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The Gaussian mother wavelet with frequency resolution $\,\alpha\equiv\omega/\sigma_\omega\,$ is given by

$$G_\alpha(t)=\left(\frac{2}{\pi\alpha^2}\right)^{1/4}\exp\left(-\frac{t^2}{\alpha^2}-it\right).$$

I'm using it to do a continuous wavelet transform (CWT) to a real singal $\,x(t)\,$ to obtain a time-dependent spectrum

$$X_\alpha(\omega,t)=\sqrt{|\omega|}\int_{-\infty}^\infty x(t')_{\,}G_\alpha[\omega(t-t')]_{\,}dt'.$$

In the limit of $\,\alpha\rightarrow\infty$, the transformation approaches the time-independent Fourier transform. Gaussian wavelets have the best time-frequency resolution as they hit the bound of the uncertainty relation. I'd like to know how one can reconstruct the signal $\,x(t)\,$ from the spectrum $X_\alpha(\omega,t)\,$ at a given $\,\alpha$. Is the wavelet basis overcomplete? Is there a reconstruction formula for the inverse CWT, or does the overcompleteness of the basis mean the reconstruction formula is not unique?

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  • $\begingroup$ have the best time-frequency resolution... Yes in the Fourier sense of frequency. Not in many other senses. $\endgroup$ – mathreadler Oct 10 '17 at 7:09
  • $\begingroup$ $X_\omega = x \ast G_\omega$ so $x(t) = \int_{-\infty}^\infty \frac{e^{i k t}}{2\pi\widehat{G}_\omega(k)} (\int_{-\infty}^\infty X_\omega(t') e^{-i k t'}dt') dk$ @mathreadler The Gaussian window minimizes $\|t h(t)\|_2^2+\|\omega \widehat{h}(\omega)\|_2^2$ $\endgroup$ – reuns Oct 10 '17 at 7:14
  • $\begingroup$ Yes. In the Fourier frequency sense. $\endgroup$ – mathreadler Oct 10 '17 at 7:16
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I figured out an inversion formula. For simplicity, let me redefine the normalization constant, so we have

$$X_\alpha(\omega,t)=\frac{|\omega|}{\sqrt{\pi}\alpha}\int_{-\infty}^\infty x(t')\exp\left[-\frac{\omega^2}{\alpha^2}(t-t')^2-i\omega(t-t')\right].$$

The new normalization makes the Gaussian window normalized rather than square-normalized, which simplifies the inversion formula. The first step is integrate over $t$ to cancel the Gaussian window. We have

$$\int_{-\infty}^\infty X_\alpha(\omega,t)\,e^{i\omega t}dt=\frac{|\omega|}{\sqrt{\pi}\alpha}\int_{-\infty}^\infty x(t')\,e^{i\omega t'}dt'\int_{-\infty}^\infty\exp\left[-\frac{\omega^2}{\alpha^2}(t-t')^2\right]dt.$$

The Gaussian integral gives a constant that happens to cancel the normalization constant in the front. So we have

$$\int_{-\infty}^\infty X_\alpha(\omega,t)\,e^{i\omega t}dt=\int_{-\infty}^\infty x(t')\,e^{i\omega t'}dt'.$$

Then we can invert the Fourier transform to obtain

$$x(t)=\int_{-\infty}^\infty dt'\int_{-\infty}^\infty\frac{d\omega}{2\pi}\,X_\alpha(\omega,t')\,e^{-i\omega(t-t')}.$$

I know the wavelet basis is overcomplete. In case you find a different inversion formula that also reconstructs the signal, I'm going to accept your answer.

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