The Convolution Theorem for Inverse Laplace Transforms

A transfer function lands on your desk: 1/(s²+1) × s/(s²+4). Neither factor is in the standard pair table as a product, and partial fractions would mean solving a four-term system, but the convolution theorem Laplace saves the inversion by turning the frequency-domain product into a time-domain integral you can evaluate directly. Set up that integral and know when it is faster than the alternatives.

The theorem is stated in Boyce & DiPrima §6.6 as ℒ⁻¹{F(s)G(s)} = (f * g)(t), where (f * g)(t) = ∫₀ᵗ f(τ)g(t−τ) dτ. The operation is commutative; you can assign either function to τ and the other to t−τ, but the integration limits are fixed: zero to t. The inverse exists and is unique for continuous f and g by Lerch's theorem, so the result you compute is the correct time-domain signal.

To invert a product F(s)G(s) you need the individual time-domain functions f(t) = ℒ⁻¹{F(s)} and g(t) = ℒ⁻¹{G(s)}. Look up each factor separately in a standard Laplace transform pair table, the one inside the front cover of Boyce & DiPrima or Kreyszig Table 6.1, and write down the two time expressions.

Choose Which Function to Shift

You will compute ∫₀ᵗ f(τ)g(t−τ) dτ. Pick whichever product gives a simpler integrand. For example, if f(τ) = e^{−aτ} and g(t−τ) = sin(ω(t−τ)), the integration is straightforward. If you reverse the assignment you get the same answer but the algebra may be messier. Try one; if it produces an intractable integral, swap the roles.

Check for a Unit Step

If either transform contains e^{-as} (a time delay), the corresponding time function is f(t−a)u(t−a). The convolution integral then runs only over the overlap interval where both arguments are non-negative. A missed unit step is the most common sign error in this method, listed as a failure mode in Boyce & DiPrima §6.6 exercises.

Let F(s)=1/(s²+1) and G(s)=s/(s²+4). From the standard pair table: f(t)=sin(t) and g(t)=cos(2t). The inverse of the product is (f * g)(t) = ∫₀ᵗ sin(τ) cos(2(t−τ)) dτ.

Use the identity sin A cos B = ½[sin(A+B) + sin(A−B)] with A=τ, B=2(t−τ). Then sin(τ) cos(2t−2τ) = ½[sin(2t−τ) + sin(3τ−2t)]. Integrate term by term:

½ ∫₀ᵗ sin(2t−τ) dτ = ½ [cos(2t−τ)]₀ᵗ = ½ [cos(t) − cos(2t)]
½ ∫₀ᵗ sin(3τ−2t) dτ = ½ [−⅓ cos(3τ−2t)]₀ᵗ = −⅙[cos(t) − cos(2t)]

Summing: (½ − ⅙)[cos(t) − cos(2t)] = ⅓[cos(t) − cos(2t)]. This is the time-domain function corresponding to the original product, no partial fractions required.

The convolution is ∫₀ᵗ e^{-2τ} e^{-3(t−τ)} dτ = e^{-3t} ∫₀ᵗ e^{τ} dτ = e^{-3t}(e^{t}−1).

Simplify: e^{-3t}(e^{t}−1) = e^{-2t} − e^{-3t}. This matches the partial-fraction result for 1/[(s+2)(s+3)], which is a routine Heaviside cover-up problem. The convolution gave the same answer without solving for coefficients, but it required an integration that is trivial here. In cases where the partial-fraction form involves complex-conjugate poles or repeated poles, the integral may be more algebra than the alternative.

Partial fractions are the default method for inverting a rational F(s). Boyce & DiPrima §6.2 and Stewart §7.4 cover the four cases. Convolution wins in three specific situations:

Three or More Distinct Factors

For a product like 1/[(s+a)(s+b)(s+c)], the partial-fraction system has three unknowns. The convolution integral, applied iteratively, replaces them with a double integral that is shorter to write and verify.

Product With a Trigonometric Factor

When one factor is s/(s²+ω²) or ω/(s²+ω²), partial fractions produce complex terms that must be recombined into sines and cosines. Convolution handles the trigonometric identities directly, as shown in Example 1.

Piecewise or Discontinuous Forcing Functions

If g(t) is defined piecewise, for instance a rectangular pulse, the convolution integral respects the unit step boundaries automatically. Partial fractions would require you to decompose the piecewise definition into separate intervals, which convolution does inside the integral limits.

The rule of thumb: if the product has more than two distinct poles, or if any pole is complex, try convolution first. Time yourself on three examples; you will find a personal threshold where switching methods saves solving a linear system.

In control engineering and signals and systems, the output Y(s) of a linear time-invariant system is the product of the transfer function H(s) and the input X(s). The time-domain output is the convolution of the impulse response h(t) with the input x(t): y(t) = (h * x)(t). This is the convolution theorem applied to control systems; the frequency-domain product becomes a time-domain integral.

For a second-order system with natural frequency ω_n and damping ratio ζ, the transfer function is ω_n²/(s²+2ζω_n s+ω_n²). If the input is a unit step, X(s)=1/s, the output transform is the product of the step and the transfer function. Convolution then produces the step response without solving the differential equation. The underdamped case (ζ < 1) gives an oscillatory response; the overdamped case (ζ > 1) does not oscillate. The same method works for the Dirac delta input δ(t), whose transform is 1, producing the impulse response directly.

The final value theorem and initial value theorem are separate checks on the result. After computing the convolution, verify that lim_{t→∞} y(t) from the theorem matches your integral's steady-state value, if the system has poles strictly in the left half-plane.