1.4 縦続接続
散乱行列
図のような散乱行列
\begin{gather}
\begin{pmatrix}
b_1 \\ b_2
\end{pmatrix}
= [\boldsymbol{S}_a]
\begin{pmatrix}
a_1 \\ a_2
\end{pmatrix}, \ \ \
[\boldsymbol{S}_a] =
\begin{pmatrix}
S_{11}^{(a)} & S_{12}^{(a)} \\
S_{21}^{(a)} & S_{22}^{(a)} \\
\end{pmatrix} \\
\begin{pmatrix}
b_2' \\ b_3
\end{pmatrix}
= [\boldsymbol{S}_b]
\begin{pmatrix}
a_2' \\ a_3
\end{pmatrix}, \ \ \
[\boldsymbol{S}_b] =
\begin{pmatrix}
S_{11}^{(b)} & S_{12}^{(b)} \\
S_{21}^{(b)} & S_{22}^{(b)} \\
\end{pmatrix}
\end{gather}
を縦続接続し（接続した端子は，\(a_2 = b_2'\)，\(b_2 = a_2'\)），
同図中の散乱行列\([\boldsymbol{S}]\)を求める．
散乱行列の縦続接続
まず，\(a_2 = b_2'\)，\(b_2 = a_2'\)より，
\begin{gather}
b_1 = S_{11}^{(a)} a_1 + S_{12}^{(a)} a_2 \label{eq:Sa-1} \\
b_2 = S_{21}^{(a)} a_1 + S_{22}^{(a)} a_2 \label{eq:Sa-2} \\
a_2 = S_{11}^{(b)} b_2 + S_{12}^{(b)} a_3 \label{eq:Sb-1} \\
b_3 = S_{21}^{(b)} b_2 + S_{22}^{(b)} a_3 \label{eq:Sb-2}
\end{gather}
上式より，\(a_2\)，\(b_2\)を消去して次式が得られればよい．
\begin{gather}
b_1 = S_{11} a_1 + S_{12} a_3 \label{eq:Sab-1} \\
b_3 = S_{21} a_1 + S_{22} a_3 \label{eq:Sab-2}
\end{gather}
そこで，式\eqref{eq:Sa-2}と式\eqref{eq:Sb-1}より，\(b_2\)を消去すると，
\begin{gather}
a_2 = S_{11}^{(b)} \left( S_{21}^{(a)} a_1 + S_{22}^{(a)} a_2 \right) + S_{12}^{(b)} a_3 \nonumber \\
\left( 1 - S_{11}^{(b)} S_{22}^{(a)} \right) a_2 = S_{11}^{(b)} S_{21}^{(a)} a_1 + S_{12}^{(b)} a_3 \nonumber \\
a_2 = \left( 1-S_{11}^{(b)} S_{22}^{(a)} \right) ^{-1} \left( S_{11}^{(b)} S_{21}^{(a)} a_1 + S_{12}^{(b)} a_3 \right)
\label{eq:a2a1a3}
\end{gather}
式\eqref{eq:Sa-1}に代入して，\(a_2\)を消去すると，
\begin{eqnarray}
b_1 &=& S_{11}^{(a)} a_1 +S_{12}^{(a)}
\left( 1-S_{11}^{(b)} S_{22}^{(a)} \right) ^{-1}
\left( S_{11}^{(b)} S_{21}^{(a)} a_1 + S_{12}^{(b)} a_3 \right)
\nonumber \\
&=& \left\{ S_{11}^{(a)}+S_{12}^{(a)} \left( 1-S_{11}^{(b)} S_{22}^{(a)} \right) ^{-1} S_{11}^{(b)} S_{21}^{(a)} \right\} a_1
\nonumber \\
&&+ \left\{ S_{12}^{(a)} \left( 1-S_{11}^{(b)} S_{22}^{(a)} \right) ^{-1} S_{12}^{(b)} \right\} a_3
\end{eqnarray}
一方，式\eqref{eq:Sa-2}と式\eqref{eq:Sb-1}より，\(a_2\)を消去すると，
\begin{align}
