2.3 マイクロストリップ素子
マイクロストリップ素子のスペクトル領域の電磁界
地導体板（\(z=0\)）をつけた誘電体基板（厚み\(d\)，大きさは無限）上にストリップ導体のあるマイクロストリップ素子について考える．
まず，\(z \geq d\) の領域\(\mathrm{(II)}\)については，半無限領域ゆえ，次のようにおくことにする．
\begin{eqnarray}
\widetilde{E}_{z,2} &=& A e^{-jk_{z,2} z} \ \ \ (z \geq d)
\\
\widetilde{H}_{z,2} &=& B e^{-jk_{z,2} z} \ \ \ (z \geq d)
\end{eqnarray}
一方，\(0 \leq z \leq d\) の領域\(\mathrm{(I)}\)については，誘電体スラブ領域ゆえ，
\begin{eqnarray}
\widetilde{E}_{z,1} &=& C \cos k_{z,1} z + D \sin k_{z,1} z \ \ \ (0 \leq z \leq d)
\\
\widetilde{H}_{z,1} &=& E \sin k_{z,1} z + F \cos k_{z,1} z \ \ \ (0 \leq z \leq d)
\end{eqnarray}
ここで，\(A\)，\(B\)，\(C\)，\(D\)，\(E\)，\(F\)は未知係数であり，境界条件によって決める．まず，\(z=0\) で完全導体ゆえ，
\begin{gather}
E_{x,1} (x,y) \big| _{z=0} = E_{y,1} (x,y) \big| _{z=0} =0
\end{gather}
より，スペクトル領域でも，
\begin{gather}
\widetilde{E}_{x,1} (k_x,k_y) \big| _{z=0} = \widetilde{E}_{y,1} (k_x,k_y) \big| _{z=0} =0
\end{gather}
ここで，
\begin{eqnarray}
\frac{\partial \widetilde{E}_{z,1}}{\partial z}
&=& \frac{\partial }{\partial z} \Big( C \cos k_{z,1} z + D \sin k_{z,1} z \Big)
\nonumber \\
&=& -C k_{z,1} \sin k_{z,1} z + Dk_{z,1} \cos k_{z,1} z
\end{eqnarray}
より，
\begin{eqnarray}
\widetilde{E}_{x,1} \big| _{z=0}
&=& \left. \frac{jk_x}{k_t^2} \frac{\partial \widetilde{E}_{z,1}}{\partial z} \right| _{z=0}
+ \left. \frac{\omega \mu k_y}{k_t^2} \widetilde{H}_{z,1} \right| _{z=0}
\nonumber \\
&=& \frac{jk_x}{k_t^2} Dk_{z,1} + \frac{\omega \mu k_y}{k_t^2} F = 0
\end{eqnarray}
\begin{eqnarray}
\widetilde{E}_{y,1} \big| _{z=0}
&=& \left. \frac{jk_y}{k_t^2} \frac{\partial \widetilde{E}_{z,1}}{\partial z} \right| _{z=0}
- \left. \frac{\omega \mu k_x}{k_t^2} \widetilde{H}_{z,1} \right| _{z=0}
\nonumber \\
&=& \frac{jk_y}{k_t^2} Dk_{z,1} - \frac{\omega \mu k_x}{k_t^2} F = 0
\end{eqnarray}
よって，
\begin{gather}
D = F = 0
\end{gather}
したがって，
\begin{eqnarray}
\widetilde{E}_{z,1} = C \cos k_{z,1} z \ \ \ (0 \leq z \leq d)
\\
\widetilde{H}_{z,1} = E \sin k_{z,1} z \ \ \ (0 \leq z \leq d)
\end{eqnarray}
また，\(z=d\)の連続条件より，電界については，
\begin{gather}
E_{{x \choose y},1} (x,y) \Big| _{z=d}
= E_{{x \choose y},2} (x,y) \Big| _{z=d}
\end{gather}
より，
\begin{gather}
