On Wed, Jul 08, 2009 at 11:43:08AM -0700, Scott Webster wrote: > I think a great deal (or perhaps all) of the confusion comes from the > continuous-discrete issue.
Part of it, but see below. > ...when you perform a discrete approximation to an integral you need to > sum the data points and multiply by the data point spacing. Since the > fourier transform is an integral shouldn't you multiply by the spacing > here too? Yes, but you also divide by the area at the end... Considering units, you need to look at the resulting units after all the operations. > Anyway, so when you take the FT of a scan line (this appears to be how > the PSDF function in the statistics tool does it, averaging in the > other direction) you get the the units of "z x" or length^2 in this > case (if you follow my above advice and multiply by the data point > spacing). Then when you square the FT you get length^4. That's what > I thought PSD should be length^4 in my previous email. It would > appear that to get the proper units for the PSD you have to divide out > by the scan size after squaring the FT. I kinda looked at the code > and I think this probably is happening. Here's the other source of confusion. Let's assume: c = ∫ w(k) dk C = ∫ w²(k) dk Denoting [x] the units of x we have: [c] = [w][k] [C] = [w]²[k] Hence, generally [C] = [c]²/[k] ≠ [c]² This means that the square of modulus density (coefficients) is not power density and vice versa. If PSDF is our central quantity we have to consider it directly instead of starting from coefficients. And this was just units, it gets more hairy when you start considering values. So, the last time I went through this I walked from the other end. We want the PSDF to have certrain properties: 1) Scale-resolution independency. That's the two properties detailed in my previous e-mail. They can be prectically verified and they uniquely determine what unit-bearing factors you need to include and how (numerical factors may still need to be decided). 2) Fitting the PSDF expressions in Graph -> Fit gives the same σ as Statistical Quantities (for surfaces that have the right statistics at least, fortunately artifical Gaussian rough surfaces can be easily generated so we can check that). If this is satisfied, maybe we differ in the number of 2s and πs in the formulas from others but at least we are consistent. Maybe Petr remembers why he chose the normalization in the PSDF expressions so, or were it came from, I don't. Anyway, if our PSDF has properties 1) and 2) we are happy for all practical purposes. I tried it again now, because bugs occur, however, it seems to me that 1) and 2) are satisfied by Gwyddion PSDF. I hope this also answers the Parseval theorem comment. Regards, Yeti ------------------------------------------------------------------------------ Enter the BlackBerry Developer Challenge This is your chance to win up to $100,000 in prizes! For a limited time, vendors submitting new applications to BlackBerry App World(TM) will have the opportunity to enter the BlackBerry Developer Challenge. See full prize details at: http://p.sf.net/sfu/Challenge _______________________________________________ Gwyddion-users mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/gwyddion-users
