After investigating signal analysis methods based on empirical mode decomposition (EMD), I found that recent developments are mostly related to the Hilbert Huang transform (HHT) and the Local Mean Decomposition (LMD) method.

I have been reading a few articles on the subject and I would like to have your opinion regarding HHT.

EMD leads to intrinsic mode functions that are particularly well suited for Hilbert transform. HHT seems to be really widely used for many distinct types of industrial or academic applications.

Am I right to think that HHT is essentially EMD + Hilbert transform? Do you consider the specificity of the HHT lies in the sifting process based on spline interpolation (to the contrary of the LMD using a moving average algorithm for instance)?


After going through the literature regarding HHT and EMD, I found that the "Huang" part of HHT comes from the fact that he is the one who proposed EMD in the first place. That explains the name of the method...

For more developments regarding EMD and HHT, I recommend the papers by Rilling et al. "On empirical mode decomposition and its algorithms". For the lucky ones who speak French, Rilling's PhD. thesis on EMD no longer seems to be available online; the document seems very complete and contains a very detailed mathematical analysis of EMD. Related articles are also available in English here on Google Scholar.

One may sum-up HHT this way:

  1. EMD: decomposes the initial signal as a list of Intrinsic Mode Functions (IMF);
  2. Hilbert tranform computes the instantaneous frequencies associated to the IMF (which are precisely well suited for such transformation)
  3. Hilbert spectrum, meaning a representation of the amplitude of the IMF in a frequency/time domain using the instantaneous frequencies.

Here is a simple example based on the time signal: $y(t)=t^2sin(t^2)$ for $t\in[0;~16]$s. This signal features a quadratic growth of its amplitude combined with a linear growth of its frequency. Here is what a HHT of the signal leads to:

signal of interest

enter image description here

IMF obtained from EMD

first IMF, very similar to <span class=$y(t)$">

second IMF, almost 0 in this example

instantaneous frequencies

instantaneous frequency associated with IMF 1

Hilbert spectrum (white to black from 0 to maximum amplitude)

enter image description here

Some peaople see the HHT as a generalized Fourier Transform in the sense that the decomposition of the signal interest by HHT leads to both amplitude and frequency varying time signals.

A signifiant drawback of HHT lies in its sensitivity to edge effects (what happens for the signal close to its left and right boundaries in 1D). Several techniques exist in order to mitigate these effects. Articles by Rilling mentioned above go with mirroring strategies while other engineering oriented strategies may involve waveform matching techniques.

| improve this answer | |
  • $\begingroup$ Welcome to DSP.se! It is great that you came back and answered your own question when you found out the answer (you can even accept it). However, it would be even more great (and benefit to others) if your answer could clearly answer the question you asked: e.g. write a 1-or-2-sentence comparison of HHT and EMD, and if in fact the specificity of HHT lies in what you assumed. Currently you offered an OK answer, I think adding a few more lines would make it a great one. $\endgroup$ – penelope Oct 4 '13 at 14:53
  • $\begingroup$ Thanks for your good information; I need Matlab code to apply EMD of signal. Would you please give me some tips? $\endgroup$ – user7571 Jan 16 '14 at 5:19
  • 1
    $\begingroup$ Well, it depends how much time you have for it... If you want to code the EMD yourself, I recommed you read the article (ncbi.nlm.nih.gov/pmc/articles/PMC1618495) by Smith. Otherwise, there is a code freely available here for Matlab : cosmostat.org/rilling/software.php $\endgroup$ – Alain Jan 17 '14 at 11:32
  • $\begingroup$ @Alain - both of your links are dead or cannot be accessed. $\endgroup$ – Duck Apr 24 '19 at 19:52
  • $\begingroup$ @SpaceDog The first link is still working for me. It seems that the second one is dead however... $\endgroup$ – Alain Apr 24 '19 at 23:04

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.