Title: Physics of Sound
1Physics 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)
2Sound
- 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.
4Wave 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 --gt, ?0 ?(x,t) )
5Wave 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
8
6Wave 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 equation v
7Uniform 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
8Uniform Tube Model (lossless)
- To find the particular solution let,
- at x 0, ug(t) u(0,t) Ug(O) ej O 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
9Uniform Tube Model (lossless)
- These are standing waves.
- The envelopes are orthogonal in space and in time
l
0
x
10Uniform Tube Model (lossless)
- The frequency response for vol. Velocity, Va(O)
- 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
O / (2?) 2000 k / 8 250, 750, 1250,...
As l decrease resonance frequencies increase
Volume velocity
O
11Uniform Tube Model (lossless)
- Acoustic impedance The ratio of presure to
volume velocity. - The frequency response can be changed to transfer
function O ? s / j - Under some restrictions it can be written as
- The poles are the resonant frequencies of the tube
12Energy Loss Due to Wall Vibration
- Let the crosssection of the tube be A(x,t), then
- Now consider the model
Wave eqns.
13Energy 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, AA0?A)
14Energy Loss Due to Wall Vibration
- Assuming again ug(t) u (0, t) Ug(O) ej O t
yields solutions of the form - These forms eliminate time dependence and the
equations become - They are solved by numerical techniques
15Formants would be at 500, 1500, 2500,..., in the
lossless case.
l17.5 cm, A05cm2, mw 0.4gr/cm2, bw
6500dyne-sec/cm3 , kw0 Bandwidth is not zero!
Viscosity (friction of air with walls) and
thermal loss included.