2.3 方形TEモードによる放射特性の例
方形TE$_{0n}$モード（$m=0$）による遠方界
TE$_{0n}$モードのとき，$m=0$とおいて，
\begin{align}
\frac{\VEC{F}_{[0n]} (\theta ,\phi) \Big|_{\Gamma=0}}{\sqrt{Z_w}}
&= -\frac{ab\sqrt{ab \epsilon _0 \epsilon _n}}{4} \frac{\lambda _{c,[0n]}}{\lambda}
\sin \theta \ \Psi _{0n} (\theta ,\phi) \sqrt{z_{[0n]}}
\nonumber \\
&\cdot \left[ \frac{1+ y_{[0n]} \cos \theta}{2} \right.
\left\{ \left( \frac{0\pi}{a} \sin \phi \right) ^2 - \left( \frac{n \pi}{b} \cos \phi \right) ^2 \right\} \VEC{a}_\theta
\nonumber \\
&\left. + \frac{\cos \theta + y_{[0n]}}{2}
k_{c,[0n]}^2 \sin \phi \cos \phi \ \VEC{a}_\phi \right]
\end{align}
ここで，
\begin{align}
&k_{c,[0n]} = \frac{n\pi}{b}\\
&\lambda_{c,[0n]} = \frac{2\pi}{k_{c,[0n]}}
= 2 \pi \frac{b}{n \pi} = \frac{2b}{n}\\
&\epsilon_0 = 1, \ \ \ \epsilon_{n(\ne 0)} = 2
\end{align}
また，
\begin{align}
&\Psi _{0n} (\theta ,\phi)
= \frac{\mbox{sinc} \left( u_x \cos \phi \right)}{u_x \cos \phi}
\cdot \frac{\mbox{sinc} \left( u_y \sin \phi + \frac{n\pi}{2} \right)}{u_y \sin \phi - \frac{n\pi}{2}}
\ e^{j\frac{(n+1)\pi}{2}}
\\
&e^{j\frac{(n+1)\pi}{2}}
= e^{j\frac{n\pi}{2}} e^{j\frac{\pi}{2}}
= j e^{j\frac{n\pi}{2}}
= j \cos \frac{n\pi}{2} -\sin \frac{n\pi}{2}
\end{align}
ここで，
\begin{gather}
u_x = \frac{\pi a}{\lambda} \sin \theta
\end{gather}
より，
\begin{align}
&\Psi _{0n} (\theta ,\phi)
= \frac{\lambda}{\pi a} \frac{\mbox{sinc} \left( u_x \cos \phi \right)}{u_x \cos \phi} \Phi_{y,n}
\\
&\Phi_{y,n} \equiv \frac{\mbox{sinc} \left( u_y \sin \phi + \frac{n\pi}{2} \right)}{u_y \sin \phi - \frac{n\pi}{2}}
e^{j\frac{(n+1)\pi}{2}}
\end{align}
これらより，
\begin{gather}
\frac{\VEC{F}_{[0n]} (\theta ,\phi) \Big|_{\Gamma=0}}{\sqrt{Z_w}}
= -\frac{ab\sqrt{ab \cdot 2}}{4} \frac{n\pi}{b} \frac{1}{\lambda} \sin \theta
\frac{\lambda}{\pi a} \frac{\mbox{sinc} \left( u_x \cos \phi \right)}{u_x \cos \phi} \Phi_{y,n} \sqrt{z_{[0n]}}
\nonumber \\
\cdot \left( \frac{n \pi}{b} \right) ^2 \left[ \frac{1+ y_{[0n]} \cos \theta}{2}
(-\cos^2 \phi ) \VEC{a}_\theta
+ \frac{\cos \theta + y_{[0n]}}{2} \sin \phi \cos \phi \ \VEC{a}_\phi \right]
\end{gather}
整理して，
\begin{align}
\frac{\VEC{F}_{[0n]} (\theta ,\phi) \Big|_{\Gamma=0}}{\sqrt{Z_w}}
&= \frac{n \pi \sqrt{ab}}{\sqrt{2}} \mbox{sinc} (u_x \cos \phi) \Phi_{y,n} \sqrt{z_{[0n]}}
\nonumber \\
&\cdot \left( \frac{1+y_{[0n]} \cos \theta}{2} \cos \phi \ \VEC{a}_\theta
- \frac{y_{[0n]}+\cos \theta}{2} \sin \phi \ \VEC{a}_\phi \right)
\end{align}