\widetilde{E}_{{x \choose y},1} (k_x,k_y) \Big| _{z=d}
= \widetilde{E}_{{x \choose y},2} (k_x,k_y) \Big| _{z=d}
\end{gather}
これより，電界について求めると，
\begin{eqnarray}
&& \widetilde{E}_{x,1} \big| _{z=d}
= \frac{jk_x}{k_t^2} \Big( -C k_{z,1} \sin k_{z,1} d \Big)
+ \frac{\omega \mu _1 k_y}{k_t^2} \cdot E\sin k_{z,1} d
\nonumber \\
&=& \widetilde{E}_{x,2} \big| _{z=d}
= \frac{jk_x}{k_t^2} \Big( -jA k_{z,2} e^{-jk_{z,2} d} \Big)
+ \frac{\omega \mu _2 k_y}{k_t^2} \cdot B e^{-jk_{z,2} d}
\end{eqnarray}
\begin{eqnarray}
&& \widetilde{E}_{y,1} \big| _{z=d}
= \frac{jk_y}{k_t^2} \Big( -C k_{z,1} \sin k_{z,1} d \Big)
- \frac{\omega \mu _1 k_x}{k_t^2} \cdot E\sin k_{z,1} d
\nonumber \\
&=& \widetilde{E}_{y,2} \big| _{z=d}
= \frac{jk_y}{k_t^2} \Big( -jA k_{z,2} e^{-jk_{z,2} d} \Big)
- \frac{\omega \mu _2 k_x}{k_t^2} \cdot B e^{-jk_{z,2} d}
\end{eqnarray}
一方，磁界については，
\(z=d\) の境界の\(x=x_0\)，\(y=y_0\)（位置ベクトル\(\boldsymbol{r}_0\)）に\(x\)方向の微小電流を考え，
\begin{gather}
\boldsymbol{u}_z \times \big( \boldsymbol{H}_2 - \boldsymbol{H}_1 \big) \Big| _{z=d} = \delta (\boldsymbol{r}-\boldsymbol{r}_0) \boldsymbol{u}_x
\end{gather}
ただし，
\begin{gather}
\boldsymbol{r}_0 = x_0 \boldsymbol{u}_x + y_0 \boldsymbol{u}_y + d \boldsymbol{u}_z
\end{gather}
これをフーリエ変換すると，
\begin{eqnarray}
&& \iint _{-\infty}^\infty \boldsymbol{u}_z \times \big( \boldsymbol{H}_2 (x,y) - \boldsymbol{H}_1 (x,y) \big) \Big| _{z=d} \ e^{-j(k_x x + k_y y)} dx dy
\nonumber \\
&=& \iint _{-\infty}^\infty \delta (\boldsymbol{r}-\boldsymbol{r}_0) \boldsymbol{u}_x \ e^{-j(k_x x + k_y y)} dx dy
\end{eqnarray}
これより，
\begin{eqnarray}
&& \boldsymbol{u}_z \times \big( \widetilde{\boldsymbol{H}}_2 (k_x,k_y)- \widetilde{\boldsymbol{H}}_1 (k_x,k_y) \big) \Big| _{z=d}
\nonumber \\
&=& \boldsymbol{u}_z \times \Big\{ \big( \widetilde{H}_{x,2} \big| _{z=d} - \widetilde{H}_{x,1} \big| _{z=d} \big) \boldsymbol{u}_x
+ \big( \widetilde{H}_{y,2} \big| _{z=d} - \widetilde{H}_{y,1} \big| _{z=d} \big) \boldsymbol{u}_y \Big\}
\nonumber \\
&=& \big( \widetilde{H}_{x,2} \big| _{z=d} - \widetilde{H}_{x,1} \big| _{z=d} \big) \boldsymbol{u}_y
