Re: [Ifeffit] pi/2deltak

[Ritimukta Sarangi]

Thank you very much for the explanations and your time.
Best Wishes,
-Riti



> On Feb 1, 2016, at 9:16 AM, Robert Gordon  wrote:
> 
> Hi Riti,
> 
> 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.
> 
> cheers,
> Robert
> 
> On 1/31/2016 6:33 PM, Matt Newville wrote:
>> Hi Riti, 
>> 
>> On Sat, Jan 30, 2016 at 7:35 PM, Ritimukta Sarangi < 
>> ritimu...@gmail.com 
>> > wrote:
>> Hello,
>> 
>> 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,
>> Best,
>> -Riti
>> 
>> 
>> 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, 
>> 
>> --Matt
>> 
>> 
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Re: [Ifeffit] pi/2deltak

[Robert Gordon]

Hi Riti,

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.


cheers,
Robert

On 1/31/2016 6:33 PM, Matt Newville wrote:

Hi Riti,

On Sat, Jan 30, 2016 at 7:35 PM, Ritimukta Sarangi 
> wrote:


Hello,

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,
Best,
-Riti


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,

--Matt


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Re: [Ifeffit] pi/2deltak

[Farquhar, Erik]

Any references to this quantity in more recent papers seem to point back to the 
1981 Lee/Citrin/Eisenberger/Kincaid Rev. Mod. Phys. Review - 
http://dx.doi.org/10.1103/RevModPhys.53.769 (see p. 794 in particular).

Erik

From: Ifeffit [mailto:ifeffit-boun...@millenia.cars.aps.anl.gov] On Behalf Of 
Ritimukta Sarangi
Sent: Saturday, January 30, 2016 8:35 PM
To: XAFS Analysis using Ifeffit
Subject: [Ifeffit] pi/2deltak

Hello,

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,
Best,
-Riti

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Re: [Ifeffit] pi/2deltak

[Avakyan L.A.]

Hello Riti,

the criterion of R-resolution of pi/2deltak can be drawn from the 
Nyquist theorem.
The amount of information in the spectra is limited by Nidp = 2 deltak 
deltaR / pi.
Then, in assumption of equal distribution of information over the 
R-interval, the minimal interval can be estimated as deltaR / Nidp = 
pi/2deltak.
The problem is that exact formula for Nidp is not known. Sometimes it 
has addition of +1 or +2 to represent the fact that single point 
(deltak=0) still contains information.
Moreover, information is not equally distributed: some k-regions are 
more rich, other are less...


And when energy interval is very short (glass, liquid) this question 
becomes vital. Sometimes, this resolution limit can even be overcome as 
it is shown in [http://dx.doi.org/10.1103/PhysRevB.82.064204].


Best regards,
Leon

Leon Avakyan
PhDr, Physics Faculty,
Southern Federal University
laavak...@sfedu.ru


On 31.01.2016 21:00, ifeffit-requ...@millenia.cars.aps.anl.gov wrote:

Message: 2
Date: Sat, 30 Jan 2016 17:35:04 -0800
From: Ritimukta Sarangi<ritimu...@gmail.com>
To: XAFS Analysis using Ifeffit<ifeffit@millenia.cars.aps.anl.gov>
Subject: [Ifeffit] pi/2deltak
Message-ID:
<camwif7ymrkedszw31pcgwczzh5tsfvfk+151gxfjrz9juar...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"

Hello,

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,
Best,
-Riti


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Re: [Ifeffit] pi/2deltak

[Matt Newville]

Hi Riti,

On Sat, Jan 30, 2016 at 7:35 PM, Ritimukta Sarangi 
wrote:

> Hello,
>
> 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,
> Best,
> -Riti
>
>
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,

--Matt
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[Ifeffit] pi/2deltak

[Ritimukta Sarangi]

Hello,

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,
Best,
-Riti
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