Can anyone explain how do I manually find F0, F1, F2, etc on these images?
Thank you
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There are others on here far more expert than me, but FWIW (1) it's much easier to identify the formants from a spectrogram (2) the centres of the dark bars you would see on a spectrogram correspond to the x-axis values of the peaks on your graphs (3) peaks get broken up by interference, so a cluster of peaks can actually represent a single formant. For this reason you really need multiple tokens, or some idea of what the formants are going to be. (4) F3 is not going to be above say 4kHz, so that helps narrow it down. (5) B, D and E do not look like vowels.– rchiversCommented Jun 18, 2020 at 14:02
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or rather, if they are vowels then you're going to have to ignore a lot of the tiny peaks and look at the underlying shape - again, difficult to do without other tokens for comparison and/or an idea of what the formants are going to be.– rchiversCommented Jun 18, 2020 at 14:22
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1 Answer
In the LPC spectrum in C you are looking for the first 3 peaks, which look to be about at 750, 1800 and 3000. The corresponding DFT spectrum in D gives you those same peaks (though the computed frequency will be somewhat different), but the problem is that you have to apply some complicated notions about bandwidth to figuring out which peak is the real peak (each harmonic is itself a peak, you only want the peak of peaks, those peaks that are stand out from other peaks). It is useful to know that the the fundamental is often "extra-prominent", which may keep you from identifying the 1st and 3rd harmonics in D as F1 and F2. A and B are more challenging in that you get 4 peaklets in A below 3000, but not 4 formants. You have to decide which ones to ignore: I am guessing that this is a vowel like [u]. Guessing formants from spectra a made easier if you have an expectation, which would tell you that 300, 800, 2700 is not appropriate for [a], which could lead you to pick different local maxima. This kind of peak-reading of spectra is pretty inaccurate in its relationship to the transfer function, which is what people are really interested in.
