1996
DOI: 10.1016/0926-9851(96)00007-9
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Analytical solution for direct current electrical potentials in an arbitrarily anisotropic half-space

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Cited by 7 publications
(4 citation statements)
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“…Furthermore, because of the small amount of data available we shall assume a constant anisotropic conductivity. With these simplifying assumptions, the problem is analytically tractable and, for the full‐space case, the electrical potential at an electrode r e produced by a unit current source point r s reads (Das & Li 1996), where is the electrical resistivity tensor with components ρ αβ =σ −1 αβ . In practice, the resistivity tensor is often given in terms of both its principal values, (ρ 1 , ρ 2 , ρ 3 ), expressed in Ω.…”
Section: Inverse Problemmentioning
confidence: 99%
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“…Furthermore, because of the small amount of data available we shall assume a constant anisotropic conductivity. With these simplifying assumptions, the problem is analytically tractable and, for the full‐space case, the electrical potential at an electrode r e produced by a unit current source point r s reads (Das & Li 1996), where is the electrical resistivity tensor with components ρ αβ =σ −1 αβ . In practice, the resistivity tensor is often given in terms of both its principal values, (ρ 1 , ρ 2 , ρ 3 ), expressed in Ω.…”
Section: Inverse Problemmentioning
confidence: 99%
“…The boundary condition corresponding to the half‐space geometry is imposed by using the image method, and the Green's potential takes the form (Das & Li 1996), where r i s is the location of the source image. Because of the electrical field distortion produced by anisotropy, the location of the image with respect to both the plane boundary and the primary source is not simple and may be found in Li & Uren (1997).…”
Section: Inverse Problemmentioning
confidence: 99%
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