Chapter 7: Problem 3
If \(\Sigma\) is internally stable, then \(\Lambda_{\Sigma}\) is UBIBO.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 7: Problem 3
If \(\Sigma\) is internally stable, then \(\Lambda_{\Sigma}\) is UBIBO.
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Let \(\Sigma_{1}\) and \(\Sigma_{2}\) be two systems, and let $$ k_{i}: y_{1} \times y_{2} \rightarrow \mathcal{U}_{i}, \quad i=1,2 $$ be two maps. We say that the interconnection of \(\Sigma_{1}\) and \(\Sigma_{2}\) through \(k_{1}\) and \(k_{2}\) is well-posed if for each \(\left(x_{1}, x_{2}\right) \in X_{1} \times \mathcal{X}_{2}\) there exist unique functions $$ \xi_{i}:[0, \infty) \rightarrow X_{i}, \quad i=1,2 $$ such that, with the notations $$ \begin{aligned} \eta_{i}(t) &:=h_{i}\left(\xi_{i}(t)\right) \\ \omega_{i}(t) &:=k_{i}\left(\eta_{1}(t), \eta_{2}(t)\right) \end{aligned} $$ it holds that \(\xi_{i}=\psi_{i}\left(x_{i}, \omega_{i}\right)\) for \(i=1,2\).
Denote by \(\mathcal{K}\) the class of functions \([0, \infty) \rightarrow[0, \infty)\) which are zero at zero, strictly increasing, and continuous, by \(\mathcal{K}_{\infty}\) the set of unbounded \(\mathcal{K}\) functions, and by \(\mathcal{K} \mathcal{L}\) the class of functions \([0, \infty)^{2} \rightarrow[0, \infty)\) which are of class \(\mathcal{K}\) on the first argument and decrease to zero on the second argument. A continuous-time time-invariant system \(\dot{x}=f(x, u)\) with state space \(X=\mathbb{R}^{n}\) and control-value space \(\mathcal{U}=\mathbb{R}^{m}\) is said to be input-to-state stable (ISS) if there exist a function \(\beta \in \mathcal{K} \mathcal{L}\) and a function \(\gamma \in \mathcal{K}_{\infty}\) so that the following property holds: For each \(T>0\), if \(\omega\) is a control on the interval \([0, T]\) and \(\xi\) is a solution of \(\dot{\xi}=f(\xi, \omega)\) on the interval \([0, T]\), then $$ \|\xi(T)\| \leq \max \left\\{\beta(\|\xi(0)\|, T), \gamma\left(\|\omega\|_{\infty}\right)\right\\} $$
For linear time-invariant continuous-time or discrete-time systems, \((A, C)\) is asymptotically observable if and only if \(\left(A^{\prime}, C^{\prime}\right)\) is asymptotically controllable.
When the system \(\Sigma\) is discrete-time and observable, one may pick a matrix \(L\) so that \(A+L C\) is nilpotent, by the Pole-Shifting Theorem. Thus, in that case a deadbeat observer results: The estimate becomes exactly equal to the state after \(n\) steps.
If \(w\) is stable, then for each \(k \neq 0\) such that \(\frac{1}{k} \notin \Gamma\), the closed-loop system with \(v=k\) is stable if and only if \(\mathrm{cw}\left(\Gamma, \frac{1}{k}\right)=0\).
What do you think about this solution?
We value your feedback to improve our textbook solutions.