Hi Tadej, ________________________________ From: ifeffit-bounces@millenia.cars.aps.anl.gov [mailto:ifeffit-bounces@millenia.cars.aps.anl.gov] On Behalf Of Tadej Rojac Sent: Thursday, December 15, 2005 6:44 AM To: Ifeffit@millenia.cars.aps.anl.gov Subject: [Ifeffit] my problem Dear collegues, my name is Tadej Rojac and I'm writing from the Institute Jozef Stefan in Ljubljana, Slovenia. I work in the Department of electronic ceramics. Actually, I'm working on a EXAFS spectra with Artemis. What I am trying to do is mainly to describe an EXAFS spectrum with a model concerning the structure of Nb2O5. I'm working togheter with prof. Iztok Arcon, who is a specialist in the field. My main problem is that I have to do some suppositions. In fact Nb2O5 structure is quite complicated. It is composed of seven different Nb positions which I took into account using 7 feffs in Artemis. I don't know the exact occupancies of the seven positions. In order to make the search easier I did at first a supposition that all the occupancies must be positive, which is realistic. I set this with the command "abs" in each feff file. Secondly, I need to supose that the sum of all occupancies is 1. In that way the search for the result is much easier, otherwise I get the sum greater than one, which, for sure, is not the case. My question is how can I do that? For example, if the occupancies are O1, O2, O3, O4, O5, O6 and O7 and I define them with a starting value (let use say the same, so 1/7) than I can do O = O1+O2+O3+O4+O5+O6+O7. Finally I need to do O =1, but here Artemis doesn't want to define the same parameter (O in this case) twice. So, how can I insert the conditions, that the sum of all the occupancies is 1, into Artemis? I hope you can help me. [Kelly, Shelly D.] In the path page for each path of each feff calculation you need to have a variable that represents the occupancy. S02xN = abs(S02 * O1) ..... Then in the variable page (guess set def) Guess O1 = 0.15 Guess O2 = 0.15 Guess O3 = 0.15 .... Def O7 = 1 - abs(O1) -abs(O2) -abs(O3) - abs(O4) - abs(O5) -abs(O6) Hence you get one occupancy for free for knowing one more piece of information. Sounds like a tough problem. I'm looking forward t hearing from you...thank you... Tadej Rojac Jozef Stefan Institute Jamova 39 1000 Ljubljana Slovenia Tel.: +386 1 477 38 34 Fax: +386 1 477 38 87 E-mail: tadej.rojac@ijs.si