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Room AcousticsAE4300/8300 by Ir Dr Siu-Kit Lau

Sound Distributions in Room

• Direct Sound Field in Room

• Diffuse Sound Field in Room

– ,

sound field is obtained and the SPL due to the reflected component

tends to be independent of position within the room:

1. The size and shape of the room is not too large and one dimension isnot significantly greater than the other two.

2. There are no surfaces of high sound absorption.

– If the assumptions are achieved, the reflected sound field isnormally referred to as the reverberant field. The SPLs of the two

, , .

W

r



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Sound Propagation Outdoor

2

2

4 /

) /(

r W

c p

I

I D orms

π

ρ

== c

p

r

DW

o

rms ρ π

2

24 = cW

p

p

p

W

W

r

D

oref

ref

ref

rms

ref ρ π

2

2

2

24 ×=×

r ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ +

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ =⎟⎟

⎠

⎞⎜⎜⎝

⎛ +

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

cW

p

p

p

r

D

W

W

oref

ref

ref

rms

ref ρ π

2

2

2

2log10log10

4log10log10

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ +=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +

cW

p L

r

D L

oref

ref

pW ρ π

2

2log10

4log10

dBcW

p

oref

ref 15.0log10

2

−=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

ρ Q

⎞⎛ D

≈0

⎟ ⎠

⎜⎝

+=2

4og

r W p

π


• Direct Sound Field Calculation

– The sound intensity radiated by a Simple Sound Source of power

W can be calculated as follows:

– A surface directivity factor D is defined and is dependent upon the

position of the source in relation to the room boundaries.

24 r

W I

π =

24 r

DW I

π =



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• Direct Sound Field Calculations

⎞⎛ =

D

– L p,direct : Direct field sound pressure level

– Lw : Sound power level of the source

– r : Distance between source and receiver

– D: Directivity factor

⎠⎝ 2,

4 r W direct p

π

– y e n on e rec soun e s e soun a ransm s rom esource to the receiver and, as such, undergoes no reflections at anyboundaries. The LP of the direct sound field is therefore totally independentof the room and is directly related to the Inverse Square Law (6 dB

reduction for each doubling of the distance).


• Reverberant Sound (High-Frequency Approx.)

– When a sound source starts up in a room, the first sound a listeners hearsis that which arrives directly from the source. This is not affected by theboundaries of the room and is attenuated only according to the inversesquare law. The next sound which is heard is that which is reflected onceand, in addition to distance attenuation, it is also affected by the soundabsorptive quality of the surface. Then there is the sound wave whicharrives after two, three or more reflections, each one arriving successivelylater and in a characteristic manner determined by the absorption of thesurfaces from which they have been reflected and the size of the room.



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• Reverberant Sound – These multiple reflected or reverberant

sounds combine together, and with the directsound, to form the resulting SPL. Close to thesource, the direct sound will predominate andwill reduce with increasing distance from thesource until the reverberant sound becomesdominant, when the SPL will remainreasonably constant, irrespective of increasingdistance. It is to be expected that, if soundenergy is continuously introduced a room, theresulting SPL will rise until an equilibriumstate is achieved. At this point the rate of

as that absorbed either through or in theboundary materials. Consequently, a roomwith hard reflective walls and little soundabsorption in it is termed ‘live’ and will have ahigher reverberant SPL than with a lot ofsound absorption which is called ‘dead’.

Sound Distribution in Room

• Reverberant/Diffuse Field Calculations – Absorption coefficient α : The proportion of incidence sound

the room.

– Mean coefficient : The average value of the absorption

coefficient of all the room boundaries.

reflectionTotal

absorptionTotal

⇒=

⇒=

0

1

α

α

nn

S

SSSS α α α α α

++++=

L332211

α

• Snαn represents the area of a particular surface multiplied by its absorption coefficient; Stotal is the totalgeometric area of all surfaces in the room.

