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I`m a 2nd year mathematics student, and I have to do a project for the course Systems Theory. The project is about a discrete time system. We have only dealt with continuous time systems so far in the course, so there is no literature I can use. I have already expressed the matrices of the discrete time system in terms of the sampling period T and the matrices of the continuous time system.

Now I have to show that the discrete time system is observable, given that the continuous time system is observable. One way to tackle this problem is to show that $\langle$kerH $\mid$ F $\rangle$ = {0}, where H = C and F = $e^{At}$ in terms of the matrix of the continuous time system, $\Sigma$, which is given by:

x' = Ax + Bu

y = Cx

I know that in order to show this, I have to show that the observability matrix [ \begin{matrix} C \\ C e^{At} \\ Ce^{2At} \\ \cdot \\ \cdot \\ \cdot \\ Ce^{(n-1)At} \end{matrix} ] has rank n. This is the dimension of the matrix A and the length of the vector x. Since $\Sigma$ is observable, I already know that the matrix
[ \begin{matrix} C \\ CA \\ CA^{2} \\ \cdot \\ \cdot \\ \cdot \\ CA^{n-1} \end{matrix} ] has rank n.

However, I have no idea how to prove this for the discrete time system! I really hope someone can help me with this, since I`ve already been working on it for almost 3 weeks right now.

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    $\begingroup$ The matrix exponential is a power-series, $e^A = \sum_{k=0}^\infty{1 \over k!}A^k $, so I think it should be essentially equivalent to what you've already done for the continuous-time system. $\endgroup$ – Steve Heim Dec 29 '16 at 6:10
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Using the same trick based on the Cayley-Hamilton theorem that is used to show that one can stop constructing the observability matrix with the term in ower $n-1$, one should be able to convert the 1st observability matrix in the question into a form similar to the 2nd. It's a good exercise!

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