Convolution Theorem

Let $f(t)$ and $f(t)$ be arbitrary functions of time $t$ with Fourier Transforms. Take

$$f(t) = {\mathcal F}^{-1}[F(\nu)]=\int_{-\infty}^\infty F(\nu)e^{2\pi i\nu t}\,d\nu$$ (1)

$$g(t) = {\mathcal F}^{-1}[G(\nu)]=\int_{-\infty}^\infty G(\nu)e^{2\pi i\nu t}\,d\nu,$$ (2)

where ${\mathcal F}^{-1}$ denotes the inverse Fourier Transform (where the transform pair is defined to have constants $A=1$ and $B=-2\pi$). Then the Convolution is

$$f*g \equiv \int_{-\infty}^\infty g(t')f(t-t')dt'$$

$$= \int_{-\infty}^\infty g(t')\left[{\int_{-\infty}^\infty F(\nu)e^{2\pi i\nu(t-t')} \,d\nu}\right]\,dt'.$$ (3)

Interchange the order of integration,

$$f*g = \int_{-\infty}^\infty F(\nu)\left[{\int_{-\infty}^\infty g(t')e^{-2\pi i\nu t'}\,dt'}\right]e^{2\pi i\nu t}\,d\nu$$

$$= \int_{-\infty}^\infty F(\nu)G(\nu)e^{2\pi i\nu t}\,d\nu$$

$$= {\mathcal F}^{-1}[F(\nu)G(\nu)].$$ (4)

So, applying a Fourier Transform to each side, we have

$${\mathcal F}[f*g] = {\mathcal F}[f]{\mathcal F}[g].$$ (5)

The convolution theorem also takes the alternate forms

$${\mathcal F}[fg] = {\mathcal F}[f]*{\mathcal F}[g]$$ (6)

$${\mathcal F}({\mathcal F}[f]{\mathcal F}[g]) = f*g$$ (7)

$${\mathcal F}({\mathcal F}[f]*{\mathcal F}[g]) = fg.$$ (8)

See also Autocorrelation, Convolution, Fourier Transform, Wiener-Khintchine Theorem

References

Arfken, G. ``Convolution Theorem.'' §15.5 in Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 810-814, 1985.

Bracewell, R. ``Convolution Theorem.'' The Fourier Transform and Its Applications. New York: McGraw-Hill, pp. 108-112, 1965.