---
title: "Advantages of closed loop with mathematical support"
subject: "Control Systems"
chapter: 1
key: "semester_3/control_systems:cl-advantages-math"
source: https://www.bctengineer.com/q/semester_3/control_systems/cl-advantages-math
site: BCT Engineering
---

# Advantages of closed loop with mathematical support

Control Systems · Chapter 1: Control System Background

- Asked in: 2079 Bhadra, 6 marks
- Video: [Closed Loop System Explained: Block Diagram, Applications, Pros, and Cons](https://www.youtube.com/watch?v=RKajJKFEooE) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/cl-advantages-math
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-1

## Question

Mention any two advantages of closed loop system in comparison to open loop system. You must justify them with any mathematical support.

## Short answer

Closed-loop reduces sensitivity to parameter changes: sensitivity S = 1/(1+GH) vs. 1 for open loop. It also speeds up response: a first-order plant's time constant becomes τ/(1+K) instead of τ, making it (1+K) times faster.

## Answer

Two important advantages of a closed-loop (feedback) system over an open-loop system are (1) reduced sensitivity to parameter variations and (2) faster response. (Better disturbance rejection is a third.)

#### Advantage 1: Reduced sensitivity to parameter changes
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%.)

#### Advantage 2: Faster response (smaller time constant)
Take a first-order plant $G(s) = \dfrac{K}{\tau s + 1}$.

**Open loop:** for a unit step input

$$
c(t) = K\left(1 - e^{-t/\tau}\right), \qquad \text{time constant} = \tau
$$

**Closed loop (unity feedback):**

$$
\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{K}{\tau s + 1 + K} = \frac{\dfrac{K}{1+K}}{\dfrac{\tau}{1+K}s + 1}
$$

$$
c(t) = \frac{K}{1+K}\left(1 - e^{-t(1+K)/\tau}\right), \qquad \tau_{cl} = \frac{\tau}{1+K}
$$

The time constant becomes $\tau/(1+K)$, so the system responds $(1+K)$ times faster. The loss of gain ($K \to K/(1+K)$) is easily recovered with a preamplifier.

Example: $\tau = 1$ s, $K = 9$. Open loop: $\tau = 1$ s, settling time $4\tau = 4$ s. Closed loop: $\tau_{cl} = 1/10 = 0.1$ s, settling time $= 0.4$ s, i.e. ten times faster.

## Check yourself

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

1. The closed-loop sensitivity of T = G/(1+GH) to G is
   A. 1/(1+GH)
   B. 1+GH
   C. G/(1+GH)^2
   D. 1
   Answer: A. S = (dT/dG)(G/T) = 1/(1+GH). (Key point: Sensitivity formula)

2. For K/(tau s+1) with unity feedback, how much faster is the closed loop?
   A. K times
   B. (1+K) times
   C. tau times
   D. Not faster
   Answer: B. The time constant drops from tau to tau/(1+K). (Key point: Speed advantage)

3. With tau = 1 s and K = 9, the closed-loop settling time (4 time constants) is
   A. 4 s
   B. 0.4 s
   C. 0.9 s
   D. 10 s
   Answer: B. tau_cl = 1/10 = 0.1 s, so settling = 0.4 s. (Key point: Example numbers)

4. How is the closed-loop gain loss K to K/(1+K) overcome?
   A. Removing feedback
   B. Adding a preamplifier in the forward path
   C. Increasing tau
   D. Lowering K
   Answer: B. An amplifier restores the lost gain. (Key point: Gain loss)

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