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I have a block of audio and I have performed an FFT on it. Now what I want to do is convert this FFT into a set of MFCCs, however while I know I need to do something with weighting the FFT samples based upon a set f triangular filters I haven't really got a clue what this involves.

Can anyone go into some good detail on it? If you are going to give me a matlab example please explain each step well as matlab allows you to cut quite a few corners and I'm trying to implement MFCC extraction on iPhone.

Goz
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1 Answers1

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There's a lot of literature on MFCCs on the web, so it would be a bit easier if you could be more specific as to which part of the processing you don't understand. But I'll give an overview of what needs to be done, hoping this is helpful for you:

  1. compute the squared magnitudes of the FFT bins
  2. weigh the bins using triangular windows; usually the windows are chosen such that the centers of the triangles are equidistant on a mel-frequency scale, and such that each triangle begins and ends at the centers of the two adjacent triangles. The mel-frequency scale is defined by $$m=2595\log_{10}(1+f/700)$$ where $f$ is the frequency in Hz. Look at the figure to see how it works: enter image description here Note that the kernels are normalized such that the sum of the weights per triangle equals 1. Usually around 20 such triangular windows are used.
  3. Take the logarithm of the weighted coefficients.
  4. Compute the DCT.
Matt L.
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  • Cheers Matt, I have read all this but where I fall down is "weigh the bins using triangular windows" I haven't got a clue exactly what this means. Should I be taking a set of bins and then weighting each bin dependent on this triangular "window" and then adding them all together and calling that a "bin"? (This is m best guess). If so, how do I calculate the triangular window? The mel frequency equation you posted ... I don't get how multiple frequencies can relate to one mel "bin" via that ... Any further information on this specific step would be hugely appreciated! :) – Goz May 10 '13 at 21:31
  • You're right about the windows: you compute weighted sums of your FFT bins. The weights are the individual values of the triangular functions. The mel scale is just for centering the triangles. Choose e.g. 20 equidistant points on the mel scale and transform back to Hertz via the formula to get the center frequencies in Hz. Of course you need to round the result to match the values to valid FFT bins. – Matt L. May 10 '13 at 21:34
  • Cheers the centering the bins part makes sense. But how do I triangularly (bad choice of word but its great to try and say ;)) weight the windows? ie How wide should they be? Also why triangular and not some other window type? Gaussian perhaps? – Goz May 10 '13 at 21:36
  • As another aside (And potentially should be asked in another questions) why do you calculate MFCCs (ignoring the weighting) so differently to a cepstrum? ie why the DCT and not inverse FFT? – Goz May 10 '13 at 21:39
  • Well, you don't weigh the windows but the FFT bins. Simple example: triangle with values 1,2,1, so your weighted output for this triangle would be 1(first FFT bin) + 2(second FFT bin) + 3*(third FFT bin). Why triangular is a good question, that's just the standard way of doing it. Try others if you like, but I doubt you'll get considerably better results. – Matt L. May 10 '13 at 21:40
  • The DCT is better at decorrelating the input values. Decorrelation is good for further processing. – Matt L. May 10 '13 at 21:41
  • One final question on the weighting, you say the whole 1 * first + 2 * second etc ... but how do i choose those weights really? Should I assume the total weighting is 1? ie If, for a given bin, I have 3 FFT bins to read should the first be 0.25, 2nd 0.5 and 3rd 0.25 (ie the weights add up to 1)? Your graph above does imply that as the wider the triangular windows get the less tall they get which would fit what I'm thinking. – Goz May 10 '13 at 21:45
  • Re: DCT, Gonna have to do more reading on DCT because I've always assumed the cepstrum transform is very similar to the autocorrelation (much the same minus the log)... And in my head the DCT is just a special case of the FFT so I would've thought you'd need an inverse DCT ... – Goz May 10 '13 at 21:45
  • Re: my last question on weighting ... should've have read your answer thoroughly ... sorry. – Goz May 10 '13 at 21:51