Doppler Shifting A Chirp Signal

Question

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?

Thanks

Answer 1

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);

Answer 2

Rather than re-sampling, this is best accomplished by re-creating the signal after it is Doppler shifted. This allows any arbitrary waveform to have Doppler imparted on it. See Mark A. Richards, "Fundamentals of Radar Signal Processing," section 2.6.1. and in particular equation 2.90 for details on why this works.

Let's say that we start with a complex chirp:

% tau is the pulse length
% fs is the sampling frequency
% B is the LFM swept bandwidth
% fc is the center RF frequency of the chirp
% tgtrng is the radial range to the target

t = (0:tau*fs-1)/fs;
chirp = exp(1j*pi*B/tau * (t - tau/2).^2);

Calculate the radial velocity of the target using vector math, then:

beta = v_radial/c;
alpha = (1+beta)/(1-beta);

Now resample the chirp by actually creating a new, time-adjusted version

t = (0:fs/PRF-1)/fs;
t_new = alpha*(t - 2*tgtrng/(1+beta)/c);
sig = exp(1j*2*pi*fc*(t_new-t)).*exp(1j*pi*B/tau*(t_new - tau/2).^2);
sig(t_new < 0)=0;
sig(t_new > tau)=0;

sig is now the sampled version of the incoming waveform with Doppler.

Answer 3

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