Vowels (one last time)

1. Vowels (one last time) March 2, 2010

2. Fun Stuff • Any questions or updates on the lab exercise? • Cardinal Vowels, revisited • Delamont (2009): Adaptive Dispersion in Tsuu T’ina • Orthographically, Tsuu T’ina makes use of the vowels /a/, /i/, /o/ and /u/ • Q: How are they phonetically realized?

3. Tsuu T’ina Vowels

4. Tsuu T’ina Vowels Note: Tsuu T’ina has ~50 speakers

5. Navajo Vowels Navajo has ~150,000 speakers

6. How Many? • Delamont rejects the hypothesis that Tsuu T’ina really has only a two-vowel system. • Minimal overlap between /u/ and /a/. • Three-vowel system? • Delamont: maybe a contrast is collapsing. • “This leads us to consider the possibility that Tsuu T’ina does indeed have a four-vowel system which is seemingly unaffected by the ideals of dispersion theory.” • “If Tsuu T’ina does in fact have a four-vowel system, why is it such a mess? Eung-Do Cook (1989) suggests that, as a {language dies}, its phonological system tends to go haywire.”

7. Theory #2 • The second theory of vowel production is the two-tube model. • Basically: • A constriction in the vocal tract (approximately) divides the tract into two separate “tubes”… • Each of which has its own characteristic resonant frequencies. • The first resonance of one tube produces F1; • The first resonance of the other tube produces F2.

8. Open up and say... • For instance, the shape of the articulatory tract while producing the vowel resembles two tubes. front tube back tube • Both tubes may be considered closed at one end... • and open at the other.

9. Resonance at Work • An open tube resonates at frequencies determined by: • fn = (2n - 1) * c • 4L • If Lf = 9.5 cm: • F1 = • 35000 / 4 * 9.5 • = 921 Hz

10. Resonance at Work • An open tube resonates at frequencies determined by: • fn = (2n - 1) * c • 4L • If Lb = 8 cm: • F1 = • 35000 / 4 * 8 • = 1093 Hz •  for : • F1 = 921 Hz • F2 = 1093 Hz

11. Check it out • Take a look at the actual F1 and F2 values of .

12. Coupling • The actual formant values are slightly different from the predictions because the tubes are acoustically coupled. • = The “closed at one end, open at the other” assumption is a little too simplistic. • The amount of coupling depends on the cross-sectional area of the open end of the small tube. • The larger the opening, the more acoustic coupling… •  the more the formant frequencies will resemble those of a uniform, open tube.

13. Coupling: Graphically • The amount of acoustic coupling between the tubes increases as the ratio of their cross-sectional area becomes closer to 1. •  Coupling shifts the formants away from each other.

14. Switching Sides • Note that F1 is not necessarily associated with the front tube; • nor is F2 necessarily determined by the back tube... • Instead: • The longer tube determines F1 resonance • The shorter tube determines F2 resonance

15. Switching Sides

16. Switching Sides

17. A Conundrum • The lowest resonant frequency of an open tube of length 17.5 cm is 500 Hz. (schwa) • In the tube model, how can we get resonant frequencies lower than 500 Hz? • One option: • Lengthen the tube through lip rounding. • But...why is the F1 of [i]  300 Hz? • Another option: • Helmholtz resonance

18. Helmholtz Resonance • A tube with a narrow constriction at one end forms a different kind of resonant system. • The air in the narrow constriction itself exhibits a Helmholtz resonance. • = it vibrates back and forth “like a piston” Hermann von Helmholtz (1821 - 1894) • This frequency tends to be quite low.

19. Some Specifics • The vocal tract configuration for the vowel [i] resembles a Helmholtz resonator. • Helmholtz frequency:

20. An [i] breakdown • Helmholtz frequency: Length(bc) = 1 cm Area(bc) = .15 cm2 Volume(ab) = 60 cm3

21. An [i] Nomogram Helmholtz resonance • Let’s check it out...

22. Slightly Deeper Thoughts • Helmholtz frequency: Length(bc) Area(bc) Volume(ab) • What would happen to the Helmholtz resonance if we moved the constriction slightly further back... • to, oh, say, the velar region?

23. Ooh! • The articulatory configuration for [u] actually produces two different Helmholtz resonators. • = very low first and second formant F1 F2

24. Size Matters, Again • Helmholtz frequency: • What would happen if we opened up the constriction? • (i.e., increased its cross-sectional area, Abc) • This explains the connection between F1 and vowel “height”...

25. Theoretical Trade-Offs • Perturbation Theory and the Tube Model don’t always make the same predictions... • And each explains some vowel facts better than others. • Perturbation Theory works better for vowels with more than one constriction ([u] and ) • The tube model works better for one constriction. • The tube model also works better for a relatively constricted vocal tract • ...where the tubes have less acoustic coupling. • There’s an interesting fact about music that the tube model can explain well…

26. Some Notes on Music • The lowest note on a piano is “A0”, which has a fundamental frequency of 27.5 Hz. • The frequencies of the rest of the notes are multiples of 27.5 Hz. • Fn = 27.5 * 2(n/12) • where n = number of note above A0 • There are 87 notes above A0 in all

27. Octaves and Multiples • Notes are organized into octaves • There are twelve notes to each octave •  12 note-steps above A0 is another “A” (A1) • Its frequency is exactly twice that of A0 = 55 Hz • A1 is one octave above A0 • Any note which is one octave above another is twice that note’s frequency. • C8 = 4186 Hz (highest note on the piano) • C7 = 2093 Hz • C6 = 1046.5 Hz • etc.

28. Frame of Reference • The central note on a piano is called “middle C” (C4) • Frequency = 261.6 Hz • The A above middle C (A4) is at 440 Hz. • The notes in most western music generally fall within an octave or two of middle C. • Recall the average fundamental frequencies of: • men ~ 125 Hz • women ~ 220 Hz • children ~ 300 Hz

29. Extremes • Not all music stays within a couple of octaves of middle C. • Check this out: • Source: “Der Rache Hölle kocht in meinem Herze”, from Die Zauberflöte, by Mozart. • Sung by: Sumi Jo • This particular piece of music contains an F6 note • The frequency of F6 is 1397 Hz. • (Most sopranos can’t sing this high.)

30. Implications • Are there any potential problems with singing this high? • F1 (the first formant frequency) of most vowels is generally below 1000 Hz--even for females • There are no harmonics below 1000 Hz for the vocal tract “filter” to amplify • a problem with the sound source •  It’s apparently impossible for singers to make F1-based vowel distinctions when they sing this high. • But they have a trick up their sleeve...

31. Singer’s Formant • Discovered by Johan Sundberg (1970) • another Swedish phonetician • Classically trained vocalists typically have a high frequency resonance around 3000 Hz when they sing. • This enables them to be heard over the din of the orchestra • It also provides them with higher-frequency resonances for high-pitched notes • Check out the F6 spectrum.

32. more info at: http://www.ncvs.org/ncvs/tutorials/voiceprod/tutorial/singer.html How do they do it? • Evidently, singers form a short (~3 cm), narrow tube near their glottis by making a constriction with their epiglottis • This short tube resonates at around 3000 Hz • Check out the video evidence.

33. Singer’s Formant Demo