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If we have OFDM system with $N$ sub-carriers, the DFT matrix can be expressed as follows:

$F = dfmtx(N)/sqrt(N)$

My question, is it possible and practical to use the matrix without Normalization, it means use the iDFT as $F = (dfmtx(N))'$, and then at the receiver end, I will multiply with that matrix and divided by $N$ (like this $dfmtx(N)/N)$ to get scalar diagonal matrix whose diagonal elements are 1?

thank you

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Yes. In fact, depending on your receiver architecture, you may not need the divide by $N$ in the receiver.

Generally, in a receiver you care about the relative phases of the bins (or possibly their strengths with respect to some other part of the same signal). So there's no reason to do that multiply if you don't want to.

If you're working with fixed-point math, just make sure that you don't experience over- or under-flow.

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  • $\begingroup$ OK thank you, but about your saying "depending on your receiver architecture, you may not need the divide by NN in the receiver." ,But, if I didn't divide by $N$, it means that $(dfmtx(N))′*(dfmtx(N))$ is a scalar diagonal matrix with diagonal elements equal to $N$ which won't be able to detect the modulated elements. $\endgroup$
    – Fatima_Ali
    Mar 1 at 5:11
  • $\begingroup$ How so? How are the relative phases of the bins changed by their amplitude? And how are the relative amplitudes changed by multiplying each and every one of them by N? And if the absolute amplitude matters, how is a receiver going to work after the signal is run through a channel with some arbitrary and unknown attenuation? $\endgroup$
    – TimWescott
    Mar 1 at 5:24
  • $\begingroup$ emmm .. you it doesn't raise any issue even if the amplitude is different, right? $\endgroup$
    – Fatima_Ali
    Mar 1 at 5:29
  • $\begingroup$ It's more that in a real system the first thing you lose any certainty of is the gain in the channel. If your receiver cannot seamlessly cope with everything being multiplied by some random but slowly changing number, then you need a different receiver -- or a different modulation scheme. $\endgroup$
    – TimWescott
    Mar 1 at 16:18

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