Pulse TF for two sampler configurations

The answer depends on whether a sampler separates $G_1$ and $G_2$. With a sampler between them, the two pulse transfer functions multiply. Without one, the two s-domain transfer functions must be multiplied first and then z-transformed.

(a) Sampler between $G_1$ and $G_2$

X(s)  _/  X*  +----+ U(s)  _/  U*  +----+ Y(s)  _/  Y*
 ---o/ o---->| G1 |------o/ o---->| G2 |------o/ o--->
      T      +----+        T      +----+        T

• $U(s) = G_1(s)X^*(s)$, so $U^*(s) = G_1^*(s)X^*(s)$, i.e. $U(z) = G_1(z)X(z)$.
• $Y(s) = G_2(s)U^*(s)$, so $Y(z) = G_2(z)U(z)$.

$$\frac{Y(z)}{X(z)} = G_1(z)\,G_2(z), \qquad G_1(z) = \mathcal{Z}[G_1(s)],\; G_2(z) = \mathcal{Z}[G_2(s)]$$

(b) No sampler between $G_1$ and $G_2$

X(s)  _/  X*  +----+ U(s)  +----+ Y(s)  _/  Y*
 ---o/ o---->| G1 |------->| G2 |------o/ o--->
      T      +----+        +----+        T

• $Y(s) = G_1(s)G_2(s)X^*(s)$, so $Y^*(s) = [G_1G_2]^*(s)\,X^*(s)$.

$$\frac{Y(z)}{X(z)} = G_1G_2(z) = \mathcal{Z}\big[G_1(s)G_2(s)\big]$$

In general, $G_1G_2(z) \neq G_1(z)G_2(z)$.

Example

Take $G_1(s) = \dfrac{1}{s}$ and $G_2(s) = \dfrac{a}{s+a}$.

Case (a):

$$\frac{Y(z)}{X(z)} = \frac{z}{z-1}\cdot\frac{az}{z-e^{-aT}} = \frac{az^2}{(z-1)(z-e^{-aT})}$$

Case (b):

$$\begin{aligned} \frac{a}{s(s+a)} &= \frac{1}{s} - \frac{1}{s+a} \\ \frac{Y(z)}{X(z)} &= \frac{z}{z-1} - \frac{z}{z-e^{-aT}} = \frac{(1-e^{-aT})z}{(z-1)(z-e^{-aT})} \end{aligned}$$

The two results are different. You cannot move or remove a sampler without changing the system.