<p>We have tried to walk through the process in this paper:</p>
<p><a href="https://marinebioacoustics.files.wordpress.com/2018/04/madsen_et_al_2007a.pdf" rel="nofollow noreferrer">https://marinebioacoustics.files.wordpress.com/2018/04/madsen_et_al_2007a.pdf</a></p>

CC BY-SA 4.0
2022-07-05T14:26:08.007Z
Hi @Marinebioacoustics Aarhus Univ, the linked paper looks extremely relevant. Could you please provide a brief description of the process in the answer itself. Hyperlink only answers are discouraged in Stack Exchange (https://meta.stackexchange.com/questions/156941/one-line-answers).
Thejasvi
CC BY-SA 4.0
2022-07-05T11:08:51.067Z
2022-07-05T11:08:51.067Z
Marinebioacoustics Aarhus Univ
<p>Yes, if you know the hydrophone and recorder calibrations, you can calculate the levels in the water and they should come out pretty much the same whatever the individual hydrophone calibrations are.
The level in your recorder is the level in the water + the hydrophone sensitivity + any other gains + the recorder sensitivity (assuming all values in dB). Personally, I try to work out the recorder sensitivity in units of dB re 'counts per volt', i.e. what numbers do you get in python, Matlab, or whatever software you use, for a 1V input. Then, you calculate the dB level in counts (which you've more or less done, but you should leave out the 1e-6 to get an rms level in dB re 1 count), then your level in water will be that value - the recorder sensitivity - any other gains - the hydrophone sensitivity.</p>

CC BY-SA 4.0
2022-06-25T10:36:34.100Z
the '-' in the last sentence mean 'minus'
WMXZ
CC BY-SA 4.0
2022-06-22T20:37:17.167Z
2022-06-22T20:37:17.167Z
Doug Gillespie
<p>I have two hydrophones to record the acoustic landscape of my study area.
The two hydrophones have different sensitivities so I get different dB values for the same acoustic events.
To get the energy I use Python :</p>
<pre><code>import numpy as np
10*np.log10(np.mean(np.square(signal))/1e-6)
</code></pre>
<p>Is there a way to use the sensitivity of the hydrophone to get comparable dB Re 1 microPa values ?</p>

<hydrophone><db>
Comparing energy levels from two different hydrophones
# Comparing energy levels from two different hydrophones

(Asked by: username_0 on 2022-06-22)

I have two hydrophones to record the acoustic landscape of my study area.
The two hydrophones have different sensitivities so I get different dB values for the same acoustic events.
To get the energy I use Python :

```
import numpy as np
10*np.log10(np.mean(np.square(signal))/1e-6)

```

Is there a way to use the sensitivity of the hydrophone to get comparable dB Re 1 microPa values ?

**Tags:** hydrophone, db


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## 2 answers

### Answer by username_1 on 2022-07-05. Score: 2 | Accepted answer

We have tried to walk through the process in this paper:

<https://username_1.files.wordpress.com/2018/04/madsen_et_al_2007a.pdf>

> **Comments:**
> **username_2**: Hi @username_1, the linked paper looks extremely relevant. Could you please provide a brief description of the process in the answer itself. Hyperlink only answers are discouraged in Stack Exchange (https://meta.stackexchange.com/questions/156941/one-line-answers).  

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### Answer by username_3 on 2022-06-22. Score: 8

Yes, if you know the hydrophone and recorder calibrations, you can calculate the levels in the water and they should come out pretty much the same whatever the individual hydrophone calibrations are.
The level in your recorder is the level in the water + the hydrophone sensitivity + any other gains + the recorder sensitivity (assuming all values in dB). Personally, I try to work out the recorder sensitivity in units of dB re 'counts per volt', i.e. what numbers do you get in python, Matlab, or whatever software you use, for a 1V input. Then, you calculate the dB level in counts (which you've more or less done, but you should leave out the 1e-6 to get an rms level in dB re 1 count), then your level in water will be that value - the recorder sensitivity - any other gains - the hydrophone sensitivity.

> **Comments:**
> **username_4**: the '-' in the last sentence mean 'minus'  

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