開口径が十分大きい場合，$z_{[0n]} \simeq 1$，$y_{[0n]} \simeq 1$と近似して，
\begin{align}
\bar{\VEC{F}}_{[0n]} (\theta ,\phi)
&= \frac{n\pi\sqrt{ab}}{\sqrt{2}} \ \frac{1+\cos \theta}{2} \
\mbox{sinc} \left( u_x \cos \phi \right)
\nonumber \\
&\cdot \ \frac{\sin \left( u_y \sin \phi + \frac{n\pi}{2} \right)}{(u_y \sin \phi )^2 - \frac{n^2\pi^2}{4}}
\ e^{j\frac{(n+1)\pi}{2}} \ \VEC{a}_\xi
\end{align}
ここで，
\begin{gather}
\VEC{a}_\xi = \cos \phi \VEC{a}_\theta - \sin \phi \VEC{a}_\phi
\end{gather}
正面方向（$\theta =0$，$\phi =0$）では，
\begin{gather}
\bar{\VEC{F}}_{[0n]} (0,0)
= \frac{n\pi\sqrt{ab}}{\sqrt{2}} \cdot 1 \cdot 1 \cdot
\ \frac{\sin \left( \frac{n\pi}{2} \right)}{- \frac{n^2\pi^2}{4}}
\ e^{j\frac{(n+1)\pi}{2}} \ \VEC{a}_x
\end{gather}
ここで，
\begin{eqnarray}
\sin \frac{n\pi}{2} e^{j\frac{(n+1)\pi}{2}}
&=& \sin \frac{n\pi}{2} \left( j \cos \frac{n\pi}{2} -\sin \frac{n\pi}{2} \right)
\nonumber \\
&=& -\left( \sin \frac{n\pi}{2} \right)^2 = -1 (\mbox{odd}), \ 0 (\mbox{even})
\end{eqnarray}
より，$n$が偶数の場合，正面でヌルになる．また，$n$が奇数の場合，
\begin{align}
&\bar{\VEC{F}}_{[0n]} (0,0) = \frac{2\sqrt{2ab}}{n\pi} \VEC{a}_x
\\
&G_{[0n]} \Big| _{\theta = 0} = 4\pi \frac{ab}{\lambda ^2} \left| \frac{2\sqrt{2}}{n\pi} \right| ^2
= 4\pi \frac{ab}{\lambda ^2} \cdot \frac{8}{n^2 \pi ^2}
\end{align}
方形TE$_{01}$モード（$m=0$，$n=1$）による遠方界
TE$_{01}$モードのとき，$m=0$，$n=1$とおいて，
\begin{align}
&A_{[01]} = \frac{1}{\pi} \sqrt{\frac{2b}{a}}
\\
&\bar{\VEC{F}}_{[01]} (\theta ,\phi)
= -\frac{\pi \sqrt{ab}}{\sqrt{2}} \ \frac{1+\cos \theta}{2} \mbox{sinc} \big( u_x \cos \phi \big)
\ \frac{\cos \big( u_y \sin \phi \big)}{
\big( u_y \sin \phi \big)^2- \frac{\pi^2}{4}} \ \VEC{a}_\xi
\end{align}
E面は$\phi = 0$のときで，$\sin \phi = 0$，$\cos \phi = 1$より，
\begin{gather}
\bar{\VEC{F}}_{[01]} (\theta ,0)
= \frac{2\sqrt{2ab}}{\pi} \ \frac{1+\cos \theta }{2} \mbox{sinc} \big( u_x \big) \ \VEC{a}_x
\end{gather}
一方，H面は$\phi = \pi /2$のときで，$\sin \phi = 1$，$\cos \phi = 0$より，
\begin{gather}
\bar{\VEC{F}}_{[01]} (\theta ,\pi /2)
= -\frac{\pi \sqrt{ab}}{\sqrt{2}} \ \frac{1+\cos \theta}{2}
\frac{\cos \big( u_y \big)}{u_y^2- \frac{\pi ^2}{4}} \ \VEC{a}_x
\end{gather}
方形TE$_{m0}$モード（$n=0$）による遠方界
同様にして，TE$_{m0}$モードのとき，$n=0$とおいて，
\begin{align}
\frac{\VEC{F}_{[m0]} (\theta ,\phi) \Big|_{\Gamma=0}}{\sqrt{Z_w}}
&= -\frac{ab\sqrt{ab \epsilon _0 \epsilon _n}}{4} \frac{\lambda _{c,[m0]}}{\lambda}
\sin \theta \ \Psi _{m0} (\theta ,\phi) \sqrt{z_{[m0]}}
\nonumber \\
&\cdot \left[ \frac{1+ y_{[m0]} \cos \theta}{2} \right.