- \big( \widetilde{H}_{y,2} \big| _{z=d} - \widetilde{H}_{y,1} \big| _{z=d} \big) \boldsymbol{u}_x
\nonumber \\
&=& \boldsymbol{u}_x e^{-j(k_x x_0 + k_y y_0)}
\end{eqnarray}
よって，
\begin{eqnarray}
\widetilde{H}_{y,2} \big| _{z=d} - \widetilde{H}_{y,1} \big| _{z=d} &=& -e^{-j(k_x x_0 + k_y y_0)}
\\
\widetilde{H}_{x,2} \big| _{z=d} - \widetilde{H}_{x,1} \big| _{z=d} &=& 0
\end{eqnarray}
また，磁界について求めると，
\begin{eqnarray}
&&\widetilde{H}_{x,1} \big| _{z=d}
= \frac{jk_x}{k_t^2} \cdot E k_{z,1} \cos k_{z,1} d
- \frac{\omega \epsilon _1 k_y}{k_t^2} \cdot C \cos k_{z,1} d
\nonumber \\
&=& \widetilde{H}_{x,2} \big| _{z=d}
= \frac{jk_x}{k_t^2} \Big( -jB k_{z,2} e^{-jk_{z,2} d} \Big)
- \frac{\omega \epsilon _2 k_y}{k_t^2} \cdot A e^{-jk_{z,2} d}
\end{eqnarray}
\begin{eqnarray}
&&\widetilde{H}_{y,1} \big| _{z=d}
= \frac{jk_y}{k_t^2} \cdot E k_{z,1} \cos k_{z,1} d
+ \frac{\omega \epsilon _1 k_x}{k_t^2} \cdot C \cos k_{z,1} d
\nonumber \\
&=& \widetilde{H}_{y,2} \big| _{z=d} +e^{-j(k_x x_0 + k_y y_0)}
\nonumber \\
&=& \frac{jk_y}{k_t^2} \Big( -jB k_{z,2} e^{-jk_{z,2} d} \Big)
+ \frac{\omega \epsilon _2 k_x}{k_t^2} \cdot A e^{-jk_{z,2} d} +e^{-j(k_x x_0 + k_y y_0)}
\end{eqnarray}
整理すると，
\begin{align}
&-jk_x k_{z,1} C \sin k_{z,1} d + \omega \mu _1 k_y E \sin k_{z,1} d
\nonumber \\
&= k_x k_{z,2} A e^{-jk_{z,2} d} + \omega \mu _2 k_y B e^{-jk_{z,2}d } \label{eq:sd1}
\end{align}
\begin{align}
&-jk_y k_{z,1} C \sin k_{z,1} d - \omega \mu _1 k_x E \sin k_{z,1} d
\nonumber \\
&= k_y k_{z,2} A e^{-jk_{z,2} d} - \omega \mu _2 k_x B e^{-jk_{z,2}d } \label{eq:sd2}
\end{align}
\begin{align}
&jk_x k_{z,1} E \cos k_{z,1} d - \omega \epsilon _1 k_y C \cos k_{z,1} d
\nonumber \\
&= k_x k_{z,2} B e^{-jk_{z,2} d} - \omega \epsilon _2 k_y A e^{-jk_{z,2}d } \label{eq:sd3}
\end{align}
\begin{align}
&jk_y k_{z,1} E \cos k_{z,1} d + \omega \epsilon _1 k_x C \cos k_{z,1} d
\nonumber \\
&= k_y k_{z,2} B e^{-jk_{z,2} d} + \omega \epsilon _2 k_x A e^{-jk_{z,2}d } +k_t^2 e^{-j(k_x x_0 + k_y y_0)} \label{eq:sd4}
\end{align}
まず，\(k_y \times \)式\eqref{eq:sd1}\(-k_x \times \)式\eqref{eq:sd2}より，\(A\)，\(C\)を消去して，
\begin{gather}
\mu _1 E \sin k_{z,1} d = \mu _2 B e^{-jk_{z,2} d}
