Filter löschen
Filter löschen

Finding the auto-correlation of the noise signal x explicitly using the FFT() and IFFT() functions

5 Ansichten (letzte 30 Tage)
Hello,
Been trying for hours to calculate the auto-correlation of a noise signal without using the xcorr operator and not sure if my results are right. My tutor wants us to find the auto-correlation by calculating the auto-spectra, the auto-covariance from the spectra and finaly the auto-correlation from the auto-covariance without using matlab operators. My plots show a peak at zero, which I think is positive, with the auto-correlation having a peak at zero with the max value of 1 in the vertical axis, which seems correct. Although my x-axis show the 0:N points and I don't know how to display the x-axis correct. From what I understand should be a time-axis. I'm also not sure my results are correct or it was just a coincidence. Would be good to have some feedback and tips. The code is underneath a bit long but I hope it's clear. thanks you so much. Code:
fs=11025; %sampling frequency
N=1024; %number of points
T=N/fs; %time
x=2*rand(N,1)-1; %generate noise signal
%creating a delay of signal x
d=zeros(64,1);
x_delay=cat(1,d,x);
%fft of signal x and delay
X=fft(x);
X_delay=fft(x_delay);
%windowing signal x
h=hanning(2*N);
hX=X.*h(1:N);
HX=fft(hX);
%windowing signal x_delay
H=hanning(2*length(X_delay));
hX_delay=X_delay.*H(1:length(X_delay));
HX_delay=fft(hX_delay);
%finding the auto-spectra
for i=1:N
Sxx(i)= (HX(i).*conj(HX_delay(i))/T);
end
%finding auto-covariance
Sxxt=ifft(Sxx);
%plotting the auto-covariance
subplot(4,1,2)
plot(abs(Sxxt));
title('auto-covariance')
%finding the boundary of the auto-covariance Sxx(0)
Sxx0=mean(Sxx);
%finding the auto-correlation
Rxx=Sxxt/Sxx0;
%plotting the auto-correlation
subplot(4,1,3);
plot(d,abs(Rxx));
title('auto-correlation')

Akzeptierte Antwort

Wayne King
Wayne King am 7 Mär. 2012
Hi, here is an example:
x = randn(16,1);
xpsdmag = abs(fft(x,2^nextpow2(2*length(x)-1))).^2;
acf = ifft(xpsdmag);
acf = acf(1:length(x));
acf = acf./acf(1);
% compare to xcorr()
[xc,lags] = xcorr(x,'coeff');
xc = xc(length(x):end);
isequal(xc,acf)
  3 Kommentare
Wayne King
Wayne King am 7 Mär. 2012
Hi Catarina, I did that just so I could demonstrate perfect agreement with xcorr()
To plot the normalized acf I would use stem()
x = randn(16,1);
xpsdmag = abs(fft(x,2^nextpow2(2*length(x)-1))).^2;
acf = ifft(xpsdmag);
acf = acf(1:length(x));
acf = acf./acf(1);
stem(0:length(acf)-1,acf); set(gca,'ylim',[-1.5 1.5]);
xlabel('Lags');
Tahariet Sharon
Tahariet Sharon am 10 Mär. 2018
Hi,
I was just wondering, if you don't square (.^2) in line 2 of your answer, would it make a difference if we are interested in computing the ACF? I have seen this formula used for ACF but without squaring the absolute value. Thanks!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by