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)

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

“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.

Wave Equation Atmospheric pressure P 0 ~ 10 5 N/m 2 Pressure : P 0 + p(x,t) p(x,t) : –0 dB, threshold of hearing ~ 2(10 -5 ) N/m 2 at 1000 Hz. –threshold of pain20 N/m 2. Particle velocity: v(x,t), m/s, (around zero average) Density of air particles:  (x,t), kg/ m 3 (around an average of  0 -->,  0 +  (x,t) ) x  x x  z  y y

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 x  x x  z  y y p p + (  p/  x)  x ∞

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) =  c 2 (  v/  x) The two can be combined as (  2 p/  x 2 ) = (1/c 2 ) (  2 p/  t 2 ) wave equation for p or (  2 v/  x 2 ) = (1/c 2 ) (  2 v/  t 2 ) wave equation v

Uniform Tube Model (lossless) 2 nd 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 l Crosssection area = A p(x,t) = 0 Piston velocity is independent of pressure x = 0x = l No air friction along the walls For convenience volume velocity is defined: u (x,t) = A v (x,t) m 3 /s

To find the particular solution let, at x = 0,u g (t) = u(0,t) = U g (Ω) e j 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 Uniform Tube Model (lossless)

These are standing waves. The envelopes are orthogonal in space and in time x l0 volume velocity pressure

The frequency response for vol. Velocity, V a (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,... Uniform Tube Model (lossless) Volume velocity Ω As l decrease resonance frequencies increase

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 Uniform Tube Model (lossless)

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

Assuming A(x,t) = A 0 (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=A 0 +  A) Energy Loss Due to Wall Vibration

Assuming again u g (t) = u (0, t) = U g (Ω) e j Ω t yields solutions of the form These forms eliminate time dependence and the equations become They are solved by numerical techniques Energy Loss Due to Wall Vibration

l=17.5 cm, A 0 =5cm 2, m w = 0.4gr/cm 2, b w = 6500dyne-sec/cm 3, k w =0 Bandwidth is not zero! Viscosity (friction of air with walls) and thermal loss included. Formants would be at 500, 1500, 2500,..., in the lossless case.