0
$\begingroup$

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?

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

$\endgroup$
1
$\begingroup$

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 $$\mathcal{F}\{f \dagger g\} = \mathcal{F}\{f\} \cdot \mathcal{F}\{-g\}$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.