Detectable System problem

Question

My question is about a proof :

We have the system

$$dx/dt=Ax\\ y=Cx .$$ If $y(t)=0$ and $(A,C)$ is detectable how can i prove that $\lim_{t\to \infty} x(t) =0$ ?

Answer 1

Hint: Start with

$y=Cx$

and differentiate

$$\dot{y}=C\dot{x}=CAx$$ $$\ddot{y} = CA^2x$$ $$\vdots $$ $$\dfrac{d^{n-1}}{dt^{n-1}}y = CA^{n-1}x$$

By this procedure, we get $n$ equations that we can rewrite in the following way $$\begin{bmatrix}y\\ \dot{y}\\ \ddot{y}\\ \vdots \\ \dfrac{d^{n-1}}{dt^{n-1}}y\end{bmatrix} = \begin{bmatrix}C\\CA\\CA^2\\ \vdots \\CA^{n-1} \end{bmatrix}x.$$

From $y(t)=0$ for all $t>t_0$ we know that the right-hand side is a zero vector. If the system $(A, C)$ is controllable we are done. But if we only have a detectable system (Definition: A linear system (continuous or discrete) is detectable if all unstable modes are observable. Source) then we will need to remove the unobservable states (they are asymptotically stable) from the system and do the same procedure with the reduced system. It will be observable such that we can use the previous procedure.

Answer 2

Detectability is often defined via the implication $$ y(t) \equiv 0 \Rightarrow \lim_{t \to \infty} x(t) = 0. $$ But anyways. Let's say that a pair $(A,C)$ is detectable if every unobservable eigenvalue of $A$ is in the left-hand side part of the plane. By contraposition, assume that $y(t) \equiv 0$ but $x(t)$ does not tend to zero as $t\to\infty$. Then $$ \exists \lambda \in \mathbb{C}, \text{ with } Re(\lambda) \ge 0, \quad \exists x_0 \in \mathbb{R}^n, \text{ with } x_0 \not = 0 \, : \quad x(t) = e^{\lambda t} x_0 \ \text{ and } \ y(t) \equiv 0. $$ This implies $$ A e^{\lambda t} x_0 = A x(t) = \dot{x}(t) = \lambda e^{\lambda t} x_0 $$ and $$ 0 \equiv y(t) = C x(t) = e^{\lambda t} C x_0 . $$ Evaluating the first at $t=0$ gives $ A x_0 = \lambda x_0 $. while the second implies $C x_0 = 0$. This means that $$ \exists \lambda \in \mathbb{C}, \text{ with } Re(\lambda) \ge 0, \quad \exists x_0 \in \mathbb{R}^n, \text{ with } x_0 \not = 0 \, : \quad \begin{bmatrix} \begin{array}{c} \lambda I - A \\ C \end{array} \end{bmatrix} x_0 = 0 . $$ which (in view of the Popov-Belevitch-Hautus test) proves the claim.