0
$\begingroup$

When I apply differerent thresholding, wavelet denosing functions to non stationary time series which has been detrended via Loess regression and demean it. I expect that when this processed series are submited to denoising / thresholding will result in a clean series with smaller values than the submited signal and not to values at any point which are above those of the signal, Is my thinking correct.

On another hand could it be thought as per those values in a procesed signal which lie above the processed signal as if the procesed signal would ought to be above those levels instead than below those levels. Of course the functions that described the signal is an aproximation so fiting erros should have to be expected.

Improve this question
$\endgroup$
1
$\begingroup$

I expect that when this processed series are submited to denoising / thresholding will result in a clean series with smaller values than the submited signal

No, you can get values that are greater. For example, consider the Fourier series of a signal. The Fourier series basis functions are wavelets. If we approximate the signal with only a few of the Fourier series coefficients you will get values that are above the original signal.

E.g. from the (image from wikipedia)

enter image description here

Improve this answer
$\endgroup$
2
  • $\begingroup$ I see, however if signal is composed of signal + noise in real terms the denoised signal should always be smaller than the signal as in the real terms, however in the aproximating function the denoised as you point out should yield results as you detail in your response that may be greater but this are basically errors in denoising. my guess is that we should aim to obtain the minimum number of coeficients above the signal+noise as to get the most acurate denoised signal $\endgroup$ – Barnaby Aug 31 '15 at 9:04
  • $\begingroup$ For unknown signal and noise quantities forming the signal , there is the posibility that the signal is above what the total noise + signal represents. This will be the theoretical possition where the signal should be. As the function will always be an aproximation. $\endgroup$ – Barnaby Jan 28 '16 at 11:05

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.