Do I need a receiver channel when I using ULA?

I'm setting a FMCW radar simulation platform with ULA.
I read example code in helperFMCWTwoRaySimulate.m and Automotive Adaptive Cruise Control Using FMCW Technology.
for m = 1:Nsweep
% Update radar and target positions
[radar_pos,radar_vel] = radarmotion(sweeptime);
[tgt_pos,tgt_vel] = carmotion(sweeptime);
% Transmit FMCW waveform
sig = waveform();
txsig = transmitter(sig);
% Propagate the signal and reflect off the target
txsig = txchannel(txsig,radar_pos,tgt_pos,radar_vel,tgt_vel); % propagate the signal
rxsig = cartarget(txsig);
rxsig = rxchannel(rxsig,radar_pos,tgt_pos,radar_vel,tgt_vel); % propagate the signal
% Dechirp the received radar return
rxsig = receiver(rxsig);
xd = dechirp(rxsig,sig);
xr_unmixed(:,m) = rxsig;
xr(:,m) = xd;
end
In % Propagate the signal and reflect off the target, it used two channels for RX and TX respectively
I noticed that if I set a ULA, I don't have to use a reciver channel after collector. Is that right?
for m = 1:Nsweep
% Update radar and target positions
[radar_pos,radar_vel] = radarmotion(waveform.SweepTime);
[tgt_pos,tgt_vel] = carmotion(waveform.SweepTime);
% Transmit FMCW waveform
sig = waveform();
txsig = transmitter(sig);
% Radiate the pulse toward the target
txsig = radiator(txsig,tgtang);
% Propagate the signal and reflect off the target
txsig = channel(txsig,radar_pos,tgt_pos,radar_vel,tgt_vel);
rxsig = cartarget(txsig);
% txsig = channel(txsig,tgt_pos,radar_pos,tgt_vel,radar_vel);
% Collect the echo from the incident angle at the antenna
rxsig = collector(rxsig,tgtang);
% Dechirp the received radar return
rxsig = receiver(rxsig);
dechirpsig = dechirp(rxsig,sig);
xr(:,:,m) = dechirpsig;
end

 Respuesta aceptada

Honglei Chen
Honglei Chen el 2 de En. de 2019

0 votos

The receiver also models the thermal noise in the receiver. If you need to include that effect, then you should not skip the receiver.
HTH

1 comentario

Wei-Min Chen
Wei-Min Chen el 2 de En. de 2019
Thanks for your answer !
I just read the documents. It's a problem about one-way propagation or two-way propagation.

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el 2 de En. de 2019

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