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SAR ATR Based on 3D Parametric Electromagnetic Scattering Model
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摘要: 合成孔径雷达目标识别是雷达数据解译中一个长期研究的难点问题。近年来，基于模型的SAR目标识别方法由于在扩展条件下的识别性能表现良好而备受关注。在联合国内多家研究单位进行攻关的基础上，该文简要阐述了对该问题的初步研究成果及思考。首先从3个方面出发梳理了散射部件模型发展的技术脉络并对其进行了补充完善；然后从正向推算和逆向反演两条技术途径提出了复杂目标电磁散射参数化建模方法；最后提出了基于复杂目标电磁散射参数化模型的目标识别新框架。论文最后对基于模型的SAR目标识别下一步研究方向进行了展望。Abstract: Automatic Target Recognition (ATR) is one of the most difficult problems in Synthetic Aperture Radar (SAR) data interpretation. In recent years, the model-based SAR target recognition method has attracted much attention because of its good performance in the extended operation condition. Based on the research of a few domestic research institutes, this paper briefly introduces the preliminary research results and gives some thoughts about SAR ATR problem. First of all, the development of parametric scattering model are discussed from three aspects. Next, two ways to model the parametric electromagnetic scattering for complex target are put forward. Finally, we propose a new framework for a Three-Dimensional (3D) parametric scattering model based SAR ATR. In the end, the future research direction of model-based SAR target recognition is prospected.
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图 5 目标电磁散射参数化模型所成图像与仿真数据对比
Figure 5. Comparison between the scattering center model generated image and the simulated image
图 7 重构的简单坦克宽角度3维属性散射中心模型
Figure 7. The rebuilt model for simplified tank target based on 3D attributed scattering center model
图 9 基于3维部件级电磁散射参数化模型的识别框架
Figure 9. Framework of 3D parametric electromagnetic part model based SAR ATR
表 1 几种不同散射中心的 ${α}_{i}$取值
Table 1. The ${α}_{i}$ for different scattering centers
ai 取值 散射中心 –1 角绕射，尖顶绕射 –1/2 边缘绕射 0 点散射，双曲面反射，直边镜面反射 1/2 单曲面反射 1 平板法向发射，二面角反射，三面角反射 
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表 2 我们研究的耦合散射中心模型的频率依赖因子
Table 2. The ${α}$ in the proposed model by our team
电磁散射中心类型 电磁散射中心名称 a取值 镜面-镜面耦合 平面-平面耦合 1 单弯曲-平面耦合 0.5 双弯曲-平面耦合 0 单弯曲-单弯曲耦合 0.5 双弯曲-单弯曲耦合 0 双弯曲-双弯曲耦合 0 镜面-边缘耦合 平面-直边耦合 0 单弯曲-直边耦合 0 双弯曲-直边耦合 –0.5 曲边-平面耦合 –0.5 曲边-单弯曲耦合 –0.5 曲边-双弯曲耦合 –0.5 边缘-边缘耦合 直边-直边耦合 –0.5 直边-曲边耦合 –1 曲边-曲边耦合 –1
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表 3 简易坦克的部件级3维电磁散射模型
Table 3. The 3D parametric electromagnetic part model for the simplified tank target
编号 形成原因 可见 范围(°) 参数化 模型形式 RCS峰值 (dB) 1 车身 俯仰: [19, 39] 方位: [75, 105] 平板 1 2 车身平板 与立方体 俯仰: [9, 30] 方位: [75, 105] 方形顶帽 32 … … … … … 28 右履带与车身 前斜平板 俯仰: [2, 30] 方位: [75, 105] 二面角 33 29 车身平板与炮筒 俯仰: [11, 30] 方位: [75, 105] 平板上 倒圆柱 10
