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Convolution — Definition, Formula & Examples

Convolution is an operation that takes two functions and produces a new function by integrating the product of one function with a shifted, reversed copy of the other. It measures how the shape of one function is modified by the other.

Given two functions ff and gg defined on R\mathbb{R}, their convolution (f∗g)(t)(f * g)(t) is defined as the integral ∫−∞∞f(τ) g(t−τ) dτ\int_{-\infty}^{\infty} f(\tau)\, g(t - \tau)\, d\tau, provided the integral exists. The operation is commutative, associative, and distributive over addition.

Key Formula

(f∗g)(t)=∫−∞∞f(τ) g(t−τ) dτ(f * g)(t) = \int_{-\infty}^{\infty} f(\tau)\, g(t - \tau)\, d\tau
Where:
  • f,gf, g = The two input functions being convolved
  • tt = The independent variable of the output function
  • τ\tau = The integration (dummy) variable

How It Works

To compute (f∗g)(t)(f * g)(t) at a specific value of tt, you reverse gg to get g(−τ)g(-\tau), then shift it by tt to get g(t−τ)g(t - \tau). Next, multiply f(τ)f(\tau) by g(t−τ)g(t - \tau) pointwise and integrate the product over all τ\tau. Repeating this for every value of tt traces out the full convolution. In practice, convolution often appears alongside the Laplace or Fourier transform, where it converts into simple multiplication: L{f∗g}=F(s)⋅G(s)\mathcal{L}\{f * g\} = F(s)\cdot G(s). This property makes convolution a powerful tool for solving differential equations and analyzing linear systems.

Worked Example

Problem: Compute the convolution (f∗g)(t)(f * g)(t) for t≥0t \geq 0, where f(t)=e−tu(t)f(t) = e^{-t}u(t) and g(t)=e−2tu(t)g(t) = e^{-2t}u(t), with u(t)u(t) the unit step function.
Set up the integral: Since both functions are zero for negative arguments, the integration limits reduce from (−∞,∞)(-\infty, \infty) to (0,t)(0, t).
(f∗g)(t)=∫0te−τ e−2(t−τ) dτ(f * g)(t) = \int_{0}^{t} e^{-\tau}\, e^{-2(t-\tau)}\, d\tau
Simplify the integrand: Factor out terms that do not depend on τ\tau and combine the exponentials.
=e−2t∫0te−τ⋅e2τ dτ=e−2t∫0teτ dτ= e^{-2t} \int_{0}^{t} e^{-\tau} \cdot e^{2\tau}\, d\tau = e^{-2t} \int_{0}^{t} e^{\tau}\, d\tau
Evaluate the integral: Integrate eτe^{\tau} from 00 to tt and simplify.
=e−2t[eτ]0t=e−2t(et−1)=e−t−e−2t= e^{-2t}\left[e^{\tau}\right]_0^t = e^{-2t}(e^{t} - 1) = e^{-t} - e^{-2t}
Answer: (f∗g)(t)=(e−t−e−2t) u(t)(f * g)(t) = (e^{-t} - e^{-2t})\, u(t)

Why It Matters

Convolution is central to solving linear ordinary and partial differential equations using Laplace and Fourier transforms. In electrical engineering and signal processing, every linear time-invariant system is characterized by convolving an input signal with the system's impulse response. It also underpins convolutional neural networks used in image recognition.

Common Mistakes

Mistake: Using incorrect integration limits when both functions have restricted domains (e.g., defined only for t≥0t \geq 0).
Correction: When both ff and gg are zero for t<0t < 0, the limits of integration collapse from (−∞,∞)(-\infty, \infty) to (0,t)(0, t). Always account for where each function is nonzero before integrating.