Inverse Laplace Transform Calculator

Calculate the inverse Laplace transform of F(s). Input the function and get f(t), ROC, and a step-by-step solution using partial fractions. An essential tool for engineering students. Informational only—consult a textbook.

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Published: October 23, 2025 | Updated: October 23, 2025 | Reviewed by: Science/Math Editor

Input Function F(s)

Enter a Laplace transform F(s). Use ^ for exponents, * for multiplication.

Preset Functions

Common Laplace Transform Pairs

F(s)f(t)ROC
1/s1Re(s) > 0
1/s²tRe(s) > 0
1/(s+a)e^(-at)Re(s) > -a
1/(s²+a²)(1/a)sin(at)Re(s) > 0
s/(s²+a²)cos(at)Re(s) > 0
1/(s²-a²)(1/a)sinh(at)Re(s) > |a|
s/(s²-a²)cosh(at)Re(s) > |a|

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How it works

L⁻¹{F(s)} = f(t). We use partial fraction decomposition to break down F(s) into simpler terms, then use a lookup table of common inverse transforms to find f(t).

The process involves:

  1. Parse the input: Identify the structure of F(s)
  2. Partial fraction decomposition: Break complex fractions into simpler forms
  3. Lookup table matching: Match each term to a known inverse transform
  4. Combine results: Sum all inverse transforms to get f(t)
  5. Determine ROC: Identify the region based on pole locations

Inputs explained

  • Input Function F(s): The Laplace transform function in the s-domain. Use standard notation: s for the complex variable, ^ for exponents, * for multiplication.
  • Time Variable: The output variable (usually t or τ) for the time-domain function.
  • Complex Variable: The Laplace domain variable (always s).

Syntax examples:

  • 1/s - Simple pole at origin
  • 1/(s+2) - Exponential decay
  • 1/(s^2 + 4) - Sinusoidal
  • s/(s^2 + 1) - Cosine
  • 1/(s(s+1)) - Requires partial fractions

Example

Input: F(s) = 1/(s² + 4)

Step 1: Recognize standard form 1/(s² + a²) where a² = 4, so a = 2

Step 2: Apply inverse transform: L⁻¹{1/(s² + a²)} = (1/a)sin(at)

Step 3: Substitute a = 2: f(t) = (1/2)sin(2t)

Result: f(t) = 0.5·sin(2t), ROC: Re(s) > 0

Tips & notes

  • The inverse Laplace transform is used to solve differential equations in engineering and physics.
  • The Region of Convergence (ROC) is important for the uniqueness of the transform.
  • Partial fraction decomposition is essential for complex rational functions.
  • Complex functions may require advanced techniques like the Bromwich integral (contour integration).
  • Always verify your result by taking the Laplace transform of f(t) to recover F(s).
  • The linearity property: L⁻¹{aF(s) + bG(s)} = aL⁻¹{F(s)} + bL⁻¹{G(s)}

FAQs

The operation that converts a function from the s-domain (frequency domain) back to the time domain. It's denoted as L⁻¹{F(s)} = f(t) and is essential for solving differential equations.

Using partial fraction expansion to break down complex fractions, then applying a table of common transforms to find the time-domain function. For simple cases, pattern matching suffices; complex cases may require the Bromwich integral.

The set of complex s-values for which the Laplace integral converges. It's crucial for determining the uniqueness of the inverse transform. For causal systems, ROC is typically Re(s) > largest pole real part.

A method to break down a complex rational function into simpler fractions that match entries in the Laplace transform table. For example, 1/(s(s+1)) = 1/s - 1/(s+1).

In solving linear differential equations, analyzing control systems, studying circuit behavior, and signal processing. It converts transfer functions back to time-domain responses.

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Disclaimer

Informational educational tool based on standard mathematical methods. Complex transforms may require advanced techniques. Consult a textbook for detailed theory.

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