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!

  • $\begingroup$ 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 $\endgroup$ – Diana Duque Mar 19 '17 at 22:02
  • $\begingroup$ 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 $\endgroup$ – dduque Mar 21 '17 at 14:38
  • $\begingroup$ @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=] $\endgroup$ – Ken Tsui Mar 22 '17 at 13:33

The hydrophone has -180 dB 1V/uPa or -60 dB V/Pa. 1 Pa produces hence 1mV or, other way around, the sensitivity is 1mV/Pa. Peak voltage is 5.64 peak to peak. That's 2.82V peak or 2V RMS.

A full scale sine wave has 2 V RMS which is 2000 Pa. dBSPL is referenced against 20uPa so 2000 Pa is equivalent to 160 dBSPL.

The clipping point of your wave file corresponds to 160 dB SPL. The exact relationship between number and dBSPL depends on the format of the wave file (fixed vs. floating, number of bits). For a 16 bit fixed point format, the clipping point corresponds to the number 32768.

  • $\begingroup$ Assume my wav file is 16 bit fixed point format. Let A0 be the amplitude at certain time. Then the SPL of A0 is, SPL=20 log(A0 *2Vrms/32768/1mV/20μPa) Is it correct? Moreover, I found that the hydrophone has a pre-amplifier gain of 18dB. How can I modify the above equation to incorporate this effect? $\endgroup$ – Ken Tsui Dec 24 '13 at 2:45
  • $\begingroup$ Sounds about right. Just subtract the 18 dB. Basically your sensitivity after the pre-amp is now 8mv/Pa. Clipping point is 2V/8mv*1Pa = 250 Pa or 142 dBSPL. So it's SPL=20 log10(A0 /32768*2V/(8mV/Pa)/20μPa) $\endgroup$ – Hilmar Dec 24 '13 at 13:34
  • $\begingroup$ Shouldn't the limit number be 32767 instead of 32768? $\endgroup$ – user19954 Mar 9 '16 at 19:24
  • $\begingroup$ @user19954 : Please do NOT leave comments as answers. Please answer / ask questions and get some rep before commenting. $\endgroup$ – Peter K. Mar 9 '16 at 20:13
  • $\begingroup$ @user19954 : 16 bit fixed point goes from -32768 to +32767, so the largest absolute value you can see 32768. In practice it makes very little difference unless you have a super high precision measurement $\endgroup$ – Hilmar Mar 10 '16 at 20:12

protected by jojek Mar 21 '17 at 17:25

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