2.4 方形TE$_{mn}$モードによるフレネル領域の放射界
積分項が $t_x \ne 0$ のとき，
\begin{gather}
I_{s1} \equiv \int_{-1}^1 \sin \frac{m\pi}{2} \big( \bar{x} +1 \big) e^{j\bar{x} u_x \cos \phi}
e^{-j 2\pi t_x \bar{x}^2} d\bar{x}
\\
I_{c1} \equiv \int_{-1}^1 \cos \frac{m\pi}{2} \big( \bar{x} +1 \big) e^{j\bar{x} u_x \cos \phi}
e^{-j 2\pi t_x \bar{x}^2} d\bar{x}
\end{gather}
また， $t_y \ne 0$ のとき，
\begin{gather}
I_{s2} \equiv \int_{-1}^1 \sin \frac{n\pi}{2} \big( \bar{y} +1 \big) e^{j\bar{y} u_y \cos \phi}
e^{-j 2\pi t_y \bar{y}^2} d\bar{y}
\\
I_{c2} \equiv \int_{-1}^1 \cos \frac{n\pi}{2} \big( \bar{y} +1 \big) e^{j\bar{y} u_y \sin \phi}
e^{-j 2\pi t_y \bar{y}^2} d\bar{y}
\end{gather}
上式は，次の不定積分が計算できればよい．
\begin{gather}
I_{s} = \int \sin B(v+1) e^{Av} e^{-j\frac{\pi}{2}Cv^2} dv
\\
I_{c} = \int \cos B(v+1) e^{Av} e^{-j\frac{\pi}{2}Cv^2} dv
\end{gather}
フレネル積分
フレネル積分$S(x)$，$C(x)$は，
\begin{gather}
S(x) = \int_0^x \sin \left( \frac{\pi}{2} t^2 \right) dt
= \sqrt{\frac{2}{\pi}} \int_0^{\sqrt{\frac{\pi}{2}}x} \sin (t^2) dt\\
C(x) = \int_0^x \cos \left( \frac{\pi}{2} t^2 \right) dt
= \sqrt{\frac{2}{\pi}} \int_0^{\sqrt{\frac{\pi}{2}}x} \cos (t^2) dt
\end{gather}
フレネル積分$S(x)$，$C(x)$の計算例
$\displaystyle{\int_0^x e^{j\frac{\pi}{2}t^2} dt = C(x) + j S(x)}$（Cornuのらせん）
不定積分を変形して，
\begin{eqnarray}
I_{s}
&=& \int \sin B(v+1) e^{Av} e^{-j\frac{\pi}{2}Cv^2} dv
\nonumber \\
&=& \frac{1}{j2} \int \Big( e^{jB(v+1)} - e^{-jB(v+1)} \Big) e^{Av} e^{-j\frac{\pi}{2}Cv^2} dv
\nonumber \\
&=& \frac{e^{jB}}{j2} \int e^{(A+jB)v} e^{-j\frac{\pi}{2}Cv^2} dv
- \frac{e^{-jB}}{j2} \int e^{(A-jB)v} e^{-j\frac{\pi}{2}Cv^2} dv
\\
I_{c}
&=& \int \cos B(v+1) e^{Av} e^{-j\frac{\pi}{2}Cv^2} dv
\nonumber \\
&=& \frac{1}{2} \int \Big( e^{jB(v+1)} + e^{-jB(v+1)} \Big) e^{Av} e^{-j\frac{\pi}{2}Cv^2} dv
\nonumber \\
&=& \frac{e^{jB}}{2} \int e^{(A+jB)v} e^{-j\frac{\pi}{2}Cv^2} dv
+ \frac{e^{-jB}}{2} \int e^{(A-jB)v} e^{-j\frac{\pi}{2}Cv^2} dv
\end{eqnarray}
いま，$jA_i \equiv A\pm jB \ (i=1,2)$とおくと，
\begin{gather}
