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Problem 11

Assume that \(\Sigma\) is observable and that $$ \delta(\lambda-\mu) \neq 2 k \pi i, \quad k=\pm 1, \pm 2, \ldots, $$ for every two eigenvalues \(\lambda, \mu\) of \(A\). Then \(\Sigma\) is also \(\delta\)-sampled observable.

Expert verified

Under the given condition \(\delta(\lambda-\mu) \neq 2 k \pi i\) for every two eigenvalues \(\lambda, \mu\) of \(A\), the continuous-time system \(\Sigma\) is not only observable but also \(\delta\)-sampled observable. This follows from the fact that the observability matrix \(O\) has full rank, and none of the eigenvalues of the system matrix \(A\) will cause the \(e^{A\delta}\) term to become singular or reducible to trivial cases after sampling, ensuring that the observability matrix associated with the discrete-time system \(O_d\) also has full rank.

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Chapter 6

For linear systems, \((x, \sigma)\) and \((z, \sigma)\) are indistinguishable on \([\sigma, \tau]\) if and only if $$ \lambda_{x-z}^{\sigma, t}(\mathbf{0})=0 $$ for all \(t \in[\sigma, \tau]\); that is, \((x-z, \sigma)\) is indistinguishable from \((0, \sigma)\) using zero controls on \([\sigma, \tau) .\) Similarly, any two states \(x, z\) are indistinguishable if and only if \(x-z\) is indistinguishable from 0 .

Chapter 6

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.)

Chapter 6

If \((A, B, C)\) realizes \(\mathcal{A}\), then $\operatorname{rank} \mathcal{H}_{s, t}(\mathcal{A}) \leq \max \left\\{\operatorname{rank} \mathbf{O}_{s}(A, C), \operatorname{rank} \mathbf{R}_{t}(A, B)\right\\} \leq n\( for all \)s, t$.

Chapter 6

Let \(\omega_{i}, i=1, \ldots, k\) be \(k\) different positive real numbers. Show that there is some continuous-time time-invariant linear system with outputs and no inputs \(\Sigma=(A, 0, C)\) such that: \- \(\Sigma\) is observable, and \- for each set of \(2 k\) real numbers \(a_{i}, \varphi_{i}, i=1, \ldots, k\), there is some initial state \(x\) so that $$ \lambda_{x}^{0, t}=\eta(t)=\sum_{i=1}^{k} a_{i} \sin \left(2 \pi \omega_{i} t+\varphi_{i}\right) $$ for all \(t \geq 0\). Conclude from the above discussion that, if $$ \frac{1}{\delta}>2 \max _{i=1, \ldots, m}\left|\omega_{i}\right| $$ then the complete function \(\eta(t)\) can be recovered from the values $$ \eta(0), \eta(\delta), \eta(2 \delta), \ldots $$ for every set of \(a_{i}\) 's and \(\varphi_{i}\) 's.

Chapter 6

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

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