Gaussian wavelet reconstruction formula 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?
 A: 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.
