Re: [Ifeffit] pi/2deltak

Hello Riti, the criterion of R-resolution of pi/2deltak can be drawn from the Nyquist theorem. The amount of information in the spectra is limited by Nidp = 2 deltak deltaR / pi. Then, in assumption of equal distribution of information over the R-interval, the minimal interval can be estimated as deltaR / Nidp = pi/2deltak. The problem is that exact formula for Nidp is not known. Sometimes it has addition of +1 or +2 to represent the fact that single point (deltak=0) still contains information. Moreover, information is not equally distributed: some k-regions are more rich, other are less... And when energy interval is very short (glass, liquid) this question becomes vital. Sometimes, this resolution limit can even be overcome as it is shown in [http://dx.doi.org/10.1103/PhysRevB.82.064204]. Best regards, Leon Leon Avakyan PhDr, Physics Faculty, Southern Federal University laavakyan@sfedu.ru On 31.01.2016 21:00, ifeffit-request@millenia.cars.aps.anl.gov wrote:

Message: 2 Date: Sat, 30 Jan 2016 17:35:04 -0800 From: Ritimukta Sarangi To: XAFS Analysis using Ifeffit Subject: [Ifeffit] pi/2deltak Message-ID: Content-Type: text/plain; charset="utf-8"

Hello,

I was recently asked about the accuracy of this formulation for obtaining EXAFS resolution and I did not have a good answer. Can someone point to a reference or explain here? Thank you for your time, Best, -Riti