FFT in discrete and continous time

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

Answer 1

[Walter Roberson]

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]

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?