スペクトル領域の電磁界
直角座標系\((x,y,z)\)において，\(xy\)面にマイクロストリップ素子がある場合を考え，電磁界
\(\boldsymbol{E}(x,y,z)\)，\(\boldsymbol{H}(x,y,z)\)
より，
空間領域\((x,y)\)からスペクトル領域\((k_x,k_y)\)へのフーリエ変換
\(\widetilde{\boldsymbol{E}}(k_x,k_y,z)\)，\(\widetilde{\boldsymbol{H}}(k_x,k_y,z)\)
を定義する．
\begin{gather}
\widetilde{\boldsymbol{E}}(k_x,k_y,z) = \iint _{-\infty}^\infty \boldsymbol{E}(x,y,z) \ e^{-j(k_x x + k_y y)} dx dy \\
\widetilde{\boldsymbol{H}}(k_x,k_y,z) = \iint _{-\infty}^\infty \boldsymbol{H}(x,y,z) \ e^{-j(k_x x + k_y y)} dx dy
\end{gather}
フーリエ変換対の関係より，
\begin{gather}
\boldsymbol{E}(x,y,z) = \frac{1}{(2\pi)^2} \iint _{-\infty}^\infty \widetilde{\boldsymbol{E}}(k_x,k_y,z) \ e^{j(k_x x + k_y y)} dk_x dk_y \\
\boldsymbol{H}(x,y,z) = \frac{1}{(2\pi)^2} \iint _{-\infty}^\infty \widetilde{\boldsymbol{H}}(k_x,k_y,z) \ e^{j(k_x x + k_y y)} dk_x dk_y
\end{gather}
これより，\(\nabla \cdot \boldsymbol{E}\) は，
\begin{eqnarray}
&&\nabla \cdot \boldsymbol{E}(x,y,z)
\nonumber \\
&=& \nabla \cdot \left[
\frac{1}{(2\pi)^2} \iint _{-\infty}^\infty \widetilde{\boldsymbol{E}}(k_x,k_y,z) \ e^{j(k_x x + k_y y)} dk_x dk_y \right]
\nonumber \\
&=& \frac{1}{(2\pi)^2} \iint _{-\infty}^\infty \nabla \cdot \big\{ \widetilde{\boldsymbol{E}} \ e^{j(k_x x + k_y y)} \big\} dk_x dk_y
\nonumber \\
&=& \frac{1}{(2\pi)^2} \iint _{-\infty}^\infty \left( jk_x\boldsymbol{u}_x + jk_y\boldsymbol{u}_y + \frac{\partial }{\partial z} \boldsymbol{u}_z \right)
\nonumber \\
&&\cdot \widetilde{\boldsymbol{E}}(k_x,k_y,z) \ e^{j(k_x x + k_y y)} dk_x dk_y
\end{eqnarray}
フーリエ変換対の関係より，
\begin{align}
&\left( jk_x\boldsymbol{u}_x + jk_y\boldsymbol{u}_y + \frac{\partial }{\partial z} \boldsymbol{u}_z \right) \cdot \widetilde{\boldsymbol{E}}
\nonumber \\
&= \iint _{-\infty}^\infty \nabla \cdot \boldsymbol{E}(x,y,z) \ e^{-j(k_x x + k_y y)} dx dy
\end{align}
ここで，
\begin{gather}
\boldsymbol{k}_t \equiv k_x\boldsymbol{u}_x + k_y\boldsymbol{u}_y
\end{gather}
とおくと，
\begin{gather}
\left( j\boldsymbol{k}_t + \frac{\partial }{\partial z} \boldsymbol{u}_z \right) \cdot \widetilde{\boldsymbol{E}}(k_x,k_y,z)
= \iint _{-\infty}^\infty \nabla \cdot \boldsymbol{E}(x,y,z) \ e^{-j(k_x x + k_y y)} dx dy
\end{gather}
同様にして，\(\nabla \cdot \boldsymbol{H}\) のフーリエ変換は，
\begin{gather}
\left( j\boldsymbol{k}_t + \frac{\partial }{\partial z} \boldsymbol{u}_z \right) \cdot \widetilde{\boldsymbol{H}}(k_x,k_y,z)
= \iint _{-\infty}^\infty \nabla \cdot \boldsymbol{H}(x,y,z) \ e^{-j(k_x x + k_y y)} dx dy
\end{gather}
また，
\(\nabla \times \boldsymbol{E}\)，\(\nabla \times \boldsymbol{H}\)
のフーリエ変換は（導出省略），
\begin{align}
