Categories: Mathematics.

Convolution theorem

The convolution theorem states that a convolution in the direct domain is equal to a product in the frequency domain. This is especially useful for computation, replacing an \(\mathcal{O}(n^2)\) convolution with an \(\mathcal{O}(n \log(n))\) transform and product.

Fourier transform

The convolution theorem is usually expressed as follows, where \(\hat{\mathcal{F}}\) is the Fourier transform, and \(A\) and \(B\) are constants from its definition:

\[\begin{aligned} \boxed{ \begin{aligned} A \cdot (f * g)(x) &= \hat{\mathcal{F}}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\ B \cdot (\tilde{f} * \tilde{g})(k) &= \hat{\mathcal{F}}\{f(x) \: g(x)\} \end{aligned} } \end{aligned}\]

Laplace transform

For functions \(f(t)\) and \(g(t)\) which are only defined for \(t \ge 0\), the convolution theorem can also be stated using the Laplace transform:

\[\begin{aligned} \boxed{(f * g)(t) = \hat{\mathcal{L}}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}} \end{aligned}\]

Because the inverse Laplace transform \(\hat{\mathcal{L}}^{-1}\) is unpleasant, the theorem is often stated using the forward transform instead:

\[\begin{aligned} \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)} \end{aligned}\]


  1. O. Bang, Applied mathematics for physicists: lecture notes, 2019, unpublished.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.