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Chapter 6: Problem 7

Consider the discrete-time time-invariant finite system with $$ x=\\{a, b, c, d\\}, \quad \mathcal{U}=\\{u, v\\}, \quad y=\\{\alpha, \beta\\}, $$ local-in-time transition function \(\mathcal{P}\) given by the following table: \begin{tabular}{|c|c|c|} \hline \(\mathcal{P}\) & \(u\) & \(v\) \\ \hline\(a\) & \(c\) & \(c\) \\ \hline\(b\) & \(c\) & \(b\) \\ \hline\(c\) & \(a\) & \(a\) \\ \hline\(d\) & \(d\) & \(a\) \\ \hline \end{tabular} and output function \(h(a)=h(b)=\alpha\) and \(h(c)=h(d)=\beta\). The pairs $$ \\{a, c\\},\\{a, d\\},\\{b, c\\},\\{b, d\\} $$ are instantaneously distinguishable. To distinguish $$ \\{a, b\\} $$ one may use any sequence starting with \(v\) and to distinguish $$ \\{c, d\\} $$ any sequence starting with \(u\). In summary, every pair of distinct states can be distinguished in time 1 , and the system is observable in time \(1 .\) There is however no single "universal" sequence \(\omega \in \mathcal{U}^{[\sigma, \tau)}\) so that for every pair \(x \neq z\) it would hold that $$ \lambda_{x}^{\sigma, t}\left(\left.\omega\right|_{[\sigma, t)}\right) \neq \lambda_{z}^{\sigma, t}\left(\left.\omega\right|_{[\sigma, t)}\right) $$ for some \(t\) (that is allowed to depend on \(x\) and \(z\) ). This is because if \(\omega\) starts with \(u\), then all outputs when starting at \(a\) or \(b\) coincide, since the state trajectories coincide after the first instant, but if \(\omega\) starts with \(v\), then \(\omega\) cannot separate \(c\) from \(d\).

Short Answer

Expert verified
In conclusion, the given discrete-time time-invariant finite system is observable in time 1. However, there is no single "universal" input sequence omega that can distinguish all distinct pairs of states in the time horizon [sigma, tau). This is because no sequence can adequately distinguish both {a, b} and {c, d} at the same time.

Step by step solution

01

Identify the observable pairs of states

According to the given information, four pairs of states are instantaneously distinguishable: {a, c}, {a, d}, {b, c}, and {b, d}. Additionally, x = {a,b} can be distinguished using any sequence starting with v, and x = {c,d} can be distinguished using any sequence starting with u. Before we move forward, let's recall that a pair of states x and z are considered to be observable if there exists some input sequence omega for which the transition function leads to outputs that are different for x and z.
02

Find if there is a universal sequence

Now we need to check if there is a single input sequence omega that can distinguish all the distinct pairs of states. We will test two cases: 1. If omega starts with 'u': * Using the input string starting with 'u', we can separate {c, d}. However, when the initial states are 'a' or 'b', the output will be the same, and therefore, state trajectories will coincide after the first instant. Hence it's not possible to separate 'a' and 'b'. 2. If omega starts with 'v': * Using the input string starting with 'v', we can separate {a, b}. However, when the initial states are 'c' or 'd', the output will be the same, and therefore, it is not possible to separate 'c' and 'd'. In conclusion, since there is no single "universal" sequence omega in U that can distinguish all distinct pairs of states in the time horizon [sigma, tau), we can conclude the system is observable in time 1 but does not have a universal sequence.

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Key Concepts

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

Observability in Discrete-Time Systems
In the context of discrete-time systems, observability is a pivotal concept. It essentially refers to the ability to infer the entire state of a system from its outputs over a period of time. Imagine you have a set of states and specific sequences of inputs, and each sequence produces certain outputs. The system is said to be observable if you can determine precisely which state you are in by looking at the outputs.

