Chapter 6: Problem 1
The Markov sequence \(\mathcal{A}\) is realizable if and only if it is recursive.
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 6: Problem 1
The Markov sequence \(\mathcal{A}\) is realizable if and only if it is recursive.
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
Assume that \(\Sigma\) is a final-state observable time-invariant discrete-time complete system for which card \(X<\infty\). Show that there exists some \(T \geq 0\) and some fixed control \(\omega\) of length \(T\) so that \(\omega\) final-state distinguishes every pair of states. (Hint: Consider any control for which the set of pairs of states \((x, z)\) that are final-state indistinguishable by \(\omega\) is of minimal possible cardinality.)
Let \(\mathcal{A}\) be any recursive Markov sequence. Consider the transposed sequence $$ \mathcal{A}^{\prime}:=\mathcal{A}_{1}^{\prime}, \mathcal{A}_{2}^{\prime}, \ldots $$ This satisfies a recursion with the same coefficients \(\alpha_{i}\) 's. For these coefficients we let \((A, B, C)\) be the observability form realization of \(\mathcal{A}^{\prime}\). Since \(C A^{i-1} B=\mathcal{A}_{i}^{\prime}\) for all \(i\), also \(B^{\prime}\left(A^{\prime}\right)^{i-1} C^{\prime}=\mathcal{A}_{i}\) for all \(i\), and the system \(\left(A^{\prime}, C^{\prime}, B^{\prime}\right)\) is a controllable realization of \(\mathcal{A}\). We have obtained the controllability form realization of \(\mathcal{A}:\) $$ A=\left(\begin{array}{ccccc} 0 & 0 & \cdots & 0 & \alpha_{1} I \\ I & 0 & \cdots & 0 & \alpha_{2} I \\ 0 & I & \cdots & 0 & \alpha_{3} I \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & I & \alpha_{n} I \end{array}\right) \quad B=\left(\begin{array}{c} I \\ 0 \\ 0 \\ \vdots \\ 0 \end{array}\right) \quad C=\left(\begin{array}{llll} \mathcal{A}_{1} & \mathcal{A}_{2} & \cdots & \mathcal{A}_{n} \end{array}\right) $$ of dimension \(m n\).
Show that, if \(m=p=1\) and if \(W_{\mathcal{A}}=P / q\) with \(P\) of degree \(\leq n-1\) and \(q\) of degree \(n\), then \(\mathcal{A}\) has rank \(n\) if and only if \(P\) and \(q\) are relatively prime. (Hint: Use Corollary \(6.6 .6\) and the fact that \(\mathcal{K}\left(\left(s^{-1}\right)\right)\) forms an integral domain.)
The data \((\mathcal{T}, \Omega, \mathcal{U}, \phi)\) define a complete time- invariant system.
Let \(\Sigma\) be a continuous-time linear system with outputs, and pick \(\sigma<\tau \in \mathbb{R}\). Consider the observability Gramian $$ W_{o}(\sigma, \tau):=\int_{\sigma}^{\tau} \Phi(s, \sigma)^{*} C(s)^{*} C(s) \Phi(s, \sigma) d s . $$ Show: \(\Sigma\) is observable in \([\sigma, \tau]\) if and only if \(W_{o}(\sigma, \tau)\) has rank \(n\), and, in that case, the operator that computes the initial state \(x\) from the observation \(\eta(t)=C(t) \Phi(t, \sigma) x\) is given by: $$ x=M_{0} \eta=W_{o}(\sigma, \tau)^{-1} \int_{\sigma}^{\tau} \Phi(s, \sigma)^{*} C(s)^{*} \eta(s) d s . $$
What do you think about this solution?
We value your feedback to improve our textbook solutions.