Inverse Laplace transform calculator (partial fractions)
Enter your values. The answer and the full worked solution update as you type.
Result
Find the inverse Laplace transform of a rational function F(s) = N(s)/D(s). Enter the coefficients of the numerator and the denominator, highest power of s first. The calculator cancels common factors, divides out an improper fraction, finds the poles, computes every partial fraction coefficient (for distinct, repeated and complex poles) and writes f(t) as a sum of exponentials, powers of t, sines and cosines.
Step-by-step solution
Write F(s) as a ratio of polynomials
Poles: roots of the denominator
Partial fractions
Transform pairs used
Inverse Laplace transform
How to find an inverse Laplace transform with partial fractions
- Make sure F(s) is proper: if the degree of the numerator is not lower than that of the denominator, divide first. The polynomial part gives impulses δ(t), δ′(t), and so on.
- Factor the denominator to find the poles: a real pole a gives a factor (s − a), a complex pair a ± bj gives (s − a)² + b².
- Write F(s) as a sum of partial fractions and find the coefficients. For a simple pole, A = [(s − a)·F(s)] at s = a (the cover-up rule); repeated poles need derivatives.
- Transform each term with the table: 1/(s − a) ↔ e^(at), 1/(s − a)ⁿ ↔ tⁿ⁻¹·e^(at)/(n − 1)! and (s − a)/((s − a)² + b²) ↔ e^(at)·cos bt.
- Add the terms to get f(t) for t ≥ 0.
Questions and answers
How do I enter F(s)?
Type the coefficients of each polynomial from the highest power of s down, separated by spaces, with zeros for missing powers. (s + 3)/(s² + 3s + 2) is numerator 1 3 and denominator 1 3 2; 1/(s³ + 4s) is numerator 1 and denominator 1 0 4 0. Fractions such as 1/2 and decimals are allowed.
What is the cover-up method?
For a simple pole at s = a, multiply F(s) by (s − a), cancel, and substitute s = a. For F(s) = (s + 3)/((s + 1)(s + 2)) that gives 2/1 = 2 at s = −1 and 1/(−1) = −1 at s = −2, so f(t) = 2e^(−t) − e^(−2t).
How are complex poles handled?
A pair of poles a ± bj gives a term (B(s − a) + C·b)/((s − a)² + b²), whose inverse is e^(at)·(B cos bt + C sin bt). The calculator finds B and C from the complex coefficient A at a + bj: B = 2·Re A and C = −2·Im A.
What if the degree of the numerator is too high?
Then F(s) is improper. Polynomial division gives F(s) = Q(s) + R(s)/D(s). The polynomial Q(s) transforms into impulses: a constant c gives c·δ(t), c·s gives c·δ′(t), and so on. The proper part R(s)/D(s) is expanded in partial fractions as usual.
How exact are the results?
Common factors and repeated factors are found exactly, with fractions. The roots and coefficients are then computed numerically to about 14 digits and shown to 4 significant figures, so irrational poles, such as those of s² + s + 1, appear as decimals.
Why is f(t) only valid for t ≥ 0?
The one-sided Laplace transform only uses the values of f for t ≥ 0. Strictly, the result is f(t)·u(t), with u the unit step, so f(t) = 0 for t < 0.
Practise it yourself
Practise Laplace transforms
A calculator gives the answer. In the exam you need the method: practise it with new problems, hints and worked solutions.