0
$\begingroup$

I'm referencing this paper here: https://arxiv.org/pdf/1203.1513.pdf

Within this paper, it states that "A wavelet transform commutes with translations, and is therefore not translation invariant". What does it mean for a transform to commute with translation?

I understand why the translation invariance is a problem, however, I don't understand why a wavelet transform would then become translation variant.

$\endgroup$
  • $\begingroup$ For any $f,g$, $f \ast g(.+a)(t) = f \ast g(t+a)$. Together with linearity, this is indeed one of the possible definition of convolution operators. The WT is a convolution operator. $\endgroup$ – reuns Apr 15 at 3:17

Your Answer

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

Browse other questions tagged or ask your own question.