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I am trying to understand the full mathematics behind FMCW. Using linear-frequency chirp $x(t)$ that has center frequency $f_c$, bandwidth $\Delta f$, duration $T_r$, and amplitude $A$, the transmitted chirp is $$x_T(t)=A\cos\left(2\pi f_c t+\pi\frac{\Delta f}{T_r}t^2+\phi_0\right)$$ The received signal reflected from a target at distance $R$ making a $t_d$ round trip time delay. $$x_R(t)=A\cos\left(2\pi f_c (t-t_d)+\pi\frac{\Delta f}{T_r}(t-t_d)^2+\phi_0\right)$$ Plotting the frequency for the two signals gives a plot like below

The beat signal (IF signal) is the lower sideband of the multiplied signal. This will have two frequency components. First, regions where Tx is greater than Rx. Second, where Tx is less than Rx. This is giving me headaches. How does the radar deal with this? The FMCW doesn't directly measure time, so the $t_d$ part cannot be discarded by timing.
I tied a method of just keeping all components. To deal with this I tried to calculate each beat frequencies and compare it. let $f_{b1}$ the beat frequency from the front side of the chrip and $f_{b2}$ the beat frequency from the $t_d$ part. $$f_{b1}=\frac{\Delta f}{T_r}t_d-f_D$$ $$f_{b2}=-\Delta f+\frac{\Delta f}{T_r}t_d-f_D$$ With this, I cannot get either $t_d$ or $f_D$. What should I do?

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The two frequencies sim to the frequency range of the chirp ($fr$) and define the region over which the distance can be uniquely determined, which is $fs/2$. For example if the chirp sweeps over $fr$ and the range to target results in a lower $fb2$, we will also get the higher $fb1=fs-fb2$.

The upper frequencies are easily excluded in the frequency domain, using FFTS, but could also be excluded with a low pass filter up to a reasonable limit below $fr/2$ allowing for filter transition.

This means that there is ambiguity for targets which are referred to as “range bins”. We cannot distinguish between a small target resulting in a $fb2$ between $f=0$ and $f=fr/2$ and a large target resulting in $fb1$ between $fr/2$ and $fr$. (Or even larger targets at further range in higher bins where the chirp basically overlaps with the next one (but can be mitigated with time spacing between chirps but that would come at the expense of receive SNR).

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  • $\begingroup$ Does it mean that only one of the two beat frequencies is used? What happens if the range is just enough to make the frequencies unfilterable? $\endgroup$
    – Moses Kim
    Dec 14, 2022 at 11:29
  • $\begingroup$ Yes only one is used, and if it is unfilterable then the range is beyond the usable range of the FMCW radar. (Unresolvable). A fast sweep results in a further range but lower SNR (less integration time). $\endgroup$ Dec 14, 2022 at 11:34
  • $\begingroup$ Oh, is then the IF signal made from the mixer is usable (unambiguously resolvable) only when the filteration step of the IF signal can discard one of the two beat frequencies? If so, assuming ideal filter, is the limit of unambiguity of $f_b$ is half the bandwidth of the chrip? Because when it passes the half point, the remaining signal swiches to the other one and I think this will blow the thing up. $\endgroup$
    – Moses Kim
    Dec 14, 2022 at 11:43
  • $\begingroup$ Yes that is correct, and easily done through an FFT but your comments on "ideal filter" is accurate in that we can't actually go all the way to half the bandwidth (similar to ADC sampling constraints). $\endgroup$ Dec 14, 2022 at 11:53

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