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 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 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.