3
$\begingroup$

Why is all pole model pretty useful in modelling room acoustics?

Is it related to reverberation?

$\endgroup$
2
  • 1
    $\begingroup$ Source please? Why do you think an all pole filter is useful in room acoustics? $\endgroup$
    – Hilmar
    Commented Nov 21, 2016 at 15:27
  • $\begingroup$ Perhaps you mean to say that an all-pole filter is useful for correcting room acoustics? $\endgroup$ Commented Nov 21, 2016 at 15:47

1 Answer 1

3
$\begingroup$

It's not because of reverberation.

When you want to model the Frequency Response of the room, it's common to simplify your approximation by using either all-pole or all-zero models. You don't want to use the full zero-pole model.

To get some intuition:

  • zeros correspond to time delays and antiresonances
  • poles correspond to resonances of your Room Response

In practice all-zero models are not being used due to various reasons, such as:

  • required filter length is comparable to the IR length, and almost 40x the length of corresponding all-pole filter
  • filter will be valid only for specific distances and positions between the source and receiver (remember: time-delays).

That is why the all-pole model is used instead. As mentioned above, poles correspond to the resonances, i.e. standing waves, which are:

  • independent of the source location (quite intuitive)
  • independent of the receiver location (except of the nodes)

Additionally the required filter length is way less than in case of all-zero models. According to Mourjopoulos, for $RT_{60}\approx0.5 \mathrm{s}$, the required order is within the range of $50 < N < 500$. The same author, concludes that all-pole filters are easier to manipulate than all-zero filters, due to their filter length. Morevover, author mentioned that all-pole filters are sufficient approximation than using raw impulse response data.


Here is some literature:

Mourjopoulos J. - On the Variation and Invertibility of Room Impulse Response Functions

Mourjopoulos J., Paraskevas M. A. - Pole and Zero Modelling of Room Transfer Functions

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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