Laplace Transform & Inverse Transform Calculator

What the Laplace transform is for

The Laplace transform converts calculus into algebra. It maps a function of time onto a function of a complex frequency variable s, and under that map differentiation becomes multiplication by s. A differential equation in t therefore becomes an ordinary algebraic equation in s, which you solve with school algebra and then transform back.

The payoff is not just convenience. Because the derivative rule is L{f′} = sF(s) − f(0), the initial condition is absorbed into the algebra from the very first line. You never solve for a general solution and then chase the arbitrary constants — the transform carries f(0) and f′(0) along automatically. That is why the method dominates circuit analysis and control theory, where every problem starts from a known state at t = 0.

The transform is also what makes discontinuous forcing tractable. A switch that closes at t = 2, a unit impulse, a square pulse — all of these have clean transforms, whereas handling them with the integrating factor method means splitting the problem into intervals and matching at every seam.

The integral, the table, and the two shifting rules

The definition is a single improper integral: multiply f(t) by the decaying kernel e^(−st) and integrate from zero to infinity. The kernel is the whole story. For large enough Re(s) it decays fast enough to beat the growth of f, and the integral converges; below that threshold it does not. The threshold is the convergence abscissa σ, and the transform is properly defined only on the half-plane Re(s) > σ.

Nobody evaluates that integral in practice. You do it once for each elementary function, tabulate the result, and then assemble everything else from four rules:

• Linearity. L{αf + βg} = αF + βG. Constants and sums pass straight through.
• First shifting (s-shift). Multiplying f(t) by e^(at) in time shifts the transform: L{e^(at)f(t)} = F(s − a). This is why every exponential in the time domain simply relocates a pole.
• Second shifting (t-shift). Delaying a signal by c multiplies its transform by e^(−cs).
• Multiplication by t. L{t·f(t)} = −F′(s), which is how 1/s becomes 1/s² and how t·sin ωt acquires a squared denominator.

Combine the first two and you get the entry that covers most engineering work: L{e^(at) sin ωt} = ω / ((s − a)² + ω²). The exponential rate a sets where the poles sit horizontally, and the frequency ω sets how far above and below the real axis they sit. Poles are the whole diagnostic content of F(s): their real part tells you the decay or growth rate, their imaginary part tells you the oscillation frequency.

Going backward, you use the same table in reverse. A rational F(s) is broken into partial fractions until every piece matches a table entry. Which decomposition you need is decided entirely by the discriminant of the denominator, exactly as it is when you solve a quadratic.

How to read the poles and the convergence abscissa

Read the poles as behaviour, not as algebra. The real part of a pole is a growth or decay rate; the imaginary part is an angular frequency in radians per unit time. A conjugate pair at −2 ± 3i means a signal that oscillates at 3 rad/s inside an envelope e^(−2t), which falls to about 2% of its initial size in two seconds.

The rightmost pole governs the long run, and its real part is the convergence abscissa σ. If σ is negative, every term decays and the signal settles. If σ is positive, at least one term grows without bound — unless the coefficient multiplying that particular pole happens to be zero, which is the pole-zero cancellation that control engineers watch for. If σ is exactly zero you sit on the boundary: a single pole at the origin integrates to a constant, a conjugate pair on the imaginary axis oscillates forever without decaying, and a repeated pole on the axis grows linearly.

The discriminant classifies the same three cases you meet in the characteristic equation of a second-order ODE, because it is literally the same quadratic. Δ > 0 gives two real poles and no oscillation — overdamped. Δ = 0 gives a repeated pole and the fastest possible non-oscillatory response — critically damped. Δ < 0 gives a conjugate pair and ringing — underdamped.

Two theorems let you check an answer without inverting anything. The initial value theorem says f(0⁺) = lim s·F(s) as s → ∞, and the final value theorem says lim f(t) as t → ∞ equals lim s·F(s) as s → 0 — but the final value theorem is valid only when every pole of s·F(s) lies strictly in the left half-plane. Applied to an unstable or purely oscillatory transform it returns a confident and completely wrong number, which makes it the most commonly misused result in the subject.

The standard one-sided transform table

Verify any row against the definition if you doubt it: for f(t) = e^(at), the integral of e^(−st)e^(at) from 0 to ∞ is 1/(s − a) whenever Re(s) > a, and diverges otherwise.

