I want to know how to convert these lines of code into c
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fs = 70e6; % sampling frequency
b = 25e6; % bandwidth
trans_freq = 9.375e9; % Carrier frequency
sweep_time = 1e-3; % sweep time
total_sweeps = 4;
t=0:1/fs:sweep_time; % sweep time matrix
c = 3e8;
range = 24000;
velocity = 50; % velocity in m/s
fd = (2*trans_freq*velocity)/c;
% fd = 0;
td2 = 2*range/c; % delay in seconds
fb = (2*b*range)/(sweep_time*c);
td = round((2*range/c)*fs); % delay in number of samples w.r.to range
beat = sin(2*pi*fd*t);
z = zeros(1,td);
for i = 1:total_sweeps;
if(mod(i,2)== 0)
tx_dn=chirp(t,b,sweep_time,0);
tx((sweep_time*fs*(i-1)+i):sweep_time*fs*i+i) = tx_dn;
rx_dn=chirp(t,b+fd,sweep_time,0+fd);
rx((sweep_time*fs*(i-1)+i):sweep_time*fs*i+i) = rx_dn;
% rx((sweep_time*fs*(i-1)+i):sweep_time*fs*i+i) = rx_dn;
rx_4mix(i,:) = [z rx_dn];
tx_4mix(i,:) = [tx_dn z];
else
tx_up=chirp(t,0,sweep_time,b);% Start @ DC, cross 25MHz at t=3 millisec
tx((sweep_time*fs*(i-1)+i):sweep_time*fs*i+i) = tx_up;
rx_up=chirp(t,0+fd,sweep_time,b+fd);% Start @ DC+fd, cross 25MHz at t=3 millisec
rx((sweep_time*fs*(i-1)+i):sweep_time*fs*i+i) = rx_up;
% rx((sweep_time*fs*(i-1)+i):sweep_time*fs*i+i) = rx_up;
rx_4mix(i,:) = [z rx_up];
tx_4mix(i,:) = [tx_up z];
end
end
rx = [z rx];
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回答(1 个)
Babak
2013-4-10
Use MATLAB coder to convert matlab code into C code.
2 个评论
Jan
2013-4-11
This is obviously not trivial. Commands like chirp require a lot of work and even 0:1/fs:sweep_time is not trivial.
I assume a professional C programmer can convert this in 40 hours. But as usual for programming projects, the doubled time is realistic also.
Why do you want to convert this to C?
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