.

Sampled at $f_s$, the FM tone is

I take $f_1=2000$ Hz, $f_\text{mod}=100$ Hz and $f_s=16$ kHz. Think of a singer’s vibrato: $f_\text{dev}$ sets how far the pitch swings, and $f_\text{mod}$ how often. Make the swing wide and fast enough, and the note stops sounding like a wobble and becomes a buzz.

Why the spectrum is a comb

Call the modulator’s own angle $\theta_n=2\pi f_\text{mod}n/f_s$. The arrow $e^{j\beta\sin\theta}$ comes back to where it started each time $\theta$ goes once round, so it is a repeating wave in $\theta$. By Fourier series coefficients (7.2), it is a sum of harmonics $e^{jk\theta}$, with coefficients

Write the integrand as a cosine plus $j$ times a sine of $k\tau-\beta\sin\tau$. The sine part is odd in $\tau$, so it adds up to zero. What remains has a name:

These are the Bessel functions of the first kind. This integral is all this page needs about them. Their shapes I read off the instrument.

As in Complex exponentials & phasors (3.4), a cosine is the real part of an arrow, so $y[n]$ is the real part of $e^{j\Omega_1n}e^{j\beta\sin\theta_n}$. Put the sum of harmonics in, and take the real part:

Term $k$ is a cosine at $f_1+kf_\text{mod}$. So the spectrum is a comb of lines, spaced $f_\text{mod}$ apart around the carrier, and line $k$ has height $\lvert J_k(\beta)\rvert$. Line $-k$ is as tall as line $k$, because $J_{-k}(\beta)=(-1)^kJ_k(\beta)$.

The picture at the top of the page draws this comb for $f_1=2000$ Hz and $f_\text{mod}=100$ Hz, with line $k$ at $2000+100k$ Hz. Drag its slider and watch three indices.

At $\beta=0.5$ the carrier keeps a height of 0.938 and the first pair of sidebands 0.242. The next pair is only 0.031, and the pair after that 0.003. A small index gives a carrier and two sidebands, much like the spectrum of AM.

At $\beta=2.405$ the carrier line has gone. That is the first zero of $J_0$, at 2.4048. The power has moved into the sidebands, at heights 0.519, 0.432 and 0.199 for the first three pairs.

At $\beta=5$ there are 17 lines above 0.01, and the tallest is not the carrier but the fourth pair, at 0.391. Drag on and the carrier vanishes a second time, near $\beta=5.52$. As the index grows, power leaves the carrier for more and more sidebands, and the comb widens.

The button “Hear it” plays one second of $y[n]$ at the current index. Every line sits on a whole multiple of 100 Hz, because $2000=20\cdot100$, so the sound repeats 100 times a second. Near $\beta=0$ you hear an almost pure 2000 Hz tone. As $\beta$ grows, more lines join in and the tone turns into a buzz.

Carson’s rule

A cosine of height 1 has power $\tfrac12$, and the lines share it: line $k$ carries $J_k(\beta)^2/2$. The arrow $e^{j\beta\sin\theta}$ always has length 1. So by Parseval’s relation, from “Where the power goes” in Fourier series coefficients (7.2),

and $J_k(\beta)^2$ is line $k$‘s share of the power.

Carson’s rule says that a band centred on the carrier and $2(\beta+1)f_\text{mod}$ wide holds most of the power. It reaches $\beta+1$ line spacings out on each side. The readout “power inside” adds the shares of the lines whose $\lvert k\rvert$ is at most $\beta+1$.

At $\beta=0.5$ the band is 300 Hz wide and holds 99.8 % of the power. At 2.405 it is 681 Hz and holds 99.1 %, and at 5 it is 1200 Hz and holds 99.4 %. For every index from 0 to 8 it holds at least 95.9 %, and 99.1 % or more when the index is a whole number.

The low values come just below a whole number. At $\beta=5.95$ the band’s edge sits 6.95 spacings out, just short of line 7, which carries 3.1 % of the power. The band holds only 96.2 %. At $\beta=6$ the edge reaches line 7, and the share jumps back to 99.3 %.

The message is the phase’s slope

Now for a real message. Mine is two slow tones, $0.6\sin(2\pi\cdot8t)+0.4\sin(2\pi\cdot20t+1)$, divided by 0.9982, its largest value, so that its peak is 1. Its lowest value is then $-0.931$. The carrier is at 1000 Hz, the deviation is $f_\text{dev}=300$ Hz, and I take 0.25 s at $f_s=8$ kHz, which is 2000 samples.

