I apologize if this is not really an EE question, but it seemed most appropriate. I am considering the following application.

Object is 3000m far away and emits a sound of 94 dBSPL at 0.5m, I, therefore, expect it to be 18.44dB loud at 3000m.

1) Using a microphone with 80dB S/N ratio and -24dB sensitivity. I should be able to hear it (albeit lost in ambient noise). Correct? (eg https://www.digikey.com/product-detail/en/pui-audio-inc/AOM-5024L-HD-R/668-1596-ND/7898328)

2) My sensors are 0.5m away from each other. if the source of noise is exactly in front of one sensor the time difference of arrival would be around 0.12 uS meaning, that to meet the Nyquist sampling theorem I would need to sample to audio at least 16.5Msps, in order to be able to at least measure the time difference, more, if I would require higher degree of accuracy. (eg. https://www.tme.eu/cz/details/ad9280arsz/prevodniky-a-d-integrovane-obvody/analog-devices/)

Does that sound reasonable? Or am I missing something?

  • \$\begingroup\$ I think you slipped a decimal place or two. 0.5 meters should be around 1.4 milliseconds. \$\endgroup\$ – JRE Feb 17 '20 at 12:29
  • 1
    \$\begingroup\$ @JRE I think you've misunderstood the problem - the 2nd sensor is not in line between or behind the 1st sensor & the object - it's 0.5m to the side. So if the 1st sensor is 3000m away from the object then the 2nd sensor is 3000.00004166m away - leading to a 0.12us difference. \$\endgroup\$ – brhans Feb 17 '20 at 12:38
  • \$\begingroup\$ I think high frequency sampling is not good strategy. I would rather try to use phase detector to measure shift in signal. \$\endgroup\$ – ufok Feb 17 '20 at 13:08
  • \$\begingroup\$ @ufok Yes, detecting phase shift is what I am hoping to do. I thought that high frequency sampling and correlation is the way to go, am I wrong? \$\endgroup\$ – Dvorkam Feb 17 '20 at 13:53
  • \$\begingroup\$ You don't need to set the sampling rate according to the time difference but rather to twice the bandwidth of your signal. If this is sound in air, sampling much above 1 MHz is pointless given the attenuation of air at those frequencies. \$\endgroup\$ – user1850479 Feb 17 '20 at 14:35

Does that sound reasonable? Or am I missing something?

Theoretically SPL halves with a doubling of distance so, the extra attenuation over the distance range of 0.5 metres to 3000 metres becomes: -

$$ 20\cdot log_{10}(6000)$$

And this is 75.56 dB hence, an SPL of 94 dB (at 0.5 m) reduces to 18.44 dB (at 3 km). This assumes a perfect scenario with no obstacles and reflections.

Using a microphone with 80dB S/N ratio and -24dB sensitivity

The sensitivity figure is in fact -24 dBV/Pa at 1 kHz. And, this means that for 1 Pa of sound pressure you get an RMS voltage from the microphone of 63.1 mV at 1 kHz.

If the received SPL is only 18.44 dB then this is a pressure of 0.000167 Pa and it takes the microphone output voltage down to 10.5 uV RMS at 1 kHz.

So your challenge is to design an amplifier that can pick out the 10.5 uV RMS signal and not introduce too much noise. The microphone chosen doesn't help that much because you are working close to its own SNR.

My sensors are 0.5m away from each other. if the source of noise is exactly in front of one sensor the time difference of arrival would be around 0.12 uS

If you are not sending a continuous sinewave of 1 kHz then this affects how the microphone works and you might need to do tests. For instance, if you are emitting a pulse of sound from the source, due to the limited bandwidth of the 2 microphones, your received voltage signal will be "smudged" significantly.

This may mean that your sampling speed might be overkill because you'll never adequately be able to determine the difference between the two microphones placed 0.5 metres apart at 300 m distant. You also need to consider that the 2 microphones will not be matched so the smudging of one could be significantly different to the smudging of the other.

If you model the microphone as a 2nd order low pass filter and then compared one with a 3 dB point at 22.5 kHz with the other with a 3 dB point at 20.5 kHz, the time difference between them might be about 1 us for exactly the same input stimulus: -

enter image description here

Captured images from here.

Given that you are trying to measure a time difference of only 120 ns, you might expect problems that require some form of test calibration method.

  • \$\begingroup\$ That brings up some issues I haven't even thought about. Thank you for the elaborated answer. \$\endgroup\$ – Dvorkam Feb 17 '20 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.