二阶微扰法随机粗糙面透射特性计算方法

摘要

本发明提供了一种二阶微扰法随机粗糙面透射特性计算方法，考虑了二阶项解对粗糙面透射特性的影响，利用微扰法给出了二阶透射场、二阶透射率和二阶双向透射系数的计算方法，扩大了适用范围，提高了精度，满足微波辐射遥感定量信息反演的精度要求。

说明书

技术领域

本发明属于电磁辐射领域，具体涉及一种随机粗糙面透射特性的微扰计算方法。

背景技术

电波传播、通信、目标识别分类、环境系统检测、遥感、生物医学诊断、工程材料测试等众多学科的应用需要，推动了随机粗糙表面电磁辐射、散射和透射研究的深入发展.在解析法求解中，最重要的两种方法是基尔霍夫近似法和微扰法，而其它方法都与这两种基本方法或多或少相关。

微扰法的适用条件是，表面标准离差小于电磁波波长的5％左右，且表面平均斜度与波数和表面标准离差之积有同一数量级，即适用于小尺度粗糙度情况。在微波遥感的许多应用都会出现小尺度粗糙度，如海面的小尺度波浪(毛细重力波)，较平坦的裸土表面，以及月球表面的局部区域等。研究表明在用微波辐射计垂直观测地物表面的辐射亮温时，这种小尺度粗糙度会对辐射亮温产生重要影响。

现有微扰法可考虑零阶、一阶和二阶的散射贡献，其中零阶解是将粗糙面视作平面时相干分量解，一阶解是最低阶的非相干分量解，而二阶解则是对相干分量的最低阶矫正项，对于保持算法的能量守恒，提高反射率、发射率的计算精度有着重要意义。现有微扰法研究多集中在粗糙面的散射研究中，而在透射方面的研究则非常欠缺，实际上，粗糙面的透射特性与散射特性一样有重要的研究价值，如在对土壤、月壤等地物的微波辐射遥感中，这些地物通常可视作表面粗糙的非等温的分层媒质，要计算这些地物的辐射亮度温度，必须掌握其粗糙表面的透射特性。现有文献仅有一阶微扰法的透射特性研究，但仅考虑零阶和一阶解将造成超过20％的误差，无法满足实际应用的需求。而对二阶微扰法，近来有国外学者基于瑞利假设，仅针对周期介质给出了二阶双站透射系数，但未推导出二阶透射率的解析公式。本发明针对随机粗糙面，基于惠更斯原理和消光定理，用微扰法推导出二阶透射场、二阶双站透射系数和二阶透射率，得到了完整的二阶透射特性并验证了二阶微扰法能量守恒情况及二阶双站透射系数。

发明内容

本发明是为解决现有随机粗糙面微扰法透射模型中忽略二阶项解这一不足之处，提供一种考虑二阶透射贡献的随机粗糙面二阶微扰法透射特性计算方法，以满足微波辐射遥感定量信息反演的精度要求。

二阶微扰法随机粗糙面透射特性计算方法，包含透射场、透射率和双向透射系数的计算步骤，具体为：

(1)所述粗糙面透射场表示为：

${\bar{E}}_t^{\left(2\right)}\left(\bar{r}\right)=\int {\overline{\mathrm{dk}}}_{\perp }{e}^{{\overline{\mathrm{ik}}}_{i\perp }·{\bar{r}}_{\perp }-{\mathrm{ik}}_{1z}z}\left\{{\hat{e}}_1(-k_{1z})\right[f_{\mathrm{ee}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)+f_{\mathrm{eh}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)]$

${+\hat{h}}_1(-k_{1z})[f_{\mathrm{he}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)+f_{\mathrm{hh}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)]$

其中，

$f_{\mathrm{ee}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })=\frac{k_{\mathrm{iz}}}{k_{\mathrm{iz}}+k_{1\mathrm{iz}}}\frac{k_1^2-k^2}{k_{1z}+k_z}\{\cos ({\phi }_k-{\phi }_i)·(k_{1\mathrm{zi}}-k_z)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })$

$-2(k_1^2-k^2)\int {\overline{\mathrm{dk}}}_{\perp }^{\prime }F({\bar{k}}_{\perp }^{\prime }-{\bar{k}}_{i\perp })F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })$

$·[-\sin ({\phi }_k-{{\phi }_k}^{\prime })\sin ({{\phi }_k}^{\prime }-{\phi }_i)\frac{k_z^{\prime }{k}_{1z}^{\prime }}{k_1^2{k}_z^{\prime }+k^2{k}_{1z}^{\prime }}+\cos ({\phi }_k-{{\phi }_k}^{\prime })\cos ({{\phi }_k}^{\prime }-{\phi }_i)\frac{1}{k_{1z}^{\prime }+k_z^{\prime }}]\}$

$f_{\mathrm{eh}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })=\frac{k_1^2-k^2}{k_{1z}+k_z}\frac{{\mathrm{kk}}_{\mathrm{iz}}}{k_1^2{k}_{\mathrm{iz}}+k^2{k}_{1\mathrm{zi}}}\{(k_1^2-k_{1\mathrm{zi}}{k}_z)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })\sin ({\phi }_k-{\phi }_i)$

$-2\int {\overline{\mathrm{dk}}}_{\perp }^{\prime }F({\bar{k}}_{\perp }^{\prime }-{\bar{k}}_{i\perp })F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })[\sin ({\phi }_k-{\phi }_k^{\prime })\frac{k_1^2{k}_{\rho }^{\prime }{k}_{\mathrm{\rho i}}}{k_{\rho }^{\prime 2}k+k_z^{\prime }{k}_{1z}^{\prime }}+\frac{(k_1^2-k^2)k_{1\mathrm{zi}}}{k_1^2{k}_z^{\prime }+k^2{k}_{1z}^{\prime }}$

$·(\sin ({\phi }_k-{{\phi }_k}^{\prime })\cos ({{\phi }_k}^{\prime }-{\phi }_i)k_z^{\prime }{k}_{1z}^{\prime }+\cos ({\phi }_k-{{\phi }_k}^{\prime })\sin ({{\phi }_k}^{\prime }-{\phi }_i)(k_{\rho }^{\prime 2}+k_z^{\prime }{k}_{1z}^{\prime })\left]\right\}$

$f_{\mathrm{he}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })=\frac{k_1^2-k^2}{k_{1\mathrm{zi}}+k_{\mathrm{iz}}}\frac{k_1{k}_{\mathrm{iz}}}{k_1^2{k}_z+k^2{k}_{1z}}\{(k^2-k_{1\mathrm{zi}}{k}_z)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })\sin ({\phi }_k-{\phi }_i)$

$+2\int {\overline{\mathrm{dk}}}_{\perp }^{\prime }F({\bar{k}}_{\perp }^{\prime }-{\bar{k}}_{i\perp })F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })[\sin ({\phi }_k^{\prime }-{\phi }_i)\frac{k_1^2{k}_{\rho }^{\prime }{k}_{\rho }}{k_{\rho }^{\prime 2}k+k_z^{\prime }{k}_{1z}^{\prime }}+\frac{(k_1^2-k^2)k_z}{k_1^2{k}_z^{\prime }+k^2{k}_{1z}^{\prime }}$

$·\left(k_z^{\prime }{k}_{1z}^{\prime }\sin ({\phi }_k-{{\phi }_k}^{\prime })\cos ({{\phi }_k}^{\prime }-{\phi }_i)+(k_{\rho }^{\prime 2}+k_z^{\prime }{k}_{1z}^{\prime })\cos ({\phi }_k-{{\phi }_k}^{\prime })\sin ({{\phi }_k}^{\prime }-{\phi }_i)\right)]$

$f_{\mathrm{hh}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })=\frac{(k_1^2-k^2)}{({k_1^{2}k}_z+{k^{2}k}_{1z})}\frac{k_1{k}_{\mathrm{iz}}}{k_1^2{k}_{\mathrm{iz}}+k^2{k}_{1\mathrm{zi}}}\{k(k_1^2{k}_z-{k^{2}k}_{1\mathrm{zi}})F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })\sin ({\phi }_k-{\phi }_i)$

$+2\int {\overline{\mathrm{dk}}}_{\perp }^{\prime }F({\bar{k}}_{\perp }^{\prime }-{\bar{k}}_{i\perp })F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })[{\mathrm{kk}}_z{k}_{1\mathrm{zi}}\frac{(k_1^2-k^2)}{k_z^{\prime }+k_{1z}^{\prime }}\sin ({\phi }_k-{{\phi }_k}^{\prime })\sin ({\phi }_k^{\prime }-{\phi }_i)$

$+\frac{1}{k_1^2{k}_z^{\prime }+k^2{k}_{1z}^{\prime }}[-k(k_1^2-k^2)k_{\rho }{k}_{\rho }^{\prime 2}{k}_{\mathrm{\rho i}}+k_{\rho }{k}_{\rho }^{\prime }{k}^3(k_z^{\prime }+k_{1z}^{\prime })k_{1\mathrm{zi}}\cos ({{\phi }_k}^{\prime }-{\phi }_i)-k_{\rho }^{\prime }{k}_{\mathrm{\rho i}}{\mathrm{kk}}_1^2$

$·(k_z^{\prime }+k_{1z}^{\prime })k_z\cos ({\phi }_k-{{\phi }_k}^{\prime })-{\mathrm{kk}}_z^{\prime }{k}_{1z}^{\prime }(k_1^2-k^2)k_z{k}_{1\mathrm{zi}}\cos ({\phi }_k-{{\phi }_k}^{\prime })\cos ({{\phi }_k}^{\prime }-{\phi }_i)\left]\right\}$

(2)所述透射率表示为：

$t(\pi -{\theta }_i,{\phi }_i)=\frac{k_{1\mathrm{zi}}}{k_{\mathrm{zi}}}|1+R_{\mathrm{ho}}{|}^2(\hat{e}(-k_{\mathrm{iz}})·{\hat{e}}_i)+\frac{k^2{k}_{1\mathrm{zi}}}{{k_1}^2{k}_{\mathrm{zi}}}|1+R_{\mathrm{vo}}{|}^2(\hat{h}(-k_{\mathrm{iz}})·{\hat{e}}_i)$

