スロット素子のスペクトル領域グリーン関数

スロット素子のスペクトル領域グリーン関数の導出

　誘電体基板（厚さ

d

，比誘電率

ϵ_{r}

）の地導体に設けたスロット素子を考える．

z = 0

の境界条件より，

\begin{matrix} (1) & {\tilde{E}}_{1} (0) = - \tilde{M} \end{matrix}

このとき，

\begin{matrix} (2) & {\tilde{H}}_{1} (0) = Y_{i n}^{(+)} {\tilde{E}}_{1} (0) = - Y_{i n}^{(+)} \tilde{M} \equiv {\tilde{Y}}^{(0)} \tilde{M} \end{matrix}

ここで，

\begin{array}{rcl} {\tilde{Y}}^{(0)} & = & - Y_{i n}^{(+)} = - Y_{1} \frac{Y_{2} + j Y_{1} \tan k_{z 1} d}{Y_{1} + j Y_{2} \tan k_{z 1} d} \\ (3) & = & - Y_{1} \frac{Y_{2} \cos k_{z 1} d + j Y_{1} \sin k_{z 1} d}{Y_{1} \cos k_{z 1} d + j Y_{2} \sin k_{z 1} d} \end{array}

これより，

\begin{array}{rcl} {\tilde{Y}}_{_{T E}}^{(0)} & = & - Y_{1_{T E}} \frac{Y_{2_{T E}} \cos k_{z 1} d + j Y_{1_{T E}} \sin k_{z 1} d}{Y_{1_{T E}} \cos k_{z 1} d + j Y_{2_{T E}} \sin k_{z 1} d} \\ = & - \frac{k_{z 1}}{ω μ_{1}} \cdot \frac{\frac{k_{z 2}}{ω μ_{2}} \cos k_{z 1} d + j \frac{k_{z 1}}{ω μ_{1}} \sin k_{z 1} d}{\frac{k_{z 1}}{ω μ_{1}} \cos k_{z 1} d + j \frac{k_{z 2}}{ω μ_{2}} \sin k_{z 1} d} \\ (4) & = & - \frac{j k_{z 1}}{ω μ_{0} T_{e}} (- j k_{z 2} \cos k_{z 1} d + k_{z 1} \sin k_{z 1} d) \end{array}

また，

\begin{array}{rcl} {\tilde{Y}}_{_{T M}}^{(0)} & = & - Y_{1_{T M}} \frac{Y_{2_{T M}} \cos k_{z 1} d + j Y_{1_{T M}} \sin k_{z 1} d}{Y_{1_{T M}} \cos k_{z 1} d + j Y_{2_{T M}} \sin k_{z 1} d} \\ = & - \frac{ω ϵ_{1}}{k_{z 1}} \cdot \frac{\frac{ω ϵ_{2}}{k_{z 2}} \cos k_{z 1} d + j \frac{ω ϵ_{1}}{k_{z 1}} \sin k_{z 1} d}{\frac{ω ϵ_{1}}{k_{z 1}} \cos k_{z 1} d + j \frac{ω ϵ_{2}}{k_{z 2}} \sin k_{z 1} d} \\ (5) & = & - \frac{j ω ϵ_{0} ϵ_{r}}{k_{z 1} T_{m}} (- j k_{z 1} \cos k_{z 1} d + k_{z 2} ϵ_{r} \sin k_{z 1} d) \end{array}

よって，磁界型ダイアディック・グリーン関数の各成分は，

\begin{array}{rcl} {\tilde{G}}_{x x}^{^{H M}} & = & \frac{1}{k_{t}^{2}} (k_{y}^{2} {\tilde{Y}}_{_{T M}}^{(0)} + k_{x}^{2} {\tilde{Y}}_{_{T E}}^{(0)}) \\ = & - \frac{j ω ϵ_{0} ϵ_{r} k_{y}^{2}}{k_{t}^{2} k_{z 1} T_{m}} (k_{z 2} ϵ_{r} \sin k_{z 1} d - j k_{z 1} \cos k_{z 1} d) \\ (6) & - \frac{j k_{z 1} k_{x}^{2}}{ω μ_{0} k_{t}^{2} T_{e}} (k_{z 1} \sin k_{z 1} d - j k_{z 2} \cos k_{z 1} d) \\ {\tilde{G}}_{x y}^{^{H M}} & = & {\tilde{G}}_{y x}^{^{H M}} = \frac{k_{x} k_{y}}{k_{t}^{2}} ({\tilde{Y}}_{_{T E}}^{(0)} - {\tilde{Y}}_{_{T M}}^{(0)}) \\ = & \frac{k_{x} k_{y}}{k_{t}^{2}} {\frac{j ω ϵ_{0} ϵ_{r}}{k_{z 1} T_{m}} (k_{z 2} ϵ_{r} \sin k_{z 1} d - j k_{z 1} \cos k_{z 1} d) \\ (7) & - \frac{j k_{z 1}}{ω μ_{0} T_{e}} (k_{z 1} \sin k_{z 1} d - j k_{z 2} \cos k_{z 1} d)} \\ {\tilde{G}}_{y y}^{^{H M}} & = & \frac{1}{k_{t}^{2}} (k_{x}^{2} {\tilde{Y}}_{_{T M}}^{(0)} + k_{y}^{2} {\tilde{Y}}_{_{T E}}^{(0)}) \\ = & - \frac{j ω ϵ_{0} ϵ_{r} k_{x}^{2}}{k_{t}^{2} k_{z 1} T_{m}} (k_{z 2} ϵ_{r} \sin k_{z 1} d - j k_{z 1} \cos k_{z 1} d) \\ (8) & - \frac{j k_{z 1} k_{y}^{2}}{ω μ_{0} k_{t}^{2} T_{e}} (k_{z 1} \sin k_{z 1} d - j k_{z 2} \cos k_{z 1} d) \end{array}

