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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 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) )
x x
z y
Wave Equation3 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
z y
p p + ( p/ x) x
∞
Wave EquationF = 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
Uniform Tube Model (lossless)
• 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
t
p
c
A
x
u
t
u
Ax
p
2
l
Crosssection area = Ap(x,t) = 0
Piston velocity is independent of pressure
x = 0 x = l
• No air friction along the walls
• For convenience volume velocity is defined:
u (x,t) = A v (x,t) m3/s
cxtucxtuA
ctxp
cxtucxtutxu
//,
//,
• 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
cxtjcxtj
cxtjcxtj
ekekA
ctxp
ekektxu
//
//
,
,
Uniform Tube Model (lossless)
g
tjgtjtj
Ukk
eUekektux ,0;0
0
0,;
//
//
cljclj
cltjcltj
ekek
ekekA
ctlplx
Uniform Tube Model (lossless)
*
//
/
//
/
;
kk
ee
eUgk
ee
eUgk
cljclj
clj
cljclj
clj
tj
g
tjg
eUcl
cxl
A
cjtxp
eUcl
cxltxu
/cos
/sin,
/cos
/cos,
• These are standing waves.
• The envelopes are orthogonal in space and in time
xl0
volume velocity
pressure
• 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,...
ag
tj
lU
g
VclU
lU
eUcl
tlu
/cos
1,
/cos
1,
,
Uniform Tube Model (lossless)
,...5,3,1;22
/ kl
ckkcl
Vol
ume
velo
city
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
cxl
A
cj
txu
txpZ A /tan
,
,
Uniform Tube Model (lossless)
cls
cls
clsclsae
e
ee
jc
slsV
/2
/
// 1
22
cos
1
V
1
*
1
kkk
a
sssssV
2,1,0;
2
12
k
l
ckjsn
Energy Loss Due to Wall Vibration• Let the crosssection of the tube
be A(x,t), then
• Now consider the model
t
A
t
pA
cx
ut
Au
x
p
2
1
/
Wave eqns.
• 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)
txpAkdt
Adb
dt
Adm www ,
2
2
Energy Loss Due to Wall Vibration
Akdt
Adb
dt
Admp
t
A
t
p
c
A
x
u
t
u
Ax
p
www
2
2
20
0
• 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
tjtjtj exAtxAexUtxuexPtxp ,ˆ,,,,,,,
Energy Loss Due to Wall Vibration
,ˆ,ˆ,ˆ,
,ˆ,,
,,
2
20
0
xAkxAbjxAmxP
xAxPc
A
x
xU
t
xU
Ax
xP
www
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.
Formants would be at 500, 1500, 2500,..., in the lossless case.