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I'm a complete beginner on signal processing, so please forgive the basicness of this question - but I'm very confused by this. Regarding the Morlet wavelet, Wikipedia says:

The wavelet is defined as a constant $\kappa_\sigma$ subtracted from a plane wave and then localised by a Gaussian window: $$ \Psi_\sigma(t) = c_\sigma\pi^{-\frac{1}{4}}e^{-\frac{1}{2}t^2}(e^{i\sigma t} - \kappa_\sigma) $$ where $\kappa_\sigma=e^{-\frac{1}{2}\sigma^2}$ is defined by the admissibility criterion, and the normalisation constant $c_\sigma$ is: $$ c_\sigma = \left( 1 + e^{-\sigma^2} - 2e^{-\frac{3}{4}\sigma^2} \right)^{-\frac{1}{2}} $$ ......... The parameter $\sigma$ in the Morlet wavelet allows trade between time and frequency resolutions. Conventionally, the restriction $\sigma>5$ is used to avoid problems with the Morlet wavelet at low $\sigma$ (high temporal resolution).

Presumably, given the comment about the typical restriction of $\sigma>5$, this trade-off parameter $\sigma$ is meant to be a dimensionless value. But then (or, in fact, whatever the dimension of $\sigma$ is), the formula for $\Psi_\sigma(t)$ is not dimensionally invariant under changes of the units in which time is measured! Shouldn't there be at least some extra parameter in front of $t$, so that the formula makes sense in terms of dimensional analysis, if it is meant to be applicable to real-world data (as suggested by comments about "conventional restriction" regarding trade-off)?

(Incidentally, I've seen elsewhere much simpler formulae for the Morlet wavelet, which do make dimensional sense, and where $\sigma$ has the dimension of time - which makes sense, as low $\sigma$ means a narrow time-window of significant relevance.)

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Okay, I think I've understood it now: The number $t$ in $\Psi_\sigma(t)$ is a dimensionless number, because in the formula for the wavelet transform it ends up being the ratio between an actual time and a time-scale.

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