Inverse Laplace Transform Calculator

How to use this calculator

1. Enter coefficients, signs includedFor (s + 4)/(s² + 5s + 6) enter p = 1, q = 4, a = 1, b = 5, c = 6. A missing s term in the numerator means p = 0.
2. Set a = 0 for a first-order transformThe calculator handles a linear denominator too, giving a single exponential.
3. Read the case before the answerDistinct, repeated or complex poles determine everything about the response. The discriminant row tells you which.
4. Check f(0)The initial value theorem gives f(0) as the limit of s·F(s), which is p ÷ a. The breakdown shows both so you can verify the decomposition.

Poles decide everything

The roots of the denominator are called poles, and they determine the entire character of the time-domain response. The numerator affects only the size of each component, never its shape.

A real pole at s = r contributes e^(rt). If r is negative that term decays; if positive it grows. The magnitude of r sets how fast — a pole at −10 decays five times faster than one at −2.

A repeated real pole contributes both e^(rt) and t·e^(rt). The factor of t makes the response rise before the exponential eventually pulls it down, which is why critically damped systems have a distinctive shape.

A complex conjugate pair at α ± βi contributes e^(αt)cos(βt) and e^(αt)sin(βt). The real part α is the envelope, controlling decay or growth. The imaginary part β is the angular frequency of the oscillation inside that envelope.

This is why control engineers talk about pole locations rather than about equations. Everything you need to know about stability and response shape is in where the poles sit on the complex plane.

The left half-plane — poles with negative real parts — means every component decays and the system is stable. The right half-plane means at least one component grows without bound, and the system is unstable regardless of what else is happening. Poles exactly on the imaginary axis oscillate forever without decaying, which is marginal stability.

Moving poles further left makes the response faster. Moving them further from the real axis makes it oscillate more. The whole business of control design is placing poles where you want them.

Worked example: (s + 4) ÷ (s² + 5s + 6)

First the poles. The discriminant is 25 − 24 = 1, positive, so two distinct real roots.

Factoring: s² + 5s + 6 = (s + 2)(s + 3), so the poles are at s = −2 and s = −3.

Now partial fractions. Write (s + 4) ÷ ((s + 2)(s + 3)) = A ÷ (s + 2) + B ÷ (s + 3).

The cover-up rule is the fastest route. To find A, cover the (s + 2) factor and evaluate what remains at s = −2: (−2 + 4) ÷ (−2 + 3) = 2 ÷ 1 = 2.

To find B, cover the (s + 3) factor and evaluate at s = −3: (−3 + 4) ÷ (−3 + 2) = 1 ÷ (−1) = −1.

So F(s) = 2 ÷ (s + 2) − 1 ÷ (s + 3).

Each simple pole inverts to an exponential, giving f(t) = 2e^(−2t) − e^(−3t).

Check it two ways. At t = 0, f(0) = 2 − 1 = 1. The initial value theorem says f(0) should equal the limit of s·F(s) as s → ∞, which is the ratio of leading coefficients, p ÷ a = 1 ÷ 1 = 1 ✓.

And the final value theorem: the limit of s·F(s) as s → 0 is 0 × 4/6 = 0, matching the fact that both exponentials decay to zero ✓.

Both poles are negative, so the response decays without oscillating — overdamped. The e^(−3t) term dies faster, so after a short time the behaviour is dominated by 2e^(−2t). The slowest pole always dominates eventually, which is why it is called the dominant pole.

The three pole cases

The last row is the special case where the damping term vanishes entirely. Poles sit on the imaginary axis and the oscillation continues at constant amplitude forever. In a real physical system some damping always exists, so this is an idealisation — but it is the reference case against which the damping ratio of every other system is measured.

Partial fractions, and the cases beyond a quadratic

Partial fraction decomposition is the whole technique, and the form of each piece depends on the type of factor.

A distinct linear factor (s − r) contributes a single term A ÷ (s − r), and the cover-up rule finds A in one step: delete that factor from the denominator and evaluate everything remaining at s = r.

A repeated linear factor (s − r)ⁿ contributes n terms, over (s − r), (s − r)², and so on up to (s − r)ⁿ. The cover-up rule finds the highest-power coefficient directly; the others need differentiation or equating coefficients.

An irreducible quadratic contributes (Ds + E) ÷ (quadratic), and the standard move is to complete the square so it matches the shifted sine and cosine pairs. Splitting the numerator into a part proportional to (s − α) and a constant part is what separates the cosine term from the sine term.

One prerequisite is easy to miss: the transform must be proper, meaning the numerator degree is strictly lower than the denominator degree. If it is not, divide out first. The polynomial part of the division inverts to impulses and their derivatives — a delta function appears, which is why an improper transform signals a system with direct feedthrough.

For higher-order denominators, factor into linear and irreducible quadratic pieces and apply the rules above to each. A cubic always has at least one real root, so it factors into a linear times a quadratic.

The convolution theorem provides an alternative when a transform is a product of two whose inverses are known individually — the inverse of the product is the convolution of the inverses. It is more work than partial fractions for rational functions and indispensable when one factor is not rational.

In practice, most inverse Laplace problems in coursework reduce to the three cases above. Recognising which case you are in from the discriminant, before doing any algebra, saves a great deal of time and prevents the most common error — trying to factor a quadratic that has no real roots.

Assumptions and limitations

• Handles a linear numerator over a quadratic or linear denominator
• Higher-order denominators must be factored into linear and quadratic pieces first
• Assumes a proper transform — an improper one produces impulse terms
• Does not perform symbolic partial fractions on an arbitrary expression
• Does not handle transforms containing e^(−as), which represent time delays
• Coefficients are treated as decimals, so results may show rounding
• Assumes the one-sided transform with zero initial conditions built into F(s)

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