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表 4 基于部件级3维电磁散射模型的识别结果
Table 4. The recognition results
操作条件 数据描述 基于目标3维参数化电磁散射模型 方法的识别率(%) 基于模板方法的识别率(%) 标准 500幅电磁计算数据和75幅暗室测量数据 99.13 91.2 扩展 62幅电磁计算数据和106幅暗室测量数据 90.48 58.2
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1 10种典型体参数化模型表达式列表
1. The parametric models for ten scatterers
典型散射部件名称 图形 参数化模型 模型适用角度范围 各参数排列顺序和含义的解释 长方体 
$\begin{array}{l}S_{\rm vv,hh}^{\rm cuboid} = {S_1} + {S_2} + {S_3} + {S_{\rm 4vv,hh}} + {S_{\rm 5vv,hh}} + {S_{\rm 6vv,hh}}\\{S_1} = - {\rm j}\frac{{kbc}}{{\sqrt {π} }}{l_x}{\rm sinc}\left( {kb{l_y}} \right){\rm sinc}\left( {kc{l_z}} \right){{\rm e}^{{\rm j}k\left( {a{l_x} + c{l_z}} \right)}}\\{S_2} = - {\rm j}\frac{{kca}}{{\sqrt {π} }}{l_y}{\rm sinc}\left( {kc{l_z}} \right){\rm sinc}\left( {ka{l_x}} \right){{\rm e}^{{\rm j}k\left( {b{l_y} + c{l_z}} \right)}}\\{S_3} = - {\rm j}\frac{{kab}}{{\sqrt {π} }}{l_z}{\rm sinc}\left( {ka{l_x}} \right){\rm sinc}\left( {kb{l_y}} \right){{\rm e}^{{ \rm j}2kc{l_z}}}\\{S_{4{\rm vv,hh}}} = \frac{{\sqrt 3 }}{9}\frac{c}{{\sqrt {π} }}\left\{ \begin{array}{l}\left[ { \mp 1 - \frac{{\sqrt 3 }}{2}\frac{{\sin \displaystyle\frac{1}{3}\left( {{π} /2 - \phi } \right)}}{{\cos \displaystyle\frac{\phi }{3}\sin \displaystyle\frac{2}{3}\left( {{π} - \phi } \right)}}} \right]\\ \cdot {\rm sinc}\left( {kc{l_z}} \right){{\rm e}^{{\rm j}k\left( {a{l_x} - b{l_y} + c{l_z}} \right)}}\\ + \left[ { \mp 1 + \frac{{\sqrt 3 }}{4}\frac{1}{{\cos \displaystyle\frac{1}{3}\left( {{π} /2 - \phi } \right)\cos \displaystyle\frac{\phi }{3}}}} \right]\\\cdot {\rm sinc}\left( {kc{l_z}} \right){{\rm e}^{{\rm j}k\left( {a{l_x} + b{l_y} + c{l_z}} \right)}}\\ + \left[ { \mp 1 - \frac{{\sqrt 3 }}{2}\frac{{\sin \displaystyle\frac{\phi }{3}}}{{\cos \displaystyle\frac{1}{3}\left( {{π} /2 - \phi } \right)\sin \displaystyle\frac{2}{3}\left( {{π} /2 + \phi } \right)}}} \right]\\\cdot {\rm sinc}\left( {kc{l_z}} \right){{\rm e}^{{\rm j}k\left( {a{l_x} - b{l_y} + c{l_z}} \right)}}\end{array} \right\}\end{array}$ $\begin{array}{l}\theta \in \left[ 0^{°},{{90}^{°}} \right]\\\phi \in \left[ 0^{°},{{90}^{°}} \right]\end{array}$ a, b, c：长方体边长 ${S_{5{\rm vv,hh}}} = \frac{{\sqrt 3 }}{9}\frac{a}{{\sqrt {π} }}\left\{ \begin{array}{l}\left[ { \pm 1 - \frac{{\sqrt 3 }}{2}\frac{{\sin \displaystyle\frac{\theta }{3}}}{{\cos \displaystyle\frac{1}{3}\left( {{π} /2 - \theta } \right)\sin \displaystyle\frac{2}{3}\left( {{π} /2 + \theta } \right)}}} \right]\\\cdot {\rm sinc}\left( {ka{l_x}} \right){{\rm e}^{{\rm j}kb{l_y}}}\\ + \left[ { \pm 1 + \frac{{\sqrt 3 }}{4}\frac{1}{{\cos \displaystyle\frac{\theta }{3}\cos\displaystyle \frac{1}{3}\left( {{π} /2 - \theta } \right)}}} \right]\\\cdot {\rm sinc}\left( {ka{l_x}} \right){{\rm e}^{{\rm j}k\left( {b{l_y} + 2c{l_z}} \right)}}\\ + \left[ { \pm 1 - \frac{{\sqrt 3 }}{2}\frac{{\sin \displaystyle\frac{1}{3}\left( {{π} /2 - \theta } \right)}}{{\cos \displaystyle\frac{\theta }{3}\sin \displaystyle\frac{2}{3}\left( {{π} - \theta } \right)}}} \right]\\\cdot {\rm sinc}\left( {ka{l_x}} \right){{\rm e}^{{\rm j}k\left( { - b{l_y} + 2c{l_z}} \right)}}\end{array} \right\}\\{S_{6{\rm vv,hh}}} = \frac{{\sqrt 3 }}{9}\frac{b}{{\sqrt {π} }}\left\{ \begin{array}{l}\left[ { \pm 1 - \displaystyle\frac{{\sqrt 3 }}{2}\frac{{\sin \displaystyle\frac{1}{3}\left( {{π} /2 - \theta } \right)}}{{\cos \displaystyle\frac{\theta }{3}\sin \frac{2}{3}\left( {{π} - \theta } \right)}}} \right]\\\cdot {\rm sinc}\left( {kb{l_y}} \right){{\rm e}^{{\rm j}k\left( { - a{l_x} + 2c{l_z}} \right)}}\\ + \left[ { \pm 1 + \displaystyle\frac{{\sqrt 3 }}{4}\frac{1}{{\cos \displaystyle\frac{1}{3}\left( {{π} /2 - \theta } \right)\cos \displaystyle\frac{\theta }{3}}}} \right]\\\cdot {\rm sinc}\left( {kb{l_y}} \right){{\rm e}^{{\rm j}k\left( {a{l_x} + 2c{l_z}} \right)}}\\ + \left[ { \pm 1 - \frac{{\sqrt 3 }}{2}\frac{{\sin \displaystyle\frac{\theta }{3}}}{{\cos \displaystyle\frac{1}{3}\left( {{π} /2 - \theta } \right)\sin \displaystyle\frac{2}{3}\left( {{π} /2 + \theta } \right)}}} \right]\\\cdot {\rm sinc}\left( {kb{l_y}} \right){{\rm e}^{{\rm j}ka{l_x}}}\end{array} \right\}$ $\begin{array}{l}\theta \in \left[ 0^{°},{{90}^{°}} \right]\\\phi \in \left[ 0^{°},{{90}^{°}} \right]\end{array}$ a, b, c：长方体边长 圆柱体 
$S_{\rm vv,hh}^{\rm cylinder} = {S_{1{\rm vv,hh}}} + {S_{2{\rm vv,hh}}} + {S_{3{\rm vv,hh}}}$ 当 $2kr\sin \theta > 2.44$且 $kh\cos \theta > 2.25$ $\begin{array}{l}{S_{1{\rm vv,hh}}} = \sqrt {\frac{{{\rm j}r}}{{3k\sin \theta }}} \left[ { \mp \frac{2}{3} - {{\left( {\frac{1}{2} + \cos \frac{{4\theta }}{3}} \right)}^{ - 1}}} \right]{{\rm e}^{{\rm j}2kr\sin \theta }}\\{S_{2{\rm vv,hh}}} = \sqrt {\frac{{{\rm j}r}}{{3k\sin \theta }}} \left[ { \mp \frac{2}{3} - {{\left( {\frac{1}{2} + \cos \frac{{4\left( {{π} - \theta } \right)}}{3}} \right)}^{ - 1}}} \right]{{\rm e}^{{\rm j}2k\left( {r\sin \theta + h\cos \theta } \right)}}\\{S_{3{\rm vv,hh}}} = {\rm j}\sqrt {\frac{{{\rm j}r}}{{3k\sin \theta }}} \left[ { \mp \frac{2}{3} - {{\left( {\frac{1}{2} + \cos \frac{{4\left( {{π} /2 - \theta } \right)}}{3}} \right)}^{ - 1}}} \right]{{\rm e}^{{\rm j}2k\left( { - r\sin \theta + h\cos \theta } \right)}}\end{array}$ 当 $kh\cos \theta \le 2.25$ ${\left( {{S_1} + {S_2} + {S_3}} \right)_{\theta \to \displaystyle\textstyle\frac{{π} }{2}}} = - \sqrt {{\rm j}kr{h^2}\sin \theta } {\rm sinc}\left( {kh\cos \theta } \right){{\rm e}^{{\rm j}k\left( {h\cos \theta + 2r\sin \theta } \right)}}$ 当 $2kr\sin \theta \le 2.44$ ${\left( {{S_1} + {S_2} + {S_3}} \right)_{\theta \to 0}} = - {\rm j}2\sqrt {π} k{r^2}\left| {\cos \theta } \right|\frac{{{J_1}\left( {2kr\sin \theta } \right)}}{{2kr\sin \theta }}{{\rm e}^{{\rm j}kh\left( {\cos \theta + \left| {\cos \theta } \right|} \right)}}$ $\begin{array}{l}\theta \in \left[ {{0^{°}},{{90}^{°}}} \right]\\\phi \in \left[ {{0^{°}},{{90}^{°}}} \right]\end{array}$ r：圆柱半径
h：圆柱的长方形顶帽 
$\begin{array}{l}S_{{\rm vv,hh}}^{{\rm cuboid - hat}} = {S_1} + {S_2} + {S_3} + {S_{4{\rm vv,hh}}} + {S_{5{\rm vv,hh}}}\\{S_1} = - {\rm j}\frac{{kbc}}{{\sqrt {π} }}{l_x}{\rm sinc}\left( {kb{l_y}} \right){\rm sinc}\left( {kc{l_z}} \right){{\rm e}^{{\rm j}k\left( {a{l_x} + c{l_z}} \right)}}\\{S_2} = - {\rm j}\frac{{kca}}{{\sqrt {π} }}{l_y}{\rm sinc}\left( {kc{l_z}} \right){\rm sinc}\left( {ka{l_x}} \right){{\rm e}^{{\rm j}k\left( {b{l_y} + c{l_z}} \right)}}\\{S_3} = - {\rm j}\frac{{kab}}{{\sqrt {π} }}{l_z}{\rm sinc}\left( {ka{l_x}} \right){\rm sinc}\left( {kb{l_y}} \right){{\rm e}^{{\rm j}2kc{l_z}}}\\{S_{4{\rm vv,hh}}} =\!\! \mp \frac{1}{{\sqrt {π} }}\!\! \frac{{\min \left\{ {c{l_x},{d_1}{l_z}} \right\}}}{{{l_y}}}\left\{ \begin{array}{l}{\!\!\! {\rm e}^{{\rm j}k\left( {b - {\delta _1}} \right){l_y}}} {\rm sinc}\left( {k{\delta _1}{l_y}} \right)\! - \!{{\rm e}^{ - {\rm j}kb{l_y}}},{\delta _1} \! \le \! b\\\frac{b}{{{\delta _1}}}\left[ {{\rm sinc}\left( {kb{l_y}} \right) - {{\rm e}^{ - {\rm j}kb{l_y}}}} \right],\quad{\delta _1} > b\end{array} \right\}{{\rm e}^{{\rm j}ka{l_x}}}\\{S_{5{\rm vv,hh}}} = \mp \frac{1}{{\sqrt {π} }}\frac{{\min \left\{ {c{l_y},{d_2}{l_z}} \right\}}}{{{l_x}}}\left\{ \begin{array}{l}{{\rm e}^{{\rm j}k\left( {a - {\delta _2}} \right){l_x}}}{\rm sinc}\left( {k{\delta _2}{l_x}} \right) - {{\rm e}^{ - {\rm j}ka{l_x}}},{\delta _2} \le a\\\frac{a}{{{\delta _2}}}\left[ {{\rm sinc}\left( {ka{l_x}} \right) - {{\rm e}^{ - {\rm j}ka{l_x}}}} \right],\quad\ \ {\delta _2} > a\end{array} \right\}{{\rm e}^{{\rm j}kb{l_y}}}\end{array}$ $\begin{array}{c}\theta \in \left[ {{{20}^{°}},{{90}^{°}}} \right]\\\!\!\!\!\phi \in \left[ {{0^{°}},{{15}^{°}}} \right]\\\quad \cup \left[ {{{75}^{°}},{{90}^{°}}} \right]\end{array}$ a, b, c：长方体边长 d1：x方向“帽沿” 宽度 d2：y方向“帽沿” 宽度 圆形顶帽 
