Download ppt - Physics of Sound Wave equation: Part. diff. equation relating pressure and velocity as a function of time and space Nonlinear contributions are not considered

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

//

//

,

,


g

tjgtjtj

Ukk

eUekektux ,0;0

0

0,;

//

//

cljclj

cltjcltj

ekek

ekekA

ctlplx


*

//

/

//

/

;

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,

,


,...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

,

,


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.