chapter 2

Sound Propagation: An Impedance Based Approach Yang-Hann Kim 2010 John Wiley & Sons (Asia) Pte Ltd

Sound PropagationAn Impedance Based Approach

Acoustic Wave Equationand Its Basic Physical Measures

Yang-Hann Kim

Chapter 2


Outline

2.1 Introduction/Study Objectives

2.2 One-dimensional Acoustic Wave Equation

2.3 Acoustic Intensity and Energy

2.4 The Units of Sound

2.5 Analysis Methods of Linear Acoustic Wave Equation

2.6 Solutions of the Wave Equation

2.7 Chapter Summary

2.8 Essentials of Wave Equations and Basic Physical Measures

2


2.1 Introduction/Study Objectives

The governing equation is the total expression of every possible wave.

This chapter explores the underlying physics and sensible physicalmeasures that are related to acoustic waves. Impedance plays a centralrole with regard to its effect on these measures.

The final objective of this chapter is to determine rational means of findingthe solutions of acoustic wave equations.

3



The simplest case is illustrated in Figure 2.1.

4

twcos

pS Sdxxpp

+

x

xD

x xx D+

+

xuu

tur=

ly.harmonical pipe, theof end one excite when we generated becan that waves thedepicts This

)fluid(kg/m ofdensity :directionin velocity particle fluid:

)(m areasection cross:

3

2

r-xu

S

0

0

twcos

pS Sdxxpp

+

x

xD

x xx D+

+

xuu

tur=

ly.harmonical pipe, theof end one excite when we generated becan that waves thedepicts This



3

2

r-xu

S

0

0

Figure 2.1 Relation between forces and motion of an infinitesimal fluid element in a pipe (expressing momentum balance:the left-hand side shows the forces and the right exhibits the change of momentum)


Equations 2.2 and 2.3 are different simply because the displacement ofthe string has to be 0 at the end, but the acoustic pressure is maximal at

.

We can also predict that the driving point impedance is governed by(wave number multiplied by its length ).


If the pipe is semi-infinitely long, then the pressure in the pipe canbe mathematically written as

(2.1)where is the pressure magnitude, and is an initial phase.

5

If the pipe is of finite length , then the waves in the pipe can be expressedby Equation 1.67. Recall that the displacement of the string, in this case, is

(2.2)

However, the possible acoustic pressure in the pipe can be written as(2.3)

( )( ),p x t

( ) ( )0, cos ,p x t P kx tw f= - +0P f

L

( ) ( ), sin cos .y x t A k L x tw= -

( ) ( )0, cos cos .p x t P k L x tw= -

x L=

kLk L


To understand what is happening in the pipe, we have to understand howpressures and velocities of the fluid particles behave and are associatedwith each other.

As illustrated in Figure 2.1, the forces acting on the fluid between andand its motion will follow the conservation of momentum principle.

That is,Sum of the forces acting on the fluid = momentum change. (2.4)

We can mathematically express this equality as(2.5)

where it has been assumed that the viscous force is small enough (relativeto the force induced by pressure) to be neglected.

6


xx x+ D

( ) ( ) ,x x xdupS pS S xdt

r+D- = D


The rate of change of the velocity can be expressed by

(2.6)where is a function of position ( ) and time ( ) and the velocity is the timerate change of the displacement.

Therefore, we can rewrite Equation 2.6 as

(2.7) If the cross-section between and is maintained constant and

becomes small , then Equation 2.5 can be expressed as

(2.8)where

(2.9), (2.10), (2.11)

7


( )/du dt,du u u x

dt t x t = +

u x t

.du u uudt t x

= +

x x x+ D xD( )0xD

,p u u Duux t x Dt

r r - = + =

0 ',r r r= + .D uDt t x

= + 0

',p p p= +


Equation 2.11 is the total derivative, and is often called the materialderivative. As can be anticipated, the second term is generally smallerthan the first.

If the static pressure ( ) and density ( ) do not vary significantly in spaceand time, then Equation 2.8 becomes

(2.12)where is acoustic pressure and is directly related to acoustic wavepropagation.

Equations 2.8 and 2.12 describe three physical parameters, pressure, fluiddensity, and fluid particle velocity. In other words, they express therelations between these three basic variables. In order to completelycharacterize the relations, two more equations are needed.

8


0p 0r

0' ,p u

x tr - =

'p


The relation between density and fluid particle velocity can be obtained byusing the conservation of mass. Figure 2.2 shows how much fluid entersthe cross-section at and how much exits through the surface at .