$\begin{array}{l}S_{{\rm vv,hh}}^{{\rm top - hat}} = {S_1} + {S_2} + {S_{3{\rm vv,hh}}}\\{S_1} = - \sqrt {{\rm j}kr{h^2}\sin \theta } {\rm sinc}\left( {kh\cos \theta } \right){{\rm e}^{{\rm j}k\left( {h\cos \theta + 2r\sin \theta } \right)}}\\{S_2} = - {\rm j}2\sqrt {π} k{r^2}\left| {\cos \theta } \right|\frac{{{J_1}\left( {2kr\sin \theta } \right)}}{{2kr\sin \theta }}{{\rm e}^{{\rm j}kh\left( {\cos \theta + \left| {\cos \theta } \right|} \right)}}\\{S_{3{\rm vv,hh}}} = \mp 2\sqrt {{\rm j}kr\sin \theta } \min \left\{ {h,\frac{d}{{\tan \theta }}} \right\}{{\rm e}^{{\rm j}2kr\sin \theta }}\end{array}$ $\begin{array}{l}\theta \in \left[ {{{0.1}^{°}},{{90}^{°}}} \right]\\\phi \in \left[ {{0^{°}},{{90}^{°}}} \right]\end{array}$ r：圆柱半径 h：圆柱的长d：“帽沿”宽度 平板上倒圆柱 
$\begin{array}{c}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! S_{{\rm vv,hh}}^{{\rm cyl - plane}} = {S_1} + {S_{2{{\rm vv,hh}}}} + {S_{3{\rm vv,hh}}}\\\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{S_1} = - \sqrt {{\rm j}kr\sqrt {1 - l_y^2} } L{\rm sinc}\left( {kL{l_y}} \right){{\rm e}^{{\rm j}2k\left( {r + h} \right){l_z}}}\\{S_{2{\rm vv,hh}}} = \pm \sqrt { - {\rm j}kr{L^2}\sin \theta } \sin \! {\rm c}\left( {kL{l_y}} \right)U\left( {\arctan \left( {\frac{d}{{r + h}}} \right) - \theta } \right){{\rm e}^{{\rm j}2kr\sin \theta }}\\\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{S_{3{\rm vv,hh}}} = \frac{{2{b_0}}}{{\sqrt {3{π} } }}\left[ { \pm \frac{1}{3} + \frac{1}{{1 + 2\cos \displaystyle\frac{{4\theta }}{3}}}} \right]{\rm sinc}\left( {k{b_0}{l_y}} \right){{\rm e}^{{\rm j}k{a_0}{l_x}}}\\ \quad\quad+ \frac{{2{b_0}}}{{\sqrt {3{π} } }}\left[ { \pm \frac{1}{3} + \frac{1}{{1 + 2\cos \displaystyle\frac{{4\left( {{π} - \theta } \right)}}{3}}}} \right]{\rm sinc}\left( {k{b_0}{l_y}} \right){{\rm e}^{{\rm j}k\left( {{a_0}{l_x} + 2{d_0}{l_z}} \right)}}\\ \quad\quad \quad\ + \frac{{2{b_0}}}{{\sqrt {3{π} } }}\left[ { \pm \frac{1}{3} + \frac{1}{{1 + 2\cos \displaystyle\frac{{4\left( {{π} /2 - \theta } \right)}}{3}}}} \right]{\rm sinc}\left( {k{b_0}{l_y}} \right){{\rm e}^{{\rm j}k\left( { - {a_0}{l_x} + 2{d_0}{l_z}} \right)}}\end{array}$ $\begin{array}{l}\theta \in \left[ {{0^{°}},{{90}^{°}}} \right]\\\phi \in \left[ {{0^{°}},{{10}^{°}}} \right]\end{array}$ r：圆柱半径 L：圆柱的长h：圆柱面到平面距离 a0：平板边长 b0：平板边长 d0：平板厚度 直二面角 