&b_2 = S_{21}^{(a)} a_1 + S_{22}^{(a)} \left( S_{11}^{(b)} b_2 + S_{12}^{(b)} a_3 \right)
\nonumber \\
&\left( 1-S_{22}^{(a)} S_{11}^{(b)} \right) b_2 = S_{21}^{(a)} a_1 + S_{22}^{(a)} S_{12}^{(b)} a_3
\nonumber \\
&b_2 = \left( 1-S_{22}^{(a)} S_{11}^{(b)} \right) ^{-1} \left( S_{21}^{(a)} a_1 + S_{22}^{(a)} S_{12}^{(b)} a_3 \right)
\label{eq:b2a1a3}
\end{align}
式\eqref{eq:Sb-2}に代入して，\(b_2\)を消去すると，
\begin{eqnarray}
b_3 &=& S_{21}^{(b)} \left( 1-S_{22}^{(a)} S_{11}^{(b)} \right) ^{-1} %\nonumber \\ \hspace{10mm} \cdot
\left( S_{21}^{(a)} a_1 + S_{22}^{(a)} S_{12}^{(b)} a_3 \right) +S_{22}^{(b)} a_3
\nonumber \\
&=& \left\{ S_{21}^{(b)} \left( 1-S_{22}^{(a)} S_{11}^{(b)} \right) ^{-1} S_{21}^{(a)} \right\} a_1
\nonumber \\
&&+ \left\{ S_{21}^{(b)} \left( 1-S_{22}^{(a)} S_{11}^{(b)} \right) ^{-1} S_{22}^{(a)} S_{12}^{(b)} + S_{22}^{(b)} \right\} a_3
\end{eqnarray}
したがって，散乱行列要素\(S_{11}\)，\(S_{12}\)，\(S_{21}\)，\(S_{22}\)は，
\begin{eqnarray}
S_{11} &=& S_{11}^{(a)}+S_{12}^{(a)} \left( 1-S_{11}^{(b)} S_{22}^{(a)} \right) ^{-1} S_{11}^{(b)} S_{21}^{(a)} \\
S_{12} &=& S_{12}^{(a)} \left( 1-S_{11}^{(b)} S_{22}^{(a)} \right) ^{-1} S_{12}^{(b)} \\
S_{21} &=& S_{21}^{(b)} \left( 1-S_{22}^{(a)} S_{11}^{(b)} \right) ^{-1} S_{21}^{(a)} \\
S_{22} &=& S_{21}^{(b)} \left( 1-S_{22}^{(a)} S_{11}^{(b)} \right) ^{-1} S_{22}^{(a)} S_{12}^{(b)} + S_{22}^{(b)}
\end{eqnarray}
縦続接続ポートの波動振幅
式\eqref{eq:a2a1a3}，\eqref{eq:b2a1a3}に示した\(a_2\)，\(b_2\)は，2つの散乱行列を縦続接続しているポート2の波動振幅であり，
縦続接続した両者の散乱行列による多重反射を考慮できる．
いま，ポート1から励振して，ポート3を整合させたとき，ポート2の波動振幅\(a_2\)，\(b_2\)は，\eqref{eq:a2a1a3}，\eqref{eq:b2a1a3}より次のようになる．
\begin{eqnarray}
a_2 \Big|_{a_3 =0} &=& \left( 1-S_{11}^{(b)} S_{22}^{(a)} \right) ^{-1} S_{11}^{(b)} S_{21}^{(a)} a_1
\\
b_2 \Big|_{a_3 =0} &=& \left( 1-S_{22}^{(a)} S_{11}^{(b)} \right) ^{-1} S_{21}^{(a)} a_1
\end{eqnarray}
逆に，ポート3から励振して，ポート1を整合させたときは，
\begin{eqnarray}
a_2 \Big|_{a_1 =0} &=& \left( 1-S_{11}^{(b)} S_{22}^{(a)} \right) ^{-1} S_{12}^{(b)} a_3
\\
b_2 \Big|_{a_1 =0} &=& \left( 1-S_{22}^{(a)} S_{11}^{(b)} \right) ^{-1} S_{22}^{(a)} S_{12}^{(b)} a_3
\end{eqnarray}