xBSM Analysis - Dan Peterson
Review of the optics elements: Pinhole (“GAP”), FZP, Coded Aperture
Extracting information from the GAP
what is the GAP width? introduction to diffraction calculations
Extracting information from the FZP
how is the in-focus image separated from the out-of-focus?
is there an intrinsic image width?
Extracting information from the Coded Aperture
how can this image be fitted?
Results from a representative set of runs
Pin Hole (GAP)
Height is about 50μm,
optimized for the minimum
image size.
Fresnel Zone Plate (FZP)
Optimized for 2.5 keV
239 rings, center opaque
1200 μm diameter,
diam. of center opaque: 77.8 μm,
width of last ring: 1.26 μm.
Focus if by diffraction; paths through
successive rings differ by one
wavelength.
Coded Aperture (CA)
Height is 310 μm.
(Yellow is transmitting.)
Transmitting slits range in size from
40μm (begin diffraction) to
10μm (very diffracting).
The image from a single 10μm slit is
much wider than that of the 40 μm
The pinhole is a simple camera – similar to what
can be used to view a solar eclipse.
I made one of these contraptions in 1959.
It was July 20, 1963 when a total eclipse
was visible from Nova Scotia.
The image size (σi) is a convolution of components due to
the source (or object) size (σS) and
the pinhole (or GAP) size (σG).
σS is the beam size
at the source (our product)
o
G is the full width of the pinhole,
σG is the pinhole size; σG =G/sqrt(12)
σi is the total image size
as observed on the detector
i
σiG pinhole image
σiG is the pinhole image size as observed on the detector
σSG is the pinhole image size “referred to the source”
σSG = σiG /M, where M is the magnification, i/o = 2.34
We have been using the formula:
σS = [(σi/M)2 - σSG2 ]1/2 (which is equivalent to σS =[ σi2 - σiG2 ]1/2 /M )
Problem: calculating σiG (and hence σSG ) is tricky.
The image is a projection or “shadow” for large pinhole size and a diffraction pattern
for small pinhole size. We have been using σSG = 16μm; I recommend 21μm.
For large pinhole size, the image is a projection: σiG ~ G.
For small pinhole size, the image is a diffraction pattern, σiG ~ 1/G.
There is a value of G that gives a minimum σiG .
We minimize the image size, σi , while varying G (actually TASGAP).
Above are the measurements for C-line from 20100803 and 20101223
where the minimum was found at TASGAP=4.55 and 4.57.
We do not know (or care) what is G,
but must rely on a calculation to determine σiG .
JimA did calculations in December 2009, using the x-ray spectrum,
where he finds σSG between 14.1 and 17.0μm.
We are using σSG = 16 μm.
This could be revisited.
I wrote a program (an Excel spreadsheet) to perform the numerical integration:
∑ exp(i 2π P(x) /λ T(x) )
where x is a position along the optics element,
P(x) is the path length from the source, to the position on the optic, to the detector pixel,
T(x) is the transparency of the optics element at that position (1 or 0).
Pulse height is calculated for 64 bins of size 25μm (1/2 pixel).
Limitations: the paths are separated by 0.5 μm at the optic.
It does not integrate over x-ray energy; I use one energy at a time.
G= 10μm
20μm
30μm
80μm
100μm
240μm
The minimum image is found with G =45μm.
However, extracting σiG from this distribution is not simple.
Projecting the pinhole, G: σSG=G/sqrt(12)*(i+o)/o /M=18.5 μm.
Simple measurement: Γ=5.1 bins; σiG =54.1; σSG = 23.1 μm.
The diffraction bumps will be interpreted as background when
fitting the data and must be treated as background when
determining the width of the central peak.
G= 45μm
RMS to the first zero crossing
of the background-subtracted image
( case of 30 μm is shown)
To define the central peak of the image,
a cut-off width must be determined.
At any candidate cut-off width,
the next 7 bins are used to define the background.
The cut-off width is chosen where
the background subtracted distribution crosses zero.
The RMS calculated within the cut-off width.
Note that the RMS of the full distribution is much
larger.
The minimum RMS image size,
for x-ray energy about 2keV, is
σiG = 49 μm.
σSG = σiG / M = 21 μm
The Fresnel Zone Plate produces a true focus.
The image, for σS=0, is so small that it can be
lost with the 25μm bins.
I model a FZP with 7 transmitting rings.
