Changing pulse width for linear FM waveform
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The help on linear FM waveforms states that increasing the pulse width improves detection capability. However, when I tried doubling or halving my pulse width, I did not notice a change in probability of detection. Am I implementing this incorrectly?
I used the basic monostatic radar example to get started and made modifications from there. The code below shows the linear FM waveform initialization followed by a loop used to get my simulated probability of detection, which will be sum(detected)/n_steps.
hWav = phased.LinearFMWaveform(...
'PulseWidth',pulse_width,...
'SweepBandwidth',pulse_bw,...
'SampleRate',fs,...
'PRF',prf); % linear fm waveform instead of rectangular waveform
n_steps = 1000;
detected = false(n_steps, 1);
%hWav.release(); % uncomment these two lines to run with modified pulse width
%hWav.PulseWidth = 1/pulse_bw/2;
for istep = 1:n_steps
x = step(hWav);
[s, tx_status] = step(hTx, x);
rSig = zeros(size(s,1),3);
[tgt_pos, tgt_vel] = step(hTargetplatform, 1/prf);
[tgt_rng, tgt_ang] = rangeangle(tgt_pos);
tSig = step(hRadiator, s, tgt_ang);
tSig = step(hTargetchannel, tSig,...
ant_pos, tgt_pos, ant_vel, tgt_vel);
rSig = step(hTarget, tSig, true);
rSig = step(hCollector, rSig, tgt_ang);
rx_pulses = step(hRx, rSig, ~(tx_status>0));
% *Detection Threshold*
npower = noisepow(noise_bw, hRx.NoiseFigure, hRx.ReferenceTemperature);
threshold = npower * db2pow(npwgnthresh(pfa));
above_inds = abs(rx_pulses).^2 > threshold;
expected_bin = 2*tgt_rng/c*fs + 1;
detected(istep) = above_inds(expected_bin);
end
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Honglei Chen
2015-11-17
0 个投票
This does not seem to have all the relevant code but from what I can see you seem to be comparing the received signal sample value with the noise sample value at the sample point that corresponds to the target location? If so then you won't see the improvement this way. The reason a longer linear FM waveform can improve the detection is because it can pack more energy in the waveform. However, at each sample, the power remains the same. In this case, you are comparing the sample value before the integration, so you are using the power of a single sample, which is the same as before. You will have to use a matched filter or something similar to integrate the power to see the difference.
HTH
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