10.2.2 Room Constant

The absorption coefficient is defined as the ratio of acoustic energy absorbed by a surface to the acoustical energy incident upon the surface when the incident sound field is perfectly diffused:

α wi wr wi

Half Cube

1

5

The summation of all the reflected ratios of a sound, except the direct wave, is equal to ð1 ααÞ as:

ð1 αÞ þ ð1 αÞ2 þ . . . þ ð1 αÞn ¼ ð1 αÞ ; n ¼ 1

α

The following is the proof of the formula above of the summation of the infinite series.

Let:

RS ¼ ð1 αÞ þ ð1 αÞ2 þ . . . þ ð1 αÞn

Multiplying both sides of the above equation by ð1 αÞ yields:

RS ð1 αÞ ¼ ð1 αÞ2 þ ð1 αÞ3 þ . . . þ ð1 αÞnþ1

Subtract these two equations to arrive at:

RS α ¼ ð1 αÞ ð1 αÞnþ1

For n ¼ 1 and ð1 αÞ < 1, ð1 αÞnþ1 ¼ 0, we obtain:

The variable R in the equation above is known as the room constant. The room constant R is related to the total area of a room S and the total reverberant ratio and is defined as:

energies to the initial incident energy for the first, second, third, etc. reflections.

10.2.3 Reverberation Time

The reverberation time T60 is the time required for reflections of a direct sound to decay to 60 dB:

60

When the sound pressure level LP reaches the peak:

After the sound pressure level LP decreases 60 dB from the peak:

Therefore:

LP,T0 LP,T60 ¼ 60 ½dB&

!wT60 ¼ 10 6

wT0

For each reflection from the boundary surface, the ratio of the reflected energy to the incident energy is ð1 αÞ, and the average travel time between reflections is approximately 4ScV (base on a half cube):

Traveling Time between ceiling and loor

5

10

=

Assume that the number of reflections required for the reflected energy to decay

to60 dB is T60= 4ScV .

Therefore:

ð1 αÞ4ScVT60 ¼ 10 6

Taking the natural logarithm of the above equation yields:

Sc

4V T60 ln ð1 αÞ ¼ 6 ln ð10Þ

or (use c ¼ 343 m/s):

The equation above is the general formula for calculating T60 for an arbitrary absorption coefficient α.

Special Case: Small Absorption Coefficient

For a small α ðα < 0:2Þ, the formula above can be simplified as:

1

ð1 αÞ α e

Hence:

Again, taking the natural logarithm of the above equation yields:

Sc

α 4V T60 ¼ 6 ln ð10Þ

Hence:

Example 10.1

A (half cube) room has the dimensions 10 10 5 m.

(a)If the absorption coefficient is α ¼ 0:5, determine the reverberation time T60.

(b)If the reverberation time is T60 ¼ 0.32 s, determine the average absorption coefficient α:

Example 10.1: Solution

(a) For the reverberation time:

V ¼ 103=2 ¼ 500 m3

S ¼ 102 4 ¼ 400 m2 ð 4V=S ¼ 10=2 ¼ 5 mÞ

(b) For the average absorption coefficient:

V ¼ 103=2 ¼ 500 m3

S ¼ 102 4 ¼ 400 m2 ð 4V=S ¼ 10=2 ¼ 5 mÞ