/*! 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 1 Let \(\mathcal{U} \subseteq \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 1

Let \(\mathcal{U} \subseteq \mathbb{R}^{m}\) and pick any two \(S, T \geq 0\). Then $$ \mathcal{R}_{u}^{T}(0)+e^{T A} \mathcal{R}_{u}^{S}(0)=\mathcal{R}_{u}^{S+T}(0) . $$

Short Answer

Expert verified
In summary, by calculating the reachable sets \(\mathcal{R}_{u}^{T}(0)\), \(\mathcal{R}_{u}^{S}(0)\), and \(\mathcal{R}_{u}^{S+T}(0)\), and proving their respective properties, we have shown that \(\mathcal{R}_{u}^{T}(0)+e^{T A} \mathcal{R}_{u}^{S}(0)=\mathcal{R}_{u}^{S+T}(0)\).

Step by step solution

01

Define the given terms

Before solving the problem, let's define the given terms. \(\mathcal{U}\) is a subset of \(\mathbb{R}^{m}\), where m is the dimension of the real number space. \(S\) and \(T\) are non-negative real numbers. \(\mathcal{R}_{u}^{T}(0)\), \(\mathcal{R}_{u}^{S}(0)\), and \(\mathcal{R}_{u}^{S+T}(0)\) represent reachable sets at times \(T\), \(S\), and \(S+T\), respectively. Matrix \(A\) is a given matrix associated with the linear system.
02

Express the reachable sets

Reachable sets can be expressed as: \[ \mathcal{R}_{u}^{T}(0) = \{ x \in \mathbb{R}^{m} : x = \int_{0}^{T} e^{A(T-s)}Bu(s) ds, u(\cdot) \in \mathcal{U}\} \] Similarly, we can express \(\mathcal{R}_{u}^{S}(0)\) and \(\mathcal{R}_{u}^{S+T}(0)\).
03

Calculate \(e^{TA}\mathcal{R}_{u}^{S}(0)\)

To calculate the term \(e^{TA}\mathcal{R}_{u}^{S}(0)\), we have to multiply the matrix exponential of \(TA\) with the reachable set \(\mathcal{R}_{u}^{S}(0)\). \[ e^{TA}\mathcal{R}_{u}^{S}(0) = \{ e^{TA}x : x \in \mathcal{R}_{u}^{S}(0)\} \]
04

Prove the equality

Now we need to show that \[ \mathcal{R}_{u}^{T}(0) + e^{TA}\mathcal{R}_{u}^{S}(0) = \mathcal{R}_{u}^{S+T}(0) \] Let \(x_1 \in \mathcal{R}_{u}^{T}(0)\) and \(x_2 \in e^{TA}\mathcal{R}_{u}^{S}(0)\). Then, we have \[ x_1 = \int_{0}^{T} e^{A(T-s)}Bu_1(s) ds \quad \text{and} \quad x_2 = e^{TA}\int_{0}^{S} e^{A(S-s)}Bu_2(s) ds \] Now let \(x = x_1 + x_2\). Then \[ x = \int_{0}^{T} e^{A(T-s)}Bu_1(s) ds + e^{TA}\int_{0}^{S} e^{A(S-s)}Bu_2(s) ds \] Using the change of variables, \(v = T-s\), we get \[ x = \int_{0}^{T} e^{A(T-s)}Bu_1(s) ds + \int_{T}^{S+T} e^{A(S+T-v)}Bu_2(v-T) dv \] Since \(u_1(\cdot) \in \mathcal{U}\) and \(u_2(\cdot) \in \mathcal{U}\), we have that \(x \in \mathcal{R}_{u}^{S+T}(0)\). Conversely, let \(x \in \mathcal{R}_{u}^{S+T}(0)\). Then, we have \[ x = \int_{0}^{S+T} e^{A(S+T-s)}Bu(s) ds \] Splitting the integral, we get \[ x = \int_{0}^{T} e^{A(S+T-s)}Bu(s) ds + \int_{T}^{S+T} e^{A(S+T-s)}Bu(s) ds \] Let \(x_1 = \int_{0}^{T} e^{A(S+T-s)}Bu(s) ds\) and \(x_2 = \int_{T}^{S+T} e^{A(S+T-s)}Bu(s) ds\). Then, we have \(x_1 \in \mathcal{R}_{u}^{T}(0)\) and \(x_2 \in e^{TA}\mathcal{R}_{u}^{S}(0)\). Thus, we have shown that \(\mathcal{R}_{u}^{T}(0) + e^{TA}\mathcal{R}_{u}^{S}(0) = \mathcal{R}_{u}^{S+T}(0)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Reachable Set
The concept of the reachable set in mathematical control theory is fundamental to understanding the behavior of dynamical systems over time. A reachable set, denoted as \(\mathcal{R}_u^T(0)\), represents all the states a system can attain, starting from the origin, at time \(T\) given the set of all possible control inputs \(\mathcal{U}\). This set is vital in control theory as it helps us determine which states are accessible and consequently, whether we can steer the system to a desired state within a specific timeframe.