$+2\mathrm{Re}\left(\frac{k_{1\mathrm{zi}}}{k_{\mathrm{zi}}}(1+R_{\mathrm{ho}})f_{\mathrm{ee}}^{t\left(2\right)}\left(k_{i\perp }\right)\right)(\hat{e}(-k_{\mathrm{iz}})·{\hat{e}}_i)+2\mathrm{Re}\left(\frac{{\mathrm{kk}}_{1\mathrm{zi}}}{k_1{k}_{\mathrm{zi}}}(1+R_{\mathrm{vo}})f_{\mathrm{hh}}^{t\left(2\right)}\left(k_{i\perp }\right)\right)$

$·(\hat{h}(-k_{\mathrm{iz}})·{\hat{e}}_i)+\int d{\bar{k}}_{\perp }\frac{k_{1\mathrm{zi}}}{k_{\mathrm{zi}}}W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })\left[\right|f_{\mathrm{ee}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })·(\hat{e}(-k_{\mathrm{iz}})·{\hat{e}}_i)+f_{\mathrm{eh}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })·(\hat{h}(-k_{\mathrm{iz}})·{\hat{e}}_i){|}^2$

$+|f_{\mathrm{he}}^{t\left(1\right)}(\hat{e}(-k_{\mathrm{iz}})·{\hat{e}}_i)+f_{\mathrm{hh}}^{t\left(1\right)}(\hat{h}(-k_{\mathrm{iz}})·{\hat{e}}_i){|}^2]$

其中是随机起伏高度的相关函数的Fourier变换，

$W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })=\frac{h^2{l}^2}{4\pi }\exp (-\frac{(k_{\rho }^2+k_{\mathrm{\rho i}}^2)l^2}{4}+\frac{k_{\rho }{k}_{\mathrm{\rho i}}{l}^2}{2}\cos ({\phi }_k-{\phi }_i));$

(3)所述双向透射系数表示为：

在入射波h极化，透射波h极化情况下的双向透射系数为：

$\frac{1}{4\pi }{\gamma }_{\mathrm{hh}}^t(\pi -{\theta }_t,{\phi }_t;\pi -{\theta }_i,{\phi }_i)=\left[\frac{k_{1\mathrm{zi}}}{k_{\mathrm{zi}}}\right|1+R_{\mathrm{ho}}{|}^2+1\mathrm{Re}\left(\frac{k_{1\mathrm{zi}}}{k_{\mathrm{zi}}}(1+R_{\mathrm{ho}})f_{\mathrm{ee}}^{t\left(2\right)}\left({\bar{k}}_{i\perp }\right)\right)]\delta ({\cos \theta }_t-\frac{k_{1\mathrm{zi}}}{k_1})\delta ({\phi }_t-{\phi }_i)$

$+\frac{k_1{k}_{1z}^2}{k_{\mathrm{iz}}}W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })|f_{\mathrm{ee}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp }){|}^2$

在入射波v极化，透射波h极化情况下，

$\frac{1}{4\pi }{\gamma }_{\mathrm{hv}}^t(\pi -{\theta }_t,{\phi }_t;\pi -{\theta }_i,{\phi }_i)=\frac{k_1{k}_{1z}^2}{k_{\mathrm{iz}}}W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })|f_{\mathrm{eh}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp }){|}^2$

在入射波h极化，透射波v极化情况下，

$\frac{1}{4\pi }{\gamma }_{\mathrm{vh}}^t(\pi -{\theta }_t,{\phi }_t;\pi -{\theta }_i,{\phi }_i)=\frac{k_1{k}_{1z}^2}{k_{\mathrm{iz}}}W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })|f_{\mathrm{he}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp }){|}^2$

在入射波v极化，透射波v极化情况下，

$\frac{1}{4\pi }{\gamma }_{\mathrm{vv}}^t(\pi -{\theta }_t,{\phi }_t;\pi -{\theta }_i,{\phi }_i)=\left[\frac{k^2{k}_{1\mathrm{zi}}}{k_1^2{k}_{\mathrm{iz}}}\right|1+R_{\mathrm{vo}}{|}^2+2\mathrm{Re}\left(\frac{{\mathrm{kk}}_{1\mathrm{zi}}}{k_1{k}_{\mathrm{iz}}}(1+R_{\mathrm{vo}})f_{\mathrm{hh}}^{t\left(2\right)}\left({\bar{k}}_{i\perp }\right)\right)]$

$·\delta (\cos {\theta }_t-\frac{k_{1\mathrm{zi}}}{k_1})\delta ({\phi }_t-{\phi }_i)+\frac{k_1{k}_{1z}^2}{k_{\mathrm{iz}}}W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })|f_{\mathrm{hh}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp }){|}^2$

其中a，b分别为v极化和h极化的两种情况，且表达镜向透射方向应以θ_t，φ_t的形式：其中时，θ_t才指向透射方向，Re()表示对括号中的数取实部；

$f_{\mathrm{ab}}^{t\left(2\right)}\left({\bar{k}}_{i\perp }\right)=<f_{\mathrm{ab}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })>a,b=v,h$

$f_{\mathrm{ee}}^{t\left(2\right)}\left({\bar{k}}_{i\perp }\right)=\frac{k_{\mathrm{iz}}(k_1^2-k^2)}{{(k_{\mathrm{iz}}+k_{1\mathrm{zi}})}^2}\{(k_{1\mathrm{zi}}-k_{\mathrm{iz}}){\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })-2(k_1^2-k^2){\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })$

$·[{\sin }^2({\phi }_k-{\phi }_i)\frac{k_z{k}_{1z}}{k_1^2{k}_z+k^2{k}_{1z}}+{\cos }^2({\phi }_k-{\phi }_i)\frac{1}{k_{1z}+k_z}]\}$

$f_{\mathrm{hh}}^{t\left(2\right)}\left({\bar{k}}_{i\perp }\right)=\frac{k_1({k_1}^2-k^2)k_{\mathrm{iz}}}{{({k_1}^2{k}_{\mathrm{iz}}+k^2{k}_{1\mathrm{zi}})}^2}\{{\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })k(k_1^2{k}_{\mathrm{iz}}-k^2{k}_{1\mathrm{zi}})+$

$2{\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })[{-\mathrm{kk}}_{\mathrm{iz}}{k}_{1\mathrm{zi}}\frac{(k_1^2-k^2)}{k_2+k_{1z}}{\sin }^2({\phi }_k-{\phi }_i)+$

$\frac{1}{k_1^2{k}_z+k^2{k}_{1z}}(-k({k_1}^2-k^2)k_{\mathrm{\rho i}}^2{k}_{\rho }^2+k_{\rho }{k}_{\mathrm{\rho i}}k(k_z+k_{1z})(k^2{k}_{1\mathrm{zi}}-k_1^2{k}_{\mathrm{iz}})$

$·\cos ({\phi }_k-{\phi }_i)-{\mathrm{kk}}_z{k}_{1z}{k}_{\mathrm{iz}}{k}_{1\mathrm{zi}}({k_1}^2-k^2){\cos }^2({\phi }_k-{\phi }_i)\left)\right]\}$

入射波矢量

入射波矢量的水平分量为

k_ix＝ksinθ_icosφ_i，k_iy＝ksinθ_isinφ_i，k_iz＝kcosθ_i，k_ρi＝ksinθ_i，

入射波水平分量的模

散射波矢量

散射波水平分量

散射波水平分量的模

散射波垂直分量的模

透射波矢量

透射波矢量的水平分量为

透射波水平分量的模

透射波垂直分量的模

$k_{1\mathrm{zi}}=\sqrt{k_1^2+k_{i\perp }^2},$

入射波的入射角θ_i和方位角透射波的透射角θ_t和方位角和分别代表原点和场点的位置矢量，

当入射场为TE极化时入射场为TM极化时

μ_o和μ₁分别代表介质0和介质1的磁导率，

ε_o和ε₁分别代表介质0和介质1的电导率，

w为角频率，

η₁为介质1的波阻抗，

粗糙面的均方根高度h和相关长度l，

$\hat{e}\left(k_z\right)=\hat{k}\times \hat{z}/|\hat{k}\times \hat{z}|=(1/k_{\rho })(\hat{x}{k}_y-\hat{y}{k}_x),$

$\hat{e}\left({-k}_z\right)=(1/k_{\rho })(\hat{x}{k}_y-\hat{y}{k}_x),$

$\hat{e}\left({-k}_{1\mathrm{zi}}\right)={\hat{k}}_1\times \hat{z}/|{\hat{k}}_1\times \hat{z}|=(1/k_{1\mathrm{\rho i}})(\hat{x}{k}_{1\mathrm{yi}}-\hat{y}{k}_{1\mathrm{xi}}),$

${\hat{e}}_1\left(k_{1z}\right)={\hat{k}}_1\times \hat{z}/|{\hat{k}}_1\times \hat{z}|=(1/k_{\rho })(\hat{x}{k}_y-\hat{y}{k}_x),$

$\hat{h}\left(k_z\right)=\frac{1}{k}\hat{e}\times \hat{k}=\frac{{-k}_z}{{\mathrm{kk}}_{\rho }}(\hat{x}{k}_x+\hat{y}{k}_y)+\frac{k_{\rho }}{k}\hat{z},$

$\hat{h}\left({-k}_z\right)=\frac{k_z}{{\mathrm{kk}}_{\rho }}(\hat{x}{k}_x+\hat{y}{k}_y)+\frac{k_{\rho }}{k}\hat{z},$

${\hat{h}}_1\left(k_{1z}\right)=\frac{1}{k_1}{\hat{e}}_1\times {\hat{k}}_1=\frac{{-k}_{1z}}{k_1{k}_{\rho }}(\hat{x}{k}_x+\hat{y}{k}_y)+\frac{k_{\rho }}{k_1}\hat{z},$

${\hat{h}}_1\left({-k}_{1z}\right)=\frac{k_{1z}}{k_1{k}_{\rho }}(\hat{x}{k}_{1x}+\hat{y}{k}_{1y})+\frac{k_{\rho }}{k_1}\hat{z},$

