The signal processing steps for a pulsed radar are: Fast time matched filter -> Slow time FFT.

I cannot understand how is the second step able to detect frequency, because the results after the matched filter will look like:

enter image description here

I.e each peak will be shifted from its prior.

How is the FFT on slow time able to detect some kind of frequency here?

My signal is:

enter image description here

Thanks alot in advance!


1 Answer 1


For each pulse a range is determined. The OP has drawn the results for the case that the velocity is so great that the range will result in a significant shift in time of the matched filter result. Consider the cases where the velocity is much slower, whereas there will be a shift in phase with each subsequent matched filter result yet the resulting matched filter outputs in magnitude will be more or less aligned in time. A change in phase with a change in time is frequency, and therefore if we use the FFT to determine the frequency over several pulse results (slow time) the resulting FFT will have a result where the frequency as depicted by the FFT is proportional to the velocity of the target.

From this we can understand how there will be a resolvable velocity (up to a maximum) based on the time between pulses and the change in phase from pulse to pulse based on the targets velocity: If the phase difference between two matched filter outputs exceeds $2\pi$ there will be ambiguity in the resolved velocity.

  • $\begingroup$ So the magnitude is more or less the same, and each point on the slow time axis is just shifted a bit in time, i.e diffrent phase? (am i to assume the wave 'riding' the magnitude of the correlation result is the carrier wave?) what is the mathematical relation between the phase change the doppler frequency? also are you familliar with a good visualization/rigorous math explanation to this phenomena? $\endgroup$ Jan 13, 2023 at 16:07
  • $\begingroup$ A phase shift between the received and transmitted signal will result in an identical phase shift in the cross-correlation. Review how each pulse from a target with constant velocity will rotate the cross-correlation by a similar phase step $\Delta \phi$, which is basically $Ae^{j\Delta \phi n}$ for each pulse $n$. The equation $e^{j \omega n}$ is a constant frequency at rate $\omega$ radians per sample. As far as fantastic visualizations along with the math, I recommenend Stimson's Introduction to Airborne Radar book: $\endgroup$ Jan 13, 2023 at 19:10
  • $\begingroup$ thriftbooks.com/w/… $\endgroup$ Jan 13, 2023 at 19:10
  • $\begingroup$ I guess my confusion arises because i dont understand at which "wave" are we looking at. we have the chirp signal which varies its frequency in time, and the carrier wave. are we strictly looking at the carrier's wave phase shift? the phase shift caused by the chirp is insignificant? I really appreciate your help. $\endgroup$ Jan 13, 2023 at 19:41
  • $\begingroup$ You transmit a chirp and then you multiply that chirp by the received chirp, which essentially dechirps in that product as part of your pulse compression process (eliminating the actual carrier frequency). The phase shift in the carrier between what was transmitted and received will be translated to a phase shift in the baseband signal at the output of your received multiply and filter operation. I assume you have a complex baseband signal which will have a magnitude and phase for each sample (as a complex waveform). $\endgroup$ Jan 13, 2023 at 21:51

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