Why is noise required to get expected Magnitude Squared Coherence (mscohere)

Question

Jerry Gregoire 2015-3-23

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/184686-why-is-noise-required-to-get-expected-magnitude-squared-coherence-mscohere

编辑： Jerry Gregoire 2015-3-24

Hi all,

I am missing something with magnitude Squared Coherence and/or its algorithm. If two signals are compared without or with little noise I get unexpected results. As an example taking from the ML help page:

Fs = 1000; t = 0:1/Fs:1-1/Fs;

x = cos(2*pi*100*t)+sin(2*pi*200*t)+0.5*randn(size(t)); y = 0.5*cos(2*pi*100*t-pi/4)+0.35*sin(2*pi*200*t-pi/2)+ ... 0.5*randn(size(t)); [Pxy,F] = mscohere(x,y,hamming(100),80,100,Fs);

gives the expected two peak response. I would have thought that with no noise the mscohere would be similar and even stronger but it is not. Run the same code without the noise

x = cos(2*pi*100*t)+sin(2*pi*200*t); y = 0.5*cos(2*pi*100*t-pi/4)+0.35*sin(2*pi*200*t-pi/2);

[Pxy,F] = mscohere(x,y,hamming(100),80,100,Fs);

and rather than getting two strong peaks and the rest near or at zero, you get unity for all frequencies.

You don't need much noise, 0.5% or -46dB will do. Below this and the results get real funky.

Furthermore, without some noise the algorithm sees harmonics very strongly even though they are not in both signals:

x = cos(2*pi*100*t)+sin(2*pi*200*t)+0.5*randn(size(t)); y = 0.5*cos(2*pi*100*t-pi/4);

still gives two strong peaks at 100 and 200 unless y has noise. Then all is as expected.

Why is this?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Greg Dionne 2015-3-23

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/184686-why-is-noise-required-to-get-expected-magnitude-squared-coherence-mscohere#answer_172460

编辑：Greg Dionne 2015-3-23

You might be thinking of CPSD instead of MSCOHERE. MSCOHERE will normalize CPSD by the PSD of each signal (i.e. Cxy = (abs(Pxy).^2)./(Pxx.*Pyy), where Pxy is the CPSD. When the spectrum of Pxy, Pxx and Pyy are very near zero, you'll be looking at division of two numbers very close to zero. Adding the noise decouples them somewhat.

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Jerry Gregoire 2015-3-24

编辑：Jerry Gregoire 2015-3-24

That was quick and makes sense. mscohere is what I want since it is normalised. I just need to be aware of the singularitiy that noiseless signals create. Hadn't thought of that. Thanks

请先登录，再进行评论。