About the Laplace Transform Calculator
This Laplace transform calculator works in both directions. In forward mode, type a function of t such as t^2 + 3sin(2t), e^(−2t)cos(3t) or t·sin(t) and it splits it into standard pieces — powers of t, exponentials, sines, cosines, sinh and cosh, and their products — and applies the table of transforms and the shift and multiplication-by-tⁿ theorems to return F(s), the region of convergence and a numerical check of the integral.
In inverse mode, type a rational function of s such as (s + 3)/(s^2 + 4s + 13) or 1/(s(s + 1)^2). The calculator finds the poles of the denominator, computes the partial-fraction (residue) coefficients — including repeated and complex poles — and writes f(t) as a combination of exponentials, powers of t, sines and cosines.
It is built for differential equations, signals and control-systems students who need to check a transform from a table, and for engineers solving circuit or vibration problems. Inverse transforms are computed numerically from the poles, so coefficients are shown as fractions when they are simple and as decimals otherwise. Use t as the variable in forward mode and s in inverse mode.
How to use the laplace transform calculator
- 1Choose Laplace transform (f(t) → F(s)) or inverse (F(s) → f(t)).
- 2Type f(t) using t, or F(s) as a fraction of polynomials in s.
- 3Read the result and the region of convergence or list of poles.
- 4Enter a value of s or t to evaluate the result there.
- 5In forward mode, compare the value with the numerical integral to confirm it.
Formula and method
The Laplace transform turns a function of time into a function of the complex frequency s. Because it is linear, a sum is transformed term by term. The core table entries are L{1} = 1/s, L{tⁿ} = n!/sⁿ⁺¹, L{sin bt} = b/(s² + b²), L{cos bt} = s/(s² + b²), L{sinh bt} = b/(s² − b²) and L{cosh bt} = s/(s² − b²). Multiplying by e^(at) shifts s to s − a (first shifting theorem), and multiplying by tⁿ gives (−1)ⁿ times the n-th derivative of F(s). Products of sines and cosines are first rewritten with product-to-sum identities.
For the inverse transform, a proper rational function N(s)/D(s) is expanded in partial fractions. Each pole p of multiplicity m contributes Σ Aⱼ tʲ⁻¹ e^(pt)/(j − 1)!, where the coefficients come from the Taylor expansion of (s − p)ᵐ F(s) at p (the residue method). A pair of complex poles α ± βi combines into e^(αt)(C cos βt + D sin βt).
- s
- Complex frequency variable
- a
- Exponential rate in e^(at) — shifts s to s − a
- b
- Angular frequency of sin/cos (rad per unit t)
- Re(s) > σ
- Region of convergence where the integral exists
Worked examples
L{t² + 3 sin 2t}
L{t²} = 2!/s³ = 2/s³ and L{sin 2t} = 2/(s² + 4), so 3 sin 2t gives 6/(s² + 4). At s = 1: 2 + 6/5 = 3.2, matching the numerical integral.
Shifted cosine: L{e^(−2t) cos 3t}
L{cos 3t} = s/(s² + 9); multiplying by e^(−2t) replaces s with s + 2, giving (s + 2)/((s + 2)² + 9). At s = 1 that is 3/18 ≈ 0.1667.
Multiplication by t: L{t sin t}
L{t·f(t)} = −F′(s). With F(s) = 1/(s² + 1), −F′(s) = 2s/(s² + 1)², which equals 4/25 = 0.16 at s = 2.
Inverse with complex poles: (s + 3)/(s² + 4s + 13)
Completing the square gives ((s + 2) + 1)/((s + 2)² + 9), so f(t) = e^(−2t)(cos 3t + ⅓ sin 3t). The poles are −2 ± 3i and f(0.5) = e^(−1)(cos 1.5 + ⅓ sin 1.5) ≈ 0.1483.
Inverse with a repeated pole: 1/(s(s + 1)²)
Partial fractions give 1/s − 1/(s + 1) − 1/(s + 1)², so f(t) = 1 − e^(−t) − t e^(−t). At t = 1 that is 1 − 2/e ≈ 0.2642.
Inverse of 1/(s² + 4)
Since L{sin 2t} = 2/(s² + 4), the inverse of 1/(s² + 4) is ½ sin 2t, which equals 0.5 at t = π/4.
Frequently asked questions
What is the Laplace transform used for?+
It converts linear differential equations with initial conditions into algebraic equations in s, which are easier to solve; the inverse transform then gives the solution in time. It is standard in circuit analysis, control systems, vibrations and signal processing.
What is the Laplace transform of 1, t and tⁿ?+
L{1} = 1/s, L{t} = 1/s² and in general L{tⁿ} = n!/sⁿ⁺¹ for whole numbers n, valid for s > 0.
How do you find an inverse Laplace transform?+
Write F(s) as partial fractions, then match each piece with a table entry: A/(s − a) → A e^(at), A/(s − a)ᵏ → A tᵏ⁻¹ e^(at)/(k − 1)!, and quadratic factors (s − α)² + β² → e^(αt) times sines and cosines.
What is the first shifting theorem?+
If L{f(t)} = F(s), then L{e^(at) f(t)} = F(s − a). For example, L{e^(−2t) sin 3t} = 3/((s + 2)² + 9).
What is the region of convergence?+
It is the set of s values for which the defining integral converges. For f(t) built from e^(at) terms, it is Re(s) > the largest a; for e^(at) times sinh or cosh of bt it is Re(s) > a + |b|.
Why does the calculator say F(s) is improper?+
If the numerator degree is at least the denominator degree, F(s) does not go to 0 as s → ∞, and its inverse contains impulse terms such as δ(t). Divide the polynomials first; the proper remainder can then be inverted.