If we apply the principle of conservation of mass to the fluid volumebetween and , the following equality can be written.

the rate of mass increase in the infinitesimal element= the decrease of mass resulting from the fluid that is entering

and exiting through the surface at and (2.13a)9


( )xSur ( ) xxSu D+r( )xSt D r

x xx D+

x0



3

2

r-xu

S

( )xSur ( ) xxSu D+r( )xSt D r

x xx D+

x0



3

2

r-xu

S

Figure 2.2 Conservation of mass in an infinitesimal element of fluid (increasing mass of the infinitesimal volume results from a net decrease of the mass through the surfaces of the volume).

x x x+ D

x x x+ D

x x x+ D


Expressing this equality mathematically leads to

(2.13b) As assumed before, if the area of the cross-section ( ) remains constant,

then Equation 2.13 can be rewritten as

(2.14) We can linearize this equation by substituting Equation 2.10 into Equation

2.14. Equation 2.14 then becomes

(2.15) Equations 2.12 and 2.15 express the relation between the sound pressure

and fluid particle velocity, as well as the relation with the fluctuatingdensity and fluid particle velocity, respectively. One more equation istherefore needed to completely describe the relations of the threeacoustic variables.

10


( ) ( ) ( ) .x x xS x uS uSt r r r +D D = -

S

( ).ut xr r = -

0' .u

t xr r = -


The other equation must describe how acoustic pressure is related to thefluctuating density. Recall that a pressure change will induce a change indensity as well as other thermodynamic variables, such as entropy. Thisleads us to postulate that the acoustic pressure is a function of density andentropy, that is

(2.16)where denotes entropy.

We can then write the change of pressure, or fluctuating pressure, or ,by modifying Equation 2.16 as follows :

(2.17)

11


( ), ,p p sr=s

dp 'p

.s

p pdp d dss r

rr = +


Equation 2.17 simply states that a pressure change causes a densitychange ( ) and entropy variation ( ). It is noticeable that the fluid obeysthe law of isentropic processes when it oscillates within the range of theaudible frequency. The second term on the right-hand side of Equation2.17 is therefore negligible.

Note that the second relation of Equation 2.18 is mostly foundexperimentally. This reduces Equation 2.17 to the form

(2.18)

where is the bulk modulus that expresses the pressure required for aunit volume change and is the speed of sound.

12


dr ds

2

0

' ,'

p B cr r

= =

Bc


We can obtain Equation 2.18 by considering wave front propagation in a duct. Suppose that we make a disturbance which induces a small volume change in the one-dimensional duct as illustrated in Figure 2.21.

We now want to find the relation between the speed of sound propagationand other physical variables, such as pressure and density.

13


Figure 2.21 is cross-sectional area of the duct, is the coordinate that measures the distance from the disturbance, and isthe disturbance velocity

S x v

c


Conservation of mass implies the identity:(2.121)

The left-hand side simply represents the amount of mass change per unittime due to the disturbance. The right-hand side is the mass flux of thefluid at rest. These two have to be balanced, and can be written as

(2.122) We next apply Newtons second law to the fluid of interest. The force

difference will be , and the corresponding momentumchange under consideration is which neglects higherorder terms induced by . We can therefore write

(2.123) Equations 2.122 and 2.123 lead to the following relation which describes

the relation between the speed of propagation and other physical variablesof fluid:

(2.124)

14


( )( ) .d c dv S cSr r r+ - =

.dv cdr r=

( ( / ) )pS p p x dx S- + {( ) }cS Sdv cS cr- -

dr.dp cdvr=

2 .dpcdr

=


Equation 2.124 states that the square of the speed of sound depends onthe rate of compression with respect to density, that is, the amount ofpressure requires to generate a unit change in density. Note, however,that the change in pressure and density of the fluid also depends ontemperature or entropy. Therefore, Equation 2.124 has to be rewritten as

(2.125)

For isentropic process, Equation 2.125 can be written as

(2.126)

15


2 .p pcsr

= +

2 .cs

pcr=


For example, if the fluid can be assumed to be an ideal gas in isentropicprocess, then we can obtain the relations between pressure and density(the ideal gas law and the isentropic relation) as

(2.127)

(2.128)where is the number of moles defined as mass ( ) per unit molarmass ( ), is the universal gas constant (= 8.314 ) instandard air), is the absolute temperature ( ) and is the heat capacityratio which is defined as the ratio of the specific heat capacity underconstant pressure to the specific heat capacity under constant volume.

Consequently, we can predict the speed of sound for an ideal gas as(2.129)

under isentropic (i.e., no change in entropy) conditions.

16


,p nRTr=

constant ,p p por ggr r r = =

n Mkg / mol / ( mol)J K

,c nRTg=

KR

T g

kg


Tables 2.1 and 2.2 summarize the speed of sound in accordance with thestate of gas.

Table 2.1 Dependency of speed of sound on temperature

17



Table 2.2 The dependency of the speed of sound on relative humidity and on frequency

18


Sound Propagation: An Impedance Based Approach Yang-Hann Kim 2010 John Wiley & Sons (Asia) Pte Ltd 19


Figure 2.3 Pictorial relation between three variables that govern acoustic wave propagation ( and express the meanpressure and static density, respectively; and denote acoustic pressure and fluctuating density, respectively; denotes thespeed of propagation, and is the velocity of the fluctuating medium)

0p 0rp r c

u


If we eliminate and from Equations 2.12, 2.15 and 2.18, then we obtain

(2.19)

Based on what we have studied so far, two conclusions can be drawn. There is an analogy between the wave propagation along a string and acoustic

waves, that is, the waves in compressible fluid. There are definite relations between three acoustic variables, which are

illustrated in Figure 2.3.

