Inverse Laplace Transform Calculator — Step-by-Step Solutions

Compute the inverse Laplace transform of rational functions with detailed steps. Free online inverse Laplace transform calculator using partial fraction decomposition and standard transform tables. Get f(t) from F(s) instantly.

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Inverse Laplace Transform Calculator

Enter a rational function F(s) to find its inverse Laplace transform f(t). Supports partial fraction decomposition and common transform pairs.

Enter F(s) and click Calculate Inverse Laplace to see f(t).

Inverse Laplace Transform Formula & Definition

The inverse Laplace transform converts a function F(s) from the complex frequency domain (s-domain) back to the time domain f(t). It is denoted as:

f(t) = L⁻¹{F(s)}

The formal definition uses the Bromwich integral (complex inversion formula):

f(t) = (1 / 2πi) ∫γ-i∞γ+i∞ est F(s) ds

In practice, most inverse Laplace transforms are computed using partial fraction decomposition combined with a table of standard transform pairs, avoiding the need for complex contour integration.

Key Variables

  • F(s) — Laplace-domain function (function of complex variable s)
  • f(t) — Time-domain function (function of real variable t, defined for t ≥ 0)
  • L⁻¹ — Inverse Laplace transform operator
  • γ — Real number greater than the real part of all singularities of F(s)

How to Calculate the Inverse Laplace Transform

Follow these steps to compute an inverse Laplace transform using the partial fraction method:

  1. Check if F(s) is a proper rational function — The degree of the numerator must be less than the degree of the denominator. If not, perform polynomial long division first.
  2. Factor the denominator — Find all roots (real and complex) of the denominator polynomial.
  3. Set up partial fractions — For each linear factor (s+a), use A/(s+a). For repeated factors (s+a)ⁿ, use A₁/(s+a) + A₂/(s+a)² + ... + Aₙ/(s+a)ⁿ. For irreducible quadratic factors (s²+bs+c), use (As+B)/(s²+bs+c).
  4. Solve for coefficients — Multiply both sides by the denominator and solve the resulting system of equations.
  5. Apply the inverse Laplace transform to each term — Use a standard transform table to convert each partial fraction term from the s-domain to the time domain.
  6. Sum all time-domain terms — Add all the individual f(t) components to obtain the final result.

Common Inverse Laplace Transform Pairs Table

Reference these standard transform pairs when computing inverse Laplace transforms:

F(s)f(t) = L⁻¹{F(s)}Condition
1/s1 (unit step u(t))t ≥ 0
1/s²tt ≥ 0
1/sⁿtⁿ⁻¹/(n-1)!n = 1,2,3,...
1/(s+a)e-atAny a
1/(s+a)ⁿtⁿ⁻¹e-at/(n-1)!n = 1,2,3,...
1/(s²+a²)sin(at)/aa ≠ 0
s/(s²+a²)cos(at)a ≠ 0
1/(s²-a²)sinh(at)/aa ≠ 0
s/(s²-a²)cosh(at)a ≠ 0
1/((s+a)²+b²)e-atsin(bt)/bb ≠ 0
(s+a)/((s+a)²+b²)e-atcos(bt)b ≠ 0
e-as/su(t-a)a ≥ 0

Inverse Laplace Transform Worked Examples

Example 1: Simple Exponential

Find L⁻¹{1/(s+3)}

F(s) = 1/(s+3) matches the form 1/(s+a) with a=3
f(t) = e-3t for t ≥ 0

Example 2: Sine Function

Find L⁻¹{1/(s²+4)}

F(s) = 1/(s²+4) matches 1/(s²+a²) with a=2
f(t) = sin(2t)/2 for t ≥ 0

Example 3: Partial Fractions Required

Find L⁻¹{1/(s²+3s+2)}

Denominator factors: s²+3s+2 = (s+1)(s+2)
1/((s+1)(s+2)) = 1/(s+1) - 1/(s+2)
f(t) = e-t - e-2t for t ≥ 0

Example 4: Damped Oscillation

Find L⁻¹{(s+1)/(s²+4s+13)}

Complete the square: s²+4s+13 = (s+2)²+3²
F(s) = (s+2-1)/((s+2)²+3²) = (s+2)/((s+2)²+9) - 1/((s+2)²+9)
f(t) = e-2tcos(3t) - (1/3)e-2tsin(3t)