${\hat{h}}_1\left({-k}_{1\mathrm{zi}}\right)=\frac{k_{1\mathrm{zi}}}{k_1{k}_{\mathrm{\rho i}}}(\hat{x}{k}_{\mathrm{xi}}+\hat{y}{k}_{\mathrm{yi}})+\frac{k_{\mathrm{\rho i}}}{k_1}\hat{z},$

傅里叶变换

$\frac{1}{{4\pi }^2}\int d{\bar{r}}_{\perp }{e}^{-i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }}{f}^2\left({\bar{r}}_{\perp }\right)=F^2\left({\bar{k}}_{\perp }\right),$

$F^2\left({\bar{k}}_{\perp }\right)={\int }_{-\infty }^{\infty }d{{\bar{k}}^{\prime }}_{\perp }F\left({{\bar{k}}^{\prime }}_{\perp }\right)F({\bar{k}}_{\perp }-{{\bar{k}}^{\prime }}_{\perp }),$

是TE波的菲涅尔透射系数，

是TM波的菲涅尔透射系数，

φ′_k，k′_ρ，k′_z和k′_1z是积分所需要的中间变量。

本发明相比现有技术具有如下优点：

针对随机粗糙面，现有的微扰法只给出了一阶透射场和一阶透射率的计算公式，一阶透射公式的适用范围是：但即便是在该范围内，计算出的透射率和反射率不符合能量守恒，将造成接近20％的误差，无法满足实际应用的需求。本发明利用微扰法，经过严格的理论推导给出了二阶透射场、二阶透射率和二阶双向透射系数的计算公式，并验证了一阶、二阶微扰法的能量守恒，结果表明二阶透射公式的适用范围比一阶的要宽，且在其适用范围内，计算出的透射率和反射率符合能量守恒，能满足实际应用的需求。

附图说明

图1是二维随机粗糙面示意图；

图2是在粗糙度在一阶适用范围内的能量守恒验证图，其中图2(a)是v极化，图2(a)是h极化。

图3是在粗糙度超出一阶适用范围的能量守恒验证图，其中图3(a)是v极化，图3(a)是h极化。

具体实施方式

首先计算二阶透射场，然后计算二阶透射率。

一、透射场求解

假设有一平面波：由介质0入射到随机粗糙面上进入介质1中，介质0的介电常数为ε₀，介质1的介电常数ε₁。其中入射波矢量入射波矢量的水平分量为对随机粗糙面建立三维直角坐标系，由z＝f(x，y)随机函数描述此粗糙面，对函数取集平均得<f(x，y)>＝0。(如图1)f_min和f_max分别代替粗糙面f(x，y)的最小和最大值。

利用惠更斯原理和消光原理知，介质0中的散射电场和磁场和介质1中透射电场和磁场满足：

${\bar{E}}_i\left(\bar{r}\right)+{\int }_{S^{\prime }}d{S}^{\prime }\{\mathrm{iw}{\mu }_o\stackrel{\overline{‾}}{G}(\bar{r},{\bar{r}}^{\prime })·[\hat{n}\times \bar{H}\left({\bar{r}}^{\prime }\right)]+▿\times \stackrel{\overline{‾}}{G}(\bar{r},{\bar{r}}^{\prime })·[\hat{n}\times \bar{E}\left({\bar{r}}^{\prime }\right)\left]\right\}=\left(\begin{array}{ccc}\bar{E}& z>f\left(\bar{r}\right)& \left(a\right)\\ 0& z<f\left(\bar{r}\right)& \left(b\right)\end{array}\right)$

${\int }_{S^{\prime }}d{S}^{\prime }\{\mathrm{iw}{\mu }_1{\stackrel{\overline{‾}}{G}}_1(\bar{r},{\bar{r}}^{\prime })·[{\hat{n}}_d\times {\bar{H}}_1\left({\bar{r}}^{\prime }\right)+▿\times {\stackrel{\overline{‾}}{G}}_1(\bar{r},{\bar{r}}^{\prime })·[{\hat{n}}_d\times {\bar{E}}_1\left({\bar{r}}^{\prime }\right)]\}=\left(\begin{array}{ccc}0& z>f\left(\bar{r}\right)& \left(a\right)\\ {\bar{E}}_1\left(\bar{r}\right)& z<f\left(\bar{r}\right)& \left(b\right)\end{array}\right)---\left(2\right)$

S′表示对整个粗糙面进行积分，是指向介质0的面元单位法向量，表示指向介质1的单位法向量。和表示介质0和介质1中的格林函数。

根据电场与磁场切向分量连续的边界条件，可以令

$d{{\bar{r}}^{\prime }}_{\perp }\bar{a}\left({{\bar{r}}^{\prime }}_{\perp }\right)=d{S}^{\prime }\eta \hat{n}\times \bar{H}\left({\bar{r}}^{\prime }\right)=d{S}^{\prime }\eta \hat{n}\times {\bar{H}}_1\left({{\bar{r}}^{\prime }}_{\perp }\right),d{{\bar{r}}^{\prime }}_{\perp }\bar{b}\left({{\bar{r}}^{\prime }}_{\perp }\right)=d{S}^{\prime }\hat{n}\times \bar{E}\left({\bar{r}}^{\prime }\right)=d{S}^{\prime }\hat{n}\times {\bar{E}}_1\left({{\bar{r}}^{\prime }}_{\perp }\right)---\left(3\right)$

$\hat{n}\left({{\bar{r}}^{\prime }}_{\perp }\right)·\bar{a}\left({{\bar{r}}^{\prime }}_{\perp }\right)=0,\hat{n}\left({{\bar{r}}^{\prime }}_{\perp }\right)·\bar{b}\left({{\bar{r}}^{\prime }}_{\perp }\right)=0$

由于可以推出：

$a_z\left({{\bar{r}}^{\prime }}_{\perp }\right)=(\hat{x}\frac{\partial f\left({{\bar{r}}^{\prime }}_{\perp }\right)}{{\partial x}^{\prime }}+\hat{y}\frac{\partial f\left({{\bar{r}}^{\prime }}_{\perp }\right)}{{\partial y}^{\prime }})·\bar{a}\left({{\bar{r}}^{\prime }}_{\perp }\right),b_z\left({{\bar{r}}^{\prime }}_{\perp }\right)=(\hat{x}\frac{\partial f\left({{\bar{r}}^{\prime }}_{\perp }\right)}{{\partial x}^{\prime }}+\hat{y}\frac{\partial f\left({{\bar{r}}^{\prime }}_{\perp }\right)}{{\partial y}^{\prime }})·\bar{b}\left({{\bar{r}}^{\prime }}_{\perp }\right)---\left(5\right)$

则由(1b)和(2a)可推得：

${\bar{E}}_i\left(\bar{r}\right)=\frac{1}{8{\pi }^2}\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }}{e}^{-{\mathrm{ik}}_{z}z}\frac{k}{k_z}\int d{{\bar{r}}^{\prime }}_{\perp }{e}^{-i{\bar{k}}_{i\perp }·{{\bar{r}}^{\prime }}_{\perp }}{e}^{i{k}_{z}f\left({{\bar{r}}^{\prime }}_{\perp }\right)}\times \left\{\right[\hat{e}(-k_z)\hat{e}(-k_z)+\hat{h}(-k_z)·\hat{h}(-k_z)]·\bar{a}\left({{\bar{r}}^{\prime }}_{\perp }\right)---\left(6\right)$

$+[-\hat{h}\left({-k}_z\right)\hat{e}\left({-k}_z\right)+\hat{e}(-k_z)·\hat{h}(-k_z)]·\bar{b}\left({{\bar{r}}^{\prime }}_{\perp }\right)\}$

$0=\frac{1}{8{\pi }^2}\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }}{e}^{{\mathrm{ik}}_{1z}z}\frac{k_1}{k_{1z}}\int d{{\bar{r}}^{\prime }}_{\perp }{e}^{-i{\bar{k}}_{i\perp }·{{\bar{r}}^{\prime }}_{\perp }}{e}^{-{\mathrm{ik}}_{1z}f\left({{\bar{r}}^{\prime }}_{\perp }\right)}\times \left\{\frac{k}{k_1}\right[{\hat{e}}_1\left(k_{1z}\right){\hat{e}}_1\left(k_{1z}\right)+{\hat{h}}_1\left(k_{1z}\right)·{\hat{h}}_1\left(k_{1z}\right)]·\bar{a}\left({{\bar{r}}^{\prime }}_{\perp }\right)---\left(7\right)$

$+[-{\hat{h}}_1\left({\mathrm{kk}}_{1z}\right){\hat{e}}_1\left(k_{1z}\right)+{\hat{e}}_1\left(k_{1z}\right)·{\hat{h}}_1\left(k_{1z}\right)]·\bar{b}\left({{\bar{r}}^{\prime }}_{\perp }\right)\}$

其中散射波矢量

$\bar{k}={\bar{k}}_{\perp }+\hat{z}{k}_z=\hat{x}{k}_x+\hat{y}{k}_y+\hat{z}{k}_z,k=\omega \sqrt{{\mu }_0{ϵ}_0}---\left(8\right)$

散射波矢量的水平分量为

${\bar{k}}_{\perp }=\hat{x}{k}_x+\hat{y}{k}_y---\left(9\right)$

其中透射波矢量

${\bar{k}}_1={\bar{k}}_{\perp }+\hat{z}{k}_{1z}=\hat{x}{k}_x+\hat{y}{k}_y+\hat{z}{k}_{1z},k_1=\omega \sqrt{{\mu }_0{ϵ}_0}---\left(10\right)$

透射波矢量的水平分量为

${\bar{k}}_{1\perp }=\hat{x}{k}_x+\hat{y}{k}_y=-{\bar{k}}_{\perp }--\left(11\right)$

水平分量的模

$k_{\rho }=\sqrt{k_x^2+k_y^2},k_z=\sqrt{k^2-k_{\rho }^2},k_{1z}=\sqrt{{k_1}^2-k_{\rho }^2}---\left(12\right)$

$\hat{e}\left(k_z\right)=\hat{k}\times \hat{z}/|\hat{k}\times \hat{z}|=(1/k_{\rho })(\hat{x}{k}_y-\hat{y}{k}_x)---\left(13\right)$

