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Vowels (one last time). March 2, 2010. 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?. - PowerPoint PPT Presentation
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Vowels (one last time)
March 2, 2010
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?
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Tsuu T’ina Vowels
Tsuu T’ina Vowels
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Note: Tsuu T’ina has ~50 speakers
Navajo Vowels
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Navajo has ~150,000 speakers
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.”
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.
Open up and say...• For instance, the shape of the articulatory tract while producing the vowel resembles two tubes.
• Both tubes may be considered closed at one end...
• and open at the other.
back tube
front tube
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
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
Check it out• Take a look at the actual F1 and F2 values of .
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.
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.
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
Switching Sides
Switching Sides
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
Helmholtz Resonance
Hermann von Helmholtz (1821 - 1894)
• 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”
• This frequency tends to be quite low.
Some Specifics• The vocal tract configuration for the vowel [i] resembles a Helmholtz resonator.
• Helmholtz frequency:
€
f =c
2π
AbcVabLbc
An [i] breakdown
• Helmholtz frequency:
€
f =c
2π
AbcVabLbc
Volume(ab) = 60 cm3
Length(bc) = 1 cm
Area(bc) = .15 cm2
€
f =35000
2π
.15
60*1≈ 280Hz
An [i] Nomogram
Helmholtz resonance
• Let’s check it out...
Slightly Deeper Thoughts
• Helmholtz frequency:
€
f =c
2π
AbcVabLbc
• What would happen to the Helmholtz resonance if we moved the constriction slightly further back...
• to, oh, say, the velar region?
Volume(ab)
Length(bc)
Area(bc)
Ooh!• The articulatory configuration for [u] actually produces two different Helmholtz resonators.
• = very low first and second formant
F1 F2
Size Matters, Again
• Helmholtz frequency:
€
f =c
2π
AbcVabLbc
• 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”...
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…
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
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.
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
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.)
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...
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.
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.
more info at: http://www.ncvs.org/ncvs/tutorials/voiceprod/tutorial/singer.html
Singer’s Formant Demo
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