20


'r u

2 2

2 2 2' 1 ' .p p

x c t =


We now extend Equations 2.12, 2.15 and 2.19 to a three-dimensional case.First, Eulers equation can be written as

(2.20)

where we use coordinate for convenience. , and denote thevelocity with respect to the coordinate system.

We may use a vector notation to express Equation 2.20, which will yield amore compact form. This gives

(2.21)

where(2.22)

21


0'p u

x tr - =

0'p v

y tr - =

0' ,p w

z tr - =

( ), ,x y z

( ), ,x y zu v w

0' ,upt

r - =

r

( ), , .u u v w=r


Equation 2.15 can also be extended to the three-dimensional form. That is,

(2.23)

The right-hand term of Equation 2.23 represent the net mass flow into theunit volume in space.

If we eliminate and using Equations 2.21, 2.23 and 2.18, then

(2.24)is obtained, which is a three-dimensional form of a wave equation.

22


0' .u

tr r = -

r

'r ur2

22 2

1 '' ppc t

=


For simplicity, we consider a one-dimensional case (Figure 2.4). Wedenote acoustic pressure ( ) as , and fluctuating density ( ) as .

23


0p pp +0

l

lD-

ll /D-

p

D-=

llpp 2

1e

0p pp +0

l

lD-

ll /D-

p

D-=

llpp 2

1e

Figure 2.4 Volume change and energy for a one-dimensional element ( is potential energy density and is for convenience)pe 'pp

'p p 'r r


As illustrated in Figure 2.4, there will be a volume change of becauseof the pressure difference along the element. The length of the elementwill be shortened by due to the small pressure change . The energystored in the unit volume, potential or elastic energy, can then be writtenas

(2.25)where has to obey the conservation of mass.

We therefore have

(2.26) Rearranging this, we obtain

(2.27)

24


l SD

l-D p

1 ,2p

lpl

e -D = lD

( )( )0 0 .lS l l Sr r r= + + D

0 0 0 .l l l l lr r r r r= + + D + D


Note that the last term on the right-hand side is much smaller than theothers. Equation 2.27 therefore reduces to

(2.28) Substituting Equation 2.28 into Equation 2.25 then gives

(2.29) Using the state equation, Equation 2.18, and changing to , then gives

(2.30)where denotes the acoustic potential energy.

The kinetic energy per unit volume can be written as

(2.31)25


0

.ll

rr

D- =

0

1 .2p

p rer

=

2

20

1 ,2p

pc

er

=

r p

pe

20

1 .2k

ue r=


If we assume that the dissipated energy in the fluid is much less than thepotential energy or kinetic energy, then the total energy has to be written:

. (2.32)

Note that the potential and kinetic energy are identical if the wave ofinterest is a plane wave in an infinite domain.

The next question then is how the acoustic energy changes with time. Wecan see that the energy per unit volume has to be balanced by the netpower flow through the surfaces that enclose the volume of interest, asillustrated in Figure 2.5.

26


22

020

1 12 2t p k

p uc

e e e rr

= + = +


This observation can be written conceptually asthe rate of increase of energy =the power entering through the surface at ( ) (2.33)

the power exiting through the surface at ( ).

This can be translated into a mathematical expression as follows:

(2.34)

27


Figure 2.5 Relation between energy and one-dimensional intensity (energy in the volume and the intensity throughthe surface at and must be balanced).

( )te ( )pu S xDx x x+ D

x x x= + Dx x=

( ) ( )

.

tx x x

S x pu S pu St

pu S xx

e+D

D = -

= - D

-


Equation 2.34 can then be reduced to

(2.35)where , which we call acoustic intensity or sound intensity.

If we simply extend Equation 2.35 to a three-dimensional case, then

(2.36)

Two major points must be noted in relation to the expression of theintensity. The intensity is a vector that has direction and magnitude. The intensity is a product of two different physical quantities.

28


0,t It xe + =

I pu=

0.t Ite + =

r


When we have two physical variables, the phase difference between themhas significant meaning. The phase between the force and velocityexpresses how well the force generates the velocity (response). In thisregard, the intensity can be classified as two different categories: activeintensity and reactive intensity.

To understand the meaning of the intensities in physical terms, we lookagain at the simplest case: the intensity of waves propagating in a one-dimensional duct.

Figure 2.6 depicts the waves in an infinite-length duct and Figure 2.7shows the waves for a finite-length duct.

