Q $3$ A. Consider the Control Process \begin{pmatrix}x\ y\ z\end{pmatrix}^' = \begin{pmatrix}0 & 1 & 0\ 0 & 0 & 1\ -2 & -4 & 1\end{pmatrix}\begin{pmatrix}x\ y\ z\end{pmatrix} + \begin{pmatrix}1 & 0\ 0 & 1\ -1 & 1\end{pmatrix}\begin{pmatrix}u_1\left(t\right)\ u_2\left(t\right)\end{pmatrix}. Is this system globally null controllable? Q $3$ B. Is this system in Q $3$ A globally null controllable under the additional contraint $||\begin{pmatrix}u_1\left(t\right)\ u_2\left(t\right)\end{pmatrix}|| \le 2$ ? Use Analysis to solve the question completely

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Grade 6

Which word problem matches the one-step equation?

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Q. Q

3

A. Consider the Control Process \begin{pmatrix}x\ y\ z\end{pmatrix}^' = \begin{pmatrix}0 & 1 & 0\ 0 & 0 & 1\ -2 & -4 & 1\end{pmatrix}\begin{pmatrix}x\ y\ z\end{pmatrix} + \begin{pmatrix}1 & 0\ 0 & 1\ -1 & 1\end{pmatrix}\begin{pmatrix}u_1\left(t\right)\ u_2\left(t\right)\end{pmatrix}. Is this system globally null controllable? Q

3

B. Is this system in Q

3

A globally null controllable under the additional contraint

||\begin{pmatrix}u_1\left(t\right)\ u_2\left(t\right)\end{pmatrix}|| \le 2

? Use Analysis to solve the question completely

Form Controllability Matrix: To determine global null controllability, we need to check the controllability matrix for the system without the constraint. The controllability matrix $C$ is formed by the matrices $A$ and $B$ from the state-space representation of the system.
Calculate $AB$ and $A^2B$ : First, let's find the controllability matrix $C$ for the system. The matrix $A$ is given by: $\newline$ $A = \begin{pmatrix}0&1&0\ 0&0&1\ -2&-4&1\end{pmatrix}$ $\newline$ And the matrix $B$ is given by: $\newline$ $B = \begin{pmatrix}1&0\ 0&1\ -1&1\end{pmatrix}$
Check Rank of C: The controllability matrix $C$ is formed by the columns $[B, AB, A^2B]$ . Let's calculate $AB$ and $A^2B$ . $AB = A \times B = \begin{pmatrix}0&1&0\ 0&0&1\ -2&-4&1\end{pmatrix} \times \begin{pmatrix}1&0\ 0&1\ -1&1\end{pmatrix} = \begin{pmatrix}0&1\ -1&1\ -3&-3\end{pmatrix}$
System Controllability: Now, let's calculate $A^2B$ . $\newline$ $A^2 = A \cdot A = \begin{pmatrix}0&1&0\ 0&0&1\ -2&-4&1\end{pmatrix} \cdot \begin{pmatrix}0&1&0\ 0&0&1\ -2&-4&1\end{pmatrix} = \begin{pmatrix}-2&-4&1\ -2&-4&1\ 0&-2&-3\end{pmatrix}$ $\newline$ $A^2B = A^2 \cdot B = \begin{pmatrix}-2&-4&1\ -2&-4&1\ 0&-2&-3\end{pmatrix} \cdot \begin{pmatrix}1&0\ 0&1\ -1&1\end{pmatrix} = \begin{pmatrix}-2&-3\ -2&-3\ 1&-1\end{pmatrix}$
Consider Additional Constraint: The controllability matrix $C$ is then: $C = [B, AB, A^2B] = \begin{pmatrix}1&0&0&1&-2&-3\ 0&1&-1&1&-2&-3\ -1&1&-3&-3&1&-1\end{pmatrix}$
Conclusion: To check if the system is controllable, we need to see if the rank of the controllability matrix $C$ is equal to the dimension of the state vector, which is $3$ in this case.
Conclusion: To check if the system is controllable, we need to see if the rank of the controllability matrix $C$ is equal to the dimension of the state vector, which is $3$ in this case.Calculating the rank of $C$ , we find that the rank( $C$ ) = $3$ , since the three columns of $C$ are linearly independent.
Conclusion: To check if the system is controllable, we need to see if the rank of the controllability matrix $C$ is equal to the dimension of the state vector, which is $3$ in this case. Calculating the rank of $C$ , we find that $\text{rank}(C) = 3$ , since the three columns of $C$ are linearly independent. Since the rank of the controllability matrix $C$ is equal to the dimension of the state vector, the system is globally null controllable without the additional constraint.
Conclusion: To check if the system is controllable, we need to see if the rank of the controllability matrix $C$ is equal to the dimension of the state vector, which is $3$ in this case. Calculating the rank of $C$ , we find that the $\text{rank}(C) = 3$ , since the three columns of $C$ are linearly independent. Since the rank of the controllability matrix $C$ is equal to the dimension of the state vector, the system is globally null controllable without the additional constraint. Now, let's consider the additional constraint $||u(t)|| \leq 2$ . This constraint limits the magnitude of the control inputs but does not affect the controllability of the system. The system's controllability is determined by the $A$ and $B$ matrices, not by the magnitude of the control inputs.
Conclusion: To check if the system is controllable, we need to see if the rank of the controllability matrix $C$ is equal to the dimension of the state vector, which is $3$ in this case.Calculating the rank of $C$ , we find that the rank( $C$ ) = $3$ , since the three columns of $C$ are linearly independent.Since the rank of the controllability matrix $C$ is equal to the dimension of the state vector, the system is globally null controllable without the additional constraint.Now, let's consider the additional constraint $||u(t)|| \leq 2$ . This constraint limits the magnitude of the control inputs but does not affect the controllability of the system. The system's controllability is determined by the $A$ and $B$ matrices, not by the magnitude of the control inputs.Therefore, the system is still globally null controllable under the additional constraint $||u(t)|| \leq 2$ .