I work with a Electrophysiologist Team and i have seen something who interrogate me (i'm a noobs in DSP).

They acquires a signal into a brain structure (an LFP to be precise) :

  • in a first time they apply a zscore(standard score) normalization on the raw data

  • then they computes a spectral estimation with a CWT(continuous wavelet transform, morlet in this case)

Is that not a bias to use a normalization like a zscore before estimate the spectrum(or to do anything else in DSP)? Or is that a common way to deal?

Thank you for a potential answer.



Most standard time-frequency-scale transforms are consistent with mean/standard-deviation corrections. In a more pedantic ways, linear transforms can cope well with affine transformations, in the shape of:

$$x' = \frac{x-a}{b}$$

of which the $z$-score is a special case. The potential bias can easily be inverted.

So as long as the representation remains linear, and somehow preserves energy, such a normalization does not sound so harmful to me. With CWT tools, you can compare shapes. However, you loose amplitude, stuff becomes relative

  • $\begingroup$ Thanks a lot for your answer! Yes you're right, we loose the amplitude and in fact they applies the same process to other record (from other subject) and they mades an average of them. But if the amplitude is not the same, it sounds a little bit strange, no? Or perhaps this is the idea behind this normalization, to don't care about differences of amplitudes between the two signal? $\endgroup$
    – nicknolt
    Feb 11 '18 at 19:10
  • $\begingroup$ With my shallow knowledge about the brain: sensors response amplitudes can vary a lot with coupling, location, stimulus. So, if you want to combine, filter, separate sources from several different sensors, normalization is convenient. Whether the $z$-score is the best, most robust, etc. remains an open question (to me) $\endgroup$ Feb 11 '18 at 19:19
  • $\begingroup$ all sounds good, thanks for your help $\endgroup$
    – nicknolt
    Feb 11 '18 at 19:34

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.