I'm having trouble with the observability gramian. Here's the question:
Given a system $\dot{x}(t) = Ax(t), \quad y(t)=Cx(t)$ with $A = \begin{bmatrix} -2 & 0 \\ 1 & -1 \\ \end{bmatrix} $ and $C= \begin{bmatrix} 1 & 1 \\ \end{bmatrix} $
a) Determine the unobservable space by means of the finite-time observability gramian $W_{O}(T)$ with $T>0$.
b) Determine the unobservable space by means of the infite-time observability Gramian $W = lim_{T \rightarrow \infty} W_{O}(T)$
What I Think:
a) Can be computed using $W_{O}(T)$ = $$\sum_{t=0}^{T-1} (A^{'})^{t}C^{'}CA^{t} $$ in which $A'$ and $C'$ denote the transpose of $A$ and $C$.
I don't know at what $t$ I should stop the summation and I don't know if $A^0$ = identity or not?
b) can be computed using $A^{'}W+WA=-C^{'}C$ solving this gives $W$=\begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \end{bmatrix} the unobservable space then equals im($W$) which is \begin{bmatrix} 1\\ 1 \\ \end{bmatrix}
Thank in advance for the help.
Kind regards,
For the system you give, the observability grammian is $\begin{bmatrix} C\\ CA\end{bmatrix}$ or $$ W_o = \begin{bmatrix} 1 & 1\\ -1 & -1\\ \end{bmatrix}. $$ $W_o$ is rank 1 < 2, so the system fails to be observable. The unobservable subspace is exactly the kernel (null space) of $W_o$, which must be of dimension $2-1 =1$. Since $[1,-1]^{tr}$ is an element in the null space of $W_o$ then $\ker W_o = \text{span}\{[1,-1]^{tr} \}$, which is exactly the unobservable subspace of the system.
Note that you never need to actually calculate $\int e^{A^*t}C^*Ce^{At}dt$ for a finite dimensional system! This is always the same as $\begin{bmatrix} C\\ CA\\\ CA^2 \\ \vdots \\ CA^{n-1}\\ \end{bmatrix}$ for a state space of dimension $n$.