Jordan canonical form of 5/((z+1)²(z+2))

When the denominator has a repeated pole, the diagonal form is not possible, and the Jordan canonical form is used. The repeated pole gives a Jordan block with 1 on the superdiagonal.

Partial fractions

$$\frac{Y(z)}{U(z)} = \frac{5}{(z+1)^2(z+2)} = \frac{c_1}{(z+1)^2} + \frac{c_2}{z+1} + \frac{c_3}{z+2}$$ $$\begin{aligned} c_1 &= \left[\frac{5}{z+2}\right]_{z=-1} = 5\\ c_2 &= \frac{d}{dz}\left[\frac{5}{z+2}\right]_{z=-1} = \left[\frac{-5}{(z+2)^2}\right]_{z=-1} = -5\\ c_3 &= \left[\frac{5}{(z+1)^2}\right]_{z=-2} = 5 \end{aligned}$$

So $\dfrac{Y(z)}{U(z)} = \dfrac{5}{(z+1)^2} - \dfrac{5}{z+1} + \dfrac{5}{z+2}$.

Choice of states

$$X_3(z) = \frac{U(z)}{z+2},\qquad X_2(z) = \frac{U(z)}{z+1},\qquad X_1(z) = \frac{X_2(z)}{z+1} = \frac{U(z)}{(z+1)^2}$$

This gives

$$\begin{aligned} x_1(k+1) &= -x_1(k) + x_2(k)\\ x_2(k+1) &= -x_2(k) + u(k)\\ x_3(k+1) &= -2x_3(k) + u(k)\\ y(k) &= 5x_1(k) - 5x_2(k) + 5x_3(k) \end{aligned}$$

Jordan canonical form

$$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \\ x_3(k+1) \end{bmatrix} = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \\ x_3(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} 5 & -5 & 5 \end{bmatrix}\mathbf{x}(k) + 0\cdot u(k)$$

Check: $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \dfrac{5}{z^3+4z^2+5z+2} = \dfrac{5}{(z+1)^2(z+2)}$.