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.