1
$\begingroup$

I'm working through "A Practical Guide to Wavelet Analysis" by Torrence and Compo, and I am confused about section 3d ("Wavelet Power Spectrum").

Let $x_n$ denote the signal, sampled at increments $\delta t$; $\hat{x}_k$ its Fourier transform; $\psi$ the (normalized) mother wavelet, so that the amplitude of the wavelet transform at scale $s$ centered at $n\delta t$ is given by $$ W_n(s) = \sum_k \hat{x}_k \hat{\psi}^*(s \omega_k)e^{i\omega_k n\delta t} $$ There's some fiddling with constants here, but I understand this so far.

Then, they write that, for a white noise signal, $\mathbb{E}|W_n(s)|^2 = N \mathbb{E}|\hat{x}_k|^2 = \sigma^2$. Trying to work through this: \begin{align} \mathbb{E}|W_n(s)|^2 &= \mathbb{E}\left|\sum \hat{x}_k \hat{\psi}^*(s \omega_k)e^{i\omega_k n\delta t}\right|^2 \end{align} I see that if the wavelets are orthogonal, this works out nicely, where you can write $$ \mathbb{E}\left|\sum \hat{x}_k \hat{\psi}^*(s \omega_k)e^{i\omega_k n\delta t}\right|^2 = \sum |\hat{\psi}(s \omega_k)|^2\mathbb{E}|\hat{x}_k|^2 = \frac{\sigma^2}{N}\sum |\hat{\psi}(s \omega_k)|^2 $$ However, the authors clearly state they are not working with orthogonal wavelets specifically here, so I am unsure how they get their white noise result.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.