---
title: "Convolution theorem and inverse Laplace"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:convolution-theorem-and-inverse-laplace"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/convolution-theorem-and-inverse-laplace
site: BCT Engineering
---

# Convolution theorem and inverse Laplace

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2078 Bhadra, 5 marks
- Video: [16 Inverse Laplace Transforms | Part 1:  ILT by Convolution Theorem](https://www.youtube.com/watch?v=6QbBNn-PWw4) — The Math Virtuoso
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/convolution-theorem-and-inverse-laplace
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

State the convolution theorem for inverse Laplace transform and use it to find the inverse Laplace transform of s/((s² + 1)(s² + 4)).

## Short answer

Convolution theorem: L^{-1}{F(s)G(s)} = ∫0^t f(u)g(t-u) du. With F(s)=s/(s^2+1), G(s)=1/(s^2+4), f=cos t, g=(sin 2t)/2, integration yields (cos t - cos 2t)/3.

## Answer

#### Convolution theorem
If L⁻¹{F(s)} = f(t) and L⁻¹{G(s)} = g(t), then

L⁻¹{F(s) G(s)} = (f * g)(t) = ∫₀^t f(u) g(t − u) du.

#### Application
Take F(s) = s/(s² + 1) and G(s) = 1/(s² + 4). Then f(t) = cos t and g(t) = (sin 2t)/2.

```text
L⁻¹{s/((s² + 1)(s² + 4))} = ∫₀^t cos u · ½ sin 2(t − u) du
```

Use cos u sin(2t − 2u) = ½[sin(2t − u) + sin(2t − 3u)]:

```text
∫₀^t sin(2t − u) du  = [cos(2t − u)]₀^t = cos t − cos 2t
∫₀^t sin(2t − 3u) du = [cos(2t − 3u)/3]₀^t = (cos t − cos 2t)/3

Result = ½ · ½ · [(cos t − cos 2t) + (cos t − cos 2t)/3]
       = ¼ · (4/3)(cos t − cos 2t)
       = (1/3)(cos t − cos 2t)
```

**Answer: L⁻¹{s/((s² + 1)(s² + 4))} = (cos t − cos 2t)/3.** (Check by partial fractions: s/((s² + 1)(s² + 4)) = (1/3)[s/(s² + 1) − s/(s² + 4)].)

## Check yourself

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

1. What is L^-1{F(s)G(s)}?
   A. integral from 0 to t of f(u) g(t-u) du
   B. f(t) g(t)
   C. integral from 0 to t of f(u) g(u) du
   D. integral from 0 to infinity of f(u) g(t-u) du
   Answer: A. The inverse of a product is the convolution (f*g)(t). (Key point: Convolution theorem)

2. With G(s) = 1/(s^2+4), what is g(t)?
   A. sin 2t
   B. (sin 2t)/2
   C. (cos 2t)/2
   D. 2 sin 2t
   Answer: B. L^-1{k/(s^2+k^2)} = sin kt, so 1/(s^2+4) = (1/2)(2/(s^2+4)) gives (sin 2t)/2. (Key point: Choice of g)

3. What is the integral from 0 to t of sin(2t - u) du?
   A. cos 2t - cos t
   B. sin t - sin 2t
   C. cos t - cos 2t
   D. 1 - cos t
   Answer: C. Substituting v = 2t - u gives the integral of sin v from t to 2t = cos t - cos 2t. (Key point: One integral)

4. What is L^-1{s/((s^2+1)(s^2+4))}?
   A. (cos t + cos 2t)/3
   B. (cos t - cos 2t)/2
   C. (sin t - sin 2t)/3
   D. (cos t - cos 2t)/3
   Answer: D. Matches the partial-fraction check (1/3)[s/(s^2+1) - s/(s^2+4)]. (Key point: Final result)

## Close variants set in other exams

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