Inverse Laplace Transform Calculator
Free inverse Laplace transform calculator with steps. Enter F(s) as a numerator and denominator and get f(t) — with the partial fraction expansion and the transform pair used for every term.
Enter F(s) as a numerator over a denominator and get f(t), with the partial fraction expansion, the coefficient solve, and the transform pair used for each term all laid out.
e.g. 2s+3
Expanded or factored, e.g. (s+1)(s^2+4)
How to find an inverse Laplace transform
- 1
Check the degrees
The numerator must have lower degree than the denominator. If not, divide first.
- 2
Factor the denominator
Split it into linear factors and irreducible quadratics; repeated factors need one term per power.
- 3
Decompose into partial fractions
Write F(s) as a sum of simple terms and solve for the unknown numerators by matching coefficients.
- 4
Invert each term
Apply the table backwards: 1/(s − a) gives e^(at), b/(s² + b²) gives sin(bt), and so on. Add the results.
Inverse transform table
| F(s) | f(t) |
|---|---|
| 1/s | 1 |
| 1/s² | t |
| 1/sⁿ⁺¹ | tⁿ/n! |
| 1/(s − a) | e^(at) |
| 1/(s − a)ⁿ⁺¹ | tⁿe^(at)/n! |
| b/(s² + b²) | sin(bt) |
| s/(s² + b²) | cos(bt) |
| b/(s² − b²) | sinh(bt) |
| s/(s² − b²) | cosh(bt) |
| b/((s − a)² + b²) | e^(at)sin(bt) |
| (s − a)/((s − a)² + b²) | e^(at)cos(bt) |
| 1/(s² + b²)² | (sin bt − bt·cos bt)/(2b³) |
| s/(s² + b²)² | t·sin(bt)/(2b) |
Worked inverse transforms
Reading a system's behaviour off the poles
The roots of the denominator — the poles of F(s) — determine everything about how f(t) behaves, before you invert anything:
- Real negative pole: a decaying exponential. The further left, the faster it decays.
- Real positive pole: exponential growth, which is an unstable system.
- Pole at the origin: a constant term, or a ramp if it is repeated.
- Complex conjugate pair: an oscillation, decaying or growing according to the real part.
- Repeated poles: each repetition multiplies the response by another factor of t.
Frequently asked questions
What is the inverse Laplace transform? ▾
It is the operation that recovers f(t) from F(s). In practice you never evaluate the defining complex contour integral — you split F(s) into partial fractions and read each piece backwards off the standard transform table.
Why does the inverse transform need partial fractions? ▾
The table only lists simple building blocks such as 1/(s − a) and b/(s² + b²). Partial fraction decomposition breaks a complicated rational F(s) into exactly those pieces, and linearity lets you invert them one at a time.
What does a repeated pole give in the time domain? ▾
A factor of t. A single pole 1/(s − a) inverts to e^(at); the repeated pole 1/(s − a)² inverts to t·e^(at); and 1/(s − a)ⁿ⁺¹ inverts to tⁿe^(at)/n!.
What do complex poles mean physically? ▾
A conjugate pair a ± bi gives e^(at)(cos bt and sin bt) — an oscillation at frequency b with an envelope e^(at). If a is negative the oscillation decays; if a is zero it persists; if a is positive it grows.
Why must the numerator have lower degree than the denominator? ▾
If it does not, F(s) does not vanish as s grows, and the inverse contains the Dirac delta and its derivatives rather than an ordinary function. Divide first and handle the polynomial part separately.
How do I enter the denominator? ▾
Either expanded (s^2+3s+2) or factored ((s+1)(s+2)) — both are read correctly, including powers like (s+1)^3.
About this Calculator
Free inverse Laplace transform calculator with steps. Enter F(s) as a numerator and denominator and get f(t) — with the partial fraction expansion and the transform pair used for every term.