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Chapter 6: Problem 8

Show that the result does not extend to analytic systems by considering the system $$ x^{+}=\frac{1}{2} x, \quad y=\sin x, $$ which is observable but not in finite time.

Short Answer

Expert verified
The given system \(x^{+}=\frac{1}{2} x, \quad y=\sin x\) is observable, as there exists a unique output y for each state x in an interval less than 2蟺. However, it is not observable in finite time, since any initial state will eventually converge towards zero with the sine function converging to sin(0) = 0, regardless of the initial state value. Thus, the result does not extend to analytic systems.

Step by step solution

01

Prove the system is observable

To prove that the system is observable, we need to show that there exists a unique solution x corresponding to each output y and input. Since the output function of the system is y = sin(x), let's first investigate if the output function can map different values of x to the same output y. Taking the derivative of sin(x) with respect to x, we get: \[\frac{d}{dx}\sin(x) = \cos(x).\] As the cosine function varies between -1 and 1, the derivative of the sin function is non-zero everywhere, which implies that the sin function is injective (one-to-one) in every interval where its derivative is non-zero. Since sin(x) and cos(x) are both periodic functions with a period of 2蟺, the sin function is one-to-one in any interval of length less than 2蟺. Thus, there exists a unique output y for each state x in an interval less than 2蟺, and by analyzing the output y over time, we can uniquely determine the initial state x. Therefore, the system is observable.
02

Prove the system is not observable in finite time

We know the dynamics of the system is given by: \[x^{+} = \frac{1}{2} x.\] For a given initial state x(0), the future state after n steps can be found as: \[x(n) = \left(\frac{1}{2}\right)^n x(0).\] As n goes to infinity, x(n) approaches zero. This implies that any initial state will eventually converge towards zero, and the sine function will converge to sin(0) = 0, regardless of the initial state value. Consequently, there is no finite n after which we can observe the state and conclude the initial state of the system based on the output alone. Therefore, the system is not observable in finite time. In conclusion, the given system is observable, but not in finite time. This demonstrates that the result does not extend to analytic systems.

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Key Concepts

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

Analytic Systems
Analytic systems are systems that are described by a set of equations involving analytic functions. An analytic function is one that is locally represented by a power series. These functions have derivatives of all orders, and they tend to be quite smooth. In the context of your exercise, the system is represented by the equations \(x^{+} = \frac{1}{2} x\) and \(y = \sin(x)\). Each part of this system - the linear difference equation and the sine function - is analytic.
Analytic systems like these can have interesting properties. For instance, they are often easier to analyze due to their smooth nature. However, they also present unique challenges because their smoothness can sometimes obscure properties like observability.
  • Analytic functions are infinitely differentiable.
  • These systems can be linear or nonlinear.
  • They may exhibit complex behaviors under certain conditions.
Understanding the nuances of analytic systems is crucial when dealing with observability in control systems.
State Observability
State observability is a fundamental concept in control theory. It refers to whether the internal state of a system can be determined from its outputs over time. In simpler terms, if you can measure certain outputs of a system, can you then "see" or "observe" what's going on inside the system? In the example from your exercise, the system was described to be observable since the output \(y = \sin(x)\) allowed for the determination of the state \(x\) in intervals where \(\sin(x)\) is injective.
When a system is observable, it means that every possible state leaves a unique "fingerprint" on the outputs. This is essential for tasks like monitoring and making decisions based on the system state.
  • An observable system provides enough information to determine its state entirely from output measurements.
  • State observability does not necessarily imply finite-time observability.
  • An injective function in the output ensures unique states for observations in certain intervals.
Observability is critical for the design and implementation of efficient control systems.
Finite-Time Observability
Finite-time observability is a stronger condition than general observability. It indicates whether the state of a system can be determined from the output in a finite amount of time. In the given exercise, even though the system is generally observable, it is not observable in finite time. This is due to the dynamics \(x^{+} = \frac{1}{2} x\), which causes the state to decay towards zero over time, making it indistinct in finite observations.
Finite-time observability is crucial when quick decisions must be made, or control actions need to be implemented promptly. If a system is not finitely observable, the initial state cannot be reconstructed from the output in a practical time frame.
  • This characteristic is necessary for systems requiring timely control actions.
  • It depends heavily on system dynamics and output characteristics.
  • No unique determination of the initial state can occur if states converge to a common point in finite time.
Understanding both finite-time observability and general observability is vital for designing responsive control systems.

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Most popular questions from this chapter

If there are any \(S, T \in \mathcal{T}_{+}, T>0\), such that $$ I^{S}=I^{S+T}, $$ then necessarily \(I^{S}=I\).

A (time-invariant, complete, initialized) system is minimal if and only if it is canonical.

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 \(m n\).

If two controllable triples are similar, then the similarity must be given by the formulas obtained in the proof of Theorem 27, so in particular similarities between canonical systems are unique. This is because if \(T\) is as in Definition \(6.5 .8\), then necessarily $$ T^{-1} A^{i} B=\widetilde{A}^{i} \widetilde{B} $$ for all \(i\), so it must hold that \(T^{-1} \mathbf{R}=\widetilde{\mathbf{R}}\), and therefore $$ T=\mathbf{R} \widetilde{\mathbf{R}}^{\\#} $$ In particular, this means that the only similarity between a canonical system and itself is the identity. In the terminology of group theory, this says that the action of \(G L(n)\) on triples is free.

The map \(\beta\) induces a bijection between the quotient space \(S_{n, m, p}^{c, o} / \sim\) and \(\mathcal{M}_{n, m, p} .\)
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