Closed-loop pulse TF and continuous output C(s)

The error is sampled, and the output of $G_1$ is sampled again before $G_2$. The feedback $H$ acts on the continuous output, so $G_2$ and $H$ are not separated by a sampler.

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

Step 1: Signal equations

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

Step 2: Sample the error equation

$$\begin{aligned} E^*(s) &= R^*(s) - (G_2H)^*(s)\,U^*(s) = R^*(s) - (G_2H)^*(s)\,G_1^*(s)E^*(s) \\ E^*(s) &= \frac{R^*(s)}{1 + G_1^*(s)(G_2H)^*(s)} \end{aligned}$$

Step 3: Output

$$C^*(s) = G_2^*(s)U^*(s) = G_2^*(s)G_1^*(s)E^*(s)$$

Closed-loop pulse transfer function

$$\frac{C^*(s)}{R^*(s)} = \frac{G_1^*(s)G_2^*(s)}{1 + G_1^*(s)(G_2H)^*(s)} \quad\Longleftrightarrow\quad \frac{C(z)}{R(z)} = \frac{G_1(z)G_2(z)}{1 + G_1(z)\,G_2H(z)}$$

with $G_2H(z) = \mathcal{Z}[G_2(s)H(s)]$.

Continuous-time output $C(s)$

The output is not sampled inside the loop. $C(s)$ is $G_2(s)$ driven by the sampled signal $U^*(s)$, so do not take the star of $G_2$ here:

$$\begin{aligned} C(s) &= G_2(s)\,U^*(s) = G_2(s)\,G_1^*(s)\,E^*(s) \\ C(s) &= \frac{G_2(s)\,G_1^*(s)\,R^*(s)}{1 + G_1^*(s)\,(G_2H)^*(s)} \end{aligned}$$

Answers:

• $\dfrac{C(z)}{R(z)} = \dfrac{G_1(z)G_2(z)}{1+G_1(z)G_2H(z)}$
• $C(s) = \dfrac{G_2(s)G_1^*(s)R^*(s)}{1+G_1^*(s)(G_2H)^*(s)}$

$C(s)$ gives the output between sampling instants as well. Its starred version $C^*(s)$ gives back $C(z)$.