RADON TRANSFORM, SCATTERING, AND IMAGING J. H´ ector Morales B´ arcenas [email protected]Departamento de Matem´ aticas Universidad Aut´ onoma Metropolitana Unidad Iztapalapa, Ciudad de M´ exico Seminario Control en Tiempos de Crisis 2020 22 de junio de 2020 J. H´ ector Morales B´ arcenas
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Itinerary
1. Preamble: Applying the Fourier Transform
Solving the 3D Structure of the DNA
2. X-Ray Tomography
Computed Tomography and the Radon Transform
3. Wave-Based Scattering Models
Reconstructive Tomography with Diffracting Wavefields
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PART I
A Prelude: Structure of DNA
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Solved Molecular Structure of Nucleic Acids
• “This structure has novel features which are of considerable biological
interest.”
Figure 1: J. D. Watson and F. H. C. Crick, April 1953 Nature, showing Photo 51.
• “We have also been stimulated by a knowledge of the general nature of
the unplublished experimental results and ideas of Dr. M. H. F. Wilkins,
Dr. R. E. Franklin and their co-workers at King’s College, London.”
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Solved Molecular Structure of Nucleic Acids
Figure 2: Rosalind E. Franklin, 1950.
• Interpretation of theX-Ray Photograph
“For a smooth single-strand helixstructure factor on the nth layer lineis given by Fn = Jn(2πR) exp in(ψ+π/2) . . . ”
• Diffraction by Helices
“. . . the intensity distribution inthe diffraction pattern of aseries of points equally spacedalong a helix is given bythe squares of Bessel functions.”
• Rosalind E. Franklin and R. G. Gosling, Molecular Configuration in SodiumThymonucleate, Nature, April 25, 1953.
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Inverse Problem solved by Cochran et al., 1952
• Uniform helix of infinite length, radius r and axial spacing P , defined by
(x, y, z) = (r cos(2πz/P ), r sin(2πz/P ), z),
T (ξ, η, ζ) =
ˆe2πi(xξ+yη+zζ)dV =
ˆ P
0
e2πi[rξ cos(2π zP )+rη sin(2π zP )+zζ]dz.
• In reciprocal coordinates with R2 = ξ2 + η2, tanψ = η/ξ, and ζ = n/P with n
integer
T (R,ψ, n/P ) =
ˆ P
0
e2πi[Rr cos(2πz/P−ψ)+nz/P ]dz.
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Inverse Problem solved by Cochran et al., 1952
• With the help of the integral´ 2π
0exp(iX cosϕ) exp(inϕ)dϕ = 2πinJn(X),
with X = 2πRr and ϕ = 2πz/P . The result is
T (R,ψ, n/P ) = Jn(2πRr) exp[in(ψ + π/2)],
where Jn denotes the nth-order Bessel function.
• “The formulæ (for scattering patterns) are given for the Fourier transforms of anumber of helical structures . . . ”
Figure 3: W. Cochran, F. H. C. Crick and V. Vand. The Structure of Synthetic Polypeptides. I.
The Transform of Atoms on a Helix. Acta Cryst. (1952). 5, 581.
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Crystallography in a Nutshell
• The relationship between the diffraction pattern (reciprocal space) and thecrystal structure (direct space) is mediated by a Fourier transform representedby the electron density function:
• The main contribution of the diffracted X-rays is that the intensity pattern inthe reciprocal space is proportional to the square of the amplitude
I(h, k, `) ∝ |F (h, k, `)|2.
There are physical factors that influence this intensity. However, a majorproblem is to determine the phases.
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Crystallography in a Nutshell
• The very first solution to the phase problem was introduced by Arthur Lindo
Patterson (1902-1966). After his training under Norbert Wiener, on Fourier
transforms convolution, Patterson introduced a new function P (u, v, w) in 1934,
which defines a new space (the Patterson space).
• To obtain Patterson’s function we cross-correlate the density function at two
different points over a unit cell
1
0
ρ(x, y, z)ρ(x+ u, y + v, z + w)dxdydz.
For continuous functions, the cross-correlation operator is the adjoint of the
convolution operator. After substituting the density function we get
P (u, v, w) =1
V
∑h,k,`
|F (h, k, `)|2 cos(2π[hu+ kv + `w]).
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Crystallography in a Nutshell
(a) (b)
Figure 4: (a) Two unit cells of a structure containing four atoms. (b) The Patterson map. The
positions at these maxima (u, v, w) represent the differences between the coordinates of each pair
of atoms in the crystal, i.e. u = x1 − x2, v = y1 − y2, w = z1 − z2 (Sands, 1975).
• The Patterson Space can be considered as the most important singledevelopment in crystal-structure analysis since the discovery of X-rays byRontgen in 1895 or X-ray diffraction by Laue in 1914.
