So I have a digital linear chirp signal:

$$ x[n] = \sqrt{\frac{2}{N}} \cdot \cos\Big( 2 \pi \frac{f_{0}}{f_{s}}n + \pi \frac{B}{N} n^{2}\Big) $$

Where $f_{0}$ is the initial frequency, $f_{s}$ is the sampling frequency, $B$ is the bandwidth and $N$ is the number of samples.

I want to simulate a Doppler shift. I am not sure exactly how to do this.

I know I need to multiply by initial frequency $f_{0}$ by the Doppler shift $d$, and do the same with the bandwidth. This will give me:

$$ x_{d}[n] = \sqrt{\frac{2}{N}} \cdot \cos\Big( 2 \pi \frac{d \cdot f_{0}}{f_{s}}n + \pi \frac{d \cdot B}{N} n^{2}\Big) $$

I know Doppler compresses or expands my waveform in time, and I know how to simulate the Doppler shift in the case of a simple sinusoid. However, I am a bit confused here.

Do I need to change the number of samples $N$ as well? If so, by how much exactly?



You need to resample the signal to simulate the Doppler induced dilation. The resampling factor is I=Td/Ts where Td = Duration of the signal after dilation Ts = Actual duration of the transmitted signal

In Matlab you can use the function resample(), but you need to find the resampling parameters P and Q from I.

Also, I=1+v/c for signal expansion and I=1-v/c for signal compression. v=speed of movement in m/s and c=speed of wave propagation in the medium. For instance sound underwater c=1500 m/s.

In Matlab, you can find P and Q as [P,Q]=rat(I);

  • $\begingroup$ Okay, but how can I calculate the duration of the signal after the Doppler shift. This is what I am confused about how to do. $\endgroup$ – The Dude Mar 31 '19 at 18:00
  • $\begingroup$ See my edited answer above. $\endgroup$ – Harris Mar 31 '19 at 23:52
  • $\begingroup$ Thanks! Could you point to a proof for this idea? ie that the time compression function takes this exact form? Ie, when $I = d$, $d$ being the Doppler shift? Because the guy below (@Digiproc) seems to claim that $I = \frac{1}{d}$ $\endgroup$ – The Dude Apr 1 '19 at 0:54
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    $\begingroup$ Look into this book for the details: A. W. Rihaczek,Principles of High-Resolution Radar: Peninsula Publishing, 1985. $\endgroup$ – Harris Apr 1 '19 at 1:06
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    $\begingroup$ Check also this paper out: ieeexplore.ieee.org/abstract/document/820736 $\endgroup$ – Harris Apr 1 '19 at 1:26

The number of samples should be divided by the Doppler shift, because the duration of the chirp decreases with an increase in Doppler shift.

  • $\begingroup$ Right, so in my equation above, I should replace $N$ by $\frac{N}{d}$, with $d$ being the Doppler shift? $\endgroup$ – The Dude Mar 31 '19 at 18:01

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