Why does periodogram varies with data length?

Hello,

I'm trying to understand better how periodogram works by using it with pure sinusoids.

Fs = 1000;
t = 0:1/Fs:1-1/Fs;
x = sin(2*pi*100*t);
[psdestx,Fxx] = periodogram(x,[],length(x),Fs);
plot(Fxx,psdestx); grid on;
xlabel('Hz');
title('Periodogram Power Spectral Density Estimate');

I took most of the above code from a MATLAB example in the documentation center: http://www.mathworks.com/help/signal/ug/psd-estimate-using-fft.html

The response, 0.5 peak at 100 Hz, seems correct to me, since the theoretical average power of a sinusoid is (A^2)/2. It's also the same answer I get when I type:

mean(x.^2)

My doubt arises when I increase data length, from the code's second line:

t = 0:1/Fs:2-1/Fs;

Why does this change my PSD estimate? Since I'm dealing with a periodic signal, shouldn't the average power remains constant, despite data length?

Thank you, Vinícius

Answer 1

[Wayne King]

Because the periodogram is strictly a PSD estimate unless you specify the 'power' option. By doubling the time interval you are integrating over twice the interval.

If you specify the 'power' option, you'll get what you expect 0.5

Fs = 1000;
 t = 0:1/Fs:2-1/Fs;
 x = sin(2*pi*100*t);
 [psdestx,Fxx] = periodogram(x,[],length(x),Fs,'power');
 plot(Fxx,psdestx); grid on;
 xlabel('Hz');
 title('Periodogram Power Estimate');

Answer 2

[Vinicius]

Thanks for the reply Wayne.

The 'power' option doesn't work in my version of MATLAB (2012a), but I could replace it by 'ms'. Do you know if it's the same thing? I can't see it described in any documentation.

Also, could you clarify the meaning of the values I see in the plot with default options? For example, let's say I see a peak of 0.5 at 100 Hz and a peak of 2 at 200 Hz, in a otherwise "clean" periodogram. I thought this would mean that, to reconstruct the original signal, I would need two sinusoids, with 100 and 200 Hz frequency, each with average power of 0.5 and 2, respectively (plus phase adjustment). Is that not correct?

Finally, is the plot with default options more meaningful, or more used? And why?

Answer 3

[Wayne King]

In R2012a, you can use spectrum.periodogram and then msspectrum

    Fs = 1000;
    t = 0:1/Fs:2-1/Fs;
    x = 0.5*cos(2*pi*100*t)+2*sin(2*pi*200*t);
    hper = spectrum.periodogram;
    hper.Windowname = 'Flat top';
    hms = msspectrum(hper,x,'Fs',Fs);
    plot(hms.Frequencies,hms.Data)

Looking at the plot (this is a one-sided power spectrum), the intepretation is this, the power at 100 Hz is 0.125 so you have a cosine (ignoring phase) with an amplitude of

sqrt(2*0.125)

The power estimate at 200 Hz is 2 so you have a sinusoid with an amplitude of

sqrt(2*2)

Answer 4

[Vinicius]

Great, this was exactly my understanding!

I just still don't get what the plot with the default option means. And why is that even the default? This second option seems much more logical and easy to interpret.