[Ifeffit] Re: Independent data poins in Artemis

Kropf, Arthur Jeremy kropf at cmt.anl.gov
Thu Jan 4 14:20:31 CST 2007

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 at cmt.anl.gov

> -----Original Message-----
> From: ifeffit-bounces at millenia.cars.aps.anl.gov 
> [mailto:ifeffit-bounces at 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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