Inverse Laplace Transform Calculator

The inverse Laplace transform turns a function F(s) in the s-domain back into a function f(t) in the time domain, and it's the key step for solving linear ODEs, analyzing RLC circuits, and working with control-system transfer functions. This tool covers both situations engineering and math students search for: entering a general rational function F(s) = N(s)/D(s) and getting the full partial fraction breakdown, or plugging values into a standard transform pair like 1/(s-a) or s/(s²+a²) to get f(t) instantly.

Quick Answer

L⁻¹{1/s} = 1 | L⁻¹{1/(s−a)} = e^at | L⁻¹{a/(s²+a²)} = sin(at) | L⁻¹{s/(s²+a²)} = cos(at)
For a general F(s) = N(s)/D(s), factor D(s), split into partial fractions, then match each piece to a known pair below and add the results (the inverse transform is linear).

Enter F(s) = N(s) ÷ D(s) as coefficient lists, highest power of s first, separated by commas. F(s) must be a proper fraction (degree of N(s) less than degree of D(s)). Supports denominators up to degree 6, with real and/or complex-conjugate poles.

Numerator N(s) coefficients:

e.g. "3, 5" means 3s + 5

Denominator D(s) coefficients:

e.g. "1, 3, 2" means s² + 3s + 2

Pick a standard form of F(s), fill in the constant K and the parameter(s), and get f(t) directly from the transform table.

Form of F(s):

K (multiplier):

f(t) =

What Is the Inverse Laplace Transform

The Laplace transform converts a time-domain function f(t) into an s-domain function F(s), and the inverse Laplace transform, written L⁻¹{F(s)} = f(t), reverses that process. It's most useful when a differential equation has already been transformed into an algebraic equation in s — once you solve for F(s), the inverse transform hands you back the actual time-domain solution f(t) you originally wanted.

The inverse Laplace transform is linear: L⁻¹{aF(s) + bG(s)} = aL⁻¹{F(s)} + bL⁻¹{G(s)}
Most F(s) that show up in practice are rational functions — a polynomial divided by a polynomial
Every rational F(s) can be rewritten as a sum of simple terms via partial fraction decomposition
Each simple term corresponds to one row of the standard transform table below

Standard Laplace Transform Pairs

This is the reference table the calculator above matches partial fraction terms against. Memorizing these rows (or bookmarking this table) is the fastest way to invert most rational functions by hand.

F(s)	f(t), t ≥ 0	Notes
1/s	1	Unit step
1/s²	t	Ramp
1/sⁿ	tⁿ⁻¹ ⁄ (n−1)!	n = 1, 2, 3, …
1/(s−a)	e^at	Real pole at s = a
1/(s−a)ⁿ	tⁿ⁻¹e^at ⁄ (n−1)!	Repeated real pole
a/(s²+a²)	sin(at)	Complex poles at ±ai
s/(s²+a²)	cos(at)	Complex poles at ±ai
a/(s²−a²)	sinh(at)	Real poles at ±a
s/(s²−a²)	cosh(at)	Real poles at ±a
b/((s−a)²+b²)	e^atsin(bt)	Complex poles at a ± bi
(s−a)/((s−a)²+b²)	e^atcos(bt)	Complex poles at a ± bi
1/(s(s+a))	(1 − e^−at) ⁄ a	Two real poles, one at 0

Partial Fraction Method, Step by Step

Step 1 — Check the fraction is proper. The degree of N(s) must be less than the degree of D(s). If not, do polynomial long division first.
Step 2 — Factor D(s). Find the roots of the denominator. These are the poles of F(s), and they can be real, repeated, or complex conjugate pairs.
Step 3 — Write the partial fraction form. Each distinct real root r contributes a term A/(s−r); each complex conjugate pair a ± bi contributes a term (Bs + C)/((s−a)²+b²); repeated roots contribute one term per power.
Step 4 — Solve for the unknown constants by clearing denominators and matching coefficients, or by plugging in convenient values of s (the "cover-up" method for simple poles).
Step 5 — Match each term to the transform table above and write down its f(t), then add every piece together — linearity guarantees the sum is the correct inverse transform of the original F(s).

