Coherence vs. Magnitude Squared Coherence

Question

currently I am writing my masther thesis. The theory part is about the turbulent wind field generation, where the coherence (not magnitude squared) is used:

$$\text{coh}(f) = \frac{|P_{xy}|}{\sqrt{P_{xx} P_{yy}}}$$

I am wondering what is the general advantage of using the magnitude squared coherence instead:

$$\text{coh}^2(f) = \frac{|P_{xy}|^2}{P_{xx} P_{yy}}$$

I think this relationship is similar to that of the correlation coefficient $R$ and coefficient of determination $R^2$. I tried to google the difference regarding the coherence, but unfortunately I don't find any (good) explanations.

Hopefully someone can help me. Thanks in advance!

Apollo3zehn

Answer 1

Your analogy between the coherence function and the correlation coefficient is correct. (The slight difference is that $-1 \leq R \leq 1$, whereas $0 \leq \operatorname{coh}(f) \leq 1$ in your definition, likely because it is possible for $P_{xy}$ to be complex-valued.

Intuitively, the cross spectral density (CSD) $P_{xy}$ can be thought of as a measure of the correlation of the spectra of the individual signals at each frequency. Normalizing the CSD by the square root of the product of the individual PSDs creates something akin to a correlation coeffcient, like the $R$ you pointed out.

Answer 2

I believe that the calculation of the magnitude squared coherence is faster for computers (multiplication is faster than square root calculations). And that's pretty much it. :P

Answer 3

I've also been a bit befuddled by which of these I should use. There is no clear standard, unfortunately. To make things worse, I've come across written work where it is unclear which is being used. I've lost a few days trying to decipher this.

Personally, I like magnitude squared coherence. The lack of a sqrt seems cleaner to me. If I were king of the world, I'd make everyone use $\mathrm{coh}^2$.

But---since I'm not ;)---you can use whichever you like. Just be sure to clearly document which one you are using and how you're calculating it. Many people don't realize that you have to do some ensemble averaging of $P$ before computing $\mathrm{coh}$ or $\mathrm{coh}^2$.

Answer 4

it might also be that the magnitude-squared has no discontinuities whereas the magnitude has a discontinuity in its derivative at zero. so the magnitude-square is easier to do calculus with.

it's like why the formula for fitting a line (or any polynomial or sum of basis functions) to data and using the mean-square error instead of the mean-absolute-value error. minimizing the mean-square error, with respect to some parameter, is easy because you can take derivatives without worry that the derivative might not exist for some value.