We are working on LTE Uplink CFO Correction for 3 MHz band. The time domain CFO correction is easy, which depends on the processing of 3840 samples step by step multiply by $\exp(-j 2\pi f_\text{offset}t)$. . What will be the equivalent frequency domain correction?
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3 Answers
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Generally, a modulation with a complex sinusoid in time domain is equivalent to a shift in frequency domain (I find that quite intuitive, since it's the very definition of frequency shift).
That means that for $f_\text{offset} = M\frac{f_\text{sample}}{N_\text{FFT}},\quad M\in\mathbb Z$ you can just correct things by shifting the frequency domain vector left or right.
However, you also have to deal with the fractional frequency offset. If you don't, you'll get Inter-Carrier-Interference (ICI).
The only way to solve that is by interpolating the correct values from the values you have with sinc interpolation.
Sinc interpolation is easiest done by taking the IDFT, then modulating with a $e^{j2\pi f_\text{offset fractional} t}$, and then DFT'ing back.
In other words: that frequency correction is done in time domain.
So, do your fine frequency correction in time domain, not in frequency domain. Doing it in frequency domain requires at least the effort of going back to time domain.
answered May 10, 2019 at 9:39
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If we process the samples after de-precoding stage, which will convert from (rxGrid)frequency to time domain samples and further processing by CFO correction as explained above in time domain. The frequency correction is not as per the transmitted data. It is still rotating for CFO above 300 Hz and working for low value of CFO.
The channel equalization method which will also resolves the issue for 300-400 Hz and for higher value of SNR.
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I believe the duality properties of Fourier transform might help you. If you multiply a function $f$ (In your case:$ exp^{−1*j*2*π*f_{offsett}}$ ) to a signal in the time domain, It is equivalent to convolving the Fourier transform of that function with the signal in the frequency domain. Here is the mathematical representation:
The Fourier transform of your function is simply is an impulse at $f_{offsett}$.
