Hi Jeremy, For a sample of thickness t in the beam, and a good measurement being I_t = I_0 * exp(-t*mu) I think having 1/2 the sample missing (and assuming uniform I_0) would be:
I_t_measured = (I_0 / 2)*exp(-0*mu) + (I_0/2) * exp(-t*mu) = I_0 * (1 + exp(-t*mu) / 2 So that t*mu_measured = -ln(I_t_measured/I_0) = -ln (1 + exp(-t*mu)/2) An ifeffit script showing such an effect would be: read_data(cu.xmu, group =good) plot good.energy, good.xmu set pinhole.energy = good.energy set pinhole.xmu = -ln(1 + exp(-good.xmu)/2) spline(pinhole.energy, pinhole.xmu, kmin=0, kweight=2, rbkg=1) spline(good.energy, good.xmu, kmin=0, kweight=2, rbkg=1) newplot(good.k, good.chi*good.k^2, xmax=18, xlabel='k (\A)', ylabel='k\u2\d\gx(k)', key='no pinholes') plot pinhole.k, pinhole.chi*pinhole.k^2, key='half pinholes' With the corresponding plot of k^2 * chi(k) attached. Corrections welcome, --Matt On Wed, Nov 24, 2010 at 2:27 PM, Kropf, Arthur Jeremy <kr...@anl.gov> wrote: > Anatoly, > > I think that may be exactly the point. If you have half the beam on a foil > and half off, even with a uniform beam, you cant get the same spectrum as > with the whole beam on the foil. > > I tried to come up with a quick proof by demonstration, but I got bogged > down on normalization. That will have to wait. > > Jeremy > > ________________________________ > From: ifeffit-boun...@millenia.cars.aps.anl.gov > [mailto:ifeffit-boun...@millenia.cars.aps.anl.gov] On Behalf Of Frenkel, > Anatoly > Sent: Wednesday, November 24, 2010 1:33 PM > To: XAFS Analysis using Ifeffit > Subject: RE: [Ifeffit] Distortion of transmission spectra due to > particlesize > > Jeremy: > > In your simulation, "(c) 1/2 original, 1/2 nothing (a large "pinhole")" it > appears that chi(k) is half intensity of the original spectrum. > Does it mean that when the pinhole is present, EXAFS wiggles are half of the > original ones in amplitude but the edge step remains the same? > Or, equivalently, that the wiggles are the same but the edge step doubled? > > Either way, I don't think it is the situation you are describing (a large > pinhole). If there is a large pinhole made in a perfect foil (say, you > removed half of the area of the foil from the footprint of the beam and it > just goes through from I0 to I detector, unaffected). > Then, if I0 is a well behaving function of energy, i.e., the flux density > is constant over the entire sample for all energies, EXAFS in the both cases > should be the same. > > Or I misunderstood your example, or, maybe, the colors? > > Anatoly > > ________________________________ > From: ifeffit-boun...@millenia.cars.aps.anl.gov on behalf of Kropf, Arthur > Jeremy > Sent: Wed 11/24/2010 1:08 PM > To: XAFS Analysis using Ifeffit > Subject: Re: [Ifeffit] Distortion of transmission spectra due to > particlesize > > It's not that I don't believe in mathematics, but in this case rather > than checking the math, I did a simulation. > > I took a spectrum of a copper foil and then calculated the following: > (a) copper foil (original edge step 1.86) > (b) 1/3 original, 1/3 with half absorption, and 1/3 with 1/4 absorption > (c) 1/2 original, 1/2 nothing (a large "pinhole") > (d) 1/4 nothing, 1/2 original, 1/4 double (simulating two randomly > stacked layers of (c)) > > Observation 1: Stacking random layers does nothing to improve chi(k) > amplitudes as has been discussed. They are identical, but I've offset > them by 0.01 units. > > Observation 2: Pretty awful uniformity gives reasonable EXAFS data. If > you don't care too much about absolute N, XANES, or Eo (very small > changes), the rest is quite accurate (R, sigma2, relative N). > > Perhaps I'll simulate a spherical particle next with absorption in the > center of 10 absorption lengths or so - probably not an uncommon > occurance. > > Jeremy > > Chemical Sciences and Engineering Division > Argonne National Laboratory > Argonne, IL 60439 > > Ph: 630.252.9398 > Fx: 630.252.9917 > Email: kr...