At each sample, FM turns the phase by the carrier’s step plus $2\pi f_\text{dev}x[n]/f_s$. The message’s part adds up, sample after sample:

To get the message back, I need the speed of the turning. In “A spinning arrow’s length and speed” of The Hilbert transform and the analytic signal (27.2), the analytic signal was one arrow. The step of its angle from $n-1$ to $n$, divided by $2\pi$, was the instantaneous frequency. For the FM signal $y$, with analytic signal $y_\text{an}$, that is

cycles per sample. Times $f_s$ that is a frequency in hertz. Subtract the carrier’s 1000 Hz and divide by 300 Hz, and what is left is the message that set that step, $x[n-1]$. This is a discriminator: it turns a change of frequency into a change of value.

Watch the middle panel: the phase drifts far from the message, but its slope, in the bottom panel, lands on the sent message.

The frequency swings between 721 Hz and 1300 Hz, as $1000+300x[n]$ says it should, since the message runs from $-0.931$ to 1. Yet the phase does not look like the message at all. It rises to 54.01 rad at 0.065 s, falls back to 2.26, rises to 47.32 and falls to $-1.61$ at 0.245 s. Its slope is what tracks the message.

The recovered message matches the sent one to within $10^{-11}$: what is left is the computer’s rounding. The analytic signal comes from the FFT method of 27.2, which has trouble near the ends of a record. Here the record holds whole periods of the carrier and of both tones, so there are no ends to cause trouble.

PM, FM and the index

Phase modulation would put the message straight into the middle panel. FM puts its running sum there instead. So FM of a message is PM of its running sum. Reading the phase itself recovers PM, and reading its slope recovers FM.

The index of this record is large. The 8 Hz tone alone swings the frequency by $300\cdot0.6/0.9982=180$ Hz, so its index is $180/8=22.5$. The 20 Hz tone swings it by 120 Hz, so its index is 6.0.

Both are far above 1: this is wide-band FM, the kind broadcast radio uses. There the deviation is 75 kHz and the audio reaches 15 kHz, so a 15 kHz tone has $\beta=5$ and Carson’s band is $2(5+1)\cdot15=180$ kHz wide.

The FM signal’s height never changes: its envelope $\lvert y_\text{an}[n]\rvert$ is 1 at every sample. The discriminator uses only angles, and multiplying the received signal by a positive number leaves every angle as it was.

Scaled by 0.01, this record still gives back the message to within $10^{-11}$. In AM the level is the message, so a change of level is a change of message. That is why FM resists fading and changes of level.

Worked example

1. Carson’s rule. Take $\beta=5$ and $f_\text{mod}=100$ Hz. The band is $2(5+1)\cdot100=1200$ Hz wide and reaches out to lines $\lvert k\rvert\le6$. The heights $\lvert J_0\rvert$ to $\lvert J_6\rvert$ are 0.1776, 0.3276, 0.0466, 0.3648, 0.3912, 0.2611 and 0.1310, and their squares 0.0315, 0.1073, 0.0022, 0.1331, 0.1531, 0.0682 and 0.0172. The carrier counts once and each other line twice, for lines $k$ and $-k$: $0.0315+2\cdot0.4810=0.9936$. The band holds 99.4 % of the power.
2. The index of the demodulation clip. The deviation is 300 Hz for the whole message, which peaks at 1. Its 8 Hz part has height $0.6/0.9982=0.601$, so it swings the frequency by 180 Hz, and $\beta=180/8=22.5$. The 20 Hz part swings it by 120 Hz, and $\beta=120/20=6.0$.

Where you’ll meet this

FM broadcast radio and two-way radios carry speech and music as FM. In 1973 John Chowning showed that a modulator at an audio frequency makes rich timbres from just two oscillators. With $f_1$ and $f_\text{mod}$ in a whole-number ratio, as in the first instrument, the lines fall on the harmonics of one pitch. Changing $\beta$ during a note brightens and darkens it, and this FM synthesis has been the sound of synthesisers since the 1980s.

Modems and Bluetooth send bits by frequency-shift keying: one frequency for a 0 and another for a 1, which is FM with a message of steps. Sending symbols as points in the complex plane comes next, in Digital modulation and OFDM (27.5).

Reference card