ECE 598: The Speech Chain Lecture 6: Vowels

Today The Three Basic Tube Terminations Hard Wall (e.g., at the Glottis) Open Space (e.g., at the Lips) Abrupt Area Change Two-Tube Models of the Vocal Tract The Vowel Space The Four Basic Admittances

The Three Basic Tube Terminations p2+ p1+ A1 A2 p1- p2- -Lb Lf x Solid wall: Air doesn’t travel into the wall: v(-Lb,t)=0 Open space: Air pressure of the world outside is unchanged: p(Lf,t)=0 Abrupt area change (0- is a small number less than zero; 0+ is a small number greater than zero) Continuity of air pressure across the boundary: p(0-,t)=p(0+,t) Conservation of mass across the boundary: A1v(0-,t)=A2 v(0+,t)

Let’s look at each of those in more detail… Is the glottis really closed all the time? (if so, how is sound created?) Is the air pressure at the lips really zero (if so, how does the acoustic wave in the room get started?) ... And what happens at an area change?

Resonance with No Excitation Air Velocities: Air at the lips moves back and forth; average velocity is zero. Air at the glottis never moves

Resonance with Excitation Glottis Air Velocities: Open Open Closed Closed Closed Pulses of air escape the glottis, then stop dead when the glottis closes. Air at the lips moves forward and backward; average forward velocity is a little higher than zero.

Is the Glottis Closed? A “slightly open glottis” doesn’t change the resonant frequencies that much Open glottis (e.g., breathy voice) raises the resonant frequencies a little (during /h/, F1 may be as high as 800Hz) Open glottis also reduces resonant amplitude So we get nearly correct results by assuming that v(-Lb,t)=0 v(x,t) = ejwt (p1+e-jkx – p1-ejkx)/rc p1+e-jk(-Lb) - p1-ejk(-Lb) = 0

Is Pressure Zero at the Lips? The Model: Forward-going wave here has amplitude and phase given by p2+e-jkLf The “inertia” of the outside world causes the total air pressure here to be 0, so… p2- Lb Lf x The “inertia of the outside world” reflects the forward-going wave backward toward the glottis; the backward-going wave exactly cancels out the forward-going wave at position x=Lf, i.e.: p2-ejkLf + p2+e-jkLf = 0

Is Pressure Zero at the Lips? The Reality: Forward-going wave here has amplitude and phase given by p2+e-jkLf Air pressure in this region rapidly decays toward zero as the wave radiates into the room: Radiated pressure is p(r,t) = (a/r) plips(t-r/c) where “a” is the radius of the lips Since the wave rapidly decays toward zero pressure after leaving the lips, the “remainder” of the pressure is reflected back into the vocal tract… after some delay. How MUCH delay?

Is Pressure Zero at the Lips? The End Correction Require p(x,t) = 0 here 0.8a The backward wave is a reflected copy of the forward wave: AS IF it had to exactly cancel the forward wave… at a distance r=0.8a outside the lips This is exactly the same reflection that we would get if we required that p(Lf+0.8a, t)=0 p2+e-jk(Lf+0.8a) + p2-ejk(Lf+0.8a) = 0 Alternatively (much simpler) we can just redefine the length of the front cavity to be Lf = Lf+0.8a (0.8a is called the “end correction”). Then p2+e-jkLf + p2-ejkLf = 0

The Three Basic Tube Terminations p2+ p1+ p1- p2- Lb Lf x Solid wall: v(-Lb,t)=0 p1+e-jk(-Lb) – p1-ejk(-Lb) = 0 Open space: p(Lf,t)=0 p2+e-jkLf + p2-ejkLf = 0 Abrupt area change: p(0-,t)=p(0+,t) A1v(0-,t)=A2v(0+,t)

Abrupt Area Change Pressure continuity across the boundary: Lb Lf x Pressure continuity across the boundary: p(0-,t)=p(0+,t) p1+ + p1- = p2+ + p2- Conservation of mass across the boundary: A1v(0-,t)=A2v(0+,t) (A1/rc)(p1+ - p1-) = (A2/rc)(p2+ - p2-)

