#### David83

##### Advanced Member level 1

what is the relationship between OFDM and DTF? I tried to figure it out as following:

In continuous time, the multicarrier symbol is formed as:

\[s(t)=\sum_{k=0}^{N-1}s_k\cos(2\pi f_kt)=\text{Re}\left\{\sum_{k=0}^{N-1}s_ke^{j2\pi f_kt}\right\}=\text{Re}\left\{\sum_{k=0}^{N-1}s_ke^{j2\pi\frac{k}{T}t}e^{j2\pi f_0t}\right\}\]

where \[f_k=f_0+\frac{k}{T}\], \[T=NT_s\] is the multicarrier symbol duration, and \[T_s\] is the symbol duration of the original stream. So, the low pass equivalent of the multicarrier symbol is:

\[\tilde{s}(t)=\sum_{k=0}^{N-1}s_ke^{j2\pi\frac{k}{T}t}\]

Now the IDFT is given by:

\[s[n]=\sum_{k=0}^{N-1}s_ke^{j\frac{2\pi}{N}kn}\]

which means that \[\{s[n]\}\] are samples of \[\tilde{s}(t)\] at \[t=nT_s\]. So, we can generate the samples \[\{s[n]\}\] using IDFT, and then pass them to D/A converter and up convert the frequency to \[f_0\].

The question is: the signal \[\tilde{s}(t)\] has duration \[NT_s\] so ideally, its bandwidth is \[W=\frac{1}{NT_s}\]. So, the sampling rate \[F_s\geq 2W\] for distortion-less recovering. Do we choose the sampling rate for the OFDM system to be \[f_s=\frac{1}{T_s}>\frac{1}{NT_s}\] so that the Nyquist criterion is met

**and**the DFT-IDFT operations can be used?

Thanks in advance