Controllable canonical form of (3z²+7z+15)/(z²+2z+4)

The numerator and denominator have the same degree, so first separate the direct term $D = b_0$. Then use the standard controllable canonical form.

Coefficients

$$X(z) = \frac{3z^2+7z+15}{z^2+2z+4}$$

Treat this as $Y(z)/U(z)$ with $b_0 = 3$, $b_1 = 7$, $b_2 = 15$, $a_1 = 2$, $a_2 = 4$:

$$\begin{aligned} b_2 - a_2b_0 &= 15 - 12 = 3\\ b_1 - a_1b_0 &= 7 - 6 = 1 \end{aligned}$$

So

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

Derivation

Let $\dfrac{Q(z)}{U(z)} = \dfrac{1}{z^2+2z+4}$, so $q(k+2) = -4q(k) - 2q(k+1) + u(k)$. Take $x_1 = q(k)$ and $x_2 = q(k+1)$:

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

Controllable canonical form

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

Check

$$[3\ \ 1]\frac{1}{z^2+2z+4}\begin{bmatrix} 1 \\ z \end{bmatrix} + 3 = \frac{z + 3 + 3z^2 + 6z + 12}{z^2+2z+4} = \frac{3z^2 + 7z + 15}{z^2+2z+4}$$