ECE 598: The Speech Chain
Lecture 4: Sound
Today
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Ideal Gas Law + Newton’s Second =
Sound
Forward-Going and Backward-Going
Waves
Pressure, Velocity, and Volume Velocity
Boundary Conditions:
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Open tube: zero pressure
Closed tube: zero velocity
Resonant Frequencies of a Pipe
Speech Units: cgs
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Distance measured in cm
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Mass measured in grams (g)
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1L/s = air flow rate during speech
Force measured in dynes (1 d = 1 g cm/s2)
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1g = mass of one cm3 of H20 or biological tissue
1g = mass of the vocal folds
Time measured in seconds (s)
Volume measured in liters (1L=1000cm3)
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1cm = length of vocal folds; 15-18cm = length vocal tract
1000 dynes = force of gravity on 1g (1cm3) of H20
Pressure measured in dynes/cm2 or cm H20
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1000 d/cm2 = pressure of 1cm H20
Lung pressure varies from 1-10 cm H20
Why Does Sound Happen?
A
x
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x+dx
Consider three blocks of air in a pipe.
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Boundaries are at x (cm) and x+dx (cm).
Pipe cross-sectional area = A (cm2)
Volume of each block of air: V = A dx (cm3)
Mass of each block of air = m (grams)
Density of each block of air: r = m/(Adx) (g/cm3)
Step 1: Middle Block Squished
A
v
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Velocity of air is v or v+dv (cm/s)
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v+dv
In the example above, dv is a negative number!!!
In dt seconds, Volume of middle block changes by
dV = A ( (v+dv) - dv ) dt = A dv dt
cm3 = (cm2) (cm/s) s
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Density of middle block changes by
dr = m/(V+dV) - m/V
≈ -r dV/V = -r dt dv/dx
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Rate of Change of the density of the middle block
dr/dt = - r dv/dx
(g/cm3)/s = (g/cm3) (m/s) / m
Step 2: The Pressure Rises
A
v
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Ideal Gas Law: p = rRT (pressure proportional to
density, temperature, and a constant R)
Adiabatic Ideal Gas Law: dp/dt = c2dr/dt
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When gas is compressed quickly, “T” and “r” both increase
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This is called “adiabatic expansion” --- it means that c2>R
“c” is the speed of sound!!
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v+dv
“c” depends on chemical composition (air vs. helium), temperature
(body temp. vs. room temp.), and atmospheric pressure (sea level
vs. Himalayas)
dp/dt = c2dr/dt = - rc2 dv/dx
Step 3: Pressure X Area = Force
A
p
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p+dp
Force acting on the air between  and :
F = pA - (p+dp)A = -A dp
dynes = (cm2) (d/cm2)
Step 4: Force Accelerates Air
A
Air velocity has changed here!
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Newton’s second law:
F = m dv/dt = (rAdx) dv/dt
-dp/dx = r dv/dt
(d/cm2)/cm = (g/cm3) (cm/s)/s
Acoustic Constitutive Equations
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Newton’s Second Law:
-dp/dx = r dv/dt
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Adiabatic Ideal Gas Law:
dp/dt = - rc2 dv/dx
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Acoustic Wave Equation:
d2p/dt2 = c2 d2p/dx2
pressure/s2 = (cm/s)2 pressure/cm2
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Things to notice:
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p must be a function of both time and space: p(x,t)
“c” is a speed (35400 cm/s, the speed of sound at body
temperature, or 34000 cm/s at room temperature)
“ct” is a distance (distance traveled by sound in t seconds)
Solution: Forward and Backward
Traveling Waves
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Wave Number k:
k = w/c
(radians/cm) = (radians/sec) / (cm/sec)
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Forward-Traveling Wave:
p(x,t) = p+ej(wt-kx) = p+ej(w(t-x/c))
d2p/dt2 = c2 d2p/dx2
-w2p+ = -(kc)2p+
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Backward-Traveling Wave:
p(x,t) = p-ej(wt+kx) = p-ej(w(t+x/c))
d2p/dt2 = c2 d2p/dx2
-w2p- = -(kc)2p-
Forward and Backward Traveling Waves
Other Wave Quantities Worth
Knowing
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Wave Number k:
k = w/c
(radians/cm) = (radians/sec) / (cm/sec)
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Wavelength l:
l = c/f = 2p/k
(cm/cycle) = (cm/sec) / (cycles/sec)
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Period T:
T = 1/f
(seconds/cycle) = 1 / (cycles/sec)
Air Particle Velocity
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Forward-Traveling Wave
p(x,t)=p+ej(w(t-x/c)), v(x,t)=v+ej(w(t-x/c))
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Backward-Traveling Wave
p(x,t)=p-ej(w(t+x/c)), v(x,t)=v-ej(w(t+x/c))
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Newton’s Second Law:
-dp/dx = r dv/dt
(w/c)p+ = wrv+
-(w/c)p- = wrv-
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Characteristic Impedance of Air: z0 = rc
v+= p+/rc
v-= -p-/rc
Volume Velocity
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In a pipe, it sometimes makes more sense
to talk about movement of all of the
molecules at position x, all at once.