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PART II
X-Ray Tomography
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X-Ray Tomography
• CT is the prime example of medical imaging.
• It comprises different techniques for (tomographic) imaging 2D
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X-Ray Tomography Pseudo-Inversion
• Actually, this filter resembles a bit a differential operator in the sense that
F [(H H)f ](σ) = |σ|2F (σ).
It turns out that |σ|2 is the (top-order) symbol of the Laplace operator −∆.
The reason is that FT replaces derivatives for multiplications.
• In the inversion formula the ‘annoying’ factor |σ| means that we differentiate
the data twice, then “change the sign and finally take the square root”.
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X-Ray Tomography Pseudo-Inversion
• In that way we don’t directly compute f but Λf which is defined through
F (Λf)(σ) = |σ|F (σ),
or Λf = (−∆)−1/2f .
• The same result is found in RN by
Λf(x) = cN∆N
ˆSN−1
(Rf)(θ,x · θ)dθ.
A deep mathematical result states that this pseudo-differential operator Λ
preserves the singular support, i. e. Λf has jumps whenever f has jumps
(Louis, 1992).
§
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X-Ray Tomography Inversion via Hilbert Transform
Theorem 5 (Inversion Formula RN). (Louis, 1992). Given Rf on the cylinder
SN−1 × R the function f is computed as back-projection
f(x) = 2−1(2π)1−NˆSN−1
g(θ,x · θ)dθ
of the function g given as
G(θ, σ) = |σ|N−1F [Rf ](θ, σ).
• If we avoid the use of the FT in g we note that, for odd N we have
|σ|N−1 = σN−1, and the multiplication of the FT with its argument correspond
to differentiation.
• In even dimensions we have |σ|N−1 = sign(σ)σN−1.
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X-Ray Tomography Inversion via Hilbert Transform
• The multiplication with the sign is the Hilbert transform
H [φ](s) =1
π
φ(t)
s− tdt.
This gives the reformulation of the inversion formula.
Lemma 1.
f(x) =1
2(2π)1−N
ˆSN−1
g(θ,x · θ)dθ,
where
g(θ, s) =
(−1)N/2−1 ∂N−1
∂sN−1g(θ, s) N odd,
(−1)(N−1)/2H ∂N−1
∂sN−1g(θ, s) N even.
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X-Ray Tomography: Regularization
• We can generalize the FBP in terms of the filter I−α in R2, such that
F [I−αf ](σ) = |σ|−αF (σ) (Natterer, 2001):
f =1
4πI−αR∗Iα−1g.
• For α > 0, the image is smoothed (high frequencies suppressed) and high
frequencies in the data g are sharpened.
• For α = 1, f = (4π)−1I−1(R∗R)f , so (R∗R)f = 4πI1f , for which we can
foresee the Tikhonov formula
fµ = (R∗R+ µ2)−1R∗g, or fµ = R∗(RR∗ + µ2)−1g.
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X-Ray Tomography: PET, SPECT and beyond
• Single-photon emission computed tomography (SPECT) is a nuclear medicine
tomographic imaging technique using gamma rays:
I =
ˆL
f(x) exp
(−ˆL(x)
µ(y)dy
)dx.
• Positron emission tomography (PET) is an imaging technique that uses
radioactive substances to visualize and measure metabolic processes in the
body:
I =
ˆL
f(x) exp
(−ˆL+(x)
µ(y)dy −ˆL−(x)
µ(y)dy
)dx.
• The attenuation coefficient µ depends on the energy E of the X-rays:
Ie
Id=
ˆT (E) exp
(−ˆL
µ(x, E)dx
)dE,
where T (E) is the energy spectrum of the source (Natterer, 2001).
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X-Ray Tomography
• The Nobel Prize in Physiology or Medicine 1979:
“for the development of computer assisted tomography”.
EARLY TWO-DIMENSIONALRECONSTRUCTIONANDRECENT TOPICS STEMMING FROM IT
Nobel Lecture, 8 December, 1979
byALLAN M . CORMACKPhysics Department, Tufts University, Medford, Mass., U.S.A.
