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# Phasors: Beating between Tones

Module by: Louis Scharf. E-mail the author

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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 Figure 3.7. Let's see if we can derive a mathematical model for the beating of tones.

We begin with two pure tones whose frequencies are ω0+νω0+ν and ω0-νω0-ν (for example, ω0=2π×103rad/secω0=2π×103rad/sec and ν=2πrad/secν=2πrad/sec). The average frequency is ω0, and the difference frequency is 2ν2ν. What you hear is the sum of the two tones:

x(t)=A1 cos[(ω0+ν)t+φ1]+A2cos[(ω0-ν)t+φ2].x(t)=A1 cos[(ω0+ν)t+φ1]+A2cos[(ω0-ν)t+φ2].
(1)

The first tone has amplitude A 1 A 1 and phase φ 1 φ 1 ; the second has amplitude A 2 A 2 and phase φ 2 φ 2 . We will assume that the two amplitudes are equal to A A. Furthermore, whatever the phases, we may write them as

φ 1 = φ + ψ and φ 2 = φ - ψ φ = 1 2 ( φ 1 + φ 2 ) and ψ = 1 2 ( φ 1 - φ 2 ) . φ 1 = φ + ψ and φ 2 = φ - ψ φ = 1 2 ( φ 1 + φ 2 ) and ψ = 1 2 ( φ 1 - φ 2 ) .
(2)

Recall our trick for representing x(t)x(t) as a complex phasor:

x ( t ) = A Re { e j [ ( ω 0 + ν ) t + φ + ψ ] + e j [ ( ω 0 - ν ) t + φ - ψ ] } = A Re { e j ( ω 0 t + φ ) [ e j ( ν t + ψ ) + e - j ( ν t + ψ ) ] } = 2 A Re { e j ( ω 0 t + φ ) cos ( ν t + ψ ) } = 2 A cos ( ω 0 t + φ ) cos ( ν t + ψ ) . x ( t ) = A Re { e j [ ( ω 0 + ν ) t + φ + ψ ] + e j [ ( ω 0 - ν ) t + φ - ψ ] } = A Re { e j ( ω 0 t + φ ) [ e j ( ν t + ψ ) + e - j ( ν t + ψ ) ] } = 2 A Re { e j ( ω 0 t + φ ) cos ( ν t + ψ ) } = 2 A cos ( ω 0 t + φ ) cos ( ν t + ψ ) .
(3)

This is an amplitude modulated wave, wherein a low frequency signal with beat frequency νν rad/sec modulates a high frequency signal with carrier frequency ω0ω0 rad/sec. Over short periods of time, the modulating term cos(νt+cos(νt+ψ)ψ) remains essentially constant while the carrier term cos(ω0t+φ)cos(ω0t+φ) turns out many cycles of its tone. For example, if t t runs from 0 to 2π10ν2π10ν (about 0.1 seconds in our example), then the modulating wave turns out just 1/10 cycle while the carrier turns out 1¯0ν¯ωΔ1¯0ν¯ωΔ cycles (about 100 in our example). Every time νtνt changes by 2π2π 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=2πt=2πν seconds ,νt=2πt=2πνseconds,
(4)

which is 1 second in our example. In this 1 second the carrier turns out 1000 cycles.

## Exercise 1

Find out the frequency of A A above middle C C on a piano. Assume two pianos are mistuned by ±1Hz(±2π rad/sec )±1Hz(±2π rad/sec ). Find their beat frequency νν and their carrier frequency ω 0 ω 0 .

## Exercise 2

(MATLAB) Write a MATLAB program to compute and plot A cos [(ω0+ν)t+φ1],AAcos[(ω0+ν)t+φ1],A cos [(ω0-ν)t+φ2][(ω0-ν)t+φ2], and their sum. Then compute and plot 2Acos(ω0t+φ)cos(νt+ψ)2Acos(ω0t+φ)cos(νt+ψ). Verify that the sum equals this latter signal.

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