Controllable, observable and diagonal forms

The three forms all realise

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

with $a_1 = 1$, $a_2 = 0.6$, $b_0 = 0$, $b_1 = 1$, $b_2 = 1$.

(i) Controllable canonical form

Let $\dfrac{Q(z)}{U(z)} = \dfrac{1}{z^2+z+0.6}$, so $Y = (z+1)Q$. With $x_1(k) = q(k)$ and $x_2(k) = q(k+1)$:

$$\begin{aligned} x_1(k+1) &= x_2(k)\\ x_2(k+1) &= -0.6x_1(k) - x_2(k) + u(k)\\ y(k) &= x_1(k) + x_2(k) \end{aligned}$$ $$\mathbf{x}(k+1) = \begin{bmatrix} 0 & 1 \\ -0.6 & -1 \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 1 & 1 \end{bmatrix}\mathbf{x}(k)$$

(ii) Observable canonical form

$Y = z^{-1}\left[U - Y + z^{-1}(U - 0.6Y)\right]$. Take $x_2 = y$ and $X_1 = z^{-1}(U - 0.6Y)$:

$$\begin{aligned} x_1(k+1) &= -0.6x_2(k) + u(k)\\ x_2(k+1) &= x_1(k) - x_2(k) + u(k)\\ y(k) &= x_2(k) \end{aligned}$$ $$\mathbf{x}(k+1) = \begin{bmatrix} 0 & -0.6 \\ 1 & -1 \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} 1 \\ 1 \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\mathbf{x}(k)$$

(iii) Diagonal canonical form

Poles:

$$z^2 + z + 0.6 = 0 \;\Rightarrow\; p_{1,2} = \frac{-1 \pm \sqrt{1-2.4}}{2} = -0.5 \pm j0.5916$$

They are distinct, but complex. Residues:

$$\begin{aligned} c_1 &= \left[\frac{z+1}{z-p_2}\right]_{z=p_1} = \frac{0.5 + j0.5916}{j1.1832} = 0.5 - j0.4226\\ c_2 &= \bar{c}_1 = 0.5 + j0.4226 \end{aligned}$$ $$\frac{Y(z)}{U(z)} = \frac{0.5 - j0.4226}{z + 0.5 - j0.5916} + \frac{0.5 + j0.4226}{z + 0.5 + j0.5916}$$

With $X_i = U/(z - p_i)$:

$$\mathbf{x}(k+1) = \begin{bmatrix} -0.5 + j0.5916 & 0 \\ 0 & -0.5 - j0.5916 \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} 1 \\ 1 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} 0.5 - j0.4226 & 0.5 + j0.4226 \end{bmatrix}\mathbf{x}(k)$$

The states are complex conjugates of each other, but the output $y(k)$ is real. Exact values: $p_{1,2} = -\tfrac12 \pm j\tfrac{\sqrt{35}}{10}$ and $c_{1,2} = \tfrac12 \mp j\tfrac{\sqrt{35}}{14}$.

Check

• $c_1 + c_2 = 1$, the coefficient of z in the numerator.
• $-(c_1p_2 + c_2p_1) = 1$, the constant term.

All three forms give $\dfrac{z+1}{z^2+z+0.6}$. The poles have $|p| = \sqrt{0.6} = 0.775 < 1$, so the system is stable.