基準反射係数と基準インピーダンスの関係
一般化散乱行列の複素基準インピーダンス\(\hat{Z}_1\)，\(\hat{Z}_2\)と，散乱行列の基準インピーダンス\(R_{01}\)，\(R_{02}\)，基準（反射）係数\(\gamma_1\)，\(\gamma_2\)の関係を行列表示すると，
\begin{eqnarray}
[\hat{Z}] &=&
\begin{pmatrix}
\hat{Z}_1 & 0 \\ 0 & \hat{Z}_2
\end{pmatrix}
=
\begin{pmatrix}
\frac{1+\gamma_1}{1-\gamma_1} R_{01} & 0 \\
0 & \frac{1+\gamma_2}{1-\gamma_2} R_{02}
\end{pmatrix}
\nonumber \\ \hspace{5mm}
&=&
\begin{pmatrix}
R_{01} & 0 \\ 0 & R_{02}
\end{pmatrix}
\begin{pmatrix}
1+\gamma_1 & 0 \\ 0 & 1+\gamma_2
\end{pmatrix}
\begin{pmatrix}
1-\gamma_1 & 0 \\ 0 & 1-\gamma_2
\end{pmatrix}^{-1}
\nonumber
\end{eqnarray}
ここで，対角行列を，
\begin{gather}
[R_0] =
\begin{pmatrix}
R_{01} & 0 \\ 0 & R_{02}
\end{pmatrix}, \ \ \ \ \
[U] =
\begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix}, \ \ \ \ \
[\gamma] =
\begin{pmatrix}
\gamma_1 & 0 \\ 0 & \gamma_2
\end{pmatrix}
\nonumber
\end{gather}
とおくと，
\begin{eqnarray}
[\hat{Z}]
&=& [R_0] \big( [U] +[\gamma] \big) \big( [U] -[\gamma] \big) ^{-1}
\nonumber \\
&=& [R_0] \big( [U] -[\gamma] \big) ^{-1} \big( [U] +[\gamma] \big)
\nonumber \\
&=& \left[ \sqrt{R_0} \right] \big( [U] -[\gamma] \big) ^{-1} \big( [U] +[\gamma] \big)
\left[ \sqrt{R_0} \right]
\label{eq:hatZR0}
\end{eqnarray}
複素共役をとると，
\begin{eqnarray}
[\hat{Z}]^*
&=& [R_0] \big( [U] +[\gamma]^* \big) \big( [U] -[\gamma]^* \big) ^{-1}
\nonumber \\
&=& [R_0] \big( [U] -[\gamma]^* \big) ^{-1} \big( [U] +[\gamma]^* \big)
\nonumber \\
&=& \left[ \sqrt{R_0} \right] \big( [U] -[\gamma]^* \big) ^{-1} \big( [U] +[\gamma]^* \big)
\left[ \sqrt{R_0} \right]
\label{eq:hatZcR0}
\end{eqnarray}
対角行列\([\hat{Z}]\)の実部の要素\(\hat{R}_i \ (i=1,2)\)を次のように変形する．
\begin{eqnarray}
\hat{R}_i &=& \frac{\hat{Z}_i + \hat{Z}_i^*}{2}
\nonumber \\
&=& \frac{1}{2} \left( \frac{1+\gamma_i}{1-\gamma_i} R_{0i}
+ \frac{1+\gamma_i^*}{1-\gamma_i^*} R_{0i} \right)
\nonumber \\
&=& \frac{R_{0i}}{2} \cdot \frac{(1+\gamma_i)(1-\gamma_i^*)+(1+\gamma_i^*)(1-\gamma_i)}{(1-\gamma_i)(1-\gamma_i^*)}
\nonumber \\
&=& R_{0i} \frac{1-\gamma_i \gamma_i^*}{(1-\gamma_i)(1-\gamma_i^*)}
\nonumber
\end{eqnarray}
これより，
\begin{gather}
\sqrt{\hat{R}_i} = \sqrt{R_{0i}} \sqrt{\frac{1-\gamma_i \gamma_i^*}{(1-\gamma_i)(1-\gamma_i^*)}}
\nonumber
\end{gather}
いま，
\begin{eqnarray}
\Lambda_i
&\equiv& (1-\gamma_i^*) \sqrt{\frac{1-\gamma_i \gamma_i^*}{(1-\gamma_i)(1-\gamma_i^*)}}
\nonumber \\
&=& (1-\gamma_i^*) \sqrt{\frac{1-|\gamma_i|^2}{|1-\gamma_i^*|^2}}
\nonumber \\
&=& \frac{1-\gamma_i^*}{|1-\gamma_i^*|} \sqrt{1-|\gamma_i|^2}
\end{eqnarray}
を定義すると，次式が成り立つ．
\begin{eqnarray}
\Lambda_i \Lambda_i^*
&=& (1-\gamma_i^*) (1-\gamma_i) \frac{1-\gamma_i \gamma_i^*}{(1-\gamma_i)(1-\gamma_i^*)}
\nonumber \\
&=& 1-\gamma_i \gamma_i^*
\end{eqnarray}
また，
\begin{gather}
\sqrt{\frac{1-\gamma_i \gamma_i^*}{(1-\gamma_i)(1-\gamma_i^*)}}
= \frac{\Lambda_i}{1-\gamma_i^*} = \frac{\Lambda_i^*}{1-\gamma_i}
\end{gather}
これより，\(\sqrt{\hat{R}_i}\)，\(1/\sqrt{\hat{R}_i}\)は，
\begin{eqnarray}
\sqrt{\hat{R}_i}
&=& \sqrt{R_{0i}} \sqrt{\frac{1-\gamma_i \gamma_i^*}{(1-\gamma_i)(1-\gamma_i^*)}}
\nonumber \\
&=& \sqrt{R_{0i}} \frac{\Lambda_i^*}{1-\gamma_i}
\\
\frac{1}{\sqrt{\hat{R}_i}}
&=& \frac{1}{\sqrt{R_{0i}}} \frac{1-\gamma_i}{\Lambda_i^*}
\nonumber \\
&=& \frac{1}{\sqrt{R_{0i}}} \frac{1-\gamma_i^*}{\Lambda_i}
\end{eqnarray}
行列表示して，
\begin{eqnarray}
\left[ \sqrt{\hat{R}} \right]
&=& \left[ \sqrt{R_0} \right] \big( [U] - [\gamma] \big) ^{-1} [\Lambda]^*
\label{eq:hatR}\\
\left[ \sqrt{\hat{R}} \right]^{-1}
&=& \left[ \sqrt{R_0} \right]^{-1} [\Lambda]^{-1} \big( [U] - [\gamma]^* \big)
\nonumber \\
&=& \left[ \sqrt{R_0} \right]^{-1} [\Lambda]^{-1*} \big( [U] - [\gamma] \big)
\label{eq:hatRi}
\end{eqnarray}