I am working on LTI system study and I am somewhat confused about what a system being observable actually means from the perspective of the application.
For LTI systems observability proofs usually involve taking the output vector $y(t)$ and $n$ derivatives and solving a system of linear equations using the fact the observability matrix is full rank.
My question is, how are the derivatives of the output vector supposed to be found? It seems like this theorem is purely of theoretical significance since I don't see a way for someone to compute the output vectors derivatives from measurements.
Usually, the derivatives of the output vector are not supposed to be found; instead, an observer is constructed. An observer is an auxiliary system that provides an estimate of the state of a given system.
Consider, for example, a LTI system \begin{equation}\tag{1} \left\{ \begin{array}{l} \dot x= Ax\\ y=Cx, \end{array}\right. \end{equation} where $x\in\mathbb R^n$ is a state vector, $y\in\mathbb R$ is a scalar output, the system (1) is observable, i.e. the observability matrix $$ V=\left( \begin{array}{c} C\\AC\\A^2C\\\vdots\\A^{n-1}C \end{array}\right) $$ is invertible. Also consider the observer system $$ \dot{\hat x}= A\hat x+F(y-C\hat x),\tag{2} $$ where $F$ is some matrix of size $n\times 1$. The input $y$ of the observer is the output of the real system (1). Denote the error vector $e(t)=x-\hat x$; $$ \dot e= Ax-A\hat x-F(y-C\hat x)= A(x-\hat x)-F(Cx-C\hat x)= (A-FC)e.\tag{3} $$ Since $V$ is invertible, it is always possible to choose (using, for example, the Ackermann's formula: $F^T= (0\; 0\; \ldots\; 1) (V^T)^{-1} P(A^T)$) the column $F$ to make the system (3) asymptotically stable, i.e. $$ \lim_{t\to\infty} e(t)= \lim_{t\to\infty} (x(t)-\hat x(t))=0. $$ Hence, $\hat x(t)$ is an estimation of the (unknown) state $x(t)$ of the system (1).