&\left( j\boldsymbol{k}_t + \frac{\partial }{\partial z} \boldsymbol{u}_z \right) \times \widetilde{\boldsymbol{E}}(k_x,k_y,z)
\nonumber \\
&= \iint _{-\infty}^\infty \nabla \times \boldsymbol{E}(x,y,z) \ e^{-j(k_x x + k_y y)} dx dy
\\
&\left( j\boldsymbol{k}_t + \frac{\partial }{\partial z} \boldsymbol{u}_z \right) \times \widetilde{\boldsymbol{H}}(k_x,k_y,z)
\nonumber \\
&= \iint _{-\infty}^\infty \nabla \times \boldsymbol{H}(x,y,z) \ e^{-j(k_x x + k_y y)} dx dy
\end{align}
さらに，
\(\nabla ^2 \boldsymbol{E}\)
は，
\begin{gather}
\nabla ^2 \boldsymbol{E}(x,y,z)
= \nabla ^2 E_x(x,y,z) \boldsymbol{u}_x + \nabla ^2 E_y(x,y,z) \boldsymbol{u}_y + \nabla ^2 E_z(x,y,z) \boldsymbol{u}_z
\end{gather}
いま，添え字 \(x\)，\(y\)，\(z\) を \(i\) で置き換え，
\begin{eqnarray}
&&\nabla ^2 E_i(x,y,z)
\nonumber \\
&=& \nabla ^2 \left[
\frac{1}{(2\pi)^2} \iint _{-\infty}^\infty \widetilde{E}_i(k_x,k_y,z) \ e^{j(k_x x + k_y y)} dk_x dk_y \right]
\nonumber \\
&=& \frac{1}{(2\pi)^2} \iint _{-\infty}^\infty \nabla ^2 \big\{ \widetilde{E}_i \ e^{j(k_x x + k_y y)} \big\} dk_x dk_y
\nonumber \\
&=& \frac{1}{(2\pi)^2} \iint _{-\infty}^\infty \left( (jk_x)^2 + (jk_y)^2 + \frac{\partial ^2}{\partial z^2} \right)
\nonumber \\
&&\cdot \widetilde{E}_i(k_x,k_y,z) \ e^{j(k_x x + k_y y)} dk_x dk_y
\nonumber \\
&=& \frac{1}{(2\pi)^2} \iint _{-\infty}^\infty \left( -k_t^2 + \frac{\partial ^2}{\partial z^2} \right)
\widetilde{E}_i \ e^{j(k_x x + k_y y)} dk_x dk_y
\end{eqnarray}
ここで，フーリエ変換対の関係より，
\begin{gather}
\left( -k_t^2 + \frac{\partial ^2}{\partial z^2} \right) \widetilde{E}_i(k_x,k_y,z)
= \iint _{-\infty}^\infty \nabla ^2 E_i(x,y,z) \ e^{-j(k_x x + k_y y)} dx dy
\end{gather}
ベクトルでも同様で，
\begin{gather}
\left( -k_t^2 + \frac{\partial ^2}{\partial z^2} \right) \widetilde{\boldsymbol{E}}(k_x,k_y,z)
= \iint _{-\infty}^\infty \nabla ^2 \boldsymbol{E}(x,y,z) \ e^{-j(k_x x + k_y y)} dx dy
\end{gather}
これより，電磁流がない場合のMaxwellの方程式をフーリエ変換すると，
\begin{gather}
\nabla \times \boldsymbol{E} = -j\omega \mu \boldsymbol{H} \ \ \to \ \
\left( j\boldsymbol{k}_t + \frac{\partial }{\partial z} \boldsymbol{u}_z \right) \times \widetilde{\boldsymbol{E}} = -j\omega \mu \widetilde{\boldsymbol{H}} \label{eq:sde} \\
\nabla \times \boldsymbol{H} = j\omega \epsilon \boldsymbol{E} \ \ \to \ \
\left( j\boldsymbol{k}_t + \frac{\partial }{\partial z} \boldsymbol{u}_z \right) \times \widetilde{\boldsymbol{H}} = j\omega \epsilon \widetilde{\boldsymbol{E}} \label{eq:sdh} \\
\nabla \cdot \boldsymbol{E} = 0 \ \ \to \ \
\left( j\boldsymbol{k}_t + \frac{\partial }{\partial z} \boldsymbol{u}_z \right) \cdot \widetilde{\boldsymbol{E}} = 0 \label{eq:sde2} \\
