---
title: "Discrete output C(z) with feedback H(s)"
subject: "Digital Control System"
chapter: 3
key: "semester_6/digital_control_system:discrete-output-feedback-h"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/discrete-output-feedback-h
site: BCT Engineering
---

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

Digital Control System · Chapter 3: Analysis of Discrete Time Control System

- Asked in: 2075 Bhadra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/discrete-output-feedback-h
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3

## Question

Find the discrete time output C(z) of the following closed loop system. [Figure: R(s) → summing junction (+) → E(s) → G₁(s) → sampler → G₂(s) → C(s) → sampler → C(z); C(s) is fed back through H(s) as B(s) to the negative input of the summing junction]

## Short answer

C(z) = G₂(z) G₁R(z) / [1 + G₁G₂H(z)], where G₁R(z)=Z[G₁(s)R(s)] and G₁G₂H(z)=Z[G₁(s)G₂(s)H(s)]. No closed-loop transfer function C(z)/R(z) exists because the input passes through G₁ before the sampler.

## Answer

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.

```text
       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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Where is the only sampler inside the loop?
   A. On the error signal E(s)
   B. After G1(s), before G2(s)
   C. On the feedback path H(s)
   D. Before the summing junction
   Answer: B. The sampler lies between G1 and G2. The output sampler only reads C(z) and is not in the loop. (Key point: Sampler location)

2. What is C(z) for this system?
   A. G1(z)G2(z) R(z) / [1 + G1(z)G2(z)H(z)]
   B. G2(z) G1R(z) / [1 + G1G2H(z)]
   C. G1G2(z) R(z) / [1 + H(z)]
   D. G2(z) R(z) / [1 + G1G2H(z)]
   Answer: B. From V* = (G1R)* / (1 + (G1G2H)*) and C* = G2* V*, we get C(z) = G2(z) G1R(z) / (1 + G1G2H(z)). (Key point: Output equation)

3. Why does no closed-loop C(z)/R(z) exist here?
   A. Because the system is unstable
   B. Because R(s) enters before G1 and cannot be separated from it, so only G1R(z) appears
   C. Because H(s) is not sampled
   D. Because C(s) is not sampled
   Answer: B. R(s) is not sampled before it reaches G1, so the product G1R(z) cannot be factored into G1(z)R(z). (Key point: No transfer function C(z)/R(z))

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