Changing pulse width for linear FM waveform

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

 採用された回答

Honglei Chen
Honglei Chen 2015 年 11 月 17 日

0 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

その他の回答 (0 件)

質問済み:

2015 年 11 月 14 日

回答済み:

2015 年 11 月 17 日

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by