Inverse Laplace Transform Calculator

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To invert a Laplace transform, split F(s) into partial fractions and read each piece from a table. For example, \(\frac{s+1}{s^{2}+4s+13} = \frac{(s+2) – 1}{(s+2)^{2}+9}\), so \(f(t) = e^{-2t}\cos 3t – \frac{1}{3}e^{-2t}\sin 3t\).

FreeStep by stepExact answers

\(F(s) = \frac{s + 1}{s^{2} + 4 s + 13}\)A ratio of polynomials in s. You may multiply by e^(-as) for a delayed signal.

Try:
Inverse Laplace transform
\(\displaystyle f(t) = e^{-2t} \cos\left(3t\right) – \frac{1}{3} e^{-2t} \sin\left(3t\right)\)
f(t) = ℒ⁻¹{F(s)} for t ≥ 0.
tf(t)12345678910−0.20−0.15−0.10−0.050.050.100.150.20

Graph of f(t) for 0 ≤ t ≤ 10.

Partial fractions \(\frac{s + 1}{s^{2} + 4s + 13}\)
Check Taking the Laplace transform of the answer gives back F(s) ✓
Show steps
  1. Start with \(F(s) = \frac{s + 1}{s^{2} + 4s + 13}\).
  2. Factor the denominator: \(s^{2} + 4s + 13 = \left(s^{2} + 4s + 13\right)\)
  3. Split into partial fractions: \(F(s) = \frac{s + 1}{s^{2} + 4s + 13}\)
  4. For quadratic factors, complete the square so each piece matches \((s-\alpha)^{2}+\omega^{2}\).
  5. Use the table entry \(\mathcal{L}^{-1}\left\{\frac{s-\alpha}{(s-\alpha)^{2}+\omega^{2}}\right\} = e^{\alpha t}\cos(\omega t),\quad \mathcal{L}^{-1}\left\{\frac{\omega}{(s-\alpha)^{2}+\omega^{2}}\right\} = e^{\alpha t}\sin(\omega t)\).
  6. Result: \(f(t) = e^{-2t} \cos\left(3t\right) – \frac{1}{3} e^{-2t} \sin\left(3t\right)\)

How to use the inverse Laplace calculator

  1. Enter F(s) as a fraction of polynomials in s, for example (3s + 5)/((s + 1)(s + 2)).
  2. Press Invert. The calculator factors the denominator, finds the partial fraction coefficients exactly, and converts each term with a table rule.
  3. The answer is checked by taking its Laplace transform numerically and comparing with F(s).

Method: partial fractions

  1. Factor the denominator into linear factors \((s – a)\) and irreducible quadratics \(s^{2} + ps + q\).
  2. Write \(F(s)\) as a sum of terms \(\frac{A}{(s-a)^{k}}\) and \(\frac{Bs + C}{(s^{2}+ps+q)^{k}}\), and solve for the constants.
  3. For quadratics, complete the square: \(s^{2} + ps + q = (s – \alpha)^{2} + \omega^{2}\).
  4. Use the table entries below.

Inverse transform table

F(s) f(t)
\(\frac{1}{s – a}\) \(e^{at}\)
\(\frac{1}{(s-a)^{n}}\) \(\frac{t^{n-1}e^{at}}{(n-1)!}\)
\(\frac{s-\alpha}{(s-\alpha)^{2}+\omega^{2}}\) \(e^{\alpha t}\cos(\omega t)\)
\(\frac{\omega}{(s-\alpha)^{2}+\omega^{2}}\) \(e^{\alpha t}\sin(\omega t)\)
\(\frac{s}{(s^{2}+\omega^{2})^{2}}\) \(\frac{t\sin(\omega t)}{2\omega}\)
\(\frac{1}{(s^{2}+\omega^{2})^{2}}\) \(\frac{\sin(\omega t) – \omega t\cos(\omega t)}{2\omega^{3}}\)
\(e^{-cs}F(s)\) \(u(t-c)\,f(t-c)\)

Worked example

\(F(s) = \frac{3s+5}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}\). Setting \(s = -1\) gives \(A = 2\); setting \(s = -2\) gives \(B = 1\). So \(f(t) = 2e^{-t} + e^{-2t}\).

Tips

  • If the numerator’s degree is at least the denominator’s, divide first; the polynomial part gives impulse terms δ(t).
  • Complex roots always produce sines and cosines; real roots produce exponentials.
  • Repeated roots add powers of t.

Inverse Laplace transform practice quiz: 10 questions

Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.

Question 1 of 10
1. ℒ⁻¹{1/(s − 3)} = ?

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2. ℒ⁻¹{2/s³} = ?

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3. ℒ⁻¹{s/(s² + 9)} = ?

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4. ℒ⁻¹{4/(s² + 16)} = ?

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5. 1/((s + 1)(s + 2)) splits as…

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6. ℒ⁻¹{1/(s + 2)²} = ?

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7. Complete the square: s² + 6s + 13 = ?

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8. ℒ⁻¹{e−2s/s} = ?

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9. Complex roots in the denominator lead to…

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10. ℒ⁻¹{(s + 1)/((s + 1)² + 4)} = ?

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Frequently asked questions

How do you find an inverse Laplace transform?

Write F(s) as partial fractions, complete the square for quadratic factors, and match each term with a table entry such as 1/(s − a) ↔ e^(at).

What kinds of F(s) can this calculator invert?

Any ratio of polynomials whose denominator factors into linear and quadratic pieces with rational coefficients, including repeated factors, plus an optional e^(−as) delay.

What does e^(−as) do?

It delays the signal: the inverse of e^(−as)F(s) is u(t − a)·f(t − a), where u is the unit step.

Why do I see δ(t) in the answer?

When the numerator’s degree is not lower than the denominator’s, the polynomial part of F(s) corresponds to impulse functions.

How is the result verified?

The calculator numerically computes the Laplace transform of its answer at two values of s and checks that it equals F(s).

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