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Very often in literature we see that effect of carrier frequency offset is seen as rotating of a constellation points. And explanation is generally that after down converting and sampling the time domain signal can be seen as:

x'[n] = x[n]*exp(2* pi * Delta_f * n)

And as we see there is a phase soft associated with time domain signal , this will what rotates the point of constellation diagram with no effect on amplitude.

BUT we know constellation diagram is plotted post fft (i.e in frequency domain) which means this phase shift actually will a freq shift in time domain. Then how is the constellation point being rotated?

Can someone please clear my doubt

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  • $\begingroup$ Why do you think the constellation diagram follows a DFT? Are you thinking of OFDM? $\endgroup$
    – MBaz
    Sep 6 at 2:30
  • $\begingroup$ Yes , I should have mentioned. I am referring to OFDM $\endgroup$
    – psb
    Sep 6 at 5:19
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Assuming no timing/sampling error, the time domain samples of the m-th OFDM symbol (index from 0) are

$$y[n] = x\left[m(N+N_p)+N_p+n\right] \exp\left(j2\pi \frac{\epsilon}{N} \big(m(N+N_p)+N_p+n\big) \right) \tag{1}$$

for $n=0,...,N-1$ where $N$ is the FFT size, $N_p$ is cyclic prefix length in sample, and $\epsilon=\frac{f_\textrm{offset}}{f_\textrm{sampling}}$ is the normalized frequency offset.

Take the FFT of (1), the demoludated sample of the m-th OFDM symbol $Y_m[k]$ is \begin{align} Y_m[k]&=\exp\left(j2\pi \frac{\epsilon}{N} \big(m(N+N_p)+N_p\big) \right) \times\\ &\sum_{n=0}^{N-1} x\left[m(N+N_p)+N_p+n\right] \exp\left(-j2\pi \frac{k-\epsilon}{N} n \right) \tag{2} \end{align}

Check the modulated $X_m[k]$, $$X_m[k]=\sum_{n=0}^{N-1} x\left[m(N+N_p)+N_p+n\right] \exp\left(-j2\pi \frac{k}{N} n \right) \tag{3}$$ you can see that $Y_m[k]$ is suffered from

  • Inter-carrier interference (ICI) at $(k-\epsilon)$ and
  • rotation $\exp\left(j2\pi \frac{\epsilon}{N} \big(m(N+N_p)+N_p\big) \right)$ that cannot be ignored unless $\epsilon=0$.
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