I am interested in the equation that depicts the resolution of an FMCW radar. I write the same below.

$\Delta d = \frac{c}{2b}$

First I would like to have an intuitive explanation to the above derivation.

Secondly, if possible, I would like to know why the chirp's gradient doesn't play a role in determining the resolution of the FMCW radar. Does it not matter how fast I sweep the $BW$? If not so, what impact does the chirp gradient have?



  1. $\tau$: round-trip delay. $\Delta \tau$ = delay resolution.
  2. $r$: range. $r=c\tau/2$. Range resolution $\Delta r = c \Delta \tau / 2$.
  3. $B$: chirp bandwidth
  4. $T$: chirp sweep time
  5. $R$: chirp rate. $R=B/T$

Basic steps:

  1. We estimate the delay by estimating the frequency shift between the received and transmitted signal. Typically the resolution of a frequency estimate is the inverse of the observation time, which in this case is $1/T$. Why? Think of taking the Fourier Transform of a tone with a rectangular window T. The window will have nulls at $f-1/T$ and $f+1/T$, where $f$ is the tone's frequency.
  2. Because the chirp rate is $R=B/T$, the time resolution corresponding to a frequency resolution of $\frac{1}{T}$ is $\frac{1/T}{B/T} = \frac{1}{B}$.
  3. Plugging this into the range resolution equation, we have $\Delta r = c/(2B)$.

Intuitively, we improve range resolution by (1) improving resolution of the frequency shift estimate (i.e., increasing sweep time), and (2) decreasing the time delay corresponding to a given frequency shift (i.e., increasing chirp rate). To do one without sacrificing the other, we need to increase the bandwidth. If we double the chirp rate but cut the sweep time in half, thus keeping the bandwidth the same, the time delay per frequency shift decreases but the frequency resolution gets worse. Likewise, if we cut the chirp rate in half but double the sweep time, the time delay per frequency shift increases while the frequency resolution improves. In either case, the range resolution does not change.


The range resolution represented by the equation that you wrote in your question describes the capability of a radar to distinguish between two closed targets, so if the BW is much large then the resolution is better. by increasing the BW we enhance radar performance in tems of range resolution.

  • $\begingroup$ In that case, what is the impact of Chirp Rate or the slope of the Chirp? $\endgroup$ – Denis Jul 10 '18 at 16:09
  • $\begingroup$ @Dina The slope of the chirp does not impact your range resolution. Your ability to resolve between two targets is dependent on the bandwidth of the chirp only, no matter how fast you chirp it. The pulse width will affect your ability to process Doppler, however. $\endgroup$ – Envidia Jul 10 '18 at 16:51

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