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


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?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.