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Beam hardening 2: the no-linearize theorem

The last post showed that beam hardening causes a nonlinearity between the log of the measurements and the A-vector. It is natural to think that we can eliminate the beam hardening artifacts by measuring the nonlinearity and then “linearizing” it with an inverse transformation. In this post, I will show that this is not possible in general. Although there are some special cases when we can linearize and a linearizing transformation may reduce the artifacts, we cannot do this for every object. I will show that this is due to the fact that we need at least a two dimension basis set to represent the attenuation coefficient.

The goal of linearizing

A conventional, non-energy selective CT system attempts to reconstruct the linear attenuation coefficient μ(r, E0) at a single average energy E0, where r is the position in the cross section. Introducing the linear decomposition of the attenuation coefficient, μ(r, E0) = a1( r)f1(E0) + a2( r)f2(E0), we can see that by linearizing the CT system attempts to reconstruct a constant linear combination of the a1 and a2
(1) μ(r, E0) = a1( r)f1(E0) + a2( r)f2(E0)
where the f1(E0) and f2(E0) are constants. Integrating Eq. 1↑ along lines, this is equivalent to reconstructing
(2) ℒ = A1f1(E0) + A2f2(E0).
From this, we see that the goal of linearizing is to compute a linear combination with constant coefficients of the A-vector components from a broad spectrum measurement.

The no-linearize theorem

I introduced and proved this theorem in Chapter 6 of my dissertation, which has other interesting results on beam hardening artifacts. Based on the goal of linearizing discussion in the previous section, we can state the no-linearize theorem as follows:
Suppose we have a measurement I with a spectrum S(E) and the object A-vector is [ A1,  A2 ]T so that
(3) I(A1, A2) = S(E)exp[ − A1f1(E) − A2f2(E)]dE.
If the object consists of more than one material and the spectrum is not monoenergetic, then an invertible function g does not exist such that
(4) g[I(A1, A2)] = A1f1(E0) + A2f2(E0)
where f1(E0) = k1 and f2(E0) = k2 are constants.
The proof has two steps. First, I will show that if g exists, then the contour lines of I in A-space, that is the set of points where it is a constant, are straight lines. Next, I show that except for special cases the contours are not straight lines. This completes the proof that g cannot exist by contradiction.
For the first part of the proof, we can reason as follows: The contour lines of I are defined by I(A1, A2) = C, where C is a constant. Applying g to both sides of this equation and using Eq. 4↑
g[I(A1, A2)] = k1A1 + k2A2 = g(C).
Since g(C) is also a constant, if g exists then the contour curve is
(5) k1A1 + k2A2 = constant.
This is the equation of a straight line, A2 =  − k1k2A1 + constant. This shows the first step of the proof.
For the second part of the proof, we need to show that the contours are not straight lines. Taking the total derivative of I(A1, A2)
dI = (I)/(A1)dA1 + (I)/(A2)dA2
On a contour of I,  dI = 0 so solving for the slope
(6) (dA2)/(dA1) =  − ((I)/(A1))/((I)/(A2)).
Differentiating Eq. 3↑ with respect to A1 and A2
(I)/(A1) =  − f1(E)S(E)exp[ − A1f1(E) − A2f2(E)]dE
(I)/(A2) =  − f2(E)S(E)exp[ − A1f1(E) − A2f2(E)]dE.
Substituting in Eq. 6↑
(dA2)/(dA1)  =  −  (f1(E)S(E)exp[ − A1f1(E) − A2f2(E)]dE)/(f2(E)S(E)exp[ − A1f1(E) − A2f2(E)]dE)
Defining u(E) = S(E)exp[ − A1f1(E) − A2f2(E)] and using bracket notation for the integrals
(dA2)/(dA1) =  − (f1, u)/(f2, u).
Multiplying across, the equation becomes
(7) f1, u + (dA2)/(dA1)f2, u = 0.
By definition, the slope of a straight line is constant. Eq. 7↑ states that if the slope dA2dA1is constant, in the space of u(E) functions, f1(E) and f2(E) are linearly dependent. Notice however that we have not said anything about the spectrum. Therefore the function u can be any function and Eq. 7↑ says that if the contours of I(A1, A2) are straight lines then f1(E) and f2(E) are linearly dependent. But we know this is not true or else we would not need a two function basis set to represent μ(E). Therefore, the contours cannot be straight lines and this completes the proof of the no-linearize theorem by contradiction.

Linearizable cases

It turns out that the linearizable cases are the same as the cases for no beam hardening artifacts discussed in the previous post. The first case is a monoenergetic spectrum. Then
u(E0)  =  S(E0)exp[ − A1f1(E0) − A2f2(E0)]  =  constant.
In this case, the function space u(E) is degenerate so the contour lines of I(A1, A2) are straight lines even though f1(E) and f2(E) are linearly independent. In the second case, the object consists of a single material. Then the A-vector is constrained to be on a straight line through the origin so the possible A-space is not two dimensional and the contour lines are not meaningful.


The no-linearize theorem does not say that we should not linearize. Linearizing does help to reduce artifacts. But it is not clear what material to use to define the linearizing curve. Is it the curve for, say, soft tissue or for bone? If the object composition does not conform to your choice then you will have artifacts so you cannot linearize perfectly for all possible materials. One of the advantages of energy-selective systems is that they can in principle eliminate beam hardening artifacts for any body material composition.
In the following posts, I will discuss another approach that is sometimes used to try to eliminate artifacts--using an iterative reconstruction algorithm to eliminate the “inconsistent” portions of the projections. That is, if you can eliminate the part of the projections that are due to the nonlinearity then the assumption is that you will eliminate beam hardening artifacts. However, I will show that nonlinear transformations of some objects are consistent so with these objects you will still have artifacts.
Last edited Jun 19, 2015
Copyright © 2015 by Robert E. Alvarez
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