Inverse Tangent Calculator
EN– Change language
New calculation

Inverse Laplace Transform Calculator

An inverse Laplace transform calculator turns F(s) into the time function f(t). Type a fraction in s to get f(t) from partial fractions, with every step, the poles, the initial and final values, a graph of f(t) and a pole-zero map. It handles repeated poles, complex poles and delays e^(−as).

What is the Inverse Laplace Transform?

The inverse Laplace transform turns a function F(s) back into the time function f(t) with 𝓛{f(t)} = F(s). For F(s) = 1/(s + 2), f(t) = e⁻²ᵗ.

f(t) = 𝓛⁻¹{F(s)}, F(s) = ∫₀^∞ f(t) e^(−st) dt
  • The forward transform turns a differential equation in t into algebra in s; the inverse brings the answer back to t.
  • Most F(s) in engineering are fractions of polynomials. Their inverse comes from partial fractions and a short table.
  • The poles of F(s), the roots of its denominator, set the shape of f(t): a real pole a gives eᵃᵗ, and a pair α ± βi gives e^(αt)·cos(βt) and e^(αt)·sin(βt).
  • Poles with a negative real part give terms that die out. Poles with a positive real part give terms that grow.

Poles decide the shape

Move the pole pair α ± βi. The top plot is the s-plane; the bottom plot is f(t).

How to Find the Inverse Laplace Transform?

Factor the denominator, split F(s) into partial fractions, and invert each fraction with the Laplace table. For F(s) = (3s + 5)/((s + 1)(s + 2)):

  1. Write F(s) = A/(s + 1) + B/(s + 2).
  2. Cover-up at s = −1: A = (3(−1) + 5)/(−1 + 2) = 2.
  3. Cover-up at s = −2: B = (3(−2) + 5)/(−2 + 1) = 1.
  4. Invert with 1/(s + a) → e^(−at): f(t) = 2e⁻ᵗ + e⁻²ᵗ.

Check at t = 0: f(0) = 2 + 1 = 3, and the initial value theorem gives lim sF(s) = 3 as s → ∞.

Partial fractions, step by step

Type F(s) in terms of s.

How to Use This Inverse Laplace Calculator?

Type F(s) and the calculator returns f(t) with the poles, the partial fractions and every step. The steps:

  1. Choose Find f(t) for the formula, f(t) at a time for one value, or Partial fractions for the split alone.
  2. Type F(s) with s as the variable: 1/(s + 2), (3s + 5)/(s^2 + 4s + 13) or e^(-2s)/s.
  3. Read f(t), the initial and final values, the poles and the stability.
  4. Open the graph of f(t) or the pole-zero map, and the steps under the result.

The calculator handles any fraction of polynomials with a denominator of degree 20 or less, repeated and complex poles, and delays e^(−as).

Try an example

Each button fills the calculator above and scrolls to the result.

Inverse Laplace Transform Table

The table lists the standard Laplace pairs that every partial-fraction answer reduces to. Click a row to load F(s) into the calculator.

  • 1/s → 1, and n!/sⁿ⁺¹ → tⁿ.
  • 1/(s − a) → eᵃᵗ, and 1/(s − a)ⁿ → tⁿ⁻¹eᵃᵗ/(n − 1)!.
  • ω/(s² + ω²) → sin(ωt), and s/(s² + ω²) → cos(ωt).
  • Replacing s by s − a multiplies f(t) by eᵃᵗ, which turns sin and cos into damped waves.
  • e^(−as) shifts f(t) right by a and switches it on at t = a.

Laplace pairs

Filter by value, or click a row to send it to the calculator.

F(s)f(t)
1δ(t)
e^(−2s)δ(t − 2)
1/s1
1/s²t
2/s³t²
1/(s+2)e⁻²ᵗ
1/(s+1)²te⁻ᵗ
3/(s²+9)sin(3t)
s/(s²+9)cos(3t)
2/((s+1)²+4)e⁻ᵗsin(2t)
(s+1)/((s+1)²+4)e⁻ᵗcos(2t)
2/(s²−4)sinh(2t)
s/(s²−4)cosh(2t)
2s/(s²+1)²t·sin(t)
1/(s(s+1))1 − e⁻ᵗ
e^(−2s)/su(t − 2)

Inverse Laplace with Repeated Poles

A pole repeated n times needs n fractions, A₁/(s − a) + A₂/(s − a)² + … + Aₙ/(s − a)ⁿ, and each one inverts to tᵏ⁻¹eᵃᵗ/(k − 1)!.

