There is a calculator with detailed steps.
Use this Laplace transform calculator to find the Laplace transformation of a function f(t) or an ordinary differential equation (ODE). Our calculator applies relevant Laplace formulas and integral operations to provide accurate results with detailed steps.
Laplace transform is a mathematical technique used to convert a time-domain function f(t) into a function of a complex variable s. It is widely used in physics, engineering, and control theory to solve ODEs.
Mathematically:
\(F(s) = \int_{0}^{\infty} f(t)e^{-st} dt\)
Where:
There are two main methods:
\(F(s) = \int_{0}^{\infty} f(t)e^{-st} dt\)
Example:
Given:
\(f(t) = 6e^{-5t} + e^{3t} + 5t^3 - 9\)
Step 1: Apply the Laplace formula
\(F(s) = \int_{0}^{\infty} (6e^{-5t} + e^{3t} + 5t^3 - 9) e^{-st} dt\)
Step 2: Solve each term individually
Step 3: Combine all terms
\(F(s) = \frac{6}{s+5} + \frac{1}{s-3} + \frac{30}{s^4} + \frac{9}{s}\)
To convert this back to the time domain, use our inverse Laplace transform calculator.
This table shows common Laplace transforms:
| Function name | Time-domain f(t) | Laplace transform F(s) |
|---|---|---|
| Constant | 1 | 1/s |
| Linear | t | 1/s² |
| Power | t^n | n!/s^(n+1) |
| Exponent | e^(at) | 1/(s-a) |
| Sine | sin(at) | a / (s² + a²) |
| Cosine | cos(at) | s / (s² + a²) |
| Hyperbolic sine | sinh(at) | a / (s² - a²) |
| Hyperbolic cosine | cosh(at) | s / (s² - a²) |
| Growing sine | t sin(at) | 2as / (s² + a²)² |
| Growing cosine | t cos(at) | (s² - a²) / (s² + a²)² |
| Decaying sine | e^(-at) sin(ωt) | ω / ((s+a)² + ω²) |
| Decaying cosine | e^(-at) cos(ωt) | (s+a) / ((s+a)² + ω²) |
| Delta function | δ(t) | 1 |
| Delayed delta | δ(t-a) | e^(-as) |
| Property | Equation |
|---|---|
| Linearity | L{f(t) + g(t)} = F(s) + G(s) |
| Time Delay | L{f(t-td)} = e^(-s td) F(s) |
| First Derivative | L{f'(t)} = s F(s) - f(0-) |
| Second Derivative | L{f''(t)} = s² F(s) - s f(0-) - f'(0-) |
| Nth Derivative | L{f^(n)(t)} = s^n F(s) - s^(n-1) f(0-) - ... - f^(n-1)(0-) |
| Integration | L{∫f(t) dt} = 1/s F(s) |
| Convolution | L{f(t) * g(t)} = F(s) G(s) |
| Initial Value Theorem | lim(s→∞) s F(s) = f(0-) |
| Final Value Theorem | lim(s→0) s F(s) = f(∞) |
Laplace Transform converts a time-domain signal into a complex frequency-domain signal. Fourier Transform converts it into the 'jw' complex plane, which is a special case of Laplace Transform when the real part is zero.
Yes. If f(t) = 0, then F(s) = 0. This follows from the linearity property of the Laplace Transform.
Wikipedia: Bilateral Laplace Transform, Inverse Laplace Transform.
Paul's Online Notes: Laplace Transforms, Solving IVPs.
Swarthmore College: Laplace Properties - Linearity, Time Delay, Derivatives, Initial and Final Value Theorems.
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