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Problem 1

The discrete-time system \(\Sigma\) is linear (over the field \(\mathbb{K}\) ) if: \- It is complete; \- \(X\) and \(\mathcal{U}\) are vector spaces; and \- \(\mathcal{P}(t, \cdot \cdot \cdot)\) is linear for each \(t \in \mathbb{Z}\). The system with outputs \(\Sigma\) is linear if in addition: \- \(y\) is a vector space; and \- \(h(t, \cdot)\) is linear for each \(t \in \mathbb{Z}\). The system is finite dimensional if both \(\mathcal{U}\) and \(X\), as well as \(y\) for a system with outputs, are finite dimensional; the dimension of \(\Sigma\) is in that case the dimension of \(\mathcal{X}\).

Expert verified

A discrete-time system Σ is linear if it is complete, and the input, state, and output spaces (\( \mathcal{U} \), \( \mathcal{X} \), and y) are vector spaces. Additionally, the operators \( \mathcal{P}(t, \cdot, \cdot) \) and \( h(t, \cdot) \) must be linear for each \(t \in \mathbb{Z}\). The system is finite-dimensional if the input, state, and output spaces have finite dimensions, with the dimension of Σ being the same as the dimension of the state space \( \mathcal{X} \).

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Chapter 2

If \(\Sigma\) is linear, then \(\Lambda_{\Sigma, 0}\) is linear.

Chapter 2

Let \(\Sigma\) be a \(\mathcal{C}^{1}\) discrete-time system over \(\mathbb{R}\), and assume that \(\Gamma=(\bar{\xi}, \bar{\omega})\) is a trajectory for \(\Sigma\) on an interval \(\mathcal{I}\). The linearization of \(\Sigma\) along \(\Gamma\) is the discrete-time linear system \(\Sigma_{*}[\Gamma]\) with local- in-time description $\left(\mathbb{R}^{n}, \mathbb{R}^{m}, \mathcal{P}_{*}\right)$, where $$ \mathcal{P}_{*}(t, \cdot, \cdot):= \begin{cases}\mathcal{P}(t, \cdot, \cdot)_{*}[\bar{\xi}(t), \bar{\omega}(t)] & \text { if } t \in \mathcal{I}, \\\ 0 & \text { if } t \notin \mathcal{I} .\end{cases} $$ If \(\Sigma\) is a system with outputs, then \(\Sigma_{*}[\Gamma]\) is the discrete-time linear system with outputs having the above $\mathcal{P}, y=\mathbb{R}^{p}$, and the readout map $$ h_{*}(t, \cdot):= \begin{cases}h(t, \cdot)_{*}[\bar{\xi}(t)] & \text { if } t \in \mathcal{I} \\ 0 & \text { if } t \notin \mathcal{I}\end{cases} $$ When \(\mathcal{I}=\mathbb{Z}\) and \(\bar{\xi}\) and \(\bar{\omega}\) are both constant, \(\bar{\xi}(t) \equiv x\) and \(\bar{\omega}(t) \equiv u\), then \(\Sigma_{*}[\Gamma]\) is the linearization of \(\Sigma\) at \((x, u)\).

Chapter 2

Let \(\mathcal{X}\) be an open subset of \(\mathbb{K}^{n}\) and let \(\mathcal{U}\) be a metric space. \(A\) right-hand side (rhs) with respect to \(X\) and \(\mathcal{U}\) is a function $$ f: \mathbb{R} \times \mathcal{X} \times \mathcal{U} \rightarrow \mathbb{K}^{n} $$ which can be obtained in the following way: There must exist another metric space \(\mathcal{S}\) as well as maps $$ \tilde{f}: \mathcal{S} \times x \times \mathcal{U} \rightarrow \mathbb{K}^{n} $$ and $$ \pi: \mathbb{R} \rightarrow \mathcal{S} $$ so that $$ f(t, x, u)=\tilde{f}(\pi(t), x, u) $$ and the following properties hold: 1\. \(\widetilde{f}(s, \cdot, u)\) is of class \(\mathcal{C}^{1}\) for each fixed \(s, u\); 2\. both \(\tilde{f}\) and the partial derivative \(\tilde{f}_{x}\) are continuous on \(\mathcal{S} \times \mathcal{X} \times \mathcal{U}\); 3\. \(\pi\) is a measurable locally essentially bounded function.

Chapter 2

Assume that \(K\) in formula (2.27) is continuous. Pick any fixed $i \in\\{1, \ldots, m\\}\( and any \)T>0\(. For each \)\varepsilon>0\(, let \)\omega_{e}$ be the control defined on the interval \([0, T]\) by $$ \omega_{\varepsilon}(t):= \begin{cases}\frac{1}{\varepsilon} e_{i} & \text { if } 0 \leq t<\varepsilon \\ 0 & \text { otherwise }\end{cases} $$ where \(e_{i}=(0, \ldots, 0,1,0, \ldots, 0)^{\prime}\) is the \(i\) th canonical basis vector. Prove that $\lim _{\varepsilon \rightarrow 0} y_{\varepsilon}(T)=K_{i}(T)$,

Chapter 2

The discrete-time i/o behavior \(\Lambda\) is linear (over the field \(\mathbb{K}\) ) if: \- It is complete; \- U and \(y\) are vector spaces; \- For every \(\sigma \leq \tau\) in $\mathbb{Z}, \lambda(\sigma, \sigma, \diamond)=0\( and the mapping \)\lambda^{\sigma, \tau}$ is linear; and - For each \(\sigma<\mu<\tau\) in \(\mathbb{Z}\) and each input $\omega \in \mathcal{U}^{[\mu, \tau)}$, it holds that $$ \lambda(\tau, \sigma, 0 \omega)=\lambda(\tau, \mu, \omega) $$ where \(\mathbf{0}\) is the input identically equal to 0 on \([\sigma, \mu) .\)

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