Inverse Laplace Calculator

The inverse Laplace transform is the operation that recovers a time function f(t) from its Laplace transform F(s). It is written L⁻¹, so f(t) = L⁻¹[F(s)]. The forward transform multiplies f(t) by e^−st and integrates from 0 to ∞; the inverse transform undoes that step.

Formally, the inverse is the Bromwich contour integral f(t) = (1/2πi) ∫ F(s) e^st ds taken along a vertical line in the complex plane. In practice almost nobody evaluates that integral directly. For rational functions, which cover circuits, control systems, and mechanical vibrations, the residue theorem reduces it to a short sum: every pole p of F(s) contributes a term proportional to e^pt.

That single fact explains the whole method:

• A real pole at s = −a gives a decaying exponential e^−at.
• A pole at s = 0 gives a constant (a step).
• A complex pair at s = σ ± iω gives an oscillation e^σtcos(ωt) or e^σtsin(ωt).
• A repeated pole multiplies its term by t, t², and so on.

To calculate an inverse Laplace transform by hand, factor the denominator, expand F(s) into partial fractions, and convert each fraction with a table entry.

1. Make the fraction proper. If the numerator degree is not lower than the denominator degree, divide first. The quotient becomes impulse terms δ(t), δ′(t), and so on.
2. Find the poles. Solve D(s) = 0. Note each root and how many times it repeats.
3. Find the coefficients. For a simple pole p, the coefficient is the residue A = lim (s − p)F(s) as s → p, the "cover-up" rule. For a pole repeated m times, differentiate (s − p)^mF(s) up to m − 1 times.
4. Invert term by term. A/(s − p) → Ae^pt, A/(s − p)^k → A·t^k−1e^pt/(k − 1)!, and conjugate complex terms combine into a damped sinusoid.
5. Add the pieces. Linearity lets you sum every term to obtain f(t).

The calculator follows exactly these steps, using a numerical root finder for the poles and a Taylor-series expansion for repeated-pole coefficients, so you can check homework answers line by line.

Five properties let you invert expressions that are not in the table directly. Each one turns an operation in the s-domain into a simple change in the time domain.

• Linearity: L⁻¹[aF + bG] = a·f(t) + b·g(t). This is what makes partial fractions work.
• First shifting (frequency shift): replacing s with s − a multiplies the answer by e^at.
• Multiplication by 1/s: dividing by s integrates f(t) from 0 to t.
• Differentiation in s: −F′(s) inverts to t·f(t), which is where the t·e^at terms come from.
• Time scaling: F(s/k)/k inverts to f(kt), stretching or squeezing the signal.

Seeing a (s + a) everywhere in place of s is the clue to use the shifting property; a lone s in the denominator is the clue to integrate.

The location of the poles of F(s) tells you how f(t) behaves before you calculate anything. Poles in the left half of the s-plane (negative real part) produce terms that die out, so the signal is stable. A single pole in the right half-plane makes the signal grow without bound.

The classic second-order system ω_n²/(s² + 2ζω_ns + ω_n²) shows all the cases with one damping ratio ζ:

• ζ = 0, undamped: poles on the imaginary axis, f(t) is a pure sine that never decays.
• 0 < ζ < 1, underdamped: a complex pair, f(t) rings and decays.
• ζ = 1, critically damped: a repeated real pole, the fastest response without overshoot.
• ζ > 1, overdamped: two distinct real poles, a slow exponential settle.

The Laplace method turns a linear differential equation into algebra, and the inverse transform turns the algebraic answer back into y(t). Take y″ + b·y′ + c·y = 0 with y(0) = y₀ and y′(0) = v₀. Transforming each term gives

s²Y − s·y₀ − v₀ + b(sY − y₀) + cY = 0

so Y(s) = ((s + b)·y₀ + v₀) / (s² + b·s + c). The initial conditions are already built in, so there are no constants left to solve for after inverting.

This is why the inverse Laplace transform is the standard tool for RLC circuits (current and voltage after a switch closes), spring-mass-damper systems, heat flow in rods, and the step response of control loops.