\nabla \cdot \boldsymbol{H} = 0 \ \ \to \ \
\left( j\boldsymbol{k}_t + \frac{\partial }{\partial z} \boldsymbol{u}_z \right) \cdot \widetilde{\boldsymbol{H}} = 0 \label{eq:sdh2}
\end{gather}
また，
\begin{align}
&\big( \nabla ^2 + k^2 \big) \boldsymbol{E} = 0
\nonumber \\
&\to \ \
\left( -k_t^2 + \frac{\partial ^2}{\partial z^2} + k^2 \right) \widetilde{\boldsymbol{E}}
= \left( \frac{\partial ^2}{\partial z^2} + k_z^2 \right) \widetilde{\boldsymbol{E}} = 0
\\
&\big( \nabla ^2 + k^2 \big) \boldsymbol{H} = 0
\nonumber \\
&\to \ \
\left( -k_t^2 + \frac{\partial ^2}{\partial z^2} + k^2 \right) \widetilde{\boldsymbol{H}}
= \left( \frac{\partial ^2}{\partial z^2} + k_z^2 \right) \widetilde{\boldsymbol{H}} = 0
\end{align}
式\eqref{eq:sde}より，
\begin{eqnarray}
\boldsymbol{u}_z \cdot \left\{\left( j\boldsymbol{k}_t + \frac{\partial }{\partial z} \boldsymbol{u}_z \right) \times \widetilde{\boldsymbol{E}} \right\}
&=& j \boldsymbol{u}_z \cdot ( \boldsymbol{k}_t \times \widetilde{\boldsymbol{E}} )
\nonumber \\
&=& -j\omega \mu \widetilde{\boldsymbol{H}} \cdot \boldsymbol{u}_z
\end{eqnarray}
\begin{gather}
k_x \widetilde{E}_y - k_y \widetilde{E}_x = - \omega \mu \widetilde{H}_z
\end{gather}
また，式\eqref{eq:sde2}より，
\begin{align}
&j\boldsymbol{k}_t \cdot \widetilde{\boldsymbol{E}} = - \boldsymbol{u}_z \cdot \frac{\partial \widetilde{\boldsymbol{E}}}{\partial z}
\nonumber \\
&k_x \widetilde{E}_x + k_y \widetilde{E}_y = j \frac{\partial \widetilde{E}_z}{\partial z}
\end{align}
これより，\(\widetilde{E}_x\)，\(\widetilde{E}_y\)について解くと，
\begin{gather}
\widetilde{E}_x = \frac{jk_x}{k_t^2} \frac{\partial \widetilde{E}_z}{\partial z} + \frac{\omega \mu k_y}{k_t^2} \widetilde{H}_z
\\
\widetilde{E}_y = \frac{jk_y}{k_t^2} \frac{\partial \widetilde{E}_z}{\partial z} - \frac{\omega \mu k_x}{k_t^2} \widetilde{H}_z
\end{gather}
同様にして，\(\widetilde{H}_x\)，\(\widetilde{H}_y\)は，
\begin{gather}
\widetilde{H}_x = \frac{jk_x}{k_t^2} \frac{\partial \widetilde{H}_z}{\partial z} - \frac{\omega \epsilon k_y}{k_t^2} \widetilde{E}_z
\\
\widetilde{H}_y = \frac{jk_y}{k_t^2} \frac{\partial \widetilde{H}_z}{\partial z} - \frac{\omega \epsilon k_x}{k_t^2} \widetilde{E}_z
\end{gather}
ただし，
\begin{gather}
k_t^2 = k_x^2 + k_y^2
\end{gather}
ここで，
\begin{gather}
\left( \frac{\partial ^2}{\partial z^2} + k_z^2 \right) \widetilde{E}_z = 0
\\
\left( \frac{\partial ^2}{\partial z^2} + k_z^2 \right) \widetilde{H}_z = 0
\end{gather}
より，\(\widetilde{E}_z\)，\(\widetilde{H}_z\) の解としては，
\(e^{-jk_z z}\)，\(e^{jk_z z}\) あるいは，\(\sin k_z z\)，\(\cos k_z z\)
のいずれかを考えればよい．