$\begin{array}{l}S_{{\rm vv,hh}}^{{\rm dihedral}} = {S_1} + {S_2} + {S_{3{\rm vv,hh}}}\\{S_1} = - {\rm j}\frac{{kah}}{{\sqrt {π} }}{l_y}{\rm sinc}\left( {kh{l_z}} \right){\rm sinc}\left( {ka{l_x}} \right){{\rm e}^{{\rm j}k\left( {a{l_x} + h{l_z}} \right)}}\\{S_2} = - {\rm j}\frac{{kbh}}{{\sqrt {π} }}{l_x}{\rm sinc}\left( {kh{l_z}} \right){\rm sinc}\left( {kb{l_y}} \right){{\rm e}^{{\rm j}k\left( {b{l_y} + h{l_z}} \right)}}\\{S_{\rm 3vv,hh}} = \left\{ {\begin{array}{*{20}{c}}{ \pm \frac{1}{{\sqrt {π} }}\frac{{\min \{ b{l_x},a{l_y}\} }}{{{l_z}}}\left\{ {{{\rm e}^{{\rm j}k\left( {h - \delta } \right){l_z}}}{\rm sinc}\left( {k\delta {l_z}} \right) - {{\rm e}^{ - {\rm j}kh{l_z}}}} \right\}{{\rm e}^{{\rm j}kh{l_z}}}},&\!\!{\delta \le h}\\\!\!\!\!\!\!\!\!\!\! { \pm \frac{1}{{\sqrt {π} }}\frac{{\min \{ b{l_x},a{l_y}\} }}{{{l_z}}}\left\{ {\frac{h}{\delta }{\rm sinc}\left( {kh{l_z}} \right) - {{\rm e}^{ - {\rm j}kh{l_z}}}} \right\}{{\rm e}^{{\rm j}kh{l_z}}}},&\!\!{\delta > h}\end{array}} \right.\\\quad\quad\ \delta = \min \left\{ {b{l_z}/{l_y},a{l_z}/{l_x}} \right\}\end{array}$ $\begin{array}{l}\theta \in \left[ {{{40}^{°}},{{90}^{°}}} \right]\\\phi \in \left[ {{0^{°}},{{90}^{°}}} \right]\end{array}$ a, b：二面角两面宽度h：二面角的长度 直三面角 
$\begin{array}{l}S_{{\rm vv,hh}}^{\rm trihedral} = {S_1} + {S_2} + {S_3} + {S_{12{\rm vv,hh}}} + {S_{\rm 23vv,hh}} + {S_{\rm 31vv,hh}} + {S_{123}}\\{S_1} = - {\rm j}\frac{{k{a^2}}}{{\sqrt {π} }}{l_x}{\rm sinc}\left( {ka{l_y}} \right){\rm sinc}\left( {ka{l_z}} \right){{\rm e}^{{\rm j}ka\left( {{l_y} + {l_z}} \right)}}\\{S_2} = - {\rm j}\frac{{k{a^2}}}{{\sqrt {π} }}{l_y}{\rm sinc}\left( {ka{l_z}} \right){\rm sinc}\left( {ka{l_x}} \right){{\rm e}^{{\rm j}ka\left( {{l_z} + {l_x}} \right)}}\\{S_3} = - {\rm j}\frac{{k{a^2}}}{{\sqrt {π} }}{l_z}{\rm sinc}\left( {ka{l_x}} \right){\rm sinc}\left( {ka{l_y}} \right){{\rm e}^{{\rm j}ka\left( {{l_x} + {l_y}} \right)}}\\{S_{\rm 12vv,hh}} = \mp {\rm j}2\frac{{k{a^2}}}{{\sqrt {π} }}\min \left\{ {{l_x},{l_y}} \right\}{\rm sinc}\left( {ka{l_z}} \right){{\rm e}^{{\rm j}ka{l_z}}}\\{S_{\rm 23vv,hh}} = \pm {\rm j}2\frac{{k{a^2}}}{{\sqrt {π} }}\min \left\{ {{l_y},{l_z}} \right\}{\rm sinc}\left( {ka{l_x}} \right){{\rm e}^{{\rm j}ka{l_x}}}\\{S_{\rm 31vv,hh}} = \pm {\rm j}2\frac{{k{a^2}}}{{\sqrt {π} }}\min \left\{ {{l_z},{l_x}} \right\}{\rm sinc}\left( {ka{l_y}} \right){{\rm e}^{{\rm j}ka{l_y}}}\\\begin{aligned}{S_{123}} = & - {\rm j}\frac{{k\sqrt 3 {a^2}}}{{\sqrt {π} }}\min \left\{ {\sin \left( {\theta + {π} /4 - \alpha } \right),\cos \left( {\theta + {π} /4 - \alpha } \right)} \right\} \\ &\cdot \min \left\{ {\sin \phi ,\cos \phi } \right\}\end{aligned}\end{array}$ $\begin{array}{l}\theta \in \left[ {{{10}^{°}},{{80}^{°}}} \right]\\\phi \in \left[ {{{10}^{°}},{{80}^{°}}} \right]\end{array}$ a：三面角边长 圆边浅凹体 