• All surfaces, e.g. furniture, screens, people, etc., provide equivalent absorbent areas in a room andtherefore should be taken into account, together with all the room boundaries. All materials haveabsorption coefficients that are different at different frequencies. It is standard practice to specifysound absorption coefficients in manufacturer literature in the octave bands 125, 250, 500, 1000,2000 and 4000 Hz.

tota



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• Reverberant / Diffuse Field Calculations – The SPL of the multiply-reflected / reverberant sound field

– It can be seen from the formula that for any given frequency,the terms involved are all constant (surface area and

absorption coefficient). Thus the formula predicts that the

α

α

−=

⎟ ⎠

⎞⎜⎝

⎛ +=

1

4log10,

S R

R L L W reflected p

R is called Room Constant

reverberant sound field will be unaltered by changing the

location of the receiver or the source. Both source andsurface directivity do not affect this ideal reverberant

field.


• Total SPL in Room (High-Freq. Approx.)

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ++= Rr

D L L W p

4

4log10

2π



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• Reverberation Time – It is the time required for the SPL to decrease by 60 dB after the

sound source has sto ed o eration. (Fre uenc De endent)

α Sc

V RT

⋅=

25.55

Sabine Formula

RT: Reverberant time in sec.V: Volume of the room

α: Average absorption coefficientC: Speed of Sound


• Reverberation Time

– In order to quantitatively evaluate reverberation, Sabine

.required fro the SPL to decrease by 60 dB after the sound

source has ceased. Sabine demonstrated that thereverberation time depends on the volume of the roomspace and on the treatment (absorption) of the room

surfaces. Sabine formula gives the relationship betweenreverberation time RT , volume V and surface S of a room,

and equivalent absorption coefficient . The formula isα

one of the most important in room acoustics and its use is

universal. Although many assumptions are made when it isderived, it is a sufficiently accurate formula in practice. Othermore accurate relationships exist but are less convenient to

use.



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• Answers: – Cubical Room

• L3 = 3000 L = 14.42 m

– Assume D =


• Example

– An equipment inside a 30 m3 plant room radiates asound of 1 W at 500Hz. A measurement gives aSPL of 116 dB at a point, which is far enough fromthe sound source.

a) If the total area of the room is 50 m2, calculate the mean absorptioncoefficient of the room.

b) If the source is suddenly stopped, how much time is needed to have a10 dB reduction?

c) If an additional material of 10 m with an α = 0.8 is put on the wall,calculate the SPL. [Assume the mean absorption coefficient of theother area are same in (a)]

d) If three other identical equipment are added to the room, what will bethe total SPL inside the room.



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• Answer (a): – Measurement is far from the source:

• It is done in reverberant field and the direct sound is

very small due to the long distance (inverse-squared law)


• Answer

– (b) Time to reduce 10 dB

• = . = . sec.

– (c) Additional material added A new

• Assume the of other areas are same.

'α

α

– (d) Original room with

• Now there are 4 equipment

– SPL =

α



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Sound Absorbing Materials

• Absorption can be determined byex erimental rocedures, includin ;

– Random incidence coefficient α sabine using areverberation chamber.

– Normal incidence coefficient α n using animpedance tube.

Sound Absorbing Materials

• Random incidence coefficient: – The absorption coefficient of a material is defined as the ratio

between the energy absorbed to that of the incident sound ray. Its

absorbing no sound energy) and 1 (the surface absorbs all incidentenergy).

– For a given material, the absorption coefficient depends onfrequency and the angle of incidence. In practice an averaged valueover all possible angles of incidence is used. However, itsvariations with frequency are not averaged and should always betaken into account.

• Reverberation Time Method: – A patch of material is placed in a large, highly reverberant room

avng a use e . α sabine s ca cu ate rom measurements osound decay (reverberation time) in the room both with and withoutthe material sample in place. It is a better approximation to realinstallations of absorptive materials, where the incidence angle canbe anything.



