State model from 20(4s+2)/(s³+5s²+8s+2)

$$\frac{Y(s)}{U(s)} = \frac{20(4s+2)}{s^3+5s^2+8s+2} = \frac{80s + 40}{s^3 + 5s^2 + 8s + 2}$$

Since the numerator has $s$ terms, use the phase-variable (controllable canonical) form with an intermediate variable $X(s)$.

Step 1: Split the transfer function

$$\frac{Y(s)}{U(s)} = \frac{X(s)}{U(s)}\cdot\frac{Y(s)}{X(s)},\qquad \frac{X(s)}{U(s)} = \frac{1}{s^3+5s^2+8s+2},\qquad \frac{Y(s)}{X(s)} = 80s + 40$$

So $\dddot{x} + 5\ddot{x} + 8\dot{x} + 2x = u$ and $y = 80\dot{x} + 40x$.

Step 2: Choose state variables

$x_1 = x$, $x_2 = \dot{x}$, $x_3 = \ddot{x}$:

$$\begin{aligned} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= x_3 \\ \dot{x}_3 &= -2x_1 - 8x_2 - 5x_3 + u \\ y &= 40x_1 + 80x_2 \end{aligned}$$

Step 3: Matrix form

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -8 & -5 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}u$$ $$y = \begin{bmatrix} 40 & 80 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$

Step 4: Characteristic equation from the matrix

The characteristic equation is $|sI - A| = 0$:

$$sI - A = \begin{bmatrix} s & -1 & 0 \\ 0 & s & -1 \\ 2 & 8 & s+5 \end{bmatrix}$$ $$\begin{aligned} |sI-A| &= s\left[s(s+5) + 8\right] + 1\left[0 + 2\right] \\ &= s^3 + 5s^2 + 8s + 2 = 0 \end{aligned}$$

This equals the denominator of the transfer function, as it must. Its roots (eigenvalues of $A$) are $s = -0.304$ and $-2.348 \pm j1.029$, all in the left half-plane, so the system is stable.

Answer: $A = \begin{bmatrix} 0&1&0\\0&0&1\\-2&-8&-5 \end{bmatrix}$, $B = \begin{bmatrix}0\\0\\1\end{bmatrix}$, $C = \begin{bmatrix}40&80&0\end{bmatrix}$; characteristic equation $s^3 + 5s^2 + 8s + 2 = 0$.