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Physics of Sound

Physics of Sound. Wave equation: Part. diff. e quation relating pressure and velocity as a function of time and space Nonlinear contributions are not considered. Valid under small part ical Velocity. Three steps Lossless uniform tube model Nonuniform, losses due to voval tract walls.

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Physics of Sound

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  1. Physics of Sound • Wave equation: Part. diff. equation relating pressure and velocity as a function of time and space • Nonlinear contributions are not considered. • Valid under small partical Velocity. • Three steps • Lossless uniform tube model • Nonuniform, losses due to voval tract walls. • Boundary effects (lip radiation)

  2. Sound • Sound is vibration of particles in a medium. • Particle velocity • Pressure • Sound wave is the propagation of disturbance of particles through a medium. • c =  f •  = 2 f • /c = 2 /  : wave number • At sea level c = 344 m/s (70º F) • At f = 50 Hz  = c / f = 6.88 m

  3. “Isothermal” processes • Slow variation (of pressure), temp. stays constant (no time for heat transfer) • “Adiabatic” processes • Fast variation (of pressure), temp. changes (time for heat transfer) • Example: Bicycle pump • Typical usage of the terms: “ Isothermal/adiabatic compression of a gas” • For most frequencies (except very low frequencies) sound is adiabatic.

  4.  y  z  x x Wave Equation • Atmospheric pressure P0 ~ 105 N/m2 • Pressure : P0 + p(x,t) • p(x,t) : • 0 dB, threshold of hearing ~ 2(10-5) N/m2 at 1000 Hz. • threshold of pain20 N/m2. • Particle velocity: v(x,t), m/s, (around zero average) • Density of air particles: (x,t), kg/ m3 (around an average of 0 -->, 0 +(x,t) )

  5. p + (p/ x) x p  y  z  x x Wave Equation 3 laws of physics, to be applied on the cubic volume of air. • F = ma • P V = Const; P: total pressure, V: volume,  = 1.4 • Conservation of mass: The cube may be deformed if pressure changes but the # of particles inside remains the same. F = - (p/ x) x (y z) net press. vol. (no frictional pressure, zero viscosity) m =  x y z ∞

  6. Wave Equation F = m a  - (p/x) x (y z) =  x y z (dv/dt) dv = (v/x) dx + v (v/t) dt dv/dt = v (v/x) + (v/t) nonlinear; can be neglected in speech production since particle velocity is small  - (p/ x) =  (v/t) Gas law and cons. of mass yields coupled wave equation pair  - (p/ t) = c2 (v/x) The two can be combined as (2p/x2) = (1/c2) (2p/t2) wave equation for p or (2v/x2) = (1/c2) (2v/t2) wave equationv

  7. l p(x,t) = 0 Crosssection area= A Piston velocity is independent of pressure x = 0 x = l Uniform Tube Model (lossless) • No air friction along the walls • For convenience volume velocity is defined: • u(x,t) = A v(x,t) m3/s • 2nd order wave equations are the same in this case except the replacement v(x,t)  u(x,t) • Coupled pair becomes • The solutions are of the form

  8. Uniform Tube Model (lossless) • To find the particular solution let, at x = 0, ug(t) = u(0,t) = Ug(Ω) ej Ω t (glottal flow) at x = l, p(l,t) = 0 (no radiation at the lips) • The general solution is • To solve for unknown constants k+ and k-, apply the boudary conditions above. and

  9. volume velocity pressure Uniform Tube Model (lossless) • These are standing waves. • The envelopes are orthogonal in space and in time 0 l x

  10. Uniform Tube Model (lossless) • The frequency response for vol. Velocity, Va(Ω) • The resonances occur at Ex: Consider a uniform tube of length l = 35 cm. For c = 350 m/s, the roots, resonances, are at f = Ω / (2) = 2000 k / 8 = 250, 750, 1250,... As l decrease resonance frequencies increase Volume velocity Ω

  11. Uniform Tube Model (lossless) • Acoustic impedance: The ratio of presure to volume velocity. • The frequency response can be changed to transfer function: Ω s / j • Under some restrictions it can be written as • The poles are the resonant frequencies of the tube

  12. Energy Loss Due to Wall Vibration Wave eqns. • Let the crosssection of the tube be A(x,t), then • Now consider the model

  13. Energy Loss Due to Wall Vibration • Assuming A(x,t) = A0(x,t)+ A(x,t), an equaton can be written for A(x,t): • Then, the three equations can be written (under some simplifications, A=A0+A)

  14. Energy Loss Due to Wall Vibration • Assuming again ug(t) = u (0, t) = Ug(Ω) ej Ω t yields solutions of the form • These forms eliminate time dependence and the equations become • They are solved by numerical techniques

  15. Formants would be at 500, 1500, 2500,..., in the lossless case. l=17.5 cm, A0=5cm2, mw= 0.4gr/cm2, bw= 6500dyne-sec/cm3 , kw=0 Bandwidth is not zero! Viscosity (friction of air with walls) and thermal loss included.

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