$\hat{h}\left(k_z\right)=\frac{1}{k}\hat{e}\times \hat{k}=\frac{{-k}_z}{{\mathrm{kk}}_{\rho }}(\hat{x}{k}_x+\hat{y}{k}_y)+\frac{k_{\rho }}{k}\hat{z}---\left(14\right)$

${\hat{e}}_1\left(k_{1z}\right)={\hat{k}}_1\times \hat{z}/|{\hat{k}}_1\times \hat{z}|=(1/k_{\rho })(\hat{x}{k}_y-\hat{y}{k}_x)---\left(15\right)$

${\hat{h}}_1\left(k_{1z}\right)=\frac{1}{k_1}{\hat{e}}_1\times {\hat{k}}_1=\frac{{-k}_{1z}}{{k_{1}k}_{\rho }}(\hat{x}{k}_x+\hat{y}{k}_y)+\frac{k_{\rho }}{k_1}\hat{z}---\left(16\right)$

用(6)和(7)我们可以确定粗糙面的表面场，即解出两个未知量和一旦我们确定这个粗糙面的表面场，在介质0的散射场和在介质1的透射场就可以由(1a)和(2b)式确定出来。

由(1a)和(2b)可推知

${\bar{E}}_s\left({\bar{r}}_{\perp }\right)=-\frac{1}{8{\pi }^2}\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }}{e}^{{\mathrm{ik}}_{z}z}\frac{k}{k_z}\int d{{\bar{r}}^{\prime }}_{\perp \perp }{e}^{-i{\bar{k}}_{\perp }·{{\bar{r}}^{\prime }}_{\perp }}{e}^{-{\mathrm{ik}}_{z}f\left({{\bar{r}}^{\prime }}_{\perp }\right)}$

$\times \left\{\right[\hat{e}\left(k_z\right)\hat{e}\left(k_z\right)+\hat{h}\left(k_z\right)\hat{h}\left(k_z\right)]·\bar{a}\left({{\bar{r}}^{\prime }}_{\perp }\right)---\left(17\right)$

$+[-\hat{h}\left(k_z\right)\hat{e}\left(k_z\right)+\hat{e}\left(k_z\right)\hat{h}\left(k_z\right)]·\bar{b}\left({{\bar{r}}^{\prime }}_{\perp }\right)\}$

${\bar{E}}_t\left({\bar{r}}_{\perp }\right)=\frac{1}{8{\pi }^2}\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }}{e}^{{\mathrm{ik}}_{1z}z}\frac{k_1}{k_{1z}}\int d{{\bar{r}}^{\prime }}_{\perp \perp }{e}^{-i{\bar{k}}_{i\perp }·{{\bar{r}}^{\prime }}_{\perp }}{e}^{-{\mathrm{ik}}_{1z}f\left({{\bar{r}}^{\prime }}_{\perp }\right)}$

$\times \left\{\frac{k}{k_1}\right[{\hat{e}}_1(-k_{1z}){\hat{e}}_1\left({-k}_{1z}\right)+{\hat{h}}_1\left({-k}_{1z}\right){\hat{h}}_1(-k_{1z})]·\bar{a}\left({\bar{r}}_{\perp }^{\prime }\right)---\left(18\right)$

$+[-{\hat{h}}_1(-k_{1z}){\hat{e}}_1(-k_{1z})+{\hat{e}}_1\left({-k}_{1z}\right){\hat{h}}_1(-k_{1z})]·\bar{b}\left({\bar{r}}_{\perp }^{\prime }\right)\}$

下面利用微扰法来进一步推导计算，分解和如下：

$\bar{a}\left({\bar{r}}_{\perp }^{\prime }\right)=\underset{m=0}{\overset{\infty }{\Sigma }}{\bar{a}}^{\left(m\right)}\left({\bar{r}}_{\perp }^{\prime }\right),\bar{b}\left({\bar{r}}_{\perp }^{\prime }\right)=\underset{m=0}{\overset{\infty }{\Sigma }}{\bar{b}}^{\left(m\right)}\left({\bar{r}}_{\perp }^{\prime }\right)---\left(19\right)$

这里代表第m阶的分量。

由泰勒级数展开可以得到：

$e^{\pm {\mathrm{ik}}_{z}f\left({\bar{r}}_{\perp }^{\prime }\right)}=\underset{m=0}{\overset{\infty }{\Sigma }}\frac{{[\pm {\mathrm{ik}}_{z}f\left({\bar{r}}_{\perp }^{\prime }\right)]}^m}{m!},e^{\pm {\mathrm{ik}}_{1z}f\left({\bar{r}}_{\perp }^{\prime }\right)}=\underset{m=0}{\overset{\infty }{\Sigma }}\frac{{[\pm {\mathrm{ik}}_{1z}f\left({\bar{r}}_{\perp }^{\prime }\right)]}^m}{m!}---\left(20\right)$

在微扰法中，及其斜率参数都要求是小尺度参数，即假利用(5)和(19)可以得到如下关系：

${\bar{a}}_z^{\left(0\right)}\left({\bar{r}}_{\perp }^{\prime }\right)={\bar{b}}_z^{\left(0\right)}\left({\bar{r}}_{\perp }^{\prime }\right)=0---\left(21\right)$

$a_z^{\left(m\right)}\left({\bar{r}}_{\perp }^{\prime }\right)=(\hat{x}\frac{\partial f\left({\bar{r}}_{\perp }^{\prime }\right)}{\partial x^{\prime }}+\hat{y}\frac{\partial f\left({\bar{r}}_{\perp }^{\prime }\right)}{\partial y^{\prime }})·{\bar{a}}_{\perp }^{(m-1)}\left({\bar{r}}_{\perp }^{\prime }\right),b_z^{\left(m\right)}\left({\bar{r}}_{\perp }^{\prime }\right)=(\hat{x}\frac{\partial f\left({\bar{r}}_{\perp }^{\prime }\right)}{\partial x^{\prime }}+\hat{y}\frac{\partial f\left({\bar{r}}_{\perp }^{\prime }\right)}{\partial y^{\prime }})·{\bar{b}}_{\perp }^{(m-1)}\left({\bar{r}}_{\perp }^{\prime }\right)---\left(22\right)$

把(19)和(20)式带入(5)，(6)和(7)并保持相同阶数，可以计算出各阶表面场，将表面场代入(17)，(18)式则可以得到各阶的散射和透射场。由于利用微扰法来研究二阶透射场的特性，所以以二阶为例来说明。

我们定义表面场的傅里叶变换：

$\bar{A}\left(\bar{k}\right)=\frac{1}{{\left(2\pi \right)}^2}\int d{\bar{r}}_{\perp }^{\prime }\bar{a}\left({\bar{r}}_{\perp }^{\prime }\right)e^{-i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }^{\prime }},\bar{B}\left(\bar{k}\right)=\frac{1}{{\left(2\pi \right)}^2}\int d{\bar{r}}_{\perp }^{\prime }\bar{b}\left({\bar{r}}_{\perp }^{\prime }\right)e^{-i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }^{\prime }}---\left(23\right)$

以下和分别是对的傅里叶变换，在谱域，(23)结合(22)式可以得到：

$A_z\left({\bar{k}}_{\perp }\right)=i{\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }^{\prime }({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime }){\bar{A}}_{\perp }\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime }),B_z\left({\bar{k}}_{\perp }\right)=i{\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }^{\prime }({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime }){\bar{B}}_{\perp }\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })---\left(24\right)$

由于推导的是二阶，所以泰勒展开也近似到二阶即：

$e^{\pm {\mathrm{ik}}_{1z}f\left({\bar{r}}_{\perp }^{\prime }\right)}=1\pm {\mathrm{ik}}_{1z}f\left({\bar{r}}_{\perp }^{\prime }\right)-\frac{k_{1z}^2}{2}{f}^2\left({\bar{r}}_{\perp }^{\prime }\right)---\left(25\right)$

再对f，f²进行傅里叶变换，

$\frac{1}{4{\pi }^2}\int d{\bar{r}}_{\perp }{e}^{-i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }}f\left({\bar{r}}_{\perp }\right)=F\left({\bar{k}}_{\perp }\right),\frac{1}{4{\pi }^2}\int d{\bar{r}}_{\perp }{e}^{-i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }}{f}^2\left({\bar{r}}_{\perp }\right)=F^2\left({\bar{k}}_{\perp }\right)---\left(26\right)$

可以证明

$F^2\left({\bar{k}}_{\perp }\right)={\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }^{\prime }F\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })---\left(27\right)$

将(23)和(24)及上述泰勒近似代入(6)(7)式可以得到：

${\bar{E}}_i\left(\bar{r}\right)=\frac{1}{2}\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }-{\mathrm{ik}}_{z}z}\frac{k}{k_z}\left\{\right[\hat{e}(-k_z)\hat{e}(-k_z)+\hat{h}(-k_z)·\hat{h}(-k_z)]·[\bar{A}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_z\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}_{\perp }\left({\bar{k}}_{\perp }^{\prime }\right)$

$·F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_z^2}{2}\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}_{\perp }\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]+[-\hat{h}(-k_z)\hat{e}(-k_z)+\hat{e}(-k_z)·\hat{h}(-k_z)]---\left(28\right)$

$·[\bar{B}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_z\int d{\bar{k}}_{\perp }^{\prime }{\bar{B}}_{\perp }\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_z^2}{2}\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}_{\perp }\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]\}$

$0=\frac{1}{2}\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }+{\mathrm{ik}}_{1z}z}\frac{k_1}{k_{1z}}\left\{\frac{k}{k_1}\right[\hat{e}\left(k_{1z}\right){\hat{e}}_1\left(k_{1z}\right)+{\hat{h}}_1\left(k_{1z}\right)·{\hat{h}}_1\left(k_{1z}\right)]·[\bar{A}\left({\bar{k}}_{\perp }\right)-{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}_{\perp }\left({\bar{k}}_{\perp }^{\prime }\right)$

$·F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_{1z}^2}{2}\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}_{\perp }\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]+[-{\hat{h}}_1\left(k_{1z}\right){\hat{e}}_1\left(k_{1z}\right)+{\hat{e}}_1\left(k_{1z}\right)·{\hat{h}}_1\left(k_{1z}\right)]---\left(29\right)$

