In 1 they say that the energy of $\psi_\lambda(t)$ is concentrated in an interval of length $2\pi Q/\lambda$. I understand the inverse proportionality between the frequency band $\lambda/Q$ and the time support but I'm missing where that formula is coming from. I thought that the precise relation depends on the specific filter/wavelet but I might be missing something.


If I consider Morlet

$\theta(t) = \frac{e^{\frac{-t^2}{2\sigma_t^2}}}{\sigma_t\sqrt{2\pi}}$


$\hat\theta(\omega) = e^{-\frac{\sigma_t^2 \omega^2}{2}} = e^{-\frac{ \omega^2}{2 \sigma^2_\omega}}$

then: $\sigma_t\sigma_\omega = 1$

If $\sigma_\omega$ is $\lambda/Q$, why $\sigma_t$ is $2\pi Q/\lambda$ instead of simply $Q/\lambda$. Where the $2\pi$ is coming from?


1 Answer 1


It's indeed only a statement on proportionality. Wavelets aren't compat, so any notion of "support" invokes a heuristic (engineered criterion).

However, it's not applicable to all wavelets: Morlets are unimodal and symmetric in time and frequency, which enables inverse proportionality (counterexample).

Kymatio defines "support" as amplitude envelope decaying to 1/1000th of its max value, for example. $Q$ only specifies the number of wavelets per octave - which, together with r_psi (redundancy factor), determines the actual quality factor (Q in CQT), which in turn determines the exact decay characteristics, so an exact formula for support must be in terms of QF and $\lambda$ - and that's for Morlets; for general, see here.

  • $\begingroup$ Thanks for the explanation, I understand the point. However, I still cannot understand where the $2\pi$ comes from for the Morlet. Please see the edited question above. $\endgroup$
    – dac
    Dec 24, 2021 at 23:12
  • 1
    $\begingroup$ @dac $2\pi$ is simply for linear units, $f$ rather than $\omega$ (latter is radian). The relations are $\sigma_t \sigma_\omega = .5$, and $\sigma_t \sigma_f = \pi / 4$ -- related. Also be wary of omitting the zero-meaning constant ($\kappa$ here), it can yield significant implementation artifacts. $\endgroup$ Dec 25, 2021 at 0:02
  • $\begingroup$ Shit. I was doing $Q/ \lambda 2\pi$. Thanks. With the definition used above, $\sigma_t =1/ \sigma_\omega$. However, I think you refer to the fact that, to calculate the spread for the Heisenberg box, I should double that value. $\endgroup$
    – dac
    Dec 25, 2021 at 0:56
  • $\begingroup$ @dac The definition is incorrect. The sigmas properly occupy the slots of "widths" in respective domains, but their product is 1 when it must be 1/2. You may find this interactive Morlet useful (check "Computed parameters"). $\endgroup$ Dec 25, 2021 at 1:21
  • $\begingroup$ From en.wikipedia.org/wiki/Fourier_transform (206) $f(t) = e^{-\alpha t^2}$; $\hat f(\omega) = \frac{1}{\sqrt{2\alpha}} e^{-\frac{\omega^2}{4\alpha}}$. Do you define $\sigma_t = \sqrt{\alpha}$ and $\sigma_\omega = \sqrt{\frac{1}{4\alpha}}$? $\endgroup$
    – dac
    Dec 25, 2021 at 3:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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