0
$\begingroup$

I am working with Bluetooth specification 5.1 where the advertisement packets can send a constant tone extension (CTE) over the baseband signal to estimate the angle of arrival with an antenna array.

Due to inaccuracy of transciever and reciever clocks for bluetooth devices, the CTE is not exact.. This leads to a phase shift in the recieved signal of the antennas.

Based on my last discussion over here, I have understand that I need to estimate the CFO and phase rotation to compensate for the inaccuracy of the recieved signal.

I have simulated the I/Q sampling for Bluetooth baseband model with CTE signal of 250 Khz with a sampling of 125 KHz.

enter image description here

As expected the phase of the I/Q samples remains fairly constant.

In this case as I am trying to understand how to compensate CFO, I have introduced a freq. deviation of 5 Khz to the signal.

Phase shift of 5Khz

Based on this I am now trying to figure out how to compensate this deviation I introduced to return back to the first figure where the phase is constant. What method do I need to apply to use the knowledge of the 5 Khz freq. deviation from the original 250 Khz signal?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

Such carrier frequency offsets are typically compensated for digitally by using a numerically controlled oscillator (NCO)- the complex signal at or near baseband is multiplied by the complex NCO output with a full complex multiplier, consisting of 4 real multipliers and 2 adders according to;

$$I_o = I_1I_2-Q_1Q_2$$ $$Q_o = I_1Q_2+I_2Q_1$$

Where $I_o+jQ_o$ is the correct red baseband signal and $I_1+jQ_1$ is the received signal with a carrier offset and $I_2+jQ_2$ is the NCO. All are functions of time as $I_o[n]$ etc with sample index $n$.

This can be done in a carrier recovery loop when combined with a phase detector to measure the phase error from symbol to symbol and a loop filter.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Would this mean I could represent the NCO signal $I_2+jQ_2$ as $$ \exp(-i \cdot 2 \pi \cdot f_{\text{dev}} \cdot t) $$ and then multiply it to recieved signal $I_1+jQ_1$ ? $\endgroup$
    – Yudop
    Commented Jul 3, 2023 at 22:23
  • $\begingroup$ Yes exactly! The sign for the NCO exponent is opposite the sign for the frequency error such that when you multiply the two, the exponent representing phase is zero. Phase vs time is frequency (a change of phase with a change in time or $d\phi/dt$) $\endgroup$
    – Dan Boschen
    Commented Jul 3, 2023 at 23:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.