Original post from yesterday here

So I did the FFT and I have my transformed data (which is a bunch of values for each frame of data. I'm unsure what the transformed data actually represents. What information do I need, and what math do I need to perform to turn the $k$ values into the appropriate frequency values and the $X[k]$ values into $\textrm{dB}$ values? With $X[k]$ being equal to the sum of $x[n]$ from $n$ to $N-1$ times [...]

I understand how the FFT algorithm works now, I'm just not real clear on what the actual results represent.

  • 1
    $\begingroup$ This is probably best answered by a text book on the matter. $\endgroup$ – AnonSubmitter85 Dec 9 '15 at 20:25
  • $\begingroup$ Liz, do you understand that the FFT is a (fast) method of computing the DFT? can you write down the equation that defines the DFT, that expresses $X[k]$ in terms of $x[n]$? then write down the equation that is the inverse DFT (that tells you what $x[n]$ is in terms of the $X[k]$ you've just computed. that last equation tells you what the output of the DFT (that is $X[k]$) fundamentally mean. $\endgroup$ – robert bristow-johnson Dec 9 '15 at 20:56
  • $\begingroup$ Do you know the sample rate at which your original samples of data were taken? $\endgroup$ – hotpaw2 Dec 9 '15 at 21:53
  • $\begingroup$ 44.100kHz. I'm processing wave files, and some are at other sample rates but I can parse that information from the file, it's listed at the beginning of ever wave file in hex. I think, after a bit more digging, that these 'bins', multiplied by sampleRate/frameLength (so in my case, 44100/4096) give the actual frequencies that correspond to the modulus power levels. A bit more math converts the frequencies to musical notes. I guess the only thing I haven't figured out is how to turn the X[k] levels into dB. It's my understanding that I need a 'reference level'? What would that be for audio? $\endgroup$ – Liz Dec 9 '15 at 22:56
  • $\begingroup$ are you trying to make a audio-to-MIDI converter? $\endgroup$ – robert bristow-johnson Dec 10 '15 at 1:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.