29


L



t

avgpu

txpu ,0)( =

x)0,(xp0P

2l

{ })cos()cos(),(

)cos(Re),(

00

0

0)(

0

kxtUkxtc

Ptxu

kxtPePtxp kxtj

-=-=

-== --

wwr

wwx

t

0U),0( tu

t

0P

2p

),0( tp

x

0,)( =txpu avgpu

x)0,(xu

0U

t

avgpu

txpu ,0)( =

x)0,(xp0P

2l

{ })cos()cos(),(

)cos(Re),(

00

0

0)(

0

kxtUkxtc

Ptxu

kxtPePtxp kxtj

-=-=

-== --

wwr

wwxx

t

0U),0( tu

t

0P

2p

),0( tp

x

0,)( =txpu avgpu

x)0,(xu

0U

Figure 2.6 The acoustic pressure and intensity in an infinite duct. Note that the pressure and velocity are in phase with each other. Also, the active intensity (or average intensity with respect to time) is constantavgpu



Figure 2.7 The acoustic pressure and intensity in a duct of finite length of . Note that the phase difference between thepressure and velocity is 90( ))

L/ 2p


When the waves propagate in an infinite duct, the pressure and velocityhave the same phase. It can be observed that the average intensity with respect to time is

constant, as can be seen in Figure 2.6. The instantaneous intensity changes with regard to the position along the duct. The excitation effectively supplies energy to the system.

If we have the same excitation at one end, the duct has a finite length of ,and a rigid boundary condition exists at the other end , then the phasedifference between the pressure and velocity will be 90( ). It is not possible to effectively put energy into the system. The intensity is always zero at the nodal point of the duct , but it

oscillates between these points where the energy vibrates and does notpropagate anywhere.

32


avgpu

x L=2p

( )1 / 4x n L= +

L


We now need to explore more specific characteristics of the soundintensity, such as how to calculate and measure the intensity.

The mathematical definition of intensity can be written as

(2.37) The one-dimensional expression is simply

(2.38) The velocity can be obtained from the Euler equation (2.12):

(2.39) To obtain the derivative with respect to space, we may use two

microphones. This means that we approximate the derivative as

(2.40)33


.I pu=r r

.I pu=

0

1 .pu dtxr

= -

1 2 .p ppx x

- D


The pressure ( ) at the position of the measurement can be approximatelyobtained as:

(2.41)where the pressure fluctuates in time and is therefore a dynamic quantity.

We now look at the intensity measurement and calculation by consideringa plane wave with a radian frequency . The pressure can then be writtenas

(2.42)where denotes the pressure magnitude which has a real value and

represents the possible phase change in space.

34


p

1 2 ,2

p pp +

w

( ) ( )( )( , ) ,pj t xx t P x e w f- +=p( )P x

( )p xf


To obtain the velocity using the linearized Euler equation a pressuregradient is needed, that is

(2.43) Equations 2.39 and 2.43 then allow us to obtain the particle velocity:

(2.44) The intensity that is generated by the real part of pressure (2.42) and the

corresponding velocity (2.44), which is in phase with the real part of thepressure, can be obtained as

(2.45)

This is normally referred to as the active component of sound intensity.35


( ).pj tpddP j P e

x dx dxw ff - + = -

p

( )0

1 .pj tpd dPP j edx dx

w ffwr

- + = - +

u

( ) ( ) ( )

( )0

2 2

0

, cos cos

1 cos .

pa p p

pp

dPI x t P t tdx

dP t

dx

fw f w f

wrf

w fwr

= + - +

= - +


The time average of this intensity is often referred to as a mean intensity,or an active intensity, and can be written:

(2.46)This intensity can effectively supply power to space.

On the other hand, the multiplication of the real part of the pressure andthe imaginary part of the velocity that has 90phase difference willgenerate the following intensity:

(2.47)

We refer to this intensity as the reactive component of sound intensity.The time average of this intensity is 0 and, therefore, there is no netenergy transport; it only oscillates.

36


( ) 20

1 .2

pavg

dI x P

dxf

wr = -

( ) ( ) ( )

( )0

2

0

1, cos sin

1 sin 2 .4

re p p

p

dPI x t P t tdx

dP tdx

w f w fwr

w fwr

= + - +

= - +


The directions of intensities The direction of the active intensity is perpendicular to the wave front where

the phase is constant. The direction of reactive intensity has to be perpendicular to the surface over

which the mean square pressure is constant.

Equations 2.45 and 2.47 are referred to as the instantaneous activeintensity and instantaneous reactive intensity in the strict sense,respectively. However, when we say active intensity, we are referring to a time average of the

instantaneous active intensity, that is,

(2.46) for the reactive intensity case, we call its amplitude

(2.48)as reactive intensity.

37


( )2

0

1 ,4r

dPI xdxwr

= -

( ) 20

1 .2

pavg

dI x P

dxf

wr = -


The instantaneous intensity expressed by Equation 2.38 is composed oftwo components : the instantaneous active intensity (2.45) andinstantaneous reactive intensity (2.47). We can therefore write them as

(2.49) Using a complex function, Equation 2.49 can be expressed in simpler form,

that is

(2.50)where . This is often referred as a complex intensity.

