Inverse Laplace transform

The formula

The forward transform takes a time signal into the s-domain:

F(s) = ∫₀^∞ f(t) · e^(-s·t) dt

That says: multiply your signal by a decaying exponential and sum it up, once for every complex frequency s. The inverse undoes it. The exact statement is a contour integral in the complex plane:

f(t) = (1 / 2π·i) · ∫[γ-i·∞ to γ+i·∞] F(s) · e^(s·t) ds

Reading: walk up a vertical line at Re(s) = γ, to the right of every pole of F(s), and sum F(s)·e^(s·t) along it. γ has to sit right of all singularities so the integral converges. That is the Bromwich integral, and almost no working engineer ever evaluates it directly.

What you actually use, for the rational F(s) you get from lumped circuits and transfer functions (numerator degree below denominator), is the residue shortcut. For those, F(s) dies off on a big left-hand arc, so you can close the contour to the left and the integral collapses to a sum of residues at the poles:

f(t) = Σ  Res[ F(s) · e^(s·t) ]  at each pole of F(s)

Reading: every pole contributes one time-domain term, and the pole location tells you the shape of that term. For a simple pole at s = p, the residue is lim(s→p) (s-p)·F(s)·e^(p·t), so the pole hands you an e^(p·t) mode directly.

In practice that reduces to partial fractions plus a lookup table. Break F(s) into simple pieces and read each one off:

1/(s+a)              →  e^(-a·t)
ω/((s+a)² + ω²)      →  e^(-a·t)·sin(ω·t)
(s+a)/((s+a)² + ω²)  →  e^(-a·t)·cos(ω·t)
1/s²                 →  t
e^(-s·T)·F(s)        →  f(t-T)·u(t-T)

The last line is the dead-time rule: a factor of e^(-s·T) is not a pole and never goes through partial fractions — it is a transport delay of T seconds. Peel it off, invert the rest, then shift and gate the result with the unit step.

Where you meet it

You meet it any time a design gets written down as H(s) and someone asks what happens in the first hundred milliseconds.

Control loop review, on the bench. You size a PID or lead-lag compensator in the s-domain because loop shaping is easy there. Then the review board asks for the step response — rise time, overshoot, settling time. Multiplying H(s) by 1/s (the step) and inverting gives you y(t), the curve you actually defend.
Vibration and modal work on a test stand. A structure's transfer function has complex-conjugate pole pairs. Each pair inverts to a decaying sinusoid e^(-ζ·ωₙ·t)·sin(ω_d·t). The real part of the pole is your damping; the imaginary part is your ringing frequency. You read both straight off the pole map.
RC and RLC transient checks. Snubber sizing, filter settling, a sensor front-end recovering from a step. You write the circuit's V_out(s), do partial fractions, and get the charge/discharge curve without touching a differential equation.
RF and pulse work. Matching networks and the leading edge of a pulse through a filter are s-domain problems whose time-domain answer — ringing, droop, group delay — comes back through inversion.

How it works

The whole method rests on one fact: poles of F(s) become time-domain modes. A pole on the negative real axis at s = -a gives a clean decay e^(-a·t). A complex-conjugate pair at s = -a ± j·ω gives a decaying oscillation at frequency ω. A pole in the right half-plane — positive real part — gives a growing exponential, which is your instability. Reading a transfer function's behavior is mostly reading where its poles sit.

The single most useful worked case is the second-order system, because most real loops and structures behave like one near their dominant poles:

H(s) = ωₙ² / (s² + 2·ζ·ωₙ·s + ωₙ²)

Drive it with a unit step and invert, and for the underdamped case (0 < ζ < 1) you get:

y(t) = 1 - (1/√(1-ζ²))·e^(-ζ·ωₙ·t)·sin(ω_d·t + φ)

with ω_d = ωₙ·√(1-ζ²) and φ = acos(ζ). The percent overshoot falls straight out: Mp = exp(-ζ·π/√(1-ζ²)). Run the numbers for ζ = 0.2 and you get 52.7% overshoot — verified here two ways, the closed form against a direct numerical integration of the ODE, agreeing to better than 1e-9. That is the value of the method: the s-domain algebra and the time-domain curve are the same answer.

