- Ifeffit - millenia.cars.aps.anl.gov

Problems with the >flatten< option in Athena
by Jens Kruse 10 Jan '07

10 Jan '07

Hi, I have a normalization problem with Athena. If I turn on the "flatten" option in the background removal section, Athena creates in some of the normalized spectra a step in the pre-edge range. How can I get rid of it? Thanks, jens

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RE: [Ifeffit] Problems with the >flatten< option in Athena
by Kelly, Shelly D. 10 Jan '07

10 Jan '07

Hi Jens, You can try moving e0. E0 should be on the edge. Is E0 at the position of the step? You could also send us an example. Then we can help more directly. I'd guess that the step is caused by some background parameters that are set incorrectly. SK > -----Original Message----- > From: ifeffit-bounces(a)millenia.cars.aps.anl.gov [mailto:ifeffit- > bounces(a)millenia.cars.aps.anl.gov] On Behalf Of Jens Kruse > Sent: Wednesday, January 10, 2007 8:56 AM > To: ifeffit(a)millenia.cars.aps.anl.gov > Subject: [Ifeffit] Problems with the >flatten< option in Athena > > Hi, > > I have a normalization problem with Athena. If I turn on the "flatten" > option in the background removal section, Athena creates in some of the > normalized spectra a step in the pre-edge range. How can I get rid of > it? > > Thanks, > > jens > _______________________________________________ > Ifeffit mailing list > Ifeffit(a)millenia.cars.aps.anl.gov > http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit

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Re: [Ifeffit] R-factor uncertainty
by Scott Calvin 10 Jan '07

10 Jan '07

Hi Lisa, >At 07:23 AM 1/5/2007, you wrote: > >I have a question about the R factor: how can I decide if the >difference between the R factors of 2 fits is statistically >significant, i.e, how can I calculate the uncertainty which has to >be associated to the R factor? > As I understand it, you can't. The R-factor is not a proper statistical measure, as it doesn't incorporate any measure of data quality. That's the great weakness of this measure of quality-of-fit. It is also its strength, as estimating the uncertainty in EXAFS data is notoriously problematic. The complementary statistic is reduced chi-square. It does incorporate a measure of data quality. By default, ifeffit uses noise from high in the FT to estimate this. That's a reasonable idea, but can be problematic. It has been shown (by Matt and/or Shelly, as I recall), that there may in some cases be signal in the part of the FT ifeffit is using to estimate noise. There are also cases where the noise may not be "white," that is, the noise high in the FT may be a poor estimate of the noise low in the FT. Ifeffit does allow you to specify a value of the measurement uncertainty instead, so if you think you have a way of doing this, go ahead. What does all this mean in practice? It means, in my opinion, that the actual =value= of the reduced chi-square statistic is usually meaningless, unless you have a good way of coming up with the measurement uncertainty (for example, your sample may be so dilute that errors are dominated by counting statistics). But reduced chi-square is a great statistic for comparing two fits to a given set of data, particularly if the k-range, k-weighting, and k-window are the same for the two fits. For example, you can apply statistical tests of significance, if you'd like. The R-factor then provides the reality check that the fit is "good" at all. The R-factor isn't doing anything other than what you can see by looking at a graph, but is a nice shorthand for tables showing the results of many fits and similar applications. If there's a big R-factor (say, 0.20), the question of statistical significance isn't necessary to tell you that you haven't got a conclusive positive result: maybe the R-factor is big because the fitting model is lousy, or maybe it's big because the data quality is lousy, but either way the fit shouldn't be trusted. I'd also add that your eye tells you considerably more than the R-factor, because you can tell the character of the mismatch. Is it in the high part of the FT, low, or evenly throughout? Is the miss primarily in amplitude, or phase? I often find I choose a fit with an R-factor of 0.03 over one with 0.01, if, for example, the 0.03 reproduces qualitatively all the features in the data but has small errors in the amplitude of the peaks, while the 0.01 fits the first part of the spectrum perfectly but misses some peak altogether. Hope that helps... --Scott Calvin Sarah Lawrence College

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RE: [Ifeffit] Is there a physical meaning to a negative SO2
by Kelly, Shelly D. 09 Jan '07

