電磁界のスペクトル領域解析

スペクトル領域の電磁界

　直角座標系

(x, y, z)

において，

x y

面にマイクロストリップ素子がある場合を考え，電磁界

E (x, y, z)

，

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

\tilde{E} (k_{x}, k_{y}, z)

，

\tilde{H} (k_{x}, k_{y}, z)

を定義する．

\begin{matrix} (1) & \tilde{E} (k_{x}, k_{y}, z) = \iint_{- \infty}^{\infty} E (x, y, z) e^{- j (k_{x} x + k_{y} y)} d x d y \\ (2) & \tilde{H} (k_{x}, k_{y}, z) = \iint_{- \infty}^{\infty} H (x, y, z) e^{- j (k_{x} x + k_{y} y)} d x d y \end{matrix}

フーリエ変換対の関係より，

\begin{matrix} (3) & E (x, y, z) = \frac{1}{(2 π)^{2}} \iint_{- \infty}^{\infty} \tilde{E} (k_{x}, k_{y}, z) e^{j (k_{x} x + k_{y} y)} d k_{x} d k_{y} \\ (4) & H (x, y, z) = \frac{1}{(2 π)^{2}} \iint_{- \infty}^{\infty} \tilde{H} (k_{x}, k_{y}, z) e^{j (k_{x} x + k_{y} y)} d k_{x} d k_{y} \end{matrix}

これより，

\nabla \cdot E

は，

\begin{array}{rcl} \nabla \cdot E (x, y, z) \\ = & \nabla \cdot [\frac{1}{(2 π)^{2}} \iint_{- \infty}^{\infty} \tilde{E} (k_{x}, k_{y}, z) e^{j (k_{x} x + k_{y} y)} d k_{x} d k_{y}] \\ = & \frac{1}{(2 π)^{2}} \iint_{- \infty}^{\infty} \nabla \cdot {\tilde{E} e^{j (k_{x} x + k_{y} y)}} d k_{x} d k_{y} \\ = & \frac{1}{(2 π)^{2}} \iint_{- \infty}^{\infty} (j k_{x} u_{x} + j k_{y} u_{y} + \frac{\partial}{\partial z} u_{z}) \\ (5) & \cdot \tilde{E} (k_{x}, k_{y}, z) e^{j (k_{x} x + k_{y} y)} d k_{x} d k_{y} \end{array}

フーリエ変換対の関係より，

\begin{aligned} (j k_{x} u_{x} + j k_{y} u_{y} + \frac{\partial}{\partial z} u_{z}) \cdot \tilde{E} \\ (6) & = \iint_{- \infty}^{\infty} \nabla \cdot E (x, y, z) e^{- j (k_{x} x + k_{y} y)} d x d y \end{aligned}

ここで，

\begin{matrix} (7) & k_{t} \equiv k_{x} u_{x} + k_{y} u_{y} \end{matrix}

とおくと，

\begin{matrix} (8) & (j k_{t} + \frac{\partial}{\partial z} u_{z}) \cdot \tilde{E} (k_{x}, k_{y}, z) = \iint_{- \infty}^{\infty} \nabla \cdot E (x, y, z) e^{- j (k_{x} x + k_{y} y)} d x d y \end{matrix}

同様にして，

\nabla \cdot H

のフーリエ変換は，

\begin{matrix} (9) & (j k_{t} + \frac{\partial}{\partial z} u_{z}) \cdot \tilde{H} (k_{x}, k_{y}, z) = \iint_{- \infty}^{\infty} \nabla \cdot H (x, y, z) e^{- j (k_{x} x + k_{y} y)} d x d y \end{matrix}

また，

\nabla \times E

，

\nabla \times H

のフーリエ変換は（導出省略），

\begin{aligned} (j k_{t} + \frac{\partial}{\partial z} u_{z}) \times \tilde{E} (k_{x}, k_{y}, z) \\ (10) & = \iint_{- \infty}^{\infty} \nabla \times E (x, y, z) e^{- j (k_{x} x + k_{y} y)} d x d y \\ (j k_{t} + \frac{\partial}{\partial z} u_{z}) \times \tilde{H} (k_{x}, k_{y}, z) \\ (11) & = \iint_{- \infty}^{\infty} \nabla \times H (x, y, z) e^{- j (k_{x} x + k_{y} y)} d x d y \end{aligned}

さらに，

\nabla^{2} E

は，

\begin{matrix} (12) & \nabla^{2} E (x, y, z) = \nabla^{2} E_{x} (x, y, z) u_{x} + \nabla^{2} E_{y} (x, y, z) u_{y} + \nabla^{2} E_{z} (x, y, z) u_{z} \end{matrix}

いま，添え字

x

，

y

，

z

を

i

で置き換え，

\begin{array}{rcl} \nabla^{2} E_{i} (x, y, z) \\ = & \nabla^{2} [\frac{1}{(2 π)^{2}} \iint_{- \infty}^{\infty} {\tilde{E}}_{i} (k_{x}, k_{y}, z) e^{j (k_{x} x + k_{y} y)} d k_{x} d k_{y}] \\ = & \frac{1}{(2 π)^{2}} \iint_{- \infty}^{\infty} \nabla^{2} {{\tilde{E}}_{i} e^{j (k_{x} x + k_{y} y)}} d k_{x} d k_{y} \\ = & \frac{1}{(2 π)^{2}} \iint_{- \infty}^{\infty} ((j k_{x})^{2} + (j k_{y})^{2} + \frac{\partial^{2}}{\partial z^{2}}) \\ \cdot {\tilde{E}}_{i} (k_{x}, k_{y}, z) e^{j (k_{x} x + k_{y} y)} d k_{x} d k_{y} \\ (13) & = & \frac{1}{(2 π)^{2}} \iint_{- \infty}^{\infty} (- k_{t}^{2} + \frac{\partial^{2}}{\partial z^{2}}) {\tilde{E}}_{i} e^{j (k_{x} x + k_{y} y)} d k_{x} d k_{y} \end{array}

