Does modified Morlet wavelet function satisfy the admissibility condition ?
Do you have any reference for the answer please.
Thanks.
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No, it doesn't (at least not in theory). You know that the admissibility condition is
$$\int_{-\infty}^{\infty}\frac{|\Psi(\omega)|^2}{|\omega|}d\omega<\infty$$
where $\Psi(\omega)$ is the Fourier transform of the mother wavelet $\psi(t)$. If $\Psi(\omega)$ decays sufficiently fast (which is the case for the Morlet wavelet) then the admissibility condition reduces to $\Psi(0)=0$, i.e. $\psi(t)$ must satisfy a zero-mean condition. However, this condition is not satisfied for the Morlet wavelet. So the Morlet wavelet is not admissible, but in practice this is no problem because $\Psi(0)$ is very small.
[1] M. Vetterli, J. Kovačević: Wavelets and Subband Coding
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$\begingroup$ I think there are 2 definitions for Morlet function. first one is $ \psi (t)= \frac{1}{ \sqrt{ \pi f_{b} } } \big( e^{j2\pi f_{c}t}- e^{-f_{b}(\pi f_{c})^{2}} \big) e^{ \frac{-t^{2}}{f_{b}} }$. The second definition is $ \psi (t)= \frac{1}{ \sqrt{ \pi f_{b} } } e^{j2\pi f_{c}t} e^{ \frac{-t^{2}}{f_{b}} }$ So I guess it's not admissible for the 2nd definition. Do you agree? @matt-l $\endgroup$ – SAH Jul 1 '14 at 8:43
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$\begingroup$ @Electricman: Yes, the first one has a correction term to make the original Morlet wavelet (your second formula) admissible. $\endgroup$ – Matt L. Jul 1 '14 at 9:10
