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# (Solved): whicj id the correct observer conical form pls im in a test Consider the following plant: $G(s)=\f ... whicj id the correct observer conical form pls im in a test Consider the following plant: \[ G(s)=\frac{s+15}{(s+1)(s+2)(s+4)}$ From the information given which one of the following is the correct observer canonical form? $\left[\begin{array}{l} \dot{x}_{1} \\ \dot{x}_{2} \\ \dot{x}_{3} \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -8 & -14 & -7 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right]+\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] u \quad, \quad y=\left[\begin{array}{lll} 15 & 1 & 0 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right]$ $$\left[\begin{array}{c}\dot{x}_{1} \\ \dot{x}_{2} \\ \dot{x}_{3}\end{array}\right]=\left[\begin{array}{ccc}-8 & 1 & 0 \\ -14 & 0 & 1 \\ -7 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]+\left[\begin{array}{l}0 \\ 1 \\ 15\end{array}\right] u \quad, \quad y=\left[\begin{array}{lll}1 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]$$ $$\left[\begin{array}{l}\dot{x}_{1} \\ \dot{x}_{2} \\ \dot{x}_{3}\end{array}\right]=\left[\begin{array}{ccc}-7 & 1 & 0 \\ -14 & 0 & 1 \\ -8 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]+\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right] u \quad y=\left[\begin{array}{lll}15 & 1 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]$$ $\left[\begin{array}{l} \dot{x}_{1} \\ \dot{x}_{2} \\ \dot{x}_{3} \end{array}\right]=\left[\begin{array}{ccc} -7 & 1 & 0 \\ -14 & 0 & 1 \\ -8 & 0 & 0 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right]+\left[\begin{array}{c} 0 \\ 1 \\ 15 \end{array}\right] u \quad, \quad y=\left[\begin{array}{lll} 1 & 0 & 0 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right]$

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