38


( ) ( ) ( ) ( ) ( ), 1 cos2 sin 2 .avg p r pI x t I x t I x tw f w f = + + + +

( ) ( ) ( ){ }2, Re 1 ,pj tI x t x e w f- + = + C( ) ( ) ( )avg rx I x jI x= +C



The units that are relevant to sound can be classified into two groups:absolute units and subjective units.

Absolute units The unit of pressure = Pascal(Pa) = N/m2

The unit of velocity = m/sec The unit of intensity = Pam/sec = watt/m2

The unit of energy = joule = wattsec

To understand subjective units, we need to understand how we hear. Thismeans that we need to study our hearing system. Figure 2.8(a) depicts thehuman hearing system.

39



40

Figure 2.8 The structure of the ear and its frequency band characteristics. (a) The structure of a human ear. (Redrawn withpermission from D. Purves et al., Neuroscience, 3rd edition, 2004, pp. 288 (Figure 12.3), Sinauer Associates, Inc.,Massachusetts, USA. 2004 Sinauer Associates, Inc.) (b) External, middle, and inner ear. (c) Basilar membrane and Cortiorgan. (d) The cross-section of the cochlea shows the sensory cells (located in the organ of Corti) surrounded by the cochlearfluids. (e) Space-frequency map: moving along the cochlea, different locations are preferentially excited by different inputacoustic frequencies. (f) Tonotopic organization. (Figure 2.8(bf) drawings by Stephan Blatrix, from Promenade around thecochlea, EDU website: http://www.cochlea.org by Re my Pujol et al., INSERM and University Montpellier.)



41Figure 2.8 (Continued)



42

Figure 2.8 (Continued)



It is well known that humans do not hear the frequency of sound inabsolute scale, but rather relatively. Due to this reason, we normally userelative units for the frequencies.

The octave band is a typical relative scale (Figure 2.9).

43

1f 0f 12 2 ff = 002/1

12

12

12/1

0

7.0222

fffffff

ff

==-=D==

-

1f 0f 131

2 2 ff = 0121

3/12

16/1

0

23.022

ffffffff

=-=D==

1f 0f 11

2 2 ff n=

121

0 2 ff n=

1

1

2 2 ff n=

scaleoctave

scaleoctave 1/3

scaleoctave 1/n

Figure 2.9 Octave, 1/3 octave, and 1/n octave

The center frequency ( ) of each band is at the geometrical center of theband.

0f



44

Table 2.3 The center frequency of octave and 1/3 octave



For the amplitude of the sound pressure, we use the sound pressure level(SPL or ). It is defined as

(2.51)and is measured in units of decibels (dB); is the reference pressure,

is the average pressure, and log10 is a log function that has a base of10. is 20 Pa(20106N/m2).

From Figure 2.10, we can see that the human can hear from about 0 dB tosomewhere in the range of 130-140 dB.

Table 2.4 collects some typical samples of sound levels that we canencounter, providing some practical references of the sound pressure level(SPL).

45

pL

refpavgp

mrefp

2

10 210logavg

pref

pSPL L

p

= =



46

Figure 2.10 Equal-loudness contour: each line shows the SPL with respect to the frequency which corresponds to a loudness(phon) of 1 kHz pure sound. (Reproduced from ISO 226 (2003), Normal equal-loudness-level contours, InternationalStandards Organization, Geneva.)



47

Table 2.4 Daily life noise level in SPL



In order to calculate SPL, we write

(2.52) where denotes the measurement time.

Equation 2.52 expresses as the sum of every frequency component,equivalent to

(2.53) We then use the well-known relation

(2.54)where * denotes the complex conjugate.

48

( )2 20

1 ,T

avgp p t dtT=

( )p t

{ } { }20

1 Re Re .m nT j t j t

avg m nn m

p e e dtT

w w- -= P P

{ } ( )1Re ,2j t j t j te e ew w w- -= + *P P P

T



(2.55)

. This is only valid if and only if Equation 2.55 has a maximum when .

When , the slowly fluctuating terms with frequency are muchgreater than those with a frequency of .

49

{ }202

1 2Re4

12

T

mm

mm

dtT

@

=

P

P

( ){ }( ){ }

0

*

1 2Re4

2Re

m n

m n

T j tm n

n m

j tm n

eT

e dt

w w

w w

- +

- -

=

+

P P

P P

n mn m=

( )m nw w-( )m nw w+

If we rearrange Equation 2.53 using Equation 2.54, then we obtain( ) ( )

( ) ( )

2 *

0

* * *

14

m n m n

m n m n

T j t j tavg m n m n

n m

j t j tm n m n

p e eT

e e dt

w w w w

w w w w

- + - -

- +

= +

+ +

P P P PP P P P



50

1 2 3

SPL

SPL

1 2 31 2 3

t transformFourier

(Hz)Frequency (Hz)Frequency

(Hz)frequency center octave 1/3

mP

PPP

M3

2

1p

2avgp 2

|| 22P

2|| 21P

2|| 23P

1 2 3

SPL

SPL

1 2 31 2 3

t transformFourier

(Hz)Frequency (Hz)Frequency

(Hz)frequency center octave 1/3

mP

PPP

M3

2

1p

2avgp 2

|| 22P

2|| 21P

2|| 23P

Figure 2.11 Total mean square pressure and the mean square pressure of each frequency band

Figure 2.11 illustrates the relation between the sound pressure level andthe mean square pressure.