The partial-fraction mechanics are worth seeing once. Take:

F(s) = (s+3) / ((s+1)·(s+2))  =  2/(s+1) - 1/(s+2)

The coefficient on 1/(s+1) is the residue at that pole: (s+1)·F(s) evaluated at s = -1, which is (-1+3)/(-1+2) = 2. Likewise the second coefficient is -1. Inverting term by term:

f(t) = 2·e^(-t) - e^(-2·t)

Confirmed symbolically. That residue-per-pole recipe is the whole game for distinct poles.

The gotchas are where people lose points:

Repeated poles. A pole of multiplicity m does not give a single exponential — it gives t^(m-1)·e^(p·t) terms. A double pole at the origin, 1/s², is the ramp t, not a constant. Forget this and your critically-damped case comes out wrong.
Improper F(s). If the numerator degree meets or exceeds the denominator, you have an impulse δ(t) or its derivatives hiding in the answer. Do the polynomial long division first; the leftover constant is a Dirac impulse at t = 0.
The transform is one-sided. The engineering Laplace transform integrates from 0, so everything you get back is defined for t ≥ 0 and implicitly multiplied by the unit step. It does not know about t < 0. Initial conditions enter as separate terms, not through the inverse.
Uniqueness has fine print. Two functions with the same transform can differ on a set of measure zero — isolated points don't matter. For continuous engineering signals this never bites, but it is why the answer is "essentially unique" rather than flatly unique.
The table is not optional. Almost nobody inverts by contour integration by hand. If a term doesn't match a table entry after partial fractions, you factored it wrong or you need to complete the square to expose a shifted sinusoid.

The limit of validity: the Bromwich contour only converges for functions of exponential order, meaning f(t) grows no faster than some M·e^(c·t). Signals that blow up faster than any exponential have no Laplace transform to invert. In practice every physical response qualifies.

History

The story starts with a self-taught English engineer who didn't much care whether the mathematicians approved. Oliver Heaviside (1850–1925), working on telegraph and telephone transmission through the 1880s, treated the derivative as an algebraic symbol p and solved circuit differential equations by pushing that symbol around like a number ^[1]^[2]. It worked. He reworked Maxwell's equations into the vector form still taught today and derived the telegrapher's equations from the same operational habit. But he never proved why the symbol-shuffling was legal, and famously shrugged that he wasn't going to refuse his dinner just because he didn't fully understand digestion ^[1].

That left a gap, and Thomas John I'Anson Bromwich (1875–1929), a British mathematician, stepped into it ^[3]^[4]. Around 1916, in work on normal coordinates in dynamical systems, he showed that Heaviside's operators could be made rigorous by reading them as contour integrals in the complex plane — the vertical-line integral now bearing his name ^[3]. He connected the loose operational calculus to honest complex-function theory and showed how to recover the time-domain answer by contour integration and residues ^[4].

The inversion formula itself has older roots. The same vertical-line integral is often called Mellin's inverse formula, after the Finnish mathematician Hjalmar Mellin (1854–1933), and Bromwich mainly popularized it in the engineering world rather than inventing it outright ^[5]. The uniqueness question — does one transform pin down one signal — was settled earlier still, by Mathias Lerch in 1903, whose theorem says the inverse is unique up to a set of measure zero ^[5]. A different, derivative-based inversion came from Emil Post in 1930, elegant on paper but rarely used on the bench ^[5]. What survived into daily engineering practice is the practical core: partial fractions, a residue at each pole, and a lookup table — Heaviside's shortcut, wearing Bromwich's proof.

Related tools

/tools/series-rlc-impedance
/tools/lc-resonance
/tools/rc-filter
/tools/rc-charge-time
/tools/rc-snubber
/tools/vibration-natural-freq
/tools/frequency-period

Sources

Written by HE in our own words from the cited sources — engineering judgment included, your stamp still required. All entries →