Laplace Transform — Mathematics Knowledge System

Overview

The Laplace transform, named after Pierre-Simon Laplace, is an integral transform that converts a function f(t) of a real variable (typically time) into a function F(s) of a complex variable s. Lowercase letters conventionally denote the original time-domain function and the corresponding uppercase letter denotes its frequency-domain transform, e.g. f(t) and F(s). Its principal use is turning differential equations in t into algebraic equations in s, since differentiation in t corresponds to multiplication by s in the transformed equation.

Formal Definition

Definition

The (unilateral) Laplace transform of f(t), defined for t ≥ 0, is:

\mathcal{L}\{f\}(s) = F(s) = \int_0^\infty f(t)e^{-st}\,dt

Laplace transform definition

\mathcal{L}\{f'\}(s) = sF(s) - f(0)

Transform of a first derivative

\mathcal{L}\{f''\}(s) = s^2F(s) - sf(0) - f'(0)

Transform of a second derivative

Worked Examples

Substitute directly into the definition.
$\mathcal{L}\{e^{at}\}(s) = \int_0^\infty e^{at}e^{-st}\,dt = \int_0^\infty e^{-(s-a)t}\,dt$
Evaluate the improper integral, which converges for Re(s) > a.
$= \left[\frac{-e^{-(s-a)t}}{s-a}\right]_0^\infty = \frac{1}{s-a}$

Answer: L{e^(at)}(s) = 1/(s-a), for Re(s) > a.

Practice Problems

Difficulty 6/10

Show that L{sin(bt)}(s) = b/(s²+b²) and L{cos(bt)}(s) = s/(s²+b²) for Re(s) > 0.

Difficulty 6/10

Why is the Laplace transform so useful for solving the differential equations of circuits and structural dynamics?

Quiz

The Laplace transform is powerful for ODEs because it turns differentiation into:

integrationmultiplication by s (converting the ODE into an algebraic equation)a trigonometric identitya matrix

A major practical advantage of the Laplace method over classical ODE methods is that it:

requires no initial conditionsbuilds initial conditions in directly and handles step/impulse (discontinuous) inputs cleanlyonly works for linear algebraavoids all integration

Summary

The Laplace transform maps f(t) to F(s) = ∫₀^∞ f(t)e^{-st}dt, converting time-domain functions into the complex frequency domain.
Differentiation in t becomes multiplication by s (plus initial-value correction terms): L{f'}(s) = sF(s) - f(0).
Standard transform pairs: L{e^{at}} = 1/(s-a), L{sin(bt)} = b/(s²+b²), L{cos(bt)} = s/(s²+b²).
This turns linear ODEs with constant coefficients into algebraic equations in s, which are solved and then inverted back to the time domain.

References

WebsiteWikipedia — Laplace transform