In (1), it's said that:

If two identical training symbols are transmitted consecutively, the corresponding signals with CFO of $\epsilon$ are related with each other as follows: $y_2[n]=y_1[n]\exp(j2\pi \epsilon)$.

How is this relation between $y_1[n]$ and $y_2[n]$ achieved? How is the exponential part associated with $y_1[n]$ obtained?

(1) Ishtiaq Akbar, A. Naveed Malik, "Performance analysis of carrier frequency offset estimation techniques for OFDM systems", Research journal of Applied Sciences, Engineering and Technology 6(11): 2041-2044, 2013


This is very simple. If there is a carrier frequency offset then there will be a phase rotation from symbol to symbol proportional to the frequency offset. For an offset of $\omega$ rad/sec, and an symbol period of $T$ the expected phase difference due to the CFO from symbol to symbol will be $\omega T$ radians. This can be expressed similar to what you've shown since the only difference between identical consecutive training symbols will be the phase due to the CFO.

The indices don't make much sense as you've shown them, but if identical symbols $y$ are received at samples $n$ and $n+1$ with CFO = $\omega$ rad/sec and symbol period $T$ seconds, then

$y(n+1) = y(n)\mathrm{exp}(j\omega T)$

  • $\begingroup$ The notation in the reference given probably refers to $y_1$ and $y_2$ as two consecutive OFDM symbols where $n$ starts from 0, respectively. In your answer shouldn't it be $y(n+N_\mathrm{S})$, where $N_\mathrm{S}$ is the length of one OFDM symbol in samples, including guard interval? $\endgroup$ – Deve Feb 3 '14 at 16:27
  • $\begingroup$ @Deve: I assume that Eric's indexing uses $n$ as the received symbol index, i.e. the sampling interval between successive values of $n$ is equal to the symbol period (including any cyclic prefix). $\endgroup$ – Jason R Feb 3 '14 at 16:51

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