\left\{ \left( \frac{m\pi}{a} \sin \phi \right) ^2 - \left( \frac{0 \pi}{b} \cos \phi \right) ^2 \right\} \VEC{a}_\theta
\nonumber \\
&\left. + \frac{\cos \theta + y_{[m0]}}{2}
k_{c,[0n]}^2 \sin \phi \cos \phi \ \VEC{a}_\phi \right]
\end{align}
ここで，
\begin{align}
&k_{c,[m0]} = \frac{m\pi}{a}\\
&\lambda_{c,[m0]} = \frac{2\pi}{k_{c,[m0]}}
= 2 \pi \frac{a}{m \pi} = \frac{2a}{m}\\
&\epsilon_0 = 1, \ \ \ \epsilon_{m(\ne 0)} = 2
\end{align}
また，
\begin{align}
&\Psi _{m0} (\theta ,\phi)
= \frac{\mbox{sinc} \left( u_x \cos \phi + \frac{m\pi}{2} \right)}{u_x \cos \phi - \frac{m\pi}{2}}
\cdot \frac{\mbox{sinc} \left( u_y \sin \phi \right)}{u_y \sin \phi}
\ e^{j\frac{(m+1)\pi}{2}}
\end{align}
ここで，
\begin{gather}
u_y = \frac{\pi b}{\lambda} \sin \theta
\end{gather}
より，
\begin{align}
&\Psi _{m0} (\theta ,\phi)
= \frac{\lambda}{\pi b} \frac{\mbox{sinc} \left( u_y \sin \phi \right)}{u_y \sin \phi} \Phi_{x,n}
\\
&\Phi_{x,n} \equiv \frac{\mbox{sinc} \left( u_x \cos \phi + \frac{m\pi}{2} \right)}{u_x \cos \phi - \frac{m\pi}{2}}
e^{j\frac{(m+1)\pi}{2}}
\end{align}
これらより，
\begin{gather}
\frac{\VEC{F}_{[m0]} (\theta ,\phi) \Big|_{\Gamma=0}}{\sqrt{Z_w}}
= -\frac{ab\sqrt{ab \cdot 2}}{4} \frac{m\pi}{a} \frac{1}{\lambda} \sin \theta
\frac{\lambda}{\pi b} \frac{\mbox{sinc} \left( u_y \sin \phi \right)}{u_y \sin \phi} \Phi_{x,n} \sqrt{z_{[m0]}}
\nonumber \\
\cdot \left( \frac{m \pi}{a} \right) ^2 \left[ \frac{1+ y_{[m0]} \cos \theta}{2}
\sin^2 \phi \VEC{a}_\theta
+ \frac{\cos \theta + y_{[m0]}}{2} \sin \phi \cos \phi \ \VEC{a}_\phi \right]
\end{gather}
整理して，
\begin{align}
\frac{\VEC{F}_{[m0]} (\theta ,\phi) \Big|_{\Gamma=0}}{\sqrt{Z_w}}
&= -\frac{m \pi \sqrt{ab}}{\sqrt{2}} \mbox{sinc} (u_y \sin \phi) \Phi_{x,n} \sqrt{z_{[m0]}}
\nonumber \\
&\cdot \left( \frac{1+y_{[m0]} \cos \theta}{2} \sin \phi \ \VEC{a}_\theta
+ \frac{y_{[m0]}+\cos \theta}{2} \cos \phi \ \VEC{a}_\phi \right)
\end{align}
開口径が十分大きい場合，$z_{[m0]} \simeq 1$，$y_{[m0]} \simeq 1$と近似して，
\begin{align}
\bar{\VEC{F}}_{[m0]} (\theta ,\phi)