ここで，フーリエ変換対の関係より，

\begin{matrix} (14) & (- k_{t}^{2} + \frac{\partial^{2}}{\partial z^{2}}) {\tilde{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)} d x d y \end{matrix}

ベクトルでも同様で，

\begin{matrix} (15) & (- k_{t}^{2} + \frac{\partial^{2}}{\partial z^{2}}) \tilde{E} (k_{x}, k_{y}, z) = \iint_{- \infty}^{\infty} \nabla^{2} E (x, y, z) e^{- j (k_{x} x + k_{y} y)} d x d y \end{matrix}

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{matrix} (16) & \nabla \times E = - j ω μ H \to (j k_{t} + \frac{\partial}{\partial z} u_{z}) \times \tilde{E} = - j ω μ \tilde{H} \\ (17) & \nabla \times H = j ω ϵ E \to (j k_{t} + \frac{\partial}{\partial z} u_{z}) \times \tilde{H} = j ω ϵ \tilde{E} \\ (18) & \nabla \cdot E = 0 \to (j k_{t} + \frac{\partial}{\partial z} u_{z}) \cdot \tilde{E} = 0 \\ (19) & \nabla \cdot H = 0 \to (j k_{t} + \frac{\partial}{\partial z} u_{z}) \cdot \tilde{H} = 0 \end{matrix}

また，

\begin{aligned} (\nabla^{2} + k^{2}) E = 0 \\ (20) & \to (- k_{t}^{2} + \frac{\partial^{2}}{\partial z^{2}} + k^{2}) \tilde{E} = (\frac{\partial^{2}}{\partial z^{2}} + k_{z}^{2}) \tilde{E} = 0 \\ (\nabla^{2} + k^{2}) H = 0 \\ (21) & \to (- k_{t}^{2} + \frac{\partial^{2}}{\partial z^{2}} + k^{2}) \tilde{H} = (\frac{\partial^{2}}{\partial z^{2}} + k_{z}^{2}) \tilde{H} = 0 \end{aligned}

式

(16)

より，

\begin{array}{rcl} u_{z} \cdot {(j k_{t} + \frac{\partial}{\partial z} u_{z}) \times \tilde{E}} & = & j u_{z} \cdot (k_{t} \times \tilde{E}) \\ (22) & = & - j ω μ \tilde{H} \cdot u_{z} \end{array}

\begin{matrix} (23) & k_{x} {\tilde{E}}_{y} - k_{y} {\tilde{E}}_{x} = - ω μ {\tilde{H}}_{z} \end{matrix}

また，式

(18)

より，

\begin{aligned} j k_{t} \cdot \tilde{E} = - u_{z} \cdot \frac{\partial \tilde{E}}{\partial z} \\ (24) & k_{x} {\tilde{E}}_{x} + k_{y} {\tilde{E}}_{y} = j \frac{\partial {\tilde{E}}_{z}}{\partial z} \end{aligned}

これより，

{\tilde{E}}_{x}

，

{\tilde{E}}_{y}

について解くと，

\begin{matrix} (25) & {\tilde{E}}_{x} = \frac{j k_{x}}{k_{t}^{2}} \frac{\partial {\tilde{E}}_{z}}{\partial z} + \frac{ω μ k_{y}}{k_{t}^{2}} {\tilde{H}}_{z} \\ (26) & {\tilde{E}}_{y} = \frac{j k_{y}}{k_{t}^{2}} \frac{\partial {\tilde{E}}_{z}}{\partial z} - \frac{ω μ k_{x}}{k_{t}^{2}} {\tilde{H}}_{z} \end{matrix}

同様にして，

{\tilde{H}}_{x}

，

{\tilde{H}}_{y}

は，

\begin{matrix} (27) & {\tilde{H}}_{x} = \frac{j k_{x}}{k_{t}^{2}} \frac{\partial {\tilde{H}}_{z}}{\partial z} - \frac{ω ϵ k_{y}}{k_{t}^{2}} {\tilde{E}}_{z} \\ (28) & {\tilde{H}}_{y} = \frac{j k_{y}}{k_{t}^{2}} \frac{\partial {\tilde{H}}_{z}}{\partial z} - \frac{ω ϵ k_{x}}{k_{t}^{2}} {\tilde{E}}_{z} \end{matrix}

ただし，

\begin{matrix} (29) & k_{t}^{2} = k_{x}^{2} + k_{y}^{2} \end{matrix}

ここで，

\begin{matrix} (30) & (\frac{\partial^{2}}{\partial z^{2}} + k_{z}^{2}) {\tilde{E}}_{z} = 0 \\ (31) & (\frac{\partial^{2}}{\partial z^{2}} + k_{z}^{2}) {\tilde{H}}_{z} = 0 \end{matrix}

より，

{\tilde{E}}_{z}

，

{\tilde{H}}_{z}

の解としては，

e^{- j k_{z} z}

，

e^{j k_{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．