Chapter 14
The Acoustical
Phenomena Governing
the Musical
Relationships of Pitch
Use of Beats for Tuning

Produce instrument tone and standard
Tuning fork or concert master
 Download NCH Tone Generator from Study Tools
page and try it





Open two instances of Tone Generator
Set one for 440 Hz and the other for 442 Hz
Adjust instrument until beat frequency is zero
Here we examine other ways of producing
and using beats
Beat Experiment
Mask one ear
of a subject so
nothing can be
heard.
In the other ear introduce a
strong, single frequency (say,
400 Hz) source and a much
weaker, adjustable frequency
sound (the search tone).
Vary the search tone from 400 Hz up.
We hear beats at multiples of 400 Hz.
Alteration of the Experiment

Produce search tones of equal amplitude but
180° out of phase.
Search tone now completely cancels single tone.
 Result is silence at that harmonic
 Each harmonic is silenced in the same way.
 How loud does each harmonic need to be to get
silence of all harmonics?

Waves Out of Phase
Waves Out of Phase
1
0.8
Displacement
0.6
Superposition
of these waves
produces zero.
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
Time
Loudness of the Beat Harmonics




400 Hz
800 Hz
1200 Hz
1600 Hz
95 SPL
75 SPL
75 SPL
75 SPL
Source Frequency
Note: harmonics are 20 dB or 100 times fainter
than source (10% as loud)
Start with a Fainter Source




400 Hz
800 Hz
1200 Hz
1600 Hz
89 SPL
63 SPL
57 SPL
51 SPL
Source – ½ loudness
¼ as loud as above
1/8 as loud as above
1/16 as loud as above
…And Still Fainter Source






400 Hz
800 Hz
1200 Hz
1600 Hz
75 SPL
55 SPL
35 SPL
15 SPL
Source
Too faint
Too faint
This example is appropriate to music.
Where do the extra tones come from?

They are not real but are produced in the ear/brain
Heterodyne Components

Consider two tones (call them P and Q)


From above we see that the ear/brain will produce harmonics at
(2P), (3P), (4P), etc.
Other components will also appears as combinations of P and Q
Original
Components
Simplest Heterodyne
Components
Next-Appearing
Heterodyne
Components
P
(2P)
(3P)
(P + Q), (P – Q)
(2P + Q), (2P – Q)
(2Q + P), (2Q – P)
(2Q)
(3Q)
Q
Heterodyne Component Example
Original
Components
Simplest Heterodyne
Components
Next-Appearing
Heterodyne
Components
400
(800)
(1200)
(1000), (200)
(1400), (800)
(1600), (800)
(1200)
(1800)
600
So the ear hears (200), 400, 600, (800), (1000), (1200), (1400), (1600), (1800).
Producing Beats


Beats can occur between closely space
heterodyne components, or between a main
frequency and a heterodyne component.
Ex. Consider three tones P at 200, Q at 396,
and R at 605 Hz.
Two of the many heterodyne components are
(Q – P) = 196 Hz and (R – Q) = 209 Hz.
 Also (Q – P) will beat with P at 4 Hz.

Mechanical Analogy to
Heterodyne Components
For small oscillations of the tip, we have
simple harmonic motion. The bar never loses
contact at A or comes into contact at B. The
graph of the motion of the tip is a pure sine
wave. Make the natural frequency 20 Hz.
Higher Amplitudes

Bar loses contact at A on upward swing



Bar touches clamp at B



Bar is momentarily longer and less stiff
Amplitude is greater than the pure sine wave.
Bar is momentarily shorter and more stiff
Amplitude is less than the pure sine wave.
The red curve on the next slide describes the
situation

But the red curve is the superposition of the two sine waves
shown.
Graph High Amplitude Motion of Tip
Fundamental pure
sine wave
1st Harmonic pure
sine wave
Resultant
Driven System

Now add the spring and drive the system at a
variety of frequencies.
We expect large amplitudes when the driver
frequency matches the natural frequency of 20
Hz.
 We also get increases in amplitude at ⅓ and ½
the natural frequency (6⅔ Hz and 10 Hz)
 See the response graph on the next slide

Driven System Response
Natural
Frequency, fo
3rd Harmonic
is fo
2nd Harmonic
is fo
Response Curve Explained



When the driver frequency becomes 6⅔ Hz,
the heterodyne component (third harmonic)
is also excited. 3 X 6⅔ Hz = 20 Hz, the
natural frequency.
When the driver frequency is 10 Hz, the
second harmonic (2 X 10 Hz = 20 Hz) is also
stimulated as a heterodyne component.
The 20 Hz frequency is self-generated
More than One Driving Source

We should expect high amplitude whenever
a heterodyne component is close to 20 Hz.
EX: Suppose two frequencies are used at
P = 9 Hz and Q = 30 Hz.
 We get a heterodyne component at (Q-P) = 21
Hz, which is close to the natural 20 Hz frequency.

