Controllable and observable forms (denominator 0.36z⁻²)

Here the constant term of the denominator is 0.36, not 0.368. The method is the same; only the entry $-a_2$ changes.

Coefficients

$$\frac{Y(z)}{U(z)} = \frac{0.368z^{-1}+0.264z^{-2}}{1-1.368z^{-1}+0.36z^{-2}} = \frac{0.368z+0.264}{z^2-1.368z+0.36}$$

(i) Controllable canonical form

Let $\dfrac{Q(z)}{U(z)} = \dfrac{1}{z^2-1.368z+0.36}$ with $x_1(k) = q(k)$, $x_2(k) = q(k+1)$:

$$\begin{aligned} x_1(k+1) &= x_2(k)\\ x_2(k+1) &= -0.36x_1(k) + 1.368x_2(k) + u(k)\\ y(k) &= 0.264x_1(k) + 0.368x_2(k) \end{aligned}$$ $$\mathbf{G} = \begin{bmatrix} 0 & 1 \\ -0.36 & 1.368 \end{bmatrix},\quad \mathbf{H} = \begin{bmatrix} 0 \\ 1 \end{bmatrix},\quad \mathbf{C} = \begin{bmatrix} 0.264 & 0.368 \end{bmatrix},\quad D = 0$$

(ii) Observable canonical form

From $Y = z^{-1}\left[0.368U + 1.368Y + z^{-1}(0.264U - 0.36Y)\right]$ with $x_2 = y$:

$$\begin{aligned} x_1(k+1) &= -0.36x_2(k) + 0.264u(k)\\ x_2(k+1) &= x_1(k) + 1.368x_2(k) + 0.368u(k)\\ y(k) &= x_2(k) \end{aligned}$$ $$\mathbf{G} = \begin{bmatrix} 0 & -0.36 \\ 1 & 1.368 \end{bmatrix},\quad \mathbf{H} = \begin{bmatrix} 0.264 \\ 0.368 \end{bmatrix},\quad \mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix},\quad D = 0$$

Check and remark

• Both give $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \dfrac{0.368z+0.264}{z^2-1.368z+0.36}$.
• The poles are $z = 0.3556$ and $z = 1.0124$. The second lies just outside the unit circle, so this system is (slightly) unstable, unlike the 0.368 version, which has a pole at exactly $z = 1$.