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Reverberation Time Method

• Sabines formula is used to calculate the equivalent absorption area ofthe material, then, by dividing this quantity by the surface area, theabsorption coefficient can be determine by:

– T s & T e is the reverberation time (RT) with the sample & the empty reverberation room

– V is the volume, in cubic meters, of the empty reverberation room

– S is the area, in square meters, covered by the test specimen

• -

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=

es T T cS

V 1125.55α

.


• Example:

– The sound absorption coefficient of a fiberglass is to be determined byreverberation chamber method. The volume of the chamber is 60 m3 andthe surface area of the fiberglass put into the chamber is 3.6 m . Determinethe sound absorption coefficients using the following measuredreverberation times.

Band (Hz) 160 200 250 315 400 500 630 800 1000 1250 1600 2000 2500 3150 4000

Empty Room, Te 0.89 1.46 1.72 1.96 1.79 2.08 2.08 1.99 2.07 2.08 2.03 2.04 1.84 1.79 1.82

With Specimen (Fibreglass), Ts 0.73 1.12 1.20 1.35 1.10 1.22 1.25 1.17 1.24 1.30 1.33 1.39 1.37 1.29 1.28

Reverberation Time(s)

1.70

1.90

2.10

2.30

0.50

0.70

0.90

1.10

1.30

1.50

160 200 250 315 400 500 630 800 1000 1250 1600 2000 2500 3150 4000 5000

Empty Room, Te

With Specimen (Fibreglass), Ts



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• Answer:Band (Hz) 160 200 250 315 400 500 630 800 1000 1250 1600 2000 2500 3150 4000

Reverberation Time(s)

0.80

0.90

1.00

Empty Room, Te 0.89 1.46 1.72 1.96 1.79 2.08 2.08 1.99 2.07 2.08 2.03 2.04 1.84 1.79 1.82

With Specimen (Fibreglass), Ts 0.73 1.12 1.20 1.35 1.10 1.22 1.25 1.17 1.24 1.30 1.33 1.39 1.37 1.29 1.28

Sound Absortpion Coefficient 0.66 0.56 0.68 0.62 0.94 0.91 0.86 0.95 0.87 0.77 0.70 0.62 0.50 0.58 0.62

0.40

0.50

0.60

.

160 200 250 315 400 500 630 800 1000 1250 1600 2000 2500 3150 4000 5000

Impedance Tube Measurement

• The normal incidence absorption coefficient is the ratio of energyabsorbed/energy incident, for a plane wave, normally incident on an absorptivesurface. It is easy to determine using a “standing wave tube” (sometimes called

“ ” .

• It uses a small sample (typically 4” diameter) and has limited validity andusefulness due to the small sample size and the dif ference between a truenormal incidence condition, and the actual incidence conditions (nearly random)seen in most real installations. But it is still useful for comparison purposes.

• The diameter of the tube must be smaller than ½ wavelength to insure planewave sound propagation. A 4” tube is good up to about 3300 Hz. For higherfrequencies, a smaller diameter tube is used.



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• Any heavy walled tube may be used for the construction of animpedance tube.

•

the material to be tested should be mounted at the other end.

• A pure tone (or band of noise) is generated using aloudspeaker.

Impedance tube with a movable microphone or probe


• When a tonal sound field is setup in the tube, the incident wave

from the speaker combines with the reflected wave from the endof the tube to form a standin wave. A attern of re ularl

spaced maxima and minima along the tube will result, which isuniquely determined by the driving frequency and the absorption

of the sample at the end of the tube.



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• We experimentally measure the maximum and minimumpressures inside the tube by sliding a microphone along thecenterline, from which we can calculate the normal incidence

absorption coefficient, α n

• L is the difference between the maximum and minimum sound

pressure levels of the standing wave measured in the tube.

2

110

1101

20

20

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

−−=

L

L

nα


• Example:

– The measured difference between adjacent maximum and

Hz. What is the normal incidence absorption coefficient for

the sample at 1000 Hz.

– Since, L = 8 dB, the normal incidence absorption coefficientis given by



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