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A(x) = cross-sectional area of the pipe at
position x (cm2)
v(x,t) = velocity of air molecules (cm/s)
u(x,t) = A(x)v(x,t) = “volume velocity” (cm3/s
= mL/s)
Acoustic Waves
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Pressure
p(x,t) = ejwt (p+e-jkx + p-ejkx)
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Velocity
v(x,t) = ejwt (1/rc)(p+e-jkx - p-ejkx)
Standing Waves:
Resonances in the Vocal
Tract
Boundary Conditions
0
L
Boundary Conditions
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Pressure=0 at x=L
0 = p(L,t) = ejwt (p+e-jkL + p-ejkL)
0 = p+e-jkL + p-ejkL
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Velocity=0 at x=0
0 = v(0,t) = ejwt (1/rc)(p+e-jk0 - p-ejk0)
0 = p+e-jk0 - p-ejk0
0 = p+ - pp+ = p-
Two Equations in Two Unknowns
(p+ and p-)
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Two Equations in Two Unknowns
0 = p+e-jkL + p-ejkL
p+ = p-
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Combine by Variable Substitution:
0 = p+(e-jkL +ejkL)
…and now, more useful
trigonometry…
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Definition of complex exponential:
ejkL = cos(kL)+j sin(kL)
e-jkL = cos(-kL)+j sin(-kL)
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Cosine is symmetric, Sine antisymmetric:
e-jkL = cos(kL) - j sin(kL)
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Re-combine to get useful equalities:
cos(kL) = 0.5(ejkL+e-jkL)
sin(kL) = -0.5 j (ejkL-e-jkL)
Two Equations in Two Unknowns
(p+ and p-)
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Two Equations in Two Unknowns
0 = p+e-jkL + p-ejkL
p+ = p-
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Combine by Variable Substitution:
0 = p+(e-jkL +ejkL)
0 = 2p+cos(kL)
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Two Possible Solutions:
0 = p+ (Meaning, amplitude of the wave is 0)
0 = cos(kL) (Meaning…)
Resonant Frequencies of a Uniform
Tube, Closed at One End, Open at
the Other End
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p+ can only be nonzero (the amplitude of
the wave can only be nonzero) at
frequencies that satisfy…
0 = cos(kL)
kL = p/2, 3p/2, 5p/2, …
k1=p/2L, w1=pc/2L, F1=c/4L
k2=3p/2L, w2=3pc/2L, F2=3c/4L
k3=5p/2L, w3=5pc/2L, F3=5c/4L
…
Resonant Frequencies of a Uniform
Tube, Closed at One End, Open at
the Other End
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For example, if the vocal tract is 17.5cm
long, and the speed of sound is 350m/s at
body temperature, then c/L=2000Hz, and
F1=500Hz
F2=1500Hz
F3=2500Hz
Closed at One End, Open at the
Other End = “Quarter-Wave
Resonator”
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The wavelength are:
l1=4L
l2=4L/3
l3=4L/5
Standing Wave Patterns
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The pressure and velocity are
p(x,t) = p+ ejwt (e-jkx+ejkx)
v(x,t) = ejwt (p+/rc)(e-jkx-ejkx)
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Remember that p+ encodes both the
magnitude and phase, like this:
p+=P+ejf
Re{p(x,t)} = 2P+cos(wt+f) cos(kx)
Re{v(x,t)} = (2P+/rc)cos(wt+f) sin(kx)
Standing Wave Patterns
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The “standing wave pattern” is the part
that doesn’t depend on time:
|p(x)| = 2P+cos(kx)
|v(x)| = (2P+/rc) sin(kx)
Standing Wave Patterns: QuarterWave Resonators
Tube Closed at the Left End, Open at the Right End
Half-Wave Resonators
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If the tube is open at x=0 and at x=L,
then boundary conditions are p(0,t)=0
and p(L,t)=0
If the tube is closed at x=0 and at x=L,
then boundary conditions are v(0,t)=0 and
v(L,t)=0
In either case, the resonances are:
F1=0, F2=c/2L, F3=c/L, , F3=3c/2L
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Example, vocal tract closed at both ends:
F1=0, F2=1000Hz, F3=2000Hz, , F3=3000Hz
Standing Wave Patterns: Half-Wave
Resonators
Tube Closed at Both Ends
Tube Open at Both Ends
Schwa and Invv
(the vowels in “a tug”)
F3=2500Hz=5c/4L
F2=1500Hz=3c/4L
F1=500Hz=c/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
Summary
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Newton’s Second Law:
-dp/dx = r dv/dt
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Adiabatic Ideal Gas Law:
dp/dt = - rc2 dv/dx
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Acoustic Wave Equation:
Solutions
Resonances
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d2p/dt2 = c2 d2p/dx2Pressure
p(x,t) = ejwt (p+e-jkx + p-ejkx)
v(x,t) = ejwt (1/rc)(p+e-jkx - p-ejkx)
Quarter-wave resonator: F1=c/4L, F2=3c/4L, F3=5c/4L
Half-wave resonator: F1=0, F2=c/2L, F3=c/L
Standing-wave Patterns
|p(x)| = 2P+cos(kx)
|v(x)| = (2P+/rc) sin(kx)