State transition matrix (G = [0 1; −0.21 −1])

The state transition matrix of $\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k) + \mathbf{H}u(k)$ is

$$\boldsymbol{\psi}(k) = \mathbf{G}^k = \mathcal{Z}^{-1}\left[(z\mathbf{I}-\mathbf{G})^{-1}z\right]$$

It depends only on $\mathbf{G}$. $\mathbf{H}$, $\mathbf{C}$ and $\mathbf{D}$ are needed only for the response.

Step 1: $(z\mathbf{I}-\mathbf{G})^{-1}$

$$z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z & -1 \\ 0.21 & z+1 \end{bmatrix},\qquad |z\mathbf{I}-\mathbf{G}| = z^2 + z + 0.21 = (z+0.3)(z+0.7)$$ $$(z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{(z+0.3)(z+0.7)}\begin{bmatrix} z+1 & 1 \\ -0.21 & z \end{bmatrix}$$

Step 2: Partial fractions of $(z\mathbf{I}-\mathbf{G})^{-1}z$

$$\begin{aligned} \frac{(z+1)z}{(z+0.3)(z+0.7)} &= \frac{1.75z}{z+0.3} - \frac{0.75z}{z+0.7}\\ \frac{z}{(z+0.3)(z+0.7)} &= \frac{2.5z}{z+0.3} - \frac{2.5z}{z+0.7}\\ \frac{-0.21z}{(z+0.3)(z+0.7)} &= \frac{-0.525z}{z+0.3} + \frac{0.525z}{z+0.7}\\ \frac{z\cdot z}{(z+0.3)(z+0.7)} &= \frac{-0.75z}{z+0.3} + \frac{1.75z}{z+0.7} \end{aligned}$$

(For example, for (1,1): at $z = -0.3$, $\frac{0.7}{0.4} = 1.75$; at $z = -0.7$, $\frac{0.3}{-0.4} = -0.75$.)

Step 3: Inverse z-transform

Using $\mathcal{Z}^{-1}\left[\dfrac{z}{z+a}\right] = (-a)^k$:

$$\boldsymbol{\psi}(k) = \begin{bmatrix} 1.75(-0.3)^k - 0.75(-0.7)^k & 2.5(-0.3)^k - 2.5(-0.7)^k \\ -0.525(-0.3)^k + 0.525(-0.7)^k & -0.75(-0.3)^k + 1.75(-0.7)^k \end{bmatrix}$$

Check

• $k = 0$: $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \mathbf{I}$
• $k = 1$: $\begin{bmatrix} -0.525+0.525 & -0.75+1.75 \\ 0.1575-0.3675 & 0.225-1.225 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -0.21 & -1 \end{bmatrix} = \mathbf{G}$
• $k = 2$: $\begin{bmatrix} -0.21 & -1 \\ 0.21 & 0.79 \end{bmatrix} = \mathbf{G}^2$

Both eigenvalues ($-0.3$, $-0.7$) are inside the unit circle, so $\boldsymbol{\psi}(k) \to \mathbf{0}$ and the system is asymptotically stable.