How to solve an ODE with the transform

1. Transform every term

   Apply L to both sides. Use L{y′} = sY(s) − y(0) and L{y″} = s²Y(s) − s·y(0) − y′(0). The initial conditions enter here, not later.

2. Solve algebraically for Y(s)

   Collect the Y(s) terms on one side and divide. You now have a rational function whose denominator is the characteristic polynomial of the original equation.

3. Decompose into table pieces

   Split Y(s) by partial fractions. Compute the discriminant of the quadratic factor first — it tells you whether to expect exponentials, a repeated root with a factor of t, or a decaying sinusoid.

4. Invert term by term

   Match each piece to a table entry and add the results. Linearity means you never need to invert the sum as a whole.

5. Check at t = 0

   Your answer must reproduce the initial conditions you fed in. Evaluating y(0) and y′(0) from the final expression catches most algebra slips immediately.

Mistakes that produce a wrong transform

• Forgetting to normalise the denominator. This calculator assumes the s² coefficient is exactly 1. If your F(s) is 15/(2s² + 8s + 26), divide numerator and denominator by 2 first.
• Using the wrong numerator for a sine. The sine pair needs ω on top, not 1. Inverting 1/((s+2)²+9) gives ⅓e^(−2t) sin 3t, not e^(−2t) sin 3t.
• Applying the shift to only part of the denominator. The first shifting rule replaces every s in F(s), so ω/(s²+ω²) becomes ω/((s−a)²+ω²) — the ω² term is untouched but the s must move.
• Using the final value theorem on an unstable transform. It requires all poles of s·F(s) in the open left half-plane. Applied to 1/(s−1) it returns 0, while f(t) = e^t is unbounded.
• Dropping the initial conditions. The −f(0) in the derivative rule is the entire reason the method handles IVPs. Omitting it silently solves a different problem.
• Ignoring the region of convergence. Two different time functions can share the same algebraic F(s) with different convergence regions; for the one-sided transform used here, causality resolves the ambiguity.
• Mixing the one-sided and two-sided transforms. Tables for the bilateral transform look similar but carry different convergence strips and no initial-condition terms.

Where the Laplace transform sits among related tools

The Laplace transform is one of a family. Set s = iω and restrict to signals defined on the whole line and you have the Fourier transform, which trades the ability to handle growth and initial conditions for a cleaner interpretation as a frequency spectrum. Replace the integral by a sum over integer times and you get the z-transform, the discrete-time equivalent used for digital filters and difference equations. The algebra of poles is the same in all three.

Within differential equations, reach for Laplace when the equation is linear with constant coefficients and you have initial conditions at t = 0, especially with a discontinuous or impulsive forcing term. For a homogeneous constant-coefficient equation with no forcing, the characteristic equation is quicker and gives the same answer. For a first-order equation with variable coefficients, the transform does not help at all and you want an integrating factor. For a nonlinear equation the transform simply does not apply — the method depends on linearity at every step — and you should be marching numerically with RK4 or Euler's method.

One limitation of this calculator is worth stating plainly: the inverse mode handles a numerator of degree at most one over a monic quadratic. That covers essentially every second-order system — RLC circuits, mass-spring-damper models, PID loops with second-order plants — but higher-order transfer functions must be split into quadratic and linear factors by hand first, then inverted one factor at a time and added by linearity.

Key terms

Pole

A value of s where F(s) becomes infinite. Poles determine the form of f(t): each one contributes an exponential whose rate is the pole's real part and whose frequency is its imaginary part.

Convergence abscissa

The real number σ such that the defining integral converges exactly for Re(s) > σ. It equals the real part of the rightmost pole for a rational transform.

Residue

The coefficient of a simple partial-fraction term. For F(s) with a simple pole at r, the residue is the limit of (s − r)F(s) as s approaches r.

Causal signal

One that is zero for all t < 0. The one-sided transform only ever sees, and only ever reconstructs, the causal part of a function.

Unit step u(t − c)

The Heaviside function: zero before time c, one after. It is the standard way to write a switch turning on inside a forcing term.

Transfer function

The ratio of output transform to input transform for a linear system at zero initial conditions. Its poles are the system's natural modes.