Key takeaways
- Laplace transform: F(s) = ∫₀^∞ f(t)·e^(−st) dt.
- L{e^(at)} = 1/(s − a), and it exists only for s > a.
- It turns y' into sY(s) − y(0), so derivatives become algebra.
- Watch the sign: L{e^(−3t)} = 1/(s + 3).
In this guide
What is the Laplace transform?
Think of it as a translator. You give it a signal f(t) that changes over time, such as a voltage or a spring's position. It gives back a new function F(s). Hard calculus in the time world (derivatives) becomes simple algebra in the s world (multiplying by s). You solve the easy algebra, then translate back.
It works by multiplying f(t) by a fading exponential e^(−st) and adding up the area from t = 0 to infinity. The fading factor tames the function so that the area is finite, as long as s is big enough.
Why does the Laplace transform matter?
Engineers use it every day to design systems that must respond smoothly: cruise control, drone stabilisers, thermostats, audio filters and power circuits. Its superpower is one rule: the transform of a derivative is just sY(s) − y(0). So a differential equation becomes an algebra problem, with the starting values already built in.
Solve for Y(s) with algebra, look up the answer in a table, and translate back. No guessing, no integrating factors.
The formula
For the exponential function f(t) = e^(at), which this calculator handles:
| Symbol | What it means | Example |
|---|---|---|
| f(t) | The original function of time t (t ≥ 0) | e^(2t) |
| a | Growth rate: a > 0 grows, a < 0 decays, a = 0 gives 1 | 2 |
| s | The transform variable (real here) | 4 |
| F(s) | The result, 1/(s − a) | 0.5 |
The calculator plugs your a and s into 1/(s − a). Always choose s > a. If s is less than a, the calculator tells you the integral does not converge, because the integrand grows without bound. At s = a exactly it shows ∞, since the area grows forever.
A short table of common transforms
| f(t) | F(s) = L{f(t)} | Valid for |
|---|---|---|
| 1 | 1/s | s > 0 |
| e^(at) | 1/(s − a) | s > a |
| tⁿ (whole n) | n!/sⁿ⁺¹ | s > 0 |
| sin(bt) | b/(s² + b²) | s > 0 |
| cos(bt) | s/(s² + b²) | s > 0 |
| y'(t) | sY(s) − y(0) | — |
For tⁿ you'll need factorials; see the factorial calculator.
How to find L{e^(at)} step by step
- 1Write the integral.
∫₀^∞ e^(at) · e^(−st) dt = ∫₀^∞ e^(−(s−a)t) dt. - 2Integrate. The antiderivative is
−e^(−(s−a)t) ÷ (s − a). - 3Take the limit as t → ∞. If s > a, the exponential fades to 0.
- 4Subtract the value at t = 0. That value is
−1/(s − a), so the result is0 − (−1/(s − a)) = 1/(s − a).
Now enter your own a and s below. The calculator checks that s > a and shows the working.
Try it yourself
Pre-filled with the example — change any value.
Common mistakes to avoid
- Getting the sign wrong.
L{e^(−3t)} = 1/(s + 3), because a = −3 gives s − (−3). - Starting the integral at −∞. The Laplace transform starts at t = 0.
- Forgetting it is linear.
L{2e^(at)} = 2/(s − a); constants just come along. - Forgetting the starting value.
L{y'} = sY(s) − y(0), not just sY(s).
Tips and tricks
- Split sums:
L{3 + e^(2t)} = 3/s + 1/(s − 2), because the transform is linear. - Solving ODEs: it handles equations like the second-order ODE of a spring, starting values included.
- Repeating signals: for those, the related tool is the Fourier series.
Where you'll use it in real life
- Control systems: cruise control, drones, robots and thermostats.
- Electronics: analysing circuits with capacitors and inductors.
- Mechanical engineering: springs, dampers and vibrating machines.
- Signal processing: designing filters for audio and communications.
Quick summary
The Laplace transform multiplies f(t) by e^(−st) and integrates from 0 to ∞, turning calculus into algebra. For e^(at) the result is 1/(s − a), which exists only when s > a. Use the calculator above to evaluate it for any a and s, with the steps shown.
Worked example
In a circuit problem, the current grows like e^(2t). Your textbook asks for its Laplace transform, evaluated at s = 4. Since 4 > 2, the transform exists. Use a = 2 and s = 4.
- 01L{e^(2t)} = 1/(s-2)
- 02= 1/(4-2)
- 03= 1/2
- 04= 0.5
These numbers are pre-filled in the calculator above.
Frequently asked questions
What is the Laplace transform of e^(at)?
It is 1/(s − a), valid for s > a. For example, L{e^(2t)} = 1/(s − 2), which equals 1/3 at s = 5.
What is the Laplace transform of 1?
It is 1/s, for s > 0. This is the e^(at) formula with a = 0, because e^(0·t) = 1.
Why is the Laplace transform useful for differential equations?
It turns the derivative y' into sY(s) − y(0), which is plain algebra. You solve for Y(s), then use a table to transform back and get y(t).
What is the difference between the Laplace and Fourier transforms?
The Fourier transform breaks a signal into pure waves. The Laplace transform adds a fading factor e^(−st) and starts at t = 0, so it also handles signals that grow and includes starting values.
