Help to generate radar chirp signal
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Hi,
i am interesting in generating a linear chirp signal for simulating radar signal. I find out there is a function in matlab called: _Y = CHIRP(T,F0,T1,F1)_This function generates samples of a linear swept-frequency signal at the time instances defined in array T. The instantaneous frequency at time 0 is F0 Hertz. The instantaneous frequency F1 is achieved at time T1. By default, F0=0, T1=1, and F1=100.
But, i am confused as i do not know which will be a suitable sample rate. Can you help me? Could you write me a simple and brief example? Do i need other data, such as bandwidth or pulsewidth?
3 comentarios
Mike
el 1 de Feb. de 2013
John Petersen
el 4 de Feb. de 2013
The shape of the chirp is based on the amplitude of each successive sample. The sampling rate would be determined by the bandwidth of your radar signal.
Respuestas (2)
Honglei Chen
el 5 de Feb. de 2013
Editada: Honglei Chen
el 5 de Feb. de 2013
0 votos
I suggest you to take look at Phased Array System Toolbox if you have access to it.
There are two issues with chirp:
- It only generates real chirp, if you want a complex band for your radar, you need to do it twice.
- It does not generate a pulse so if you are simulating a pulse radar, you need to pad zero yourself.
If you want to use chirp, the sample rate is defined in the first argument, T. The bandwidth is defined by the difference of F0 and F1 and the pulse width is defined in T1.
14 comentarios
Mike
el 6 de Feb. de 2013
Mike
el 6 de Feb. de 2013
Youssef Khmou
el 6 de Feb. de 2013
hi Mike, the proper sample rate Fs must satisfy : Fs>= 2* F2
in this case : Fs >= 24 GHz
Honglei Chen
el 7 de Feb. de 2013
Let's say your pulse width is 1 second and your duty cycle is 50% so your signal has a PRF of 0.5 Hz. If you can simulate things in the baseband, then all you need is to sample at the Nyquist rate defined by the bandwidth. So in your case, you need the sample rate to be 2*bw=2000Hz. So as an example, using your numbers, you can generate the signal as
bw = 1000;
Fs = 2*bw;
pw = 1;
prf = 0.5;
t = 0:1/Fs:1/prf-1/Fs;
Nsamp = numel(t);
x = zeros(Nsamp,1);
x(1:Nsamp/2)=chirp(t(1:Nsamp/2),0,pw,bw);
plot(t,x)
Mike
el 7 de Feb. de 2013
Honglei Chen
el 7 de Feb. de 2013
T1 is where you want to reach the frequency F1, i.e., the end of your bandwidth. If you don't need the zero after that, then simply set prf as 1 in my previous example. Alternatively, if you really liked your previous t vector, just use that t and set T1 to t(end)
Honglei Chen
el 7 de Feb. de 2013
use t(end), like
chirp(t,0,t(end),1000)
Honglei Chen
el 8 de Feb. de 2013
Yes, this is up to you, I'm really just giving an example here since in one of your comment you used 1000 Hz for bandwidth.
Mike
el 11 de Feb. de 2013
Jaya Srivastava
el 4 de Jun. de 2019
Editada: Jaya Srivastava
el 4 de Jun. de 2019
I have generated chirp signal in matlab and i have to transmit it through an antenna in USRP 2920. But for that I will have to convert the chirp into a complex baseband signal for my radar. How shall i accomplish that?
Honglei Chen
el 4 de Jun. de 2019
You can try hilbert which will return the analytical signal.
Mary Lanzerotti
el 12 de Jul. de 2019
Honglei Chen,
We were given a complex baseband signal, below. How can we create a pulsed waveform from this signal with 50% duty cycle? Also how can we output I(t) and Q(t) (as ascii files)?
% Waveform parameters
pulse_width = 100e-6;
bandwidth = 500e6;
center_frequency = 10e9;
% Receiver parameters
adc_sample_rate = 2e9;
lo_freq = 10e9; % For baseband, lo = center_freq of waveform
% FM Slope
lfm_slope = bandwidth / pulse_width;
% Number of ADC samples in a pulse
nsamp = pulse_width * adc_sample_rate;
% Sample time grid
t = ((0 : nsamp - 1) - 0.5 * (nsamp + 1)) / adc_sample_rate;
% Calculate transmit phase
phi_tx = pi * lfm_slope * t.^2 ...
+ 2*pi * (center_frequency) * t;
% Calculate receiver LO phase
phi_rcv = 2*pi * lo_freq * t;
% Calcuate phase of received signal
phi = phi_tx - phi_rcv;
% Calculate signal
s = exp(1i * phi); % Baseband signal is complex
Youssef Khmou
el 6 de Feb. de 2013
Editada: Youssef Khmou
el 6 de Feb. de 2013
Hi Mike , i suggest to take a look at this topic :
Here is the code :
Fs=1000; % sample rate
tf=2; % 2 seconds
t=0:1/Fs:tf-1/Fs;
f1=100;
f2=400; % start @ 100 Hz, go up to 400Hz
semi_t=0:1/Fs:(tf/2-1/Fs);
sl=2*(f2-f1/2);
f1=f1*semi_t+(sl.*semi_t/2);
f2=f1(end)+f2*semi_t-sl.*semi_t/2;
f=[f1 f2];
y=1.33*cos(2*pi*f.*t);
plot(t,y)
You alter the code by changing f1,f2 and Fs>2*f2 .
Or you can tape "chirp" in the Help Mat, to see the 5 EXAMPLES provided there with a good way to visualize the frequency via spectrogram .
5 comentarios
Mike
el 6 de Feb. de 2013
Youssef Khmou
el 6 de Feb. de 2013
1.because there must no discontinuity in the dynamic frequency,
2.yes Fs>2*Max([f1 f2])
Youssef Khmou
el 7 de Feb. de 2013
Hi, im afraid i can not tell to much about that, you can try to work with snapshots like getting 500 snapshots (small duration) from the source and sample with 24e+9. but an advice is needed from real-time specialist .
Mike
el 7 de Feb. de 2013
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