\end{gather}
また，\(k_x \times \)式\eqref{eq:sd3}\(+k_y \times \)式\eqref{eq:sd4}より，\(A\)，\(C\)を消去して，
\begin{gather}
jk_{z,1} E \cos k_{z,1} d = k_{z,2} B e^{-jk_{z,2}d} + k_y e^{-j(k_x x_0 + k_y y_0)}
\end{gather}
これより，\(B\)を消去すると\(E\)は，
\begin{gather}
E = \frac{-j\mu _2 k_y e^{-j(k_x x_0 + k_y y_0)}}{
k_{z,1} \mu _2 \cos k_{z,1} d + j k_{z,2} \mu _1 \sin k_{z,1} d}
\end{gather}
これより，\(B\)は，
\begin{eqnarray}
B &=& \frac{\mu _1}{\mu _2} E \sin k_{z,1}d \ e^{jk_{z,2}d}
\nonumber \\
&=& \frac{-j\mu _1 k_y \sin k_{z,1}d \ e^{jk_{z,2}d} \ e^{-j(k_x x_0 + k_y y_0)} }{
k_{z,1} \mu _2 \cos k_{z,1} d + j k_{z,2} \mu _1 \sin k_{z,1} d}
\end{eqnarray}
いま，\(\mu _i = \mu _0 \mu _{r,i} \ (i=1,2)\)，
\begin{gather}
T_e \equiv k_{z,1} \mu _{r,2} \cos k_{z,1} d + j k_{z,2} \mu _{r,1} \sin k_{z,1} d
\end{gather}
とおくと，\(\widetilde{H}_{z,1}\)，\(\widetilde{H}_{z,2}\)は次のようになる．
\begin{eqnarray}
\widetilde{H}_{z,1}
&=& E \sin k_{z,1} z
\nonumber \\
&=& \frac{-j \mu _{r,2} k_y \sin k_{z,1} z}{T_e} e^{-j(k_x x_0 + k_y y_0)}
\end{eqnarray}
\begin{eqnarray}
\widetilde{H}_{z,2}
&=& B e^{-jk_{z,2} z}
\nonumber \\
&=& \frac{-j \mu _{r,1} k_y \sin k_{z,1}d \ e^{-jk_{z,2}(z-d)} }{
T_e} \ e^{-j(k_x x_0 + k_y y_0)}
\end{eqnarray}
同様にして，
\(k_y \times \)式\eqref{eq:sd1}\(+k_x \times \)式\eqref{eq:sd2}より，\(B\)，\(E\)を消去して，
\begin{gather}
-j k_{z,1} C \sin k_{z,1} d = k_{z,2} A e^{-jk_{z,2} d}
\end{gather}
また，\(k_x \times \)式\eqref{eq:sd3}\(-k_y \times \)式\eqref{eq:sd4}より，\(B\)，\(E\)を消去して，
\begin{gather}
-\omega \epsilon _1 C \cos k_{z,1} d = -\omega \epsilon _2 A e^{-jk_{z,2}d} - k_x e^{-j(k_x x_0 + k_y y_0)}
\end{gather}
これより，\(A\)を消去すると\(C\)は，
\begin{gather}
C = \frac{k_x k_{z,2} e^{-j(k_x x_0 + k_y y_0)}}{
\omega \big( k_{z,2} \epsilon _1 \cos k_{z,1} d + j k_{z,1} \epsilon _2 \sin k_{z,1} d \big)}
\end{gather}
よって，\(A\)は，
\begin{eqnarray}
A &=& \frac{-jk_{z,1}}{k_{z,2}} C \sin k_{z,1}d \ e^{jk_{z,2}d}
\nonumber \\
&=& \frac{-jk_x k_{z,1} \sin k_{z,1}d \ e^{jk_{z,2}d} \ e^{-j(k_x x_0 + k_y y_0)} }{
\omega \big( k_{z,2} \epsilon _1 \cos k_{z,1} d + j k_{z,1} \epsilon _2 \sin k_{z,1} d \big)}
\end{eqnarray}
