State transition matrix of a 2×2 discrete system

The state transition matrix of $\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k) + \mathbf{H}u(k)$ is $\boldsymbol{\psi}(k) = \mathbf{G}^k$. It is found as

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

$\mathbf{H}$ and $\mathbf{C}$ are not needed for $\boldsymbol{\psi}(k)$.

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

$$z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z & -2 \\ 0.2 & z+1.5 \end{bmatrix},\qquad |z\mathbf{I}-\mathbf{G}| = z^2 + 1.5z + 0.4$$ $$(z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{z^2+1.5z+0.4}\begin{bmatrix} z+1.5 & 2 \\ -0.2 & z \end{bmatrix}$$

Step 2: Eigenvalues

$$z = \frac{-1.5 \pm \sqrt{2.25 - 1.6}}{2} = -0.75 \pm 0.4031 \;\Rightarrow\; \lambda_1 = -0.3469,\quad \lambda_2 = -1.1531$$

$\lambda_1 - \lambda_2 = 0.8062$.

Step 3: Partial fractions of each element of $(z\mathbf{I}-\mathbf{G})^{-1}z$

Each element has the form $\dfrac{N(z)\,z}{(z-\lambda_1)(z-\lambda_2)} = \dfrac{N(\lambda_1)}{\lambda_1-\lambda_2}\dfrac{z}{z-\lambda_1} + \dfrac{N(\lambda_2)}{\lambda_2-\lambda_1}\dfrac{z}{z-\lambda_2}$.

Using $\mathcal{Z}^{-1}\left[\dfrac{z}{z-\lambda}\right] = \lambda^k$:

Result

$$\boldsymbol{\psi}(k) = \begin{bmatrix} 1.4303(-0.3469)^k - 0.4303(-1.1531)^k & 2.4807(-0.3469)^k - 2.4807(-1.1531)^k \\ -0.2481(-0.3469)^k + 0.2481(-1.1531)^k & -0.4303(-0.3469)^k + 1.4303(-1.1531)^k \end{bmatrix}$$

Check

• $k = 0$: $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
• $k = 1$: $\begin{bmatrix} 0 & 2 \\ -0.2 & -1.5 \end{bmatrix} = \mathbf{G}$
• $k = 2$: $\begin{bmatrix} -0.4 & -3 \\ 0.3 & 1.85 \end{bmatrix} = \mathbf{G}^2$

Exact form: $\lambda_{1,2} = -0.75 \pm \sqrt{65}/20$ and $1.4303 = 0.5 + 7.5/\sqrt{65}$, $2.4807 = 20/\sqrt{65}$.

Note: $|\lambda_2| = 1.153 > 1$, so $\boldsymbol{\psi}(k)$ grows without bound and the system is unstable.