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I am currently writing an essay on Wavelet transforms, and as part of such, I am trying to show that the Morlet wavelet satisfies the standard criteria:

$$ \int^\infty_{-\infty} \psi(t) dt = 0 $$ $$ \int^\infty_{-\infty} |\psi(t)|^2 dt = 1 $$

Using the following definition from this article (as cited on the Wikipedia page):

$$ \psi_{\sigma }(t)=c_{\sigma }e^{-{\frac {1}{2}}t^{2}}\left(e^{-i\sigma t}-e^{-{\frac {1}{2}}\sigma ^{2}}\right) $$ With $\sigma$ our "reference frequency" parameter and $c_\sigma$ is our normalisation constant: $$ c_{\sigma }=\pi^{-\frac{1}{4}}\left(1+e^{-\sigma ^{2}}-2e^{-{\frac {3}{4}}\sigma ^{2}}\right)^{-{\frac {1}{2}}} $$

However, when computing the integral for the "unit energy" condition, I obtain:

$$ \int^\infty_{-\infty} |\psi_\sigma(t)|^2 dt = \int^\infty_{-\infty} \left|c_{\sigma }e^{-{\frac {1}{2}}t^{2}}(e^{-i\sigma t}-e^{-{\frac {1}{2}}\sigma ^{2}})\right|^2 dt = c_{\sigma}^2\int^\infty_{-\infty}e^{-t^{2}}\left(e^{-i\sigma t}-e^{-{\frac {1}{2}}\sigma ^{2}}\right)^2 dt = c_\sigma^2 \left[2\sqrt{{\pi}}\left(e^{-{\sigma}^2}-e^{-\frac{3}{4}\sigma^2}\right)\right] = \frac{\sqrt{{\pi}}\left(2e^{-{\sigma}^2}-2e^{-\frac{3}{4}\sigma^2}\right)} {\sqrt{\pi}\left(1+e^{-\sigma ^{2}}-2e^{-{\frac {3}{4}}\sigma ^{2}}\right)} \neq 1 $$

For this to be one, our normalisation constant would have to be:

$$ c_\sigma = \pi^{-\frac{1}{4}}\left(2e^{-\sigma ^{2}}-2e^{-{\frac {3}{4}}\sigma ^{2}}\right)^{-{\frac {1}{2}}} $$

I am inclined to believe the source is correct based on plotting this function - simultaneously I cannot find the error in my integration:

My Solution:

