Loading in 5 sec....

ECE 598: The Speech ChainPowerPoint Presentation

ECE 598: The Speech Chain

- 81 Views
- Uploaded on
- Presentation posted in: General

ECE 598: The Speech Chain

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

ECE 598: The Speech Chain

Lecture 4: Sound

- Ideal Gas Law + Newton’s Second = Sound
- Forward-Going and Backward-Going Waves
- Pressure, Velocity, and Volume Velocity
- Boundary Conditions:
- Open tube: zero pressure
- Closed tube: zero velocity

- Resonant Frequencies of a Pipe

- Distance measured in cm
- 1cm = length of vocal folds; 15-18cm = length vocal tract

- Mass measured in grams (g)
- 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)
- 1L/s = air flow rate during speech

- Force measured in dynes (1 d = 1 g cm/s2)
- 1000 dynes = force of gravity on 1g (1cm3) of H20

- Pressure measured in dynes/cm2 or cm H20
- 1000 d/cm2 = pressure of 1cm H20
- Lung pressure varies from 1-10 cm H20

A

- Consider three blocks of air in a pipe.
- 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)

x

x+dx

A

- Velocity of air is v or v+dv (cm/s)
- 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

- Density of middle block changes by
dr = m/(V+dV) - m/V

≈ -r dV/V = -r dt dv/dx

- Rate of Change of the density of the middle block
dr/dt = -r dv/dx

(g/cm3)/s = (g/cm3) (m/s) / m

v

v+dv

A

- Ideal Gas Law: p = rRT (pressure proportional to density, temperature, and a constant R)
- Adiabatic Ideal Gas Law: dp/dt = c2dr/dt
- When gas is compressed quickly, “T” and “r” both increase
- This is called “adiabatic expansion” --- it means that c2>R

- “c” is the speed of sound!!
- “c” depends on chemical composition (air vs. helium), temperature (body temp. vs. room temp.), and atmospheric pressure (sea level vs. Himalayas)

- When gas is compressed quickly, “T” and “r” both increase
- dp/dt = c2dr/dt = -rc2 dv/dx

v

v+dv

A

- Force acting on the air between and :
F = pA - (p+dp)A = -A dp

dynes = (cm2) (d/cm2)

p

p+dp

A

- Newton’s second law:
F = m dv/dt = (rAdx) dv/dt

-dp/dx = r dv/dt

(d/cm2)/cm = (g/cm3) (cm/s)/s

Air velocity has changed here!

- Newton’s Second Law:
-dp/dx = r dv/dt

- Adiabatic Ideal Gas Law:
dp/dt = -rc2 dv/dx

- Acoustic Wave Equation:
d2p/dt2= c2 d2p/dx2

pressure/s2 = (cm/s)2 pressure/cm2

- Things to notice:
- 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)

- Wave Number k:
k = w/c

(radians/cm) = (radians/sec) / (cm/sec)

- Forward-Traveling Wave:
p(x,t) = p+ej(wt-kx) = p+ej(w(t-x/c))

d2p/dt2= c2 d2p/dx2

-w2p+ = -(kc)2p+

- Backward-Traveling Wave:
p(x,t) = p-ej(wt+kx) = p-ej(w(t+x/c))

d2p/dt2= c2 d2p/dx2

-w2p- = -(kc)2p-

- Wave Number k:
k = w/c

(radians/cm) = (radians/sec) / (cm/sec)

- Wavelength l:
l = c/f = 2p/k

(cm/cycle) = (cm/sec) / (cycles/sec)

- Period T:
T = 1/f

(seconds/cycle) = 1 / (cycles/sec)

- Forward-Traveling Wave
p(x,t)=p+ej(w(t-x/c)), v(x,t)=v+ej(w(t-x/c))

- Backward-Traveling Wave
p(x,t)=p-ej(w(t+x/c)), v(x,t)=v-ej(w(t+x/c))

- Newton’s Second Law:
-dp/dx = r dv/dt

(w/c)p+ = wrv+

-(w/c)p- = wrv-

- Characteristic Impedance of Air: z0 = rc
v+= p+/rc

v-= -p-/rc

- In a pipe, it sometimes makes more sense to talk about movement of all of the molecules at position x, all at once.
- 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)

- Pressure
p(x,t) = ejwt (p+e-jkx + p-ejkx)

- Velocity
v(x,t) = ejwt (1/rc)(p+e-jkx - p-ejkx)

L

0

- Pressure=0 at x=L
0 = p(L,t) = ejwt (p+e-jkL + p-ejkL)

0 = p+e-jkL + p-ejkL

- 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+- p-

p+= p-

- Two Equations in Two Unknowns
0 = p+e-jkL + p-ejkL

p+= p-

- Combine by Variable Substitution:
0 = p+(e-jkL +ejkL)

- Definition of complex exponential:
ejkL = cos(kL)+j sin(kL)

e-jkL = cos(-kL)+j sin(-kL)

- Cosine is symmetric, Sine antisymmetric:
e-jkL = cos(kL) - j sin(kL)

- 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
0 = p+e-jkL + p-ejkL

p+= p-

- Combine by Variable Substitution:
0 = p+(e-jkL +ejkL)

0 = 2p+cos(kL)

- Two Possible Solutions:
0 = p+ (Meaning, amplitude of the wave is 0)

0 = cos(kL) (Meaning…)

- 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

…

- 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

- The wavelength are:
l1=4L

l2=4L/3

l3=4L/5

- The pressure and velocity are
p(x,t) = p+ ejwt (e-jkx+ejkx)

v(x,t) = ejwt (p+/rc)(e-jkx-ejkx)

- 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)

- The “standing wave pattern” is the part that doesn’t depend on time:
|p(x)| = 2P+cos(kx)

|v(x)| = (2P+/rc) sin(kx)

Tube Closed at the Left End, Open at the Right End

- 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

- Example, vocal tract closed at both ends:
F1=0, F2=1000Hz, F3=2000Hz, , F3=3000Hz

Tube Closed at Both Ends

Tube Open at Both Ends

F3=2500Hz=5c/4L

F2=1500Hz=3c/4L

F1=500Hz=c/4L

/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

- Newton’s Second Law:
-dp/dx = r dv/dt

- Adiabatic Ideal Gas Law:
dp/dt = -rc2 dv/dx

- Acoustic Wave Equation:
d2p/dt2= c2 d2p/dx2Pressure

- Solutions
p(x,t) = ejwt (p+e-jkx + p-ejkx)

v(x,t) = ejwt (1/rc)(p+e-jkx - p-ejkx)

- Resonances
- 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)