Help with Peig function
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Karin Norén-Cosgriff
el 30 de Ag. de 2018
Comentada: Karin Norén-Cosgriff
el 4 de Sept. de 2018
Hi, I have problems understanding the meaning of nwin and nooverlaps in Peig. In the documentation it is stated that nwin is the window length and noverlap is the number of input sample points by which successive windows overlap. But the defualt values are 2xp(1) and nwin-1 respectively. Having a 18000 sample signal with one expected embedded sinusoid, I use p(1) = 2 in Peig. I would like to compare the results using a window length equal to the number of samples (and no everlaps), and dividing the signal in blocks with 6000 samples each and using 50 % overlap. How do I do this? Can someone explain and provide me with an example? Karin
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Vishal Bhutani
el 3 de Sept. de 2018
By my understanding you want to compare results of peig function using default values of nwin and noverlap. So you can find the sample example hope it will help:
fs = 100;
t = 0:1/fs:1-1/fs;
s = 2*sin(2*pi*25*t)+sin(2*pi*35*t)+randn(1,100);
X = corrmtx(s,12,'mod');
figure(1)
peig(X,2,512,fs) % default values of nwin = p(1) = 2 and noverlap = nwin-1.
figure(2)
peig(X,2,512,fs,20,10); % specifying nwin = 20, noverlap = 10.
For your case you can specify nwin = 6000 and noverlap = 3000.
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