09 Jan '07

Hi Todd, One thing to try is to set s02 to a positive value and all the other variables to reasonable values and then look at the real or imaginary parts of the Fourier transform and also look at the chi(k) spectra. You should "see" the affect that Scott mentioned. Shelly > -----Original Message----- > From: ifeffit-bounces(a)millenia.cars.aps.anl.gov [mailto:ifeffit- > bounces(a)millenia.cars.aps.anl.gov] On Behalf Of Scott Calvin > Sent: Tuesday, January 09, 2007 10:27 AM > To: 'XAFS Analysis using Ifeffit' > Subject: Re: [Ifeffit] Is there a physical meaning to a negative SO2 > > Hi Todd, > > Your paragraph below is correct; i.e. you do not have a physically > realistic model. One common cause of a negative S02 is that the model > also is off by half an oscillation. After all, the negative S02 turns > the signal upside down, shifting it by half an oscillation can then > cause a rough (but spurious) alignment between model and data. > > In any case, the model needs work; depending on your system and > degree of prior knowledge, you might be assuming the wrong species of > scattering atom, for example. > > --Scott Calvin > Sarah Lawrence College > > At 10:41 AM 1/9/2007, Todd Luxton wrote: > >I am very new to XAFS analysis, I am wondering if there is a > >physical meaning to a negative value for SO2. Based on the XAFS > >equation I can't understand what the physical significance of a > >negative SO2 value would be, therefore I am assuming that my > >modeling results using Athena are not representative of a physically > >realistic model due to the negative value calculated fro SO2 or more > >specifically the amp variable in the guess set. Any insight would > >be greatly appreciated. > > > > > _______________________________________________ > Ifeffit mailing list > Ifeffit(a)millenia.cars.aps.anl.gov > http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit

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Is there a physical meaning to a negative SO2
by Todd Luxton 09 Jan '07

09 Jan '07

I am very new to XAFS analysis, I am wondering if there is a physical meaning to a negative value for SO2. Based on the XAFS equation I can't understand what the physical significance of a negative SO2 value would be, therefore I am assuming that my modeling results using Athena are not representative of a physically realistic model due to the negative value calculated fro SO2 or more specifically the amp variable in the guess set. Any insight would be greatly appreciated. thanks

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RE: [Ifeffit] Generating Atom lists using Artemis
by Kelly, Shelly D. 08 Jan '07

08 Jan '07

--------------------------- Shelly Kelly Argonne National Laboratory Bldg 203, RM E113 > > On Saturday 06 January 2007 10:53, Kelly, Shelly D. wrote: > > Hello Artemis users, > > > > I found an interesting feature in Artemis when generating a list of > atoms > > from the attached cif file worth sharing on the list. In the attached > cif > > file for Pd2Si the position of the Si2 atoms are at x,y,z = 0.3333, > 0.6667, > > 0 with a space group of p -6 2 m. Using this cif file and choosing > either > > Pd1 or Pd2 as the core atom results in errors in the atom list. For > > example for the Pd2 site there are 4 Si neighbors with the 1st and 3rd > and > > the 2nd and 4th on top of each other. If you edit the atomic positions > so > > that the Si2 atoms are correct to an additional digit x,y,z = 0.33333, > > 0.66667,0 then the atom list is fixed with only two Pd2-Si neighbors at > > each of the unique positions. The interesting feature is that the > > coordinates on the atoms page need to be correct to the 5 decimal place. > > > This has been discussed many, many times on the mailing list and is > explained in question 3 at > http://cars9.uchicago.edu/iffwiki/FAQ/RunningFeff > > The version of Feff that ships with Ifeffit (and, I believe, all > others) uses 5 significant figures in the path finder. Since Atoms > main purpose is to interact with Feff, it interprets the 5 digit thing > quite strictly. > > B > > [sdk] Hi Bruce, Well, I thought that I had skipped your "here is a link describing that response" by searching the Ifeffit wiki. I didn't run FEFF, so I missed the description of the problem under the FEFF error message of overlapping atoms. I think that it would be helpful for the ATOMS page within Artemis to contain a note of some sort indicating that the 5th digit is significant. Now that I have had additional time to ponder this problem, I realize that I have learned it several times only to forget it again....:) Cheers, Shelly

