# Sound - Attenuation and the Directivity Coefficient

## The attenuation in a room depends on the location of the sound source and the receiver - and the room constant.

For a continuing sound source the sound level in a room is the sum of direct and reverberant sound. The sound pressure for a receiver can be expressed as

L_{p}= L_{N}+ log (D / (4 π r^{2}) + 4 / R)(1)

where

L_{p}= received sound pressure level (dB)

L_{N}= sound power level from source (dB)

D= directivity coefficient

R= room constant (m^{2}Sabine)

π= 3.14 .....

r= distance from source (m)

### Directivity coefficient - *D*

The figure below can be used to approximate the** Directivity coefficient - D **- for typical locations of the receiver and the sound source

*(1)* can be transformed to express the difference between the received sound pressure level and emitted sound power level - the attenuation - as:

L_{p}- L_{N}= 10 log (D / (4 π r^{2}) + 4 / R)(2)

### Room Sound Attenuation Calculator

D - directivity coefficient

R -room constant- (m^{2}Sabine)

r - distance from source (m)Room Sound Attenuation

(L:_{p}- L_{N}) - (dB)

The attenuation *(L _{p} - L_{N}*) can also be estimated from the diagram below by modifying

*(2)*to be expressed as the ratio between "distance

*r*between source and receiver" and "square of the directivity coefficient

*D"*- and the total room sound absorption in

*m*

^{2}Sabine.#### Example - Sound Attenuation in a Room

For a room the acoustic property total room sound absorption is *1000 m ^{2} Sabine*. For a listener in the middle of the room with a directivity constant

*1*and at distance

*1 m*from the sound source - the ratio

*r / D ^{1/2} = 1 / (1 m)^{1/2}*

* = 1*

From the diagram above the attenuation can estimated to approximately

*10 dB*