Laplace Transform Calculator
Interactive Laplace transform calculator for common functions: exponentials, powers, sine, cosine, damped sinusoids, step shifts and impulses. See formulas in LaTeX, region of convergence (ROC) and numeric evaluation.
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Laplace Transform Calculator
Compute Laplace transforms for standard time-domain functions, see the formulas and region of convergence (ROC), and get numeric values at a chosen real s0.
Interactive Laplace transform tool
Choose a standard time-domain signal \(f(t)\), set its parameters, and let the calculator produce the symbolic Laplace transform \(F(s)\), with a short explanation and region of convergence. Optionally, evaluate \(F(s)\) at a specific real value \(s_0\).
All signals are assumed causal (zero for \(t < 0\)) unless explicitly shifted by \(a_0\).
If provided and within the ROC, the tool evaluates \(F(s_0)\) numerically.
Real gain (can be negative).
Real \(a\). For a stable decaying exponential, use \(a < 0\).
\(f(t) = K e^{a t} u(t)\) → \(F(s) = \dfrac{K}{s - a}\), ROC: \(\mathrm{Re}(s) > \mathrm{Re}(a)\).
Definition of the Laplace transform
The unilateral Laplace transform of a time-domain function \(f(t)\), assumed zero for \(t < 0\), is defined as
\(\displaystyle \mathcal{L}\{f(t)\}(s) = F(s) = \int_{0^-}^{\infty} f(t)\,e^{-st}\,\mathrm{d}t\)
Here \(s\) is a complex variable, \(s = \sigma + j\omega\). For some functions the integral converges only for a subset of the complex plane, called the region of convergence (ROC).
In control, circuits, and mechanical systems, the Laplace transform is a standard tool to turn differential equations into simple algebra in the \(s\)-domain, where transfer functions and poles/zeros are easier to interpret.
Standard Laplace transform pairs
The table below collects a few core Laplace transform pairs used most often in introductory courses, signal processing, and control engineering.
| \(f(t)\) | \(F(s) = \mathcal{L}\{f(t)\}\) | ROC |
|---|---|---|
| \(u(t)\) | \(\dfrac{1}{s}\) | \(\mathrm{Re}(s) > 0\) |
| \(e^{at} u(t)\) | \(\dfrac{1}{s - a}\) | \(\mathrm{Re}(s) > \mathrm{Re}(a)\) |
| \(t^n u(t)\), \(n \in \mathbb{N}_0\) | \(\dfrac{n!}{s^{n+1}}\) | \(\mathrm{Re}(s) > 0\) |
| \(\sin(bt) u(t)\) | \(\dfrac{b}{s^2 + b^2}\) | \(\mathrm{Re}(s) > 0\) |
| \(\cos(bt) u(t)\) | \(\dfrac{s}{s^2 + b^2}\) | \(\mathrm{Re}(s) > 0\) |
| \(e^{at}\sin(bt) u(t)\) | \(\dfrac{b}{(s-a)^2 + b^2}\) | \(\mathrm{Re}(s) > \mathrm{Re}(a)\) |
| \(e^{at}\cos(bt) u(t)\) | \(\dfrac{s-a}{(s-a)^2 + b^2}\) | \(\mathrm{Re}(s) > \mathrm{Re}(a)\) |
| \(u(t-a)f(t-a)\) | \(e^{-as} F(s)\) | Same ROC as \(F(s)\) |
| \(\delta(t-a)\) | \(e^{-as}\) | All \(s\) |
ROC and stability intuition
The region of convergence encodes how quickly the signal decays (or grows) and determines whether the Laplace transform exists. For example, a causal exponential \(e^{at}u(t)\) only has a finite transform if \(e^{(a-\sigma)t}\) decays as \(t \rightarrow \infty\), which requires \(\mathrm{Re}(s) = \sigma > \mathrm{Re}(a)\).
In systems theory, the ROC relates directly to stability: if all poles of a causal LTI system lie in the left half-plane (\(\mathrm{Re}(s) < 0\)), then the ROC includes the imaginary axis and the system is BIBO stable.
Laplace transform – FAQ
Why is the Laplace transform so useful for differential equations?
The Laplace transform converts differentiation into multiplication by \(s\): \(\mathcal{L}\{f'(t)\} = sF(s) - f(0^-)\). Higher-order derivatives lead to higher powers of \(s\). This turns linear constant-coefficient differential equations into algebraic equations in \(s\), which are much easier to solve; you then apply the inverse transform to return to \(f(t)\).
What is the difference between unilateral and bilateral Laplace transforms?
The unilateral (one-sided) Laplace transform integrates from \(0^-\) to \(\infty\) and is tailored to causal signals and initial-value problems. The bilateral (two-sided) transform integrates from \(-\infty\) to \(\infty\) and is useful when signals have non-zero values for negative time. This calculator follows the common unilateral convention.
Can I combine results for sums of signals?
Yes. The Laplace transform is linear:
\(\mathcal{L}\{a f(t) + b g(t)\} = a F(s) + b G(s)\).
You can decompose a complicated \(f(t)\) into a sum of standard building blocks, apply the table or the calculator to each term, then add the corresponding \(F(s)\) expressions.
How accurate are the numeric evaluations?
The numeric values use standard double-precision floating point, comparable to scientific calculators and engineering software. For extreme parameter values (very large or small magnitudes) rounding and overflow may occur; these limitations are inherent to finite-precision arithmetic rather than the transform formulas themselves.
Frequently Asked Questions
Do I always need the Laplace transform, or is the Fourier transform enough?
The Fourier transform assumes signals are sufficiently well-behaved over all time and often focuses on steady-state sinusoidal behaviour. The Laplace transform generalizes Fourier by including exponential growth/decay and transient behaviour through the real part of s. In control and transient analysis, the Laplace transform is usually the more natural tool.
Does this calculator compute inverse Laplace transforms?
This version focuses on forward Laplace transforms (from \(f(t)\) to \(F(s)\)) with clear ROC descriptions. For many engineering problems, forward transforms combined with partial fraction decompositions and standard tables are sufficient. A future update may include a guided inverse-transform helper for simple rational \(F(s)\).
How should I document Laplace transform steps in reports or lab work?
When you use transform methods in formal work, always show the original differential equation, the initial conditions, the expression for \(F(s)\), and the corresponding inverse transform step. This calculator can help you verify each transform pair numerically, but your report should still include the full reasoning and references to standard tables.
Formula (LaTeX) + variables + units
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\(\displaystyle \mathcal{L}\{f(t)\}(s) = F(s) = \int_{0^-}^{\infty} f(t)\,e^{-st}\,\mathrm{d}t\)
\(\mathcal{L}\{a f(t) + b g(t)\} = a F(s) + b G(s)\).
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Last code update: 2026-01-19
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