Non-Linear Response

At small amplitude the system acts like a
Hooke’s Law spring (deflection [x]  load [F])



A graph of F vs. x will give a straight line (linear)
At higher amplitude the F vs. x curve
becomes curved (non-linear)
See graphs below.
Load
Load vs. Deflection
Black is linear (Hooke’s Law)
Colored is non-linear
Deflection
Notes on Non-linear Systems



In a non-linear system, the whole response
is not simply the sum of its parts.
Non-linear systems subject to sinusoidal
driving forces generate heterodyne
components, no matter what the nature of
the non-linearity.
The amplitudes of the heterodyne
components depend on the nature of the
non-linearity and the amplitude of the driver.
The Musical Tone


Special Properties of Sounds Having
Harmonic Components
Imagine a single sinusoidal frequency
produced from a speaker
At low volume the single tone is all you hear.
 At higher volumes the room and our hearing
system may produce harmonics.

Change the Source

Now have the source composed of the same
frequency, a weak second harmonic, and a
still weaker third harmonic.

The added harmonics will probably not be
noticed, but the listener may say the tone is
louder.

Reason is that the additional harmonics is exactly what
happens with the single tone at higher volume.
Almost Harmonic Components


Suppose the tones introduced are at 250 Hz
(X), a second partial at 502 Hz (Y), and a
third at 747 Hz (Z).
Heterodyne components include:
(Y-X)
 (Z-Y)
 (Z-X)
 (X+Y)
 2X

(252)
(245)
(497)
(752)
(500)
I have color-coded frequencies
which form “clumps.” These are
heard as musical tones, but may
be called “unclear.”
Frequency - Pitch



Frequency is a physical quantity
Pitch is a perceived quantity
Pitch may be affected by whether…



the tone is a single sinusoid or a group of
partials
heterodyne components are present, or
noise is a contributor
Frequency Assignments

The Equal-Tempered Scale



Each octave is divided into 12 equal parts
(semitones)
Since each octave is a doubling of the frequency,
each semitone increases frequency by 12 2
Ex. G4 has a frequency of 392 Hz
G4# has a frequency of 415.3 Hz
Cents


Each semitone is further divided into 100 parts called
cents.
The difference between G4 and G4# above is 23.3
Hz and thus in this part of the scale each cent is
0.233 Hz.
A tone of 400 Hz can be called
[G4 + (400-392)/0.233] cents, or (G4 + 34 cents).
 500 Hz falls between B4 (493.88 Hz) and C5 (525.25 Hz).
We could label 500 Hz as (B4 + 20 cents)

Calculating Cents


The fact that one octave is equal to 1200
cents leads one to the power of 2
relationship:
Or,
 f2 
ln  
f1 

