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
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
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,
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
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) /
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
You are right about the amplitude factor: it should change. The ln(1+x/2) is
not the full story, since x is the exponent: x= exp(-mu*t) and when mu*t is
small (thin samples), x is not, and vice versa. More accurate expansion of
ln(1+x/2) in the limit of thick and thin samples shows that