I_{s} = \frac{e^{jB}}{j2} \int e^{jA_1 v} e^{-j\frac{\pi}{2}Cv^2} dv
- \frac{e^{-jB}}{j2} \int e^{jA_2 v} e^{-j\frac{\pi}{2}Cv^2} dv
\\
I_{c} = \frac{e^{jB}}{2} \int e^{jA_1 v} e^{-j\frac{\pi}{2}Cv^2} dv
+ \frac{e^{-jB}}{2} \int e^{jA_2 v} e^{-j\frac{\pi}{2}Cv^2} dv
\end{gather}
積分項をまとめると，
\begin{gather}
\int e^{jA_i v} e^{-j\frac{\pi}{2}Cv^2} dv
= \int e^{j(A_i v - \frac{\pi}{2}Cv^2)} dv
\equiv \int e^{j \Theta_i(v)} dv
\end{gather}
位相項$\Theta_i(v)$を変形して，
\begin{eqnarray}
\Theta_i(v)
&=& A_i v - \frac{\pi}{2}Cv^2
\nonumber \\
&=& -\frac{\pi}{2} \left( C v^2 - \frac{2}{\pi} A_i v \right)
\nonumber \\
&=& -\frac{\pi}{2} \left\{ \left( \sqrt{C} v - \frac{A_i}{\pi \sqrt{C}} \right)^2
- \frac{A_i^2}{\pi^2 C} \right\}
\nonumber \\
&=& -\frac{\pi}{2} \left( \sqrt{C} v - \frac{A_i}{\pi \sqrt{C}} \right)^2 + \frac{A_i^2}{2\pi C}
\end{eqnarray}
ここで，
\begin{gather}
t_i(v) \equiv \sqrt{C} v - \frac{A_i}{\pi \sqrt{C}}
\end{gather}
とおくと，
\begin{gather}
dt_i = \sqrt{C} dv
\end{gather}
積分範囲は，
\begin{gather}
t_i(\mp 1) = \mp \sqrt{C} - \frac{A_i}{\pi \sqrt{C}} \equiv t_{i \mp}
\end{gather}
これより積分項は，
\begin{eqnarray}
\int_{-1}^1 e^{j \Theta_i(v)} dv
&=& \int_{t_{i-}}^{t_{i+}} e^{-j\frac{\pi}{2} t_i^2}e^{j\frac{A_i^2}{2 \pi C}} \frac{dt_i}{\sqrt{C}}
\nonumber \\
&=& \frac{e^{j\frac{A_i^2}{2 \pi C}}}{\sqrt{C}} \int_{t_{i-}}^{t_{i+}} e^{-j\frac{\pi}{2} t_i^2} dt_i
\end{eqnarray}
よって，
\begin{eqnarray}
\int_{t_{i-}}^{t_{i+}} e^{-j\frac{\pi}{2}t_i^2} dt_i
&=& \int_{t_{i-}}^{t_{i+}} \cos \left( \frac{\pi}{2} t_i^2 \right) dt_i
- j \int_{t_{i-}}^{t_{i+}} \sin \left( \frac{\pi}{2} t_i^2 \right) dt_i
\nonumber \\
&=& C(t_{i+}) - C(t_{i-}) -j \big\{ S(t_{i+}) - S(t_{i-}) \big\}
\end{eqnarray}
いま，
\begin{gather}
v \to \bar{x}, \ \ \ \ \
A \to j u_x \cos \phi, \ \ \ \ \
B \to \frac{m \pi}{2}, \ \ \ \ \
C \to 4 t_x
\nonumber
\end{gather}
とすると（$jA_i \equiv A\pm jB \ (i=1,2)$），
\begin{eqnarray}
t_{1 \mp}
&=& \mp \sqrt{C} - \frac{A_1}{\pi \sqrt{C}}
\nonumber \\
&=& \mp \sqrt{C} - \frac{-j A + B}{\pi \sqrt{C}}
\nonumber \\
