Consider an OFDM signal which has $N=8$ subcarriers and is band-limited to $[-B/2,B/2]$, $\sum_k x_k e^{j2\pi kf_0t}$ where $f_0=\frac{B}{N}$. For this case the Nyquist rate is $2\times B/2=B$, i.e., the sampling interval $T$ (distant between two consecutive samples) is $T=\frac{1}{B}$ and the $l$'th sample is obtained as

$x(lT)=\large\sum_k x_k e^{j2\pi kf_0 lT}=\sum_k x_k e^{j\frac{2\pi}{N}kl}$

which is the same as IDFT formulation and the frequency components $x_K$ can be obtained using these $x(lT)$ and DFT. $x_k$'s are the symbols in frequency subcarriers $[-4f_0,-3f_0,-2f_0,-f_0,0,f_0,2f_0,3f_0]$

Now consider a sampling frequency $B^{'}$ bigger than Nyquist ($B^{'}>B$). In other words, a sampling interval $T^{'}<T$. In this case, we will obtain $N$ new samples, let's call them $x(lT^{'})$'s:

$x(lT^{'})=\large\sum_k x_k e^{j2\pi kf_0 lT^{'}}=\sum_k x_k e^{j\frac{2\pi}{N}\frac{B}{B^{'}}kl}$.

My question is:

Is there a way to obatin $x_k$ (symbols in subcarriers $[-4f_0,-3f_0,-2f_0,-f_0,0,f_0,2f_0,3f_0]$) from these new set of samples $x(lT^{'})$.

  • $\begingroup$ You can read about oversampling and fractional sampling, they have the same idea. that should give better results, but don't forget in very high frequency sampling, the complexity will increase too. $\endgroup$ – Zeyad_Zeyad Dec 26 '18 at 8:06
  • $\begingroup$ A way to interpolate in the time domain is to add zero coefficients in the frequency domain. The bandwidth $B$ is increased ant the product $BT$ remains constant $\endgroup$ – Damien Dec 26 '18 at 8:11

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