@anl.gov > > >> -----Original Message----- >> From: ifeffit-boun...@millenia.cars.aps.anl.gov >> [mailto:ifeffit-boun...@millenia.cars.aps.anl.gov] On Behalf >> Of Scott Calvin >> Sent: Wednesday, November 24, 2010 10:41 AM >> To: XAFS Analysis using Ifeffit >> Subject: Re: [Ifeffit] Distortion of transmission spectra due >> to particlesize >> >> Matt, >> >> Your second simulation confirms what I said: >> >> > The standard deviation in thickness from point to point in >> a stack of >> > N tapes generally increases as the square root of N (typical >> > statistical behavior). >> >> Now follow that through, using, for example, Grant Bunker's >> formula for the distortion caused by a Gaussian distribution: >> >> (mu x)eff = mu x_o - (mu sigma)^2/2 >> >> where sigma is the standard deviation of the thickness. >> >> So if sigma goes as square root of N, and x_o goes as N, the >> fractional attenuation of the measured absorption stays >> constant, and the shape of the measured spectrum stays >> constant. There is thus no reduction in the distortion of the >> spectrum by measuring additional layers. >> >> Your pinholes simulation, on the other hand, is not the >> scenario I was describing. I agree it is better to have more >> thin layers rather than fewer thick layers. My question was >> whether it is better to have many thin layers compared to >> fewer thin layers. For the "brush sample on tape" method of >> sample preparation, this is more like the question we face >> when we prepare a sample. Our choice is not to spread a given >> amount of sample over more tapes, because we're already >> spreading as thin as we can. Our choice is whether to use >> more tapes of the same thickness. >> >> We don't have to rerun your simulation to see the effect of >> using tapes of the same thickness. All that happens is that >> the average thickness and the standard deviation gets >> multiplied by the number of layers. >> >> So now the results are: >> >> For 10% pinholes, the results are: >> # N_layers | % Pinholes | Ave Thickness | Thickness Std Dev | >> # 1 | 10.0 | 0.900 | 0.300 | >> # 5 | 10.0 | 4.500 | 0.675 | >> # 25 | 10.0 | 22.500 | 1.500 | >> >> For 5% pinholes: >> # N_layers | % Pinholes | Ave Thickness | Thickness Std Dev | >> # 1 | 5.0 | 0.950 | 0.218 | >> # 5 | 5.0 | 4.750 | 0.485 | >> # 25 | 5.0 | 23.750 | 1.100 | >> >> For 1% pinholes: >> # N_layers | % Pinholes | Ave Thickness | Thickness Std Dev | >> # 1 | 1.0 | 0.990 | 0.099 | >> # 5 | 1.0 | 4.950 | 0.225 | >> # 25 | 1.0 | 24.750 | 0.500 | >> >> As before, the standard deviation increases as square root of >> N. Using a cumulant expansion (admittedly slightly funky for >> such a broad >> distribution) necessarily yields the same result as the Gaussian >> distribution: the shape of the measured spectrum is >> independent of the number of layers used! And as it turns >> out, an exact calculation (i.e. >> not using a cumulant expansion) also yields the same result >> of independence. >> >> So Lu and Stern got it right. But the idea that we can >> mitigate pinholes by adding more layers is wrong. >> >> --Scott Calvin >> Faculty at Sarah Lawrence College >> Currently on sabbatical at Stanford Synchrotron Radiation Laboratory >> >> >> >> On Nov 24, 2010, at 6:05 AM, Matt Newville wrote: >> >> > Scott, >> > >> >> OK, I've got it straight now. The answer is yes, the >> distortion from >> >> nonuniformity is as bad for four strips stacked as for the single >> >> strip. >> > >> > I don't think that's correct. >> > >> >> This is surprising to me, but the mathematics is fairly clear. >> >> Stacking >> >> multiple layers of tape rather than using one thin layer >> improves the >> >> signal to noise ratio, but does nothing for uniformity. So there's >> >> nothing wrong with the arguments in Lu and Stern, Scarrow, >> etc.