For F(s) = (s + 3)/(s + 1)³:

  1. Write s + 3 = (s + 1) + 2.
  2. Split: F(s) = 1/(s + 1)² + 2/(s + 1)³.
  3. Invert with 1/(s + 1)ⁿ → tⁿ⁻¹e⁻ᵗ/(n − 1)!: f(t) = te⁻ᵗ + t²e⁻ᵗ.

The general coefficient formula differentiates (s − a)ⁿF(s): A_(n−k) = 1/k! · dᵏ/dsᵏ[(s − a)ⁿF(s)] at s = a.

Repeated poles

Type F(s) with a repeated factor, such as (s + 2)^2.

Inverse Laplace with Complex Poles

For a quadratic factor with no real roots, complete the square to (s − α)² + β², then use the damped cosine and sine pairs.

(s − α)/((s − α)² + β²) → e^(αt)·cos(βt)

For F(s) = (3s + 5)/(s² + 4s + 13):

  1. Complete the square: s² + 4s + 13 = (s + 2)² + 3².
  2. Rewrite the top: 3s + 5 = 3(s + 2) − 1.
  3. Split: F(s) = 3·(s + 2)/((s + 2)² + 9) − (1/3)·3/((s + 2)² + 9).
  4. Invert: f(t) = e⁻²ᵗ(3cos(3t) − (1/3)sin(3t)).

The poles are −2 ± 3i: the real part −2 sets the decay rate, and the imaginary part 3 sets the angular frequency.

Complex poles

Type F(s) with a quadratic factor, such as s^2 + 2s + 5.

Laplace Transform Properties for the Inverse

6 properties cover almost every inverse: linearity, the s-shift, the t-shift, division by s, products and the derivative in s.

  • Linearity: invert a sum term by term, and keep the constants.
  • s-shift: F(s − a) → eᵃᵗf(t). 1/(s + 2)² → te⁻²ᵗ.
  • t-shift: e^(−as)F(s) → u(t − a)f(t − a). e^(−3s)/s² → u(t − 3)(t − 3).
  • Division by s: F(s)/s → the integral of f from 0 to t. 1/(s(s + 1)) → 1 − e⁻ᵗ.
  • Convolution: F(s)G(s) → (f ∗ g)(t), the integral of f(τ)g(t − τ).
  • Derivative in s: F′(s) → −t·f(t).

Each rule, checked by the calculator

Pick a property. The f(t) line comes from the calculator, not from a lookup.

Time Delays and Step Functions

A factor e^(−as) in F(s) delays the matching part of f(t) by a and switches it on with the unit step u(t − a).

𝓛⁻¹{e^(−as)G(s)} = u(t − a)·g(t − a)

For F(s) = e^(−2s)/(s(s + 1)):

  1. Invert G(s) = 1/(s(s + 1)) = 1/s − 1/(s + 1): g(t) = 1 − e⁻ᵗ.
  2. Shift by 2: f(t) = u(t − 2)·[1 − e^(−(t − 2))].

f(t) is 0 until t = 2, then rises towards 1. A pulse of height 1 from t = 0 to t = 1 is (1 − e^(−s))/s.

Delays

Type F(s) with a delay, such as e^(-2s)/(s + 1).

Initial and Final Value Theorems

f(0⁺) = lim sF(s) as s → ∞, and f(∞) = lim sF(s) as s → 0, without inverting F(s).

  • For F(s) = 10/(s(s² + 2s + 10)): sF(s) = 10/(s² + 2s + 10). As s → ∞ it goes to 0, so f(0⁺) = 0. At s = 0 it is 10/10 = 1, so f(∞) = 1.
  • The final value theorem holds only when every pole of sF(s) has a negative real part. For 1/(s² + 1), f(t) = sin(t) keeps oscillating and has no final value, though s · 1/(s² + 1) → 0.
  • The initial value theorem needs F(s) to be strictly proper; s/(s + 1) contains a δ(t) at t = 0.

Theorems against the real f(t)

Type F(s). The last line evaluates f(t) at a late time to confirm f(∞).

Solving Differential Equations with the Inverse Laplace Transform

Transform the equation, solve for Y(s) with algebra, and invert Y(s) to get y(t). For y″ + 3y′ + 2y = 0 with y(0) = 1 and y′(0) = 0:

  1. Transform with 𝓛{y′} = sY − y(0) and 𝓛{y″} = s²Y − sy(0) − y′(0): s²Y − s + 3(sY − 1) + 2Y = 0.
  2. Solve: Y(s) = (s + 3)/(s² + 3s + 2) = (s + 3)/((s + 1)(s + 2)).
  3. Partial fractions: Y(s) = 2/(s + 1) − 1/(s + 2).
  4. Invert: y(t) = 2e⁻ᵗ − e⁻²ᵗ. Check: y(0) = 2 − 1 = 1.

