[Ifeffit] merging data

George Sterbinsky GeorgeSterbinsky at u.northwestern.edu
Fri Dec 14 13:00:35 CST 2012


The unexpected result I demonstrated persists for real step-like XAS data.
The data in the project I attached are actually just monitor counts, and
are only being used here to illustrate the unexpected behavior. It is not
actually data that I am trying to analyze.

George


On Fri, Dec 14, 2012 at 12:12 PM, Bruce Ravel <bravel at bnl.gov> wrote:

>
> I agree that you have uncovered an edge case where my merging
> algorithm does something unexpected.  Figuring this out and fixing it
> is nowhere near the top of my list of priorities.  I think I would be
> happier flagging an error when the user sets a zero-length range for
> the pre- or post-edge lines.
>
> Athena is not intended to be a generic data processing tool.  Its
> algorithms are fairly well tested and expected to work when you have
> data that nominally resembles XAS data -- that is, some kind of
> step-like function, possibly with wiggles above the step.  It has an
> algorithm for normalization that makes those same assumptions.
>
> We all eventually go to the beamline and measure something that needs
> interpretation, but which is not a step-like function with wiggles.
> The solution to that problem is not to expect Athena to miraculously
> do what you want.  The solution is to go find the find the right tool
> for the job.  Or, if it doesn't exist, write it yourself.
>
> B
>
> On Thursday, December 06, 2012 10:59:55 AM George Sterbinsky wrote:
> > Hello,
> >
> > I am observing some behavior in (D)Athena that I do not understand, and I
> > was hoping someone could explain it to me. I have attached a project file
> > to illustrate my question. The first two groups in the file have been
> > normalized such that mu(E) and norm(E) are identical. I then merged these
> > groups in mu(E) and norm(E), also in the file. Since mu(E) and norm(E)
> are
> > the same for the groups, I would expect the data merged in mu(E) and
> > norm(E) to also be identical, but it is not. While it is clear that
> merging
> > in mu(E) is averaging the groups, it seems merging in norm(E) is not. So
> > what is merging in norm(E) doing?
> >
> > Thank you,
> > George
>
>
> --
>
>  Bruce Ravel  ------------------------------------ bravel at bnl.gov
>
>  National Institute of Standards and Technology
>  Synchrotron Methods Group at NSLS --- Beamlines U7A, X24A, X23A2
>  Building 535A
>  Upton NY, 11973
>
>  Homepage:    http://xafs.org/BruceRavel
>  Software:    https://github.com/bruceravel
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