In 1955 I was a Lecturer in Physics at the University of Cape Town whenthe Hospital Physicist at the Groote Schuur Hospital resigned. SouthAfrican law required that a properly qualified physicist supervise the useof any radioactive isotopes, and since I was the only nuclear physicist in
Cape Town, I was asked to spend 1 1 / 2 days a week at the hospitalattending to the use of isotopes, and I did so for the first half of 1956. I wasplaced in the Radiology Department under Dr. J. Muir Grieve, and in thecourse of my work I observed the planning of radiotherapy treatments. Agirl would superpose isodose charts and come up with isodose contourswhich the physician would then examine and adjust, and the processwould be repeated until a satisfactory dose-distribution was found. Theisodose charts were for homogeneous materials, and it occurred to me thatsince the human body is quite inhomogeneous these results would be quitedistorted by the inhomogeneities - a fact that physicians were, of course,well aware of. It occurred to me that in order to improve treatmentplanning one had to know the distribution of the attenuation coefficient oftissues in the body, and that this distribution had to be found by measure-ments made external to the body. It soon occurred to me that this informa-tion would be useful for diagnostic purposes and would constitute atomogram or ser ies of tomograms , though I did not learn the word“tomogram” for many years.
At that time the exponential attenuation of X- and gamma-rays had beenknown and used for over sixty years with parallel sided homogeneous slabsof material. I assumed that the generalization to inhomogeneous materialshad been made in those sixty years, but a search of the pertinent literaturedid not reveal that it had been done, so I was forced to look at the problemab initio. It was immediately evident that the problem was a mathematicalone which can be seen from Fig. 1. If a fine beam of gamma-rays ofintensity I, is incident on the body and the emerging intensity is I, then themeasurable quantity g = In(I0/ I) = SLfds, where f is the variable absorp-
tion coefficient along the line L. Hence if f is a function in two dimensions,and g is known for all lines intersecting the body, the question is: “Can f bedetermined if g is known ?“. Again this seemed like a problem which would
55 I
COMPUTED MEDICAL IMAGING
Nobel Lecture, 8 December, 1979
BYGODFREY N . HOUNSFIELDThe Medical Systems Department of Central Research Laboratories EMI,London, England
In preparing this paper I realised that I would be speaking to a generalaudience and have therefore included a description of computed tomo-graphy (CT) and some of my early experiments that led up to the develop-ment of the new technique. I have concluded with an overall picture of theCT scene and of projected developments in both CT and other types ofsystems, such as Nuclear Magnetic Resonance (NMR).
Although it is barely 8 years since the first brain scanner was construct-ed, computed tomography is now relatively widely used and has beenextensively demonstrated. At the present time this new system is operatingin some 1000 hospitals throughout the world. The technique has succesful-ly overcome many of the limitations which are inherent in conventional X-ray technology.
When we consider the capabilities of conventional X-ray methods, threemain limitations become obvious. Firstly, it is impossible to display withinthe framework of a two-dimensional X-ray picture all the informationcontained in the three-dimensional scene under view. Objects situated indepth, i. e. in the third dimension, superimpose, causing confusion to theviewer.
Secondly, conventional X-rays cannot distinguish between soft tissues. Ingeneral, a radiogram differentiates only between bone and air, as in thelungs. Variations in soft tissues such as the liver and pancreas are notdiscernible at all and certain other organs may be rendered visible onlythrough the use of radio-opaque dyes.
Thirdly, when conventional X-ray methods are used, it is not possible tomeasure in a quantitative way the separate densities of the individualsubstances through which the X-ray has passed. The radiogram recordsthe mean absorption by all the various tissues which the X-ray has penetrat-ed. This is of little use for quantitative measurement.
Computed tomography, on the other hand, measures the attenuation ofX-ray beams passing through sections of the body from hundreds ofdifferent angles, and then , from the evidence of these measurements, a
computer is able to reconstruct pictures of the body’s interior.Pictures are based on the separate examination of a series of contiguous
cross sections, as though we looked at the body separated into a series ofthin “slices”. By doing so, we virtually obtain total three-dimensional infor-
mation about the body.
568
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X-Ray Tomography
• Despite major advances in X-ray sources, detector arrays, gantry mechanical
design and especially computer performance, one component of CT scanners
has remained virtually constant for the past 25 years—the reconstruction
algorithm1.
• CT is a global business with several major manufacturers and many minor
providers, especially of niche systems. Worldwide sales of CT scanners is more
than $2.3 billion per year, despite economic slowdown.
• “Global Market for Medical Imaging Equipment Worth $11.4 Billion by
1X. Pan, E. Y. Sidky and M. Vannier. Why do commercial CT scaners still employ traditional, filtered back-projection for imagereconstruction? Inverse Problems 25 (2009) 123009 (36pp)
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PART III
Wave-Based Scattering Models
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Wave-Based Scattering Models
• In diffraction tomography the goal is to form an image of the interior of a
body from the echo of a wavefield.
• An image is a mapping of the locus of the singularities of the wave speed.
Figure 11: Typical experiment of travel-time tomography (Tarantola, 2005).
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Seismic Imaging
It is the process through which Earth’s interior properties are mapped
into seismograms recorded on its surface.
Figure 12: Seismogram showing accompanying ray paths (Stein and Wysession, 2003).