In the given exercise, we have a discrete system with states \(\{a, b, c, d\}\) and inputs \(\{u, v\}\). Various pairs of these states can be instantly distinguished from one another through specific input sequences. For example:
  • Pairs \(\{a, c\}\), \(\{a, d\}\), \(\{b, c\}\), and \(\{b, d\}\) are distinguishable at once based on their outputs.
  • Pair \(\{a, b\}\) is distinguishable using an input sequence starting with \(v\).
  • Pair \(\{c, d\}\) is distinguishable using an input sequence starting with \(u\).
This insight implies that after a very short period (specifically one time unit), it is possible to observe and determine exactly which state the system was in. Therefore, the system is observable in time 1, meaning that all state pairs can be distinguished almost immediately.
State Distinguishability
State distinguishability is the essence of determining system states through outputs. It focuses on ensuring each state within a system produces a unique output or changes in outputs when a specific input is given.

In the presented problem, we see how specific states can be told apart:
  • Pairs \(\{a, c\}\), \(\{a, d\}\), \(\{b, c\}\), and \(\{b, d\}\) are instantly distinguishable through their outputs without ambiguity.
  • The challenge comes with pairs such as \(\{a, b\}\) and \(\{c, d\}\), which need a strategic input sequence to be distinguished.
The core challenge arises when trying to find a universal input sequence that can differentiate any and all state pairs. In many cases, no such universal sequence exists that works for all state pairs due to the overlap in outputs for different states when a particular sequence is used. Thus, while state distinguishability is achievable for various pairs, the absence of a universal sequence means that different strategies or sequences might be needed to distinguish different pairs.
Transition Functions
The concept of transition functions is fundamental in understanding how a system moves from one state to another based on input actions. In discrete-time systems, a transition function defines these state changes clearly and concisely.

For the provided system, the transition function \(\mathcal{P}\) informs us how each state evolves with the application of inputs \(u\) and \(v\):
  • State \(a\) transitions to \(c\) with either \(u\) or \(v\).
  • State \(b\) remains \(b\) with \(v\) but changes to \(c\) with \(u\).
  • State \(c\) transitions back to \(a\) with any input.
  • State \(d\) stays the same with \(u\) but moves to \(a\) if \(v\) is applied.
These transitions dictate how observable the system is and how easily one can distinguish between different states. By thoroughly analyzing transition functions, you can predict or determine the system's behavior over time, further aiding in understanding its observability and state distinguishability.

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Most popular questions from this chapter

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.

One can in fact obtain many of the results for the linear theory as consequences of the above abstract considerations. For instance, take the proof of the fact that canonical realizations of linear behaviors must be unique up to a linear isomorphism. Assume that two systems are given, with the same spaces \(\mathcal{U}, y\), which are assumed to be vector spaces over a field \(\mathbb{K}\), that the state spaces \(x_{1}, x_{2}\) are also vector spaces, and that the maps \(\phi_{i}\) and \(h_{i}\) are linear in \((x, \omega)\) and \(x\), respectively. Furthermore, we assume that the initial states are zero. Then we claim that the set \(G\) is a linear subspace, which implies that the unique morphism \(T\) must correspond to a linear map \(x_{1} \rightarrow x_{2}\). Indeed, if \(\left(x_{1}, x_{2}\right)\) and \(\left(z_{1}, z_{2}\right)\) are in \(G\) and if \(k \in \mathbb{K}\), then linearity gives that $$ \begin{aligned} h\left(\phi_{1}\left(\tau, \sigma, x_{1}+k z_{1}, \omega\right)\right) &=h\left(\phi_{1}\left(\tau, \sigma, x_{1}, \mathbf{0}\right)\right)+k h\left(\phi_{1}\left(\tau, \sigma, z_{1}, \omega\right)\right) \\ &=h\left(\phi_{2}\left(\tau, \sigma, x_{2}, \mathbf{0}\right)\right)+k h\left(\phi_{2}\left(\tau, \sigma, z_{2}, \omega\right)\right) \\ &=h\left(\phi_{2}\left(\tau, \sigma, x_{2}+k z_{2}, \omega\right)\right) \end{aligned} $$ which implies that \(\left(x_{1}+k z_{1}, x_{2}+k z_{2}\right) \in G\). In particular, if both systems are canonical, then \(G\) is the graph of a linear isomorphism. Arbitrary linear isomorphisms are not very interesting when considering infinite dimensional linear systems. For example, in the context of systems whose state spaces \(X_{i}\) are Banach spaces, the output value set \(y\) is a normed space, and the maps \(\phi_{i}\) and \(h_{i}\) are assumed to be bounded (continuous) linear operators, one may want to conclude that the morphism \(T\) given by the Theorem is bounded, too. This is an easy consequence of the above proof: It is only necessary to notice that \(G\) must be closed (because of the assumed continuities), so by the Closed Graph Theorem (see, for instance, [399], Theorem 4.2-I) the operator \(T\) must indeed be continuous. If both systems are canonical, \(T^{-1}\) is also bounded, by the same argument.