\begin{align*} \int^\infty_{-\infty} |\psi_\sigma(t)|^2 dt &= \int^\infty_{-\infty} \left|c_{\sigma }e^{-{\frac {1}{2}}t^{2}}(e^{-i\sigma t}-e^{-{\frac {1}{2}}\sigma ^{2}})\right|^2 dt = c_{\sigma}^2\int^\infty_{-\infty}e^{-t^{2}}\left(e^{-i\sigma t}-e^{-{\frac {1}{2}}\sigma ^{2}}\right)^2 dt\\ &= c_{\sigma }^2\left(\int^\infty_{-\infty} e^{-2i\sigma t-t^{2}}dt -2e^{-{\frac {1}{2}}\sigma ^{2}}\int^\infty_{-\infty}e^{-i\sigma t-t^{2}} dt + e^{-\sigma ^{2}}\int^\infty_{-\infty}e^{-t^{2}} dt\right) \end{align*} Consider each of these integrals separately: \begin{align} \int^\infty_{-\infty} e^{-2i\sigma t-t^{2}} dt &= \int^\infty_{-\infty}e^{-(t+i\sigma)^2-\sigma^2} dt = \frac{\sqrt{\pi}e^{-\sigma^2}}{2}\int^\infty_{-\infty}\frac{2}{\sqrt{\pi}}e^{-(t+i\sigma)^2} dt&\\ &= \frac{\sqrt{\pi}e^{-\sigma^2}}{2} \:\left[erf(t+i\sigma)\right]^\infty_{-\infty} = \sqrt{\pi}e^{-\sigma^2}?\nonumber& \end{align} \begin{align} \int^\infty_{-\infty}e^{-i\sigma t-t^{2}} dt &= \int^\infty_{-\infty}e^{-\left(t+\frac{i\sigma}{2}\right)^2-\frac{\sigma^2}{4}} dt = \frac{\sqrt{\pi}e^{-\frac{\sigma^2}{4}}}{2}\int^\infty_{-\infty}\frac{2}{\sqrt{\pi}}e^{-\left(t+\frac{i\sigma}{2}\right)^2} dt&\\ &= \frac{\sqrt{\pi}e^{-\frac{\sigma^2}{4}}}{2} \:\left[erf\!\left(t+\frac{i\sigma}{2}\right)\right]^\infty_{-\infty} = \sqrt{\pi}e^{-\frac{\sigma^2}{4}}?\nonumber& \end{align} \begin{align} &\int^\infty_{-\infty}e^{-t^{2}} dt = \frac{\sqrt{\pi}}{2}\int^\infty_{-\infty}\frac{2e^{-t^{2}}}{\sqrt{\pi}}dt = \frac{\sqrt{\pi}}{2} \:[erf(t)]^\infty_{-\infty} = \sqrt{\pi}& \end{align} Substituting these back into our equation: \begin{align*} \int^\infty_{-\infty} |\psi_\sigma(t)|^2 dt &= c_{\sigma }^2\left(\sqrt{\pi}e^{-\sigma^2} -2e^{-{\frac {1}{2}}\sigma ^{2}}\sqrt{\pi}e^{-\frac{\sigma^2}{4}} + e^{-\sigma ^{2}}\sqrt{\pi}\right)\\ &= 2\sqrt{\pi}c_{\sigma }^2\left(e^{-\sigma^2} -e^{-{\frac {3}{4}}\sigma ^{2}}\right)\\ &= \frac{2\sqrt{\pi}\left(e^{-\sigma^2} -e^{-{\frac {3}{4}}\sigma ^{2}}\right)} {\left(\pi^{\frac{1}{4}}\left(1+e^{-\sigma ^{2}}-2e^{-{\frac {3}{4}}\sigma ^{2}}\right)^{{\frac {1}{2}}}\right)^2}\\ &= \frac{2\sqrt{\pi}\left(e^{-\sigma^2} -e^{-{\frac {3}{4}}\sigma ^{2}}\right)} {\sqrt{\pi}\left(1+e^{-\sigma ^{2}}-2e^{-{\frac {3}{4}}\sigma ^{2}}\right)} \end{align*}

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1 Answer 1

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You didn't correctly compute the squared magnitude of a complex-valued function. You just took the square instead of the squared magnitude:

\begin{align*} |\psi_{\sigma}(t)|^2 &= c_{\sigma }^2e^{-t^{2}}\left|e^{-i\sigma t}-e^{-\sigma ^{2}/2}\right|^2 \\ &= c_{\sigma }^2e^{-t^{2}}\left(1-2e^{-\sigma^2/2}\cos(\sigma t)+e^{-\sigma^2}\right) \end{align*}

Hence,

\begin{align*} \int_{-\infty}^{\infty}|\psi_{\sigma}(t)|^2 dt &= c_{\sigma }^2\int_{-\infty}^{\infty}e^{-t^{2}}\left(1-2e^{-\sigma^2/2}\cos(\sigma t)+e^{-\sigma^2}\right)dt \\ &=c_{\sigma }^2\left(1+e^{-\sigma^2}\right)\int_{-\infty}^{\infty}e^{-t^2}dt-2c_{\sigma }^2e^{-\sigma^2/2}\int_{-\infty}^{\infty}e^{-t^2}\cos(\sigma t)dt\\ &= c_{\sigma }^2\left(1+e^{-\sigma^2}\right)\sqrt{\pi}-2c_{\sigma }^2e^{-\sigma^2/2}e^{-\sigma^2/4}\sqrt{\pi}\\&=c_{\sigma}^2\sqrt{\pi}\left(1+e^{-\sigma^2}-2e^{-3\sigma^2/4}\right) \end{align*}

which results in the correct value of $c_{\sigma}$. Note that the integrals above are standard integrals which can be found in integration tables or which can be solved with a little help from WolframAlpha.

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  • $\begingroup$ Thank you very much :) $\endgroup$ Commented Mar 28 at 11:38

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