Laplace Transform
Common Laplace transforms table and the definition L{f(t)} = integral of e^(-st)f(t)dt.
Used to solve differential equations.
The Formula
The Laplace transform converts a function of time f(t) into a function of the complex variable s. This transforms differential equations into algebraic equations, which are much easier to solve.
After solving in the "s-domain," you apply the inverse Laplace transform to get back to the time domain. This technique is fundamental in control theory, circuit analysis, and signal processing.
Variables
| Symbol | Meaning |
|---|---|
| f(t) | Original function in the time domain |
| F(s) | Transformed function in the s-domain (frequency domain) |
| s | Complex frequency variable (s = σ + jω) |
| t | Time variable (t ≥ 0) |
Common Laplace Transform Pairs
| f(t) | F(s) = L{f(t)} |
|---|---|
| 1 (unit step) | 1/s |
| t | 1/s² |
| tⁿ | n!/sⁿ⁺¹ |
| eᵃᵗ | 1/(s − a) |
| sin(ωt) | ω/(s² + ω²) |
| cos(ωt) | s/(s² + ω²) |
| t·eᵃᵗ | 1/(s − a)² |
| eᵃᵗ sin(ωt) | ω/((s − a)² + ω²) |
| δ(t) (impulse) | 1 |
Example 1
Find the Laplace transform of f(t) = 3t² + 2eᵗ.
Use linearity: L{3t² + 2eᵗ} = 3·L{t²} + 2·L{eᵗ}
From the table: L{t²} = 2!/s³ = 2/s³
From the table: L{eᵗ} = 1/(s − 1)
F(s) = 6/s³ + 2/(s − 1)
Example 2
Solve the differential equation y' + 2y = 6, with y(0) = 1, using the Laplace transform.
Take the Laplace transform of both sides: L{y'} + 2·L{y} = L{6}
sY(s) − y(0) + 2Y(s) = 6/s
sY(s) − 1 + 2Y(s) = 6/s
(s + 2)Y(s) = 6/s + 1 = (6 + s)/s
Y(s) = (s + 6) / (s(s + 2))
Partial fractions: Y(s) = 3/s − 2/(s + 2)
Inverse transform: y(t) = 3 − 2e⁻²ᵗ
y(t) = 3 − 2e⁻²ᵗ (verified: y(0) = 3 − 2 = 1 ✓)
When to Use It
The Laplace transform is a powerful tool in mathematics and engineering.
- Solving ordinary differential equations with initial conditions
- Analyzing electrical circuits (RLC circuits, transfer functions)
- Control systems design and stability analysis
- Signal processing and communication systems
- Mechanical vibration analysis