24 Nov
2010
24 Nov
'10
6:08 p.m.
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@anl.gov
> -----Original Message-----
> From: ifeffit-bounces@millenia.cars.aps.anl.gov
> [mailto:ifeffit-bounces@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
> >
>
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