Hi Bindu,
> > Why the peak in the Fourier transform of the experimental > > spectrum is less than the actual one. > The EXAFS equation is: > N S02 F(K) > chi(k) = ----------- exp(-2k^2 sigma^2) sin(2kR + phi(k)) > 2kR^2 > In the oscillatory term, there is a piece that is the phase shift > associated with the scattering of the photoelectron. When you do a > Fourier transform, the peak of the sin wave is not at R, rather it is > shifted by an amount that depends on the size of phi(k). That should be, of course, "...the peak of the *Fourier transform* of the sine wave..." B -- Bruce Ravel ----------------------------------- [EMAIL PROTECTED] National Institute of Standards and Technology Synchrotron Methods Group at Brookhaven National Laboratory Building 535A Upton NY, 11973 My homepage: http://cars9.uchicago.edu/~ravel EXAFS software: http://cars9.uchicago.edu/~ravel/software/exafs/
<<winmail.dat>>
_______________________________________________ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
