Hi all,
I would like to know what are the right formulas to extract the frequency and amplitude values from the data stored in a 1GB5 SDIF (which are encoded as complex numbers). Thanks in advance for your kind concern.
Best regards,
Francesco Vitale
Dear Francesco,
For the amplitude this is relatively simple:
A = sqrt(I^2 + R^2)
where I and R are the real and imaginary parts of the complex number. The result is the magnitude of the FFT of the windowed signal, this means the size of the window affects the amplitude. For sinusoidal analysis/resynthesis you would need to normalize by the sum of the analysis window which you can do in supervp by means of adding the -ns flag. If you use this flag you can use the amplitude values of the spectral peaks that come out of the formula to resynthesize sinusoids.
The notion of frequency is somewhat ambiguous. Do you want the frequency of the STFT bin, or the frequency corresponding to the phase vocoder frequency that is related to the phase increment with respect to the previous frame (which is what you get from the 1GB4 SDIF format)?
The first is simply
F = B/FFTSIZE*samplerate
where B is the index of the bin (start counting with DC having B=0).
The second is a bit more complicated and requires to take into account the phase of the previous frame. If you want this please read through sections 3.1 and 3.2 of my lecture on signal processing with the phase vocoder in http://recherche.ircam.fr/anasyn/roebel/amt_audiosignale/VL3.pdf. The formula you need is eq. 18.
Best
Axel
Dear Axel,
thanks for your reply. As you correctly guessed, I need to extract the frequency data that you can also get with a 1GB4 SDIF. As regards the equation
$latex {{\Theta }_{k}}=\frac{[arg(X((l+\text{ }1)I,k))-arg(X(lI,k))-I\frac{2\pi k}{N}]2\pi }{I}$,
I understood that $latex I$ is the step size; $latex (X(lI,k)$ and $latex (X((l+\text{ }1)I,k)$ are two successive frames (is it right?), but I’m not sure what N and arg mean… Sorry for the silly question.
Best,
Francesco
Hi Francesco,
yes, and N is the FFT size, and arg means the principle argument, that is the phase of the complex number in ]-pi, pi].
You get that from the function atan2.
Best
Axel
Dear Axel,
so, if I understand well, (as regards what lies within the square brackets of the equation) one must calculate the arctangent of a frame l at bin k, minus the arctangent of its previous frame l-1 at bin k-1, and then subtract from the obtained result the step size multiplied with 2πk/N (where N is the FFT size). But what is exactly 2πk/N?
P.S. I tried in my previous post to write the equation in LaTex translating it for WordPress, but, as I can see, it did not work as expected…
Best,
Francesco
Hi Francesco,
Did you read the document sections I mentioned? Could you please read additionally appendix 6.2 starting on page 57, this appendix is linked to from the section I mentioned earlier, and it explains the whole procedure over three pages. I somehow feel it would be a waste of time to repeat this given one needs formulas and as you have seen those do not work here.
Best
Axel
Thanks Axel,
following your initial indications, I read only sections 3.1 and 3.2. Adding the appendix makes all much clearer.
P.S. It’s a pity that LaTeX didn’t work. In theory, WordPress should support LaTeX formulas without problems.
Best,
Francesco