This seemingly simple question is actually quite fundamental. To answer the question we need to consider sound intensity. Now sound intensity is defined as “*the average rate of flow of energy through a unit area normal to the direction of wave propagation*”. The average rate of flow of energy is energy per second which we recognise as power in Watts. Intensity then has units of Watts per square metre [latex]W/m^2[/latex] as it has dimensions of power transmitted per unit area. We could also define in old fashioned units like [latex]erg/cm^{2}sec[/latex] and so on, but it is better to stay with Watts. Just in case you want a modern reference note that [latex]1 Joule = 10^{7} erg[/latex]

Now the reference for Watts in air was defined as [latex]10^{-12}W[/latex], which at the time was considered as the intensity of a just barely audible 1kHz tone to a normal human ear. This seems a reasonable choice as we are usually dealing with sound which is definitely audible, and often annoyingly so.

Moving on if we consider either plane or spherical acoustic waves in perfect free field conditions, that is a situation where no reflections occur, then it is straight forward to show that the intensity is given by

[latex]I = \frac{p^2}{2{\rho}c}[/latex]

where $p$ is the peak pressure of the sound wave, [latex]\rho[/latex] is the air density and [latex]c[/latex] is the speed of sound in air. Now for a sine wave

[latex]\frac{p^2}{2} = {p_{rms}^2}[/latex]

so we have

[latex]I = \frac{p_{rms}^2}{{\rho}c}[/latex]

If we take the dB level we have

[latex]dB_I = 10log\left(\frac{{p_{rms}^2}/{{\rho}c}}{10^{-12}}\right)[/latex]

Now in air [latex]{\rho}c[/latex] is approximately 400 and we have [latex]10^{-12} = (2*10^{-5})^2/400[/latex]

So we can write

[latex]dB_I = 10log\left(\frac{{p_{rms}^2}/{{\rho}c}}{10^{-12}}\right) = 10log\left(\frac{p_{rms}}{2*10^{-5}}\right)^2 = 20log\left(\frac{p_{rms}}{2*10^{-5}}\right)[/latex]

That is by using 2 x 10^{-5} as the reference we are relating the rms pressure to the barely audible intensity in a free field, or conversely a pressure of 2 x 10^{-5} is the sound level we can just hear at 1kHz

In normal measurements we are not in perfect free field conditions, so the dB level of the pressure is referred to as the Sound Pressure Level, namely

[latex]dB_{SPL} = 20 log \left(\frac{p_{rms}}{2*10^{-5}}\right)[/latex]

If the rms pressure used is over the entire frequency range then we have the **Overall Sound Pressure Level**, often just called the **Overall Level**. If the rms pressure is the output of say a third octave filter then we call it the **dB Band Level** at that third octave centre frequency.

#### Dr Colin Mercer

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