[Ifeffit] Distortion of transmission spectra due to particlesize
Kropf, Arthur Jeremy
kropf at anl.gov
Wed Nov 24 15:47:18 CST 2010
That looks right except for a minor quibble. Your calc is for 1/3 foil and 2/3 "pinhole". I think the equation should have an extra set of parenthesis:
set pinhole.xmu = -ln((1 + exp(-good.xmu))/2)
Jeremy
PS. That is a whole lot easier than what I did for the plot - export data to another program, do some calculations, and then reimport the results to Athena. I really need to learn to use ifeffit better. Thanks for the demonstration. :)
> -----Original Message-----
> From: ifeffit-bounces at millenia.cars.aps.anl.gov
> [mailto:ifeffit-bounces at millenia.cars.aps.anl.gov] On Behalf
> Of Matt Newville
> Sent: Wednesday, November 24, 2010 3:01 PM
> To: XAFS Analysis using Ifeffit
> Subject: Re: [Ifeffit] Distortion of transmission spectra due
> to particlesize
>
> 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
> <kropf at 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-bounces at millenia.cars.aps.anl.gov
> > [mailto:ifeffit-bounces at 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-bounces at 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: kropf at anl.gov
> >
> >
> >> -----Original Message-----
> >> From: ifeffit-bounces at millenia.cars.aps.anl.gov
> >> [mailto:ifeffit-bounces at 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 at millenia.cars.aps.anl.gov
> >> http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
> >>
> >
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