Transforms and Convolutions

There are many theorems of the form

{\cal T}(f*g) = {\cal T}(f){\cal T}(g)

where f and g are functions, T is an integral transform, and * is a kind of convolution. In words, the transform of a convolution is the product of transforms.

When the transformation changes, the notion of convolution changes.

Here are three examples.

Fourier transform and convolution

With the Fourier transform defined as

{\cal T}(f)(s) = \int_{-\infty}^\infty \exp(-isx) f(x)\, dx

convolution is defined as

(f* g)(x) = \int_{-\infty}^\infty f(x - y) g(y)\, dy

Note: There are many minor variations on the definition of the Fourier transform. See these notes.

Laplace transform and convolution

With the Laplace transform defined as

{\cal T}(f)(s) = \int_0^\infty \exp(-sx) f(x)\, dx

convolution is defined as

(f* g)(x) = \int_0^x f(x - y) g(y)\, dy

Mellin transform and convolution

With the Mellin transform defined as

{\cal T}(f)(s) = \int_0^\infty x^{s-1} f(x)\, dx

convolution is defined as

(f* g)(x) = \int_0^\infty f(y) \, g\left( \frac{x}{y} \right)\, \frac{dy}{y}

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One thought on “Transforms and Convolutions

  1. While you’re at it, you might well add moment generating functions for probability distributions.

    On the discrete side there is the z-transform and the probability generating function.

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