# Comparison of PSD calculating methods.

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Alex Bobrovnik 2011 年 11 月 20 日
Hello everyone. I have some problems with my Matlab code. I need to compare 3 methods of calculating PSD:
1. Welch-method;
2. FFT of autocorrelation function;
3. Averaging of frequency bands;
close all;
clear all;
clc;
fs=16000;
t=0.1;
b1=rand(1,fs*t);
1)
% PSD method 1 (welch)
figure(1)
h=spectrum.welch;
psd(h,b1)
% it works correctly, here i use some different windows and so on
figure(2)
wlen=256 ;
h=spectrum.welch('hamming',wlen,0.5)
psd(h,b1)
2)
%PSD method 2 (xcorr)
rxx=xcorr(b1); %getting ACF
PSD=abs(fft(rxx));
figure(3);
plot(PSD);
• i don't think that it's right, because of strange shape of PSD;
• how to make ACF symmetrically of "0"?
• if i want to "cut" ACF by the "triangular" window, what should i do?
3) ...

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### 採用された回答

Wayne King 2011 年 11 月 21 日
Fs = 1000;
t = 0:1/Fs:1-1/Fs;
x = cos(2*pi*100*t)+randn(size(t));
Rxx = xcorr(x,'biased');
Rxxdft = abs(fftshift(fft(Rxx)));
freq = -Fs/2:Fs/length(Rxx):Fs/2-(Fs/length(Rxx));
plot(freq,Rxxdft);

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### その他の回答 (3 件)

Alex 2011 年 11 月 20 日
xcorr adds some padding numbers to either side. I forget how to manually truncate the series to take out the side bands (and I don't have Matlab at home).
So, here is one idea: calculate the auto correlation yourself. The function is pretty easy to implement with a few for loops. However, this method will take more processing time (Matlab's xcorr function using the fft & ifft to calculate it, which is much faster).
Second, do you know if your auto-corr should be biased or not? that will give you slightly different vairiations.
Third, the plot((abs(fft)) does not plot 0 at the center.
lets say x = -4:5, then plot(abs(fft(ifft(x))) will end up plotting (1,2,3,4,5,-4,-3,-2,-1).
to get the plot to come out right, use something like this:
rez = abs(fft(xcorr))
rez = circshift(rez',round(length(rez)/2))
plot(rez)
##### 5 件のコメント表示非表示 4 件の古いコメント
Alex Bobrovnik 2011 年 11 月 21 日
and about windowing... "cutting" ACF gives us smoother estimation, isn't it? I want to show this effect. I mean if i'll use 64 points of ACF or 128 points it'll give different shape.

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Wayne King 2011 年 11 月 21 日
Hi Alex, If the time series is zero-mean, the periodogram is equivalent to Fourier transform of the biased autocorrelation sequence.
##### 4 件のコメント表示非表示 3 件の古いコメント
Wayne King 2011 年 11 月 21 日
yes, it's fine to have the psd of a nonzero mean process, I never said it wasn't. All I said was, if you are comparing PSD methods, then you can easily make your process zero mean, that's not going to affect any comparison of PSD methods, and then you can use the periodogram which is the same as the Fourier transform of the biased autocorrelation sequence.

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Alex Bobrovnik 2011 年 11 月 21 日
Today I tried to improve my code:
... % % PSD method 2 (xcorr) % Calculate the Power Spectral Density of Original Signal%
Rxx=xcorr(b1,b1,length(b1)-1);
Pxx=abs(fftshift(fft(Rxx)));
figure(2)
subplot(211)
plot(Rxx)
title('Autocorrelation function');
subplot(212)
plot(Pxx)
title('PSD');
...
but i'm not sure... i can't understand why ACF isn't symmetric on 0
##### 2 件のコメント表示非表示 1 件の古いコメント
Alex 2011 年 11 月 21 日
I just ran your code, and I'm getting what I expect, the spikes centered at 0.
Going back to what you mentioned in commenting on my post earlier, you can see the differences between using different # of points with:
Rxx64 = xcorr(b1,64,'biased');
Rxx128 = xcorr(b1,128,'biased');

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