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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3 thoughts 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.

  2. Another similar example is Dirichlet convolution and Dirichlet series for arithmetic functions.

  3. Why does the Laplace case of convolution generalize to groups even though the it is the Fourier transform which holds generally?

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