25 μm bins
2 μm bins
Image at the optimization x-ray energy, 2.5keV
The distance, zero-to-zero, is 26 μm.
Ex-ray = 2.1
2.3
2.4
one pixel
Ex-ray = 2.5
2.6
2.7
Chromatic aberration is severe, bandwidth ≈ 0.1keV.
Off energy, but within the bandwidth, the peak does not broaden; it get smaller.
The “wide part” is due to the off-energy x-rays and will be relatively insensitive to beam size.
The width of the central peak is only 5μm; the width “referred to the source” is negligible.
The only contribution that that can be removed from the observed image is due to the pixel size:
σSG = 50μm / 121/2 / 2.34 = 6.2 μm.
2.9
The coded aperture is an array of slits ranging
is width from 10μm to 40μm.
25 μm bins
Shown is the image for σS=0. (blue points)
With a goal of developing a parameter-based,
rather than template-based fitting procedure,
the red line is the
sum of 12 gaussian distributions:
F0 = ∑J=1,12 AJ exp ( -½ ( x-x0J)2/(σJ2) )
The 36 parameters: AJ , x0J , σJ , are fixed.
This is the red line.
A coded aperture, with template fitting, can be used solve more complicated problems:
multiple sources and unknown energy in addition to beam size.
xBSM must solve only beam size, location, and intensity;
there is a single source, the energy spectrum is fixed.
The variation of F with beam size, location, and intensity is straightforward.
A parameter-based fitting function with
12 gaussian distributions:
From the previous page:
σS= 0
F0 = ∑J=1,12 AJ exp ( -½ ( x-xJ0)2/(σJ2) )
The 36 parameters: AJ , x0J , σJ , are fixed.
For non-zero source size ,
contributions from the source add incoherently.
There are 3 floating variables:
C, an overall normalization,
X0, an overall position offset, and
σi, the image width due to the beam size.
F = C ∑J=1,12 AJ (σJ) /(σJ2 + σi2)½
exp ( -½ ( x-x0J - X0)2/(σJ2 + σi2 ) )
Shown at right is the calculation and function
for σS= 15 μm.
σS= 15 μm
So, now there is a function for fitting each of the images.
PinHole
1 gaussian + background,
4 parameters
Extract the beam size: σS = [(σi/M)2 – 21μm 2 ]1/2
FZP
2 gaussians + background, 4 parameters
The wide component has fixed width
The wide part area fraction is fixed at 88%.
The central values are tied.
The narrow component is the measurement.
Extract the beam size: σS = [(σi/M)2 – 6μm 2 ]1/2
Coded Aperture
F = C ΣJ=1,12 AJ (σJ) /(σJ2 + σi2)½
exp ( -½ ( x-x0J - X0)2/(σJ2 + σi2 ) )
Extract the beam size: σS = σi/M ;
there are no other corrections.
The beam position is measured for individual turns.
The peak is located. Then the average is measured within a window about the peak.
(The window is different for each optics element.)
The individual turn is shifted to the average beam position and accumulated in an
average distribution. Only the average distribution is fitted.
And this is a comparison of a set of runs, using the 3 optics elements,
taken during beam dynamics studies, 2010-12-20
(sigma of position is at the image, not source)
In conclusion…
Have parameter based fitting of all three optics: pinhole, FZP, coded aperture.
Improvements can be made to
the pinhole size correction and
coded aperture function at σS=0
by introducing the energy distribution to the optics calculation.
(It is straight forward to average the result of 5 energies.)
Work is needed to better understand the FZP function:
improved parameterization of the wide part, including the asymmetry,
improved estimate of the fraction of the signal in the narrow part.
Very low beam size data is needed.
Have not tested the beam center finding for the coded aperture for large motion.
The original reason for creating the numerical integration
was to look at the image KEK Coded Aperture.
The KEK Coded Aperture has all features reduced to half
size. The image for 2keV, σS= 15 μm, does not have wellresolved peaks. This is why we replaced it with the
Canadian Coded Aperture in the C-line.
At 4keV, the half size Coded Aperture produces resolved
peaks.
The peaks of the Coded Aperture are not
direct images of individual slits. This is
very much a diffractive device.
The first 10μm slit , alone, produces a
broad diffraction peak.
The first two 10μm slits produce a
diffraction pattern with 4 easily visible
peaks.
The widest slit, at 40μm, produces a peak
similar to the optimized GAP image, but
does not directly correspond to the large
feature in the image.