4.7 SOUND TRANSMISSION THROUGH A WALL

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One of the more important problems in noise control is the determination of

the energy transmitted through a wall, as shown in Fig. 4-8. The following

analysis is valid if the wall is not too thin, in which case, vibrations of the

wall as a whole can occur. Also, the analysis is valid if the frequency is not

high enough that energy dissipation can occur. These effects will be examined

later in this chapter. The sound wave is considered to strike the wall at

normal incidence.

The expressions for the acoustic pressure in each of the three media

may be written as follows:

Transmission of Sound 101

FIGURE 4-8 Sound transmission from one material through a second material into

a third material for normal incidence.

Copyright © 2003 Marcel Dekker, Inc.

p1рx; tЮ ј A1 ejр!t_k1xЮ ю B1 ejр!tюk1xЮ (4-107)

(Incident wave) ю (Reflected wave)

p2рx; tЮ ј A2 ejр!t_k2xЮ ю B2 ejр!tюk2xЮ (4-108)

p3рx; tЮ ј A3 ejЅ!t_k3рx_LЮ_ (4-109)

(Transmitted wave)

The constants A1; B1; etc., are complex quantities in this case.

For a plane wave at normal incidence, the instantaneous particle velocity

in each material may be written as follows:

u1рx; tЮ ј р1=Z1ЮЅA1 ejр!t_k1xЮ _ B1 ejр!tюk1xЮ_ (4-110)

u2рx; tЮ ј р1=Z2ЮЅA2 ejр!t_k2xЮ _ B2 ejр!tюk2xЮ_ (4-111)

u3рx; tЮ ј р1=Z3ЮA3 ejЅ!t_k3рx_LЮ_ (4-112)

At the first interface рx ј 0Ю, the pressure in medium 1 and the pressure

in medium 2 are equal, and the particle velocities in mediums 1 and 2

are also the same at the interface. Using these conditions in Eqs (4-107),

(4-108), (4-110), and (4-111), we find the following relations:

A1 ю B1 ј A2 ю B2 р4-113)

A1 _ B1

Z1 ј

A2 _ B2

Z2

(4-114)

At the second interface рx ј LЮ, the pressures and particle velocities are also

equal. Using this condition in Eqs (4-108), (4-109), (4-111), and (4-112), we

obtain a second set of relationships between the coefficients:

A2 e_jk2L ю B2 ejk2L ј A3 (4-115)

A2 e_jk2L _ B2 ejk2L

Z2 ј

A3

Z3

(4-116)

We may combine Eqs (4-113) through (4-116) to obtain the expression for

the following complex number ratio:

A1

A3 ј

1

4

1 ю

Z1

Z2

_ _

1 ю

Z2

Z3

_ _

ejk2L ю

1

4

1 _

Z1

Z2

_ _

1 _

Z2

Z3

_ _

e_jk2L

(4-117)

The exponential terms may be written as follows:

ejk2L ј cosрk2LЮ ю j sinрk2LЮ (4-118)

e_jk2L ј cosрk2LЮ _ j sinрk2LЮ (4-119)

102 Chapter 4

Copyright © 2003 Marcel Dekker, Inc.

Substituting the results from Eqs (4-118) and (4-119) into Eq. (4-117), the

following expression is obtained:

A1

A3 ј

1

2

1 ю

Z1

Z3

_ _

cosрk2LЮ ю j

1

2

Z1

Z2 ю

Z2

Z3

_ _

sinрk2LЮ (4-120)

For any complex quantity, the magnitude is given by Eq. (4-23). The

magnitude of the ratio A1=A3 may be written from Eq. (4-120):

A1

A3

____

____

ј

1

2

1 ю

Z1

Z3

_ _2

cos2рk2LЮ ю

Z1

Z2 ю

Z2

Z3

_ _2

sin2рk2LЮ

" #1=2

(4-121)

The sound power transmission coefficient for transmission of acoustic

energy from medium 1 through medium 2 into medium 3 is given by:

at ј

Itr

Iin ј j p3j2=Z3

j pinj2=Z1 ј

A3

A1

____

____

2Z1

Z3

(4-122)

Eliminating the ratio jA3=A1j by using Eq. (4-121), we obtain the final

expression for the sound power transmission coefficient:

at ј

4рZ1=Z3Ю

1 ю

Z1

Z3

_ _2

cos2рk2LЮ ю

Z1

Z2 ю

Z2

Z3

_ _2

sin2рk2LЮ

(4-123)

Note that when the trigonometric terms are evaluated numerically, the term

k2L must be expressed in radians.

