ECE 598: The Speech Chain

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ECE 598 The Speech Chain

• Lecture 6 Vowels

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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

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The Three Basic Tube Terminations
p2
p1
A2
A1
p1-
p2-
x
Lf
0
-Lb

• Solid wall
• Air doesnt 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)

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Lets 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?

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Resonance with No Excitation

• Air Velocities

Air at the lips moves back and forth average
velocity is zero.
Air at the glottis never moves
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Resonance with Excitation

• Air Velocities

Glottis
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.
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Is the Glottis Closed?

• A slightly open glottis doesnt 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 (p1e-jkx p1-ejkx)/rc
• p1e-jk(-Lb) - p1-ejk(-Lb) 0

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Is Pressure Zero at the Lips? The Model
Forward-going wave here has amplitude and phase
given by p2e-jkLf
The inertia of the outside world causes the
total air pressure here to be 0, so
p2-
x
Lf
0
Lb
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 xLf, i.e.
p2-ejkLf p2e-jkLf 0
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Is Pressure Zero at the Lips? The Reality
Forward-going wave here has amplitude and phase
given by p2e-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?
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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 r0.8a outside the lips
• This is exactly the same reflection that we would
  get if we required that
• p(Lf0.8a, t)0
• p2e-jk(Lf0.8a) p2-ejk(Lf0.8a) 0
• Alternatively (much simpler) we can just redefine
  the length of the front cavity to be Lf Lf0.8a
  (0.8a is called the end correction). Then
• p2e-jkLf p2-ejkLf 0

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The Three Basic Tube Terminations
p2
p1
p1-
p2-
x
Lf
0
Lb

• Solid wall v(-Lb,t)0
• p1e-jk(-Lb) p1-ejk(-Lb) 0
• Open space p(Lf,t)0
• p2e-jkLf p2-ejkLf 0
• Abrupt area change
• p(0-,t)p(0,t)
• A1v(0-,t)A2v(0,t)

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Abrupt Area Change
p2
p1
p1-
p2-
x
Lf
0
Lb

• 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-)

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Abrupt Area Change
p2
p1
p1-
p2-
x
Lf
0
Lb

• 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 (1g)p1 - gp2-
• Reflection coefficient g
• g (A1-A2)/(A1A2)

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Hard Wall, Open Space are Special Cases of the
Abrupt Area Change
p2
p1
p1-
p2-
x
Lf
0
Lb

• Reflection coefficient g
• g (A1-A2)/(A1A2)
• p1- gp1 (1-g)p2-
• Open space
• As A2 ? 8, g ? -1
• p1- -p1
• Hard Wall
• As A2 ? 0, g ? 1
• p1- p1

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The Three Basic Tube Terminations
p2
p1
p1-
p2-
x
Lf
0
Lb

• Solid wall v(-Lb,t)0
• p1e-jk(-Lb) p1-ejk(-Lb) 0
• Open space p(Lf,t)0
• p2e-jkLf p2-ejkLf 0
• Abrupt area change
• p1- gp1 (1-g)p2-
• p2 (1g)p1 - gp2-

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Two-Tube Models of the Vocal Tract A2A1
p2
p1
p1-
p2-
x
Lf
0
Lb

• Pretend that g -1 (A2 A1)
• p1- -p1 (like an open space termination)
• p2 p2- (like a hard wall termination)

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Two-Tube Models of the Vocal Tract A2A1
p2
p1
p1-
p2-
Lf
0
-Lb
0

• Pretend that g -1 (A2 A1)
• p1- -p1 (like an open space termination)
• p2 p2- (like a hard wall termination)

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Two-Tube Models of the Vocal Tract A2A1
p2
p1
p1-
p2-
Lf
0
-Lb
0

• 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,

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Example Vowel /a/
p2
p1
p1-
p2-
Lf
0
-Lb
0

• Lb 8cm
• f 1100Hz, 3300Hz,
• Lf 9cm
• f 983Hz, 2950Hz,
• Formant frequencies F1983, F21100, F32950

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Example Vowel /ae/
p2
p1
p1-
p2-
Lf
0
-Lb
0

• Lb 2cm
• f 4425Hz,
• Lf 15cm
• f 590Hz, 1770Hz, 2950Hz, 4130Hz,
• Formant frequencies F1590, F21770, F32950

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Two-Tube Models of the Vocal Tract A1A2
p1
p2
p2-
p1-
x
Lf
0
Lb

• Pretend that g 1 (A1 A2)
• p1- p1 (like a hard wall termination)
• p2 -p2- (like an open space termination)

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Two-Tube Models of the Vocal Tract A1A2
p1
p2
p2-
p1-
Lf
0
-Lb
0

• Pretend that g 1 (A1 A2)
• p1- -p1 (like an open space termination)
• p2 p2- (like a hard wall termination)

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Two-Tube Models of the Vocal Tract A1A2
p1
p2
p2-
p1-
Lf
0
-Lb
0

• Resonant frequencies of the back cavity
• f 0, c/2Lb, c/Lb, 3c/2Lb,
• Resonant frequencies of the back cavity
• f 0, c/2Lf, c/Lf, 3c/2Lf,

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Example Vowel /i/
p1
p2
p2-
p1-
Lf
0
-Lb
0

• Lb 9cm
• f 0, 1966Hz, 3933Hz,
• Lf 8cm
• f 0, 2212Hz, 4425Hz,
• Formant frequencies F10, F21966, F32212

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Example Vowel /u/
p1
p2
p2-
p1-
Lf
0
-Lb
0

• Lb 16cm
• f 0, 1106Hz, 2212Hz,
• Lf 0cm
• f 0, 17700Hz,
• Formant frequencies F10, F21106, F32212

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The Vowel Quadrangle
2000
i
e
ae
F2 (Hz) Degree of tongue fronting
?
1500
o
a
1100
u
1000
0
500
F1 (Hz) 1000 Tongue height
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Wait a Minute --- F10Hz??!!

• F1 of /i/ and /u/ is not really 0Hz. Its 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.
• Lets learn about low-frequency coupling.

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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)

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The Four Basic Impedances
v0
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) 8
• Tube closed at the opposite end
• pe-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)

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The Four Basic Impedances
v(w)
p0
v(w)
p(w)

• Open space
• Air pressure p(w)0 regardless of w, therefore
• Admittance y(w) v(w)/p(w) 8
• Impedance z(w) p(w)/v(w) 0
• Tube open at the opposite end
• pe-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)

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Matching Admittances
vb
vf
Af
Ab
pb
pf
Lf
0
-Lb
0

• 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

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Low-Frequency Approximation
vb
vf
A2
A1
pb
pf
Lf
0
-Lb
0

• tan(q) q for small enough q. (q
• 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

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Helmholtz Resonance of the Vocal Tract
vb
vf
A2
A1
pb
pf
Lf
0
-Lb
0

• Helmholtz resonant frequency
• f (c/2p) (Af/VbLf)1/2
• (35400 cm/s/2p) (0.5cm2/(40cm3 ? 6cm))1/2
• 250 Hz

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Summary

• Abrupt area change
• p1- gp1 (1-g)p2-
• p2 (1g)p1 - gp2-
• If g1 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

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