# How to obtain the Sound Pressure Level from .wav file based on these given parameters?

I have a wav file recorded from a hydrophone. I am not quite familiar with signal processing so I really need some advice from you.

I have gathered the information which looks useful and related.

sample rate = 250 000 Hz

Receive sensitivity of hydrophone = -180 db re 1V/μ Pa

Peak-to-peak voltage range = 5.64 V

Pre-amplifier gain = 18 dB (Modified)

Let A be the amplitude saved in the wav file. What is the formula to transform A from amplitude to Sound Pressure Level?

If the given information is not sufficient, please tell me what more do I need?

Any helps are appreciated!

• I don't seem to be able to add a comment to the only answer. I have a question. So the sound pressure for A and -A is the same?. (A is a given amplitude). BTW, thank you for your answer. I almost didn't find one really helpful – Diana Duque Mar 19 '17 at 22:02
• I am new here, so I can't comment but I have two questions: 1. Does that mean that the SPL is the same for A and -A (A being a given amplitude) 2. Shouldn't it be the clipping point at Vp, (2/0.707) instead of Vrms? Thank you – dduque Mar 21 '17 at 14:38
• @DIANACAROLINADUQUEMONTOYA Hi, Diana. This topic is quite long ago to me but I will still try my best to answer. 1) A and -A should give the same dB. 2) I don't know the exact definition of the so called "clipping point". I would say the amplitude A can be converted to Vp or Vrms. The reason to use Vrms in the equation is that the sensitivity is defined as Vrms per Pa. So we have to use Vrms. Feel free to let me know if u have more questions=] – Ken Tsui Mar 22 '17 at 13:33