Abrupt Area Change Continuity of pressure and mass: Lb Lf x Continuity of pressure and mass: p1+ + p1- = p2+ + p2- A1 (p1+ - p1-) = A2 (p2+ - p2-) Re-arrange to get the outgoing waves (p1-, p2+) as functions of the incoming waves (p1+, p2-): p1- = gp1+ + (1-g)p2- p2+ = (1+g)p1+ - gp2- Reflection coefficient g: g = (A1-A2)/(A1+A2)

Hard Wall, Open Space are Special Cases of the “Abrupt Area Change” Lb Lf x Reflection coefficient g: g = (A1-A2)/(A1+A2) p1- = gp1+ + (1-g)p2- Open space: As A2 → ∞, g → -1 p1- = -p1+ Hard Wall: As A2 → 0, g → 1 p1- = p1+

The Three Basic Tube Terminations p2+ p1+ p1- p2- Lb Lf x Solid wall: v(-Lb,t)=0 p1+e-jk(-Lb) – p1-ejk(-Lb) = 0 Open space: p(Lf,t)=0 p2+e-jkLf + p2-ejkLf = 0 Abrupt area change: p1- = gp1+ + (1-g)p2- p2+ = (1+g)p1+ - gp2-

Two-Tube Models of the Vocal Tract: A2>>A1 p2+ p1+ p1- p2- Lb Lf x Pretend that g ≈ -1 (A2 >> A1): p1- ≈ -p1+ (like an “open space” termination) p2+ ≈ p2- (like a “hard wall” termination)

Two-Tube Models of the Vocal Tract: A2>>A1 p2+ p1+ p1- p2- -Lb Lf Pretend that g ≈ -1 (A2 >> A1): p1- ≈ -p1+ (like an “open space” termination) p2+ ≈ p2- (like a “hard wall” termination)

Two-Tube Models of the Vocal Tract: A2>>A1 p2+ p1+ p1- p2- -Lb Lf Resonant frequencies of the back cavity: f = c/4Lf, 3c/4Lf, 5c/4Lf, … Resonant frequencies of the front cavity: f = c/4Lb, 3c/4Lb, 5c/4Lb, …

Example: Vowel /a/ Lb ≈ 8cm: Lf ≈ 9cm: Lf Lb ≈ 8cm: f ≈ 1100Hz, 3300Hz, … Lf ≈ 9cm: f ≈ 983Hz, 2950Hz, … Formant frequencies: F1≈983, F2≈1100, F3≈2950

Example: Vowel /ae/ Lb ≈ 2cm: Lf ≈ 15cm: Lf Lb ≈ 2cm: f ≈ 4425Hz, … Lf ≈ 15cm: f ≈ 590Hz, 1770Hz, 2950Hz, 4130Hz, … Formant frequencies: F1≈590, F2≈1770, F3≈2950

Two-Tube Models of the Vocal Tract: A1>>A2 p1+ p2+ p2- p1- Lb Lf x Pretend that g ≈ 1 (A1 >> A2): p1- ≈ p1+ (like a “hard wall” termination) p2+ ≈ -p2- (like an “open space” termination)

Two-Tube Models of the Vocal Tract: A1>>A2 p1+ p2+ p2- p1- -Lb Lf Pretend that g ≈ 1 (A1 >> A2): p1- ≈ -p1+ (like an “open space” termination) p2+ ≈ p2- (like a “hard wall” termination)

Two-Tube Models of the Vocal Tract: A1>>A2 p1+ p2+ p2- p1- -Lb Lf Resonant frequencies of the back cavity: f = 0, c/2Lb, c/Lb, 3c/2Lb, … f = 0, c/2Lf, c/Lf, 3c/2Lf, …

Example: Vowel /i/ Lb ≈ 9cm: Lf ≈ 8cm: Lf Lb ≈ 9cm: f ≈ 0, 1966Hz, 3933Hz, … Lf ≈ 8cm: f ≈ 0, 2212Hz, 4425Hz, … Formant frequencies: F1≈0, F2≈1966, F3≈2212

