How to go from continuous-time OFDM to discrete-time?

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David83

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Hello all,

I am interested in understanding how do we obtain the discrete-time OFDM signal from its continuous-time counter part. In particular, the received OFDM signal over time-invariant channel of length L is given by:

\[y_n=\sum_{l=0}^{L-1}h_lx_{n-l}+w_n\]

where {h_l} are the channel gains, {x_n} are the N-point IFFT of transmitted symbols, and {w_n} are AWGN samples. Then we take the FFT of the above received samples as:

\[Y_k=\sum_{n=0}^{N-1}y_n\,e^{j\frac{2\pi}{N}nk}=H_kX_k+W_K\]

where H_k, X_k, and W_k are the frequency counterparts of the the channel, signal, and noise, respectively, and N is the number of subcarriers.

Now in the continuous-time, the transmitted signal is given by:

\[x(t)=\Re\left\{\sum_{k=0}^{N-1}X_ke^{j 2 pi f_kt}\right\},\,\,\,t\in[-T_g, T]\]

where T_g is the length of cyclic prefix in seconds, and T is the OFDM symbol length. The LTI channel impulse response is given by:

\[h(\tau;t)=\sum_{p=1}^{P}h_p\delta(\tau-\tau_p)\]

where h_p and tau_p are the channel gain and delay pf the pth path. The received baseband signal is given by:

\[y(t)=\sum_{k=0}^{N-1}X_k\sum_{p=1}^Ph_pe^{j2pi f_k \tau_p}e^{j2pi\frac{k}{T}t}+w(t)\]

I assume we do sampling next at t=nTs, but then what?

Thanks
 

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