Discretization and state transition matrix

Discretization of a continuous-time state model

Consider $\dot{\mathbf{x}}(t) = \mathbf{A}\mathbf{x}(t) + \mathbf{B}\mathbf{u}(t)$, $\mathbf{y}(t) = \mathbf{C}\mathbf{x}(t) + \mathbf{D}\mathbf{u}(t)$, where the input comes from a zero-order hold, so $\mathbf{u}(t) = \mathbf{u}(kT)$ for $kT \le t < (k+1)T$.

The solution from $t = kT$ to $(k+1)T$ is

$$\mathbf{x}((k+1)T) = e^{\mathbf{A}T}\mathbf{x}(kT) + \left(\int_0^T e^{\mathbf{A}\lambda}\,d\lambda\right)\mathbf{B}\,\mathbf{u}(kT)$$

Hence

$$\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k) + \mathbf{H}\mathbf{u}(k),\qquad \mathbf{G} = e^{\mathbf{A}T},\quad \mathbf{H} = \left(\int_0^T e^{\mathbf{A}\lambda}d\lambda\right)\mathbf{B}$$

$\mathbf{C}$ and $\mathbf{D}$ are unchanged. $e^{\mathbf{A}t}$ is found from $\mathcal{L}^{-1}[(s\mathbf{I}-\mathbf{A})^{-1}]$. If $\mathbf{A}$ is invertible, $\mathbf{H} = \mathbf{A}^{-1}(e^{\mathbf{A}T}-\mathbf{I})\mathbf{B}$.

State transition matrix of the given system

$$\boldsymbol{\psi}(k) = \mathbf{G}^k = \mathcal{Z}^{-1}\left[(z\mathbf{I}-\mathbf{G})^{-1}z\right],\qquad \mathbf{G} = \begin{bmatrix} 0 & 2 \\ -0.2 & -1.5 \end{bmatrix}$$ $$|z\mathbf{I}-\mathbf{G}| = z^2 + 1.5z + 0.4 = 0 \;\Rightarrow\; \lambda_1 = -0.3469,\ \lambda_2 = -1.1531$$

For distinct eigenvalues, $\mathbf{G}^k = \lambda_1^k\mathbf{M}_1 + \lambda_2^k\mathbf{M}_2$ with $\mathbf{M}_1 = \dfrac{\mathbf{G}-\lambda_2\mathbf{I}}{\lambda_1-\lambda_2}$ and $\mathbf{M}_2 = \dfrac{\mathbf{G}-\lambda_1\mathbf{I}}{\lambda_2-\lambda_1}$, where $\lambda_1-\lambda_2 = 0.8062$:

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

Check: $k = 0$ gives $\mathbf{I}$, and $k = 1$ gives $\mathbf{G}$. Since $|\lambda_2| > 1$, the system is unstable.