Example: Vowel /u/ Lb ≈ 16cm: Lf ≈ 0cm: Lf Lb ≈ 16cm: f ≈ 0, 1106Hz, 2212Hz, … Lf ≈ 0cm: f ≈ 0, 17700Hz, … Formant frequencies: F1≈0, F2≈1106, F3≈2212

The Vowel Quadrangle 2000 i e ae F2 (Hz) ≈ Degree of tongue fronting 1500 ə o a 1100 u 500 1000 F1 (Hz) ≈ 1000 – Tongue height

Wait a Minute --- F1=0Hz??!! F1 of /i/ and /u/ is not really 0Hz. It’s really about 250Hz. 250Hz is the “Helmholtz resonance” of the vocal tract. “Helmholtz resonance” is caused by coupling between the back cavity and front cavity at very low frequencies. Let’s learn about low-frequency coupling.

Admittance/Impedance vb vf pb pf The far end of a tube specifies a relationship, called “impedance,” between pressure and velocity at the near end of the tube. Impedance: z(w) = p(w)/v(w) Admittance: y(w) = v(w)/p(w) = 1/z(w)

The Four Basic Impedances v=0 v(w) p(w) Hard wall: Air velocity v(w)=0 regardless of w, therefore Admittance: y(w) = v(w)/p(w) = 0 Impedance: z(w) = p(w)/v(w) = ∞ Tube closed at the opposite end: p+e-jkL – p-ejkL = 0, so u(w) ~ 2j sin(kL) p(w) ~ 2 cos(kL) Admittance: y(w) = j sin(kL)/cos(kL) = j tan(kL) Impedance: z(w) = 1/y(w) = 1/j tan(kL)

The Four Basic Impedances v(w) p=0 v(w) p(w) Open space: Air pressure p(w)=0 regardless of w, therefore Admittance: y(w) = v(w)/p(w) = ∞ Impedance: z(w) = p(w)/v(w) = 0 Tube open at the opposite end: p+e-jkL + p-ejkL = 0, so v(w) ~ 2 cos(kL) p(w) ~ 2j sin(kL) Admittance: y(w) = cos(kL)/jsin(kL) = 1/j tan(kL) Impedance: z(w) = 1/y(w) = j tan(kL)

Matching Admittances Pressure continuity: pb = pf vb vf Ab Af pb pf -Lb Lf Pressure continuity: pb = pf Conservation of mass: Abvb = -Afvf zb/Ab = -zf/Af 1/jAbtan(kLb) = -j tan(kLf)/Af 1/Abtan(kLb) = tan(kLf)/Af

Low-Frequency Approximation vb vf A1 A2 pb pf -Lb Lf tan(q) ≈ q for small enough q. (q << p/2) 1/(AbkLb) ≈ kLf/Af 1/Vb ≈ (w/c)2 Lf/Af Same as a spring-mass system!! Lf/Af is the “mass per unit area” of the air in the front tube 1/Vb is the “stiffness” of the air in the back tube Helmholtz resonant frequency: w = (k/m)1/2 = c(Af/VbLf)1/2 f = (c/2p) (Af/VbLf)1/2

Helmholtz Resonance of the Vocal Tract vb vf A1 A2 pb pf -Lb Lf Helmholtz resonant frequency: f = (c/2p) (Af/VbLf)1/2 ≈ (35400 cm/s/2p) (0.5cm2/(40cm3  6cm))1/2 ≈ 250 Hz

Summary Abrupt area change: If g≈1 or g≈-1 we can “decouple” the tubes p1- = gp1+ + (1-g)p2- p2+ = (1+g)p1+ - gp2- If g≈1 or g≈-1 we can “decouple” the tubes “Vowel quadrangle:” /i/-/u/-/a/-/ae/ Decoupling fails at very low frequencies – we need to replace the 0Hz resonance with a Helmholtz resonance at w = (k/m)1/2 = c (Af/VbLf)1/2

The Vowel Quadrangle 2000 i e ae F2 (Hz) ≈ Degree of tongue fronting 1500 ə o a 1100 u 250 500 1000 F1 (Hz) ≈ 1000 – Tongue height