いま，\(\epsilon _i = \epsilon _0 \epsilon _{r,i} \ (i=1,2)\)，
\begin{gather}
T_m \equiv k_{z,2} \epsilon _{r,1} \cos k_{z,1} d + j k_{z,1} \epsilon _{r,2} \sin k_{z,1} d
\end{gather}
とおくと，\(\widetilde{E}_{z,1}\)，\(\widetilde{E}_{z,2}\)は次のようになる．
\begin{eqnarray}
\widetilde{E}_{z,1}
&=& C \cos k_{z,1} z
\nonumber \\
&=& \frac{k_x k_{z,2} \cos k_{z,1} z}{\omega \epsilon _0 T_m} e^{-j(k_x x_0 + k_y y_0)}
\end{eqnarray}
\begin{eqnarray}
\widetilde{E}_{z,2}
&=& A e^{-jk_{z,2} z}
\nonumber \\
&=& \frac{-jk_x k_{z,1} \sin k_{z,1}d \ e^{-jk_{z,2}(z-d)}}{
\omega \epsilon _0 T_m} \ e^{-j(k_x x_0 + k_y y_0)}
\end{eqnarray}
微小電流のかわりに，\(x\)成分のみもつ面電流分布\(J_x(x',y')\)を考えても同様に，
\begin{eqnarray}
\widetilde{E}_{z,2}
&=& \frac{-jk_x k_{z,1} \sin (k_{z,1}d) }{\omega \epsilon _0 T_m} e^{-jk_{z,2}(z-d)} \widetilde{J}_x(k_x,k_y)
\\
\widetilde{H}_{z,2}
&=& \frac{-j \mu _{r,1} k_y \sin (k_{z,1}d) }{T_e} e^{-jk_{z,2}(z-d)} \widetilde{J}_x(k_x,k_y)
\\
\widetilde{E}_{z,1}
&=& \frac{k_x k_{z,2} \cos (k_{z,1} z)}{\omega \epsilon _0 T_m} \widetilde{J}_x(k_x,k_y)
\\
\widetilde{H}_{z,1}
&=& \frac{-j \mu _{r,2} k_y \sin (k_{z,1} z)}{T_e} \widetilde{J}_x(k_x,k_y)
\end{eqnarray}
ただし，\(\widetilde{J}_x(k_x ,k_y)\)は\(J_x(x',y')\)のフーリエ変換を示し，次のようになる．
\begin{gather}
\widetilde{J}_x (k_x,k_y)
= \iint _{-\infty}^\infty J_x(x',y') \ e^{-j(k_x x' + k_y y')} dx' dy'
\end{gather}
したがって，\(z=d\) におけるスペクトル領域の電界成分
\(\widetilde{E}_x\)，\(\widetilde{E}_y\)，\(\widetilde{E}_z\)は次のようになる．
\begin{eqnarray}
\widetilde{E}_{x}(k_x,k_y,d)
&=& \left\{ \frac{jk_x}{k_t^2} \cdot (-jk_{z,2}) \frac{-jk_x k_{z,1} \sin (k_{z,1}d) }{\omega \epsilon _0 T_m} \right.
\nonumber \\
&&\left. + \frac{\omega \mu _2 k_y}{k_t^2} \cdot \frac{-j \mu _{r,1} k_y \sin (k_{z,1}d) }{T_e} \right\} \widetilde{J}_x
\nonumber \\
&=& \frac{-j}{\omega \epsilon _0} \cdot \frac{\sin (k_{z,1}d)}{k_t^2} \left\{ \frac{k_x^2 k_{z,1} k_{z,2}}{T_m}
+ \frac{\mu _{r,1} \mu _{r,2} k_y^2 k_0^2}{T_e}
\right\} \widetilde{J}_x(k_x,k_y)
\end{eqnarray}
\begin{eqnarray}
\widetilde{E}_{y}(k_x,k_y,d)
&=& \left\{ \frac{jk_y}{k_t^2} \cdot (-jk_{z,2}) \frac{-jk_x k_{z,1} \sin (k_{z,1}d) }{\omega \epsilon _0 T_m} \right.