&=& \mp 2 \sqrt{t_x} - \frac{u_x \cos \phi + \frac{m \pi}{2}}{2\pi \sqrt{t_x}}
\\
t_{2 \mp}
&=& \mp \sqrt{C} - \frac{A_2}{\pi \sqrt{C}}
\nonumber \\
&=& \mp \sqrt{C} - \frac{-j A - B}{\pi \sqrt{C}}
\nonumber \\
&=& \mp 2 \sqrt{t_x} - \frac{u_x \cos \phi - \frac{m \pi}{2}}{2\pi \sqrt{t_x}}
\end{eqnarray}
このとき，
\begin{eqnarray}
\frac{e^{j\frac{A_1^2}{2 \pi C}}}{\sqrt{C}}
&=& \frac{1}{2\sqrt{t_x}} e^{j\frac{(u_x \cos \phi + \frac{m \pi}{2})^2}{8\pi t_x}}
\\
\frac{e^{j\frac{A_2^2}{2 \pi C}}}{\sqrt{C}}
&=& \frac{1}{2\sqrt{t_x}} e^{j\frac{(u_x \cos \phi - \frac{m \pi}{2})^2}{8\pi t_x}}
\end{eqnarray}
同様にして，
\begin{gather}
v \to \bar{y}, \ \ \ \ \
A \to j u_y \sin \phi, \ \ \ \ \
B \to \frac{n \pi}{2}, \ \ \ \ \
C \to 4 t_y
\nonumber
\end{gather}
として，
\begin{eqnarray}
t'_{1 \mp} &=& \mp 2 \sqrt{t_y} - \frac{u_y \sin \phi + \frac{n \pi}{2}}{2\pi \sqrt{t_y}}
\\
t'_{2 \mp} &=& \mp 2 \sqrt{t_y} - \frac{u_y \sin \phi - \frac{n \pi}{2}}{2\pi \sqrt{t_y}}
\end{eqnarray}
このとき，
\begin{eqnarray}
\frac{e^{j\frac{A_1^2}{2 \pi C}}}{\sqrt{C}}
&=& \frac{1}{2\sqrt{t_y}} e^{j\frac{(u_y \sin \phi + \frac{n \pi}{2})^2}{8\pi t_y}}
\\
\frac{e^{j\frac{A_2^2}{2 \pi C}}}{\sqrt{C}}
&=& \frac{1}{2\sqrt{t_y}} e^{j\frac{(u_y \sin \phi - \frac{n \pi}{2})^2}{8\pi t_y}}
\end{eqnarray}
よって，
\begin{eqnarray}
I_{s1}
&=& \frac{e^{jB}}{j2} \int_{-1}^1 e^{j\Theta_1 \bar{x}} d\bar{x}
- \frac{e^{-jB}}{j2} \int_{-1}^1 e^{j\Theta_2 \bar{x}} d\bar{x}
\nonumber \\
&=& \frac{1}{j4 \sqrt{t_x}}
\left( e^{j\frac{m\pi}{2}} e^{j\frac{(u_x \cos \phi + \frac{m \pi}{2})^2}{8\pi t_x}}
\int_{t_{1-}}^{t_{1+}} e^{-j\frac{\pi}{2} t^2} dt \right.
\nonumber \\
&&\left. - e^{-j\frac{m\pi}{2}} e^{j\frac{(u_x \cos \phi - \frac{m \pi}{2})^2}{8\pi t_x}}
\int_{t_{2-}}^{t_{2+}} e^{-j\frac{\pi}{2} t^2} dt \right)
\nonumber \\
&=& \frac{1}{j4 \sqrt{t_x}}
\nonumber \\
&&\cdot
\left( e^{j\frac{m\pi}{2}} e^{j\frac{(u_x \cos \phi + \frac{m \pi}{2})^2}{8\pi t_x}}
\Big[ C(t_{1+}) - C(t_{1-}) - j \big\{ S(t_{1+}) - S(t_{1-}) \big\} \Big] \right.
