$$ x = \begin{bmatrix} 0 & 6 & -5\\ 1 & 0 & 2\\ 3&2&4 \end{bmatrix} \space x + \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \space u $$ $$ y = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \space x$$

Topic : State Variable Models

Difficulty : High

Marks : 10M

Given:

$A=\begin{bmatrix} 0 & 6 & -5 \\ 1 & 0 & 2 \\ 3 & 2 & 4 \end{bmatrix}-----(1)$

$B=\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}-----(2)$

$C=\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}-----(3)$

Since A is a 3x3 matrix, so n=3.

Controllability: The necessary and sufficient condition for controllability is

${ Q }_{ C }=\begin{bmatrix} B & AB & ----- & { A }^{ n-1 }B \end{bmatrix}$

${ Q }_{ C }=\begin{bmatrix} B & AB & { A }^{ 2 }B \end{bmatrix}-----(4)$

From equation (2),

$B=\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}-----(5)$

$A\cdot B=\begin{bmatrix} 0 & 6 & -5 \\ 1 & 0 & 2 \\ 3 & 2 & 4 \end{bmatrix}\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$

$A\cdot B=\begin{bmatrix} -4 \\ 4 \\ 10 \end{bmatrix}-----(6)$

${ A }^{ 2 }B=A\cdot (AB)=\begin{bmatrix} 0 & 6 & -5 \\ 1 & 0 & 2 \\ 3 & 2 & 4 \end{bmatrix}\begin{bmatrix} -4 \\ 4 \\ 10 \end{bmatrix}$

${ A }^{ 2 }B=\begin{bmatrix} -26 \\ 16 \\ 36 \end{bmatrix}-----(7)$

Put equation (5), (6), (7) in equation (4),

${ Q }_{ C }=\begin{bmatrix} 0 & -4 & -26 \\ 1 & 4 & 16 \\ 2 & 10 & 36 \end{bmatrix}$

Now, find the determinant of $Q_{C}$

$|{ Q }_{ C }|=\begin{vmatrix} 0 & -4 & -26 \\ 1 & 4 & 16 \\ 2 & 10 & 36 \end{vmatrix}$

$|Q_{C}|=-36$

$\therefore |Q_{C}| \neq 0$

Since the determinant of $Q_{C}$ is non-zero, therefore of $Q_{C}=n=3$. Hence, the system is completely controllable.

Observability:

Given,

$B=\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$

$B'=\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}$

$C=\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$

$C'=\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

$A=\begin{bmatrix} 0 & 6 & -5 \\ 1 & 0 & 2 \\ 3 & 2 & 4 \end{bmatrix}$

$A'=\begin{bmatrix} 0 & 1 & 3 \\ 6 & 0 & 2 \\ -5 & 2 & 4 \end{bmatrix}$

The necessary and sufficient condition for observability.

${ Q }_{ C }=\begin{bmatrix} C^{T} & A^{T}C^{T} & ---- & { (A^{T}) }^{ n-1 }C^{T} \end{bmatrix}$

For n=3,

${ Q }_{ C }=\begin{bmatrix} C^{T} & A^{T}C^{T} & { (A^{T}) }^{ 2 }C^{T} \end{bmatrix}-----(8)$

$C^{T}=\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}-----(9)$

$ A^{T}C^{T}=\begin{bmatrix} 0 & 1 & 3 \\ 6 & 0 & 2 \\ -5 & 2 & 4 \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

$ A^{T}C^{T}=\begin{bmatrix} 11 \\ 12 \\ 11 \end{bmatrix}-----(10)$

$ (A^{T})^{2}C^{T}=A^{T} \cdot (A^{T}C^{T})=\begin{bmatrix} 0 & 1 & 3 \\ 6 & 0 & 2 \\ -5 & 2 & 4 \end{bmatrix}\begin{bmatrix} 11 \\ 12 \\ 11 \end{bmatrix}$

$ (A^{T})^{2}C^{T}=\begin{bmatrix} 45 \\ 88 \\ 13 \end{bmatrix}-----(11)$

Put equations (9), (10), (11) in equation (8),

${ Q }_{ C }=\begin{bmatrix} 1 & 11 & 45 \\ 2 & 12 & 88 \\ 3 & 11 & 13 \end{bmatrix}$

Now, find the determinant of $Q_{C}$

$|{ Q }_{ C }|=\begin{vmatrix} 1 & 11 & 45 \\ 2 & 12 & 88 \\ 3 & 11 & 13 \end{vmatrix}$

$|Q_{C}|=1176$

Since, $|Q_{C}| \neq 0$, the rank of $Q_{C}$ is n=3.

$\therefore$ The system is completely observable.