\nonumber \\
&&\left. - \frac{\omega \mu _2 k_x}{k_t^2} \cdot \frac{-j \mu _{r,1} k_y \sin (k_{z,1}d) }{T_e} \right\} \widetilde{J}_x
\nonumber \\
&=& \frac{-j}{\omega \epsilon _0} \cdot \frac{k_x k_y \sin (k_{z,1}d)}{k_t^2} \left\{ \frac{k_{z,1} k_{z,2}}{T_m}
- \frac{\mu _{r,1} \mu _{r,2} k_0^2}{T_e}
\right\} \widetilde{J}_x(k_x,k_y)
\end{eqnarray}
\begin{eqnarray}
\widetilde{E}_z(k_x,k_y,d)
&=& \frac{-j}{\omega \epsilon _0} \cdot \frac{k_x k_{z,1} \sin (k_{z,1}d) }{T_m} \widetilde{J}_x(k_x,k_y)
\end{eqnarray}
同様にして\(y\)成分のみをもつ面電流分布\(J_y(x',y')\)について求めると，
\begin{gather}
\widetilde{E}_{x}(k_x,k_y,d)
= \frac{-j}{\omega \epsilon _0} \cdot \frac{k_x k_y \sin (k_{z,1}d)}{k_t^2} \left\{ \frac{k_{z,1} k_{z,2}}{T_m}
- \frac{\mu _{r,1} \mu _{r,2} k_0^2}{T_e}
\right\} \widetilde{J}_y(k_x,k_y) \\
\widetilde{E}_{y}(k_x,k_y,d)
= \frac{-j}{\omega \epsilon _0} \cdot \frac{\sin (k_{z,1}d)}{k_t^2} \left\{ \frac{k_y^2 k_{z,1} k_{z,2}}{T_m}
+ \frac{\mu _{r,1} \mu _{r,2} k_x^2 k_0^2}{T_e}
\right\} \widetilde{J}_y(k_x,k_y)
\end{gather}
行列表示すると，
\begin{gather}
{\widetilde{E}_x \choose \widetilde{E}_y}
= {\widetilde{Z}_{xx}^{EJ} \ \ \widetilde{Z}_{xy}^{EJ} \choose \widetilde{Z}_{yx}^{EJ} \ \ \widetilde{Z}_{yy}^{EJ}}
{\widetilde{J}_x \choose \widetilde{J}_y}
\end{gather}
ただし，
\begin{eqnarray}
\widetilde{Z}_{xx}^{EJ}
&=& \frac{-j}{\omega \epsilon _0} \cdot \frac{\sin (k_{z,1}d)}{k_t^2} \left\{ \frac{k_x^2 k_{z,1} k_{z,2}}{T_m}
+ \frac{\mu _{r,1} \mu _{r,2} k_y^2 k_0^2}{T_e} \right\}
\\
\widetilde{Z}_{yx}^{EJ}
&=& \frac{-j}{\omega \epsilon _0} \cdot \frac{k_x k_y \sin (k_{z,1}d)}{k_t^2} \left\{ \frac{k_{z,1} k_{z,2}}{T_m}
- \frac{\mu _{r,1} \mu _{r,2} k_0^2}{T_e} \right\}
\nonumber \\
&=& \widetilde{Z}_{xy}^{EJ}
\\
\widetilde{Z}_{yy}^{EJ}
&=& \frac{-j}{\omega \epsilon _0} \cdot \frac{\sin (k_{z,1}d)}{k_t^2} \left\{ \frac{k_y^2 k_{z,1} k_{z,2}}{T_m}
+ \frac{\mu _{r,1} \mu _{r,2} k_x^2 k_0^2}{T_e} \right\}
\end{eqnarray}
また，ベクトル表示すると，\(xy\)面内の電界成分
