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

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time

Dis

pla

cem

ent Superposition

of these waves produces zero.

Loudness of the Beat Harmonics

400 Hz 95 SPL Source Frequency 800 Hz 75 SPL 1200 Hz 75 SPL 1600 Hz 75 SPL

Note: harmonics are 20 dB or 100 times fainter than source (10% as loud)

Start with a Fainter Source

400 Hz 89 SPL Source – ½ loudness 800 Hz 63 SPL ¼ as loud as above 1200 Hz 57 SPL 1/8 as loud as above 1600 Hz 51 SPL 1/16 as loud as above

…And Still Fainter Source

400 Hz 75 SPL Source 800 Hz 55 SPL 1200 Hz 35 SPL Too faint 1600 Hz 15 SPL 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

OriginalComponents

Simplest HeterodyneComponents

Next-AppearingHeterodyneComponents

P (2P) (3P)

(P + Q), (P – Q)(2P + Q), (2P – Q)(2Q + P), (2Q – P)

Q (2Q) (3Q)

Heterodyne Component Example

OriginalComponents

Simplest HeterodyneComponents

Next-AppearingHeterodyne

Components

400 (800) (1200)

(1000), (200)(1400), (800)(1600), (800)

600 (1200) (1800)

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 toHeterodyne 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 is momentarily longer and less stiff Amplitude is greater than the pure sine wave.

Bar touches clamp at B 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 puresine wave

1st Harmonic puresine 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

2nd Harmonic is fo

3rd Harmonic is fo

Natural Frequency, 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 vs. Deflection

Deflection

Lo

ad

Black is linear (Hooke’s Law)

Colored is non-linear

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) (252) (Z-Y) (245) (Z-X) (497) (X+Y) (752) 2X (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:

ln(2)

ff

ln

1200 cents 1

2

Or,

Advantage of the Cents Notation

Bbo 29.135 Hz Bb

4 466.16 Hz Bb7 3729.3 Hz

Ao 27.5 A4 440.0 A7 3520

f 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

  Note0 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

Intervals (Hz) for the Equal-Tempered Scale

Octave

Note0 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

Frequency Value of CentThrough the Keyboard

0.0

0.5

1.0

1.5

2.0

2.5

3.0

0 1000 2000 3000 4000 5000

Frequency

Hz/cent

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 Pitch match occurs if the sine wave is made sharp.

Present the two together 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

Beat Frequency 2 4 6 8

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   Additive Components  

244 246 248 254 256 258        

496 498 506 508 502  

748 758         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   Additive Components  

157 188 219 312 343 374      

438 469 593 624     531    

719 874         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   Additive Components  

248.5 249 249.5 251 251.5 252      

499 499.5 501.5 502     500.5    

749.5 752         750.5 751  

            1000.5 1001 1001.5

Results

A collection of beats may be heard.

Achieving unison is well-defined.

Here are the eight components near 250 Hz sounded together.

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 200 400 600 800

Tone Q 401 802 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

Frequencies above 1600 Hz are few in number and amplitude 1803 1804

2003 2004

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 Additive Components

99 101 103    

198 202    

299 303    

402 404    

499 503 501  

604      

703   701  

804   802  

    901  

1004   1002  

1101 1103

1202

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 1200

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).

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 CentsNumbers of

Frequencies inclumps

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

8/5 Minor sixth 814 (800) 3 groups of two

7/6 267 3 groups of two