# FFT in discrete and continous time

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S. David on 29 Jul 2013
Hello all,
I have in my code for OFDM system something like fft(x,N)/N. Does division by N in the discrete time equivalent to division by T (OFDM symbol time) in the continuous time?
Thanks

Walter Roberson on 29 Jul 2013
No. If you were to keep expanding the sampling frequency, even without adding any more information to the signal, then that would correspond to N getting larger and larger. The limit of increasing N is N -> infinity. But limit(fft(x,N)/N, N->infinity) is going to be 0 (provided that x is not a constant spectrum.) So you are asking if 0 (limit of discrete) is going to be the same as division of the continuous time signal (infinitely dense) by the OFDM symbol time (time of a bit of information contained in the signal), which is obviously not going to be the case (unless T itself approaches infinity.)
You need to distinguish between sampling frequency needed for correct analysis (Nyquist frequency) and sampling frequency of the A/D convertor: your formula should not tend to 0 as the actual sampling frequency increases beyond the minimum necessary to decode the data properly.
S. David on 29 Jul 2013
I do the following:
b=rand(N,1);
d=2.*b-1;
x=ifft(d,N).*N;
%%transmit x over the channel which gives the noise-free signal z
Z=fft(z,N)/N;
ZEq=conj(H).*Z;
H is the fft of the channel. Theoretically, ZEq=T.*abs(H).^2.*d, but it is actually ZEq=abs(H).^2.*d. Why then?

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