The range calculation for FMCW is much simpler than this and quite intuitive- given a ramp rate in frequency and a stationary object, the round trip delay between the transmit signal and receive signal results in a constant frequency offset after the received chirp is multiplied with the transmitted chirp and low pass filtered, as depicted in the diagram below.

[![FMCW][1]][1]

The round trip delay $\tau$ is calculated directly from the measured offset frequency $f_{\Delta}$ for a given target and the ramp rate $R$ as $f_{\Delta}/R$. For a simple example, if the chirp was sweeping at a rate of R = 10 MHz/us and the round trip delay was 100 ns, the frequency difference between transmit and receive would be 1 MHz:

$$f_{\Delta}= \tau R = (1E-9)(10E6/1E-6) = 1E6$$  

In free space, propagation is approximately 1 ns/foot as given by the speed of light $c$, so the round trip delay of 100 ns would be associated with a target 50 feet away. Or as calculated for the rest of the world:

$$d = \frac{c \tau}{2} = \frac{(3E8) (100E-9)}{2} = 15 \text{ meters}$$


For the case of a moving target, consider first the target with a constant velocity moving toward the trasmitter. Assuming the frequency change over the ramp is significantly smaller than the carrier frequency transmitted (as would be typical for FMCW), the Doppler offset in this case would be constant and independent of the frequency ramp rate, shifting the frequency of the received signal higher as depicted in the graphic below and given by the formula below that. 

[![Doppler Offset][2]][2]

$$f_d = \frac{2v_r f_{tx}}{c}$$

Where 

$f_d$: Doppler frequency offset

$v_r$: velocity between transmitter and target

$f_{tx}$: carrier frequency of transmitter (assumed >> frequency ramp)

$c$: speed of light.

In addition to this offset, and not necessarily visible in the scale of the graphic above, the measured frequency difference after the low pass filter ($f_{\Delta}$) will be changing with time (as itself a ramp if velocity is constant). Note that there is no cross-dependence in the ramp due to the time change in range, and the offset due the Doppler frequency due to the target velocity; therefore these two effects can be computed independently and summed.

Consider our example with the transmitted frequency as 10 GHz, with a relative velocity between transmitter and receiver of 20 m/s (target moving toward transmitter) the Doppler offset would then shift the received chirp upward in frequency by:

$$f_d = \frac{2 (20) 10E9 }{3E8} = 1.33 \text{ KHz}$$

As the target is moving toward the transmitter, the frequency difference due to the range alone would be decreasing. If we assumed a case where we use all measured data was for the target from 15m range as originally derived, up to a 14 m range, or 1 meter corresponding to a time of 1/20 second (given the velocity of 20 m/s), the frequency after the low pass filter without accounting for Doppler would be 1 MHz at 15m corresponding to a 100 ns delay as computed above, and for 14m:

$$\tau = \frac{2d}{c} = \frac {2(14)}{3E8} = 93.33 \text{ ns}$$

$$f_{\Delta} = \tau R = (93.33E-9)(10E6/1E-6) = 933.3 \text{ KHz}$$ 

Thus the frequency difference $f_{\Delta}$ before accounting for Doppler went from 1 MHz to 933.3 KHz over the 50 ms measurement interval. As noted by the graphic above, the effect of the increase in Doppler is to decrease this measured frequency difference by a constant amount, specifically 1.33 KHz in this case, and does not change the rate of change for the measured frequency; so the actual measured $f_{\Delta}$ in this case would be a frequency ramp starting at 998.67 KHz and ending at 932 KHz.  Independent of this frequency shift, we can still derive the velocity directly from the rate of change in the measured frequency.

The above is to show that the ramp of the chirp itself can be removed from the equation with Doppler and velocity determination based on the derived waveform for $f_{\Delta}$, simplifying the computation. 

Combining the above to a final fomulation, we get:

$$d = \frac{c \tau}{2} = \frac{c (f_{\Delta} + f_d)}{2R} =\frac{c f_{\Delta} + 2v_r f_{tx}}{2R}$$ 


Where 

$R$: Ramp rate of chirp

$f_d$: Doppler frequency offset

$f_{\Delta}$: Measured frequency after low pass filter in receiver

$f_{tx}$: carrier frequency of transmitter (assumed >> frequency ramp)

$c$: speed of light.

And $v_r$ is derived directly from the time rate of change in the frequency after the low pass filter, which is not affected by the Doppler or other parameters as described above.

With the above computations we are assuming the receiver bandwidth is wide enough and time resolution for all processing small enough to support the rate of change due to the velocity of the target, otherwise these effects can also significantly impact the result.


  [1]: https://i.sstatic.net/i3FrT.png
  [2]: https://i.sstatic.net/73B7M.png