Imagine you're a pilot of an airplane: the reachable set would encapsulate all the possible locations you could fly to, starting from the airport, in a given amount of time considering the range of maneuvers the plane can perform. In mathematical terms, the reachable set is commonly expressed through an integral involving the state-transition matrix \(e^{A(T-s)}\) and a control matrix \(B\), compounded by the control function \(u(s)\) for \(s\) in the range from \(0\) to \(T\). It maps the trajectory of the system states influenced by the control inputs over time.

Understanding Reachable Sets Intuitively

Reachable sets can often be visualized as geometric shapes (such as ellipsoids or polytopes) in state space, illustrating the span of possible states at a given instance. This visualization aids learners in grasping the extent of control they exert over a system, as well as the system's flexibility and limitations. By analyzing these shapes, one can identify if certain conditions make a state unreachable, which is a crucial aspect when designing control laws or verifying the performance of a system.
Matrix Exponential
The matrix exponential, noted as \(e^{TA}\), plays a key role in both the theoretical and practical aspects of control systems analysis. It is analogous to the scalar exponential function but is extended to matrices, particularly in the context of solving linear systems of differential equations.

In simple terms, if you think of regular exponential growth, like how money earns interest, the matrix exponential describes more complex growth patterns that can be applied to vectors and not just single numbers. This growth pattern can be influenced by various directions and magnitudes, much like how an ecosystem can grow and evolve in multiple dimensions and not just in population size.

Computing Matrix Exponentials

Computing the matrix exponential involves summing an infinite series similar to the Taylor series for the scalar exponential function but involves matrix powers instead of scalar powers. When you look at \(e^{TA}\), it's essentially a way to describe how a system evolves from time \(0\) to time \(T\), effectively transforming the state of the system according to the dynamics dictated by matrix \(A\).

This tool allows us to predict how a linear system evolves through time and answer questions such as: 'If we know the state of the system now, where will it be after \(T\) seconds?' When used in conjunction with the reachable sets, it provides a powerful method to translate states over time, giving us a clear view of the system's behavior as it moves through its state space.
Linear Systems
Linear systems form the bedrock upon which much of control theory is built. These systems are defined by linear differential equations, meaning that the change in the system's state is proportional to the current state and the input. Due to their linearity, these systems are more manageable to analyze and control compared to nonlinear systems.

A classic example of a linear system might be a spring-mass-damper system. It can be described using linear equations that account for the forces involved鈥攍ike the spring's resistance and the damper's friction鈥攁nd from these equations, we can predict how the system will respond to various inputs, like being compressed or extended.

Properties and Solutions of Linear Systems

One of the most important properties of linear systems is the principle of superposition, which states that the response caused by two different inputs is the sum of the responses that would be caused by each input independently. This principle facilitates understanding and predicting the behavior of linear systems, especially when multiple forces or controls are at play.

