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Inverse Laplace transform calculator

Type a fraction of two polynomials in s, such as (s + 3) / ((s + 1)(s + 2)), and get f(t), with the poles and the steps of the partial fraction decomposition.

A ratio of two polynomials in s, such as 1/(s^2 + 4) or 5/(s − 2)^2.

Fill in the fields; the answer appears here straight away.

How it works

  1. Enter the values

    Type the numbers or the formula. Decimals may use a point or a comma. No idea? Click Fill in an example.

  2. Instant answer

    The answer appears as you type, with the most important intermediate values.

  3. See the working

    Under "How it's worked out" you see the steps, handy for checking your own calculation.

How it works

  1. Find the poles: the zeros of the denominator.
  2. Split F(s) into partial fractions, one term per pole p, with (s − p)ᵏ in the denominator for repeated poles.
  3. Transform every term back with the table: L⁻¹{1 / (s − p)ᵏ} = tᵏ⁻¹ / (k − 1)! · e^(pt).

Two complex poles a ± bi together give a term e^(at)·(… cos(bt) + … sin(bt)). For the example, f(t) = 2e^(−t) − e^(−2t).

Limits

The degree of the numerator must be lower than that of the denominator (a proper fraction); otherwise f(t) would contain a delta function. The forward direction is the Laplace transform calculator.

Worked examples

F(s) f(t)
1 / (s + 1) e^(−t)
1 / (s² + 4) ½ · sin(2t)
5 / (s − 2)² 5t · e^(2t)
1 / (s² + 2s + 5) ½ · e^(−t) · sin(2t)
(s + 3) / ((s + 1)(s + 2)) 2e^(−t) − e^(−2t)

The partial fractions of the example

For (s + 3) / ((s + 1)(s + 2)) write A/(s + 1) + B/(s + 2). The coefficient A is the numerator evaluated at s = −1, divided by the other factor at −1: A = (−1 + 3) / (−1 + 2) = 2. In the same way B = (−2 + 3) / (−2 + 1) = −1. So F(s) = 2/(s + 1) − 1/(s + 2), and f(t) = 2e^(−t) − e^(−2t).

Complex poles

s² + 2s + 5 equals (s + 1)² + 4, with poles −1 ± 2i. That gives the damped oscillation e^(−t) · sin(2t) divided by 2. The forward step is the Laplace transform calculator, and to find where a quadratic denominator is zero you can use the quadratic equation solver.

Frequently asked questions

What are poles?

The zeros of the denominator of F(s). Each pole p gives a term e^(pt) in f(t): real poles give exponentials, and complex poles a ± bi give oscillations e^(at)·sin(bt) or cos(bt).

What happens with a repeated pole?

A pole p that appears k times gives terms with 1/(s − p)ᵏ, and each one transforms back to tᵏ⁻¹/(k − 1)! · e^(pt). So 1/(s − 2)² becomes t · e^(2t).

Why do I get an error for some fractions?

The numerator must have a lower degree than the denominator. An improper fraction like s² / (s + 1) would give a delta function in f(t), which this table does not cover.

Is f(t) valid for negative t?

No. The one-sided Laplace transform only describes t ≥ 0, so the result is the function for t ≥ 0.