Controllable and observable forms of a 0.368/0.264 system

Both forms come from the same coefficients. The controllable form puts the denominator in the last row of $\mathbf{G}$, and the observable form puts it in the last column.

Coefficients

$$\frac{Y(z)}{U(z)} = \frac{0.368z^{-1}+0.264z^{-2}}{1-1.368z^{-1}+0.368z^{-2}} = \frac{0.368z + 0.264}{z^2 - 1.368z + 0.368}$$

$a_1 = -1.368$, $a_2 = 0.368$, $b_0 = 0$, $b_1 = 0.368$, $b_2 = 0.264$. Since $b_0 = 0$: $b_2 - a_2b_0 = 0.264$ and $b_1 - a_1b_0 = 0.368$.

(i) Controllable canonical form

Let $Q(z) = \dfrac{U(z)}{z^2-1.368z+0.368}$, $x_1 = q(k)$, $x_2 = q(k+1)$, $y = 0.264x_1 + 0.368x_2$:

$$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -0.368 & 1.368 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} 0.264 & 0.368 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}$$

(ii) Observable canonical form

Write $Y = z^{-1}\left[0.368U + 1.368Y + z^{-1}(0.264U - 0.368Y)\right]$ and take $x_2 = y$, $x_1(k+1) = -0.368x_2(k) + 0.264u(k)$:

$$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & -0.368 \\ 1 & 1.368 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0.264 \\ 0.368 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}$$

Check

For both forms, $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \dfrac{0.368z+0.264}{z^2-1.368z+0.368}$. The observable form is the transpose of the controllable one: $\mathbf{G}_o = \mathbf{G}_c^T$, $\mathbf{H}_o = \mathbf{C}_c^T$, $\mathbf{C}_o = \mathbf{H}_c^T$. (The denominator factors as $(z-1)(z-0.368)$.)