# Conversion of power estimation in Audacity to real Sound Pressure Level in ear phones

Axel Hutt

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## To cite this version:

Axel Hutt. Conversion of power estimation in Audacity to real Sound Pressure Level in ear phones. INRIA Nancy Grand Est. 2023. hal-04177151

HAL Id: hal-04177151

https://inria.hal.science/hal-04177151

Submitted on 4 Aug 2023

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# Conversion of power estimation in Audacity to real Sound Pressure Level in ear phones

A. Hutt,

Team MIMESIS, INRIA Nancy Grand Est, France

August 4, 2023

In Audacity (version 3.3.3), it is possible to tune the amplitude of the output signal. In the menu under *Effect.../Volume and Compression/Amplify*, the signal amplitude can be defined by the root mean squared error (RMSE) of the signal. However, Audacity does not provide this value **rmse**, but gives this value in decibel (dB) with respect to an internal reference value $p_o$. Let us call this amplitude value (in dB) as amplitude $p_{Au}$, which then relates to the real RMSE of the output by

$$p_{Au}^{1} = 20 \log \left( \frac{\text{rmse}^{1}}{p_{o}} \right) \qquad (1)$$

$$p_{Au}^{ref} = 20 \log \left( \frac{\text{rmse}^{ref}}{p_{o}} \right) \qquad (2)$$

Here **rmse**¹ is the RMSE of the signal under study and **rmse**¹*ref* is the RMSE of the signal at the audition threshold, i.e. the audible signal with the lowest intensity. We note that then Eqs. (1), (2) yield

$$\text{rmse}^{1} = p_{o}10^{\frac{p_{Au}^{1}}{20}} \qquad (3)$$

$$\text{rmse}^{ref} = p_{o}10^{\frac{p_{Au}^{ref}}{20}}. \qquad (4)$$

The sound pressure level $p_{ear}$ generated in the ear phones can be supposed (in a good approximation) to be proportional to the amplitude in the
generated signal, i.e. the RMSE, and hence<fnref n="1" />

$$p_{\text{ear}}^{1} = a \cdot \text{rmse}^{1} \qquad (5)$$

$$p_{\text{ear}}^{\text{ref}} = a \cdot \text{rmse}^{\text{ref}} \qquad (6)$$

where $a \neq 0$ is any proportionality factor capturing the properties of the devices and cables between the Audacity output signal and the ear-phone signal.

The Sound Pressure Level (SPL) is defined as

$$SPL = 20 \log \left( \frac{p_{\text{ear}}^{1}}{p_{\text{ear}}^{\text{ref}}} \right) \text{dB}, \qquad (7)$$

where $p_{\text{ear}}^{1}$ and $p_{\text{ear}}^{\text{ref}}$ is the sound pressure of the heard signal under study and the sound pressure of the heard reference signal with the lowest intensity, respectively. Then inserting Eqs. (3), (4), (5) and (6) into Eq. (7) yields

$$ 
\begin{aligned}
SPL &= 20 \log \left( \frac{\text{armse}^1}{\text{armse}^{\text{ref}}} \right) \text{dB} \\
&= 20 \log \left( \frac{\text{rmse}^1}{\text{rmse}^{\text{ref}}} \right) \text{dB} \\
&= 20 \log \left( \frac{p_o 10^{\frac{p_{Au}^1}{20}}}{p_o 10^{\frac{p_{Au}^{\text{ref}}}{20}}} \right) \text{dB} \\
&= 20 \log \left( 10^{\left(\frac{p_{Au}^1}{p_{Au}^{\text{ref}}}\right)/20} \right) \text{dB} \\
&= 20(p_{Au}^1 - p_{Au}^{\text{ref}})/20 \text{dB} \\
SPL &= (p_{Au}^1 - p_{Au}^{\text{ref}}) \text{dB.}
\end{aligned}
$$

For instance, for a single subject $p_{Au}^1 = -18.177$ and $p_{Au}^{\text{ref}} = -93.177$ yielding to SPL = 75dB.

<footnote n="1">¹Bies, David A., and Hansen, Colin. (2003). Engineering Noise Control.</footnote>