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Convolution Theorem — Definition, Formula & Examples

The Convolution Theorem states that the Laplace (or Fourier) transform of a convolution of two functions equals the product of their individual transforms. This lets you turn a difficult integral operation into simple multiplication in the transform domain.

If F(s)=L{f(t)}F(s) = \mathcal{L}\{f(t)\} and G(s)=L{g(t)}G(s) = \mathcal{L}\{g(t)\}, then L{(f∗g)(t)}=F(s)⋅G(s)\mathcal{L}\{(f * g)(t)\} = F(s) \cdot G(s), where the convolution is defined as (f∗g)(t)=∫0tf(τ) g(t−τ) dτ(f * g)(t) = \int_0^t f(\tau)\,g(t - \tau)\,d\tau. Equivalently, L−1{F(s)⋅G(s)}=(f∗g)(t)\mathcal{L}^{-1}\{F(s)\cdot G(s)\} = (f * g)(t).

Key Formula

L−1{F(s)⋅G(s)}=∫0tf(τ) g(t−τ) dτ\mathcal{L}^{-1}\{F(s)\cdot G(s)\} = \int_0^t f(\tau)\,g(t - \tau)\,d\tau
Where:
  • F(s)F(s) = Laplace transform of f(t)
  • G(s)G(s) = Laplace transform of g(t)
  • τ\tau = Dummy integration variable
  • tt = Time variable in the original domain

How It Works

When you encounter a product F(s)⋅G(s)F(s) \cdot G(s) in the Laplace domain and need the inverse transform, you can split it into two recognizable transforms and compute their convolution integral instead. First, identify F(s)F(s) and G(s)G(s) separately, then find their inverse transforms f(t)f(t) and g(t)g(t). Finally, evaluate the convolution integral ∫0tf(τ) g(t−τ) dτ\int_0^t f(\tau)\,g(t-\tau)\,d\tau to obtain the time-domain result. This approach is especially useful when F(s)⋅G(s)F(s) \cdot G(s) does not match a standard transform table entry but each factor does.

Worked Example

Problem: Find the inverse Laplace transform of H(s)=1s(s−1)H(s) = \dfrac{1}{s(s-1)} using the Convolution Theorem.
Split into factors: Write the product as two known transforms:
F(s)=1s,G(s)=1s−1F(s) = \frac{1}{s}, \quad G(s) = \frac{1}{s-1}
Find inverse transforms: From standard Laplace tables:
f(t)=1,g(t)=etf(t) = 1, \quad g(t) = e^{t}
Compute the convolution integral: Apply the convolution formula with these functions:
h(t)=∫0t1⋅et−τ dτ=et∫0te−τ dτ=et[−e−τ]0t=et(1−e−t)=et−1h(t) = \int_0^t 1 \cdot e^{t-\tau}\,d\tau = e^t \int_0^t e^{-\tau}\,d\tau = e^t\left[-e^{-\tau}\right]_0^t = e^t(1 - e^{-t}) = e^t - 1
Answer: h(t)=et−1h(t) = e^t - 1

Why It Matters

The Convolution Theorem is central to solving differential equations with Laplace transforms, which appear throughout engineering and physics courses. In signal processing, it explains why filtering a signal in the frequency domain reduces to pointwise multiplication, making real-time audio and image processing computationally feasible.

Common Mistakes

Mistake: Using integration limits from −∞-\infty to ∞\infty for the Laplace convolution.
Correction: The Laplace convolution integrates from 00 to tt, not over the entire real line. The (−∞,∞)(-\infty, \infty) limits apply to the Fourier convolution instead.