Amplitude modulation

A slow message can ride on a fast wave, whose peaks trace the message out. Watch the dashed outline as the message is pressed in harder: at first it follows the message, then it folds up off zero.

Lift the message, and the envelope carries it

Say I want to send a slow message, such as a voice, over a channel that only passes high frequencies. Radio is the classic case. An antenna sends a signal well only when it is not much smaller than the signal’s wavelength, and a voice’s wavelengths are tens to thousands of kilometres long.

The trick is to multiply the message by a fast cosine, the carrier. In “Multiply by a cosine and the spectrum splits in two” of Properties of the Fourier transform (8.2), that product moved the spectrum. It made two copies at half height, one centred on each of $\pm\omega_1$.

In discrete time, “Slide the spectrum round the circle” of Properties of the DTFT (12.3) gave the same shift for one arrow $e^{j\Omega_1n}$. A cosine is two half arrows, so:

Here $x[n]$ is the message. I keep it slow, and I scale it so its largest value is 1. The carrier’s frequency is $\Omega_1$ in rad/sample, or $f_1=\Omega_1f_s/2\pi$ in hertz. On the circle of 12.3 the copies wrap round, so I keep $\Omega_1$ plus the message’s highest frequency below $\pi$.

Look at the copy centred on $+\Omega_1$. Its part above $\Omega_1$ is the message’s positive frequencies, the upper sideband. Its part below is their mirror, the lower sideband.

So this product has two sidebands and nothing at the carrier’s own frequency. It is called DSB, for double sideband.

AM adds the carrier itself. I lift the message by 1 before I multiply:

The number $\mu$ is the modulation index: how deeply the message moves the carrier’s size. It is local to this page. It is not the position between samples of Resampling by any factor (22.3), nor the step size of Adaptive filters: LMS (26.3).

Multiply it out and $y[n]=\cos\Omega_1n+\mu\,x[n]\cos\Omega_1n$. That is the carrier plus a DSB signal scaled by $\mu$. So AM’s spectrum is DSB’s two sidebands with a line at the carrier between them.

In “Multiplying by an envelope” of Operations on amplitude (2.2), a slow signal times a fast wave gave a product whose outline followed the slow one. Here the slow factor is $1+\mu x[n]$. Where it is positive, the carrier’s peaks lie on it. That outline is the envelope.

The picture at the top of the page draws all three: $y[n]$, the dotted $1+\mu x$, and the dashed envelope over it. Its message is two tones, $0.7\sin(2\pi10t)+0.3\sin(2\pi25t+0.5)$, divided by its largest value, 0.9967, so that its peak is 1. Its pattern repeats every 0.2 s, the width of the panel.

Watch the dashed envelope as $\mu$ goes from 0.5 to 1.5. At $\mu=1$, $1+\mu x$ dips to 0.064 but stays positive, and the envelope still follows it. At 1.5 it is below zero at 240 of the 1600 samples, and there the envelope folds up.

Where the envelope folds

Why does the picture break at 1.5? Where $1+\mu x[n]$ goes below zero, $y[n]$ is a carrier of size $\lvert1+\mu x[n]\rvert$ turned upside down. Its peaks then trace $\lvert1+\mu x[n]\rvert$, which folds up off the zero line instead of following the message down. This is overmodulation.

So AM keeps the message in its envelope while $1+\mu x[n]$ never goes negative. That needs $1+\mu\min x\ge0$, or $\mu\le1/\lvert\min x\rvert$. It is tempting to say ”$\mu$ up to 1”. But this message dips only to −0.936, so it overmodulates only from $\mu=1/0.936=1.068$.

Try it on the slider. At 1.06 no sample is folded. At 1.07, the first step past 1.068, ten are.

To read the envelope, a receiver needs no carrier of its own. An envelope detector is a rectifier and a smoother. It takes the size $\lvert y[n]\rvert$ of each sample, then a low-pass filter smooths away the carrier’s ripple. What is left is $\lvert1+\mu x[n]\rvert$, scaled by the average of $\lvert\cos\rvert$ over a cycle, $2/\pi$.

DSB is the case without the added 1. Its envelope is $\lvert x[n]\rvert$, which folds at every zero of the message. This message crosses zero four times in its 0.2 s, so an envelope detector would hand back $\lvert x\rvert$, not $x$. DSB needs another way back, coherent demodulation, which the next instrument shows.

Where the power goes

The added carrier costs power. Average power is the mean of the square, as in “Average per sample” of How big is a signal (1.3). A cosine of size 1 has power $\tfrac12$. The message is slow and averages to zero, so the carrier and the sidebands add their powers:

Here $P_x$ is the message’s power, 0.292 for this one. The carrier’s $\tfrac12$ carries no message at all. At $\mu=0.5$, only 6.8 % of the AM signal’s power is in the sidebands, and at $\mu=1$ it is 22.6 %.

So AM spends most of its power on a line that tells the receiver nothing. What it buys is a receiver as simple as a rectifier and a smoother.

Mix it down, filter it out

Coherent demodulation

The other way back multiplies by the carrier again. Take $2y[n]\cos\Omega_1n$. Since $2\cos^2\theta=1+\cos2\theta$:

The first part is the lifted message, back at low frequencies. The second is a copy around $2\Omega_1$. A low-pass filter keeps the first part and removes the second.