People Also Ask About Inverse Laplace Transforms

The inverse Laplace transform converts frequency-domain functions back to the time domain, making it essential for solving linear differential equations, analyzing control systems, signal processing, and electrical circuit analysis. It bridges the s-domain analysis with real-world time-domain behavior.
Decompose F(s) into a sum of simpler rational functions using partial fraction decomposition. Then apply the inverse Laplace transform to each term individually using a standard transform table. Finally, sum all the resulting time-domain functions to obtain f(t).
The inverse Laplace transform of 1/s is the unit step function u(t), which equals 1 for all t > 0. This is the most fundamental transform pair and represents a constant input that turns on at t=0 in control systems and circuit analysis.
The convolution theorem states that L⁻¹{F(s)G(s)} = f(t) * g(t) = ∫₀ᵗ f(τ)g(t-τ)dτ. This is useful when F(s) is a product of two functions whose individual inverse transforms are known, allowing computation through convolution integration rather than partial fractions.
L⁻¹{1/(s²+1)} = sin(t). This comes from the standard pair L{sin(at)} = a/(s²+a²) with a=1. More generally, L⁻¹{1/(s²+a²)} = sin(at)/a for any nonzero constant a.

Frequently Asked Questions About Inverse Laplace Transforms

Yes. When the denominator has complex conjugate roots (discriminant less than zero), the calculator handles them using the damped sine/cosine transform pairs: 1/((s+a)²+b²) transforms to e⁻ᵃᵗsin(bt)/b, and (s+a)/((s+a)²+b²) transforms to e⁻ᵃᵗcos(bt).
For improper rational functions where the numerator degree ≥ denominator degree, polynomial long division must be performed first. The result is a polynomial (whose inverse transform involves Dirac delta functions and their derivatives) plus a proper rational remainder that can be handled with partial fractions.
Use the caret symbol: s^2 means s-squared, s^3 means s-cubed, etc. For example, enter 1/(s^2+4) for 1/(s²+4) or (s+2)/(s^2+4s+13) for more complex expressions. The calculator automatically parses these exponential notations.
The current version focuses on rational functions F(s) without explicit exponential factors. For expressions with e^(-as) multipliers, the second shifting theorem applies: L⁻¹{e^(-as)F(s)} = f(t-a)u(t-a). You can first compute the inverse of F(s), then apply the time shift manually.
The one-sided Laplace transform integrates from t=0 to ∞ and is used for causal systems where f(t)=0 for t<0. The two-sided (bilateral) Laplace transform integrates from -∞ to ∞. This calculator uses the standard one-sided Laplace transform, which is most common in engineering and differential equations.
All calculations use exact symbolic methods (partial fraction decomposition and algebraic manipulation) rather than numerical approximations. Results are mathematically exact for rational functions. Decimal coefficients are displayed with up to 6 significant figures for readability.

Inverse Laplace Transform Glossary

Laplace Transform

An integral transform that converts a time-domain function f(t) into a complex frequency-domain function F(s) = ∫₀∞ e⁻ˢᵗf(t)dt.

Inverse Laplace Transform

The operation L⁻¹ that recovers f(t) from F(s), denoted f(t) = L⁻¹{F(s)}.

s-Domain

The complex frequency domain where s = σ + jω. Laplace transforms map time functions into this domain for algebraic manipulation.

Partial Fraction Decomposition

A technique for breaking a complex rational function into a sum of simpler fractions, each of which has a known inverse Laplace transform.

Unit Step Function

Denoted u(t), equals 0 for t < 0 and 1 for t ≥ 0. Its Laplace transform is 1/s, making it fundamental to causal system analysis.

Convolution Theorem

States that L⁻¹{F(s)G(s)} = (f * g)(t) = ∫₀ᵗ f(τ)g(t-τ)dτ, allowing inverse transforms of products via convolution integrals.

Bromwich Integral

The formal complex inversion formula for the inverse Laplace transform: f(t) = (1/2πi)∫ eˢᵗF(s)ds along a vertical contour in the complex plane.

Heaviside Expansion

A formula for the inverse Laplace transform of rational functions with distinct poles, expressing f(t) as a sum of exponential terms using residues.

Editorial Review & Methodology

This inverse Laplace transform calculator was built and reviewed by the NumbrWiz Editorial Team. The inverse Laplace transform is a cornerstone of advanced engineering mathematics, verified against standard textbooks including Oppenheim & Willsky, Kreyszig's Advanced Engineering Mathematics, and standard differential equations curricula.

  • Transform pair verification: All standard pairs cross-checked against multiple authoritative sources in signals and systems literature.
  • Partial fraction algorithm: Tested with real distinct roots, repeated roots, and complex conjugate pairs.
  • Edge case handling: Verified with improper rational functions, zero-pole cancellations, and high-degree denominators.

Transparency note: All calculations run entirely client-side in your browser. No input data is collected, stored, or transmitted. Results are for educational purposes; always verify critical engineering calculations independently.

Page last reviewed: May 2026 · NumbrWiz Editorial Team