Let us begin with two sound pressures that have different frequencies,and . According to Equation 2.51, the sound pressure level of each

individual tone can then be written as

(2.56)

(2.57) If these two tones occur at the same time, the SPL can be written

(2.58) If we generalize this result to different pure tone cases, the SPL is

(2.59)

51

1f 2f

1

21,

10 21: 10logavg

pref

ptone SPL L

p= =

2

22,

10 22 : 10log .avg

pref

ptone SPL L

p= =

( )1 2

21

2 21, 2,

1 2 10 2

/10 /1010

10log

10log 10 10 .p p

avg avgp

ref

L L

p pSPL L

p+++

= =

= +

( )11 2 /10/101 2 1010log 10 10 .pp NN LLN pSPL L + + ++ + + = = + +LL LN


For example, if each tone has an SPL of 80dB, that is, =80dB and=80dB, then the sum of these two must be . This

simply means that the SPL increases by 3dB.

If we have two sounds of SPL 75dB and 80dB, the resulting SPL of thesounds is 81.2dB.

52


1pL

2pL8 8

1 2 1010log (10 10 )pL + = +



As illustrated in Figure 2.10, our hearing system depends strongly onfrequency band. Therefore, SPL has to properly consider its effect. Figure2.12 shows typical weightings or scales : often used as dB(A,B,C)

53

Figure 2.12 Various weighting curves. A-weighting: 40 phon curve (SPL < 55 dB); B-weighting: 70 phon curve (SPL=5585 dB); C-weighting: 100 phon curve (SPL > 85 dB).



54

Table 2.5 Measurement standards


2.5 Analysis methods of linear Acoustic Wave Equation

This section addresses how we mathematically predict or describe soundin space and time.

Let us begin with the case of which we have a volume source in one-dimensional infinite space. The volume velocity source makes the masschange by the velocity excitation. That is

(2.60)where is the volume velocity at .

55

0 0 ,u q

t xr r r = - +

x( ),q q x t=



Substituting Equations 2.18 and 2.12 into this new mass law Equation 2.60,we obtain the governing equation that includes the acoustic source:

(2.61) We first attempt a harmonic solution,

(2.62) Equation 2.61 can then be written as

(2.63)where represents the right-hand side of Equation 2.61.

Equation 2.63 is strictly only valid where the sound source exists;otherwise a homogeneous equation is valid.

56

2 2

02 2 21 .p p q

x c t tr - = -

( ) ( ), .j tx t x e w-=p P

( )2

22 ,

d k xdx

+ = -P P f

( )xf



For example, if there is a point source at , then Equation 2.63 can berewritten as

(2.64)where is a Dirac delta function, that is

(2.65) If the source exists only in the region , then we can write the governing

equation as

(2.66)

57

0x

( ) ( )2

20 02 ,

d k x x xdx

d+ = - -P P f

( )0x xd -( )0 1;x xd

-- =

( )0 00; .x x x xd - = 0L

( ) ( )0

22

0 0 02 .Ld k x x x dxdx

d+ = - -P P f



Expanding this equation to a three-dimensional case yields

(2.67)where and express the source position and the volume where the source is, respectively.

We now look at how to mathematically express the boundary condition.

We first study the one-dimensional case. As already expressed (1.26), the boundary condition can generally be written as

(2.68)where the subscript 0 and represent boundary at .

58

( ) ( ) ( )0

2 20 0 0 ,Vk r r r dV rd + = - -r r r rP P f

0rr

0V

0, 0, 0, , 0, ,L L L x L+ = =P Ua b g

L 0,x L=



To understand the boundary conditions that are expressed by Equation2.68, let us investigate several typical cases.

First, when , the condition takes the form

(2.69)This type of boundary condition is generally known as the Dirichletboundary condition.

On the other hand, if , then the equation becomes

(2.70)This type of boundary condition is called the Neumann boundary condition.

59

0=b

.=P ga

0=a

.=U gb



If , then Equation 2.68 reduces to

(2.71)and

(2.72)The impedance is described on the boundary.

More generally, the three-dimensional case of Equation 2.71 can bewritten as

(2.73)where is the surface that encloses the space of interest, as depicted inFigure 2.13 where is the particle velocity that is normal to the surface.