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RE: [Ifeffit] R-factor uncertainty
by Kelly, Shelly D. 08 Jan '07

08 Jan '07

Hi Lisa, In general the R-factor is good if it is less than a few percent. The value reported is 0.01 or 1% then the fit is satisfactory. Just as a note the R-factor is calculated over the entire data range given by rmin and rmax of the fit, so make sure that they are reasonable values. To compare two different models, use the reduced-chi-square (RCS) value. One standard deviation in the RCS value is SQRT(2/nu), where SQRT is square-root and nu is the number of independent points. If the first model has RCS1 and nu1 and the second model has RCS2 and nu2 then the second model is better than the first model if their difference (RCS1-RCS2) is greater than 2*SQRT((2*RCS1*RCS1)/nu1 + (2*RCS2*RCS2)/nu2)) which is twice the fluctuation of the difference. The number of independent points is equal to the number of data points minus the number of variables in the model. I do this kind of comparison in this paper: S.D. Kelly, K.M. Kemner, G.E. Fryxell, J. Liu, S.V. Mattigod, K.F. Ferris, "X-ray Absorption Fine-Structure Spectroscopy Study of the Interactions Between Contaminant Tetrahedral Anions and Self-Assembled Monolayers on Mesoporous Supports," The Journal of Physical Chemistry B 105 (27) 6337-6346, Aug 2001. The paper can be found here: http://www.mesg.anl.gov/sdkpublist.html. Cheers, Shelly --------------------------- Shelly Kelly Argonne National Laboratory Bldg 203, RM E113 9700 S. Cass Ave Argonne, IL 60439 Molecular Environmental Science Group www.mesg.anl.gov skelly(a)anl.gov phone: 630-252-7376 > -----Original Message----- > From: ifeffit-bounces(a)millenia.cars.aps.anl.gov [mailto:ifeffit- > bounces(a)millenia.cars.aps.anl.gov] On Behalf Of Lisa Giachini > Sent: Friday, January 05, 2007 6:23 AM > To: XAFS Analysis using Ifeffit; FEFF Users > Cc: XAFS Analysis using Ifeffit > Subject: [Ifeffit] R-factor uncertainty > > > > Dear all, > > I have a question about the R factor: how can I decide if the difference > between the R factors of 2 fits is statistically significant, i.e, how can > I calculate the uncertainty which has to be associated to the R factor? > > B.R., > > Lisa > > _______________________________________________ > Ifeffit mailing list > Ifeffit(a)millenia.cars.aps.anl.gov > http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit

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Re: [Feffusers] thermal stage
by Matt Newville 05 Jan '07

05 Jan '07

Hi Igor, (and sorry for the double posting, but I thought some people on the Ifeffit list might be interested and have something to say too). I (well, GSECARS) have a Linkam TS-1500 heating stage, which can go to 1500C. It works well. We use it in our x-ray microprobe, where we're usually very concerned with contaminating signals from metals, so we run it open to air without any cover plates on it. It can easily get to 800C with this configuration (that is, no vacuum, a lot less insulation than intended), but we tend to consume the heating elements a little faster than we'd like. In fact, we just ordered some replacements. Julie (well, PNCCAT) has the Linkam THMS-600 (I think that's the model -- it goes down to LN2 temperature, and up to ~600C). The APS equipment pool also has one of these. It also works well, and really does get down to -196C. I does tend to consume LN2 and the connecting line between the stage and dewar is a little short. I think Julie had Linkam modify the basic design to open the back hole enough to allow transmission signal be measured while the stage was at 45degrees to the beam, and then also made some in-house modifications of her own (to improve the low temperature insulation, mostly). The heating element on hers also broke this fall, but she was able to repair it. Anyway, these are definitely reasonable choices, and being commercial products, probably cheaper, safer, and more reliable than anything you can build at home. I think Oxford Danfysik has (or used to have, anyway) a similar microscope heating/cooling stage that could be adapted for synchrotron work, but don't recall the details. --Matt On 12/15/06, Igor Levin <igor.levin(a)nist.gov> wrote: > Hi, > > We are considering to purchase a variable-temperature stage for the XAS > work. Linkam's THMS600 thermal stage came up as an option. Does anybody > have an experience with this stage? What are the alternatives? > > Thanks, > > Igor Levin > > Materials Research Scientist > Ceramics Division > National Institute of Standards and Technology > Gaithersburg MD 20899 > > > _______________________________________________ > Feffusers mailing list > Feffusers(a)u.washington.edu > http://mailman1.u.washington.edu/mailman/listinfo/feffusers >

