<p>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 <a href="https://rs.figshare.com/articles/journal_contribution/Supplementary_materials_from_Mosquito_sound_communication_are_male_swarms_loud_enough_to_attract_females_/14354689?backTo=/collections/Supplementary_material_from_Mosquito_sound_communication_are_male_swarms_loud_enough_to_attract_females_/5361985" rel="nofollow noreferrer">study</a> p.11 to remove low-frequency noise in the SPL measurements).</p>
<p>For e.g. 1/3-octave bands, each band filter of central frequency <span class="math-container">$f_c$</span> has a lower limit of <span class="math-container">$2^{-1/6}f_c$</span> and an upper limit of <span class="math-container">$2^{1/6}f_c$</span> . You can add up the sound levels of the different bands of interest with the following <a href="http://www.sengpielaudio.com/calculator-octave.htm" rel="nofollow noreferrer">formulae</a>:</p>
<p><span class="math-container">$$SPL = 10\log_{10}( \sum_{i} 10^{0.1*L_i} )$$</span></p>
<p>with <span class="math-container">$i$</span>  the <span class="math-container">$i$</span>-th frequency band.</p>
<p>Don't forget to report the frequency bands that you included with your results.</p>
<p>If possible, do the same without any communication signals to have a noise floor to compare with.</p>

CC BY-SA 4.0
2022-07-13T15:48:58.927Z
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.
Marcelo Araya-Salas
CC BY-SA 4.0
2022-07-18T14:49:26.390Z
OK. I made a new answer, with the same idea but working from the sound signal instead of different measurements on a meter.
Noil
CC BY-SA 4.0
2022-07-13T10:11:11.133Z
2022-10-19T12:04:04.393Z
2022-10-19T12:04:04.393Z
Noil
<p>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.</p>
<p>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.</p>

CC BY-SA 4.0
2022-07-18T16:20:17.960Z
could you please elaborate on how to treat noise as an error?
Marcelo Araya-Salas
CC BY-SA 4.0
2022-07-13T05:21:57.973Z
2022-07-18T17:28:05.200Z
2022-07-18T17:28:05.200Z
WMXZ
<p>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 <span class="math-container">$L_i$</span> of the different bands <span class="math-container">$i$</span> of interest with the following [formulae][2]:</p>
<p><span class="math-container">$$SPL = 10\log_{10}( \sum_i 10^{0.1L_i} )$$</span></p>
<p>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 <a href="http://www.sengpielaudio.com/calculator-octave.htm" rel="nofollow noreferrer">here</a> (scroll down up to the middle of the page). Octave or 1/3-octave bands are pretty standard.</p>
<p>EDIT:</p>
<p><strong>If you were able to record the sound background without the signal of interest</strong>, you can subtract the measured <strong>energy</strong> of the background from the <strong>energy</strong> of signal+background, which gives in term of sound levels:</p>
<p><span class="math-container">$$SPL_{signal} = 10\log_{10}(  10^{0.1SPL_{signal+background}} -  10^{0.1SPL_{background}}  )$$</span></p>
<p>In any cases, try to report exactly how you got your final SPL value.</p>

CC BY-SA 4.0
2022-07-25T01:20:54.327Z
thanks! what about measuring the SPL of the background noise and then subtract it from the SPL of the signal + noise?
Marcelo Araya-Salas
CC BY-SA 4.0
2022-07-27T09:28:09.877Z
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.
Noil
CC BY-SA 4.0
2022-07-18T14:47:36.327Z
2022-10-19T12:08:43.077Z
2022-10-19T12:08:43.077Z
Noil
<p>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).</p>

<spl><background-noise>
When measuring sound pressure level (SPL) on a signal, how can I exclude SPL coming from background noise?
# When measuring sound pressure level (SPL) on a signal, how can I exclude SPL coming from background noise?

(Asked by: username_0 on 2022-07-12)

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).

**Tags:** spl, background-noise


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## 3 answers

### Answer by username_1 on 2022-07-18. Score: 4 | Accepted answer

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](http://www.sengpielaudio.com/calculator-octave.htm) (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.

> **Comments:**
> **username_0**: thanks! what about measuring the SPL of the background noise and then subtract it from the SPL of the signal + noise?  
> **username_1**: 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.  

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### Answer by username_1 on 2022-07-13. Score: 11

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](https://rs.figshare.com/articles/journal_contribution/Supplementary_materials_from_Mosquito_sound_communication_are_male_swarms_loud_enough_to_attract_females_/14354689?backTo=/collections/Supplementary_material_from_Mosquito_sound_communication_are_male_swarms_loud_enough_to_attract_females_/5361985) 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](http://www.sengpielaudio.com/calculator-octave.htm):

$$SPL = 10\log\_{10}( \sum\_{i} 10^{0.1\*L\_i} )$$

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.

> **Comments:**
> **username_0**: 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.  
> **username_1**: OK. I made a new answer, with the same idea but working from the sound signal instead of different measurements on a meter.  

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### Answer by username_2 on 2022-07-13. Score: 8

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.

> **Comments:**
> **username_0**: could you please elaborate on how to treat noise as an error?  

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