$\begin{array}{l}\quad\quad\quad\quad\quad\quad\quad\quad\ S_{\rm vv,hh}^{\rm pan} = \pm {10^{{p_{\rm vv,hh}}\left( \theta \right)/20}}{{\rm e}^{ - {\rm j}2ka\sin \theta }}\\\quad\quad\quad\quad\quad\quad\quad\quad\ {p_{\rm vv}}\left( \theta \right) = \sum\limits_{i = 1}^{{N_{\rm vv}}} {c_i^{\rm vv}{{\left( {90 - \theta } \right)}^i}} \\\quad\quad\quad\quad\quad\quad\quad\quad\ {p_{\rm hh}}\left( \theta \right) = \sum\limits_{i = 1}^{{N_{\rm hh}}} {c_i^{\rm hh}{{\left( {90 - \theta } \right)}^i}} \end{array}$ $\begin{array}{l}\theta \in \left[ {20}^{°},{{90}^{°}} \right]\\\phi \in \left[ 0^{°},{{90}^{°}} \right]\end{array}$ ${N_{\rm vv,hh}}$：拟合函数阶数 $c_j^{\rm vv,hh}$：拟合系数 a：圆盘半径 kc：中心波数 矮顶帽 
$\begin{array}{l}S_{\rm vv,hh}^{\rm short - tophat} = \pm {10^{{p_{\rm vv,hh}}\left( \theta \right)/20}}{{\rm e}^{{\rm j}2ka\sin \theta }}\\{p_{\rm vv}}\left( \theta \right) = \sum\limits_{i = 1}^{{N_{\rm vv}}} {c_i^{\rm vv}{{\left( {90 - \theta } \right)}^i}} \\{p_{\rm hh}}\left( \theta \right) = \sum\limits_{i = 1}^{{N_{\rm hh}}} {c_i^{\rm hh}{{\left( {90 - \theta } \right)}^i}}\end{array}$ $\begin{array}{l}\theta \in \left[ {{{20}^{°}},{{90}^{°}}} \right]\\\phi \in \left[ {{0^{°}},{{90}^{°}}} \right]\end{array}$ ${N_{\rm vv,hh}}$：拟合函数阶数 $c_j^{\rm vv,hh}$：拟合系数a：顶部半径kc：中心波数 狭长二面角 
$\begin{array}{l}\quad\quad\quad S_{\rm vv,hh}^{\rm dihedral\_smallsurf}\\\quad\quad\quad\quad = \pm {{\rm e}^{ - \frac{2}{3}{{\left( {{π} fL\sin \theta /{\rm c}} \right)}^2}{{\left( {\phi - {π} } \right)}^2}}}{10^{{p_{\rm vv,hh}}\left( \theta \right)/20}}{{\rm e}^{{\rm j}2ka\sin \theta }}\\\quad\quad\quad{p_{\rm vv}}\left( \theta \right) = \sum\limits_{i = 1}^{{N_{\rm vv}}} {c_i^{\rm vv}{{\left( {90 - \theta } \right)}^i}} \\\quad\quad\quad{p_{\rm hh}}\left( \theta \right) = \sum\limits_{i = 1}^{{N_{\rm hh}}} {c_i^{\rm hh}{{\left( {90 - \theta } \right)}^i}} \end{array}$ $\begin{array}{l}\theta \in \left[ {{{20}^{°}},{{90}^{°}}} \right]\\\phi \in \left[ {{0^{°}},{{90}^{°}}} \right]\end{array}$ ${N_{\rm vv,hh}}$：拟合函数阶数 $c_j^{\rm vv,hh}$：拟合系数 d：狭长立板偏置距离 L：狭长立板长度kc：中心波数 说明： ${l_x} = \sin \theta \cos \phi, \;{l_y} = \sin \theta \sin \phi, \;{l_z} = \cos \theta $ ${J_1}\left( \cdot \right)$为一阶贝塞尔函数；min{a,b}指取a,b中的较小值；U(x)为单位阶跃函数，即x大于等于0时取1，小于0时取0； $k = \frac{{2{π} f}}{\rm c}$为波数。
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