How can we justify doing coherent radar when the phase of the return is unknown?

For instance, after a reflection, assuming a stationary target, we should get something akin to:

$$ x_{rx}(t) = |\alpha| e^{\angle\alpha} x_{tx}(t) = |\alpha| e^{j 2 \pi f_{tx} t + \phi_{tx} + \angle \alpha} $$

however $\alpha$ is generally unknown. It is highly dependent on the material of the target, the angle of the waveform's impingement, and the frequency bands used.

So why do we not always do incoherent radar? Why do all these methods for coherent radar exist? Where are they useful?

  • 2
    $\begingroup$ Consider FMCW radar where we transmit a chirp and determine range to target based on the frequency difference between the local oscillator and the reflected signal. How would you do that non-coherently and how would the range accuracy compare? $\endgroup$ Dec 5, 2019 at 18:27

1 Answer 1


A lot of radar systems are interested in measuring Doppler, one example of which is pulse Doppler (PD) radar. PD radars are capable of measuring both range and velocity (with ambiguities, but that’s a different topic).

The basic idea behind a PD radar is to transmit many pulses of the same waveform and observe how the phase of a target return changes. To measure phase, we need a coherent radar system. Nominally, we utilize in-phase and quadrature (IQ) data, which gives us the maximum likewise estimate of the phase of a signal. When we want to measure velocity of a target signal, we simply take the DFT across all of the pulses we’ve collected, which can be viewed as a matched filtering operation that is “looking” for specific Doppler speeds.

Without using IQ data, there would be an “ambiguity” in phase, and therefore the method wouldn’t really work well at all.

  • $\begingroup$ I see, so everyone knows the phase is changing and they actively look for this change. I was just hoping it wasnt some theoretical concept developed by academics who have never run a real radar. $\endgroup$ Dec 6, 2019 at 11:32

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.