A constant input c on the right side adds c/s to the top of Y(s), which adds a constant c/b to y(t) once the transients die out.

Solve y″ + ay′ + by = c

Type the coefficients and the starting values.

Inverse Laplace in MATLAB, SymPy and Wolfram

MATLAB has ilaplace, SymPy has inverse_laplace_transform and Wolfram Language has InverseLaplaceTransform.

  • MATLAB: syms s t; ilaplace((3*s + 5)/((s + 1)*(s + 2))) returns 2*exp(-t) + exp(-2*t).
  • SymPy: inverse_laplace_transform((3*s + 5)/((s + 1)*(s + 2)), s, t) returns the same answer times Heaviside(t).
  • Wolfram Language: InverseLaplaceTransform[(3 s + 5)/((s + 1) (s + 2)), s, t].
  • Maxima: ilt((3*s + 5)/((s + 1)*(s + 2)), s, t).
  • Python without SymPy: scipy.signal.residue(b, a) returns the partial-fraction coefficients and poles.

MATLAB, SymPy and Maxima need * between a number and s, such as 3*s.

The same inverse in software

Type F(s), then pick a tool.

Inverse Laplace Transforms in Real Life

Engineers use the inverse Laplace transform to turn a transfer function into a time response: how a circuit, a spring or a control loop moves after a change.

  • Spring–damper and RLC circuits: the step response of ω²/(s² + 2ζωs + ω²) overshoots when the damping ratio ζ is below 1 and creeps up without overshoot above 1.
  • Control systems: the poles of the closed loop set the settling time, about 4/|Re p| for the slowest pole.
  • Heat and drug models: 1/(s(τs + 1)) gives 1 − e^(−t/τ), the rise of a temperature or a blood level towards its final value.
  • Signal processing: the impulse response h(t) = 𝓛⁻¹{H(s)} describes a filter completely.

Step response of a damped system

Move ζ and ω. The dashed line is the final value 1.

Inverse Laplace FAQ

What is the inverse Laplace transform?

The operation that turns F(s) back into the time function f(t) with 𝓛{f(t)} = F(s). For F(s) = 1/(s + 2), f(t) = e⁻²ᵗ.

How do you find the inverse Laplace transform?

Factor the denominator, split F(s) into partial fractions, and invert each fraction with the Laplace table.

What is the inverse Laplace transform of 1/s?

1, the unit step u(t), for t ≥ 0.

What is the inverse Laplace transform of 1?

The Dirac delta δ(t), an impulse at t = 0.

What is the inverse Laplace transform of 1/(s + a)?

e^(−at). For a = 2, it is e⁻²ᵗ.

What is the inverse Laplace transform of 1/s²?

t. In general, n!/sⁿ⁺¹ inverts to tⁿ.

What is the inverse Laplace transform of 1/(s² + a²)?

sin(at)/a. For a = 3, 1/(s² + 9) inverts to sin(3t)/3.

What is the inverse Laplace transform of s/(s² + a²)?

cos(at).

How do you handle complex poles?

Complete the square to (s − α)² + β² and use e^(αt)cos(βt) and e^(αt)sin(βt). (3s + 5)/(s² + 4s + 13) gives e⁻²ᵗ(3cos(3t) − (1/3)sin(3t)).

What does e^(−as) do in F(s)?

It delays f(t) by a: e^(−as)G(s) inverts to u(t − a)g(t − a). e^(−2s)/s inverts to u(t − 2).

What if the numerator degree is not lower than the denominator degree?

Divide first. The quotient gives δ(t) and its derivatives: s²/(s² + 1) = 1 − 1/(s² + 1) inverts to δ(t) − sin(t).

What is the final value theorem?

f(∞) = lim sF(s) as s → 0, valid when every pole of sF(s) has a negative real part. 10/(s(s² + 2s + 10)) settles at 1.

Is the inverse Laplace transform unique?

Yes, for continuous f(t) on t ≥ 0. Functions that differ only at isolated points share the same F(s).

How do I find the inverse Laplace transform in MATLAB?

syms s t; ilaplace(F). For F = 1/(s + 2), ilaplace returns exp(−2t).

Can the calculator handle repeated roots?

Yes. (s + 3)/(s + 1)³ gives te⁻ᵗ + t²e⁻ᵗ, with the coefficient of each fraction shown in the steps.