## Converting hydrophone voltage time series into pressure time series

Started by 6 years ago●9 replies●latest reply 6 years ago●1535 viewsHi all

I am trying to record underwater noise in terms of pressure(upa) time series using voltage output hydrophone having receive sensitivity of -196 dB ref vrms/upa. I have some confusion how to convert hydrophone voltage time series into sound pressure time series using Rx sensitivity?

In frequency domain, i think i can convert easily by using FFT output of hydrophone voltage time series. I calculate absolute of each complex output frequency bin and multiply this absolute value by 1.414 to get Vrms for each frequency bin. Then taking 20log of this vrms and adding it with Rx sensitivity, i get sound pressure in terms of dB ref upa.

Question is how to get sound pressure in terms of time series using Rx sensitivity?

Thanks & Regards

Assuming that the transducer is specified at its center frequency then that's where the -196 dB ref vrms/upa applies. Other frequency outputs would have to be discounted according the the transducer frequency response. So, if the input frequency is 3dB down from the peak/center frequency response then the sensitivity would be -196-3=-199 dB ref vrms/upa and so forth....

Just sticking with the center frequency, 1 upa will yield -196dBv and you go from there.

I'd be a little careful about the FFT process you describe. It appears that you are assuming that the bin magnitude outputs are measured in volts. How is that justified?

Also, to convert from vrms to v-peak one would multiply by 1.414 which seems the opposite of what you're proposing. But then, I don't know how to interpret your FFT outputs anyway.

The transducer sensitivity will deal with any voltage/pressure measure - so peak, rms, etc. don't matter. You can deal with that once the rest of the process is solid.

Thanks for help.

The transducer sensitivity is almost flat across desired frequencies.

The confusion is about that 1upa will yield "-196 dBVrms" as literature says and what i have at hydrophone output is instantaneous voltage at a time instant. Should i ignore Vrms notation in sensitivity and use the relation 1upa ==>0.16 nV instead of 0.16 nVrms?

As far as FFT output is concerned, i have verified it using MATLAB simulation. I generated a time domain sinusoidal tone with some amplitude (of some frequency) and using 1.414*abs(Complex FFT_Out@desired freq bin), i always get value equivalent to Vrms of signal i generated

Best Regards

vrms or vpeak are just observation/measurement choices and have nothing to do with sensitivity, amplification, etc. as long as all those parts are *linear*, e.g. don't saturate. And, of course we are talking in terms of pure sinusoids if we are going to use a conversion of rms to peak of 1.414.

So, if you make an instantaneous measurement and don't believe that vrms is an appropriate measure, then don't use it. The transducer doesn't know any better and its sensitivity doesn't change by virtue of something that you arbitrarily chose.

So, transducer sensitivity expressed in vrms / upa is an expression of the sensitivity to a purely sinusoidal input. It raises the question: "Is the sound pressure expressed in uparms as well?" Likely so. That would be common usage.

So, consider this:

-196 vrms per 1 uparms yield -196 v/upa which is dimensionless regarding peak, rms, etc. just v/upa. But, because transducers tend to be resonant devices, it would not be a good idea to try to measure the impulse response, eh? A steady-state sinusoid (or pulse of reasonable length) is by far the most common measurement approach.

This is kind of a reiteration of what Fred's saying, but here's a step-by-step.

First, decide whether you want to pretend that the frequency response is flat, correct for the frequency response before you convert to pressure, or correct for the frequency response afterward. None of these is the right answer for everything -- it depends on your problem and your inclinations. Second, you'll need to decide if you want to correct for the frequency response in the time domain, or in the frequency domain. Again, neither is the right answer for all cases. Finally, correcting for frequency response by more than a few dB can get *very* inaccurate -- if you can get the data you need while pretending that the frequency response is flat, do so.

Putting all that aside, here's how to do it assuming a flat response:

- Find the gain of your ADC, in counts/volt. Call it \(A_{ADC}\).
- Find the gain of the amplifier from the hydrophone to the ADC, probably in volts/volt. Call it \(A_A\).
- Calculate a correction gain to apply to the raw numbers coming out of the ADC. Define \(A_c = \frac{1}{A_{ADC} A_A}\).
- Calculate the voltage from the ADC numbers: \(v_p(t) = A_c x_{adc}\).
- You're given the gain of the hydrophone. Figure out what "dB ref" means, and then calculate the gain as a multiplier (rather than something in dB). Call it \(A_p\). Now the estimated pressure at the hydrophone is equal to \(\hat p_p(t) = \frac{v_p(t)}{A_p}\).

I leave any compensation of the phone's frequency response for other posts, because doing such a thing correctly can get *very* complicated if you need more than a few dB of correction, and it can get *very* inaccurate if you try too hard.

Thanks Tim Wescott for your help.

The sensitivity is flat in desired frequency band. The actual sensitivity was -206 dB ref upa and i have included 10 dB gain of preamplifier (which also has flat gain in desired frequency band) in it.

I am using National Instruments DAQ cards at preamplifier end which gives me output in terms of voltage time series.

The confusion is that sensitivity is in terms of dBVrms i.e. 1upa will generate -196 dBVrms and what i have at hydrophone output is "instantaneous voltage" at a time instant.