---
title: "Verify commutative law of convolution"
subject: "Signals and Systems"
chapter: 5
key: "semester_4/signals_and_systems:verify-commutative-law"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/verify-commutative-law
site: BCT Engineering
---

# Verify commutative law of convolution

Signals and Systems · Chapter 5: Continuous-Time Systems

- Asked in: 2070 Bhadra, 5 marks
- Video: [Properties of Convolution](https://www.youtube.com/watch?v=mzYbd44X8Z0) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/verify-commutative-law
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5

## Question

Let x(t) be the input to an LTI system with unit impulse response h(t), where x(t) = e^(−at)u(t), a > 0 and h(t) = u(t). Verify commutative law of LTI system.

## Short answer

Computing x*h and h*x for x(t)=e^(-at)u(t), h(t)=u(t) both give (1/a)(1-e^(-at))u(t), confirming x*h = h*x. This verifies the commutative property of convolution for LTI systems.

## Answer

The **commutative law** of convolution states $x(t) * h(t) = h(t) * x(t)$: the output of an LTI system is the same whether we convolve the input with the impulse response or the other way round.

**Given:** $x(t) = e^{-at}u(t)$, $a > 0$; $h(t) = u(t)$.

#### LHS: $y_1(t) = x(t) * h(t)$

$$
y_1(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau = \int_{-\infty}^{\infty} e^{-a\tau}u(\tau)\,u(t-\tau)\,d\tau
$$

Non-zero for $0 \le \tau \le t$ (needs $t \ge 0$):

$$
y_1(t) = \int_0^{t} e^{-a\tau}d\tau = \left[\frac{e^{-a\tau}}{-a}\right]_0^t = \frac{1}{a}\left(1 - e^{-at}\right)u(t)
$$

#### RHS: $y_2(t) = h(t) * x(t)$

$$
y_2(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau = \int_{-\infty}^{\infty} u(\tau)\,e^{-a(t-\tau)}u(t-\tau)\,d\tau
$$

Non-zero for $0 \le \tau \le t$ (needs $t \ge 0$):

$$
\begin{aligned}
y_2(t) &= e^{-at}\int_0^{t} e^{a\tau}d\tau \\
&= e^{-at}\cdot\frac{e^{at} - 1}{a} \\
&= \frac{1}{a}\left(1 - e^{-at}\right)u(t)
\end{aligned}
$$

#### Result

$$
y_1(t) = y_2(t) = \frac{1}{a}\left(1 - e^{-at}\right)u(t)
$$

so $x(t) * h(t) = h(t) * x(t)$ and the commutative law is verified.

**Meaning:** an input $e^{-at}u(t)$ applied to an integrator ($h = u(t)$) gives the same output as a unit step applied to a system with impulse response $e^{-at}u(t)$. In general the roles of input and impulse response can be interchanged.

## Check yourself

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

1. What does the commutative law of convolution state?
   A. x(t) * h(t) = h(t) * x(t)
   B. x * (h1 + h2) = x*h1 + x*h2
   C. (x*h1)*h2 = x*(h1*h2)
   D. x(t) * delta(t) = 0
   Answer: A. Convolution is commutative: input and impulse response can be interchanged. (Key point: Commutative law)

2. For x(t) = e^(-at)u(t) and h(t) = u(t), over what range of tau is x(tau)h(t - tau) non-zero?
   A. tau > t
   B. -inf < tau < inf
   C. tau < 0
   D. 0 <= tau <= t (needs t >= 0)
   Answer: D. u(tau) requires tau >= 0 and u(t - tau) requires tau <= t. (Key point: Limits of integration)

3. What is x(t) * h(t)?
   A. e^(-at) u(t)
   B. a(1 - e^(-at)) u(t)
   C. (1/a)(1 - e^(-at)) u(t)
   D. (1/a) e^(-at) u(t)
   Answer: C. The integral of e^(-a tau) from 0 to t is (1 - e^(-at))/a. (Key point: LHS result)

4. How is h(t) * x(t) evaluated, and what is the outcome?
   A. e^(-at) times the integral of e^(a tau) from 0 to t, giving the same (1/a)(1 - e^(-at)) u(t)
   B. The integral of e^(-a tau) from 0 to t, giving a different answer
   C. e^(at) times the integral of e^(-a tau), giving (1/a)(e^(at) - 1)
   D. The integral of e^(-a(t - tau)) from -inf to inf, giving 1/a
   Answer: A. e^(-at)(e^(at) - 1)/a = (1 - e^(-at))/a, equal to the LHS, so the law is verified. (Key point: RHS evaluation)

## Related in Signals and Systems

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