---
title: "Closed loop reduces internal disturbances"
subject: "Control Systems"
chapter: 1
key: "semester_3/control_systems:cl-reduces-disturbance"
source: https://www.bctengineer.com/q/semester_3/control_systems/cl-reduces-disturbance
site: BCT Engineering
---

# Closed loop reduces internal disturbances

Control Systems · Chapter 1: Control System Background

- Asked in: 2081 Bhadra, 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/cl-reduces-disturbance
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-1

## Question

Justify that the closed loop system reduces effect of internal disturbances in the system.

## Short answer

Closed-loop reduces internal disturbances because feedback divides the effect of parameter variations and internal noise by (1+GH). For a disturbance D entering between G1 and G2, the output becomes G2D/(1+G1G2H), which is much smaller than the open-loop output G2D when loop gain is high.

## Answer

**Internal disturbances** are changes inside the system itself: variation of plant parameters (ageing, temperature, wear, supply changes) and internal noise. Feedback reduces their effect because the output is measured and the error is corrected.

#### 1. Parameter variation
The sensitivity of the overall gain $T$ to a change in $G$ is

$$
S_G^T = \frac{\partial T/T}{\partial G/G} = \frac{\partial T}{\partial G}\cdot\frac{G}{T}
$$

- Open loop: $T = G$, so $S_G^T = 1$ (a 10% change in $G$ gives a 10% change in output).
- Closed loop: $T = \dfrac{G}{1+GH}$, so $\dfrac{\partial T}{\partial G} = \dfrac{1}{(1+GH)^2}$ and

$$
S_G^T = \frac{1}{(1+GH)^2}\cdot\frac{G(1+GH)}{G} = \frac{1}{1+GH}
$$

Since $|1+GH| \gg 1$, the effect of a change in $G$ is reduced by the factor $(1+GH)$. (With $GH = 99$, a 10% change in $G$ changes the output by only 0.1%.)

#### 2. Disturbance generated inside the plant
Let the disturbance $D(s)$ enter between two forward blocks $G_1$ (controller) and $G_2$ (plant).

- Open loop: $C_D(s) = G_2(s)D(s)$.
- Closed loop (feedback $H$), with $R = 0$ (superposition):

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

The disturbance output is divided by $(1+G_1G_2H)$. If $|G_1G_2H| \gg 1$, $C_D \approx \dfrac{D}{G_1H}$, which is very small for large $G_1$.

#### Example
A DC motor's armature resistance rises as it heats. In open loop the speed falls in proportion. With tacho feedback and loop gain $K = 49$, the change in speed is only $1/(1+49) = 2\%$ of the open-loop change.

Hence the closed-loop system is far less affected by internal disturbances, as long as the loop gain is high and the sensor itself is accurate (the sensitivity to $H$ is $S_H^T = -GH/(1+GH) \approx -1$).

## Check yourself

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

1. In open loop T = G, so the sensitivity S of T to G is
   A. 0
   B. 1
   C. 1/(1+GH)
   D. GH
   Answer: B. A 10% change in G gives a 10% change in output, so S = 1. (Key point: Open-loop sensitivity)

2. With GH = 99, a 10% change in G changes the closed-loop output by about
   A. 10%
   B. 1%
   C. 0.1%
   D. 99%
   Answer: C. Sensitivity is 1/(1+99) = 0.01, so 10% x 0.01 = 0.1%. (Key point: Closed-loop sensitivity)

3. With disturbance D between G1 and G2 and R = 0, the closed-loop output due to D is
   A. G2 D
   B. G2 D/(1+G1G2H)
   C. G1 D/(1+G1G2H)
   D. D(1+G1G2H)
   Answer: B. The open-loop effect G2 D is divided by (1+G1G2H). (Key point: Disturbance inside the plant)

4. What limitation applies to this disturbance reduction?
   A. It works only when the loop gain is high and the sensor is accurate
   B. It works only for open loop
   C. It requires H = 0
   D. It needs the loop gain to be less than 1
   Answer: A. Reduction requires |1+GH| large, and sensor errors (S_H is about -1) pass directly to the output. (Key point: Condition)

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