[Ifeffit] Rebinning

Carlo U. Segre segre at iit.edu
Wed Mar 12 15:19:09 CST 2003


Matt:

I think that I didn't explain clearly.  First of all, I misused the name
boxcar.  In the most stringent of definitions we are not doing a boxcar
average because we make no assumptions of where the energy position is to
be.  Let me try to restate it.

1. There is no effort to put the data on a uniform energy grid.  The date
from a continuous scan is never on an even grid anyway, just
approximately.  This is because, at least at MR-CAT, we just count the
encoder steps and use that to determine the angular position and thence
the energy.  Since we can only se our monochromator to move at a uniform
angular speed (and even the spacing in angle is not always perfectly
uniform), the energy spacing will have a siusoidal dependence at the best.

2. There is no effort to put the data on an even grid in k-space.  The
purpose of the delta-k is to set the window over which we average the
data.  The algorithm determines which points in energy space are within
the desired box in k-space and then it finds the center of mass of the
points by averaging the counts AND averaging the energy too.

The resultant spectrum will have datapoints which have the desired density
but which will not be on any kind of regular grid.  The idea is to let the
analysis programs such as IFEFFIT interpolate and regrid.

Carlo

On Wed, 12 Mar 2003, Matt Newville wrote:

>Hi Carlo,
>
>On Tue, 11 Mar 2003, Carlo U. Segre wrote:
>
>> Matt:
>>
>> Jeff Terry mentioned that perhaps I did not explain myself adequately with
>> the last message.
>>
>> The algorithm averages BOTH counts and Energy.  We do not just place the
>> average number of counts at the center of the bin.  Because of this, there
>> should be no distortion and no need to weight.
>>
>> Carlo
>
>I'm not sure I see that at first glance, but I'll take your word
>for it.  That isn't what I'm concerned about.
>
>Let me explain where I get stuck: let's say you have data
>collected in energy steps of 0.25eV -- could be QEXAFS, could be
>step scan.  For the sake of argument, lets' set the energy
>resolution to 1eV.  You have to get the data on the 0.25eV grid
>to an even k-grid for the analysis (at least in ifeffit).  For
>the sake of argument, we'll say dk = 0.05Ang^-1, No matter what
>the details are, you need to get values of mu(E) for the data
>onto a gridded set E={E_i}.
>
>At k=4, E~=61.0, and k=0.05 corresponds to 1.5eV.  A boxcar
>average will average the original data between 60.25 and 61.75
>and call that the new data for E=61.  Seems reasonable, though
>giving equal weight to the data at 61.0 and 61.75 could be
>questioned for 1eV resolution.
>
>At k=16, E~=975.0 eV, k=0.05 corresponds to 6eV.  So here, you
>average of data between E=972.0 and 978.0eV and call that the
>data for E=975.0eV.  Giving equal weight to the data at 972.0 and
>975.0 when the resolution is 1eV is what worries me.  The simple
>average of mu(E) between 972.0 and 978.0 is definitely not the
>same as mu(E) at E=975.0. It would probably be better to average
>the original data between 974.0 and 976.0 for the new data at
>975.0, and might be preferrable to just do a convolution with a
>1eV point spread function for all the data.
>
>Certainly, if you were to re-bin data collected in 0.25eV steps
>to a grid of 10eV there would be real problems.  By 'using all
>the data', one can use too much data and spoil the resolution.
>
>I don't claim that the boxcar average is wrong, or that
>convolution is definitely the right thing to do, just that I'm
>confused by this.
>
>--Matt
>
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-- 
Carlo U. Segre -- Professor of Physics
Associate Dean for Research, Armour College
Illinois Institute of Technology
Voice: 312.567.3498            Fax: 312.567.3494
Carlo.Segre at iit.edu    http://www.iit.edu/~segre



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