Beats as amplitude modulation

Two sources of equal amplitude give y = A cos(2πf₁t) + A cos(2πf₂t). The sum-to-product identity cos X + cos Y = 2 cos((X − Y)/2) cos((X + Y)/2) rewrites it as y = 2A cos(π Δf t) cos(2πf₀t), with f₀ = (f₁ + f₂)/2 and Δf = f₂ − f₁, the higher tone minus the lower; cosine is even, so the slow factor does not care which way round the difference is taken. Take f₁ = 440.0 Hz and f₂ = 444.0 Hz: then f₀ = 442.0 Hz and Δf = 4.0 Hz, so y = 2A cos(2π × 2.0 t) cos(2π × 442.0 t). Read that as an oscillation at 442.0 Hz whose amplitude is the slowly changing number 2A cos(2π × 2.0 t). Nothing was approximated; the identity holds for any pair of frequencies. What makes the reading useful is the ratio of scales: 110 carrier cycles fit inside one beat here, far more than an ear can resolve one at a time, so it registers the amplitude and not the cycles.

The slow factor cos(π Δf t) has frequency Δf/2, which is 2.0 Hz in the example above. The beat rate is 4.0 Hz. The factor of two is not a slip: loudness depends on the size of the amplitude and not on its sign, so the envelope is 2A|cos(π Δf t)|, and the vertical bars halve the period. Follow the zeros. The cosine vanishes when π Δf t = π/2, 3π/2, 5π/2, …, that is at t = 1/(2Δf), 3/(2Δf), 5/(2Δf), …, spaced 1/Δf apart. With Δf = 4.0 Hz the silences fall at 0.125 s, 0.375 s, 0.625 s and 0.875 s: four of them per second. Between consecutive silences the tone swells to 2A and dies again. The carrier flips sign at each silence, and no detector of loudness can tell.

Decompose the sum into frequencies and you find exactly two, f₁ and f₂. The Fourier content of A cos(2πf₁t) + A cos(2πf₂t) has nothing at Δf, and no linear device — an ideal amplifier, a linear medium, a spectrum analyser — can produce one, because a linear system returns only the frequencies it is given. Squaring breaks that rule. Square the equal-amplitude pair at 500.0 Hz and 506.0 Hz: the cross term 2A² cos(2π × 500.0 t) cos(2π × 506.0 t) becomes A²[cos(2π × 6.0 t) + cos(2π × 1006.0 t)], while the two squared terms contribute a constant plus lines at 1000.0 Hz and 1012.0 Hz. The 6.0 Hz line is new. Any energy-sensitive detector — the ear, a microphone feeding a power meter, a photodiode — performs that squaring, which is why the beat is audible at all, and why a heterodyne receiver needs a nonlinear mixer rather than a better amplifier.

Silence at a beat minimum needs exact cancellation, so it needs equal amplitudes. In general the resultant amplitude is Aₑₙᵥ = √(A₁² + A₂² + 2A₁A₂ cos 2πΔf t), which swings between A₁ + A₂ and |A₁ − A₂|. Take A₁ = 1.00 and A₂ = 0.80 in the same unit: the envelope runs from 1.80 down to 0.20 and never reaches zero. Intensity goes as amplitude squared, so the loud-to-quiet intensity ratio is (1.80/0.20)² = 81, a level swing of 10 log₁₀ 81 = 19.1 dB — dramatic, but finite. Averaging over a few carrier cycles gives an intensity proportional to A₁² + A₂² + 2A₁A₂ cos 2πΔf t, so the fractional swing about its mean is m = 2A₁A₂/(A₁² + A₂²) = 1.60/1.64 = 0.976. Equal amplitudes give m = 1 and true silence; a badly mismatched pair gives a beat you have to strain to notice.

Beats give the sharpest frequency comparison available to an unaided ear. Judging pitch directly is good to roughly half a per cent; counting beats against a reference is limited only by how long you will listen, because a mismatch Δf needs 1/Δf to show one swell. Hold the note 20 s with no swell and the mismatch is bounded at 0.050 Hz, which is 1.1 × 10⁻⁴ of 440 Hz. The catch is that a beat rate carries |Δf| and not its sign: 3.0 beats per second against a 440.0 Hz fork means 437.0 Hz or 443.0 Hz. Resolve it with a deliberate perturbation. Tighten the string slightly; since f ∝ √F, a fractional tension change gives Δf/f = ΔF/2F, so 3.0 Hz at 440 Hz is 1.4% of tension. If the beat rate rises, the string was already sharp and you must go back. Nulling from 443.0 Hz needs the tension cut by 1 − (440.0/443.0)² = 1.35%.

The product form is exact algebra, but carrier and envelope are physical names that need Δf ≪ f₀. Count carrier cycles per beat, N = f₀/Δf. At 500.0 Hz and 506.0 Hz, N = 83.8, an unmistakable slow throb. At 500.0 Hz and 300.0 Hz the same algebra returns f₀ = 400.0 Hz and Δf = 200.0 Hz, so N = 2: the envelope completes half a cycle while the carrier completes one, and nobody hears a single throbbing tone. Hearing crosses over in stages — below about 15 Hz of separation the pair fuses into one tone whose loudness fluctuates, above that it turns rough, and once Δf exceeds the critical bandwidth, of order 10 to 20 per cent of the centre frequency, the two tones are heard apart. A related limit binds any instrument: resolving two frequencies Δf apart takes an observation of at least about 1/Δf, so a 0.10 Hz mismatch cannot be found inside a 2 s note.