$[\bar{B}\left({\bar{k}}_{\perp }\right)-{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{B}}_{\perp }\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_{1z}^2}{2}\int d{\bar{k}}_{\perp }^{\prime }{\bar{B}}_{\perp }\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]\}$

同理可以得到近似到二阶的透射场为：

${\bar{E}}_t\left(\bar{r}\right)=\frac{1}{2}\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }}{e}^{-i{k}_{1z}z}\frac{k_1}{k_{1z}}\left\{\frac{k}{k_1}\right[{\hat{e}}_1(-k_{1z}){\hat{e}}_1(-k_{1z})+{\hat{h}}_1(-k_{1z}){\hat{h}}_1\left(k_{1z}\right)]$

$[\bar{A}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }\bar{A}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_{1z}^2}{2}\int d{\bar{k}}_{\perp }^{\prime }\bar{A}\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]---\left(30\right)$

$+[-{\hat{h}}_1(-k_{1z}){\hat{e}}_1(-k_{1z})+{\hat{e}}_1(-k_{1z}){\hat{h}}_1(-k_{1z})][\bar{B}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }\bar{B}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })$

$-\frac{k_{1z}^2}{2}\int d{\bar{k}}_{\perp }^{\prime }\bar{B}\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })\left]\right\}$

对(5)式进行傅里叶变换可以推出粗糙面的水平分量与z轴分量之间的关系：

$A_z^{\left(m\right)}\left({\bar{k}}_{\perp }\right)=i{\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }^{\prime }F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime }){\bar{A}}_{\perp }^{(m-1)}\left({\bar{k}}_{\perp }^{\prime }\right)---\left(31\right)$

$B_z^{\left(m\right)}\left({\bar{k}}_{\perp }\right)=i{\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }^{\prime }F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime }){\bar{B}}_{\perp }^{(m-1)}\left({\bar{k}}_{\perp }^{\prime }\right)---\left(32\right)$

也就是说m阶粗糙面的z轴分量可以由m-1阶粗糙面的水平分量计算得到。则(6)式可以写成：

${\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }-{\mathrm{ik}}_{z}z}{\hat{e}}_i\delta ({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })=\frac{i}{8{\pi }^2}\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }-{\mathrm{ik}}_{z}z}\frac{k}{k_z}\int d{{\bar{r}}^{\prime }}_{\perp }{e}^{-i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }^{\prime }}(1+{\mathrm{ik}}_{z}f\left({\bar{r}}_{\perp }^{\prime }\right)-\frac{k_z^2}{2}{f}^2\left({\bar{r}}_{\perp }^{\prime }\right))$

$\times \left\{\right[\hat{e}(-k_z)\hat{e}(-k_z)+\hat{h}(-k_z)·\hat{h}(-k_z)]·\int d{\bar{k}}_{\perp }^{\prime }\bar{A}\left({\bar{k}}_{\perp }^{\prime }\right)e^{+i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }^{\prime }}---\left(33\right)$

$+[-\hat{h}(-k_z)\hat{e}(-k_z)+\hat{e}(-k_z)·\hat{h}(-k_z)]·\int d{\bar{k}}_{\perp }^{\prime }\bar{B}\left({\bar{k}}_{\perp }^{\prime }\right)e^{+i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }^{\prime }}\}$

化简(33)可以得到：

${\hat{e}}_i\delta ({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })=\frac{1}{2}\frac{k}{k_z}[\hat{e}(-k_z)\hat{e}(-k_z)+\hat{h}(-k_z)·\hat{h}(-k_z)]$

$·[\bar{A}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_z\int d{\bar{k}}_{\perp }^{\prime }\bar{A}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_z^2}{2}\int d{\bar{k}}_{\perp }^{\prime }\bar{A}\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]---\left(34\right)$

$+\frac{1}{2}\frac{k}{k_z}[-\hat{h}(-k_z)\hat{e}(-k_z)+\hat{e}(-k_z)\hat{h}(-k_z)]+\frac{1}{2}\frac{k}{k_z}[-\hat{h}(-k_z)\hat{e}(-k_z)+\hat{e}(-k_z)\hat{h}(-k_z)]$

$·[\bar{B}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_z\int d{\bar{k}}_{\perp }^{\prime }\bar{B}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_z^2}{2}\int d{\bar{k}}_{\perp }^{\prime }\bar{B}\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]$

对(7)式用相同的化简方式可以得到

$0=\frac{1}{2}\frac{k}{k_{1z}}[{\hat{e}}_1\left(k_{1z}\right)\hat{e}\left(k_{1z}\right)+{\hat{h}}_1\left(k_{1z}\right)·{\hat{h}}_1\left(k_{1z}\right)]$

$·[\bar{A}\left({\bar{k}}_{\perp }\right)-{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }\bar{A}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_{1z}^2}{2}\int d{\bar{k}}_{\perp }^{\prime }\bar{A}\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]---\left(35\right)$

$+\frac{1}{2}\frac{k_1}{k_{1z}}[-{\hat{h}}_1\left(k_{1z}\right){\hat{e}}_1(-k_{1z})+{\hat{e}}_1\left(k_{1z}\right){\hat{h}}_1\left(k_{1z}\right)]$

$·[\bar{B}\left({\bar{k}}_{\perp }\right)-{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }\bar{B}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_{1z}^2}{2}\int d{\bar{k}}_{\perp }^{\prime }\bar{B}\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]$

下面分阶求解(34)和(35)方程，并给出一阶和二阶透射场的推导过程。

1、零阶与一阶透射场求解

(1)零阶场

由(30)式可以推导出零介透射场为：

${\bar{E}}_t^{\left(0\right)}\left(\bar{r}\right)=\frac{1}{2}d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }-{\mathrm{ik}}_{1z}z}\frac{k_1}{k_{1z}}\left\{\frac{k}{k_1}\right[{\hat{e}}_1(-k_{1z}){\hat{e}}_1(-k_{1z})+{\hat{h}}_1(-k_{1z}){\hat{h}}_1\left(k_{1z}\right)]·{\bar{A}}^{\left(0\right)}\left({\bar{k}}_{\perp }\right)---\left(36\right)$

$+[-{\hat{h}}_1(-k_{1z}){\hat{e}}_1(-k_{1z})+{\hat{e}}_1(-k_{1z}){\hat{h}}_1(-k_{1z})]·{\bar{B}}^{\left(0\right)}\left({\bar{k}}_{\perp }\right)\}$

0阶透射解对应平面波入射到一个平面上的透射，(34)和(35)这两个消光定理的表达式近似到零阶可得：

${\hat{e}}_i\delta ({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })=\frac{k}{2k_z}[\hat{e}(-k_z)\hat{e}(-k_z)+\hat{h}(-k_z)\hat{h}(-k_z)]·{\bar{A}}_{\perp }^{\left(0\right)}---\left(37\right)$

$+\frac{k}{k_z}[-\hat{h}(-k_z)\hat{e}(-k_z)+\hat{e}(-k_z)\hat{h}(-k_z)]·{\bar{B}}_{\perp }^{\left(0\right)}$

$0=\frac{k}{2k_{1z}}[{\hat{e}}_1\left(k_{1z}\right){\hat{e}}_1\left(k_{1z}\right)+{\hat{h}}_1\left(k_{1z}\right){\hat{h}}_1\left(k_{1z}\right)]·{\bar{A}}_{\perp }^{\left(0\right)}---\left(38\right)$

$+\frac{k}{2k_{1z}}[-{\hat{h}}_1\left(k_{1z}\right){\hat{e}}_1(-k_{1z})+{\hat{e}}_1\left(k_{1z}\right){\hat{h}}_1\left(k_{1z}\right)]·{\bar{B}}_{\perp }^{\left(0\right)}$

由(36)和(37)可求出零介表面场和代入(36)式可推出零阶透射场的表达式：

${\bar{E}}_t^{\left(0\right)}=\left\{(1+R_{h0})\right[\hat{e}(-k_{\mathrm{iz}})·{\hat{e}}_i]{\hat{e}}_1(-k_{1\mathrm{zi}})+\frac{k}{k_1}(1+R_{v0})[\hat{h}(-k_{\mathrm{iz}})·{\hat{e}}_i\left]{\hat{h}}_1(-k_{1\mathrm{zi}})\right\}{e}^{i{\bar{k}}_{i\perp }{\bar{r}}_{\perp }-{\mathrm{ik}}_{1\mathrm{zi}}z}---\left(39\right)$

其中R_h0是TE波的菲涅尔反射系数，R_v0是TM波的菲涅尔反射系数，

$R_{h0}=\frac{k_{\mathrm{iz}}-k_{1\mathrm{iz}}}{k_{\mathrm{iz}}+k_{1\mathrm{iz}}};$$R_{v0}=\frac{k_1^2{k}_{\mathrm{iz}}-k^2{k}_{1\mathrm{iz}}}{k_1^2{k}_{\mathrm{iz}}+k^2{k}_{1\mathrm{iz}}}$

(2)一阶透射场的推导

由(30)式可以推导出一阶透射场为

${\bar{E}}_t^{\left(1\right)}\left(\bar{r}\right)=\frac{1}{2}\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{\perp }·{\bar{r}}_{\perp }-i{k}_{1z}z}\frac{k_1}{k_{1z}}\left\{\frac{k}{k_1}\right[{\hat{e}}_1(-k_{1z}){\hat{e}}_1(-k_{1z})+{\hat{h}}_1(-k_{1z}){\hat{h}}_1\left(k_{1z}\right)]$

$·[{\bar{A}}^{\left(1\right)}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}^{\left(0\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]+[-{\hat{h}}_1(-k_{1z}){\hat{e}}_1(-k_{1z})+{\hat{e}}_1(-k_{1z}){\hat{h}}_1(-k_{1z})]---\left(40\right)$

$·[{\bar{B}}^{\left(1\right)}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{B}}^{\left(0\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]$

由(34)和(35)这两个消光定理的表达式近似到一阶可得：

${\hat{e}}_i\delta ({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })=\frac{k}{2k_z}[\hat{e}(-k_z)\hat{e}(-k_z)+\hat{h}(-k_z)\hat{h}(-k_z)]$

$·[{\bar{A}}^{\left(1\right)}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}^{\left(0\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]---\left(41\right)$

