Could you provide the mathmatical proof that multiplication in Fourier domain is only convolution , when the flipping (of one of the signals/functions) occurs. So, that multiplication is not convolution, when there's no flipping.

See the minus sign:

$$(f * g )(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^\infty f(\tau)\, g(t-\tau)\, d\tau$$ So, if the minus sign wouldn't be there, multiplication in Fourier domain isn't convolution?

$$\mathcal{F}\{f*g\} = \mathcal{F}\{f\} \cdot \mathcal{F}\{g\}$$


So you're asking what the theorem for $$(f \dagger g )(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^\infty f(\tau)\, g(\tau-t)\, d\tau$$ is?

Then just use $f\dagger g = f * -g$ and we have $$\mathcal{F}\{f \dagger g\} = \mathcal{F}\{f\} \cdot \mathcal{F}\{-g\}$$

  • $\begingroup$ Hmm. Isn't this confusing $-g(x)$ with $g(-x)$? $\endgroup$ – knedlsepp Jan 22 '15 at 17:37

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