Using fft() to calculate PSD.

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2one
2one 2012 年 6 月 21 日
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

採用された回答

Wayne King
Wayne King 2012 年 6 月 21 日
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.
  1 件のコメント
2one
2one 2012 年 6 月 21 日
the signal is from 0 to 12 sec, sampled every 0.2 sec (61 points total).
the signal has period of 4 s so frequency peak should be at 0.25 hZ but graph is showing peak at 0.05 Hz (which would mean 20 s period).

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その他の回答 (1 件)

Wayne King
Wayne King 2012 年 6 月 21 日
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])

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