ragordon at alumni.sfu.ca
Mon Feb 1 11:16:18 CST 2016
A couple of caveats to Matt's answer:
The EXAFS equation is not simply a sum of sine functions. I can think of
two ways that the resolution criteria
can be worked around:
1. Consider the case of two close near neighbours in a single-crystal
environment whose close distances are
geometrically distinct such that, if one were to conduct
polarisation-dependent measurements, one
could turn off one of the distances in one orientation (and perhaps the
other in another), one could fit the
two sets to extract the two close distances. (i.e. you aren't trying to
resolve the two distances in one measurement)
2. If there is a pronounced chemical difference between the neighbours,
they can be resolved even if the
distance separation is small, provided the k-space range being used
provides enough independent parameters
to conduct the fit.
If you wish to convince yourself that these two caveats work, I suggest
creating a model using the AuCu
structure. If you contract or expand the c-axis, and substitute either
Cu for Au (i.e. make Cu but in lower symmetry)
or substitute a different atom on one of the Cu sites, you can test both
Matt's explanation and my caveats above.
On 1/31/2016 6:33 PM, Matt Newville wrote:
> Hi Riti,
> On Sat, Jan 30, 2016 at 7:35 PM, Ritimukta Sarangi
> <ritimukta at gmail.com <mailto:ritimukta at gmail.com>> wrote:
> 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,
> The deltaR = pi / 2DeltaK follows from general Fourier analysis and
> formulas like it can be found in many signal processing textbooks.
> For on-line resources, googling "Frequency resolution Fourier
> transform" gives several good references.
> The idea is that (using sound-waves as an example) in order to
> distinguish two close frequencies (say 440 Hz from 441 Hz, so a
> different of 1 Hz), you have to sample many periods (pi seconds) to be
> able to do this.
> For EXAFS, if there are contributions from two neighbors that are very
> closely spaced, you would have to sample enough oscillations (go high
> enough in k) to see the effect of these two different distances
> beating against each other. If you don't go out far enough in k,
> you can't tell that these two contributions are actually from
> different distances.
> Hope that helps,
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