The tangent of the phase angle between the transmitted wave and the

incident wave is found from Eq. (4-120), using Eq. (4-24):

tan        ј ЅрZ1=Z2Ю ю рZ2=Z3Ю_ tanрk2LЮ

1 ю рZ1=Z3Ю

(4-124)

There are several special cases of practical importance for Eq. (4-123).

First, suppose the materials are the same on both sides of the wall, i.e.,

materials 1 and 3 are the same. This corresponds to the transmission of

sound from air (1) through a solid wall (2) into air (3) on the other side

of the wall. For this special case, Z1 ј Z3, and Eq. (4-123) reduces to:

at ј

4

4 cos2рk2LЮ ю ЅрZ1=Z2Ю ю рZ2=Z1Ю_2 sin2рk2LЮ

(4-125)

Next, we observe that the characteristic impedance of most solids is

much larger than that of air. For example,

Transmission of Sound 103

Copyright © 2003 Marcel Dekker, Inc.

concrete: Z2 ј 7,440,000 rayl

air р258CЮ: Z1 ј 409:8 rayl

Z1=Z2 ј 0:0000551 and Z2=Z1 ј 18,200

For this special case, we may neglect the term рZ1=Z2Ю in Eq. (4-125).

For the frequency range of interest in analysis of transmission of

sound through walls, the term k2L is often small. For example, for a

100mm (3.94 in) thick wall of concrete рc2 ј 3100 m/s) at a frequency of

1000 Hz, we find the following numerical value:

k2L ј

2_fL

c2 ј р2_Юр1000Юр0:100Ю

р3100Ю ј 0:203 rad

Using this value, we find:

sinрk2LЮ ј 0:201 _ k2L ј 0:203 (within 1%)

cosрk2LЮ ј 0:980 _ 1 (within 2%)

Based on this observation, we see that for рk2LЮ _ 0:25 rad, we may approximate:

sinрk2LЮ _ k2L and cosрk2LЮ _ 1

within about 3% error. With these approximations and for Z1 ј Z3, Eq.

(4-125) reduces to the following:

at ј

1

1 ю рZ2=2Z1Ю2рk2LЮ2 (4-126)

If we make the substitution for the wave number, k2 ј 2_f =c2, we obtain

the following important relationship:

1

at ј 1 ю

__2Lf

_1c1

_ _2

(4-127)

If we introduce the quantity, Ms ј _2L, called the specific mass, Eq. (4-127)

may be written in a form often called the mass law:

1

at ј 1 ю

_Msf

_1c1

_ _2

(4-128)

Another special case of interest is when k2L ј n_, where n ј an

integer (1, 2, 3, . . . ). For this case, cos2рk2LЮ ј 1 and sinрk2LЮ ј 0. If we

make these substitutions into Eq. (4-123), we find the following expression

for the sound power transmission coefficient:

at ј

4Z1Z3

рZ1 ю Z3Ю2 (4-129)

104 Chapter 4

Copyright © 2003 Marcel Dekker, Inc.

If we have the same material on both sides of the wall рZ1 ј Z3Ю, then the

sound power transmission from Eq. (4-129) becomes unity, i.e., at ј 1. The

sound is transmitted through the wall with no attenuation!

The wall is also transparent to sound waves having a frequency given

by the following relationship, obtained from sinрk2LЮ ј 0:

k2L ј

2_fL

c2 ј n_ or f ј

nc2

2L

(4-130)

This condition from Eq. (4-130) may also be written in the following form:

2_L=_ ј n_ or L ј 1

2 n_ (4-131)

When the thickness of the wall is a half-integer multiple of the wavelength,

the sound wave is transmitted directly through the wall. This principle has

been used in the design of free-flooding streamlined domes for housing

sonar transducers (Kinsler et al., 1982). For other applications, the condition

described by Eq. (4-130) may not be practical to achieve. For example,

for a 100mm(3.94in) thick concrete wall and with n ј 1, the corresponding

frequency is as follows:

f ј р1Юр3100Ю

р2Юр0:100Ю ј 15,500Hz ј 15:5kHz

At this high frequency, dissipation effects within the material and bending

wave effects tend to become significant, and Eq. (4-123) is no longer valid.

