Pulse transfer function with first order hold

There is a sampler between $1/(s+3)$ and the FOH, so the system splits into two cascaded pulse transfer functions:

$$\frac{Y(z)}{X(z)} = G_1(z)\,G_2(z), \qquad G_1(s) = \frac{1}{s+3}, \quad G_2(s) = G_{h1}(s)\frac{1}{s+2}$$

Here $Y(z)$ is the output at the sampling instants (a fictitious output sampler).

Step 1: $G_1(z)$

$$G_1(z) = \mathcal{Z}\left[\frac{1}{s+3}\right] = \frac{z}{z-e^{-3T}} = \frac{z}{z-0.0498}\quad(T = 1)$$

Step 2: $G_2(z)$, FOH + plant

$$\begin{aligned} G_2(z) &= \mathcal{Z}\left[\frac{Ts+1}{T}\left(\frac{1-e^{-Ts}}{s}\right)^2\frac{1}{s+2}\right] = (1-z^{-1})^2\,\mathcal{Z}\left[\frac{s+1}{s^2(s+2)}\right]\quad(T=1) \end{aligned}$$

Partial fractions:

$$\frac{s+1}{s^2(s+2)} = \frac{1/4}{s} + \frac{1/2}{s^2} - \frac{1/4}{s+2}$$ $$\mathcal{Z}\left[\frac{s+1}{s^2(s+2)}\right] = \frac{0.25z}{z-1} + \frac{0.5z}{(z-1)^2} - \frac{0.25z}{z-e^{-2}}$$

Multiply by $(1-z^{-1})^2 = (z-1)^2/z^2$:

$$\begin{aligned} G_2(z) &= \frac{1}{z}\left[0.25(z-1) + 0.5 - \frac{0.25(z-1)^2}{z-e^{-2}}\right] \\ &= \frac{(3-e^{-2})z - (1+e^{-2})}{4z(z-e^{-2})} \\ &= \frac{0.7162z - 0.2838}{z(z-0.1353)} \end{aligned}$$

Check: at $z = 1$, $G_2(1) = 0.4324/0.8647 = 0.5$, which equals the DC gain of $1/(s+2)$. ✔

Step 3: Overall pulse transfer function

$$\frac{Y(z)}{X(z)} = \frac{z}{z-0.0498}\cdot\frac{0.7162z - 0.2838}{z(z-0.1353)} = \frac{0.7162z - 0.2838}{(z-0.0498)(z-0.1353)}$$

Expanding the denominator:

$$\frac{Y(z)}{X(z)} = \frac{0.7162z - 0.2838}{z^2 - 0.1851z + 0.00674} = \frac{0.7162z^{-1} - 0.2838z^{-2}}{1 - 0.1851z^{-1} + 0.00674z^{-2}}$$

Answer: $\dfrac{Y(z)}{X(z)} = \dfrac{0.7162z - 0.2838}{(z-0.0498)(z-0.1353)}$