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I am struggling to understand how linear phase shift can be determined with a transform whose main purpose is to determine frequencies rather than rate of change of a function.

In FMCW radar, received signal corresponding to a chirp is passed through a mixer and the resulting beat frequency is sampled by means of ADC. The samples are stored in a row of a matrix. Similarly, samples from other chirps are stacked up in corresponding rows. Different objects at different locations produce different frequencies (thanks to Doppler). So a FFT of each row yields dominant frequency bins. The frequency bins directly relate to object range. Absolutely fine so far.

Assuming that these objects move (and slowly), they must continue to appear in the same frequency bin (column)for all of the chirps. But since they are moving, the phase of the frequency component must change. For simplicity, we assume that the magnitude of all frequency components in a column is constant.

Thanks to Shift Theorem, a signal delayed/advanced in time results in a phase shift of the frequency components of the time domain signal.

Question-1: Am I right in assuming that a constant velocity results in a LINEAR phase shift of the frequency component across chirps?

Question-2: If the phase shift is linear, how can FFT along a column determine the same? FT/DFT essentially performs decompositions in terms of basis sinusoids. But in our case, we are interested in calculating the slope of the line (or rate of change or derivative) representing the phase-shift change across chirps. Derivative is not same as frequency. I think we need a Non-sinusoidal decomposition hich can yield various slopes. But second-stage FFT along the column seems to work.

What have I misunderstood?

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Your understanding is mostly correct but no need of non-sinusoidal decomposition and all.

Am I right in assuming that a constant velocity results in a LINEAR phase shift of the frequency component across chirps?

Yes, the change in phase due to time change $\Delta \tau$ is $\Delta\phi = 2\pi F (2\Delta \tau)$, $2\Delta\tau$ because signal has to travel back so double the distance covered by object. $F_c$ is the chirp frequency. $\Delta \phi = 4\pi F\frac{\Delta d}{c}$, where $\Delta d$ is the distance increment due to velocity of object and $c$ is velocity of light. Since wavelength $\lambda=F/c$, and $\Delta d=vT_c$ (where $T_c$ is chirp duration), phase change between 2 chirps is $\Delta \phi = 4\pi \frac{vT_c}{\lambda}$. So you can see that relative phase of peak bin of FFT output is linearly changing with respect to velocity $v$.

If the phase shift is linear, how can FFT along a column determine the same?

Once you have identified the column corresponding to peak bin, take the FFT of phases of these peak bins column wise. If phase changes linearly with time, the FFT of phases would show up as a peak bin when you take FFT along the column.

From the previous formula, we have $\Delta \phi$ as the change in phase between two consecutive chirps. So $\Delta \phi$ can be denoted as angular velocity $\omega$. Hence $v = \frac{\lambda \omega}{4\pi T_c}$. So if you take FFT of phases, if you see a peak at $\omega=\omega_0$, the velocity is $\frac{\lambda \omega_0}{4\pi T_c}$. If you see 2 distinct peaks, implies there are 2 equidistant objects approaching or retreating at different velocities at that point of time.

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  • $\begingroup$ Thank you, Jithin. Can you please explain how the phase difference is 2.Pi over Tc? It is not clear to me. I have accepted your answer though. $\endgroup$ – Raj Apr 30 at 16:08
  • $\begingroup$ Oh I think I made a mistake there. It need not change over $2\pi$ for chirp duration $T_c$. It changes because of velocity but need not be $2\pi$ change. The angular velocity $\omega$ will be how much the phase changes between two chirps. Let me correct it in answer. $\endgroup$ – jithin Apr 30 at 16:34

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