---
title: "Impulse and ramp response from first order step response"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:first-order-impulse-ramp"
source: https://www.bctengineer.com/q/semester_3/control_systems/first-order-impulse-ramp
site: BCT Engineering
---

# Impulse and ramp response from first order step response

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2074 Chaitra, 4 marks
- Video: [Introduction to First Order Systems in Control: steady state response under step, impulse and ramp](https://www.youtube.com/watch?v=VLJDxZoO0_E) — Mechanical Engineering with Dr. Sanei
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/first-order-impulse-ramp
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Suppose that the step response of a first order system is c(t) = 5(1 − e^(−t/5)). What are impulse and ramp responses?

## Short answer

Impulse response is the derivative of step response: e^{-t/5}. Ramp response is the integral: 5t - 25 + 25e^{-t/5} for t ≥ 0.

## Answer

For a linear time-invariant system, the impulse response is the **derivative** of the step response, and the ramp response is the **integral** of the step response.

#### Transfer function
$$
C(s) = \mathcal{L}\{5(1 - e^{-t/5})\} = 5\left[\frac{1}{s} - \frac{1}{s + 0.2}\right] = \frac{1}{s(s+0.2)}
$$
Since $R(s) = 1/s$:
$$
G(s) = \frac{1}{s + 0.2} = \frac{5}{5s + 1}
$$
This is a first-order system with gain 5 and time constant $T = 5$ s.

#### Impulse response
$$
c_\delta(t) = \frac{d}{dt}\left[5(1 - e^{-t/5})\right] = 5\cdot\frac{1}{5}e^{-t/5} = e^{-t/5},\quad t \ge 0
$$
Check: $\mathcal{L}^{-1}\{1/(s+0.2)\} = e^{-0.2t}$.

#### Ramp response (unit ramp)
$$
\begin{aligned}
c_r(t) &= \int_0^t 5(1 - e^{-\tau/5})\,d\tau \\
&= 5\left[\tau + 5e^{-\tau/5}\right]_0^t \\
&= 5t + 25e^{-t/5} - 25 \\
&= 5t - 25 + 25e^{-t/5},\quad t \ge 0
\end{aligned}
$$
Check: $\dfrac{1}{s^2(s+0.2)} = \dfrac{5}{s^2} - \dfrac{25}{s} + \dfrac{25}{s+0.2}$, which gives the same result.

**Answer:** impulse response $= e^{-t/5}$; ramp response $= 5t - 25 + 25e^{-t/5}$ for $t \ge 0$.

## Check yourself

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

1. How is the impulse response related to the step response of an LTI system?
   A. It is the integral of the step response
   B. It is the derivative of the step response
   C. It is the square of the step response
   D. It equals the step response
   Answer: B. Impulse is the derivative of the step input, so the response is the derivative. (Key point: Impulse from step)

2. What is G(s) for c(t) = 5(1 - e^(-t/5))?
   A. 5/(s + 5)
   B. 1/(s + 5)
   C. 1/(s + 0.2) = 5/(5s + 1)
   D. 5/(s + 0.2)
   Answer: C. C(s) = 1/[s(s + 0.2)] and R(s) = 1/s, so G = 1/(s + 0.2). (Key point: Transfer function)

3. What is the impulse response?
   A. e^(-t/5)
   B. 5e^(-t/5)
   C. 5 - 5e^(-t/5)
   D. e^(-5t)
   Answer: A. d/dt of 5(1 - e^(-t/5)) = e^(-t/5). (Key point: Impulse response)

4. What is the unit ramp response?
   A. 5t + 25 - 25e^(-t/5)
   B. 5t - 5e^(-t/5)
   C. t - 5 + 5e^(-t/5)
   D. 5t - 25 + 25e^(-t/5)
   Answer: D. Integrating the step response gives 5t + 25e^(-t/5) - 25. (Key point: Ramp response)

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