I am trying to simulate the SC-FDMA transmission and I read H Myung's paper and research on the topic but I am not sure exactly on what should be the size of FFT at the transmitter.

For a 20 MHz channel, with 1200 available subcarriers, what should be the size of the FFT block?

The research talks about IFFT at transmitter sized N = M·Q where Q is the number of users and M is the no. of subcarriers in each block of Q.

So do we perform a M pt. FFT on each Q block or on the entire 1200 symbols at the transmitter before IFFT?

Link to paper: https://ieeexplore.ieee.org/abstract/document/4099344

  • $\begingroup$ you'll need to link to that paper. Also, it's likely that you'll just do an N-point FFT, because that's the only way you can find the M carriers. $\endgroup$ – Marcus Müller Aug 1 '19 at 11:20
  • $\begingroup$ So a 2048 FFT before a 2048 IFFT? Wouldnt they just cancel out each other that way ? $\endgroup$ – samz12 Aug 1 '19 at 12:18

We perform M point FFT on each Q block, then map the FFT results to intended positions of 1200 available subcarriers (subcarrier mapping), which are finally the input of a 2048-iFFT operation. See the figure below.


Note that this is just one of possible implementations of SC-FDMA which is called DFT spread OFDM. The another is time domain approach. Check this book (Chapter 14) for more details.

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  • $\begingroup$ Thanks for the info @Alex. :) $\endgroup$ – samz12 Aug 5 '19 at 8:52

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