2
$\begingroup$

I went through the Matlab tutorial on Formant Estimation using LPC Coefficients. Though I vaguely understand the details, it's not entirely clear why we need to do this. From http://person2.sol.lu.se/SidneyWood/praate/whatform.html:

A formant is a concentration of acoustic energy around a particular frequency in the speech wave

Why is it not enough to take the DFT of the audio signal (after some pre-processing if necessary)? In the frequency-domain, the peaks correspond to these concentrations, correct?

$\endgroup$
  • $\begingroup$ the peaks (or line spectra) are where sinusoidal components go. the formants are like an envelope to those spectral lines in the frequency domain. not the same thing. $\endgroup$ – robert bristow-johnson Oct 23 '16 at 1:30
3
$\begingroup$

The frequency resolution of DFT is limited to the number of time samples. On the other hand proper LPC can have high resolution.

$\endgroup$
  • $\begingroup$ But would the resolution not fade given more predictions? $\endgroup$ – Bob Burt Oct 3 '17 at 12:45
  • $\begingroup$ @BobBurt Really do not understand what do you mean. what does more prediction means: More data? If so: we are assuming model is correct so more data means better only as data follows the model. $\endgroup$ – Creator Oct 3 '17 at 20:34
0
$\begingroup$

DFT is limited to a large number of samples that are an even power of 2. LPCs can be abstracted locally from a small number of samples, even the order M + 3 or so [so for a 12th-order filter could get by with 15 samples], and are not restricted to N being a power of 2. That said, the general idea is pretty much correct; you can think of the formants as being fat peaks in the frequency domain if it helps you to visualize what's going on.

$\endgroup$

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.