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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.

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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.
See the step by step solution

## Step 1: Define observability and $$\delta$$-sampled observability

Observability is a property of a system that determines whether the system's internal states can be determined by observing its output. In the context of linear systems, a system is observable if we can uniquely determine the initial state of the system from the output. $$\delta$$-sampled observability is a property related to discrete-time systems that concerns whether the system's states can be determined by observing a sampled output, with $$\delta$$ being the sampling interval. In other words, instead of having continuous outputs, we only observe the system outputs at specific, discrete time instants.

## Step 2: Show that the given condition ensures $$\delta$$-sampled observability

To show that $$\Sigma$$ is $$\delta$$-sampled observable under the given condition, we will first recall the result from observability theory that states: A continuous-time LTI (Linear Time-Invariant) system with state-space representation: $\begin{cases} \dot{x}(t) = Ax(t) + Bu(t) \\ y(t) = Cx(t) + Du(t) \end{cases},$ is observable if and only if the observability matrix $$O$$ has full rank, where $$O = \begin{bmatrix} C \\ CA \\ \vdots \\ CA^{n-1} \end{bmatrix}$$, and $$n$$ is the dimension of the state vector $$x(t)$$. To show $$\delta$$-sampled observability, we need to consider the discrete-time version of the system, by sampling the output at discrete time intervals $$\delta$$: $\begin{cases} x[k+1] = e^{A\delta} x[k] + B_d u[k] \\ y[k] = Cx[k] + Du[k] \end{cases},$ where $$x[k] = x(k\delta)$$, $$y[k] = y(k\delta)$$, and $$B_d = \int_0^{\delta} e^{As}B ds$$. Now, $$\Sigma$$ is $$\delta$$-sampled observable if and only if the observability matrix associated with the discrete-time system: $$O_d = \begin{bmatrix} C \\ Ce^{A\delta} \\ \vdots \\ Ce^{A(n-1)\delta} \end{bmatrix}$$, has full rank. Given that the continuous-time system is observable, which implies that $$O$$ has full rank, and the condition $$\delta(\lambda-\mu) \neq 2 k \pi i$$ for every two eigenvalues $$\lambda, \mu$$ of $$A$$, we can conclude that $$O_d$$ will also have full rank. This is because 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. Therefore, $$\Sigma$$ is also $$\delta$$-sampled observable under the given conditions.

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