Transcript 16-Tube-Model

Vowels, Tubes and Music
March 22, 2016
Pragmatic Considerations
• The Formant/Fourier Analysis homework is due in two
days!
• I still owe you a lot of grading!
• I’m setting aside a big chunk of time between now
and Thursday to get your last course project reports
graded.
• I got a couple of correct answers to the mystery
spectrogram exercise…
F3 and
• English
, revisited
has pharyngeal, palatal and labial constrictions
• These constrictions conspire to drastically lower F3
F3 and
, revisited
Retroflex Vowels
• Retroflexion is a feature which may be superimposed on
other vowel articulations.
• Retroflexion is contrastive in vowels in Badaga, a
language spoken in southern India.
Retroflex Vowel Spectrograms
[be]
F3 and [y]
• [y] has both labial and palatal constrictions
• What effect would these constrictions have on F3?
[i] vs. [y]
[li]
[ly]
Overrounded Vowels
• Note: there is typically more rounding on [u] than [o]
• and on [o] than
• all the way down the line...
• It is possible to have [u]-like rounding on lower vowels
• “over-rounding” in Assamese
• Assamese is spoken in Bangladesh.
Overrounded Vowel
Spectrograms
Overrounded Vowel
Spectrograms
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.
front tube
back tube
• Both tubes may be considered closed at one end...
• and open at the other.
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 crosssectional 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
Let’s Try It Out
• What would the first five formant frequencies be for an
articulatory configuration where:
• The length of the front tube is 7 cm…
• and the length of the back tube is 10.5 cm?
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
• 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.
Hermann von Helmholtz
(1821 - 1894)
Some Specifics
• The vocal tract configuration for the vowel [i] resembles
a Helmholtz resonator.
• Helmholtz frequency:
c
Abc
f 
2 Vab Lbc
An [i] breakdown
• Helmholtz frequency:
c
Abc
f 
2 Vab Lbc
Length(bc) = 1 cm

Area(bc) =
.15 cm2
Volume(ab) = 60 cm3
35000 .15
f 
 280Hz
2
60 *1
An [i] Nomogram
Helmholtz
resonance
• Let’s check it out...
Slightly Deeper Thoughts
• Helmholtz frequency:
c
Abc
f 
2 Vab Lbc
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?
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:
c
Abc
f 
2 Vab Lbc
• What would happen if we opened up the constriction?
• (i.e., increased its cross-sectional area)

• 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
• In western music, each note is at a specific frequency
• Notes have letter names: A, B, C, D, E, F, G
• Some notes in between are called “flats” and “sharps”
261.6 Hz
440 Hz
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
Harmony
• Notes are said to “harmonize” with each other if the
greatest common denominator of their frequencies is
relatively high.
• Example: note A4 = 440 Hz
• Harmonizes well with (in order):
• A5 = 880 Hz
(GCD = 440)
• E5 ~ 660 Hz
(GCD = 220)
(a “fifth”)
• C#5 ~ 550 Hz
(GCD = 110)
(a “third”)
....
• A#4 ~ 466 Hz
(GCD = 2)
• A major chord: A4 - C#5 - E5
(a “minor second”)
Extremes
• Not all music stays within a couple of octaves of middle C.
• Check this out:
• Source: “Der Hölle Rache kocht in meinem Herzen”, 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.