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Joint reconstruction of x-ray fluorescence and transmission tomography

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Abstract

X-ray fluorescence tomography is based on the detection of fluorescence x-ray photons produced following x-ray absorption while a specimen is rotated; it provides information on the 3D distribution of selected elements within a sample. One limitation in the quality of sample recovery is the separation of elemental signals due to the finite energy resolution of the detector. Another limitation is the effect of self-absorption, which can lead to inaccurate results with dense samples. To recover a higher quality elemental map, we combine x-ray fluorescence detection with a second data modality: conventional x-ray transmission tomography using absorption. By using these combined signals in a nonlinear optimization-based approach, we demonstrate the benefit of our algorithm on real experimental data and obtain an improved quantitative reconstruction of the spatial distribution of dominant elements in the sample. Compared with single-modality inversion based on x-ray fluorescence alone, this joint inversion approach reduces ill-posedness and should result in improved elemental quantification and better correction of self-absorption.

© 2017 Optical Society of America

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Figures (11)

Fig. 1
Fig. 1 Top view of the geometry used in x-ray fluorescence microscopy. The x-ray beam is treated as a pencil beam in the z direction that is raster-scanned across the specimen in 1D in the x direction, and the specimen is then rotated before another image is acquired (successive 2D planes are imaged by motion of the 3D specimen in the y direction, into/out of the plane of this top view). The x-ray transmission signal (absorption) is recorded, and the x-ray fluorescence (XRF) signal is recorded over an angular range of Ωv by using an energy dispersive detector located at 90° to the beam, in the direction of the elastic scattering minimum for a horizontally polarized x-ray beam. The grid overlay on the specimen shows its discretization with a pixel size of Lv; the set of pixels (in 2D; voxels in 3D) through which the XRF signal might undergo self-absorption in the specimen is indicated in orange.
Fig. 2
Fig. 2 Illustration of the x-ray fluorescence self-absorption effect, and how x-ray transmission can be used to recognize and correct for it. We show here a specimen composed of cylinders, or circles in this top view. The largest is of borosilicate glass (composition described in Sec. 2) with 200 μm diameter, followed by tungsten (W) with 10 μm diameter, and gold (Au) with 10 μm diameter. As 1D scans in beamlet positions τ are collected at successive specimen rotation angles θ, one builds up sinograms or (τ, θ) views of elemental x-ray fluorescence (XRF) signals such as the Si XRF signal shown in the middle, as well as 12.1 keV x-ray transmission (XRT) sinograms as shown at right (based on absorption contrast). If there is no self-absorption of the fluorescence signal, one obtains a Si XRF sinogram as shown in the top row, where the incident x-ray beam is partially absorbed in the small W and Au wires as they rotate into positions to intercept the incident beam before it reaches the glass cylinder. However, the 200 μm diameter glass cylinder is large compared to the 1.66 μm absorption length μ−1 of Si 1 x-rays in the glass as shown in Table 1, so that a fraction 1 − exp[−200/1.66] ≃ 1 – 5 × 10−53 of the Si XRF signal will be self-absorbed in the rod. As a result, the Si XRF signal will be detected mainly when the incident beam is at the left side of the scan; this leads to the Si XRF sinogram shown in the middle row (the sinogram also shows absorption of the Si XRF signal in the W and Au wires as they rotate through positions where they partly obscure the XRF detector’s view of the Si cylinder). In the bottom row we show the case where the glass cylinder is hollow, with a wall thickness of 30 μm that is nevertheless large compared to the 1.66 μm absorption length of Si XRF photons; in this case the Si XRF sinogram is almost unchanged, but the XRT sinogram is clearly different. By using the combined information of the fluorescence (XRF) and transmission (XRT) sinograms, one can in principle obtain a better reconstructed image of the specimen in the case of strong fluorescence self-absorption.
Fig. 3
Fig. 3 Relative elemental concentrations obtained from a MAPS-based fit of the raw x-ray fluorescence data for the glass rod sample. Due to the imperfection of fitting and background rejection (which might be able to be corrected with additional expert input), the decomposed elemental concentrations show certain artifacts, where certain elemental sinograms pick up other elements’ signals. For example, according to our knowledge of the sample, we know that Si exists only in the rod part with a cylinder shape; but its corresponding sinogram shows that it also exists in the two wires, which is caused by imperfect data fitting. Those two extra curves in the sinogram are actually picked up from Au and W signals because certain emission lines of Au and W overlap those of Si. Similar artifacts happen to Au and W sinograms as well.
Fig. 4
Fig. 4 Experimental sinograms. Left: mean (across energy channels) value of XRF raw spectrum; Right: XRT optical density. Based on the different magnitudes of these two datasets, we choose β2 = 100 as the scaling parameter to balance the measurement variability of the two data sources, so that both measurements have maxima near 15 in their respective units. As a result, the relative variability of the two detectors between the data sources is mitigated.
Fig. 5
Fig. 5 Method for choosing the parameter β1 that appears in the cost function of Eq. (8): XRF objective value ϕ ˜ R versus XRT objective value ϕ ˜ T given different values β1, with fixed β2 = 100. The curve displays the tradeoff between these two modalities.
Fig. 6
Fig. 6 Method for choosing the parameter β1 that appears in the cost function of Eq. (8): The curvature from successive points in Fig. 5; the point with maximum curvature occurs at β1 = 1.
Fig. 7
Fig. 7 Comparison of reconstruction results for MAPS+TomoPy, XRF alone, and joint reconstruction, respectively, given an initial guess of all zeros, β1 = 1, and β2 = 100. Every elemental map is rescaled to the range [0, 0.5]. It is clear that the joint reconstruction returns the best result from two perspectives: first, the glass rod is filled with Si; and second, the “elemental crosstalk” is dramatically mitigated for the reconstruction of Si and W.
Fig. 8
Fig. 8 Convergence of TN for each inner iteration j, given a maximum number of inner iterations as 52, β1 = 1, and β2 = 100. We can see that along the iterations, TN is reducing the objective function so that the forward model fits better and better to the given data.
Fig. 9
Fig. 9 Solution for each outer iteration i, given (β1, β2) = (1, 100). At iteration i = 3, Alg. 1 reaches its stopping criterion in the sense that the solution does not change anymore. The results also indicate that our alternating algorithm requires very few outer iterations.
Fig. 10
Fig. 10 Reconstruction results (given β2 = 100) for different β1 values. Apart from the two extreme cases (β1 = 0.001, where XRF dominates, and β1 = 100, where XRT dominates), the reconstructions do not show clear difference in terms of quality. Therefore, for a broad range of β1 values, the joint reconstruction is able to dramatically improve upon the single-modality reconstructions.
Fig. 11
Fig. 11 Example x-ray fluorescence spectrum. In this case, an incident beam with a photon energy of 12.1 keV is used to excite x-ray fluorescence from a specimen consisting of a borosilicate glass cylinder comprised mainly of SiO2 but with other elements present, and tungsten (W) and gold (Au) wires. The experimental spectrum is averaged over all positions of a sinogram from one scan row. The simulated spectrum based on the reconstructed elemental map is generated by the forward model described in Sec. 3; it includes tabulated [33] x-ray fluorescence lines for all elements present in the specimen along with the Gaussian energy response of the fluorescence detector, plus the background spectrum from non-specimen areas. Some additional background is present in the 4–7 keV energy range due to the materials in the experimental apparatus as indicated at specific fluorescence peaks; because this background does not change whether or not a specimen region is illuminated, it does not affect our analysis.