• Seismograms are representations of PDE’s solutions of scattering problems in
variable-parameter elastic medium.
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Synthetic-Aperture Radar Imaging
Teledetection and Monitoring: Ocean waves, soil humidity, forest ecology,
planets’ cartography, etc.
~γ(s)
x1
x2
h
x3
~xv
εT
• Moving antenna transmitting and collecting electromagnetic pulses.
• Synthesis of the antenna aperture (measurements).
• Signal processing for imaging (inverse problem).
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Acoustic Scattering Models
• We are given the wave number k ∈ C\0 with Re k ≥ 0, Im k ≥ 0, a
bounded region Ω ⊂ RN with bdry ∂Ω ∈ C 2, and solution (incident wave) uI
of the Helmholtz equation
∆u+ k2u = 0
in a region containing Ω in its interior. We assume that RN\Ω is connected.
Problem 1 (Direct Scattering Problem). Determine u ∈ C 2(RN\Ω) ∩ C 2(Ω) ∩C 1(RN) with
(a) ∆u+ k2n(x)u = 0 in RN with n = n(x) ∈ C2(RN) such that the support of
n− 1 is contained in Ω (penetrable scatterer).
(b) uS = u − uI uniformly satisfies, with respect to x = x/|x|, Sommerfeld’s
radiation condition
lim|x|→+∞
|x|(N−1)/2
(∂uS(x)
∂|x|− ikuS(x)
)= 0.
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Acoustic Scattering Models
• From the Representation Theorem (fundamental solution), one derives the
Lippmann-Schwinger integral equation for the total wavefield u:
u(x) = uI(x) + k2
¨Ω
(1− n(y))u(y)g(x,y)dy, x ∈ Ω.
This is a Fredholm integral equation of the second kind.
Problem 2 (Inverse Scattering Problem (ISP)). To find an obstacle, characterized
by its index of refraction n(x), from partial knowledge of data uS(x, y; k).
• The scattered, radiating or outgoing wavefield uS is the measured echo in x
after the monochromatic waves (fixed k) are scattered from the direction y.
• We are interested in weak singularities of the wave speed c(x) represented by
a small perturbation to the index of refraction n(x) = 1 + T (x).
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Acoustic Scattering Models
• This function T plays the role of a potential (Schrodinger Eqn.), a reflectivity
(Radar imaging), an obstacle (ultrasound tomography), an attenuation
coefficient (CT tomography):
k2n(x) =ω2
c20n(x) = k2(1 + T (x)),
where c0 is the background wave speed.
• We will transform the Lippmann-Schwinger equation as follows:
1. Decompose the total field u = uI + uS.
2. Substitute n(x) = 1 + T (x).
3. Use the far-field approximation, |x − y| = |x| − x · y + O(|x|−1), where x
is an unitary vector that denotes the direction of the incident wavefield.
4. Linearize using the Born approximation: |TuS| |TuI|.
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Acoustic Scattering Models
• Rearranging the expression in terms of new variables for sources and receivers.
Aditionally, all wave variables depend on the frequency ω:
uS(xg,xs, ω, T ) =ω2
c20
ˆΩ
T (y)uI(y,xs, ω)g(y,xg, ω)dy.
xs
source
xg1xg2
receivers
F (ω)
n
Ω
Figure 13: Geometry of one source xs and multiple receiver positions xgi.
• Now we reshape this formula introducing Kirchhoff Approximation to represent
the upward scattered data from a single reflector T .
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Acoustic Scattering Models
• By definition, g is the fundamental solution and satisfies
(∆ + ω2/c20)g(y,xg, ω) = −δ(y − xg),
therefore uI(y,xs, ω) = F (ω)g(y,xs, ω) and g is given by its WKBJ
approximation (geometrical optics):
g(x0,xs, ω) = A(x0,xs)eiωτS(x0,xs),
with τS being the travel-time in heterogeneous medium, and A being the
corresponding WKBJ amplitude (rays).
• We finally get
uS(xg,xs, ω, T ) ≈ ω2
c20F (ω)
ˆΩ
T (y)a(y, ξ)eiωΦ(y,ξ)dydξ,
where we introduced a phase function Φ(y, ξ) = τ(y,xs(ξ)) + τ(xg(ξ),y) and
an amplitude a(y, ξ) = A(y,xs(ξ))A(xg(ξ),y).
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Acoustic Scattering Models
• To solve the inverse scattering problem means that we are able to estimate the
reflector T .
• We denote by T our model for the inverse imaging map. It is suggested by
the expression
TQ(x) =
¨Q(x, ξ)e−iωΦ(x,ξ)uS(xg,xs, ω, T )dωdξ,
where the filter Q will become a regularization term.
• The complex-conjugate phase recalls the adjoint of uS.