\(\widetilde{\boldsymbol{E}}_t (= \widetilde{E}_x \boldsymbol{u}_x + \widetilde{E}_y \boldsymbol{u}_y)\)
は，
\begin{eqnarray}
\widetilde{\boldsymbol{E}}_t
&=& (\widetilde{Z}_{xx}^{EJ} \widetilde{J}_x + \widetilde{Z}_{xy}^{EJ} \widetilde{J}_y ) \boldsymbol{u}_x
+ (\widetilde{Z}_{yx}^{EJ} \widetilde{J}_x + \widetilde{Z}_{yy}^{EJ} \widetilde{J}_y ) \boldsymbol{u}_y
\nonumber \\
&\equiv& \widetilde{\bar{\bar{\boldsymbol{Z}}}}^{EJ} \cdot \widetilde{\boldsymbol{J}}_t
\end{eqnarray}
ここで，
\begin{eqnarray}
\widetilde{\bar{\bar{\boldsymbol{Z}}}}^{EJ} &=& \widetilde{Z}_{xx}^{EJ} \boldsymbol{u}_x \boldsymbol{u}_x + \widetilde{Z}_{xy}^{EJ} \boldsymbol{u}_x \boldsymbol{u}_y
+ \widetilde{Z}_{yx}^{EJ} \boldsymbol{u}_y \boldsymbol{u}_x + \widetilde{Z}_{yy}^{EJ} \boldsymbol{u}_y \boldsymbol{u}_y
\\
\widetilde{\boldsymbol{J}}_t &=& \widetilde{J}_x \boldsymbol{u}_x + \widetilde{J}_y \boldsymbol{u}_y
\end{eqnarray}
マイクロストリップ素子のスペクトル領域の電界型ダイアディック・グリーン関数の表示式
いま，\(0\leq z \leq d\) の領域\(\mathrm(I)\)が比誘電率
\(\epsilon _{r,1} \equiv \epsilon _r\)
の誘電体基板（比透磁率\(\mu _{r,1}=1\)），
\(z\geq d\) の領域\(\mathrm(II)\)が空気（真空とみなし，\(\epsilon _{r,2}=1\)，\(\mu _{r,2}=1\)）のとき，
\(k_{z,1} \equiv k_1\)，\(k_{z,2} \equiv k_2\)，自由空間波数を\(k_0\)とおくと，
\begin{align}
&k_0 ^2 \epsilon _r = k_t^2+k_1^2
\\
&k_0^2 = k_t^2+k_2^2
\\
&k_t^2 = k_x^2+k_y^2
\end{align}
これより，
\begin{eqnarray}
T_e &=& k_{z,1} \mu _{r,2} \cos k_{z,1} d + j k_{z,2} \mu _{r,1} \sin k_{z,1} d
\nonumber \\
&=& k_1 \cos k_1 d + j k_2 \sin k_1 d
\end{eqnarray}
\begin{eqnarray}
T_m &=& k_{z,2} \epsilon _{r,1} \cos k_{z,1} d + j k_{z,1} \epsilon _{r,2} \sin k_{z,1} d
\nonumber \\
&=& k_2 \epsilon _1 \cos k_1 d + j k_1 \sin k_1 d
\end{eqnarray}
よって，マトリクス要素\(\widetilde{Z}_{xx}^{EJ}\)は，
\begin{eqnarray}
&&\widetilde{Z}_{xx}^{EJ}
= \frac{-j}{\omega \epsilon _0} \cdot \frac{\sin k_1 d}{k_t^2} \left( \frac{k_x^2 k_1 k_2}{T_m}
+ \frac{k_y^2 k_0^2}{T_e} \right)
\nonumber \\
&=& \frac{-j}{\omega \epsilon _0} \cdot \frac{\sin k_1 d}{k_t^2} \cdot