Solving linear systems often involves finding a function that describes the system's state over time, which can typically be done using matrix exponentials. These solutions help us comprehend the system's dynamic behavior and are essential for designing control strategies to guide the system toward desired objectives. By mastering linear systems, students can build a strong foundation that will serve them well when they encounter more complex systems in their studies and professional careers.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\Sigma\) be a continuous-time linear system, and pick \(\sigma<\tau \in \mathbb{R}\). Consider the controllability Gramian $$ W_{c}(\sigma, \tau):=\int_{\sigma}^{\tau} \Phi(\sigma, s) B(s) B(s)^{*} \Phi(\sigma, s)^{*} d s . $$ Show: \(\Sigma\) is controllable in \([\sigma, \tau]\) if and only if \(W_{c}(\sigma, \tau)\) has rank \(n\), and, in that case, the unique control of minimum square norm that steers \(x\) to 0 is given by the formula \(\omega(t)=-B(t)^{*} \Phi(\sigma, t)^{*} W_{c}(\sigma, \tau)^{-1} x\).

Consider a discrete-time bilinear system \(x(t+1)=u(t) E x(t)\), with \(x=\mathbb{R}^{2}, \mathcal{U}=\mathbb{R}\), and $$ E=\left(\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right) . $$ Show that $$ \mathcal{R}^{T}\left(\begin{array}{l} 1 \\ 0 \end{array}\right)=\left\\{\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) \mid T x_{1}=x_{2}\right\\} $$ for each positive integer \(T\). This is a subspace for each such \(T\), but \(R^{T}\left(\begin{array}{l}1 \\ 0\end{array}\right) \neq R\left(\begin{array}{l}1 \\ 0\end{array}\right)\) for all \(T\). Does this contradict Corollary 3.2.7?

One could also define a class of systems as in (3.28) with other choices of \(\theta\). Theorem 8 may not be true for such other choices. For instance, the theorem fails for \(\theta=\) identity (why?). It also fails for \(\theta=\arctan\) : Show that the 4-dimensional, single-input system $$ \begin{aligned} \dot{x}_{1} &=\arctan \left(x_{1}+x_{2}+x_{3}+x_{4}+2 u\right) \\ \dot{x}_{2} &=\arctan \left(x_{1}+x_{2}+x_{3}+x_{4}+12 u\right) \\ \dot{x}_{3} &=\arctan (-3 u) \\ \dot{x}_{4} &=\arctan (-4 u) \end{aligned} $$ satisfies that \(B \in \mathbf{B}_{n, m}\) but is not controllable. Explain exactly where the argument given for \(\theta=\tanh\) breaks down.

Let \(\Sigma\) be a controllable continuous-time linear system, and let \(Q\) be a real symmetric positive definite \(m \times m\) matrix. Pick any \(x, z \in X\), and \(\sigma, \tau \in \mathbb{R}\). Find a formula for a control \(\omega \in \mathcal{L}_{m}^{\infty}(\sigma, \tau)\) which gives \(\phi(\tau, \sigma, x, \omega)=z\) while minimizing $$ \int_{\sigma}^{\tau} \omega(s)^{*} Q \omega(s) d s . $$ Prove that for each pair \(x, z\) there is a unique such control. Do this in two alternative ways: (1) Applying again the material about pseudoinverses, but using a different inner product in the set \(\mathcal{L}_{m}^{2}\). (2) Factoring \(Q=Q_{1}^{*} Q_{1}\) and observing that the same result is obtained after a change of variables.

Assume that \((A, B)\) is controllable and \(\mathcal{U} \subseteq \mathbb{R}^{m}\) is a neighborhood of 0 . Then \(J_{k}^{\mathrm{R}} \subseteq \mathcal{R}_{u}(0)\) for all \(k\).
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.