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.

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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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Chapter 7: Chapter 7

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Step 1: Recalling Definitions

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