# Which transform makes correllation a multiplication?

In Fourier analysis, a central theorem for the Fourier Transform states:

$$\mathcal F\{(f*g)(t)\}(\omega)=\mathcal F \{f(t)\}(\omega)\cdot \mathcal F\{g(t)\}(\omega)$$

In other words, convolution turns into multiplication.

In turn convolution is defined as $$(f*g)(t)=\int_{-\infty}^\infty f(\tau)g(t-\tau)d\tau$$

A correllation can be defined similarly:

$$corr(f,g) =\int_{-\infty}^\infty f(\tau)g(\tau-t)d\tau$$

How can we figure out which (if any) integral transform which has a similar rule for correllation as the Fourier Transform has for convolution?

• There is the so called cross correlation theorem archive.lib.msu.edu/crcmath/math/math/c/c778.htm – Davide Morgante Jul 29 '18 at 20:08
• @DavideMorgante that is of interest but what does $f^*(t)$ mean there? – mathreadler Jul 29 '18 at 20:11
• The asterisk usually means the complex conjugate! Your definition of cross correlation is valid if $f$ is real, because usually cross correlation is defined as $$(f\star g) =\int_{-\infty}^\infty f^*(\tau)g(\tau+t)d\tau = \int_{-\infty}^\infty f^*(\tau-t)g(\tau)d\tau \overset{\mbox{ }f\mbox{ real }}{=} \int_{-\infty}^\infty f(\tau-t)g(\tau)d\tau$$ – Davide Morgante Jul 29 '18 at 20:17
• ok good it is as I suspected, thank you. feel free to add as an answer if you will. – mathreadler Jul 29 '18 at 20:17