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

# Convolution theorem and inverse Laplace

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2074 Asoj, 1+4 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-two
- 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 + 4)(s² + 9)).

## Short answer

Convolution theorem: L⁻¹[F(s)G(s)] = ∫_0^t f(u)g(t-u) du. For s/(( s²+4)(s²+9)), take F(s)=s/(s²+4) (f=cos2t), G(s)=1/(s²+9) (g=(1/3)sin3t). Convolution yields (cos2t - cos3t)/5.

## Answer

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

```text
L⁻¹[F(s) G(s)] = f * g = ∫₀ᵗ f(u) g(t − u) du
```

That is, the inverse transform of a product is the convolution of the inverse transforms.

#### Inverse of s/((s² + 4)(s² + 9))
Take F(s) = s/(s² + 4), G(s) = 1/(s² + 9). Then f(t) = cos 2t, g(t) = (1/3) sin 3t.

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

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

```text
cos 2u sin(3t − 3u) = ½ [ sin(3t − u) + sin(3t − 5u) ]
```

```text
∫₀ᵗ sin(3t − u) du  = [cos(3t − u)]₀ᵗ = cos 2t − cos 3t
∫₀ᵗ sin(3t − 5u) du = [cos(3t − 5u)/5]₀ᵗ = (cos 2t − cos 3t)/5
```

```text
L⁻¹ = (1/3)(1/2) (cos 2t − cos 3t)(1 + 1/5) = (1/5)(cos 2t − cos 3t)
```

**Answer: (cos 2t − cos 3t)/5**

(Check by partial fractions: s/((s²+4)(s²+9)) = (1/5)[s/(s²+4) − s/(s²+9)].)

## Check yourself

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

1. What is L^-1{FG}?
   A. f(t) g(t)
   B. integral from 0 to t 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. The convolution. (Key point: Statement)

2. F = s/(s^2+4), G = 1/(s^2+9). What are f, g?
   A. f = cos 2t, g = (1/3) sin 3t
   B. f = sin 2t, g = cos 3t
   C. f = cos 2t, g = sin 3t
   D. f = (1/2) sin 2t, g = cos 3t
   Answer: A. Standard forms. (Key point: Choice)

3. cos 2u sin(3t-3u) equals?
   A. (1/2)[sin(3t-u) - sin(3t-5u)]
   B. (1/2)[sin(3t-u) + sin(3t-5u)]
   C. (1/2)[cos(3t-u) + cos(3t-5u)]
   D. (1/2)[sin(3t-5u) - sin(3t-u)]
   Answer: B. cosA sinB = (1/2)[sin(A+B) - sin(A-B)]. (Key point: Product to sum)

4. What is the inverse?
   A. (cos 2t - cos 3t)/3
   B. (cos 3t - cos 2t)/5
   C. (cos 2t - cos 3t)/5
   D. (cos 2t + cos 3t)/5
   Answer: C. Matches the partial fractions check. (Key point: Final result)

## Close variants set in other exams

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