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Chapter 14

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Chapter 14

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

  2. 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

  3. Beat Experiment In the other ear introduce a strong, single frequency (say, 400 Hz) source and a much weaker, adjustable frequency sound (the search tone). Mask one ear of a subject so nothing can be heard. Vary the search tone from 400 Hz up. We hear beats at multiples of 400 Hz.

  4. 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?

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

  6. 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)

  7. 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

  8. …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

  9. 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

  10. Heterodyne Component Example So the ear hears (200), 400, 600, (800), (1000), (1200), (1400), (1600), (1800).

  11. 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.

  12. 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.

  13. 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.

  14. Graph High Amplitude Motion of Tip

  15. 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

  16. Natural Frequency, fo 3rd Harmonic is fo 2nd Harmonic is fo Driven System Response

  17. 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

  18. 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.

  19. 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.

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

  21. 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.

  22. 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.

  23. 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.

  24. 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.”

  25. 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

  26. 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

  27. Cents • Each semitone is further divided into 100 parts called cents. • The difference between G4 andG4# 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)

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

  29. Advantage of the Cents Notation The same interval in different octaves will be difference frequency differences, but the interval in cents is always the same.

  30. Frequencies (Hz) for Equal-Tempered Scale("Middle C" is C4 )

  31. Intervals (Hz) for the Equal-Tempered Scale

  32. 3.0 2.5 2.0 1.5 1.0 Hz/cent 0.5 0.0 Frequency 0 1000 2000 3000 4000 5000 Frequency Value of CentThrough the Keyboard

  33. 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.

  34. 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).

  35. 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.

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

  37. Almost Unison Tones • Consider two tones constructed from partials as below. Neglect heterodyne effects for the time being.

  38. 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.

  39. 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

  40. Complete set of Heterodyne Components Can you find the differences and sums that result in these frequencies?

  41. 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.

  42. 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.

  43. Next - Separate the Tones More The spread of the clumps is quite large and the resulting sound is “nondescript.”

  44. Approaching Unison – Pitch Matching

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

  46. 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

  47. Heterodyne Components

  48. 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

  49. 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. • Now every third harmonic of M is close to a harmonic of N

  50. Heterodyne Components