Controllable and observable forms of (z+1)/(z²+1.3z+0.4)

Both forms are found from the same coefficients: $a_1 = 1.3$, $a_2 = 0.4$, $b_0 = 0$, $b_1 = 1$, $b_2 = 1$.

$$\frac{Y(z)}{U(z)} = \frac{z+1}{z^2+1.3z+0.4}$$

Controllable canonical form

Let $\dfrac{Q(z)}{U(z)} = \dfrac{1}{z^2+1.3z+0.4}$, so $Y = (z+1)Q$. With $x_1(k) = q(k)$ and $x_2(k) = q(k+1)$:

$$\begin{aligned} x_1(k+1) &= x_2(k)\\ x_2(k+1) &= -0.4x_1(k) - 1.3x_2(k) + u(k)\\ y(k) &= x_1(k) + x_2(k) \end{aligned}$$ $$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -0.4 & -1.3 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 1 & 1 \end{bmatrix}\mathbf{x}(k)$$

Observable canonical form

Divide by $z^2$ and cross-multiply:

$$Y(z) = z^{-1}\Big[U - 1.3Y + z^{-1}\big(U - 0.4Y\big)\Big]$$

Let $X_2(z) = Y(z)$ and $X_1(z) = z^{-1}(U - 0.4Y)$:

$$\begin{aligned} x_1(k+1) &= -0.4x_2(k) + u(k)\\ x_2(k+1) &= x_1(k) - 1.3x_2(k) + u(k)\\ y(k) &= x_2(k) \end{aligned}$$ $$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & -0.4 \\ 1 & -1.3 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\mathbf{x}(k)$$

Check

• Controllable form: $\mathbf{C}\,\text{adj}(z\mathbf{I}-\mathbf{G})\mathbf{H} = [1\ \ 1]\begin{bmatrix} 1 \\ z \end{bmatrix} = z+1$.
• Observable form: $[0\ \ 1]\begin{bmatrix} z+1.3 & -0.4 \\ 1 & z \end{bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = 1 + z$.
• Both have denominator $z^2 + 1.3z + 0.4$.

The two forms are duals: $\mathbf{G}_o = \mathbf{G}_c^T$, $\mathbf{H}_o = \mathbf{C}_c^T$, $\mathbf{C}_o = \mathbf{H}_c^T$.