---
title: "Suppressing disturbance by increasing G(s) or H(s)"
subject: "Control Systems"
chapter: 1
key: "semester_3/control_systems:disturbance-suppress-gain"
source: https://www.bctengineer.com/q/semester_3/control_systems/disturbance-suppress-gain
site: BCT Engineering
---

# Suppressing disturbance by increasing G(s) or H(s)

Control Systems · Chapter 1: Control System Background

- Asked in: 2071 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/disturbance-suppress-gain
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-1

## Question

Effect of disturbance in case of feedback control system can be suppressed by increasing the gain G(s) and/or H(s). Justify.

## Short answer

Disturbance output is C_D = [G2/(1+G1G2H)]D; for large loop gain it approximates D/(G1H). Increasing G1 and/or H reduces this, suppressing disturbance, provided the system remains stable.

## Answer

**The statement is justified**, provided that the gain increased is the one in the loop **before** the point where the disturbance enters, and that the system stays stable.

Let the disturbance $D(s)$ enter between $G_1(s)$ (controller) and $G_2(s)$ (plant), with feedback $H(s)$.

```text
 R  +      +----+  +     +----+  C
 -->(Σ)--->| G1 |-->(Σ)->| G2 |--+-->
     ^ -   +----+    ^ D +----+  |
     |       +----+              |
     +-------| H  |<-------------+
             +----+
```

With $R = 0$ (superposition), the output due to the disturbance is

$$
C_D(s) = \frac{G_2(s)}{1+G_1(s)G_2(s)H(s)}\,D(s)
$$

For large loop gain, $|G_1G_2H| \gg 1$:

$$
C_D(s) \approx \frac{G_2}{G_1G_2H}D(s) = \frac{D(s)}{G_1(s)H(s)}
$$

So increasing $G_1$ and/or $H$ makes $C_D$ smaller, and the disturbance is suppressed. In open loop, $C_D = G_2D$, with no reduction at all.

Output due to the reference: $C_R = \dfrac{G_1G_2}{1+G_1G_2H}R \approx \dfrac{R}{H}$, which is almost independent of the gains, so tracking is not harmed.

#### Limits
- Raising $G_2$ alone does not help much: the result tends to $D/(G_1H)$, which does not depend on $G_2$.
- Raising $H$ also lowers the closed-loop gain ($\approx 1/H$), so it must be chosen with the required output level in mind.
- Very high gain can reduce damping or cause instability in higher-order systems, and it amplifies sensor noise.

## Check yourself

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

1. With R = 0 the output due to D is
   A. G2 D/(1+G1G2H)
   B. G1 D/(1+G1G2H)
   C. G1G2 D
   D. D/(G2 H)
   Answer: A. Superposition gives C_D = G2 D/(1+G1G2H). (Key point: Disturbance output formula)

2. For large loop gain, C_D is approximately
   A. D G2
   B. D/(G1 H)
   C. D H
   D. D/G2
   Answer: B. G2/(G1G2H) = 1/(G1H). (Key point: Large-gain approximation)

3. Raising G2 alone
   A. Does not help much since C_D tends to D/(G1H), independent of G2
   B. Eliminates the disturbance
   C. Makes it unstable always
   D. Doubles the disturbance
   Answer: A. The limit does not depend on G2. (Key point: Which gain helps)

4. Increasing H also
   A. Raises closed-loop gain
   B. Lowers closed-loop gain (about 1/H)
   C. Has no effect
   D. Removes noise
   Answer: B. C/R is approximately 1/H, so H must suit the needed output level. (Key point: Limits of raising H)

## Related in Control Systems

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- [True/false: higher forward gain reduces disturbance](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-disturbance-forward-gain) — Control Systems, chapter 1
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