Partial Fraction Decomposition for Inverse Laplace Transforms

The inverse Laplace transform recovers a time-domain function f(t) from its frequency-domain representation F(s). The real work is decomposing F(s) into standard forms via partial-fraction decomposition. Almost every rational function you meet in differential equations, signals and systems, or control engineering arrives as a ratio of polynomials in s. A table lookup only works after you have split that ratio into terms whose inverse you know.

Skip this step and you get the wrong time-domain answer. Do it correctly and the table gives you exponentials, sines, cosines, and powers of t in minutes. The method is standard: Stewart 'Calculus' §7.4 defines four cases (distinct linear factors, repeated linear factors, distinct irreducible quadratic factors, repeated irreducible quadratic factors). Boyce & DiPrima apply them directly to Laplace transforms in their Chapter 6. The sections below work through each case with a real example you can replicate.

You need partial fractions whenever the degree of the numerator of F(s) is less than the degree of the denominator, a proper fraction. If the numerator degree equals or exceeds the denominator degree, you must divide first. The type of denominator factor determines which template to use.

When all denominator factors are linear and none repeat, use the Heaviside cover-up method. It gives you each coefficient by covering the corresponding factor in the denominator and evaluating what remains at the root of that factor. This is the fastest method for distinct real poles.

Template Form for Distinct Real Roots

For F(s) = N(s) / [(s - p₁)(s - p₂)(s - p₃) ... ], the partial-fraction expansion is:

A₁/(s - p₁) + A₂/(s - p₂) + A₃/(s - p₃) + ...

Each Aₖ = N(s) / (product of all factors except (s - pₖ)), evaluated at s = pₖ.

Worked Example

Find ℒ⁻¹{ (5s - 3) / (s² - 3s + 2) }.

Factor the denominator: s² - 3s + 2 = (s - 1)(s - 2). The poles are p₁ = 1 and p₂ = 2, both distinct real roots.

Write the form: A/(s - 1) + B/(s - 2).

For A, cover (s - 1) in the denominator and evaluate (5s - 3)/(s - 2) at s = 1: (5 - 3)/(1 - 2) = 2/(-1) = -2.

For B, cover (s - 2) and evaluate (5s - 3)/(s - 1) at s = 2: (10 - 3)/(2 - 1) = 7/1 = 7.

So F(s) = -2/(s - 1) + 7/(s - 2).

Inverse: ℒ⁻¹{ -2/(s - 1) } = -2eᵗ, ℒ⁻¹{ 7/(s - 2) } = 7e²ᵗ.

Failure case: Using the cover-up method on repeated roots gives missing terms (see next section). The method only works for distinct linear factors.

When a denominator factor appears more than once, the partial-fraction expansion must include terms for each power of that factor. The cover-up method still works for the highest-power coefficient, but the lower coefficients require a system of equations or limit-based methods. Boyce & DiPrima §6.2 covers the repeated real root case.

Template Form for Repeated Real Roots

For a factor (s - p) repeated m times, include:

A₁/(s - p) + A₂/(s - p)² + ... + Aₘ/(s - p)ᵐ

The numerator of each term is a constant, not a polynomial. Common error: using A/(s - p)² alone, that gives one term where two are needed.

Worked Example

Find ℒ⁻¹{ (2s + 5) / (s - 1)² }.

Denominator (s - 1)² is a repeated linear factor with multiplicity 2. Write: A/(s - 1) + B/(s - 1)².

Clear denominators: 2s + 5 = A(s - 1) + B.

Substitute s = 1: 2(1) + 5 = A(0) + B → B = 7.

Compare coefficients of s: 2 = A → A = 2.

So F(s) = 2/(s - 1) + 7/(s - 1)².

Failure case: Using the cover-up method on the repeated root and stopping at A = 2 misses the B term entirely. Always write the full form for the multiplicity.

