I am measuring SPL of playback bird songs as the RMS of amplitude envelopes. SPL is being measured on the waveform, not using a sound level meter. I want to compare SPL for sounds within the same recording. They were played from a speaker at a fix distance so comparing SPL should still be meaningful. However, for some signals I think that most SPL comes from background noise. Background noise is pretty constant across the recording. I didn't use a sound level meter as the one I have is not fast enough to measure SPL for very short signals (~100 ms).
3 Answers
You could remove the noise present in the frequency bands where there is no signal, instead of computing the sound level over the full frequency range. You can add up the sound levels $L_i$ of the different bands $i$ of interest with the following [formulae][2]:
$$SPL = 10\log_{10}( \sum_i 10^{0.1L_i} )$$
To compute the lower and higher cut-off frequencies of your i-th bandpass filter, it is better to use standard bands (EN ISO 266) as defined here (scroll down up to the middle of the page). Octave or 1/3-octave bands are pretty standard.
EDIT:
If you were able to record the sound background without the signal of interest, you can subtract the measured energy of the background from the energy of signal+background, which gives in term of sound levels:
$$SPL_{signal} = 10\log_{10}( 10^{0.1SPL_{signal+background}} - 10^{0.1SPL_{background}} )$$
In any cases, try to report exactly how you got your final SPL value.
-
$\begingroup$ thanks! what about measuring the SPL of the background noise and then subtract it from the SPL of the signal + noise? $\endgroup$ Commented Jul 25, 2022 at 1:20
-
1$\begingroup$ If the background noise is independent of your relevant signal, you cannot subtract the background from the background+signal. You can only subtract/add energies (~ pressure^2). I edited my post to take into account the case where you was able to record the background noise without the signal of interest. $\endgroup$– NoilCommented Jul 27, 2022 at 9:28
If you have a sound level meter, it is possible to measure the SPL per band (e.g. third-octave band) instead of measuring over the full frequency range of the meter microphone. This will not remove all the noise, but depending of your noise, this can remove good amount of noise easily from the frequency bands where you don't have any signal of interest (as done in e.g. this study p.11 to remove low-frequency noise in the SPL measurements).
For e.g. 1/3-octave bands, each band filter of central frequency $f_c$ has a lower limit of $2^{-1/6}f_c$ and an upper limit of $2^{1/6}f_c$ . You can add up the sound levels of the different bands of interest with the following formulae:
$$SPL = 10\log_{10}( \sum_{i} 10^{0.1*L_i} )$$
with $i$ the $i$-th frequency band.
Don't forget to report the frequency bands that you included with your results.
If possible, do the same without any communication signals to have a noise floor to compare with.
-
$\begingroup$ Thanks! I want to compare SPL for sounds within the same recording. They were played from a speaker at a fix distance so comparing SPL should still be meaningful. However, for some signals I think that most SPL comes from background noise. Background noise is pretty constant across the recording. I didn't use a sound level meter as the one I have is not fast enough to measure very short signals (~100 ms). I will add this to my question. $\endgroup$ Commented Jul 13, 2022 at 15:48
-
$\begingroup$ OK. I made a new answer, with the same idea but working from the sound signal instead of different measurements on a meter. $\endgroup$– NoilCommented Jul 18, 2022 at 14:49
While you could simply subtract the noise intensity from the signals intensity, I would not do it. Noise is always measured at different times than the signal. IMO, a better way is to consider the noise as an error in your measurements and treat it like that.
Edit: Question is how to consider noise are measurement error? Let the noise estimate (without signal) be N_rms measured in uPa then it can be considered as STD of your measurements. So, whatever you measure as signal S in uPa can be expressed as S +/- N_rms uPa.
-
$\begingroup$ could you please elaborate on how to treat noise as an error? $\endgroup$ Commented Jul 18, 2022 at 16:20