Worked Examples

EXAMPLE — distinct real poles
F(s) = (3s + 5) ⁄ (s² + 3s + 2) = (3s + 5) ⁄ ((s+1)(s+2))
Partial fractions: A ⁄ (s+1) + B ⁄ (s+2), solving gives A = 2, B = 1
f(t) = 2e^−t + e^−2t

EXAMPLE — complex conjugate poles
F(s) = 4 ⁄ (s² + 4) — poles at s = ±2i, so a = 0, b = 2
Matches the row a/(s²+a²) with a = 2, scaled by 2
f(t) = 2 sin(2t)

EXAMPLE — mixed poles with a damping term
F(s) = 1 ⁄ ((s+1)((s+1)²+4)) — one real pole at s = −1 and a complex pair at −1 ± 2i
Decomposing and matching against the damped sine/cosine rows gives a sum of an e^−t term and an e^−tsin(2t)/e^−tcos(2t) combination — exactly the kind of case the rational-function mode above solves automatically.

Key Properties

Linearity — L⁻¹{aF(s) + bG(s)} = a·f(t) + b·g(t), so you can invert term by term.
First shifting theorem — if L⁻¹{F(s)} = f(t), then L⁻¹{F(s−a)} = e^atf(t); this is why poles shifted by "a" always carry an e^at factor.
Derivative property — L{f'(t)} = sF(s) − f(0), which is exactly why Laplace transforms turn differential equations into algebraic ones.
Uniqueness — for continuous, well-behaved functions, the inverse Laplace transform is unique, so partial fractions always give the one correct f(t).

Common Mistakes

COMMON MISTAKE
Forgetting that a complex conjugate pair a ± bi must be kept as one quadratic term (Bs + C)/((s−a)²+b²) in the partial fraction setup — splitting it into two separate linear terms with complex coefficients still works mathematically, but it's far easier to make an error, and the final f(t) must still come out real.

COMMON MISTAKE
Applying partial fractions to an improper fraction (degree of N(s) ≥ degree of D(s)) without first doing polynomial long division — this silently gives a wrong decomposition.

Real-Life Uses

Control systems — converting a transfer function's step or impulse response back to the time domain to check overshoot, settling time, and stability
Circuit analysis — finding the actual current or voltage waveform in an RLC circuit after solving the circuit equations in the s-domain
Mechanical vibrations — recovering displacement or velocity of a damped spring-mass system over time
Signal processing — moving between frequency-domain filter descriptions and their time-domain impulse response
Differential equations — solving linear ODEs with initial conditions without ever finding the general homogeneous + particular solution by hand

Glossary

Laplace Transform: An integral transform that converts a time-domain function f(t) into an s-domain function F(s).
Inverse Laplace Transform: The operation L⁻¹{F(s)} = f(t) that recovers the original time-domain function from F(s).
Pole: A value of s where F(s) becomes infinite — the roots of the denominator D(s) of a rational F(s).
Partial Fraction Decomposition: Rewriting a rational function as a sum of simpler fractions, one per pole (or repeated pole power), so each piece matches a known transform.
Proper Rational Function: A fraction N(s)/D(s) where the degree of the numerator is strictly less than the degree of the denominator.
Damped Sinusoid: A time-domain term of the form e^atsin(bt) or e^atcos(bt), produced by a complex conjugate pole pair a ± bi.

Frequently Asked Questions

Q: What is the inverse Laplace transform?

A: It converts a function F(s) in the s-domain back into a time-domain function f(t), written L⁻¹{F(s)} = f(t).

Q: How do you find the inverse Laplace transform of a rational function?

A: Factor the denominator, decompose F(s) into partial fractions, match each term to the standard transform table, and add the results together.

Q: What is the inverse Laplace transform of 1/s?

A: f(t) = 1, for t ≥ 0.

Q: What is the inverse Laplace transform of 1/(s−a)?

A: f(t) = e^at.

Q: How do complex conjugate poles affect the inverse Laplace transform?

A: A pole pair a ± bi produces one real term combining e^atcos(bt) and e^atsin(bt), instead of two separate complex terms.

Q: Why must F(s) be a proper fraction before using partial fractions?

A: Because partial fractions only decompose a proper fraction directly — an improper F(s) needs polynomial long division first to split off a polynomial part.