3.3: Beating between Tones
Perhaps you have heard two slightly mistuned musical instruments play pure tones whose frequencies are close but not equal. If so, you have sensed a beating phenomenon wherein a pure tone seems to wax and wane. This waxing and waning tone is, in fact, a tone whose frequency is the average of the two mismatched frequencies, amplitude modulated by a tone whose “beat” frequency is half the difference between the two mismatched frequencies. The effect is illustrated in the Figure. Let's see if we can derive a mathematical model for the beating of tones.
We begin with two pure tones whose frequencies are \(ω_0+ν\) and \(ω_0−ν\) (for example, \(ω_0=2π×10^3\mathrm{rad}/\mathrm{sec}\) and \(ν=2π \mathrm{rad}/\mathrm{sec}\)). The average frequency is \(ω_0\), and the difference frequency is \(2ν\). What you hear is the sum of the two tones:
\[x(t)=A_1\cos[(ω_0+ν)t+φ_1]+A_2\cos[(ω_0−ν)t+φ_2] \nonumber \]
The first tone has amplitude \(A_1\) and phase \(φ_1\); the second has amplitude \(A_2\) and phase \(φ_2\). We will assume that the two amplitudes are equal to \(A\). Furthermore, whatever the phases, we may write them as
\[φ_1=φ+ψ;\mathrm{and};φ_2=φ−ψ \nonumber \]
\[φ=\frac 1 2 (φ_1+φ_2);\mathrm{and}ψ=\frac 1 2 (φ_1−φ_2) \nonumber \]
Recall our trick for representing \(x(t)\) as a complex phasor:
\[x(t)=A\mathrm{Re}\{e{j[(ω_0+ν)t+φ+ψ]},+,e^{j[(ω_0−ν)t+φ−ψ]}\} \nonumber \]
\[=A\mathrm{Re}\{e^{j(ω_0t+φ)},[e^{j(νt+ψ)}+e^{−j(νt+ψ)]}\} \nonumber \]
\[=2A\mathrm{Re}\{e^{j(ω_0t+φ)},\cos(νt+ψ)\} \nonumber \]
\[=2A\cos(ω_0t+φ)\cos(νt+ψ) \nonumber \]
This is an amplitude modulated wave, wherein a low frequency signal with beat frequency \(ν\) rad/sec modulates a high frequency signal with carrier frequency \(ω_0\) rad/sec. Over short periods of time, the modulating term \(\cos{νt+ψ}\) remains essentially constant while the carrier term \(\cos{ω_0t+φ}\) turns out many cycles of its tone. For example, if \(t\) runs from 0 to \(\frac {2π} {10ν}\) (about 0.1 seconds in our example), then the modulating wave turns out just 1/10 cycle while the carrier turns out \(10νω_Δ\) cycles (about 100 in our example). Every time \(νt\) changes by \(2π\) radians, then the modulating term goes from a maximum (a wax) through a minimum (a wane) and back to a maximum. This cycle takes
\[νt=2π⇔t=\frac {2π} {ν} \mathrm{seconds} \nonumber \]
which is 1 second in our example. In this 1 second the carrier turns out 1000 cycles.
Find out the frequency of A above middle C on a piano. Assume two pianos are mistuned by \(±1\mathrm{Hz}(±2π/mathrm{rad/sec})\). Find their beat frequency \(ν\) and their carrier frequency \(ω_0\).
(MATLAB) Write a MATLAB program to compute and plot
\(A\cos[(ω_0+ν)t+φ_1],A\cos[(ω_0−ν)t+φ_2]\), and their sum. Then compute and plot \(2A\cos(ω_0t+φ)\cos(νt+ψ)\).
Verify that the sum equals this latter signal.