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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