60

0=g

0, 0, 0L L+ =P Ua b

.= -PU

ba

0S

00, on ,S+ =P Ua b

U



We will consider the problem that is governed by the inhomogeneousgoverning equation and homogeneous boundary condition. That is ,

(2.66)and

(2.71)

61

022 =+ PP k

+ =P U

0

0V

rr

0S

0rr

022 =+ PP k

+ =P U

0

0V

rr

0S

0rr

( ) ( )0

22

0 0 02 L

d k x x x dxdx

d+ = - -P P f

0, 0, 0.L L+ =P Ua b

Figure 2.13 General boundary value problem ( is complex amplitude, is complex velocity, is the wave number, and Sexpresses the boundary; and indicate the observation position and boundary, respectively)

0Srr 0rr

P U k



One very well-known method for obtaining the solutions which satisfyEquations 2.66 and 2.71 uses eigenfunctions to express the solution. Thismeans, especially, that we first try to find the function which satisfies

(2.74)and also satisfies the boundary condition of Equation 2.71; that is

(2.75) The function which satisfies Equations 2.74 and 2.75 is the eigenfunction

or eigenmode, and the constant is the eigenvalue.

To shed more light on this problem, we consider the special case when. In this case, we have a rigid-wall boundary condition and the

eigenmode can be found, intuitively, as:

(2.76)62

nY2

22 0

nn n

ddx

+ =kY Y

0,0,

0

0.L nL nd

j dxr w+ =

b Ya Y

nk

0=a

( ) cos .n nx Lp=Y



If , which is the case for the pressure release boundary condition,then the solution has to take the form

(2.77) We generally call this method, which attempts to obtain the solution by

superimposing the eigenfunctions, a modal analysis. The advantage of thismethod is that a linear combination of the eigenmodes also satisfies thegiven boundary condition.

For the one-dimensional case, the pressure can be written as

(2.78)

63

0=b

( ) sin .n nx xLp=Y

( ) ( )0

.n nn

x x

==P aY



The main obstacle in finding the solution by a linear combination ofeigenfunctions is finding each modes contribution, or weighting, on thesolution. In other words, we must attempt to find (2.78) that satisfiesEquation 2.66.

For example, if we have one source at a point where Equation 2.64 is thegoverning equation, then we can attempt to construct the solution as givenby Equation 2.78. The coefficients can be found by using the property ofthe eigenfunctions (orthogonality condition), that is

(2.79)and

(2.80)

64

na

( ) ( )*0

1 Lm n m mnx x dxL

d= L Y Y

1 if,

0 ifmnm nm n

d==

( ) 20

1 .L

n nx dxL= L Y



Using Equations 2.66, 2.74, 2.78, 2.79 and 2.80, we can obtain theweighting as

(2.81)where denotes the complex conjugate.

Alternatively, we can try to obtain the solution that satisfies the boundarycondition by introducing Greens function.

If we denote the sound pressure due to a unit point source at as, then has to satisfy the equation:

(2.82)

65

( ) ( ) ( )*

2 2 0

1 ,L

n nn n

x x dxL k

= -L - a fk Y

*

0x x=( )0|x xG G

( )2

202 .

d k x xdx

d+ = - -G G



Multiplying by Equation 2.63 and by Equation 2.82, subtracting theformer from the latter and finally integrating with respect to lead us to

(2.83) Then, integration by parts yields:

(2.84) Changing the variable to reduces Equation 2.84 to the form

(2.85) We now investigate how to apply Equation 2.85 when we have a unit

amplitude sound source at , as illustrated in Figure 2.14. This specificcase reduces Equation 2.85 to

(2.86)

66

G Px

( ) ( )( )2 2

02 20 0.

L Ld d dx x x x dxdx dx

d - = - -

G PP G G f P

( ) ( )000 0

.L L

Ld d x dx xdx dx

- = - G PP G G f P

x 0x

( ) ( )0 0

000 0

.L

L L

d dx x dxdx dx

= + -

G PP G f P G

0x x=

( ) ( )0 0

00 0

| .L L

d dx x xdx dx

= + -

G PP G P G



67

Figure 2.14 One-dimensional and three-dimensional boundary value problems



If the velocity at is 0 (rigid-wall boundary condition), or the pressureis 0(pressure release boundary condition), then Equation 2.86 becomes

(2.87)or

(2.88) Equation 2.86 states that the sound pressure at consists of two

components: one is a direct effect from the sound source and the other isdue to the reflection from the boundary.

Expanding Equation 2.86 to a three-dimensional form yields the integralequation

(2.89)

68

0,x L=

( ) ( )0

00

|L

dx x xdx

= +

GP G P

( ) ( )0

00

| .L

dx x xdx

= -

PP G G

x

( ) ( ) ( )0 0

0 0 0 0 .V Sr r dV n dS= + - r r rP G f P G G P



If we do not have the sound source in the integral volume (Figure2.14(b)), then Equation 2.89 becomes

(2.90)

Equations 2.89 and 2.90 are referred to as Kirchhoff-Helmholtz integralequations.