Example 4-3. A sound wave having a frequency of 250Hz and an intensity

level of 90dB strikes a wooden (oak) door (material 2) at normal incidence,

as shown in Fig. 4-9. The air in which the incident wave moves (material 1)

is at 08C (328F), and the air on the other side of the door (material 3) is at

258C (778F). The thickness of the door is 40mm (1.575in). Determine the

sound pressure level of the transmitted wave.

The properties of the materials are found in Appendix B:

air at 08C _1 ј 1:292 kg=m3; c1 ј 331:3m=s; Z1 ј 428:1 rayl

oakwood _2 ј 770 kg=m3; c2 ј 4300m=s; Z2 ј 3:30 _ 106 rayl

air at 258C _3 ј 1:184 kg=m3; c3 ј 346:1m=s; Z3 ј 409:8 rayl

The wave number for the wood is:

k2 ј

2_f

c2 ј р2_Юр250Ю

р4300Ю ј 0:3653m_1

k2L ј р0:3653Юр0:040Ю ј 0:01461 rad

Transmission of Sound 105

Copyright © 2003 Marcel Dekker, Inc.

Let us evaluate the sound power transmission coefficient from the

general expression, Eq. (4-123):

at ј р4Юр428:1=409:8Ю

1 ю

428:1

409:8

_ _2

cos2р0:01461Ю ю

428:1

3:30 _ 106 ю

3:30 _ 106

409:8

!2

sin2р0:01461Ю

at ј р4Юр1:0447Ю

4:18 ю р0:000130 ю 8052:7Ю2р0:01461Ю2 ј 3:017 _ 10_4

The transmission loss is:

TL ј 10 log10р1=atЮ ј 10 log10р1=3:017 _ 10_4Ю ј 35:2dB

The intensity for the incident wave is given by:

Iin ј р10_12Ю1090=10 ј 0:0010W=m2 ј 1:00mW=m2

The intensity of the transmitted wave is found from the definition of the

sound power transmission coefficient:

Itr ј atIin ј р3:017Юр10_4Юр0:0010Ю ј 0:3017 _ 10_6W=m2

ј 0:3017 mW=m2

The intensity level of the transmitted wave is:

LI;tr ј 10 log10р0:3017 _ 10_6=10_12Ю ј 54:8dB

106 Chapter 4

FIGURE 4-9 Physical system for Example 4-3.

Copyright © 2003 Marcel Dekker, Inc.

We could also have calculated the transmitted wave intensity level from:

LI;tr ј LI;in _ TL ј 90 _ 35:2 ј 54:8dB

The acoustic pressure for a plane wave may be evaluated from:

ptr ј рZ3ItrЮ1=2 ј Ѕр409:8Юр0:3017Юр10_6Ю_1=2 ј 0:01112 Pa ј 11:12 mPa

The sound pressure level for the transmitted wave is:

Lp;tr ј 20 log10р0:01112=20 _ 10_6Ю ј 54:9dB

The phase angle between the transmitted wave and the incident wave

may be found from Eq. (4-124):

tan        ј Ѕр428:1=3:30 _ 106Ю ю р3:30 _ 106=409:8Ю_ tanр0:01461Ю

1 ю р428:1=409:8Ю

tan        ј р3938:4Юр0:01461Ю ј 57:54

             ј 1:553 rad ј 89:08

The transmitted wave is almost 908 out of phase with the incident wave.

Let us check the accuracy of the approximate equation, Eq. (4-127),

for this problem. We note that this expression is strictly valid only if

Z1 ј Z3; however, in this example, Z1=Z3 ј р428:1=409:8Ю ј 1:045. Using

Eq. (4-127), we find:

1

at ј 1 ю р_Юр770Юр0:040Юр250Ю

р428:1Ю

_ _2

ј 1 ю 3192:9 ј 3193:9

at ј 3:131 _ 10_4

The error in using Eq. (4-127) instead of the general expression for the

sound transmission loss is approximately the same as the error in assuming

the characteristic impedances are the same on both sides of the door:

р3:131 _ 3:017Ю=р3:017Ю ј 0:038 ј 3:8%

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