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So I know about the basics of doppler shift which relates the velocity of a moving object with the frequency carrier shift seen at the reciever. But, I don't understand the importance of the transmitted signal bandwidth to the process? I can use the same the same carrier frequency but with different BW... Any clarification please? Thanks

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Bandwidth affects the "gate/cell width" of your measurements. If you think about an LFM waveform reflecting back off an object and going through a matched filter to determine the range, a signal with a very low bandwidth is going to have a more ambiguous return time (range) than a more aggressive chirp. The rough calculation for the smallest range resolution for a certain bandwidth is $\frac{c}{2B}$.

I imagine the relationship is reversed for doppler processing. A signal with low bandwidth would provide a lot of doppler resolution because there isn't a great deal of ambiguity as to the return's center frequency, so the doppler gate width should increase proportional to bandwidth. That is a guess, though.

Radar and other signal processing disciplines involve these tradeoffs a lot. Certainty in frequency versus certainty in time and vice versa, maximum unambiguous range versus improvement factor from pulse repetition frequency, etc.

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  • $\begingroup$ Thank you very much for the reply. Why is it that "a signal with a very low bandwidth is going to have a more ambiguous return time (range) than a more aggressive chirp" ? $\endgroup$
    – user3921
    Dec 22 '20 at 20:09
  • $\begingroup$ Because the front and end of the chirp will be more similar. If you imagine a pulse as being just a sine wave (effectively no bandwidth) then it's entirely unclear when you're matching whether a portion of the return is the front, middle, or end of a pulse. $\endgroup$
    – Keegs
    Dec 22 '20 at 20:13

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