Using fft() to calculate PSD.
7 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
I used the code below perform spectral analysis of signal with known period of 4 s:
load lujan.dat
data_out.time = lujan(:,1);
data_out.disp = lujan(:,2);
Y = fft(data_out.disp,1000);
Y(1)=[];
NY = length(Y);
Pyy = abs(Y(1:floor(NY/2))).^2;
nyquist = 1/2;
f = (1:NY/2)/(NY/2)*nyquist;
plot(f,Pyy,'-b','LineWidth',3);
title('Power spectral density');
xlabel('Frequency (Hz)');
However the output graph shows a peak at 0.05 Hz which is not correct! IS there something obvious i'm doing wrong?
thanks
0 Kommentare
Akzeptierte Antwort
Wayne King
am 21 Jun. 2012
Are you sure the sampling frequency is 1 ? If so, how far off is the peak from where you expect it to be? Is it almost correct or way off? You also don't tell us how long the time series is, so we don't know why you are adding the length input argument to fft() . Your frequency increment of 1/NY (0.0025) is correct.
Weitere Antworten (1)
Wayne King
am 21 Jun. 2012
Then your sampling frequency is specified incorrectly.
Fs = 1/0.2;
% here I'll create a signal just to show you
t = 0:1/Fs:12-1/Fs;
x = cos(2*pi*0.25*t)+randn(size(t));
Y = fft(x,256);
Pyy = abs(Y(1:floor(256/2)+1)).^2;
freq = 0:Fs/256:5/2;
plot(freq,10*log10(Pyy))
grid on
set(gca,'xtick',[0 0.25 0.5 1 1.5 2 2.5])
0 Kommentare
Siehe auch
Kategorien
Mehr zu Spectral Measurements finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!