2. Chapter 14 The Acoustical Phenomena Governing the Musical Relationships of Pitch

3. 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

4. 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.

5. 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?

6. Waves Out of Phase Superposition of these waves produces zero.

7. Loudness of the Beat Harmonics 400 Hz95 SPLSource Frequency 800 Hz75 SPL 1200 Hz75 SPL 1600 Hz75 SPL Note: harmonics are 20 dB or 100 times fainter than source (10% as loud)

8. Start with a Fainter Source 400 Hz89 SPLSource – ½ loudness 800 Hz63 SPL¼ as loud as above 1200 Hz57 SPL 1 / 8 as loud as above 1600 Hz51 SPL 1 / 16 as loud as above

9. …And Still Fainter Source 400 Hz75 SPLSource 800 Hz55 SPL 1200 Hz35 SPLToo faint 1600 Hz15 SPLToo 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

10. 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) Q(2Q)(3Q)

11. Heterodyne Component Example Original Components Simplest Heterodyne Components Next-Appearing Heterodyne 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).

12. 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.

13. 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.

14. 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.

15. Graph High Amplitude Motion of Tip

16. 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

17. Driven System Response 2 nd Harmonic is f o 3 rd Harmonic is f o Natural Frequency, f o

18. 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

19. 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.

20. 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.

21. Load vs. Deflection Black is linear (Hooke’s Law) Colored is non-linear

22. 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.

23. 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.

24. 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.

25. 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.”

26. 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

27. 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  Ex. G4 has a frequency of 392 Hz G4 # has a frequency of 415.3 Hz

28. Cents Each semitone is further divided into 100 parts called cents. The difference between G4 and G 4 # 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)

29. Calculating Cents The fact that one octave is equal to 1200 cents leads one to the power of 2 relationship: Or,

30. Advantage of the Cents Notation BboBbo 29.135 HzBb4Bb4 466.16 HzBb7Bb7 3729.3 Hz AoAo 27.5A4A4 440.0A7A7 3520 ff 1.63526.16209.3 Interval100 cents The same interval in different octaves will be difference frequency differences, but the interval in cents is always the same.

31. Frequencies (Hz) for Equal-Tempered Scale ("Middle C" is C4 ) Octave Note 012345678 C 16.3532.765.41130.81261.63523.251046.520934186.01 C # /D b 17.3234.6569.3138.59277.18554.371108.732217.464434.92 D 18.3536.7173.42146.83293.66587.331174.662349.324698.64 D # /E b 19.4538.8977.78155.56311.13622.251244.512489.024978.03 E 20.641.282.41164.81329.63659.261318.512637.02 F 21.8343.6587.31174.61349.23698.461396.912793.83 F # /G b 23.1246.2592.5185369.99739.991479.982959.96 G 24.54998196392783.991567.983135.96 G # /A b 25.9651.91103.83207.65415.3830.611661.223322.44 A 27.55511022044088017603520 A # /B b 29.1458.27116.54233.08466.16932.331864.663729.31 B 30.8761.74123.47246.94493.88987.771975.533951.07

32. Intervals (Hz) for the Equal-Tempered Scale Octave Note 012345678 C#/Db – C0.971.953.897.7815.5531.1262.23124.46248.91 D - C # /D b 1.032.064.128.2416.4832.9665.93131.86263.72 D#/Eb - D1.12.184.368.7317.4734.9269.85139.7279.39 E - D # /E b 1.152.314.639.2518.537.0174148 F - E1.232.454.99.819.639.278.4156.81 F#/Gb - F1.292.65.1910.3920.7641.5383.07166.13 G - F # /G b 1.382.755.51122.014488176 G#/Ab - G1.462.915.8311.6523.346.6293.24186.48 A - G # /A b 1.543.096.1712.3524.749.3998.78197.56 A#/Bb - A1.643.276.5413.0826.1652.33104.66209.31 B - A # /B b 1.733.476.9313.8627.7255.44110.87221.76

33. Frequency Value of Cent Through the Keyboard 0.0 0.5 1.0 1.5 2.0 2.5 3.0 010002000300040005000 Frequency Hz/cent

34. 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.

35. 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 (C 4 ).

36. 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.

37. Practical Application In music only the first few partials have appreciable amplitude  Pitch matching for tones presented alternately and together gives the same result.

38. Almost Unison Tones Consider two tones constructed from partials as below. Neglect heterodyne effects for the time being. Harmonic1234 Tone J2505007501000 Tone K2525047561008 Beat Frequency2468

39. 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.

40. 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

41. Complete set of Heterodyne Components Tone J2505007501000 Tone K2525047561008 Subtractive Components AdditiveComponents 244246248254256258 496498506508502 748758 752754 100210041006 Can you find the differences and sums that result in these frequencies?

42. 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.

43. 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.

44. Next - Separate the Tones More Tone J2505007501000 Tone K2815628431124 Subtractive Components AdditiveComponents 157188219312343374 438469593624 531 719874 781812 103110621093 The spread of the clumps is quite large and the resulting sound is “nondescript.”

45. Approaching Unison – Pitch Matching Tone J2505007501000 Tone K250.5501751.51002 Subtractive Components AdditiveComponents 248.5249249.5251251.5252 499499.5501.5502 500.5 749.5752 750.5751 1000.510011001.5

46. Results A collection of beats may be heard.  Achieving unison is well-defined.  Here are the eight components near 250 Hz sounded together.

47. 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 P200400600800 Tone Q40180212031604

48. Heterodyne Components SubtractiveComponentsAdditiveComponents 199201202 399402403 601602603 803804 801 1003100410011002 120412011202 140414021403 16021603 Frequencies above 1600 Hz are few in number and amplitude 18031804 20032004 2204 2404

49. 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

50. 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 M200400600800 Tone N3016029031204 Now every third harmonic of M is close to a harmonic of N

51. Heterodyne Components Subtractive ComponentsAdditiveComponents 99101103 198202 299303 402404 499503501 604 703 701 804 802 901 1004 1002 11011103 1202

52. Results We get clusters of frequencies separated by 100 Hz. When the two are in tune, we will have the partials… 2003004006008009001200 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).

53. 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.

54. No Special Relationship among the Tones Consider two tones and their partials Tone V200400600800 Tone W2735468191092 Heterodyne components includes 19 [W 3 – V 4 ], 54 [V 3 – W 2 ], 73 [W 1 – V 1 ], 127 [V 2 – W 1 ], 146 [W 2 – V 2 ], 219 [W 3 – V 3 ], 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.

55. Other Harmonic Sequences Another harmonic series produced at 473 Hz by the additive heterodyne components  The series is 473 [V 1 + W 1 ], 946 [V 2 + W 2 ], 1419 [V 3 + W 3 ], and 1819 [V 4 + W 4 ]. 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 W 1 – V 1 and the other at W 1 + V 1. 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.

56. Other Special Relationships RatioMusical IntervalCents Numbers of Frequencies in clumps 1/1Unison0001 group of five 1 group of six 1 group of seven 1 group of eight 2/1Octave12001 group of three 4 groups of four 3 groups of five 3/2Fifth702 (700)3 groups of two 9 groups of three

57. Other Special Relationships (Cont’d) 4/3Fourth498 (500)12 groups of two 1 group of three 5/3Major sixth884 (900)14 groups of two 5/4Major third386 (400)10 groups of two 6/5Minor third316 (300)6 groups of two 7/49696 groups of two 7/55834 groups of two 8/5Minor sixth814 (800)3 groups of two 7/62673 groups of two