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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 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 ω0νω0ν
(for example, ω0=2π×103rad/secω0=2π×103rad/sec and ν=2πrad/secν=2πrad/sec). The average
frequency is ω_{0}, 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[(ω0ν)t+φ2].x(t)=A1 cos[(ω0+ν)t+φ1]+A2cos[(ω0ν)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=2π⇔t=2πν
seconds
,νt=2π⇔t=2πνseconds,
(4)
which is 1 second in our example. In this 1 second the carrier turns out 1000
cycles.
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
.
(MATLAB) Write a MATLAB program to compute and plot
A
cos
[(ω0+ν)t+φ1],AAcos[(ω0+ν)t+φ1],A cos [(ω0ν)t+φ2][(ω0ν)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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