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

# Convolution theorem and inverse Laplace

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2079 Baishakh, 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-four
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

What do you mean by convolution of two functions f(t) and g(t)? Hence or otherwise find the inverse Laplace transform of s²/((s² + 4)(s² + 9)).

## Short answer

Convolution is (f*g)(t)=∫_0^t f(u)g(t-u)du. With F(s)=s/(s^2+4) and G(s)=s/(s^2+9) yielding cos2t and cos3t, the inverse Laplace transform is (3/5)sin3t - (2/5)sin2t.

## Answer

#### Convolution
The convolution of f(t) and g(t) (both defined for t ≥ 0) is

```text
(f * g)(t) = ∫₀ᵗ f(u) g(t − u) du
```

Convolution theorem: L[f * g] = F(s)·G(s), so L⁻¹[F(s)G(s)] = f * g. It is commutative: f * g = g * f.

#### L⁻¹[s²/((s² + 4)(s² + 9))]
Take F(s) = s/(s² + 4), G(s) = s/(s² + 9). Then f(t) = cos 2t, g(t) = cos 3t.

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

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

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

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

```text
L⁻¹ = ½ [ (sin 2t + sin 3t)/5 + sin 3t − sin 2t ]
    = ½ [ −(4/5) sin 2t + (6/5) sin 3t ]
```

**Answer: (3/5) sin 3t − (2/5) sin 2t**

(Check by partial fractions: s²/((s²+4)(s²+9)) = (9/5)/(s²+9) − (4/5)/(s²+4), whose inverse is (3/5) sin 3t − (2/5) sin 2t ✓.)

## Check yourself

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

1. (f * g)(t) =
   A. f(t) g(t)
   B. integral from 0 to t of f(u) g(u) du
   C. integral from 0 to t of f(u) g(t-u) du
   D. f(t) + g(t)
   Answer: C. Convolution. (Key point: Definition)

2. L[f * g] =
   A. F(s) + G(s)
   B. F(s)/G(s)
   C. F(s) G(1/s)
   D. F(s) G(s)
   Answer: D. Product in s-domain. (Key point: Theorem)

3. s^2/((s^2+4)(s^2+9)) is written as product of
   A. s/(s^2+4) and s/(s^2+9)
   B. 1/(s^2+4) and s^2/(s^2+9)
   C. s/(s^2+4) and 1/(s^2+9)
   D. s^2/(s^2+4) and 1/(s^2+9)
   Answer: A. Gives cos 2t and cos 3t. (Key point: Choice)

4. The inverse is
   A. (2/5) sin 3t - (3/5) sin 2t
   B. (3/5) sin 3t - (2/5) sin 2t
   C. (3/5) sin 3t + (2/5) sin 2t
   D. (3/5) cos 3t - (2/5) cos 2t
   Answer: B. Check by partial fractions (9/5)/(s^2+9) - (4/5)/(s^2+4). (Key point: Answer)

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## Related in other subjects

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