\frac{k_x^2 k_1 k_2 T_e + k_y^2 k_0^2 T_m}{T_e T_m}
\nonumber \\
&=& \frac{-j}{\omega \epsilon _0} \cdot \frac{\sin k_1 d}{k_t^2}
\nonumber \\
&&\cdot \frac{k_2 (k_x^2 k_1^2+k_y k_0^2 \epsilon _r ) \cos k_1d + k_1(k_x^2 k_2^2+k_y^2 k_0^2 ) j \sin k_1 d}{T_e T_m}
\nonumber \\
&=& \frac{-j}{\omega \sqrt{\epsilon _0 \mu _0}} \sqrt{\frac{\mu _0}{\epsilon _0}} \frac{\sin k_1 d}{k_t^2}
\nonumber \\
&&\cdot \frac{k_2 k_t^2 (\epsilon _r k_0^2 - k_x^2 ) \cos k_1d + k_1 k_t^2 (k_0^2 -k_x^2) j \sin k_1 d}{T_e T_m}
\nonumber \\
&=& \frac{-j Z_0}{k_0} \cdot
\frac{k_2 (\epsilon _r k_0^2 - k_x^2 ) \cos k_1d + j k_1 (k_0^2 -k_x^2) \sin k_1 d}{T_e T_m} \sin k_1d
\label{eq:zejxx}
\end{eqnarray}
ただし，\(Z_0\)は自由空間の波動インピーダンスを示す．また，
\(\widetilde{Z}_{xy}^{EJ}\)，\(\widetilde{Z}_{yx}^{EJ}\)
は，
\begin{eqnarray}
&&\widetilde{Z}_{yx}^{EJ} = \widetilde{Z}_{xy}^{EJ}
= \frac{-j}{\omega \epsilon _0} \cdot \frac{k_x k_y \sin k_1 d}{k_t^2} \left( \frac{k_1 k_2}{T_m}
- \frac{k_0^2}{T_e} \right)
\nonumber \\
&=& \frac{-j}{\omega \epsilon _0} \cdot \frac{k_x k_y \sin k_1 d}{k_t^2} \cdot \frac{k_1 k_2 T_e - k_0^2 T_m}{T_e T_m}
\nonumber \\
&=& \frac{-j}{\omega \sqrt{\epsilon _0 \mu _0}} \sqrt{\frac{\mu _0}{\epsilon _0}} \frac{k_x k_y \sin k_1 d}{k_t^2}
\nonumber \\
&&\cdot \frac{k_2 (k_1^2 - k_0^2 \epsilon _r) \cos k_1d + k_1 (k_2^2-k_0^2) j \sin k_1d }{T_e T_m}
\nonumber \\
&=& \frac{-j}{\omega \epsilon _0} \cdot \frac{k_x k_y \sin k_1 d}{k_t^2}
\nonumber \\
&&\cdot \frac{k_2 (-k_t^2) \cos k_1d + j k_1 (-k_t^2) \sin k_1d }{T_e T_m}
\nonumber \\
&=& \frac{j Z_0}{k_0} \cdot
\frac{k_x k_y (k_2 \cos k_1d + j k_1 \sin k_1d)}{T_e T_m} \sin k_1 d
\label{eq:zejyx}
\end{eqnarray}
また，\(\widetilde{Z}_{yy}^{EJ}\)は，\(x\)と\(y\)を入れ換えれた式となり，
\begin{gather}
\widetilde{Z}_{yy}^{EJ}
= \frac{-j Z_0}{k_0} \cdot
\frac{k_2 (\epsilon _r k_0^2 - k_y^2 ) \cos k_1d + j k_1 (k_0^2 -k_y^2) \sin k_1 d}{T_e T_m} \sin k_1d
\label{eq:zejyy}
\end{gather}