Jordan canonical form of (2z³+6z+5)/((z+2)²(z+3))

The pole $z=-2$ is repeated, so the Jordan canonical form is used. The numerator and denominator have the same degree, so there is a direct term $D = b_0$.

Separate the direct term

Denominator: $(z+2)^2(z+3) = z^3 + 7z^2 + 16z + 12$.

$$\frac{2z^3+6z+5}{z^3+7z^2+16z+12} = 2 + \frac{-14z^2 - 26z - 19}{(z+2)^2(z+3)}$$

So $D = b_0 = 2$.

Partial fractions

$$\frac{Y(z)}{U(z)} = 2 + \frac{c_1}{(z+2)^2} + \frac{c_2}{z+2} + \frac{c_3}{z+3}$$

Let $N(z) = 2z^3 + 6z + 5$, so $N(-2) = -16 - 12 + 5 = -23$ and $N(-3) = -54 - 18 + 5 = -67$.

$$\begin{aligned} c_1 &= \left[\frac{N(z)}{z+3}\right]_{z=-2} = \frac{-23}{1} = -23\\ c_2 &= \frac{d}{dz}\left[\frac{N(z)}{z+3}\right]_{z=-2} = \left[\frac{(6z^2+6)(z+3) - N(z)}{(z+3)^2}\right]_{z=-2} = \frac{30 + 23}{1} = 53\\ c_3 &= \left[\frac{N(z)}{(z+2)^2}\right]_{z=-3} = \frac{-67}{1} = -67 \end{aligned}$$ $$\frac{Y(z)}{U(z)} = 2 - \frac{23}{(z+2)^2} + \frac{53}{z+2} - \frac{67}{z+3}$$

States

$X_2 = \dfrac{U}{z+2}$, $X_1 = \dfrac{X_2}{z+2}$, $X_3 = \dfrac{U}{z+3}$:

$$\begin{aligned} x_1(k+1) &= -2x_1(k) + x_2(k)\\ x_2(k+1) &= -2x_2(k) + u(k)\\ x_3(k+1) &= -3x_3(k) + u(k)\\ y(k) &= -23x_1(k) + 53x_2(k) - 67x_3(k) + 2u(k) \end{aligned}$$

Jordan canonical form

$$\mathbf{x}(k+1) = \begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -3 \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} -23 & 53 & -67 \end{bmatrix}\mathbf{x}(k) + 2u(k)$$

Check: $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + 2$ gives back $\dfrac{2z^3+6z+5}{(z+2)^2(z+3)}$.