$+\frac{k}{k_z}[-\hat{h}(-k_z)\hat{e}(-k_z)+\hat{e}(-k_z)\hat{h}(-k_z)]$

$·[{\bar{B}}^{\left(1\right)}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{B}}^{\left(0\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]$

$0=\frac{k}{2k_{1z}}[{\hat{e}}_1\left(k_{1z}\right){\hat{e}}_1\left(k_{1z}\right)+{\hat{h}}_1\left(k_{1z}\right){\hat{h}}_1\left(k_{1z}\right)]$

$·[{\bar{A}}^{\left(1\right)}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}^{\left(0\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]---\left(42\right)$

$+\frac{k_1}{2k_{1z}}[-{\hat{h}}_1\left(k_{1z}\right){\hat{e}}_1(-k_{1z})+{\hat{e}}_1\left(k_{1z}\right){\hat{h}}_1\left(k_{1z}\right)]$

$·[{\bar{B}}^{\left(1\right)}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{B}}^{\left(0\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]$

由(41)和(42)可以求解出一阶的表面场和再将和代入(40)可得出一阶透射场表达式：

${\bar{E}}_t^{\left(1\right)}\left(\bar{r}\right)=\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }-{\mathrm{ik}}_{1z}z}\mathrm{iF}({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })\left\{{\hat{e}}_1(-k_{1z})\right[f_{\mathrm{ee}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)$

$+f_{\mathrm{he}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)]+{\hat{h}}_1\left({-k}_{1z}\right)[f_{\mathrm{he}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)---\left(43\right)$

$+f_{\mathrm{hh}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{h}}_1\left({-k}_{1\mathrm{zi}}\right)·{\hat{e}}_1)]$

当入射场为TE极化时入射场为TM极化时我们定义并由坐标关系推出：

$f_{\mathrm{ee}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })=\frac{(k_1^2-k^2)}{k_z+k_{1z}}\left(\frac{2k_{\mathrm{iz}}}{k_{\mathrm{iz}}+k_{1\mathrm{zi}}}\right)\cos ({\phi }_k-{\phi }_i)---\left(44\right)$

同理：

$f_{\mathrm{he}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })=\frac{k_z{k}_1(k_1^2-k^2)}{k_1^2{k}_z+{k^{2}k}_{1z}}\left(\frac{2k_{\mathrm{iz}}}{k_{\mathrm{iz}}+k_{1\mathrm{zi}}}\right)\sin ({\phi }_k-{\phi }_i)---\left(45\right)$

$f_{\mathrm{eh}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })=\frac{k_{\mathrm{iz}}(k_1^2-k^2)}{k_z+k_{1z}}\left(\frac{2{\mathrm{kk}}_{1\mathrm{zi}}}{{k_1^{2}k}_{\mathrm{iz}}+{k^{2}k}_{1\mathrm{zi}}}\right)\sin ({\phi }_k-{\phi }_i)---\left(46\right)$

$f_{\mathrm{hh}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })=\frac{(k_1^2-k^2)}{k_1^2{k}_z+{k^{2}k}_{1z}}\left(\frac{2k{k}_1{k}_{\mathrm{iz}}}{k_1^2{k}_{\mathrm{iz}}+{k^{2}k}_{1\mathrm{zi}}}\right)\{k_{\rho }{k}_{\mathrm{\rho i}}+k_{1\mathrm{zi}}{k}_z\cos ({\phi }_k-{\phi }_i)\}---\left(47\right)$

φ_k，φ_i分别指透射方位角与入射方位角。

2、二阶透射场求解

由(30)式可以推导出二阶透射场为

${\bar{E}}_t^{\left(2\right)}\left(\bar{r}\right)=\frac{1}{2}\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }-{\mathrm{ik}}_{1z}z}\frac{k_1}{k_{1z}}\left\{\frac{k}{k_1}\right[{\hat{e}}_1(-k_{1z}){\hat{e}}_1(-k_{1z})+{\hat{h}}_1(-k_{1z}){\hat{h}}_1\left(k_{1z}\right)]·[{\bar{A}}^{\left(2\right)}\left({\bar{k}}_{\perp }\right)$

$+{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}^{\left(1\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_{1z}^2}{2}d{\bar{k}}_{\perp }^{\prime }{\bar{A}}^{\left(0\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]---\left(48\right)$

$+[-{\hat{h}}_1(-k_{1z}){\hat{e}}_1\left({-k}_{1z}\right)+{\hat{e}}_1(-k_{1z}){\hat{h}}_1(-k_{1z})]·[{\bar{B}}^{\left(2\right)}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{B}}^{\left(1\right)}\left({\bar{k}}_{\perp }^{\prime }\right)$

$·F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_{1z}^2}{2}\int d{\bar{k}}_{\perp }^{\prime }{\bar{B}}^{\left(0\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]$

由(34)和(35)式可以写出二阶的消光定理的表达式：

${\hat{e}}_i\delta ({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })=\frac{k}{2k_z}[\hat{e}(-k_z)\hat{e}(-k_z)+\hat{h}(-k_z)\hat{h}(-k_z)]·[{\bar{A}}^{\left(2\right)}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_{\mathrm{iz}}\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}^{\left(1\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })$

$-\frac{k_{1z}^2}{2}\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}^{\left(0\right)}\left({\bar{k}}_{\perp }^{\prime }\right){F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })}^{}]+\frac{k}{k_z}[-\hat{h}\left({-k}_z\right)\hat{e}\left({-k}_z\right)+\hat{e}\left({-k}_z\right)\hat{h}(-k_z)]---\left(49\right)$

$·[{\bar{B}}^{\left(2\right)}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{B}}^{\left(1\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_{1z}^2}{2}d{\bar{k}}_{\perp }^{\prime }{\bar{B}}^{\left(0\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]$

$0=\frac{k}{2k_{1z}}[{\hat{e}}_1\left(k_{1z}\right){\hat{e}}_1\left(k_{1z}\right)+{\hat{h}}_1\left(k_{1z}\right){\hat{h}}_1\left(k_{1z}\right)]·[{\bar{A}}^{\left(2\right)}\left({\bar{k}}_{\perp }\right)+{\mathrm{ik}}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}^{\left(1\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })$

$-\frac{k_{1z}^2}{2}\int d{\bar{k}}_{\perp }^{\prime }{\bar{A}}^{\left(0\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]+\frac{k_1}{2k_{1z}}[-{\hat{h}}_1\left(k_{1z}\right){\hat{e}}_1(-k_{1z})+{\hat{e}}_1\left(k_{1z}\right){\hat{h}}_1\left(k_{1z}\right)]---\left(50\right)$

$[{\bar{B}}^{\left(2\right)}\left({\bar{k}}_{\perp }\right)+i{k}_{1z}\int d{\bar{k}}_{\perp }^{\prime }{\bar{B}}^{\left(1\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })-\frac{k_{1z}^2}{2}\int d{\bar{k}}_{\perp }^{\prime }{\bar{B}}^{\left(0\right)}\left({\bar{k}}_{\perp }^{\prime }\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })]$

由(49)，(50)两式可以求解出二阶的表面场和再将和代入(48)，可推出出二阶透射场

${\bar{E}}_t^{\left(2\right)}\left(\bar{r}\right)=\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }-{\mathrm{ik}}_{1z}z}\left\{{\hat{e}}_1(-k_{1z})\right[f_{\mathrm{ee}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)+f_{\mathrm{eh}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)]---\left(51\right)$

$+{\hat{h}}_1(-k_{1z})[f_{\mathrm{he}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)+f_{\mathrm{hh}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)]$

$f_{\mathrm{ee}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })=\frac{k_{\mathrm{iz}}}{k_{\mathrm{iz}}+k_{1\mathrm{iz}}}\frac{k_1^2-k^2}{k_{1z}+k_z}\{\cos ({\phi }_k-{\phi }_i)·(k_{1\mathrm{zi}}-k_z)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })$

$-2(k_1^2-k^2)\int d{\bar{k}}_{\perp }^{\prime }F({\bar{k}}_{\perp }^{\prime }-{\bar{k}}_{i\perp })F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })---\left(52\right)$

$·[-\sin ({\phi }_k-{{\phi }_k}^{\prime })\sin ({{\phi }_k}^{\prime }-{\phi }_i)\frac{k_z^{\prime }{k}_{1z}^{\prime }}{k_1^2{k}_z^{\prime }+k^2{k}_{1z}^{\prime }}+\cos ({\phi }_k-{{\phi }_k}^{\prime })\cos ({{\phi }_k}^{\prime }-{\phi }_i)\frac{1}{k_{1z}^{\prime }+k_z^{\prime }}]\}$

$f_{\mathrm{eh}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })=\frac{k_1^2-k^2}{k_{1z}+k_z}\frac{k{k}_{\mathrm{iz}}}{k_1^2{k}_{\mathrm{iz}}+{k^{2}k}_{1\mathrm{zi}}}\{\left({k_1^2-k}_{1\mathrm{zi}}{k}_z\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })\sin ({\phi }_k-{\phi }_i)$

$-2\int d{\bar{k}}_{\perp }^{\prime }F({\bar{k}}_{\perp }^{\prime }-{\bar{k}}_{i\perp })F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })[\sin ({\phi }_k-{\phi }_k^{\prime })\frac{k_1^2{k}_{\rho }^{\prime }{k}_{\mathrm{\rho i}}}{k_{\rho }^{\prime 2}k+k_z^{\prime }{k}_{1z}^{\prime }}+\frac{(k_1^2-k^2)k_{1\mathrm{zi}}}{k_1^2{k}_z^{\prime }+k^2{k}_{1z}^{\prime }}---\left(53\right)$

$·(\sin ({\phi }_k-{{\phi }_k}^{\prime })\cos ({{\phi }_k}^{\prime }-{\phi }_i)k_z^{\prime }{k}_{1z}^{\prime }+\cos ({\phi }_k-{{\phi }_k}^{\prime })\sin ({{\phi }_k}^{\prime }-{\phi }_i)(k_{\rho }^{\prime 2}+k_z^{\prime }{k}_{1z}^{\prime })\left]\right\}$

