The other day I was bored so I implemented a Canonical Correlation Analysis (CCA) for Blind Source Separation (BSS).
According to wikipedia it is possible to show that $$c \text{ is an eigenvector of } \Sigma_{XX}^{-1/2}\Sigma_{XY}\Sigma_{YY}^{-1}\Sigma_{YX} \Sigma_{XX}^{-1/2}$$
where $\Sigma_{AB}$ is correllation matrix between $A$ and $B$.
Now if we choose $Y$ to be some small shift of $X$, we can achieve a BSS effect.
This works fine for example for members of the exponential family $$f_k(t) = Re\left(\exp\left(-\frac{t^n}{\sigma^n}+2\pi i (n_k(t+ \varphi_k))\right)\right)$$
We generate as linear combination: $X_k(t) = \sum_{\forall i} w_{ki} \cdot f_k(t)$ and are able to find $w_{ki}$ by the procedure described at wikipedia. Subsequently we can then find $f_k(t)$ by inverse linear transform.
As long as $n_k$ are unique. But as soon as there are two or more $n_k$ which are equal, then we do not manage to find the correct $f_k(t)$ from the $X_k(t)$ by this method.
Example where it works reasonably well. First picture is random mixture. Second picture is the original functions with found solution that is overlapped. We use Gaussian window, e.g. $n = 2$.
Own work I suspect that the reason for this in the case of $n=0$ exponent is that we can always find $$c_1\sin(kt+\varphi_1) + c_2\sin(kt+\varphi_2) = c_3\sin(kt) + c_4\cos(kt)$$
But I am not sure (why) this must hold for exponential envelopes with $n>0$.
It feels like it should be an easy exercise in functional analysis to show whether two arbitrary $\varphi_1,\varphi_2$ will fill the same space also for other exponents, but I don't remember details.