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.)

In terms of the infinite Hankel matrix which is expressed in block form as $$ \mathcal{H}(\mathcal{A})=\left(\begin{array}{ccccc} \mathcal{A}_{1} & \mathcal{A}_{2} & \cdots & \mathcal{A}_{t} & \cdots \\ \mathcal{A}_{2} & \mathcal{A}_{3} & \cdots & \mathcal{A}_{t+1} & \cdots \\ \vdots & \vdots & \cdots & \vdots & \vdots \\ \mathcal{A}_{s} & \mathcal{A}_{s+1} & \cdots & \mathcal{A}_{s+t-1} & \cdots \\\ \vdots & \vdots & \cdots & \vdots & \vdots \end{array}\right) $$ one may restate the definition of rank as follows. The rank of an infinite matrix such as \(\mathcal{H}(\mathcal{A})\) is, by definition, the dimension of the column space of \(\mathcal{H}(\mathcal{A})\), which is seen as a subspace of the space \(\mathbb{K}^{\infty}\) consisting of infinite column vectors \(\left(x_{1}, x_{2}, x_{3}, \ldots\right)^{\prime}\) with entries over \(\mathbb{K}\), with pointwise operations. When this rank is finite and less than \(n\), all columns are linear combinations of at most \(n-1\) columns, and therefore all submatrices of \(\mathcal{H}(\mathcal{A})\) must have rank less than \(n\); this implies that \(\mathcal{A}\) must have rank less than \(n\). Conversely, we claim that if \(\mathcal{A}\) has rank less than \(n\), then the rank of the infinite matrix \(\mathcal{H}(\mathcal{A})\) is less than \(n\). If this were not to be the case, there would be a set of \(n\) independent columns \(c_{1}, \ldots, c_{n}\) of \(\mathcal{H}(\mathcal{A})\). If so, let \(M_{i}\) be the \(i \times n\) matrix obtained by truncating each column \(c_{j}\) at the first \(i\) rows. Then some \(M_{i}\) has rank \(n\) : Consider the nonincreasing sequence of subspaces $$ Q_{i}:=\operatorname{ker} M_{i} \subseteq \mathbb{K}^{n} $$ and let \(Q:=\cap Q_{i}\); by dimensionality, there is some \(k\) so that \(Q_{k}=Q\), and if \(x \in Q\) then \(M_{i} x=0\) for all \(i\) means that \(x\) is a linear relation between the \(c_{j}\) 's, so \(x=0\). This means that \(Q_{k}=0\) and so \(M_{k}\) has rank \(n\) as desired. Let \(s, t\) be such that \(M_{k}\) is a submatrix of \(\mathcal{H}_{s, t}\). Then $$ \operatorname{rank} \mathcal{A} \geq \operatorname{rank} \mathcal{H}_{s, t} \geq \operatorname{rank} M_{k}=n, $$ contradicting \(\operatorname{rank} \mathcal{A}

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