また，

\begin{array}{rcl} {\tilde{H}}_{1} (d) & = & - {\tilde{E}}_{1} (0) j Y_{1} \sin k_{z 1} d + {\tilde{H}}_{1} (0) \cos k_{z 1} d \\ = & - (- j Y_{1} \sin k_{z 1} d + Y_{i n}^{(+)} \cos k_{z 1} d) \tilde{M} \\ (9) & \equiv & {\tilde{Y}}^{(d)} \tilde{M} \end{array}

\begin{array}{rcl} {\tilde{E}}_{1} (d) & = & Z_{2} {\tilde{H}}_{1} (d) \\ = & Z_{2} {\tilde{Y}}^{(d)} \tilde{M} \\ (10) & \equiv & {\tilde{Q}}^{(d)} \tilde{M} \end{array}

ここで，

\begin{array}{rcl} {\tilde{Y}}^{(d)} & = & j Y_{1} \sin k_{z, 1} d - Y_{i n}^{(+)} \cos k_{z 1} d \\ = & (j \frac{Y_{1}}{Y_{i n}^{(+)}} \sin k_{z 1} d - \cos k_{z 1} d) Y_{i n}^{(+)} \\ = & \frac{- Y_{2}}{Y_{2} \cos k_{z 1} d + j Y_{1} \sin k_{z 1} d} Y_{i n}^{(+)} \\ = & \frac{- Y_{2}}{Y_{2} \cos k_{z 1} d + j Y_{1} \sin k_{z 1} d} Y_{1} \frac{Y_{2} \cos k_{z 1} d + j Y_{1} \sin k_{z 1} d}{Y_{1} \cos k_{z 1} d + j Y_{2} \sin k_{z 1} d} \\ (11) & = & \frac{- Y_{1} Y_{2}}{Y_{1} \cos k_{z 1} d + j Y_{2} \sin k_{z 1} d} \end{array}

また，

\begin{matrix} (12) & {\tilde{Q}}^{(d)} = Z_{2} {\tilde{Y}}^{(d)} = - \frac{Y_{1}}{Y_{1} \cos k_{z 1} d + j Y_{2} \sin k_{z 1} d} = - {\tilde{P}}^{(- d)} \end{matrix}

これより，

\begin{array}{rcl} (13) & {\tilde{Q}}_{_{T E}}^{(d)} & = & - {\tilde{P}}_{_{T E}}^{(- d)} = - \frac{k_{z 1}}{T_{e}} \\ (14) & {\tilde{Q}}_{_{T M}}^{(d)} & = & - {\tilde{P}}_{_{T M}}^{(- d)} = - \frac{k_{z 2} ϵ_{r}}{T_{m}} \end{array}

よって，電界型ダイアディック・グリーン関数の各成分は，

\begin{array}{rcl} {\tilde{G}}_{x x}^{^{E M}} & = & - {\tilde{G}}_{y y}^{^{E M}} = \frac{k_{x} k_{y}}{k_{t}^{2}} ({\tilde{Q}}_{_{T E}}^{(d)} - {\tilde{Q}}_{_{T M}}^{(d)}) \\ = & \frac{k_{x} k_{y}}{k_{t}^{2}} (- \frac{k_{z 1}}{T_{e}} + \frac{k_{z 2} ϵ_{r}}{T_{m}}) \\ (15) & = & - \frac{j k_{x} k_{y} (ϵ_{r} - 1) \sin k_{z 1} d}{T_{e} T_{m}} = - {\tilde{G}}_{x x}^{^{H J}} = {\tilde{G}}_{y y}^{^{H J}} \end{array}

\begin{array}{rcl} {\tilde{G}}_{x y}^{^{E M}} & = & \frac{1}{k_{t}^{2}} (k_{x}^{2} {\tilde{Q}}_{_{T M}}^{(d)} + k_{y}^{2} {\tilde{Q}}_{_{T E}}^{(d)}) \\ = & - \frac{1}{k_{t}^{2}} (\frac{ϵ_{r} k_{x}^{2} k_{z 2}}{T_{m}} + \frac{k_{y}^{2} k_{z 1}}{T_{e}}) \\ (16) & = & \frac{k_{z 1}}{T_{e}} - \frac{j k_{x}^{2} (ϵ_{r} - 1) \sin k_{z 1} d}{T_{e} T_{m}} = - {\tilde{G}}_{y x}^{^{H J}} \end{array}

\begin{array}{rcl} {\tilde{G}}_{y x}^{^{E M}} & = & - \frac{1}{k_{t}^{2}} (k_{y}^{2} {\tilde{Q}}_{_{T M}}^{(d)} + k_{x}^{2} {\tilde{Q}}_{_{T E}}^{(d)}) \\ = & \frac{1}{k_{t}^{2}} (\frac{ϵ_{r} k_{y}^{2} k_{z 2}}{T_{m}} + \frac{k_{x}^{2} k_{z 1}}{T_{e}}) \\ (17) & = & - \frac{k_{z 1}}{T_{e}} + \frac{j k_{y}^{2} (ϵ_{r} - 1) \sin k_{z 1} d}{T_{e} T_{m}} = - {\tilde{G}}_{x y}^{^{H J}} \end{array}