# Is Morlet wavelet a wavelet?

As far as I know (according to "Wavelet Methods for Time Series Analysis" by Percival & Waldren), by definition a function $\psi$ is a wavelet, if (and only if?) it integrates to zero and its square integrates to one, that is

$\int_{-\infty}^{\infty} \psi(u) \ du = 0 \\ \int_{-\infty}^{\infty} \psi^2(u) \ du = 1$

However the Morlet wavelet (builtin in Wolfram Mathematica) doesn't satisfy these conditions:

$\psi(x)_{morlet} = \frac{e^{-\frac{x^2}{2}} \cos \left(\pi x \sqrt{\frac{2}{\log (2)}}\right)}{\sqrt[4]{\pi }}$

$\int_{-\infty}^{\infty} \psi_{morlet}(x) \ dx = \sqrt{2} \sqrt[4]{\pi } e^{-\frac{\pi ^2}{\log (2)}}$

$\int_{-\infty}^{\infty} \psi^2_{morlet}(x) \ dx = \frac{1}{2} \left(1+e^{-\frac{2 \pi ^2}{\log (2)}}\right) = 0.5$

Then how can it be a wavelet?

• the definition in your post and link don't seem to match entirely? – πr8 Mar 16 '17 at 19:11

## 1 Answer

As I figured out, the original Morlet wavelet is indeed not a perfect wavelet, since as I also showed in my question it doesn't satisfy the admissibility condition.

For this reason there is a corrected (sometimes called complete) version of the Morlet wavelet, which incorporates a correction term to satisfy the admissibility condition.

It looks like the requirements of a wavelet is not that strict that I thought.

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