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## ECE 598: The Speech Chain

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**ECE 598: The Speech Chain**Lecture 4: Sound**Today**• 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**Speech Units: cgs**• 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**Why Does Sound Happen?**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**Step 1: Middle Block Squished**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**Step 2: The Pressure Rises**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) • dp/dt = c2dr/dt = -rc2 dv/dx v v+dv**Step 3: Pressure X Area = Force**A • Force acting on the air between and : F = pA - (p+dp)A = -A dp dynes = (cm2) (d/cm2) p p+dp**Step 4: Force Accelerates Air**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!**Acoustic Constitutive Equations**• 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)**Solution: Forward and Backward Traveling Waves**• 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-**Other Wave Quantities Worth Knowing**• 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)**Air Particle Velocity**• 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**Volume Velocity**• 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)**Acoustic Waves**• Pressure p(x,t) = ejwt (p+e-jkx + p-ejkx) • Velocity v(x,t) = ejwt (1/rc)(p+e-jkx - p-ejkx)**Boundary Conditions**• 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 (p+ and p-)**• Two Equations in Two Unknowns 0 = p+e-jkL + p-ejkL p+= p- • Combine by Variable Substitution: 0 = p+(e-jkL +ejkL)**…and now, more useful trigonometry…**• 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 (p+ and p-)**• 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…)**Resonant Frequencies of a Uniform Tube, Closed at One End,**Open at the Other End • 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 • 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” • The wavelength are: l1=4L l2=4L/3 l3=4L/5**Standing Wave Patterns**• 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)**Standing Wave Patterns**• 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: Quarter-Wave Resonators**Tube Closed at the Left End, Open at the Right End**Half-Wave Resonators**• 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**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**• 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)