Laplace Transform of e^(-at)
The Laplace transform of e^(-at) is 1/(s + a). See the table rule behind it, then edit the function to transform anything else.
L{e^(-at)} = 1/(s + a). A decaying exponential gives a pole in the left half plane, 1/(s + a).
Use t as the variable. Examples: t^3, e^(-2t), sin(3t), t*cos(2t), 4t^2+3. Write δ(t) as delta(t).
The result
L{e^(-at)} = 1/(s + a)
Worked example with numbers: L{e^(-3t)} = 1/(s + 3)
- L{e^(-3t)} = 1/(s + 3) Table rule: L{e^(at)} = 1/(s − a)
A decaying exponential gives a pole in the left half plane, 1/(s + a).
More exponential transforms
Other transforms
Frequently asked questions
What is the Laplace transform of e^(-at)? ▾
L{e^(-at)} = 1/(s + a). A decaying exponential gives a pole in the left half plane, 1/(s + a).
Which rule gives the transform of e^(-at)? ▾
L{e^(at)} = 1/(s − a) applied to e^(-3t).
How do I get back from F(s) to f(t)? ▾
Use the inverse Laplace transform. Split F(s) into partial fractions and invert each piece with the same table read backwards — the inverse Laplace transform calculator does this step by step.
About this Calculator
The Laplace transform of e^(-at) is 1/(s + a). See the table rule behind it, then edit the function to transform anything else.