69

( )0rrf 0V

( ) ( )0

0 0 .Sr dS= - rP P G G P



We will start with a one-dimensional, planar acoustic wave at positionand time , . This can be written as

(2.91) A wave in a certain direction in space can be expressed as

(2.92)where is a complex amplitude.

The plane wave 2.92, as the name implies, has all the same physicalproperties at the plane perpendicular to at (Figure 2.15). Note that itsimpedance at any position and time is

(2.93)

70

xt ( ),x tp

( ) ( ) ( ), .j t kxj tx t x e e ww - --= =p P A

( ) ( ) ( ), ,j t k rj tr t r e e ww - - -= =r rr rp P A

A

kr rr

0 .p cr=Z



The plane wave in an unbounded fluid propagates in the wave numbervector direction, independent of the position, frequency, wave number, andwavelength.

Intensity, specifically the average intensity (active intensity), can beexpressed as follows :

(2.94)where is the velocity in the direction of propagation and

71

rv

kv

0

) (),( rktjetrrrr --= wAp

Planes of constant phase

Figure 2.15 A plane wave ( is normal to the planes of constant phase)

{ } 2*0

1 1Re ,2 2avg

Icr

= =PU P

U 0 .cr=U P /

kr



Therefore, the intensity (2.94) can be written as

(2.95) The governing equation can also be written in terms of the spherical

coordinate.

We assume that the pressure is independent of the polar and azimuthangles and only depends on the distance from the origin ( ). Equation 2.24then becomes

(2.96) Its solution will be

(2.97)where is a complex amplitude.

72

20

1 .2avg

I cr= U

r

( ) ( )2 2

2 2 21 .rp rp

r c t =

( ) ,j t k rr e w- - =r r

p A

A



Equation 2.97 can be rewritten as

(2.98) To assess the velocity, consider Eulers equation in the spherical

coordinate:(2.99)

where is the velocity in the radial direction.

Equations 2.98 and 2.99 allow us to calculate the velocity in the radialdirection, that is

(2.100) Therefore, the impedance at can be written as

(2.101) This is the monopole radiation impedance.

73

( ) ( ), .j t k rr t er

w- - =r rAp

0 ,rup

r tr - =

ru

r

( )0

1 1 .j t k rrj e

r c krw

r- - = +

r rAu

( )( ) ( )

2

0 2 2 .1 1r

r

kr krc jkr kr

r = = -

+ +

pZu



74

Figure 2.16 depicts Equation 2.101.

Figure 2.16 Monopole radiation. (a) The monopoles radiation impedance where is wave number, indicates the Note isnoteworthy that it behaves as a plane wave, as the observation position is far from the origin. (b) Pressure and particle velocityin near field ( is small), magnitude (left) and phase (right) of pressure (top) and particle velocity (bottom); arrows indicateintensity. (c) As for (b) for far field case

l

kr

k



In the near field, the reactive part dominates the acoustic behavior in sucha way that the waves do not propagate well in the vicinity of the origin.

In the far field, the active part dominates. Therefore, the wave propagatesas if it is a planer wave.

The monopole sound source is defined by Equations 2.98 and 2.100 andhas a singularity at . This simple solution satisfies the linear waveequation. This implies that superposition of this type of solution alsosatisfies the governing wave equation. We can therefore attempt toconstruct any type of wave by using the monopole. This concept isillustrated in Figure 2.17, a graphical expression of Huygens principle.

75

0r =



76

Figure 2.17 Huygens principle. The wave front constructed by many monopole sound sources: (a) graphical illustration and (b) shallow ripple tank



If the two monopoles are close together with the opposite phase, then adipole is formed.

77

Figure 2.18 Dipole and quadrupole distributions and their characteristics where indicates an arbitrary point in spherical coordinate, is wave number, represents the dipole-moment amplitude vector, and represents the amplitude of quadrupole:(a) pressure of the spatial pattern of dipole sound; (b) impedence of a dipole at ; (c) magnitude (left) and phase (right) of particle velocity of a dipole in near field (top) and far field (bottom) (arrows indicate intensity); (d) pressure of a quadrupole pattern in space; (e) impedance of a quadrupole at and ( ) magnitude (left) and phase (right) of particle velocity of aquadrupole in near field (top) and far field (bottom) (Section 2.8.4)

( , , )r q fk D Q

r

r f



78

Figure 2.18 (Continued )



79Figure 2.18 (Continued )


2.7 Chapter Summary

We have attempted to understand how acoustic waves are generated andpropagated in a compressible fluid.

Conservation of mass and the state equation of fluid, together withNewtons law, provide three relations between density, fluid velocity, andpressure.

Acoustic intensity expresses the direction of acoustic power flow as well asits magnitude.

We studied a way to measure the associated acoustic variables inaccordance with human perception.

We have investigated possible solution methods that predict how soundwaves propagate in space and time.

80

chapter 2

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