Observable canonical form of (4z²+3z+5)/(5z²+z−4)

The leading coefficient of the denominator is 5, so divide the numerator and denominator by 5 first. Because the degrees are equal, there is a direct term $D = b_0$.

Normalise

$$G(z) = \frac{4z^2+3z+5}{5z^2+z-4} = \frac{0.8z^2 + 0.6z + 1}{z^2 + 0.2z - 0.8}$$ $$\begin{aligned} b_2 - a_2b_0 &= 1 - (-0.8)(0.8) = 1.64\\ b_1 - a_1b_0 &= 0.6 - (0.2)(0.8) = 0.44 \end{aligned}$$

Derivation

Separate the direct term:

$$G(z) = 0.8 + \frac{0.44z + 1.64}{z^2 + 0.2z - 0.8}$$

For the strictly proper part, $W = z^{-1}\left[0.44U - 0.2W + z^{-1}(1.64U + 0.8W)\right]$. With $x_2 = w$:

$$\begin{aligned} x_1(k+1) &= 0.8x_2(k) + 1.64u(k)\\ x_2(k+1) &= x_1(k) - 0.2x_2(k) + 0.44u(k)\\ y(k) &= x_2(k) + 0.8u(k) \end{aligned}$$

Observable canonical form

$$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 0.8 \\ 1 & -0.2 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 1.64 \\ 0.44 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + 0.8u(k)$$

Check

$$\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + D = \frac{0.44z + 1.64}{z^2+0.2z-0.8} + 0.8 = \frac{0.8z^2 + 0.6z + 1}{z^2+0.2z-0.8} = \frac{4z^2+3z+5}{5z^2+z-4}$$

(Poles: $z = 0.8$ and $z = -1$. The pole at $-1$ is on the unit circle, so the system is only marginally stable.)