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Problem 12

# A continuous-time time-invariant system with outputs (but no inputs) $$\dot{x}=f(x), y=h(x)$$, having state space $$x=\mathbb{R}^{n}$$, and output- value space $$y=\mathbb{R}^{p}$$, is said to be output-to-state stable (OSS) if there exist a function $$\beta \in \mathcal{K} \mathcal{L}$$ and a function $$\gamma \in \mathcal{K}_{\infty}$$ (terminology is as in Exercise 7.3.11) so that the following property holds: For each $$T>0$$, if $$\xi$$ is a solution of $$\dot{\xi}=f(\xi)$$ on the interval $$[0, T]$$, and $$\eta(t)=h(\xi(t))$$ for $t \in[0, T]$, then $$\|\xi(T)\| \leq \max \left\\{\beta(\|\xi(0)\|, T), \gamma\left(\|\eta\|_{\infty}\right)\right\\} .$$ Show: 1\. A continuous-time time-invariant linear system is OSS if and only if it is detectable. 2\. In general, a system may be detectable (with respect to $$x^{0}=0$$ and $$y^{0}=0$$ ) yet not be OSS.

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In summary, we proved that a continuous-time time-invariant linear system is output-to-state stable (OSS) if and only if it is detectable. We also showed that, in general, a system may be detectable but not be OSS by providing an example of a detectable system for which a Lyapunov function does not exist, demonstrating that the system is not OSS.
See the step by step solution

## Step 1: Recalling Definitions

Before proceeding, let's recall the definitions of detectability, $$\mathcal{K}\mathcal{L}$$, and $$\mathcal{K}_{\infty}$$ functions. 1. A continuous-time, time-invariant linear system $$\dot{x} = Ax + Bu$$, $$y = Cx + Du$$ is said to be detectable if all the eigenvalues of A have negative real parts, or the eigenvalues with non-negative real parts are associated with eigenmodes that can be detected by the output $$y$$. 2. A function $$\beta: (0, \infty) \times [0, \infty) \rightarrow (0, \infty)$$ belongs to the class $$\mathcal{K\mathcal{L}}$$ if it is continuous, strictly increasing in its first argument, decreasing in its second argument, satisfies $$\lim_{s \rightarrow \infty} \beta(s, t) = \infty$$ for each fixed $$t \geq 0$$, and $$\lim_{t \rightarrow \infty} \beta(s, t) = 0$$ for each fixed $$s > 0$$. 3. A function $$\gamma: (0, \infty) \rightarrow (0, \infty)$$ belongs to the class $$\mathcal{K}_{\infty}$$ if it is continuous, strictly increasing, and satisfies $$\lim_{s \rightarrow \infty} \gamma(s) = \infty$$.

## Step 2: Proving Statement 1

We need to show that a continuous-time time-invariant linear system is OSS if and only if it is detectable. ($$\Rightarrow$$): Assume that the linear system is OSS. Then, there exist $$\beta \in \mathcal{K}\mathcal{L}$$ and $$\gamma \in \mathcal{K}_{\infty}$$ functions such that $\|\xi(T)\| \leq \max \left\{\beta(\|\xi(0)\|, T), \gamma\left(\|\eta\|_{\infty}\right)\right\}.$ Since the system is linear, we can rewrite it as $$\dot{x} = Ax$$, $$y = Cx$$. By Lyapunov stability theory, if the system is OSS, then there exists a positive definite matrix P such that the Lyapunov function $$V(x) = x^T P x$$ is decreasing along the system trajectories, which implies the Lyapunov equation $A^T P + PA < 0$ has a positive definite solution P. According to the detectability condition, this ensures the system is detectable. ($$\Leftarrow$$): Conversely, assume that the linear system is detectable. By Lyapunov stability theorem, there exists a positive definite matrix P such that the Lyapunov equation $A^T P + PA < 0$ has a solution P. Therefore, the Lyapunov function $$V(x) = x^T P x$$ is decreasing along the system trajectories, which implies the system is OSS, as required.

## Step 3: Proving Statement 2

Now, we need to show that in general, a system may be detectable (with respect to $$x^{0}=0$$ and $$y^{0}=0$$) yet not be OSS. Consider the following example of a continuous-time, time-invariant system: $\dot{x} = \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}x,\quad y = \begin{bmatrix} 0 & 1 \end{bmatrix}x$ This system is detectable since all the eigenvalues of the matrix A have negative real parts. However, the Lyapunov equation $A^T P + PA < 0$ has no solution P. Therefore, the Lyapunov function $$V(x) = x^T P x$$ does not exist, so the system is not OSS. We have shown an example where a system is detectable yet not OSS, which concludes the proof.

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