--it's >> >> the notion I had that we use multiple layers of tape to improve >> >> uniformity that's mistaken. >> > >> > Stacking multiple layers does improve sample uniformity. >> > >> > Below is a simple simulation of a sample of unity thickness with >> > randomly placed pinholes. First this makes a sample that >> is 1 layer >> > of N cells, with each cell either having thickness of 1 or >> 0. Then it >> > makes a sample of the same size and total thickness, but made of 5 >> > independent layers, with each layer having the same fraction of >> > randomly placed pinholes, so that total thickness for each >> cell could >> > be 1, 0.8, 0.6, 0.4, 0.2, or 0. Then it makes a sample with 25 >> > layers. >> > >> > The simulation below is in python. I do hope the code is >> > straightforward enough so that anyone interested can >> follow. The way >> > in which pinholes are randomly selected by the code may not be >> > obvious, so I'll say hear that the "numpy.random.shuffle" >> function is >> > like shuffling a deck of cards, and works on its array argument >> > in-place. >> > >> > For 10% pinholes, the results are: >> > # N_layers | % Pinholes | Ave Thickness | Thickness Std Dev | >> > # 1 | 10.0 | 0.900 | 0.300 | >> > # 5 | 10.0 | 0.900 | 0.135 | >> > # 25 | 10.0 | 0.900 | 0.060 | >> > >> > For 5% pinholes: >> > # N_layers | % Pinholes | Ave Thickness | Thickness Std Dev | >> > # 1 | 5.0 | 0.950 | 0.218 | >> > # 5 | 5.0 | 0.950 | 0.097 | >> > # 25 | 5.0 | 0.950 | 0.044 | >> > >> > For 1% pinholes: >> > # N_layers | % Pinholes | Ave Thickness | Thickness Std Dev | >> > # 1 | 1.0 | 0.990 | 0.099 | >> > # 5 | 1.0 | 0.990 | 0.045 | >> > # 25 | 1.0 | 0.990 | 0.020 | >> > >> > Multiple layers of smaller particles gives a more uniform thickness >> > than fewer layers of larger particles. The standard deviation of the >> > thickness goes as 1/sqrt(N_layers). In addition, one can >> see that 5 >> > layers of 5% pinholes is about as uniform 1 layer with 1% pinholes. >> > Does any of this seem surprising or incorrect to you? >> > >> > Now let's try your case of 1 layer of thickness 0.4 with 4 >> layers of >> > thickness 0.4, with 1% pinholes. In the code below, the simulation >> > would look like >> > # one layer of thickness=0.4 >> > sample = 0.4 * make_layer(ncells, ph_frac) >> > print format % (1, 100*ph_frac, sample.mean(), sample.std()) >> > >> > # four layers of thickness=0.4 >> > layer1 = 0.4 * make_layer(ncells, ph_frac) >> > layer2 = 0.4 * make_layer(ncells, ph_frac) >> > layer3 = 0.4 * make_layer(ncells, ph_frac) >> > layer4 = 0.4 * make_layer(ncells, ph_frac) >> > sample = layer1 + layer2 + layer3 + layer4 >> > print format % (4, 100*ph_frac, sample.mean(), sample.std()) >> > >> > and the results are: >> > # N_layers | % Pinholes | Ave Thickness | Thickness Std Dev | >> > # 1 | 1.0 | 0.396 | 0.040 | >> > # 4 | 1.0 | 1.584 | 0.080 | >> > >> > The sample with 4 layers had its average thickness increase by a >> > factor of 4, while the standard deviation of that thickness only >> > doubled. The sample is twice as uniform. >> > >> > OK, that's a simple model and of thickness only. Lu and >> Stern did a >> > more complete analysis and made actual measurements of the >> effect of >> > thickness on XAFS amplitudes. They *showed* that many thin >> layers is >> > better than fewer thick layers. >> > >> > Perhaps I am not understanding the points you're trying to >> make, but I >> > think I am not the only one confused by what you are saying. >> > >> > --Matt >> > >> >> _______________________________________________ >> Ifeffit mailing list >> Ifeffit@millenia.cars.aps.anl.gov >> http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit >> > > _______________________________________________ > Ifeffit mailing list > Ifeffit@millenia.cars.aps.anl.gov > http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit > >
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