Observable canonical form of (0.3z⁻¹+0.6z⁻²)/(1−0.2z⁻¹+0.5z⁻²)

In the observable canonical form the denominator coefficients sit in the last column of $\mathbf{G}$, the numerator terms form $\mathbf{H}$, and $\mathbf{C} = [0\ \ 1]$.

Coefficients

$$\frac{Y(z)}{U(z)} = \frac{0.3z^{-1}+0.6z^{-2}}{1-0.2z^{-1}+0.5z^{-2}}$$

$a_1 = -0.2$, $a_2 = 0.5$, $b_0 = 0$, $b_1 = 0.3$, $b_2 = 0.6$.

Derivation

Cross-multiplying:

$$Y(z) = 0.2z^{-1}Y - 0.5z^{-2}Y + 0.3z^{-1}U + 0.6z^{-2}U$$ $$Y(z) = z^{-1}\Big[0.3U + 0.2Y + z^{-1}\big(0.6U - 0.5Y\big)\Big]$$

Define $X_2(z) = Y(z)$ and $X_1(z) = z^{-1}(0.6U - 0.5Y)$. Then $X_2 = z^{-1}(0.3U + 0.2X_2 + X_1)$:

$$\begin{aligned} x_1(k+1) &= -0.5x_2(k) + 0.6u(k)\\ x_2(k+1) &= x_1(k) + 0.2x_2(k) + 0.3u(k)\\ y(k) &= x_2(k) \end{aligned}$$

Observable canonical form

$$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & -0.5 \\ 1 & 0.2 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0.6 \\ 0.3 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}$$

Check

$$|z\mathbf{I}-\mathbf{G}| = \begin{vmatrix} z & 0.5 \\ -1 & z-0.2 \end{vmatrix} = z^2 - 0.2z + 0.5$$ $$\mathbf{C}\,\text{adj}(z\mathbf{I}-\mathbf{G})\,\mathbf{H} = \begin{bmatrix} 1 & z \end{bmatrix}\begin{bmatrix} 0.6 \\ 0.3 \end{bmatrix} = 0.3z + 0.6$$

So $F(z) = \dfrac{0.3z+0.6}{z^2-0.2z+0.5} = \dfrac{0.3z^{-1}+0.6z^{-2}}{1-0.2z^{-1}+0.5z^{-2}}$, as given.

 u --+----------[0.3]----------+
     |                         v
     +-[0.6]->(+)->[z^-1]-x1->(+)->[z^-1]--x2--+--> y
               ^                ^              |
               +----[-0.5]------+-----[0.2]----+