Landmark-Based Speech Recognition: Spectrogram Reading, Support Vector Machines, Dynamic Bayesian Networks, and Phonology Mark Hasegawa-Johnson University of Illinois at Urbana-Champaign, USA

Lecture 2: Acoustics of Vowel and Glide Production One-Dimensional Linear Acoustics –The Acoustic Wave Equation –Transmission Lines –Standing Wave Patterns One-Tube Models –Schwa –Front cavity resonance of fricatives Two-Tube Models –The vowel /a/ –Helmholtz Resonator –The vowels /u,i,e/ Perturbation Theory –The vowels /u/, /o/ revisited –Glides

1. One-Dimensional Acoustic Wave Equation and Solutions

Acoustics: Constitutive Equations

Acoustic Plane Waves: Time Domain

Acoustic Plane Waves: Frequency Domain Tex

Solution for a Tube with Constant Area and Hard Walls

2. One-Tube Models

Boundary Conditions L 0

Resonant Frequencies

Standing Wave Patterns

Standing Wave Patterns: Quarter- Wave Resonators Tube Closed at the Left End, Open at the Right End

Standing Wave Patterns: Half- Wave Resonators Tube Closed at Both Ends Tube Open at Both Ends

Schwa and Invv (the vowels in “a tug”) F1=500Hz=c/4L F2=1500Hz=3c/4L F3=2500Hz=5c/4L

Front Cavity Resonances of a Fricative /s/: Front Cavity Resonance = 4500Hz 4500Hz = c/4L if Front Cavity Length is L=1.9cm /sh/: Front Cavity Resonance = 2200Hz 2200Hz = c/4L if Front Cavity Length is L=4.0cm

3. Two-Tube Models

Conservation of Mass at the Juncture of Two Tubes A1A1 A 2 = A 1 /2 U 1 (x,t) U 2 (x,t)= 2U 1 (x,t) Total liters/second transmitted = (velocity) X (tube area)

Two-Tube Model: Two Different Sets of Waves Incident Wave P1+ Reflected Wave P1-Incident Wave P2- Reflected Wave P2+

Two-Tube Model: Solution in the Time Domain

Two-Tube Model in the Frequency Domain

Approximate Solution of the Two- Tube Model, A 1 >>A 2 Approximate solution: Assume that the two tubes are completely decoupled, so that the formants include - F(BACK CAVITY) = c/4 L BACK - F(FRONT CAVITY) = c/4L FRONT L BACK L FRONT

The Vowels /AA/, /AH/ L BACK L FRONT L BACK =8.8cm  F2= c/4L BACK = 1000Hz L FRONT =12.6cm  F1= c/4L FRONT = 700Hz

Acoustic Impedance Z( ,j  )  0  0

Low-Frequency Approximations of Acoustic Impedance

Helmholtz Resonator -Z 1 ( ,j  ) =  0 Z 2 ( ,j  )  0 

The Vowel /i/ Back Cavity = Pharynx Resonances: 0Hz, 2000Hz, 4000Hz Front Cavity = Palatal Constriction Resonances: 0Hz, 2500Hz, 5000Hz Back Cavity Volume = 70cm 3 Front Cavity Length/Area = 7cm -1  1/2  √MC = 250Hz Helmholtz Resonance replaces all 0Hz partial-tube resonances. 250Hz 2000Hz 2500Hz

The Vowel /u/: A Two-Tube Model Back Cavity = Mouth + Pharynx Resonances: 0Hz, 1000Hz, 2000Hz Front Cavity = Lips Resonances: 0Hz, 18000Hz, … Back Cavity Volume = 200cm 3 Front Cavity Length/Area = 2cm -1  1/2  √MC = 250Hz Helmholtz Resonance replaces all 0Hz partial-tube resonances. 250Hz 1000Hz 2000Hz

The Vowel /u/: A Four-Tube Model Two Helmholtz Resonators = Two Low-Frequency Formants! F1 = 250Hz F2 = 500Hz F3 = Pharynx resonance, c/2L = 2000Hz 250Hz 500Hz 2000Hz Pharynx Velar Tongue Body Constriction Mouth Lips

4. Perturbation Theory

Perturbation Theory (Chiba and Kajiyama, The Vowel, 1940) A(x) is constant everywhere, except for one small perturbation. Method: 1. Compute formants of the “unperturbed” vocal tract. 2. Perturb the formant frequencies to match the area perturbation.

Conservation of Energy Under Perturbation

“Sensitivity” Functions

Sensitivity Functions for the Quarter- Wave Resonator (Lips Open)  L /AA//ER/ /IY//W/

Sensitivity Functions for the Half- Wave Resonator (Lips Rounded)  L /L,OW//UW/

Formant Frequencies of Vowels From Peterson & Barney, 1952

Summary Acoustic wave equation easiest to solve in frequency domain, for example: –Solve two boundary condition equations for P+ and P-, or –Solve the two-tube model (four equations in four unknowns) Quarter-Wave Resonator: Open at one end, Closed at the other –Schwa or Invv (“a tug”) –Front cavity resonance of a fricative or stop Half-Wave Resonator: Closed at the glottis, Nearly closed at the lips –/uw/ Two-Tube Models –Exact solution: use reflection coefficient –Approximate solution: decouple the tubes, solve separately Helmholtz Resonator –When the two-tube model seems to have resonances at 0Hz, use, instead, the Helmholtz Resonance frequency, computed with low-frequency approximations of acoustic impedance –/iy/: F1 is a Helmholtz Resonance –/uw/ and /ow/: Both F1 and F2 are Helmholtz Resonances Perturbation Theory –Perturbed area  Perturbed formants –Sensitivity function explains most vowels and glides in one simple chart