Closed-loop pulse TF with two samplers and H(s)

There is a sampler on the error (before $G_1$) and another between $G_1$ and $G_2$. Because $G_2$ and $H$ are not separated by a sampler, they appear together as $G_2H(z)$.

      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

$$\frac{C(z)}{R(z)} = \frac{G_1(z)\,G_2(z)}{1 + G_1(z)\,G_2H(z)}$$

where

• $G_1(z) = \mathcal{Z}[G_1(s)]$
• $G_2(z) = \mathcal{Z}[G_2(s)]$
• $G_2H(z) = \mathcal{Z}[G_2(s)H(s)]$, which is in general not equal to $G_2(z)H(z)$

Special case: with unity feedback ($H = 1$), this becomes $\dfrac{G_1(z)G_2(z)}{1+G_1(z)G_2(z)}$.