\nonumber \\
&&\left. - e^{-j\frac{m\pi}{2}} e^{j\frac{(u_x \cos \phi - \frac{m \pi}{2})^2}{8\pi t_x}}
\Big[ C(t_{2+}) - C(t_{2-}) - j \big\{ S(t_{2+}) - S(t_{2-}) \big\} \Big] \right)
\\
I_{c1}
&=& \frac{e^{jB}}{2} \int_{-1}^1 e^{j\Theta_1 \bar{x}} d\bar{x}
+ \frac{e^{jB}}{2} \int_{-1}^1 e^{j\Theta_2 \bar{x}} d\bar{x}
\nonumber \\
&=& \frac{1}{4 \sqrt{t_x}}
\nonumber \\
&&\cdot \left( e^{j\frac{m\pi}{2}} e^{j\frac{(u_x \cos \phi + \frac{m \pi}{2})^2}{8\pi t_x}}
\Big[ C(t_{1+}) - C(t_{1-}) - j \big\{ S(t_{1+}) - S(t_{1-}) \big\} \Big] \right.
\nonumber \\
&&\left. + e^{-j\frac{m\pi}{2}} e^{j\frac{(u_x \cos \phi - \frac{m \pi}{2})^2}{8\pi t_x}}
\Big[ C(t_{2+}) - C(t_{2-}) - j \big\{ S(t_{2+}) - S(t_{2-}) \big\} \Big] \right)
\end{eqnarray}
同様にして，
\begin{eqnarray}
I_{s2}
&=& \frac{1}{j4 \sqrt{t_y}}
\nonumber \\
&&\cdot
\left( e^{j\frac{n\pi}{2}} e^{j\frac{(u_y \sin \phi + \frac{n \pi}{2})^2}{8\pi t_y}}
\Big[ C(t'_{1+}) - C(t'_{1-}) - j \big\{ S(t'_{1+}) - S(t'_{1-}) \big\} \Big] \right.
\nonumber \\
&&\left. - e^{-j\frac{n\pi}{2}} e^{j\frac{(u_y \cos \phi - \frac{n \pi}{2})^2}{8\pi t_y}}
\Big[ C(t'_{2+}) - C(t'_{2-}) - j \big\{ S(t'_{2+}) - S(t'_{2-}) \big\} \Big] \right)
\\
I_{c2}
&=& \frac{1}{4 \sqrt{t_y}}
\nonumber \\
&&\cdot
\left( e^{j\frac{n\pi}{2}} e^{j\frac{(u_y \sin \phi + \frac{n \pi}{2})^2}{8\pi t_y}}
\Big[ C(t'_{1+}) - C(t'_{1-}) - j \big\{ S(t'_{1+}) - S(t'_{1-}) \big\} \Big] \right.
\nonumber \\
&&\left. + e^{-j\frac{n\pi}{2}} e^{j\frac{(u_y \sin \phi - \frac{n \pi}{2})^2}{8\pi t_y}}
\Big[ C(t'_{2+}) - C(t'_{2-}) - j \big\{ S(t'_{2+}) - S(t'_{2-}) \big\} \Big] \right)
\end{eqnarray}
ここで，
\begin{eqnarray}
\bar{\VEC{F}}_{[mn]}
&=& \frac{1+\cos \theta}{2}
\Big( \bar{N}_{x[mn]} \VEC{a}_\xi + \bar{N}_{y[mn]} \VEC{a}_\eta \Big)
\nonumber \\
&=& A_{[mn]} \frac{ab}{4} \frac{1+\cos \theta}{2}
\left( \frac{n \pi}{b} I_{c1} I_{s2} \VEC{a}_\xi
+ \frac{m \pi}{a} I_{s1} I_{c2} \VEC{a}_\eta \right)
\end{eqnarray}
また，
\begin{eqnarray}
\bar{N}_{x[mn]} &=& A_{[mn]} \frac{n\pi a}{4} I_{c1} I_{s2}
\\
\bar{N}_{y[mn]} &=& -A_{[mn]} \frac{m\pi b}{4} I_{s1} I_{c2}
\end{eqnarray}