How to use this calculator
Type F(s) using s as the variable. The calculator combines it into one fraction, finds the poles (roots of the denominator), works out the partial fraction coefficients and inverts each term. To solve an ODE, choose the second problem type and enter the coefficients, initial conditions and forcing.
Formulas
Simple poleA/(s − p) → Aept, A = [(s − p)F(s)]s=p
Repeated poleA/(s − p)k → Atk−1ept/(k − 1)!
Complex pairα ± βi → eαt(A cos βt + B sin βt)
Derivativesℒ{y′} = sY − y(0), ℒ{y″} = s²Y − sy(0) − y′(0)
Worked example
F(s) = 10/[s(s² + 2s + 5)]
- Poles: s = 0 and s = −1 ± 2i.
- At s = 0: A = 10/5 = 2. The pair contributes −2(s + 1) − 1·2 over (s + 1)² + 4.
- Invert: f(t) = 2 − e−t(2 cos 2t + sin 2t).
- At t = 1, f = 1.972.
Frequently asked questions
Can it handle e^(−as) time delays?
Not directly. Invert the rational part, then shift: ℒ⁻¹{e−asF(s)} = f(t − a)u(t − a).
What if the numerator degree is too high?
The fraction is improper. Long division leaves a polynomial in s, which inverts to impulses δ(t), δ′(t)…; the calculator lists those and inverts the proper remainder.
Why do the poles tell me about stability?
Each pole p gives a term ept. Poles with negative real parts decay; any pole with a positive real part grows without bound.