Observable canonical form of (3z²+7z+1)/(z²+2z+4)

The numerator and denominator have equal degree, so there is a direct term $D = b_0 = 3$. The observable form then follows the standard pattern.

Coefficients

$$\frac{Y(z)}{U(z)} = \frac{3z^2+7z+1}{z^2+2z+4}$$

$b_0 = 3$, $b_1 = 7$, $b_2 = 1$, $a_1 = 2$, $a_2 = 4$:

$$\begin{aligned} b_2 - a_2b_0 &= 1 - 12 = -11\\ b_1 - a_1b_0 &= 7 - 6 = 1 \end{aligned}$$

So

$$\frac{Y(z)}{U(z)} = 3 + \frac{z - 11}{z^2 + 2z + 4}$$

Derivation

For the strictly proper part $W(z) = \dfrac{z-11}{z^2+2z+4}U(z)$:

$$W = z^{-1}\Big[U - 2W + z^{-1}\big(-11U - 4W\big)\Big]$$

Let $x_2 = w$ and $X_1 = z^{-1}(-11U - 4W)$:

$$\begin{aligned} x_1(k+1) &= -4x_2(k) - 11u(k)\\ x_2(k+1) &= x_1(k) - 2x_2(k) + u(k)\\ y(k) &= x_2(k) + 3u(k) \end{aligned}$$

Observable canonical form

$$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & -4 \\ 1 & -2 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} -11 \\ 1 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + 3u(k)$$

Check

$$\mathbf{C}\,\text{adj}(z\mathbf{I}-\mathbf{G})\,\mathbf{H} = \begin{bmatrix} 1 & z \end{bmatrix}\begin{bmatrix} -11 \\ 1 \end{bmatrix} = z - 11$$ $$\frac{z-11}{z^2+2z+4} + 3 = \frac{3z^2 + 7z + 1}{z^2+2z+4}$$