$f_{\mathrm{he}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })=\frac{k_1^2-k^2}{k_{1\mathrm{zi}}+k_{\mathrm{iz}}}\frac{k_1{k}_{\mathrm{iz}}}{k_1^2{k}_z+{k^{2}k}_{1z}}\{\left({k^2-k}_{1\mathrm{zi}}{k}_z\right)F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })\sin ({\phi }_k-{\phi }_i)$

$+2\int d{\bar{k}}_{\perp }^{\prime }F({\bar{k}}_{\perp }^{\prime }-{\bar{k}}_{i\perp })F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })[\sin ({\phi }_k^{\prime }-{\phi }_i)\frac{k_1^2{k}_{\rho }^{\prime }{k}_{\rho }}{k_{\rho }^{\prime 2}+k_z^{\prime }{k}_{1z}^{\prime }}+\frac{(k_1^2-k^2)k_z}{k_1^2{k}_z^{\prime }+k^2{k}_{1z}^{\prime }}---\left(54\right)$

$·(k_z^{\prime }{k}_{1z}^{\prime }\sin ({\phi }_k-{{\phi }_k}^{\prime })\cos ({{\phi }_k}^{\prime }-{\phi }_i)+(k_{\rho }^{\prime 2}+k_z^{\prime }{k}_{1z}^{\prime })\cos ({\phi }_k-{{\phi }_k}^{\prime })\sin ({{\phi }_k}^{\prime }-{\phi }_i))]$

$f_{\mathrm{hh}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })=\frac{({k_1}^2-k^2)}{({k_1}^2{k}_z+k^2{k}_{1z})}\frac{k_1{k}_{\mathrm{iz}}}{k_1^2{k}_{\mathrm{iz}}+k^2{k}_{1\mathrm{zi}}}\{k(k_1^2{k}_z-k^2{k}_{1\mathrm{zi}})F^{\left(2\right)}({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })\cos ({\phi }_k-{\phi }_i)$

$+2\int d{\bar{k}}_{\perp }^{\prime }F({\bar{k}}_{\perp }^{\prime }-{\bar{k}}_{i\perp })F({\bar{k}}_{\perp }-{\bar{k}}_{\perp }^{\prime })[{\mathrm{kk}}_z{k}_{1\mathrm{zi}}\frac{(k_1^2-k^2)}{k_z^{\prime }+k_{1z}^{\prime }}\sin ({\phi }_k-{{\phi }_k}^{\prime })\sin ({{\phi }_k}^{\prime }-{\phi }_i)---\left(55\right)$

$+\frac{1}{k_1^2{k}_z^{\prime }+k^2{k}_{1z}^{\prime }}[-k({k_1}^2-k^2)k_{\rho }{k^{\prime }}_{\rho }^2{k}_{\mathrm{\rho i}}+k_{\rho }{k}_{\rho }^{\prime }{k}^3(k_z^{\prime }+k_{1z}^{\prime })k_{1\mathrm{zi}}\cos ({{\phi }_k}^{\prime }-{\phi }_i)-k_{\rho }^{\prime }{k}_{\mathrm{\rho i}}{\mathrm{kk}}_1^2$

$·(k_z^{\prime }+k_{1z}^{\prime })k_z\cos ({\phi }_k-{{\phi }_k}^{\prime })-{\mathrm{kk}}_z^{\prime }{k}_{1z}^{\prime }({k_1}^2-k^2)k_z{k}_{1\mathrm{zi}}\cos \left({\phi }_k{{-\phi }_k}^{\prime }\right)\cos ({{\phi }_k}^{\prime }-{\phi }_i)\left]\right\}$

二、双向透射系数与透射率

由以上0阶、一阶和二阶的公式可以推知精确至二阶的总的透射电场为：

${\bar{E}}_t\left(\bar{r}\right)={\bar{E}}_t^{\left(0\right)}\left(\bar{r}\right)+{\bar{E}}_t^{\left(1\right)}\left(\bar{r}\right)+{\bar{E}}_t^{\left(2\right)}\left(\bar{r}\right)$

$=\left\{(1+R_{h0})\right[\hat{e}(-k_{\mathrm{iz}})·{\hat{e}}_i]{\hat{e}}_1(-k_{1\mathrm{zi}})+\frac{k}{k_1}(1+R_{v0})[\hat{h}(-k_{\mathrm{iz}})·{\hat{e}}_i\left]{\hat{h}}_1(-k_{1\mathrm{zi}})\right\}{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }-i{k}_{1\mathrm{zi}}z}$

$+\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }-i{k}_{1z}z}\mathrm{iF}({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })\left\{{\hat{e}}_1(-k_{1z})\right[f_{\mathrm{ee}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)+f_{\mathrm{ee}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })---\left(56\right)$

$·({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)]+{\hat{h}}_1(-k_{1z})[f_{\mathrm{he}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)+f_{\mathrm{hh}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)\left]\right\}$

$+\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }-{\mathrm{ik}}_{1z}z}\left\{{\hat{e}}_1(-k_{1z})\right[f_{\mathrm{ee}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)+f_{\mathrm{eh}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)]$

$+{\hat{h}}_1(-k_{1z})[f_{\mathrm{he}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)+f_{\mathrm{hh}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)]$

再由公式η₁为波阻抗，可以推知：

${\bar{H}}_t\left(\bar{r}\right)={\bar{H}}_t^{\left(0\right)}\left(\bar{r}\right)+{\bar{H}}_t^{\left(1\right)}\left(\bar{r}\right)+{\bar{H}}_t^{\left(2\right)}\left(\bar{r}\right)$

$=e^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }-i{k}_{1\mathrm{zi}}z}\{-{\hat{h}}_1(-k_{1\mathrm{zi}})(1+R_{h0})[\hat{e}(-k_{\mathrm{iz}})·\hat{e}]+{\hat{e}}_1(-k_{1\mathrm{iz}})\frac{k}{k_1}(1+R_{v0})[\hat{h}(-k_{\mathrm{iz}})·{\hat{e}}_i\left]\right\}$

$+\frac{1}{{\eta }_1}\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }-{\mathrm{ik}}_{1z}z}\mathrm{iF}({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })\{-{\hat{h}}_1(-k_{1z})[f_{\mathrm{ee}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)+f_{\mathrm{he}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })---\left(57\right)$

$·({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)]+{\hat{e}}_1(-k_{1z})[f_{\mathrm{he}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)+f_{\mathrm{hh}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)\left]\right\}$

$+\frac{1}{{\eta }_1}\int d{\bar{k}}_{\perp }{e}^{i{\bar{k}}_{i\perp }·{\bar{r}}_{\perp }-{\mathrm{ik}}_{1z}z}\{-{\hat{h}}_1(-k_{1z})[f_{\mathrm{ee}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_1)+f_{\mathrm{eh}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)]$

$+{\hat{e}}_1(-k_{1z})[f_{\mathrm{he}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{e}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)+f_{\mathrm{hh}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })({\hat{h}}_1(-k_{1\mathrm{zi}})·{\hat{e}}_i)]$

单位面积上入射波轴分量的功率分别为：

${\bar{S}}_i·\hat{z}=-\frac{\cos {\theta }_i}{2\eta }---\left(58\right)$

$<{\bar{S}}_t·\hat{z}>=\frac{1}{2}\mathrm{Re}<{\bar{E}}_t\times {\bar{H}}_t^{*}>·\hat{z}$

$=\frac{1}{2}\mathrm{Re}{\bar{E}}_t^{\left(0\right)}\times {\bar{H}}_t^{\left(0\right)*}·\hat{z}+\frac{1}{2}\mathrm{Re}{\bar{E}}_t^{\left(0\right)}\times <{\bar{H}}_t^{\left(2\right)*}>·\hat{z}+\frac{1}{2}\mathrm{Re}<{\bar{E}}_t^{\left(2\right)}>\times {\bar{H}}_t^{\left(0\right)*}·\hat{z}---\left(59\right)$

$+\frac{1}{2}\mathrm{Re}<{\bar{E}}_t^{\left(1\right)}\times {\bar{H}}_t^{\left(1\right)*}>·\hat{z}$

其中为入射波平均坡延亭矢量，是介质1中的透射波的平均坡延亭矢量。

$t(\pi -{\theta }_i,{\phi }_i)=\frac{<{\bar{S}}_t·\hat{z}>}{{\bar{S}}_i·\hat{z}}$

$=\frac{k_{1\mathrm{zi}}}{k_{\mathrm{zi}}}{|1+R_{\mathrm{ho}}|}^2(\hat{e}(-k_{\mathrm{iz}})·{\hat{e}}_i)+\frac{k^2{k}_{1\mathrm{zi}}}{{k_1}^2{k}_{\mathrm{zi}}}{|1+R_{\mathrm{vo}}|}^2(\hat{h}(-k_{\mathrm{iz}})·{\hat{e}}_i)$

$+2\mathrm{Re}\left(\frac{k_{1\mathrm{zi}}}{k_{\mathrm{zi}}}(1+R_{\mathrm{ho}})f_{\mathrm{ee}}^{*t\left(2\right)}\left(k_{i\perp }\right)\right)(\hat{e}(-k_{\mathrm{iz}})·{\hat{e}}_i)+2\mathrm{Re}\left(\frac{{\mathrm{kk}}_{1\mathrm{zi}}}{k_1{k}_{\mathrm{zi}}}(1+R_{\mathrm{vo}})f_{\mathrm{hh}}^{*t\left(2\right)}\left(k_{i\perp }\right)\right)$

$·(\hat{h}(-k_{\mathrm{iz}})·{\hat{e}}_i)+\int d{\bar{k}}_{\perp }\frac{k_{1\mathrm{zi}}}{k_{\mathrm{zi}}}W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })[{|f_{\mathrm{ee}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })·(\hat{e}(-k_{\mathrm{iz}})·{\hat{e}}_i)+f_{\mathrm{eh}}^{t\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })·(\hat{h}(-k_{\mathrm{iz}})·{\hat{e}}_i)|}^2$

$+{|f_{\mathrm{he}}^{t\left(1\right)}(\hat{e}(-k_{\mathrm{iz}})·{\hat{e}}_i)+f_{\mathrm{hh}}^{t\left(1\right)}(\hat{h}(-k_{\mathrm{iz}})·{\hat{e}}_i)|}^2---\left(60\right)$

