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Convolution Theorem (Laplace) Calculator

States that the Laplace transform of a convolution of two functions is the product of their individual transforms.

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L{f * g}

Formula first

Overview

This theorem provides a powerful method for finding inverse Laplace transforms of products of functions by using the convolution integral.

Symbols

Variables

F(s)G(s) = L{f * g}, F(s) = F(s), G(s) = G(s)

F(s)G(s)
L{f * g}
Transform of the convolution
F(s)
F(s)
Transform of the first function
G(s)
G(s)
Transform of the second function

Apply it well

When To Use

When to use: Essential for solving non-homogeneous differential equations and analyzing linear time-invariant (LTI) systems.

Why it matters: It converts the complex operation of convolution in the time domain into simple algebraic multiplication in the frequency (s) domain.

Avoid these traps

Common Mistakes

  • Confusing convolution f*g with the pointwise product f(t)g(t).
  • Forgetting that the theorem only applies if the transforms F(s) and G(s) exist for the same region of convergence.

One free problem

Practice Problem

Given the individual transforms F(s) = 4 and G(s) = 8, calculate the Laplace transform of the convolution (f * g)(t).

F(s)4 Transform of the first function
G(s)8 Transform of the second function

Solve for: result

Hint: According to the theorem, the transform of the convolution is simply the product of the individual transforms.

The full worked solution stays in the interactive walkthrough.

References

Sources

  1. Advanced Engineering Mathematics
  2. Wikipedia: Laplace transform
  3. Differential Equations with Boundary-Value Problems by Dennis G. Zill
  4. Dennis G. Zill, Warren S. Wright. Differential Equations with Boundary-Value Problems.
  5. Erwin Kreyszig. Advanced Engineering Mathematics.
  6. Wikipedia: Convolution theorem
  7. Kreyszig, Advanced Engineering Mathematics
  8. Boyce, DiPrima, and Meade, Elementary Differential Equations and Boundary Value Problems