Hi All Greetings. I have a generic question. Does the Number of independent points in the data set depend on the step size (in k or E) of the scan during data collection? As I see Nidp = (2/pi)*(deltak * deltaR), there is no reference to the scan step. In a scan of small step size, I would expect a lot more data points than a scan with larger step size. But, as per the formula I am left with the same number of idp's irrespective of the scan step. Am I missing something? Regards Shan Shantanu Behera http://www.lehigh.edu/~skb204
Hi Shan, Nidp does not (to first order, anyway) depend on the step size used the data collection. The step size in k does dictate the highest frequency (R) that can be measured. This could, in principle, change the effective R range that could be used in the Nidp calculation, but it should never have a serious impact on it. Usually (at least in my experience), step sizes in k are around 0.05Angstroms-1 when collecting EXAFS data, which would give Rmax = pi / (2 * step_k) = 31.4 Angstroms. Of course, that is significantly larger R than we can hope to measure, but it does allow us to do "crude signal processing" steps like linear interpolation, re-binning of data as we shift E0 and so on without ruining the data below, say 15.7 Angstroms. So, for EXAFS it's probably dangerous to have a step size larger than 0.2Angstroms^-1, and not a great help to have step sizes smaller than 0.05Angstroms. For what it's worth, IFEFFIT/Artemis more or less insist on interpolating chi(k) onto a k-grid of 0.05Ang^-1. All that being said, I have not carefully tested how large the step size has to be to really impact the EXAFS below, say, 5Angstroms^-1. Cheers, --Matt
Hi Shan, The aspect you're missing is that the extra data points are not independent. Suppose you collect data points, with good statistics (lots of counts per point), at 8.05 and 8.10 inverse angstroms. Actual EXAFS features don't vary a whole lot over that interval, in general. So to a good approximation you could interpolate the values of chi(k) at 8.06, 8.07, 8.08, and 8.09 inverse angstroms. Collecting data at those points doesn't really tell you much you couldn't figure out anyway. (To make it a fair comparison, consider collecting 6 seconds of data each at 8.05 and 8.10 inverse angstroms to 2 seconds of data each at 8.05, 8.06, 8.07, 8.08, 8.09, and 8.10, so that it's the same total time of collection.) --Scott Calvin Sarah Lawrence College At 05:57 PM 3/18/2007, you wrote:
Hi All
Greetings. I have a generic question. Does the Number of independent points in the data set depend on the step size (in k or E) of the scan during data collection? As I see Nidp = (2/pi)*(deltak * deltaR), there is no reference to the scan step. In a scan of small step size, I would expect a lot more data points than a scan with larger step size. But, as per the formula I am left with the same number of idp's irrespective of the scan step.
Am I missing something?
participants (3)
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Matt Newville -
Scott Calvin -
Shan Behera