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?

Tsuu T’ina Vowels

Note: Tsuu T’ina has ~50 speakers

Navajo Vowels 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: f n = (2n - 1) * c 4L If L f = 9.5 cm: F 1 = / 4 * 9.5 = 921 Hz

Resonance at Work An open tube resonates at frequencies determined by: f n = (2n - 1) * c 4L If L b = 8 cm: F 1 = / 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

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 ( ) 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:

An [i] breakdown Helmholtz frequency: Volume(ab) = 60 cm 3 Length(bc) = 1 cm Area(bc) =.15 cm 2

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

Slightly Deeper Thoughts Helmholtz frequency: 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 F1F2

Size Matters, Again Helmholtz frequency: What would happen if we opened up the constriction? (i.e., increased its cross-sectional area, A bc ) 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. F n = 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 = Hz etc.

Frame of Reference The central note on a piano is called “middle C” (C4) Frequency = 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:

Singer’s Formant Demo