May I ask from the linear control theory community whether the solution of the following problem are available in the literature?
My system is $E \dot x=Ax+Bu, Y=Cx+Du$, $E,A$ are $m\times n$, $m\ne n$ matrices and there is no special assumption is made on the $\lambda E-A$ matrix pencil.
Detectability is necessary but not sufficient condition for the existence of Luenberger observer.
Causal Detectability is necessary and sufficient condition for the existence of Luenberger observer.
It will be great if I get some references/texts regarding these observer design line which shows the light on my asked questions. Thanks for helping.
This is more of an intuition based approach, you need to verify if this makes sense. You could multiply the equation with $E^T$ from the left to obtain
$$E^TE\dot{x}=E^TAx+E^TBu,$$
if you are lucky you will obtain an invertible matrix $E^TE$ such that you can multiply with the inverse of this matrix to obtain
$$\dot{x}=(E^TE)^{-1}E^TAx+(E^TE)^{-1}E^TBu.$$
Now, you would have obtained the standard form $$\dot{x}=\hat{A}x+\hat{B}u,$$ in which
$$\hat{A}=(E^TE)^{-1}E^TA~\text{and}$$ $$\hat{B}=(E^TE)^{-1}E^TB.$$
If $E^TE$ is not invertible, then you could try the same with $A^T$ from the left to obtain $$A^TE\dot{x}=A^TAx+A^TBu.$$
If $A^TE$ is invertible, then you can do the same as in the previous case.