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Phasors: Beating between Tones

Module by: Louis Scharf. E-mail the author

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This module is part of the collection, A First Course in Electrical and Computer Engineering. The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.

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 1. 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)
Figure 1: Beating between Tones
Figure one is a set of three sinusoidal functions graphed along a horizontal axis. Two are identical in wavelength and amplitude, but inverted so that their peaks and troughs look like they are enclosing the third function, and they meet four times at the horizontal axis when they progress in the opposite directions. The third sinusoidal function is bound by these two larger first functions, changing in amplitude based on the boundaries of those functions, but retaining a short, constant wavelength throughout the graph.

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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