I'm a french student in engineering and I'm given a kind of mandatory technical project to work out on until February.

I must investigate the following article : https://www.osapublishing.org/DirectPDFAccess/C41EF419-AD7D-EDF3-A173E8B93A0B610C_350066/oe-24-19-22110.pdf?da=1&id=350066&seq=0&mobile=no and implement the same kind of iterative algorithm on MATLAB.

Actually I'm not good at signal processing either that much, only have the basics, however I don't understand why they say "the inherent frequency interval limit (of fFT) can be overcome using discrete Fourier transform". In my opinion that is non-sense because fFT is basically just the same as dFT, you just have to sum the sampled values an other way with their odd and even indexes. Therefore I don't really understand what do they apply dFT for...

If someone could help me please he would be more than welcome !

Thank you


closed as unclear what you're asking by jojek Nov 22 '17 at 12:01

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ can't load that link. $\endgroup$ – Marcus Müller Nov 21 '17 at 23:04
  • $\begingroup$ Cannot open the file either. DFT and FFT sre different in the basic thing is that FFT requires the length of discrete data to be multiple of 2, thus the frequency resolution must also be multiple of 2. I guess that is what they mean "inherent freq interval limit". DFT is not limited by this length constraint. I cannot say anything further without the paper. $\endgroup$ – AlexTP Nov 21 '17 at 23:29
  • $\begingroup$ Hi, I am closing the question since the link is dead and you didn't provide any details about it. $\endgroup$ – jojek Nov 22 '17 at 12:02

You are right. DFT and FFT are the same things; they produce the same outputs and they operate on the same samples. FFT is just a fast implementation of DFT with much less artihmetic operations. They only have roundoff error difference which is due to the different number of arithmetic operations involved. Therefore you either misunderstood the article or what they say is wrong.


Not the answer you're looking for? Browse other questions tagged or ask your own question.