/*! 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} Problem 10 Let \(\Sigma\) be a complete tim... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 10

Let \(\Sigma\) be a complete time-invariant system with outputs and card \(y \geq 2\). Consider the system $$ \tilde{\Sigma}:=(\mathcal{T}, \tilde{\mathcal{X}}, \mathcal{U}, \tilde{\phi}) $$ defined as follows. Its state space \(\widetilde{\mathcal{X}}\) is the set of all maps \(x \rightarrow y\), and for each \(\omega \in \mathcal{U}^{[\sigma, \tau)}\) and each state \(\alpha: X \rightarrow y\) of \(\widetilde{X}\) $$ \widetilde{\phi}(\tau, \sigma, \alpha, \omega) $$ is the map $$ x \mapsto \alpha(\phi(\tau, \sigma, x, \widetilde{\omega})) $$ where \(\widetilde{\omega}\) is the time-reversed control \(\widetilde{\omega}(t):=\omega(\sigma+\tau-t) .\) Note that the output map \(h\) can be seen as an element of \(\widetilde{X}\). Prove that if \(\widetilde{\Sigma}\) is reachable from \(h\), i.e., $$ \mathcal{R}(h)=\widetilde{X} $$ then \(\Sigma\) is observable.

Short Answer

Expert verified
In conclusion, if \(\tilde{\Sigma}\) is reachable from \(h\) (\(\mathcal{R}(h)=\widetilde{X}\)), then the state of the system \(\Sigma\) can be uniquely determined based on its outputs, indicating that \(\Sigma\) is observable. This is proven through the relationships between \(\Sigma\) and \(\tilde{\Sigma}\) and the properties of reachability and observability.

Step by step solution

01

Define observability and reachability

In the context of this problem, observability refers to the ability to uniquely determine the state of a system, \(\Sigma\), based on its output, while reachability refers to the ability of a system, \(\tilde{\Sigma}\), to reach a given state from its initial state.
02

Provide definitions of the systems and their respective properties

\(\Sigma\) is a complete time-invariant system with outputs \(y \geq 2\). \(\tilde{\Sigma}\) is defined as having state space \(\widetilde{\mathcal{X}}\), which is the set of all maps \(x \rightarrow y\). The output map \(h\) can be seen as an element of state space \(\widetilde{\mathcal{X}}\). Reachability of \(\tilde{\Sigma}\) would mean that \(\widetilde{\mathcal{X}} = \mathcal{R}(h)\).
03

Express the relationship between \(\Sigma\) and \(\tilde{\Sigma}\)

Given the systems \(\Sigma\) and \(\tilde{\Sigma}\), we have the relationship: $$ \widetilde{\phi}(\tau, \sigma, \alpha, \omega) = x \mapsto \alpha(\phi(\tau, \sigma, x, \widetilde{\omega})), $$ where \(\widetilde{\omega}\) is the time-reversed control \(\widetilde{\omega}(t):=\omega(\sigma+\tau-t)\).
04

Prove that if \(\tilde{\Sigma}\) is reachable from \(h\), then \(\Sigma\) is observable

To prove the observability of \(\Sigma\), we need to show that the state of the system can be uniquely determined by its outputs. In the context of this exercise, we are given that the reachability of \(\tilde{\Sigma}\) is given by \(\mathcal{R}(h) = \widetilde{\mathcal{X}}\). Since the output map \(h\) can be seen as an element of the state space \(\widetilde{\mathcal{X}}\), the uniqueness of the outputs is guaranteed which leads to the observability of \(\Sigma\). We know that: $$ \widetilde{\phi}(\tau, \sigma, \alpha, \omega) = x \mapsto \alpha(\phi(\tau, \sigma, x, \widetilde{\omega})), $$ This relationship implies that the state of \(\Sigma\), determined by \(\phi(\tau, \sigma, x, \widetilde{\omega})\), can be mapped to the state of \(\tilde{\Sigma}\) using the state \(\alpha\). Since \(\mathcal{R}(h)=\widetilde{X}\), it implies that the state of \(\Sigma\), as determined by \(\phi(\tau, \sigma, x, \widetilde{\omega})\), can reach any other state in the state space \(\widetilde{\mathcal{X}}\). Thus, the state of \(\Sigma\) is uniquely determined based on its outputs, hence \(\Sigma\) is observable. Therefore, if \(\tilde{\Sigma}\) is reachable from \(h\), then \(\Sigma\) is observable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reachability of Systems
Imagine you're faced with a labyrinth; your goal is to determine whether you can reach its end from the starting point. In the context of control theory, this concept is translated to the reachability of systems; reachability describes the possibility of steering a system from any initial state to any desired final state in finite time, using permissible controls. For a system to be fully reachable, one should be able to reach any state within its state space.

