Direct time‐domain calculation of the transient response for a rectangular loop over a two‐layer medium
Direct time‐domain calculation of the transient response for a rectangular loop over a two‐layer medium
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DOI：
10.1190/1.1442368
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发表时间：
1987-07
期刊：
影响因子：
3.3
通讯作者：
M. Goldman;D. Fitterman
中科院分区：
文献类型：
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作者：
M. Goldman;D. Fitterman
通过将留数定理应用于解析场表达式，在时域中直接计算两层半空间表面上电偶极子引起的垂直磁场的时间导数。第二层必须是绝缘的（σ2=0）或完全导电的（σ2=∞）。第一种情况可用于估计采矿勘探问题的导电覆盖层的响应。第二种情况有助于解释当导电基底位于电阻性第一层之下时，瞬态测深电压视电阻率曲线中出现的过冲。在后期，垂直磁场的时间导数随t-4衰减，后期视电阻率随t增大（σ2=0），而当σ2=∞时，这些量分别表现为t-4 e-αt和te 2 αt/3，其中α=π2/4μ0σ 1h 12，σ1，为第一层电导率，h1为第一层厚度，μ0=4π10- 7 H/m。对电偶极子的表达式进行积分，得到矩形线圈的解。
The time derivative of the vertical magnetic field due to an electric dipole on the surface of a two‐layer half‐space is computed directly in the time domain by applying the residue theorem to the analytic field expressions. The second layer must be either insulating (σ2=0) or perfectly conducting (σ2=∞). The first case can be used to estimate the response of a conductive overburden for mining exploration problems. The second case is useful in explaining the overshoot seen in transient sounding voltage apparent‐resistivity curves when a conductive basement underlies a resistive first layer. In the late stage, the time derivative of the vertical magnetic field decays as t-4 and the late‐stage apparent resistivity increases as t for σ2=0, while for σ2=∞, these quantities behave as t-4e-αt and te2αt/3, respectively, where α=π2/4μ0σ1h12, σ1, is the first‐layer conductivity, h1 is the first‐layer thickness, and μ0=4π10-7H/m. The electric dipole expressions are integrated to obtain solutions for rectangular loo...