Inverse Laplace Transforms

The inverse Laplace transform does the opposite of the Laplace transform: it takes a function F(s) in the s-domain and recovers the original time function f(t). It is written f(t) = L⁻¹{F(s)}. Because the Laplace transform is one-to-one for the functions we use, every F(s) comes from exactly one f(t), and this reverse step is what lets us finish solving differential equations with the Laplace transform. It builds directly on the introduction to the Laplace transform.

Most inverse transforms are done by recognizing a standard form and reading it off a table — the same pairs used for calculating Laplace transforms, read right to left.

When F(s) is a ratio of polynomials that does not match a table entry, split it with partial fractions first. For example, 1/[(s−1)(s−2)] breaks into 1/(s−2) − 1/(s−1), and each piece is a standard form whose inverse is an exponential. So the inverse transform is e^2t − e^t.

The inverse transform is linear: the inverse of a sum is the sum of the inverses, and constants factor out. That is why you can break F(s) into simple pieces, invert each with the table, and add the results.

Find the inverse transform of F(s) = 3/(s² + 9). The table shows a/(s² + a²) maps to sin(at), and here a = 3, so f(t) = sin(3t).

Related lessons

• Differential Equations