3.6 An Anisotropic Inverse Problem for Measuring FAWT
73
Table 3.2 Result of FAWT inversion for a 1.5 mm-thick slab: σ y = σ z = [25, 75, 125]
Φ
σ y /sensit
σ z /sensit
No. pts.
0.212(3)
101.64/0.798
101.64/0.798
500
Table 3.3 Result of FAWT inversion for a 1.5 mm-thick slab: σ y = [25, 75, 125], σ z =
[25, 75, 125]
Φ
σ y /sensit
σ z /sensit
No. pts.
0.206(3)
101.53/0.781
125.0/164.9
14
Table 3.4 Result of FAWT inversion for a 12.7 mm-thick slab: σ y = σ z = [25, 75, 125]
Φ
σ y /sensit
σ z /sensit
No. pts.
0.239(3)
97.18/2.126
97.18/2.126
500
Table 3.5 Result of FAWT inversion for a 12.7 mm-thick slab: σ y = [25, 75, 125], σ z =
[25, 75, 125]
Φ
σ y /sensit
σ z /sensit
No. pts.
0.183(3)
99.09/1.834
75.92/13.05
500
It seems reasonable to treat the transverse conductivities as a package in which
σ y = σ z . If, however, we wish to consider problems in which they may not be
equal, perhaps due to the presence of fiber tows, then the corresponding inverse
problem treats σ y and σ z as independent unknowns, so the inverse problem is twodimensional. If we use the same one-dimensional nodes as before, the interpolation
table becomes the Cartesian product [25, 75, 125] ⊗ [25, 75, 125]. Using the same
data, over the same frequency range as before, the result of the inversion is shown
in Table 3.3. While σ y is reliably reconstructed, σ z is not. It seems that the 1.5mm
thick sample is too thin to generate a significant E z field that would be necessary
to interrogate the structure to produce σ z . The challenge in measuring σ z has been
discussed in [130].
To test this idea, we reran the inverse problem for a 12.7 mm-thick sample, using
the same data as before. The result for the restricted case of σ y = σ z , i.e., a onedimensional problem, is shown in Table 3.4. The results are still quite good, with
perhaps a little less certainty because of the larger sensitivity coefficient.
Finally, we consider the two-dimensional case with σ y = [25, 75, 125], σ z =
[25, 75, 125], independently. The result of this inversion is shown in Table 3.5. The
improvement here, over Table 3.3, is in the estimation of σ z and its reliability, as
indicated by the much smaller sensitivity coefficient. This lends credence to the
hypothesis that there must be sufficient thickness in the sample to allow a significant
value of E z to evolve. The manner in which such a component is derived can be
argued by the following simple model. In an infinite, homogeneous, isotropic plate,
the induced electric field has a vanishing z-component, where the z-direction is
normal to the surface, so the resulting eddy-currents flow in planes parallel to the
surface.
73
Table 3.2 Result of FAWT inversion for a 1.5 mm-thick slab: σ y = σ z = [25, 75, 125]
Φ
σ y /sensit
σ z /sensit
No. pts.
0.212(3)
101.64/0.798
101.64/0.798
500
Table 3.3 Result of FAWT inversion for a 1.5 mm-thick slab: σ y = [25, 75, 125], σ z =
[25, 75, 125]
Φ
σ y /sensit
σ z /sensit
No. pts.
0.206(3)
101.53/0.781
125.0/164.9
14
Table 3.4 Result of FAWT inversion for a 12.7 mm-thick slab: σ y = σ z = [25, 75, 125]
Φ
σ y /sensit
σ z /sensit
No. pts.
0.239(3)
97.18/2.126
97.18/2.126
500
Table 3.5 Result of FAWT inversion for a 12.7 mm-thick slab: σ y = [25, 75, 125], σ z =
[25, 75, 125]
Φ
σ y /sensit
σ z /sensit
No. pts.
0.183(3)
99.09/1.834
75.92/13.05
500
It seems reasonable to treat the transverse conductivities as a package in which
σ y = σ z . If, however, we wish to consider problems in which they may not be
equal, perhaps due to the presence of fiber tows, then the corresponding inverse
problem treats σ y and σ z as independent unknowns, so the inverse problem is twodimensional. If we use the same one-dimensional nodes as before, the interpolation
table becomes the Cartesian product [25, 75, 125] ⊗ [25, 75, 125]. Using the same
data, over the same frequency range as before, the result of the inversion is shown
in Table 3.3. While σ y is reliably reconstructed, σ z is not. It seems that the 1.5mm
thick sample is too thin to generate a significant E z field that would be necessary
to interrogate the structure to produce σ z . The challenge in measuring σ z has been
discussed in [130].
To test this idea, we reran the inverse problem for a 12.7 mm-thick sample, using
the same data as before. The result for the restricted case of σ y = σ z , i.e., a onedimensional problem, is shown in Table 3.4. The results are still quite good, with
perhaps a little less certainty because of the larger sensitivity coefficient.
Finally, we consider the two-dimensional case with σ y = [25, 75, 125], σ z =
[25, 75, 125], independently. The result of this inversion is shown in Table 3.5. The
improvement here, over Table 3.3, is in the estimation of σ z and its reliability, as
indicated by the much smaller sensitivity coefficient. This lends credence to the
hypothesis that there must be sufficient thickness in the sample to allow a significant
value of E z to evolve. The manner in which such a component is derived can be
argued by the following simple model. In an infinite, homogeneous, isotropic plate,
the induced electric field has a vanishing z-component, where the z-direction is
normal to the surface, so the resulting eddy-currents flow in planes parallel to the
surface.