In the given problem, by proving \(\widetilde{\mathcal{X}} = \mathcal{R}(h)\), it is established that all states in the state space of system \(\tilde{\Sigma}\) can be reached from the output map \(h\), signifying that the system is completely reachable. This property is critical, as reachability is a precondition for controllability, which is the ability to guide a system to a desired state through appropriate inputs.
Time-Invariant Control Systems
As we step into the world of time-invariant control systems, we are essentially talking about systems that do not change their behavior over time. The rules governing these systems remain constant, which implies that the system's responses to control signals are predictable and do not depend on when they are applied. The time-invariance of a system ensures that if an input signal produces a certain output today, the same input signal will produce the same output tomorrow or any day thereafter, as long as the initial conditions do not change.

The exercise provided pertains to a complete time-invariant system \(\Sigma\), indicating that the system's dynamics, represented through \(\phi\), are constant over time. This means that a time-reversed control signal, such as \(\widetilde{\omega}\), still elicits a response that can be analyzed within the framework of the system dynamics without considering the effects of time on the system's properties.
State Space Representation
Diving into the state space representation of systems is akin to examining a map that illustrates all possible states 鈥� or 'locations' 鈥� within a system and how the inputs can move the system from one state to another. This form of representation is a powerful mathematical model that describes a system's dynamics in terms of a set of input, output, and state variables interrelated by differential equations. By expressing a control system in state space form, we can analyze complex systems in a structured manner, often leading to insights about system behavior that would be difficult to discern otherwise.

In the context of the exercise, \(\tilde{\Sigma}\) has its state space defined as \(\widetilde{\mathcal{X}}\), the set of all maps \(x \rightarrow y\). This presentation is essential because it articulates the way inputs are mapped to outputs and how the system evolves over time. Understanding the state space is key to analyzing both the behavior of the system under various inputs and the responses it can generate, providing a comprehensive view of the system's potential dynamics.

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Most popular questions from this chapter

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.)

Calculate a canonical realization, and separately calculate the rank of the Hankel matrix, for each of these examples with \(m=p=1\) : 1\. The sequence of natural numbers \(1,2,3, \ldots\) 2\. The Fibonacci sequence \(0,1,1,2,3,5,8, \ldots\)

We consider again the parity check example discussed in Example 2.3.3. In particular, we shall see how to prove, using the above results, the last two claims in Exercise 2.3.4. The behavior to be realized is \(\lambda(\tau, 0, \omega)=\) $$ \begin{cases}1 & \text { if } \omega(\tau-3)+\omega(\tau-2)+\omega(\tau-1) \text { is odd and } 3 \text { divides } \tau>0 \\ 0 & \text { otherwise }\end{cases} $$ and we take the system with $$ x:=\\{0,1,2\\} \times\\{0,1\\} $$ and transitions $$ \mathcal{P}((i, j), l):=(i+1 \bmod 3, j+l \bmod 2) $$ for \(i=1,2\) and $$ \mathcal{P}((0, j), l):=(1, l) \text {. } $$ The initial state is taken to be \((0,0)\), and the output map has \(h(i, j)=1\) if \(i=0\) and \(j=1\) and zero otherwise. (The interpretation is that \((k, 0)\) stands for the state " \(t\) is of the form \(3 s+k\) and the sum until now is even," while states of the type \((k, 1)\) correspond to odd sums.) This is clearly a realization, with 6 states. To prove that there is no possible (time-invariant, complete) realization with less states, it is sufficient to show that it is reachable and observable. Reachability follows from the fact that any state of the form \((0, j)\) can be obtained with an input sequence \(j 00\), while states of the type \((1, j)\) are reached from \(x^{0}\) using input \(j\) (of length one) and states \((2, j)\) using input \(j 0\). Observability can be shown through consideration of the following controls \(\omega_{i j}\), for each \((i, j):\) $$ \omega_{01}:=0, \omega_{00}:=100, \omega_{10}:=10, \omega_{11}:=00, \omega_{21}:=0, \omega_{20}:=0 . $$ Then, \(\omega_{01}\) separates \((0,1)\) from every other state, while for all other pairs \((i, j) \neq\) \((0,1)\), $$ \lambda_{(i, j)}\left(\omega_{\alpha \beta}\right)=1 $$ if and only if \((i, j)=(\alpha, \beta)\).

Assume that \(\Sigma\) is a linear system of dimension \(n<\infty\). Then: (a) If \(\Sigma\) is discrete-time, \(I=I^{n-1}\). (b) If \(\Sigma\) is continuous-time, \(I=I^{\delta}\) for all \(\delta>0\). In particular, observable finite dimensional discrete-time systems are observable in \(n-1\) steps, and observable finite dimensional continuous-time systems are observable in arbitrarily small time.

Assume that \(\Sigma\) is a complete discrete-time system and \(1<\) \(\operatorname{card}(X)=n<\infty\). Then $$ I=I^{n-2}, $$ and in particular \(\Sigma\) is observable if and only if it is observable in time \(n-2\).
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