Then I remove the carrier’s level, the 1, and divide by $\mu$. This is coherent demodulation. For DSB the same steps give $x[n]$ directly, with no level to remove, so it works for DSB as well.

The multiplier $2\cos\Omega_1n$ is made in the receiver, so it is called the local oscillator. Multiplying by it is mixing. In hertz I write its frequency $f_\text{LO}$, and the multiplier is $2\cos(2\pi f_\text{LO}n/f_s)$.

One filter, many stations

Mixing does more than demodulate. Many stations share the air, each at its own carrier. Mixing at $f_\text{LO}$ moves a component at frequency $f$ to $\lvert f-f_\text{LO}\rvert$ and to $f+f_\text{LO}$, which folds back if it passes $f_s/2$. Set $f_\text{LO}$ on the wanted station, and that station lands at 0 Hz.

Every other station lands somewhere else. So one fixed low-pass filter at 0 Hz picks out whichever station the oscillator is set to. I tune by moving the oscillator, not the filter. This is the idea of the superheterodyne receiver.

The instrument’s filter is the window method of “Cut the ideal, and the edge ripples” of Window-method FIR design (19.1). It is an ideal low-pass at 800 Hz, cut to 129 taps and shaped by a Hamming window, with its taps scaled to add up to 1. At 16 kHz, 800 Hz is $\Omega_c=0.1\pi$. Its delay is $(129-1)/2=64$ samples.

Below 500 Hz its gain stays within 0.013 dB of 1. From 1.5 kHz up it is at most −56.1 dB. At 800 Hz itself the gain is one half, as for every window-method cutoff.

The three stations each send a short tune. Station A sends a 330 Hz tone. Station B alternates notes of 262 and 392 Hz every quarter second, close to middle C and the G above it. Station C sends a 440 Hz tone that pulses four times a second.

Each station is AM with $\mu=0.8$, at 2, 4 and 6 kHz. Their messages reach 444 Hz, so after mixing the nearest unwanted sideband sits at 1556 Hz, inside the stop band.

Watch station B slide down to 0 Hz as the oscillator mixes, and the filter keep only what lands below 800 Hz: B’s tune comes back with an SNR of 28.4 dB. Then choose another station and press “Hear it”. The same filter now gives a different tune, A at 28.5 dB and C at 23.9 dB.

The recovered SNR is the sent message’s power over the power of the error, both over the middle 0.75 s. It is in dB as in “Signal and noise, in decibels” of How big is a signal (1.3). Station C scores lower because its pulsing tune has less power, 0.19 against 0.50 for A, while the noise is the same.

Real radios, and the oscillator’s phase

A broadcast radio does not mix to 0 Hz. It mixes every station to a fixed intermediate frequency, 455 kHz in AM broadcast receivers, and filters there. Its sharp filters never move; only the oscillator does. That is the superheterodyne.

Mixing straight to 0 Hz, as the instrument does, needs the oscillator’s phase as well as its frequency. Suppose the oscillator is $2\cos(\Omega_1n+\theta)$. The low-pass part becomes $\big(1+\mu x[n]\big)\cos\theta$, so a phase error $\theta$ scales the output by $\cos\theta$.

At $\theta$ = 90° the message vanishes. That is why plain AM radios use the envelope detector after their mixer: it needs no phase at all.

Worked example

1. The overmodulation limit. The message’s lowest value is −0.936. For $1+\mu x[n]\ge0$ at that sample I need $1-0.936\mu\ge0$, so $\mu\le1/0.936=1.068$. At $\mu=1$ the lowest point is $1-0.936=0.064$: still positive, as the clip showed.

2. The spectrum at μ = 0.5. The message’s two tones have sizes $0.7/0.9967=0.702$ and $0.3/0.9967=0.301$. In $\mu x[n]\cos\Omega_1n$ each tone becomes two lines at half its size times $\mu$.

So the AM signal has lines at 225, 240, 250, 260 and 275 Hz. The carrier’s line at 250 Hz has size 1. The lines at 240 and 260 Hz have size $0.5\times0.702/2=0.176$, and those at 225 and 275 Hz have size $0.5\times0.301/2=0.075$.

3. Where the stations go. Mixing at 4 kHz sends station A, at 2 kHz, to $\lvert2-4\rvert=2$ kHz and to $2+4=6$ kHz. It sends station C, at 6 kHz, to 2 kHz and to 10 kHz. At 16 kHz sampling, 10 kHz folds to $16-10=6$ kHz.

So A and C land on top of each other, at 2 and 6 kHz, and the filter removes both. Station B goes to 0 Hz and to 8 kHz, half the sampling rate.

4. An oscillator out of phase. With $\theta$ = 30° the output is scaled by $\cos\theta=0.866$, and with $\theta$ = 60° by 0.5. Tuned to B, the carrier’s level after the filter reads 0.865 and 0.499, against 1.000 in phase.

Where you’ll meet this

AM broadcasting on long, medium and short wave sends this signal, carrier and all, so that cheap receivers can use an envelope detector. Aircraft talk to the tower in AM on VHF.

Mixing is in nearly every receiver, for radio, television and phones. Many phones and software radios mix straight to 0 Hz, as the instrument does, and solve the phase problem with two mixers; that is I/Q and complex baseband (27.3).

DSB wastes bandwidth: both sidebands carry the same message. SSB keeps one sideband only, and amateur and marine radio use it. Building it takes the Hilbert transform of The Hilbert transform and the analytic signal (27.2).

Reference card