State model, controllability and observability of 2/(s³+6s²+11s+6)

State model (phase-variable form)

Split the transfer function with an intermediate variable $X(s)$:

$$\frac{X(s)}{U(s)} = \frac{1}{s^3 + 6s^2 + 11s + 6},\qquad Y(s) = 2X(s)$$

So $\dddot{x} + 6\ddot{x} + 11\dot{x} + 6x = u$ and $y = 2x$. Choose $x_1 = x$, $x_2 = \dot{x}$, $x_3 = \ddot{x}$:

$$\dot{x}_1 = x_2,\qquad \dot{x}_2 = x_3,\qquad \dot{x}_3 = -6x_1 - 11x_2 - 6x_3 + u,\qquad y = 2x_1$$ $$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -11 & -6 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}u,\qquad y = \begin{bmatrix} 2 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$

Controllability (Kalman test)

$Q_c = [\,B \;\; AB \;\; A^2B\,]$:

$$AB = \begin{bmatrix} 0 \\ 1 \\ -6 \end{bmatrix},\qquad A^2B = A(AB) = \begin{bmatrix} 1 \\ -6 \\ -6 \cdot 0 - 11 + 36 \end{bmatrix} = \begin{bmatrix} 1 \\ -6 \\ 25 \end{bmatrix}$$ $$Q_c = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & -6 \\ 1 & -6 & 25 \end{bmatrix},\qquad |Q_c| = -1 \neq 0$$

Rank $= 3 = n$, so the system is completely state controllable.

Observability

$Q_o = \begin{bmatrix} C \\ CA \\ CA^2 \end{bmatrix}$:

$$C = [2\;\;0\;\;0],\qquad CA = [0\;\;2\;\;0],\qquad CA^2 = [0\;\;0\;\;2]$$ $$Q_o = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix},\qquad |Q_o| = 8 \neq 0$$

Rank $= 3$, so the system is completely observable.

This agrees with the transfer function: $s^3 + 6s^2 + 11s + 6 = (s+1)(s+2)(s+3)$ and the numerator is a constant, so there is no pole–zero cancellation.

Answer: the system is both controllable and observable.