Power spectral density of a under sampled signal

Question

DIPAN KAR 2020-5-9

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https://ww2.mathworks.cn/matlabcentral/answers/524225-power-spectral-density-of-a-under-sampled-signal

编辑： DIPAN KAR 2020-5-19

Hi ,

I have a under sampled signal x. which is a 2k Hz signal. This same signal is under sampled at 50Hz. I tried to extract to freq response of this signal but as the siganl is sampled below the Nyquest rate I am not able to get it. I also tried with pwelch() function but it is also not giving me correct reasult.

How should I calculate the power spectral density of this signal x?

fs = 50
t = 0:1/fs:2^22- 1/fs;                 
x =  sin(2*pi*2000*t) ;
plot(t,x);
y = fft(x);     
f = (0:length(y)-1)*50/length(y)*80;
plot(f,abs(y))
title('Magnitude')
figure(2)
num_segments = 32;
window_length = floor(length(x)/num_segments);
[PSDphase ,f] =pwelch(x ,window_length ,[] ,[] ,50 ,'twosided');
h=semilogx( f, (10* log10(PSDphase)), 'b'); %hold on;
grid on;

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Answer 1

Navya Seelam 2020-5-12

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Increase the sampling frequency of the signal to get expected results. When a signal is undersampled , aliasing happens and hence you are not able to get correct result.

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Navya Seelam 2020-5-14

Hi,

One thing to notice here is the signal x is of the form sin(2*pi*n) with n being an integer. So x is actually a zero signal or nearly zero signal (due to precision limitations) and we cannot perform any operation on a zero signal.

DIPAN KAR 2020-5-19

编辑：DIPAN KAR 2020-5-19

Thank you for pointing out the error. But I wanted to do fft of the phase and extract the PSD from it. Is this is the correct way?

fs = 50
t = 0:1/fs:2^22- 1/fs;                 
x =  2*pi*2000*t;
plot(t,x);
y = fft(x);     
f = (0:length(y)-1)*50/length(y)*80;
plot(f,abs(y))
title('Magnitude')
figure(2)
num_segments = 32;
window_length = floor(length(x)/num_segments);
[PSDphase ,f] =pwelch(x ,window_length ,[] ,[] ,50 ,'twosided');
h=semilogx( f, (10* log10(PSDphase)), 'b'); %hold on;
grid on;