Re: Independent data poins in Artemis
Hi! Bruce,
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.
----- Original Message -----
From: "Bruce Ravel"
Hi! Bruce
I have a question of the independent data points. I calculate the indenpent data points using Stern's idea, N = 2(Kmax - Kmin)*(Rmax-Rmin)/3.14 + 2. As you can see the attached file, the result of the Artemis is different from this method. How do you determine the independent data points?
Hi Han, I am taking the liberty of CCing my response to the Ifeffit mailing list since this is a very common question and quite approriate for discussion in that forum. The issue of enumerating independent points is one that has been discussed at some length already on the mailing list. Here are some links. In this one, Matt summary is quite consistent with my view on this issue: http://millenia.cars.aps.anl.gov/pipermail/ifeffit/2005-December/002222.html Here Scott makes reference to a paper applying Baysing approaches to EXAFS data analysis and to the fact that the EXAFS measurement is not, in fact, an ideally packed signal. http://millenia.cars.aps.anl.gov/pipermail/ifeffit/2005-December/002221.html It seems to me that if your analysis requires the "+2", then you are almost certainly placing excessive demand on your data. Almost certainly, you are using parmaters that are highly correlated and are seeing quite large error bars. If that is the case, you will need to think about how to use constraints to simplify your fitting problem. See question 3 at http://cars9.uchicago.edu/iffwiki/FAQ/FeffitModeling. Regards, B P.S. It would be very helpful if someone could make am entry on that same FAQ page discussing the issue of enumerating Nidp. -- Bruce Ravel ---------------------------------------------- bravel@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/
Hi Han, Although I think the basic principle at work here is to be conservative with your estimate of the number of independent points, I can still give a couple of replies to this last post. --Is Dk properly zero in your example? You don't really have a delta-function, as your measurement of k is not arbitrarily sharp. With a realistic range of data, that's not an issue of great importance, but it does impact your argument. --In a qualitative sense, does a single point in k-space yield any EXAFS information? I think not. EXAFS is an oscillatory phenomenon. You can make no estimate of periodicity from a single point, and thus the Fourier transform, while mathematically computable, doesn't really have any information that can be used in an EXAFS analysis. (Another way to put this is to point out that Fourier transforms on finite intervals always have "truncation effects." The Fourier transform of a delta function is in some sense entirely truncation effect, with no information on the signal.) --Scott Calvin Sarah Lawrence College At 12:48 PM 1/4/2007, you 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.
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 Chemical Engineering Division Argonne National Laboratory Argonne, IL 60439 Ph: 630.252.9398 Fx: 630.252.9373 kropf@cmt.anl.gov
-----Original Message----- From: ifeffit-bounces@millenia.cars.aps.anl.gov [mailto:ifeffit-bounces@millenia.cars.aps.anl.gov] On Behalf Of Scott Calvin Sent: Thursday, January 04, 2007 1:58 PM To: XAFS Analysis using Ifeffit Subject: Re: [Ifeffit] Re: Independent data poins in Artemis
Hi Han,
Although I think the basic principle at work here is to be conservative with your estimate of the number of independent points, I can still give a couple of replies to this last post.
--Is Dk properly zero in your example? You don't really have a delta-function, as your measurement of k is not arbitrarily sharp. With a realistic range of data, that's not an issue of great importance, but it does impact your argument.
--In a qualitative sense, does a single point in k-space yield any EXAFS information? I think not. EXAFS is an oscillatory phenomenon. You can make no estimate of periodicity from a single point, and thus the Fourier transform, while mathematically computable, doesn't really have any information that can be used in an EXAFS analysis. (Another way to put this is to point out that Fourier transforms on finite intervals always have "truncation effects." The Fourier transform of a delta function is in some sense entirely truncation effect, with no information on the signal.)
--Scott Calvin Sarah Lawrence College
At 12:48 PM 1/4/2007, you 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.
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participants (3)
-
"Sang-Wook Han" -
Kropf, Arthur Jeremy -
Scott Calvin