/*! 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 5 Let \(\mathcal{A}\) be any recur... [FREE SOLUTION] | 91影视

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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 \(m n\).

Short Answer

Expert verified
The controllable realization of the given recursive Markov sequence 饾挏 can be represented using the matrices A, B, and C as follows: $$ A=\left(\begin{array}{ccccc} 0 & 0 & \cdots & 0 & \alpha_{1} I \\\ I & 0 & \cdots & 0 & \alpha_{2} I \\\ 0 & I & \cdots & 0 & \alpha_{3} I \\\ \vdots & \vdots & \ddots & \vdots & \vdots \\\ 0 & 0 & \cdots & I & \alpha_{n} I \end{array}\right) \quad B=\left(\begin{array}{c} I \\\ 0 \\\ 0 \\\ \vdots \\\ 0 \end{array}\right) \quad C=\left(\begin{array}{llll} \mathcal{A}_{1} & \mathcal{A}_{2} & \cdots & \mathcal{A}_{n} \end{array}\right) $$ with dimensions \(m n\).

Step by step solution

01

Obtain the transposed sequence

First, let's find the transposed sequence 饾挏' as follows: $$ \mathcal{A}^{\prime}:=\mathcal{A}_{1}^{\prime}, \mathcal{A}_{2}^{\prime}, \ldots $$
02

Determine the observability form realization

We are given that the observability form realization of 饾挏' is (A, B, C). And we know: $$ C A^{i-1} B=\mathcal{A}_{i}^{\prime} \quad \text{for all} \quad i $$
03

Find the controllable realization of 饾挏

The system (A', C', B') can be found as an alternative realization of 饾挏. Use the given relationship for B', A' and C': $$ B^{\prime}\left(A^{\prime}\right)^{i-1} C^{\prime}=\mathcal{A}_{i} \quad \text{for all} \quad i $$
04

Write the controllability form realization

Now using the given formula, we can write the controllable realization of 饾挏 in the form of matrices A, B, and C as follows: $$ A=\left(\begin{array}{ccccc} 0 & 0 & \cdots & 0 & \alpha_{1} I \\\ I & 0 & \cdots & 0 & \alpha_{2} I \\\ 0 & I & \cdots & 0 & \alpha_{3} I \\\ \vdots & \vdots & \ddots & \vdots & \vdots \\\ 0 & 0 & \cdots & I & \alpha_{n} I \end{array}\right) \quad B=\left(\begin{array}{c} I \\\ 0 \\\ 0 \\\ \vdots \\\ 0 \end{array}\right) \quad C=\left(\begin{array}{llll} \mathcal{A}_{1} & \mathcal{A}_{2} & \cdots & \mathcal{A}_{n} \end{array}\right) $$ The dimension of these matrices is \(m n\). With these matrices for A, B, and C, we have obtained the controllable realization of the recursive Markov sequence 饾挏.

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Key Concepts

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

Recursive Markov Sequence
A Recursive Markov Sequence is a sequence of random variables where each variable depends only on a fixed number of preceding variables and certain coefficients. This dependency maintains the Markov property, implying that the future is independent of the past given the present. This concept is crucial in understanding control systems, as it lends itself to a structured way of predicting future states based solely on the current state and well-defined recursion rules.
In the original exercise, \(\mathcal{A}\) represents such a sequence. When dealing with Markov sequences, it becomes easier to handle transformations and manipulations, such as finding the transposed sequence \(\mathcal{A}^{\prime}\), because the recursive nature dictates fixed patterns of behavior regardless of variations in the sequence length or specific values. Recursive Markov Sequences form the basis for more sophisticated structures like state-space representations used extensively in control theory.
Observability Form Realization
Observability is a measure of how well internal states of a system can be inferred by observing its outputs. In control theory, observability form realization involves constructing a system where each state can be identified uniquely by the system's outputs over a finite time span. That's essentially what we do in the exercise by using a set of matrices \(A, B, C\) which encapsulate the dynamics of the sequence \( \mathcal{A}^\prime \).
When given the matrix equations like \( C A^{i-1} B = \mathcal{A}_{i}^{\prime} \), it shows how the output \( \mathcal{A}_{i}^{\prime} \) is directly related to the states \( A^{i-1} \) and the initial state through \( B \). This relation gives us full insight into how each sequence element can be predicted based on initial conditions and the matrix representations. Observability form realization guarantees that the state variables are recoverable from the outputs, which is essential for accurate system modeling.
Controllable Realization
Controllable Realization relates to the ability of a control input鈥攁pplied through matrices \( B' \), \( A' \), and \( C' \)鈥攖o move the system from any initial state to a desired final state in a finite amount of time. In the exercise, this is captured by rearranging the original Markov sequence via matrices to obtain the system \((A'^\prime, C'^\prime, B'^\prime)\).
Through the equation \( B'\left(A'\right)^{i-1} C'=\mathcal{A}_{i} \), controllability is achieved, allowing transitions through system inputs to result in the required state configurations. Controllable realization is pivotal for intervention in automated or dynamic systems, ensuring complete command over state transitions and predictability. It reinforces maintaining control over the entire range of possible states the system might endure during operation.
Matrix Representation in Control Systems
Matrices have a crucial role in representing control systems, streamlining complex calculations and highlighting the intrinsic properties of systems like stability and observability. In control theory, using matrices \( A, B, C \) provides a compact and efficient means of modeling system dynamics, allowing straightforward application of mathematical tools.
In the exercise, you encountered matrices set to form controllable and observable realizations that transform recursive sequences into understandable forms. By employing properties of matrices, such as their ability to be multiplied and transposed, we encode sequences' dynamics elegantly and solve equations that describe system behavior like \( A \) for transformation dynamics, \( B \) for input control, and \( C \) for capturing outputs.
The dimension \( mn \) of such matrices ensures their applicability to broader systems, scaling effectively with system complexity and size. This matrix-centric approach is fundamental in control systems design, analysis, and implementation across various engineering fields.

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Most popular questions from this chapter

Give an example of a system of class \(\mathcal{C}^{\infty}\) with \(x=\mathcal{U}=y=\mathbb{R}\) which is observable but for which \(\mathcal{O}\) does not separate.

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.)

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 \(n-2\).

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 \((i, j)=(\alpha, \beta)\).

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.)

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