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In 2014 someone asked here the Fourier transform of the Morlet wavelet; link below:

Fourier Transform of Morlet wavelet Function?

However, it was the approximated Morlet wavelet not written with the canonical Gaussian function.

Can somebody help to find the Fourier Transform of the full scaled Morlet wavelet below:

$$\psi_{\mu}(\tfrac{t}{\mu}) = \frac{1}{\mu} \times \frac{1}{\sqrt{2\pi} \, \sigma} \, e^{\frac{ -\left(\frac{t}{\mu} \right)^2}{2\sigma^2}} \big[e^{j\omega_c \frac{t}{\mu}}-e^{-\frac{1}{2} \omega_c^2}\big]$$

where $\omega_c = 2\pi f_c$ and $f_c$ is the centre or carrier frequency of the wavelet (the frequency to which the wavelet oscillates in the temporal domain). Also, $\mu$ is the scaling factor defined as $\mu = \frac{\omega_c}{\omega_{\mu}}$, where $\omega_{\mu}$ is the analysing frequency. Indeed, when $\omega_{\mu} = \omega_{c} \Rightarrow \mu = 1$ we have the mother Morlet wavelet.

In addition, the second term in the brackets is the correction term necessary to enforce zero mean to the Morlet wavelet.

I'd much appreciate a detailed step-by-step demonstration if possible to check where I'm getting stuck in my own demonstration. It could be done in a paper if quicker and a photo sent to me; I'll then publish here the full demonstration here.

Thanks in advance for any help.

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  • $\begingroup$ Welcome to SE.DSP. You have more chance to get an answer if you first detail your own demonstration, so that we can help detect the weaker spots $\endgroup$ – Laurent Duval Dec 20 '19 at 14:16
  • $\begingroup$ Hi, I've tried several times but got stuck. $\endgroup$ – Jean Dec 20 '19 at 14:35
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    $\begingroup$ Please show your work and where exactly you are getting stuck. Thanks! $\endgroup$ – Dan Boschen Dec 20 '19 at 15:24
  • $\begingroup$ I've split the integral into two parts and try to take the Fourier of each separately for simplicity. I'm currently finishing writing up a paper and will show my work in a couple of days hopefully. Thanks $\endgroup$ – Jean Dec 20 '19 at 20:54
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if you wanna do this with ordinary frequency $f$ (rather than angular frequency $\omega$) youcan start with

$$ \mathscr{F}\Big\{ x(t) \Big\} \triangleq \int\limits_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \, dt $$

$$ \mathscr{F}\left\{ e^{-\pi t^2} \right\} = e^{-\pi f^2} $$

$$ \mathscr{F}\left\{ e^{-\pi \alpha t^2} \right\} = \frac{1}{\sqrt{\alpha}} e^{-\frac{\pi}{\alpha} f^2} \qquad \alpha > 0 $$

$$ \mathscr{F}\left\{ e^{-\pi \alpha (t-\tau)^2} \right\} = \frac{1}{\sqrt{\alpha}} e^{-\frac{\pi}{\alpha} f^2} e^{-j 2 \pi f \tau} $$

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  • $\begingroup$ and this answer has a general gaussian Fourier transform. $\endgroup$ – robert bristow-johnson Dec 21 '19 at 6:05
  • $\begingroup$ Hi Robert, thank you so much for your great help. I've downloaded your 2001 paper, and read the comments related to the general gaussian Fourier transform. I think the latter should answer my question. But, I'd like to come back to you, later on, to understand how the Fourier of the general gaussian is obtained; I'm curious about this indeed. Once again thank you very much, and I hope to soon put the answer on this page. So great!!!! $\endgroup$ – Jean Dec 24 '19 at 16:41
  • $\begingroup$ sure, come back when you're ready. i think the trick for computing the Fourier Transform of the gaussian might be in completing the square. $\endgroup$ – robert bristow-johnson Dec 26 '19 at 5:59

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