&= -\frac{m\pi\sqrt{ab}}{\sqrt{2}} \ \frac{1+\cos \theta}{2} \
\mbox{sinc} \left( u_y \sin \phi \right)
\nonumber \\
&\cdot \ \frac{\sin \left( u_x \cos \phi + \frac{m\pi}{2} \right)}{(u_x \cos \phi )^2 - \frac{m^2\pi^2}{4}}
\ e^{j\frac{(m+1)\pi}{2}} \ \VEC{a}_\eta
\end{align}
ここで，
\begin{gather}
\VEC{a}_\eta = \sin \phi \VEC{a}_\theta + \cos \phi \VEC{a}_\phi
\end{gather}
正面方向（$\theta =0$，$\phi =0$）では，
$n$が偶数の場合，正面でヌルになる．また，$n$が奇数の場合，
\begin{align}
&\bar{\VEC{F}}_{[m0]} (0,0) = -\frac{2\sqrt{2ab}}{m\pi} \VEC{a}_y
\\
&G_{[m0]} \Big| _{\theta = 0} = 4\pi \frac{ab}{\lambda ^2} \left| \frac{2\sqrt{2}}{m\pi} \right| ^2
= 4\pi \frac{ab}{\lambda ^2} \cdot \frac{8}{m^2 \pi ^2}
\end{align}
方形TE$_{10}$モード（$m=1$，$n=0$）による遠方界
TE$_{10}$モードのとき，$m=1$，$n=0$とおいて，
\begin{gather}
\bar{\VEC{F}}_{[10]} (\theta ,\phi)
= -\frac{\pi \sqrt{ab}}{\sqrt{2}} \ \frac{1+\cos \theta}{2}
\ \frac{\cos \big( u_x \cos \phi \big)}{\big( u_x \cos \phi \big)^2- \frac{\pi^2}{4}}
\mbox{sinc} \big( u_y \sin \phi \big)
\ \VEC{a}_\eta
\end{gather}
E面は$\phi = \pi /2$のときで，$\sin \phi = 1$，$\cos \phi = 0$より，
\begin{gather}
\bar{\VEC{F}}_{[10]} (\theta ,\pi/2)
= \frac{2\sqrt{2 ab}}{\pi} \ \frac{1+\cos \theta }{2} \mbox{sinc} \big( u_y \big) \ \VEC{a}_y
\end{gather}
一方，H面は$\phi = 0$のときで，$\sin \phi = 0$，$\cos \phi = 1$より，
\begin{gather}
\bar{\VEC{F}}_{[10]} (\theta ,0)
= -\frac{\pi \sqrt{ab}}{\sqrt{2}} \ \frac{1+\cos \theta}{2}
\frac{\cos \big( u_x \big)}{u_x^2- \frac{\pi ^2}{4}} \ \VEC{a}_y
\end{gather}
ピーク利得
TE$_{01}$モードのピーク利得$G_{[01]} \Big| _{\theta = 0}$は，
\begin{gather}
G_{[01]} \Big| _{\theta = 0} = 4\pi \frac{ab}{\lambda ^2} \left| \frac{2\sqrt{2}}{\pi} \right| ^2
= 4\pi \frac{ab}{\lambda ^2} \cdot \frac{8}{\pi ^2}
\end{gather}
また，
TE$_{10}$モードのピーク利得$G_{[10]} \Big| _{\theta = 0}$は，
\begin{gather}
G_{[10]} \Big| _{\theta = 0} = 4\pi \frac{ab}{\lambda ^2} \left| \frac{2\sqrt{2}}{\pi} \right| ^2
= 4\pi \frac{ab}{\lambda ^2} \cdot \frac{8}{\pi ^2}
= G_{[01]} \Big| _{\theta = 0}
\end{gather}