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

# Convolution theorem and inverse Laplace

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2071 Shrawan, 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-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Using the convolution theorem, find the inverse Laplace transform of 3s/((s² + 4)(s² + 1)).

## Short answer

Using convolution theorem with F(s)=s/(s^2+4), G(s)=1/(s^2+1), f=cos2t, g=sin t. The integral 3 ∫_0^t cos2u sin(t-u) du simplifies to cos t - cos2t. Thus inverse is cos t - cos2t.

## Answer

**Convolution theorem:** L⁻¹[F(s)G(s)] = ∫₀ᵗ f(u) g(t − u) du, where f = L⁻¹[F], g = L⁻¹[G].

Write 3s/((s² + 4)(s² + 1)) = 3 · [s/(s² + 4)] · [1/(s² + 1)].
So f(t) = cos 2t and g(t) = sin t.

```text
L⁻¹ = 3 ∫₀ᵗ cos 2u sin(t − u) du
```

Use cos A sin B = ½[sin(A + B) − sin(A − B)] with A = 2u, B = t − u:

```text
cos 2u sin(t − u) = ½ [ sin(t + u) + sin(t − 3u) ]
```

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

```text
L⁻¹ = (3/2) [ (cos t − cos 2t) − (cos t − cos 2t)/3 ]
    = (3/2)(2/3)(cos t − cos 2t)
```

**Answer: cos t − cos 2t**

(Check by partial fractions: 3s/((s²+1)(s²+4)) = s/(s²+1) − s/(s²+4).)

## Check yourself

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

1. L^-1[F G] =
   A. f(t) g(t)
   B. integral of f(u) g(u) du
   C. f'(t) g(t)
   D. integral from 0 to t of f(u) g(t-u) du
   Answer: D. Convolution. (Key point: Theorem)

2. 3s/((s^2+4)(s^2+1)) is split with
   A. f = cos 2t, g = sin t
   B. f = sin 2t, g = cos t
   C. f = cos 2t, g = cos t
   D. f = sin 2t, g = sin t
   Answer: A. s/(s^2+4) -> cos 2t, 1/(s^2+1) -> sin t. (Key point: Choice of f, g)

3. The inverse is
   A. cos 2t - cos t
   B. cos t - cos 2t
   C. (cos t - cos 2t)/3
   D. sin t - sin 2t
   Answer: B. Matches partial fractions s/(s^2+1) - s/(s^2+4). (Key point: Result)

## Close variants set in other exams

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