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What is the definition of far-field and near-field in acoustic sensor / microphone array? Does it depend on the distance or the power of the sound?

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What is the definition of far-field and near-field in acoustic sensor / microphone array?

The voltage that the microphone (or array) produces is a 3 dimensional function of the spatial location of the source which is typical expressed in spherical co-ordinates, i.e.

$$ V = f(r,\phi,\theta) $$

You are in the "far field" if the dependency on distance can be approximated by that of a spherical wave, i.e.

$$ V(r,\phi,\theta) \approx V(r_0,\phi,\theta)\cdot \frac{r_0}{r} e^{-j k(r_0-r)}$$

where k is the wave number. In other words: it can be treated as a 2-dimensonal spatial problem instead of a 3-dimensional one.

Does it depend on the distance or the power of the sound?

It does depend on a lot of things, but power is NOT one of them. It depends on

  1. Wavelength of the sound
  2. Distance between source and receiver
  3. Physical dimensions of the array
  4. Physical dimensions of the sensor it self (if it has a large diaphragm)
  5. Type & directivity of the microphone (first or second order)
  6. Directivity and physical extension of the source.
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  • $\begingroup$ Notice that 1. is often the most problematic there: audible acoustic waves can have huge wavelengths (40 Hz at 300 m/s is about 8 m wavelength). $\endgroup$ – Marcus Müller Jun 6 at 15:35
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    $\begingroup$ Hence building a microphone array with high directivity at 40 Hz requires some serious real estate :-) $\endgroup$ – Hilmar Jun 6 at 16:08
  • $\begingroup$ Nice answer. But I think power is one of the determinants. Only under small amplitude assumption the sound wave equation can be linearlized. Once the nonlinearity can't be omitted, the spherical wave solution no longer satisfies the wave equation and thus it shouldn't be regarded as far field. $\endgroup$ – ZR Han Jun 7 at 4:46
  • $\begingroup$ @ZRHan I think once we stop assuming linearity, the terms "near and far field" have no meaning anymore, I'd agree. But that's just outside of the model, I'd say. I'd agree, for acoustics, nonlinearity is an important problem, but when looking at literature that divides the world into near and far field, we can't really have nonlinearities. $\endgroup$ – Marcus Müller Jun 7 at 7:46
  • $\begingroup$ @MarcusMüller When the sound intensity is large enough to cause nonlinearities, what if i keep the receiver far away from the source? This should make it linear again and then get a "far field". (Although it seems go back to the distance, but power is still one of the main characters) $\endgroup$ – ZR Han Jun 7 at 8:03

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