...
首页> 外文期刊>Radio Science >Conductivity and resistivity tensor rotation for surface impedance modeling of an anisotropic half-space
【24h】

Conductivity and resistivity tensor rotation for surface impedance modeling of an anisotropic half-space

机译:电导率和电阻率张量旋转,用于各向异性半空间的表面阻抗建模

获取原文
获取原文并翻译 | 示例

摘要

The electromagnetic surface impedance of a half-space with inclined conductivity anisotropy can be derived from the isotropic half-space solution provided the conductivity term used in the expressions is the effective horizontal conductivity. For a TM-mode plane wave incidence, the effective horizontal conductivity must be derived from the tensor rotation of the resistivity tensor and not from the tensor rotation of the conductivity tensor. For a TE-mode incident with the same geometry as the TM-mode, the surface impedance is independent upon the inclined anisotropy. This same formulation can then be extended to a multiple layered half-space where each layer has an inclined anisotropy. For an anisotropic half-space with coefficient of anisotropy of 4, with a horizontal conductivity of 0.001 S/m inclined at 45° with respect to the horizontal plane, the magnitude of the surface impedance calculated using the resistivity tensor rotation is approximately 53% larger than the magnitude of the surface impedance calculated using the conductivity tensor rotation.
机译:如果表达式中使用的电导率项是有效水平电导率,则可以从各向同性的半空间解中得出具有倾斜的电导率各向异性的半空间的电磁表面阻抗。对于TM模式平面波入射,有效水平电导率必须从电阻率张量的张量旋转中得出,而不是从电导率张量的张量旋转中得出。对于与TM模式具有相同几何形状的TE模式入射,表面阻抗与倾斜各向异性无关。然后可以将相同的配方扩展到多层半空间,其中每个层都具有倾斜的各向异性。对于各向异性系数为4的各向异性半空间,水平电导率相对于水平面倾斜0.001 S / m呈45°时,使用电阻率张量旋转计算的表面阻抗的大小大约要大53%使用电导率张量旋转计算出的表面阻抗的大小

著录项

相似文献

  • 外文文献
  • 中文文献
  • 专利
获取原文

客服邮箱:kefu@zhangqiaokeyan.com

京公网安备:11010802029741号 ICP备案号:京ICP备15016152号-6 六维联合信息科技 (北京) 有限公司©版权所有
  • 客服微信

  • 服务号