Acoustic localization
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: -



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.