cents  1200
ln(2)
Advantage of the Cents Notation
Bbo
29.135 Hz
Bb4
466.16 Hz
Bb7
3729.3 Hz
Ao
27.5
A4
440.0
A7
3520
Df
1.635
26.16
209.3
Interval
100 cents
100 cents
100 cents
The same interval in different octaves will be
difference frequency differences, but the interval in
cents is always the same.
Frequencies (Hz) for Equal-Tempered Scale
("Middle C" is C4 )
Octave
0
1
2
3
4
5
6
7
8
C
16.35
32.7
65.41
130.81
261.63
523.25
1046.5
2093
4186.01
C#/Db
17.32
34.65
69.3
138.59
277.18
554.37
1108.73
2217.46
4434.92
D
18.35
36.71
73.42
146.83
293.66
587.33
1174.66
2349.32
4698.64
D#/Eb
19.45
38.89
77.78
155.56
311.13
622.25
1244.51
2489.02
4978.03
E
20.6
41.2
82.41
164.81
329.63
659.26
1318.51
2637.02
F
21.83
43.65
87.31
174.61
349.23
698.46
1396.91
2793.83
F#/Gb
23.12
46.25
92.5
185
369.99
739.99
1479.98
2959.96
G
24.5
49
98
196
392
783.99
1567.98
3135.96
G#/Ab
25.96
51.91
103.83
207.65
415.3
830.61
1661.22
3322.44
A
27.5
55
110
220
440
880
1760
3520
A#/Bb
29.14
58.27
116.54
233.08
466.16
932.33
1864.66
3729.31
B
30.87
61.74
123.47
246.94
493.88
987.77
1975.53
3951.07
Note
Intervals (Hz) for the Equal-Tempered Scale
Octave
0
1
2
3
4
5
6
7
8
C#/Db – C
0.97
1.95
3.89
7.78
15.55
31.12
62.23
124.46
248.91
D - C#/Db
1.03
2.06
4.12
8.24
16.48
32.96
65.93
131.86
263.72
D#/Eb - D
1.1
2.18
4.36
8.73
17.47
34.92
69.85
139.7
279.39
E - D#/Eb
1.15
2.31
4.63
9.25
18.5
37.01
74
148
F-E
1.23
2.45
4.9
9.8
19.6
39.2
78.4
156.81
F#/Gb - F
1.29
2.6
5.19
10.39
20.76
41.53
83.07
166.13
G - F#/Gb
1.38
2.75
5.5
11
22.01
44
88
176
G#/Ab - G
1.46
2.91
5.83
11.65
23.3
46.62
93.24
186.48
A - G#/Ab
1.54
3.09
6.17
12.35
24.7
49.39
98.78
197.56
A#/Bb - A
1.64
3.27
6.54
13.08
26.16
52.33
104.66
209.31
B - A#/Bb
1.73
3.47
6.93
13.86
27.72
55.44
110.87
221.76
Note
Frequency Value of Cent
Through the Keyboard
3.0
2.5
2.0
Hz/cent
1.5
1.0
0.5
0.0
0
1000
2000
3000
Frequency
4000
5000
Frequency Matching vs.
Pitch Matching

Most cases these give the same result


Can use frequency standards to match pitch
May produce different results

Recall the difficulty of assigning pitch with bell
tones from Chapter 5.
Buzz Tone Made from Harmonic
Partials

Consider forming a
“buzz” sound by adding
25 partials of equal
amplitude and a
fundamental of 261.6
Hz (C4).
Compare the Buzz Tone to a Pure Sine
Wave of Same Frequency

Present the two alternately


Present the two together


Pitch match occurs if the sine wave is made sharp.
No frequency changes required
The physicist’s idea of matching frequency by
achieving a zero beat condition agrees with the
musician’s idea of matching pitch when the tones are
presented together, as long as the tones are
harmonic partials.
Practical Application

In music only the first few partials have
appreciable amplitude

Pitch matching for tones presented alternately
and together gives the same result.
Almost Unison Tones

Consider two tones constructed from
partials as below. Neglect heterodyne
effects for the time being.
Harmonic
1
2
3
4
Tone J
250
500
750
1000
Tone K
252
504
756
1008
2
4
6
8
Beat Frequency
Matching Pitch


As the second tone is adjusted to the first,
the beat frequency between the
fundamentals becomes so slow that it can
not easily be heard.
We now pay attention to the beats of the
higher harmonics.

Notice that a beat frequency of ¼ Hz in the
fundamental is a beat frequency of 1 Hz in the
fourth harmonic.
Now Add Heterodyne Components






(J2 – K1) = (500 – 252) = 248 Hz
(K2 – J1) = (504 – 250) = 254 Hz
(J3 – K1) = (750 – 252) = 498 Hz
(K3 – J1) = (756 – 250) = 506 Hz
Now we have frequencies near the fundamentals
and the second harmonic
Recall that heterodyne components arise from
differences between the harmonics of the two tones
Complete set of Heterodyne
Components
Tone J
250
500
750
1000
Tone K
252
504
756
1008
Subtractive Components
244
246
248
254
496
498
506
508
748
758
256
Additive Components
258
502
752
754
1002
1004
1006
Can you find the differences and sums that result in these frequencies?
Results

In the vicinity of the original partials, clumps
of beats are heard, which tends to muddy the
sound.
Eight frequencies near 250 Hz
 Seven near 500 Hz
 Six near 750 Hz
 Five near 1000 Hz.

