# How does slow-time FFT detect velocity?

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:

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:

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.
• 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: Jan 13, 2023 at 19:10