Using fft() to calculate PSD.
7 ビュー (過去 30 日間)
古いコメントを表示
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 件のコメント
採用された回答
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.
その他の回答 (1 件)
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])
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Spectral Measurements についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!