Complex-conjugate poles appear as irreducible quadratic factors in the denominator. Their partial-fraction terms have linear numerators. The next step, completing the square, rewrites the quadratic into the form (s + a)² + ω² so you can match sine and cosine transform pairs. Boyce & DiPrima ch. 6 covers complex conjugate roots in their example case types.

Template Form for Distinct Irreducible Quadratics

For a factor (s² + bs + c) that does not factor over the reals, use:

(As + B) / (s² + bs + c)

For repeated irreducible quadratics, the form extends with increasing power denominators and linear numerators, per Stewart §7.4 case 4.

Worked Example

Find ℒ⁻¹{ (s + 3) / (s² + 2s + 5) }.

First complete the square: s² + 2s + 5 = (s + 1)² + 2². The quadratic is irreducible because the discriminant (4 - 20 = -16) is negative.

Write the numerator to match the shift: s + 3 = (s + 1) + 2. So F(s) = (s + 1)/[(s + 1)² + 2²] + 2/[(s + 1)² + 2²].

Use the frequency-shift theorem: ℒ⁻¹{ (s + 1)/[(s + 1)² + 2²] } = e⁻ᵗ cos(2t) and ℒ⁻¹{ 2/[(s + 1)² + 2²] } = e⁻ᵗ sin(2t).

Add: f(t) = e⁻ᵗ[cos(2t) + sin(2t)] for t ≥ 0.

Failure case: Forgetting to complete the square and trying to match a non-standard form directly. Always complete the square first.

When the numerator degree equals or exceeds the denominator degree, the fraction is improper. Partial-fraction templates require the numerator degree to be lower. Perform polynomial long division first to extract a polynomial part plus a proper rational remainder. Boyce & DiPrima ch. 6 examples all start from proper fractions, but real transfer functions often have equal-degree numerator and denominator, especially when the DC gain is finite.

Worked Example: Find ℒ⁻¹{ (s² + 4s + 5) / (s + 1) }.

Degree of numerator (2) exceeds degree of denominator (1). Divide: (s² + 4s + 5) ÷ (s + 1) = s + 3 + 2/(s + 1).

Inverse: ℒ⁻¹{s} = δ'(t) (derivative of Dirac delta, often omitted in practice unless the problem expects it), ℒ⁻¹{3} = 3δ(t), ℒ⁻¹{2/(s+1)} = 2e⁻ᵗ. For most engineering contexts, the polynomial part corresponds to impulses and their derivatives.

Failure case: Trying to apply the partial-fraction template directly to an improper fraction produces algebraic nonsense. Always check degrees first.

Before looking up the inverse, verify the partial-fraction decomposition. Recombine the terms over a common denominator and compare with the original F(s). If they match numerically, the decomposition is correct. This step catches arithmetic errors that would otherwise produce a wrong time-domain answer.

Quick check: Multiply both sides of the expansion by the original denominator. The resulting equation must hold for all s, so substitute two or three distinct s values (avoid the poles themselves) and confirm both sides give the same number. This is faster than recombining fully when you have three or more terms.

Using software: SymPy's inverse_laplace_transform and Wolfram's InverseLaplaceTransform both perform partial fractions internally. Use them to cross-check a manual result, not to bypass the learning. If your manual coefficients differ, trace the error back to the expansion step.

Case	Denominator Factor	Partial Fraction Form	Source
1: Distinct Linear	(s-p₁)(s-p₂)...	A₁/(s-p₁) + A₂/(s-p₂) + ...	Stewart §7.4 Case 1
2: Repeated Linear	(s-p)ᵐ	A₁/(s-p) + A₂/(s-p)² + ... + Aₘ/(s-p)ᵐ	Stewart §7.4 Case 2
3: Distinct Irreducible Quadratic	(s²+bs+c)	(As+B)/(s²+bs+c)	Stewart §7.4 Case 3
4: Repeated Irreducible Quadratic	(s²+bs+c)ᵐ	(A₁s+B₁)/(s²+bs+c) + ... + (Aₘs+Bₘ)/(s²+bs+c)ᵐ	Stewart §7.4 Case 4