Why does the transmitted LFM signal decrease in amplitude at higher frequencies once it passes through the "phased.FreeSpace" object, even when sampling at the proper Nyquist frequency?
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MathWorks Support Team
2022-2-1
回答: MathWorks Support Team
2022-2-24
I was asking why the transmitted LFM signal decreases in amplitude at higher frequencies once it passes through the "phased.FreeSpace" object, even when sampling at the proper Nyquist frequency (twice the signal bandwidth). The example plots the signal amplitude as a function of time:

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MathWorks Support Team
2022-2-1
The reason why the result for the "phased.FreeSpace" propagation model is not linear with amplitude is due to the fact that the delayed signal does not start on an exact sampled time instant, which results in the distortion in the signal. This is an artifact of performing signal processing with numbers directly and not just using difference equations.
To further explain this effect, try running this code where the signal is sampled at twice the highest frequency (theoretically Nyquist is happy)
fs = 1000;
t = 0:1/fs:100/fs;
x = cos(2*pi*(fs/2)*t);
%Plot looks good
plot(t,x)
%Wait what , where is the signal ? ?
x = sin(2*pi*(fs/2)*t);
plot(t,x)
You will notice that the cosine plot looks perfect, but the sine plot is distorted and close to zero amplitude. But if you add a delay of 1/4th cycle, the sine wave looks perfect again:
%Its sine wave again just by adding a small delay
x = sin(2*pi*(fs/2)*(t-1/(2*fs)));
plot(t,x)
To avoid this effect, try increasing the sampling rate to significantly larger value e.g. fs = 50*temp*prf; you will notice only a linear path loss attenuation in the amplitude of the delayed signal. The path loss is given by the formula (4*pi*r)^2/lambda^2, shown in more detail at the link below:
https://www.mathworks.com/help/phased/ref/phased.freespace-system-object.html#bs7tj7q-1
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