---
title: "Closed vs open loop step response analytically"
subject: "Control Systems"
chapter: 1
key: "semester_3/control_systems:cl-vs-ol-step-response"
source: https://www.bctengineer.com/q/semester_3/control_systems/cl-vs-ol-step-response
site: BCT Engineering
---

# Closed vs open loop step response analytically

Control Systems · Chapter 1: Control System Background

- Asked in: 2074 Asoj, 6 marks; 2081 Baisakh, 3 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/cl-vs-ol-step-response
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-1

## Question

How would a closed loop system differ from open loop one on its step response? Explain with the required mathematical expression.

## Short answer

For a first-order plant G(s)=K/(tau s+1), open-loop step response is c(t)=K(1-e^(-t/tau)); closed loop with unity feedback yields c(t)=[K/(1+K)](1-e^(-t(1+K)/tau)), so the time constant decreases to tau/(1+K) and steady-state error is 1/(1+K). In second-order systems, feedback provides a bounded final value, and increasing gain raises natural frequency but reduces damping.

## Answer

In open loop the step response depends only on the plant; in closed loop the feedback changes the gain, speed, steady-state error and (for second-order systems) the damping.

#### First-order plant
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):**

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

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

So in closed loop: the time constant falls from $\tau$ to $\tau/(1+K)$ (faster response), and the steady-state gain falls from $K$ to $K/(1+K)$. The steady-state error for unit step is $e_{ss} = \dfrac{1}{1+K}$, which becomes small for large $K$.

#### Second-order case
For $G(s) = \dfrac{K}{s(s+a)}$, the open-loop step response grows without limit (a ramp after a transient), because of the pole at $s = 0$. With unity feedback:

$$
\frac{C(s)}{R(s)} = \frac{K}{s^2 + as + K}, \quad \omega_n = \sqrt{K}, \quad \zeta = \frac{a}{2\sqrt{K}}
$$

The output now settles at 1. Increasing $K$ raises $\omega_n$ (faster) and lowers $\zeta$ (more overshoot $M_p = e^{-\zeta\pi/\sqrt{1-\zeta^2}}$).

#### Summary
| Feature | Open loop | Closed loop |
|---|---|---|
| Time constant | $\tau$ | $\tau/(1+K)$ |
| Steady-state gain | $K$ | $K/(1+K)$ |
| Final value with plant change | changes fully | almost unchanged |
| Overshoot | none for 1st order | may appear if gain high |
| Effect of disturbance | full | reduced by $1+GH$ |

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. For G(s) = K/(tau s + 1) with unity feedback, the closed-loop step response has time constant
   A. tau
   B. tau/(1+K)
   C. tau(1+K)
   D. tau/K
   Answer: B. C/R = [K/(1+K)]/[(tau/(1+K))s + 1]. (Key point: First-order closed-loop time constant)

2. The closed-loop unit-step response of the first-order plant settles at
   A. K
   B. 1
   C. K/(1+K)
   D. 1/K
   Answer: C. The steady-state gain is K/(1+K), leaving a steady-state error of 1/(1+K). (Key point: Final value)

3. For G(s) = K/[s(s+a)] with unity feedback, how do wn and zeta depend on K?
   A. wn = sqrt(K), zeta = a/(2 sqrt(K))
   B. wn = K, zeta = a/K
   C. wn = sqrt(K), zeta = sqrt(K)/a
   D. wn = a, zeta = K
   Answer: A. The characteristic equation s^2 + as + K = 0 gives wn = sqrt(K), zeta = a/(2 sqrt(K)). (Key point: Second-order parameters)

4. What happens to the open-loop step response of K/[s(s+a)] compared with closed loop?
   A. Both settle at 1
   B. Open loop grows without limit; closed loop settles at 1
   C. Open loop settles at 1; closed loop grows
   D. Both oscillate forever
   Answer: B. The pole at s = 0 makes the open-loop response a ramp, while feedback makes the output settle at 1. (Key point: Open loop with pole at origin)

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