Analytical Solution of Electrical Impedance Tomography Problem (Laplace's Equation) for a Homogeneous Sphere1
Saeed Babaeizadeh
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1 Mathematical Model of Electrical Impedance Tomography (EIT) Problem
Electromagnetic phenomena are governed by Maxwell's equation for the electric field E and the magnetic field H. These equations are:
Ñ×E = -
¶B
¶t
(1)
Ñ×H = J+
¶D
¶t
,
(2)
where J denotes the current density, B the magnetic flux density (also called the magnetic induction), D the electric displacement, and the partial derivative is with respect to time t.
The frequencies used in EIT systems are low enough so that the quasi-static approximation holds, and thus we can ignore capacitive, propagation, and inductive effects [2]. The ACT3 system, for example, operates at 28.8 kHz, and is applied to bodies smaller than one meter in which the conductivity is generally less than one (Ohm-meter)-1 [3]. Under such quasi-static conditions, since the displacement current ¶D/ ¶t is approximately zero, by taking the divergence of (2) we get
Ñ·J = 0.
(3)
For current density J we have
J = sE+Js,
(4)
where s is the conductivity of the medium. This equation states that the total current density J is the sum of the conduction current density sE and the "impressed current" density Js due to the internal sources [2]. In regular EIT systems all the sources are on the outer boundary of the object, so can be modeled as the boundary condition; therefore, Js=0. Taking the divergence of both sides of (4), and substituting (3) as well as E=-Ñf (f is the scalar potential), we arrive at Laplace's equation:
Ñ·( sÑf) = 0,
(5)
which is the equation that governs the electric potential f inside the object.
In practice, we apply currents to electrodes on the surface S of the body. These currents produce a current density on the surface whose inward pointing normal component is denoted by g. Thus
s
¶f
¶n
= g on S.
(6)
Mathematically, the EIT problem can then be described by Laplace's equation for the potential given by (5) with Neumann boundary conditions given by (6).
For a homogeneous sphere, Laplace's equation (5) can be written as Ñ2 f = 0. A solution of this equation over a sphere is given below where the potential is written as a sum of spherical harmonics [4]:
f =
¥ å
n=0
(An rn + Bn r-(n+1) ) Pn(cosq).
(7)
Here Pn are the Legendre Polynomials, defined in the interval [-1 ¼1]. Differentiating this equation with respect to r, we get
¶f
¶r
=
¥ å
n=1
(nAnr(n-1) + (-(n+1))Bnr-(n+2) ) Pn(cosq).
(8)
Since [(¶f)/(¶r)] cannot be infinite at r=0, Bn must be zero for all n. Therefore, we need to use the boundary conditions to find only An.
Having a current electrode around q = 0 is the same as applying a block type boundary condition at the outer boundary [5]. It is 0 everywhere except at an interval around q = 0, where it is 1 (see Fig. 1). Since the total current leaving the sphere must be zero, we need to add another electrode to the model with the same block type function but with the opposite sign. Since the forward problem is linear, the resulting potential is the sum of the two potential sets computed for these two boundary conditions, separately. Hence, since sphere is a symmetric shape, that would be easier to first compute the potentials for the boundary condition being just one of these block type functions, and then rotate the sphere and subtract the new set of potentials from the previous one. Using this method, we can easily add to the model as many as electrodes we want. Below, we discuss how the potential f can be calculated when the boundary condition is one block type function.
Figure 1: A block type boundary condition.
The block type function can be expanded in Legendre Polynomials as well:
f(cosq) =
¥ å
n=0
an Pn(cosq),
(9)
where
an=
1
2
(2n+1)
ó õ
1
-1
f(cosq)Pn(cosq) d(cosq).
(10)
Using this equation and the property of Legendre Polynomials:
ó õ
Pn(u) du =
Pn+1-Pn-1
2n+1
(11)
leads to
an
=
1
2
(2n+1)
ó õ
qEndBlock
qStartBlock
Pn(u) du
=
[Pn+1(cosqEndBlock))-[Pn-1(cosqEndBlock))]
- [Pn+1(cosqStartBlock))-[Pn-1(cosqStartBlock))]
(12)
Applying the boundary condition s[(¶f)/(¶r)] = f( cosq), by equating (8) and (9) we get
snAnRn-1 = an,
(13)
where R and s denote the radius and conductivity of the sphere, respectively.
Substituting the values of An into (7), we have the final solution for f as
f =
1
s
¥ å
n=0
(
an
nRn-1
) Pn(cosq).
(14)
Fig. 2 shows the Boundary Element Methods (BEM)-based solution [6] and its difference with the analytical solution for a homogeneous sphere with R=1, s = 1 (in arbitrary units). Two electrodes with the width of p/100 are placed on the two opposite poles of the sphere. The amplitude of the injected current is ±10 (again in arbitrary units). We cut the infinite sum in (14) at the 100th element. The triangular mesh used for the BEM-based solution has 1819 nodes and 3634 triangles. We observe the magnitude of the difference between the two solutions is on the order of 3%. Part of this difference may be because of the approximations made in the analytical solution, in particular, cutting an infinite sum showed up in the solution at a finite number. Moreover, the electrodes used for the analytical solution have a non-zero width, but the point electrodes are used for the BEM-based solution. Discretization error due to representing the sphere with a limited number of points required for the BEM-based solution may also be another source of this difference.
Figure 2: EIT forward solution for a sphere with two electrodes.
S. Babaeizadeh, "3-D electrical impedance tomography for domains with
piecewise constant conductivity," Ph.D. dissertation, Northeastern
University, January 2006.
S. Babaeizadeh, D. H. Brooks, and D. Isaacson, "3-D electrical impedance
tomography for piecewise constant domains with known internal boundaries,"
IEEE Transactions on Biomedical Engineering, vol. 54, no. 1, pp.
2-10, January 2007.
Footnotes:
1To cite this article, please refer to [1] (Section 2.4.1)
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