Results (cont’d)



The multitude of beats produced by tones having
only a few partials makes a departure from equal
frequencies very noticeable.
The clumping of heterodyne beats near the harmonic
frequencies may make the beat unclear and confuse
the ear.
These two conclusions are contradictory and either
may happen depending on the relative amplitudes of
the partials.
Next - Separate the Tones More
Tone J
250
500
750
1000
Tone K
281
562
843
1124
Subtractive Components
157
188
219
312
438
469
593
624
719
874
343
Additive Components
374
531
781
812
1031
1062
1093
The spread of the clumps is quite large and the resulting sound is “nondescript.”
Approaching Unison – Pitch Matching
Tone J
250
500
750
1000
Tone K
250.5
501
751.5
1002
Subtractive Components
248.5
249
249.5
251
499
499.5
501.5
502
749.5
752
251.5
Additive Components
252
500.5
750.5
751
1000.5
1001
1001.5
Results

A collection of beats may be heard.
Here are the eight components near 250 Hz
sounded together.

 Achieving
unison is well-defined.
The Octave Relationship

We can make two tones separated by close to one
octave.


Tone P has a fundamental at 200 Hz and three harmonic
partials.
Tone Q has a fundamental at 401 Hz and three harmonic
partials
Tone P
Tone Q
200
400
401
600
800
802
1203
1604
Heterodyne Components
Subtractive Components
Additive Components
199
201
202
399
402
403
601
602
603
803
804
801
1003
1004
1001
1002
1204
1201
1202
1404
1402
1403
1602
1603
1803
1804
2003
2004
Frequencies above 1600 Hz are few in number and amplitude
2204
2404
Results


As the second tone is tuned to match the
first, we get harmonics of tone P, separated
by 200 Hz.
Only tone P is heard
The Musical Fifth



A musical fifth has two tones whose fundamentals
have the ratio 3:2.
Again consider an almost tuned fifth and look at the
heterodyne components produced.
Tone M
200
400
600
800
Tone N
301
602
903
1204
Now every third harmonic of M is close to a
harmonic of N
Heterodyne Components
Subtractive Components
99
101
198
202
299
303
402
404
499
503
Additive Components
103
501
604
703
701
804
802
901
1004
1002
1101
1202
1103
Results


We get clusters of frequencies separated by 100 Hz.
When the two are in tune, we will have the partials…
200



300
400
600
800
900
This is very close to a harmonic series of 100 Hz
The heterodyne components will fill in the missing
frequencies.
The ear will invariably hear a single 100 Hz tone
(called the implied tone).
1200
Curious Effects

If one of the tones (say tone N) is turned off
and then back on, we will hear two tones
even though the situation is the same as the
original.

Turning off tone N eliminates the frequencies at
300, 900, and 1200 and weakens the 600 Hz
tone. Turning N back on emphasizes those
partials again, making them distinct as a separate
tone.
No Special Relationship among the
Tones


Consider two tones and their partials
Tone V
200
400
600
800
Tone W
273
546
819
1092
Heterodyne components includes 19 [W3 – V4], 54
[V3 – W2], 73 [W1 – V1], 127 [V2 – W1], 146 [W2 – V2],
219 [W3 – V3], etc.

Three heterodyne components [73, 146, 219] are harmonics
of 73 Hz. Thus a 73 Hz tone (tone T) will be heard with the
tones V and W.
Other Harmonic Sequences

Another harmonic series produced at 473 Hz by the additive
heterodyne components



The series is 473 [V1 + W1], 946 [V2 + W2], 1419 [V3 + W3], and
1819 [V4 + W4]. Call this tone S.
Upward masking and the confusion of unrelated frequencies may
make this hard to hear.
Two heterodyne harmonic series are produced – one with a
fundamental at W1 – V1 and the other at W1 + V1. Tone T is
referred to as the difference tone. Tone S is called the
summation tone.

As tones V and W are moved toward a harmonic relationship, the
difference and summation tones realign to become the implied
tone.
Other Special Relationships
Ratio
Musical Interval
Cents
Numbers of
Frequencies in
clumps
1/1
Unison
000
1 group of five
1 group of six
1 group of seven
1 group of eight
2/1
Octave
1200
1 group of three
4 groups of four
3 groups of five
3/2
Fifth
702 (700)
3 groups of two
9 groups of three
Other Special Relationships (Cont’d)
4/3
Fourth
498 (500)
12 groups of two
1 group of three
5/3
Major sixth
884 (900)
14 groups of two
5/4
Major third
386 (400)
10 groups of two
6/5
Minor third
316 (300)
6 groups of two
7/4
969
6 groups of two
7/5
583
4 groups of two
814 (800)
3 groups of two
267
3 groups of two
8/5
7/6
Minor sixth