其中是随机起伏高度的相关函数的Fourier变换。则

$W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })=\frac{h^2{l}^2}{4\pi }\exp (-\frac{(k_{\rho }^2+k_{\mathrm{\rho i}}^2)l^2}{4}+\frac{k_{\rho }{k}_{\mathrm{\rho i}}{l}^2}{2}\cos ({\phi }_k-{\phi }_i)).$

在应用中我们更多的关注双向透射系数，所以将透射率t(π-θ_i，φ_i)写成双向透射系数的形式：

$t_b(\pi -{\theta }_i,{\phi }_i)=\frac{1}{4\pi }{\int }_0^{\pi /2}d{\theta }_t\sin {\theta }_t{\int }_0^{2\pi }d{\phi }_t({\gamma }_{\mathrm{ab}}^t(\pi -{\theta }_t,{\phi }_t;\pi -{\theta }_i,{\phi }_i)+{\gamma }_{\mathrm{bb}}^t(\pi -{\theta }_t,{\phi }_t;\pi -{\theta }_i,{\phi }_i))---\left(61\right)$

其中a，b分别为v极化和h极化的两种情况，且表达镜向透射方向应以θ_t，φ_t的形式：其中θ_t才指向透射方向。

经过推导可知在入射波h极化，透射波h极化情况下的双向透射系数为：

$\frac{1}{4\pi }{\gamma }_{\mathrm{hh}}^t(\pi -{\theta }_t,{\phi }_t;\pi -{\theta }_i,{\phi }_i)=[\frac{k_{1\mathrm{zi}}}{k_{\mathrm{zi}}}{|1+R_{\mathrm{ho}}|}^2+2\mathrm{Re}\left(\frac{k_{1\mathrm{zi}}}{k_{\mathrm{zi}}}(1+R_{\mathrm{ho}})f_{\mathrm{ee}}^{t^{*}\left(2\right)}\left({\bar{k}}_{i\perp }\right)\right)]\delta (\cos {\theta }_t-\frac{k_{1\mathrm{zi}}}{k_1})\delta ({\phi }_t-{\phi }_i)$

$+\frac{k_1{k}_{1z}^2}{k_{\mathrm{iz}}}W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp }){\left|f_{\mathrm{ee}}^{t^{*}\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })\right|}^2---\left(62\right)$

同理，在入射波v极化，透射波h极化情况下，

$\frac{1}{4\pi }{\gamma }_{\mathrm{hv}}^t(\pi -{\theta }_t,{\phi }_t;\pi -{\theta }_i,{\phi }_i)=\frac{k_1{k}_{1z}^2}{k_{\mathrm{iz}}}W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp }){\left|f_{\mathrm{eh}}^{t^{*}\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })\right|}^2---\left(63\right)$

在入射波h极化，透射波v极化情况下，

$\frac{1}{4\pi }{\gamma }_{\mathrm{vh}}^t(\pi -{\theta }_t,{\phi }_t;\pi -{\theta }_i,{\phi }_i)=\frac{k_1{k}_{1z}^2}{k_{\mathrm{iz}}}W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })|f_{\mathrm{he}}^{t^{*}\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp }){|}^2---\left(64\right)$

在入射波v极化，透射波v极化情况下，

$\frac{1}{4\pi }{\gamma }_{\mathrm{vv}}^t(\pi -{\theta }_t,{\phi }_t;\pi -{\theta }_i,{\phi }_i)=\left[\frac{k^2{k}_{1\mathrm{zi}}}{k_1^2{k}_{\mathrm{iz}}}\right|1+R_{\mathrm{vo}}{|}^2+2\mathrm{Re}\left(\frac{{\mathrm{kk}}_{1\mathrm{zi}}}{k_1{k}_{\mathrm{iz}}}(1+R_{\mathrm{vo}})f_{\mathrm{hh}}^{t^{*}\left(2\right)}\left({\bar{k}}_{i\perp }\right)\right)]---\left(65\right)$

$·\delta (\cos {\theta }_t-\frac{k_{1\mathrm{zi}}}{k_1})\delta ({\phi }_t-{\phi }_i)+\frac{k_1{k}_{1z}^2}{k_{\mathrm{iz}}}W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })|f_{\mathrm{hh}}^{t^{*}\left(1\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp }){|}^2$

其中$f_{\mathrm{ab}}^{t\left(2\right)}\left({\bar{k}}_{i\perp }\right)=<f_{\mathrm{ab}}^{t\left(2\right)}({\bar{k}}_{\perp },{\bar{k}}_{i\perp })>a,b=v,h$

$f_{\mathrm{ee}}^{t\left(2\right)}\left({\bar{k}}_{i\perp }\right)=\frac{k_{\mathrm{iz}}(k_1^2-k^2)}{{(k_{\mathrm{iz}}+k_{1\mathrm{zi}})}^2}\{(k_{1\mathrm{zi}}-k_{\mathrm{iz}}){\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })-2(k_1^2-k^2){\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })---\left(66\right)$

$·\left[{\sin }^2({\phi }_k-{\phi }_i)\frac{k_z{k}_{1z}}{k_1^2{k}_z+k^2{k}_{1z}}{\cos }^2({\phi }_k-{\phi }_i)\frac{1}{k_{1z}+k_z}\right]\}$

$f_{\mathrm{hh}}^{t\left(2\right)}\left({\bar{k}}_{i\perp }\right)=\frac{k_1({k_1}^2-k^2)k_{\mathrm{iz}}}{{({k_1}^2{k}_{\mathrm{iz}}+k^2{k}_{1\mathrm{zi}})}^2}\{{\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })k(k_1^2{k}_{\mathrm{iz}}-k^2{k}_{1\mathrm{zi}})+$

$2{\int }_{-\infty }^{\infty }d{\bar{k}}_{\perp }W({\bar{k}}_{\perp }-{\bar{k}}_{i\perp })[-{\mathrm{kk}}_{\mathrm{iz}}{k}_{1\mathrm{zi}}\frac{(k_1^2-k^2)}{k_z+k_{1z}}{\sin }^2({\phi }_k-{\phi }_i)+---\left(67\right)$

$\frac{1}{k_1^2{k}_z+k^2{k}_{1z}}(-k({k_1}^2-k^2)k_{\mathrm{\rho i}}^2{k}_{\rho }^2+k_{\rho }{k}_{\mathrm{\rho i}}k(k_z+k_{1z})(k^2{k}_{1\mathrm{zi}}-k_1^2{k}_{\mathrm{iz}}))$

$·\cos ({\phi }_k-{\phi }_i)-{\mathrm{kk}}_z{k}_{1z}{k}_{\mathrm{iz}}{k}_{1\mathrm{zi}}({k_1}^2-k^2){\cos }^2({\phi }_k-{\phi }_i)\left]\right\}$

下面对本发明方案进行验证：

对于如图1所示的二维粗糙面，入射波从自由空间(介质0)入射到该粗糙面上，其中已知介质0的μ₀＝4π×10^-7，同时需要设定参数为入射波的入射角θ_i和方位角下半部分介质的介电常数ε₁，粗糙面的标准离差σ和相关长度l。

首先利用入射角θ_i和方位角计算出：k_ix＝ksinθ_icosφ_i，k_iy＝ksinθ_isinφ_i，k_iz＝kcosθ_i，k_ρi＝ksinθ_i，再利用公式(56)即可计算出透射率。公式(56)中各变量的含义和取值，详见相关公式。

微扰法计算的二阶散射场及二阶反射率已由前人推导得到，为验证本文所推导的二阶透射场及二阶透射率，在不同的参数设置下，按本文方法计算二阶透射率(发射率)，按给出的公式计算二阶反射率，将反射率和透射率相加，若等于1则认为符合能量守恒，即本文二阶透射场及二阶透射率的推导结果是正确的。用于对比，还仿真了一阶微扰法透射率和反射率，及其能量守恒情况。

1、当表面粗糙度符合一阶微扰法的适用范围。

令介电常数ε＝6，均方根高度h＝0.0398λ，相关长度l＝λ，频率f＝37Ghz，其中λ为该频率下自由空间的波长。该参数设置满足一阶微扰法的适用范围。图2显示了透射率(二阶、一阶)、反射率(二阶、一阶)，及透射率+反射率(二阶、一阶)，随入射角变化而改变的情况。

图2(a)对应h极化情况，看出虽然表面粗糙度符合一阶微扰法的适用范围，但一阶微扰法并不符合能量守恒，在0度入射角时，误差为18％；而二阶微扰法在h极化时，对所有入射角，包括掠入射情况，都保持能量守恒；

图2(b)对应v极化情况，一阶微扰法在v极化，对所有入射角能量都不守恒。二阶微扰法在v极化，掠入射时能量不守恒，而在其他大部分入射角时是守恒的。

对于二阶微扰法，在掠入射时，只有h极化是符合能量守恒的，而v极化是不符合能量守恒。这一点与已有文献关于二阶微扰法的适用范围研究结论是一致的。

从这个例子看出，二阶微扰法的透射率计算精度较一阶微扰法的透射率计算精度有较大的提高。

在图3(a)，3(b)中，介电常数为ε＝6，均方根高度σ＝0.07λ，相关长度l＝0.15λ，频率f＝37Ghz，λ为该频率下自由空间的波长。该表面粗糙度及相关参数已超出了一阶微扰法的适用范围。图2(a)，2(b)，3(a)，3(b)中，r1，r2分别表示一阶和二阶发射率；t1，t2分别代表一阶和二阶的透射率。

2、当表面粗糙度一阶微扰法的适用范围。

图3(a)对应h极化情况，看出当表面粗糙度超出一阶微扰法的适用范围时，一阶微扰法能量不守恒的情况均较显著，尤其在0度时，误差已接近40％；；而二阶微扰法在h极化时，对所有入射角，包括掠入射情况，都保持能量守恒；

图3(b)对应v极化情况，一阶微扰法在v极化，对所有入射角能量都不守恒，误差多在40％左右。二阶微扰法在v极化，掠入射时能量不守恒，误差小于20％，而在其他大部分入射角时能量是守恒的。

从这个例子看出，二阶微扰法透射率计算公式较一阶而言，适用范围扩大了，精度提高了。