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(Solved):   Consider the following plant: \[ G(s)=\frac{s+15}{(s+1)(s+2)(s+4)} \] From the information ...



Consider the following plant:
\[
G(s)=\frac{s+15}{(s+1)(s+2)(s+4)}
\]
From the information given which one of the following i

 

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? a. \( \left[\begin{array}{l}\dot{x}_{1} \\ \dot{x}_{2} \\ \dot{x}_{3}\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 1 \\ -8 & -14 & -7\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]+\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] u, \quad y=\left[\begin{array}{lll}15 & 1 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right] \) b. \( \left[\begin{array}{l}\dot{x}_{1} \\ \dot{x}_{2} \\ \dot{x}_{3}\end{array}\right]=\left[\begin{array}{ccc}-7 & 1 & 0 \\ -14 & 0 & 1 \\ -8 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]+\left[\begin{array}{c}0 \\ 1 \\ 15\end{array}\right] u \quad, \quad y=\left[\begin{array}{lll}1 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right] \) c. \( \left[\begin{array}{l}\dot{x}_{1} \\ \dot{x}_{2} \\ \dot{x}_{3}\end{array}\right]=\left[\begin{array}{ccc}-7 & 1 & 0 \\ -14 & 0 & 1 \\ -8 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]+\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right] u \quad, \quad y=\left[\begin{array}{lll}15 & 1 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right] \) d. \( \left[\begin{array}{l}\dot{x}_{1} \\ \dot{x}_{2} \\ \dot{x}_{3}\end{array}\right]=\left[\begin{array}{ccc}-8 & 1 & 0 \\ -14 & 0 & 1 \\ -7 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]+\left[\begin{array}{c}0 \\ 1 \\ 15\end{array}\right] u \quad, \quad y=\left[\begin{array}{lll}1 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right] \)


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overshoot = 10%. % overshoot = ?? 0.1 * 6/²1-2² = l h (0-1)=-^&X1?- A briken 2.30: X A x & /?1-4² ) × 100. 2.30?1-²- 5-3 (1-²1: * €² (1-²) = 1.86 E² 2
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