Discrete output C(z) with feedback H(s)

There is no sampler on the error signal. The only sampler is after $G_1(s)$, so let its output be $V^*(s)$, the sampled version of $V(s) = G_1(s)E(s)$. The output sampler only reads $C(z)$; it is not inside the loop.

       E(s)  +-----+ V   _/ V*  +-----+  C(s)   _/
R(s)-->(S)-->| G1  |---o/ o--->| G2  |---+---o/ o--> C(z)
       ^ -   +-----+     T     +-----+   |    T
       |            +-----+              |
       +------------|  H  |<-------------+
            B(s)    +-----+

Step 1: Write the signal equations

$$\begin{aligned} C(s) &= G_2(s)V^*(s) \\ E(s) &= R(s) - H(s)C(s) = R(s) - G_2(s)H(s)V^*(s) \\ V(s) &= G_1(s)E(s) = G_1(s)R(s) - G_1(s)G_2(s)H(s)V^*(s) \end{aligned}$$

Step 2: Take starred (sampled) transforms

Sampling both sides, a starred quantity can be taken outside: $[G_1G_2HV^*]^* = (G_1G_2H)^*V^*$. This gives

$$\begin{aligned} V^*(s) &= (G_1R)^*(s) - (G_1G_2H)^*(s)V^*(s) \\ V^*(s) &= \frac{(G_1R)^*(s)}{1 + (G_1G_2H)^*(s)} \end{aligned}$$

Step 3: Output

$$C^*(s) = G_2^*(s)V^*(s)$$

In z-notation:

$$C(z) = \frac{G_2(z)\,G_1R(z)}{1 + G_1G_2H(z)}$$

where

• $G_2(z) = \mathcal{Z}[G_2(s)]$
• $G_1R(z) = \mathcal{Z}[G_1(s)R(s)]$
• $G_1G_2H(z) = \mathcal{Z}[G_1(s)G_2(s)H(s)]$

Note: The input $R(s)$ reaches the first sampler only after passing through $G_1(s)$, so it cannot be separated from $G_1$. You can find the output $C(z)$ for a given input, but no closed-loop pulse transfer function $C(z)/R(z)$ exists for this system.