Discrete state equations and pulse TF (T = 1)

With a zero-order hold, the discrete model is $\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k) + \mathbf{H}u(k)$, $y(k) = \mathbf{C}\mathbf{x}(k)$, where $\mathbf{G} = e^{\mathbf{A}T}$ and $\mathbf{H} = \left(\int_0^T e^{\mathbf{A}\lambda}d\lambda\right)\mathbf{B}$.

Step 1: State transition matrix of the continuous system

$$(s\mathbf{I}-\mathbf{A})^{-1} = \begin{bmatrix} s & -1 \\ 0 & s+2 \end{bmatrix}^{-1} = \begin{bmatrix} \dfrac{1}{s} & \dfrac{1}{s(s+2)} \\[2mm] 0 & \dfrac{1}{s+2} \end{bmatrix}$$ $$e^{\mathbf{A}t} = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2t}) \\ 0 & e^{-2t} \end{bmatrix}$$

Step 2: G and H for T = 1 s

$e^{-2} = 0.1353$:

$$\mathbf{G} = e^{\mathbf{A}} = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2}) \\ 0 & e^{-2} \end{bmatrix} = \begin{bmatrix} 1 & 0.4323 \\ 0 & 0.1353 \end{bmatrix}$$ $$\begin{aligned} \mathbf{H} &= \int_0^1\begin{bmatrix} \tfrac{1}{2}(1-e^{-2\lambda}) \\ e^{-2\lambda} \end{bmatrix}d\lambda = \begin{bmatrix} \tfrac{1}{2}\left[1 - \tfrac{1}{2}(1-e^{-2})\right] \\ \tfrac{1}{2}(1-e^{-2}) \end{bmatrix} = \begin{bmatrix} 0.2838 \\ 0.4323 \end{bmatrix} \end{aligned}$$

Discrete state and output equations

$$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 1 & 0.4323 \\ 0 & 0.1353 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0.2838 \\ 0.4323 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}$$

Step 3: Pulse transfer function

$$(z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{(z-1)(z-0.1353)}\begin{bmatrix} z-0.1353 & 0.4323 \\ 0 & z-1 \end{bmatrix}$$ $$\begin{aligned} F(z) &= \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{0.2838(z-0.1353) + 0.4323\times0.4323}{(z-1)(z-0.1353)}\\ &= \frac{0.2838z + 0.1485}{(z-1)(z-0.1353)} = \frac{0.2838z + 0.1485}{z^2 - 1.1353z + 0.1353} \end{aligned}$$

Check by direct transformation

$$(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+2)}\right] = \frac{(2T-1+e^{-2T})z + (1-e^{-2T}-2Te^{-2T})}{4(z-1)(z-e^{-2T})} = \frac{1.1353z + 0.5940}{4(z-1)(z-0.1353)}$$

This is the same result.

Answer: $F(z) = \dfrac{0.2838z + 0.1485}{(z-1)(z-0.1353)} = \dfrac{0.2838z^{-1} + 0.1485z^{-2}}{1 - 1.1353z^{-1} + 0.1353z^{-2}}$