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This question is from a Systems Theory test without answers or solutions.

Consider the folowing continuous-time state-space system $\dot{x}=Ax+Bu, \quad y=Cx.$

The continuous-time system given above is sampled at times $kh, k\in N$, with sampling time $h = 3$. We assume that the input function $u(t)$ is constant between two subsequent sampling times. To be more precise, $u(t)=u_{k}$, when $t \in [kh, (k+1)h]$. The result of this exact discretization is given by: $x_{k+1}=A_dx_k+B_dx_k$

$y_k=C_dx_k$

with matrices

$A_d = \begin{bmatrix} e^3 & 3e^3 \\ 0 & e^3 \end{bmatrix}, \quad B_d = \begin{bmatrix} e^3-1 \\ 0 \end{bmatrix}, \quad C_d = \begin{bmatrix} e^3 & e^3 \end{bmatrix}$

Which of the following continuous-time state-space systems yields the discrete time state space system after exact discretization is applied?

A) $ \left[ \begin{array}{c|c} A & B \\ \hline C & \\ \end{array} \right] =\left[ \begin{array}{cc|c} 1 & 1 & 0 \\ 0 & 1 & 1 \\ \hline 1 & 1 \\ \end{array} \right]$

B) $ \left[ \begin{array}{c|c} A & B \\ \hline C & \\ \end{array} \right] =\left[ \begin{array}{cc|c} 1 & 1 & 1 \\ 0 & 1 & 0 \\ \hline 1 & 1 \\ \end{array} \right]$

C) $ \left[ \begin{array}{c|c} A & B \\ \hline C & \\ \end{array} \right] =\left[ \begin{array}{cc|c} 1 & 1 & 1 \\ 0 & 1 & 0 \\ \hline e^3 & e^3 \\ \end{array} \right]$

D) $ \left[ \begin{array}{c|c} A & B \\ \hline C & \\ \end{array} \right] =\left[ \begin{array}{cc|c} 1 & 3 & 1 \\ 0 & 1 & 0 \\ \hline 1 & 1 \\ \end{array} \right]$

E) $ \left[ \begin{array}{c|c} A & B \\ \hline C & \\ \end{array} \right] =\left[ \begin{array}{cc|c} 1 & 1 & e^3-1 \\ 0 & 1 & 0 \\ \hline e^3 & e^3 \\ \end{array} \right]$

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When the input is constant during one time period is also known as zero-order hold discretization and can be calculated with

$$ \begin{bmatrix} A_d & B_d \\ 0 & 1 \end{bmatrix} = \exp\left( \begin{bmatrix} A & B \\ 0 & 0 \end{bmatrix} h \right), $$

and $C_d = C$. The matrix exponential can be calculated with

$$ \exp M = \sum_{n=0}^\infty \frac1{n!} M^n. $$

In am not sure if the question is also considering similarity transformations, otherwise every option where $C \neq C_d$ can already be omitted. When assuming that similarity transformations are not used then one only has to check two options, namely C) and E).

Calculating the discretization for C) gives

$$ \begin{bmatrix} e^3 & 3\,e^3 & e^3 - 1 \\ 0 & e^3 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \exp\left( \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} 3 \right), $$

and calculating the discretization for E) gives

$$ \begin{bmatrix} e^3 & 3\,e^3 & e^6 - 2\,e^3 + 1 \\ 0 & e^3 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \exp\left( \begin{bmatrix} 1 & 1 & e^3 - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} 3 \right), $$

from this it can be concluded that C) is the answer.

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  • $\begingroup$ Thanks Kwin. I just noticed your answer and have accepted it. $\endgroup$ – user463102 Sep 4 '18 at 9:03

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