Tables (3)

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Table 1 X-ray absorption lengths μ−1 for silicon in a borosilicate glass, tungsten and gold at the energies of selected x-ray fluorescence lines and 12.1 keV as incident x-ray energy.

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Algorithm 1 Algorithm for Solving Joint Inversion with Linearized XRF Term.

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Algorithm 2 Backtracking Line Search

Equations (18)

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F ˜ θ , τ T ( μ ˜ E ) = I 0 exp { v L v θ , τ μ ˜ v E } ,
F ¯ θ , τ T ( W ) = I 0 exp { v , e L v θ , τ μ v E W v , e } .
F θ , τ T ( W ) = v , e L v θ , τ μ v E W v , e .
I e , , s = I 0 c e ω e , ( 1 1 r e , s ) μ e E ,
[ 1 E e x ] i : = { 1 if | x i E e | = min j ( | x j E e | ) and x i 2 E e x i 1 0 otherwise .
M e , , s = 1 ( ( I e , l , s x ) ( 1 2 π σ exp { x 2 2 σ 2 } ) )
A v E , θ , τ ( W ) = exp { v μ ˜ v E L v θ , τ I v U v θ , τ } = exp { v e W v , e μ e E L v θ , τ I v U v θ , τ } ,
P v , e θ , τ ( W ) = exp { v Ω v e W v , e μ e E e a ( Ω v ) | { v : v Ω v } | } ,
F θ , τ R ( W ) = e ( v L v θ , τ A v E , θ , τ ( W ) P v , e θ , τ ( W ) W v , e ) M e .
min W 0 ϕ ( W ) ,
ϕ ( W ) = ϕ ˜ R ( W ) + ϕ ˜ T ( W ) = θ , τ ( F θ , τ R ( W ) ln ( F θ , τ R ( W ) ) D θ , τ R ) + β 1 θ , τ ( F θ , τ T ( W ) ln ( F θ , τ T ( W ) ) β 2 D θ , τ R ) ;
min W 0 ϕ i ( W ) ,
ϕ i ( W ) = θ , τ e , v L v θ , τ A v E , θ , τ ( W i ) P v , e θ , τ ( W i ) W v , e M e θ , τ ln ( e , v L v θ , τ A v E , θ . τ ( W i ) P v , e θ , τ ( W i ) W v , e M e ) D θ , τ R + β 1 θ , τ ( F θ , τ T ( W ) ln ( F θ , τ T ( W ) ) β 2 D θ , τ T ) ,
W i + 1 = TN ( ϕ i ( W ) , W i , k ) ,
f ( D j ; F j ( W ) ) = Pr ( X = D j ) = ( F j ( W ) ) D j exp { F j ( W ) } D j ! .
max W ln ( j f ( D j ; F j ( W ) ) ) = j ln ( f ( D j ; F j ( W ) ) ) = j ln ( F j ( W ) D j exp { F j ( W ) } D j ! ) = j ( ln ( F j ( W ) D j ) + ln ( exp { F j ( W ) } ) ln ( D j ! ) ) .
max W ψ ( W ) = j ( D j ln ( F j ( W ) ) F j ( W ) ) .
W v , e ψ ( W ) = j ( D j F j ( W ) 1 ) W v , e F j ( W ) .
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