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}

Hilbert Transform

The Hilbert transform is defined by

f_H(x) = \frac{1}{\pi} \int_{-\infty}^\infty \frac{f(t)}{t - x}\, dt

where the integral is interpreted as the Cauchy principal part. More on that here.

Convolution in the context of Hilbert transforms is defined as in the context of Fourier transforms. However, Hilbert transforms satisfy a different form of convolution theorem:

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

More integral transform posts

4 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?

  4. Hi John, is there a deep/insightful reason why the Laplace Transform’s convolution has x instead of infinity as its upper bound?

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