0
$\begingroup$

The DFT size in OFDM demodulation is usually matched to the DFT size in the modulation process. This makes modulation/demodulation simple where each FFT bin (i.e., sample) represents one symbol. What is the impact of taking an arbitrary larger FFT size during the demodulation process (assuming properly sampled based on new DFT size)? Intuition seems to lead you to think that the underlying symbol is now spread across multiple bins. Can you use a larger FFT size and still demodulate the data?

EDIT: sample rate clarification

Improve this question
$\endgroup$
1
  • $\begingroup$ addressed your edit, too. $\endgroup$ – Marcus Müller Jul 22 '20 at 13:28
0
$\begingroup$

What is the impact of taking an arbitrary larger FFT size during the demodulation process?

It doesn't work.

This gets pretty intuitive if you take an arbitrary OFDM introduction and look at the spectrum of OFDM, consisting of overlapping sinc functions: you need to hit exactly that raster, or you don't see the individual channels, but an undisambiguable mishmash of carriers.

Using a larger FFT at the same sample rate can't work: don't forget that the FFT is just a DFT. A DFT maps vectors of a specific length N from time domain to frequency domain vectors of exactly the same size. So, a larger FFT length simply needs more time domain samples. But your OFDM symbol is as as many samples long as the transmit FFT was.

So, no, you can't use any other FFT size, sensibly.

You'd have to increase the sample rate by the same integer factor that you've increased the FFT size.

But that would only yield an FFT where all but the original bins are only noise, so you'd have won nothing.

(This is actually done in some OFDM systems to have more headroom for frequency synchronization, but it's not desirable if you don't have a frequency offset larger than your outer guard carriers.)

Improve this answer
$\endgroup$
2
  • $\begingroup$ So assuming an integer multiple DFT size and proper sample rate, which bin would you take the symbol from? Would you average them? $\endgroup$ – BigBrownBear00 Jul 22 '20 at 13:12
  • $\begingroup$ no, no averaging. just take the right bins, throw away the rest. The thing about OFDM is the orthogonality of all other carriers: they contain nothing of the transmitted carriers at all. $\endgroup$ – Marcus Müller Jul 22 '20 at 13:29

Your Answer

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

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