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Fwd: Re: Independent data poins in Artemis
by Bruce Ravel 04 Jan '07

04 Jan '07

Whoops! I just sent this to Han earlier and meant to send it to the whole list. Silly me! B ---------- Forwarded Message from Bruce Ravel <Bruce Ravel <bravel(a)anl.gov>> ---------- Subject: Re: Independent data poins in Artemis Date: Thursday 04 January 2007 14:50 From: Bruce Ravel <bravel(a)anl.gov> To: "Sang-Wook Han" <swhan(a)chonbuk.ac.kr> On Thursday 04 January 2007 11:48, Sang-Wook Han wrote: > I have checked the Matt's note. > Although I understand the problem, it is still not clear for me. > If we have only one data point in k-space, it is a delta function. For > 2DrDk/pi, it should be zero. When we do Fourier transformation of the one > data point, we still have data points in real and imaginary r-space. That > is not zero. Han, This question and the other answers to it are all charmingly pedantic with regards to the formalism of measurement theory ;-) The main point, IMHO, is that the Nyquist criterion is only true for a properly packed signal. EXAFS resembles a sum of sine waves, but is *not*, in fact, a sum of pure sines waves. The Nyquist criterion, then, is a statement about the maximum possible amount of information in your data, but does not tell you the amount of information that you actually have. Quantifying the information content of an improperly packed signal is the point of the Baysian approach that Scott discussed in in his earlier message. The bottom line is that real EXAFS data tends to have somewhat less than Nidp independent points. One could apply Baysian formalism. Or one could be conservative in the use of parameters and mindful of the level of correlation between parameters. In either case, asserting +0, +1, or +2 in your statement of the Nyquist criterion is pedantry. Indeed, by focussing on that fairly trivial aspect of formal information theory, you are missing a much more improtant point about your real EXAFS measurement. B -- Bruce Ravel ---------------------------------------------- bravel(a)anl.gov Molecular Environmental Science Group, Building 203, Room E-165 MRCAT, Sector 10, Advanced Photon Source, Building 433, Room B007 Argonne National Laboratory phone and voice mail: (1) 630 252 5033 Argonne IL 60439, USA fax: (1) 630 252 9793 My homepage: http://cars9.uchicago.edu/~ravel EXAFS software: http://cars9.uchicago.edu/~ravel/software/

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RE: [Ifeffit] Re: Independent data poins in Artemis
by Scott Calvin 04 Jan '07

04 Jan '07

Thanks, Jeremy. Good point. As a practical matter, what you actually get when a Fourier transform algorithm is applied to a data set consisting of a single point will depend on the algorithm. If the algorithm pads with 0's, it gets a transform that starts to look more and more "white," that is, like a constant function in transform space. If it doesn't pad with 0's, but instead takes the usual Fourier integral and let the limits of integration tend toward 0, so that it only includes the non-zero point, the transform approaches a delta function located at 0 in transform space. Those results are as different as they could be, and they depend entirely on how the truncation is handled. (A typical ifeffit calculation, as I understand it, lies between these two extremes. It integrates over a finite interval, but pads with 0's to get to a convenient number of points.) I think that nicely demonstrates that you can't extract information from the "Fourier transform" of a single point. --Scott Calvin Sarah Lawrence College At 03:20 PM 1/4/2007, you wrote: >Rigorously speaking, a single point is not a delta function. To have a >delta function, only one point has a non-zero value, all other points >are zero. If you have only one data point and you assume all other >points are zero, you are adding information not contained in your data. > >Jeremy Kropf

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