I have a great concern about the observability inequality in systems without output equation. I learned (in engineering) that in a state representation of linear systems the equations are:
State equation: \begin{equation} \dot{x} = Ax + Bu \end{equation}
Output equation \begin{equation} y = Cx + Du \end{equation}
In this systems the controllability is check with Kalman Condition, that says if $\text{rank}([A \ AB \cdots A^{n-1}B]) = n$ then the system is controllabe. For the observability the condition is with if $\text{rank}([C \ CA \cdots C^{n-1}A]^T) = n$, then the system is observable.
On the other hand, we have the observability inequality.
\begin{equation}\label{eq13} \int_0^T|B^*\varphi|^2dt\geq c|\varphi(0)|^2, \end{equation}
Now in some books of control (specially in math) the output equations does not appear, and the observability only is determined for the inequality and the implication that controllability is equivalent to observability. I don't know how find a relationship between this concepts.
Thanks. (this is my first question here!!)
From the given information, I am understanding that $\varphi(t)=\exp(A^*t)\varphi(0)$ and that
$$ \int_0^T|B^*\varphi(t)|^2dt=\int_0^T\varphi(t)^*BB^*\varphi(t)dt=\varphi(0)^*\left(\int_0^T\exp(At)BB^*\exp(A^*t)dt\right)\varphi(0), $$
where we can observe that the middle matrix is nothing else but the finite-time controllability Gramian of the system $(A,B)$, which we will denote by $W_T$. Therefore, we have that
$$ \int_0^T|B^*\varphi|^2dt\geq c|\varphi(0)|^2 $$
is equivalent to saying that $W_T-cI$ is positive semidefinite. This means that if $c>0$, the above condition is equivalent to saying that $W_T$ is positive definite, which is equivalent, in turn, to say that the system is controllable at time $T$. In the case of LTI systems, this means that it is also controllable at all $T>0$. This is not true in general for LTV systems.