Discretize a continuous state model; pulse TF matrix

Discretization with a zero-order hold gives $\mathbf{G} = e^{\mathbf{A}T}$ and $\mathbf{H} = \left(\int_0^T e^{\mathbf{A}\lambda}d\lambda\right)\mathbf{B}$, with $\mathbf{C}$ and $\mathbf{D}$ unchanged. The sampling period is not given, so the result is found for a general T and then evaluated for T = 1 s.

Step 1: State transition matrix $e^{\mathbf{A}t}$

$$\mathbf{A} = \begin{bmatrix} 0 & 1 \\ 0 & -2 \end{bmatrix},\quad \mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix},\quad \mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix}$$ $$(s\mathbf{I}-\mathbf{A})^{-1} = \begin{bmatrix} s & -1 \\ 0 & s+2 \end{bmatrix}^{-1} = \begin{bmatrix} \dfrac{1}{s} & \dfrac{1}{s(s+2)} \\[2mm] 0 & \dfrac{1}{s+2} \end{bmatrix}$$ $$e^{\mathbf{A}t} = \mathcal{L}^{-1}\left[(s\mathbf{I}-\mathbf{A})^{-1}\right] = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2t}) \\ 0 & e^{-2t} \end{bmatrix}$$

Step 2: G and H

$$\mathbf{G} = e^{\mathbf{A}T} = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2T}) \\ 0 & e^{-2T} \end{bmatrix}$$ $$\mathbf{H} = \int_0^T \begin{bmatrix} \tfrac{1}{2}(1-e^{-2\lambda}) \\ e^{-2\lambda} \end{bmatrix}d\lambda = \begin{bmatrix} \tfrac{1}{4}\left(2T - 1 + e^{-2T}\right) \\ \tfrac{1}{2}(1-e^{-2T}) \end{bmatrix}$$

Discretized model

$$\mathbf{x}(k+1) = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2T}) \\ 0 & e^{-2T} \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} \tfrac{1}{4}(2T-1+e^{-2T}) \\ \tfrac{1}{2}(1-e^{-2T}) \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\mathbf{x}(k)$$

For T = 1 s ($e^{-2} = 0.1353$):

$$\mathbf{G} = \begin{bmatrix} 1 & 0.4323 \\ 0 & 0.1353 \end{bmatrix},\qquad \mathbf{H} = \begin{bmatrix} 0.2838 \\ 0.4323 \end{bmatrix}$$

Step 3: Pulse transfer function

$$(z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{(z-1)(z-e^{-2T})}\begin{bmatrix} z-e^{-2T} & \tfrac{1}{2}(1-e^{-2T}) \\ 0 & z-1 \end{bmatrix}$$

With $\mathbf{C} = [0\ \ 1]$ only the second row is needed:

$$\begin{aligned} F(z) &= \frac{[0\ \ z-1]\,\mathbf{H}}{(z-1)(z-e^{-2T})} = \frac{(z-1)\tfrac{1}{2}(1-e^{-2T})}{(z-1)(z-e^{-2T})}\\ &= \frac{\tfrac{1}{2}(1-e^{-2T})}{z-e^{-2T}} \end{aligned}$$

For T = 1 s:

$$F(z) = \frac{0.4323}{z - 0.1353}$$

Remark

The output $y = x_2$ is the velocity, so the integrator pole at $z = 1$ cancels. The result matches the direct ZOH transform of $1/(s+2)$: $(1-z^{-1})\mathcal{Z}\left[\frac{1}{s(s+2)}\right] = \frac{0.5(1-e^{-2T})}{z-e^{-2T}}$.

Answer: $F(z) = \dfrac{0.5(1-e^{-2T})}{z-e^{-2T}}$, which is $\dfrac{0.4323}{z-0.1353}$ for T = 1 s.