\begin{enumerate} \item (1) 図2のようにループを考えると次の関係式が導かれる。 \begin{eqnarray} E_{1} &=& j\omega L I_{1} + \frac{1}{j\omega C}(I_{1}-I_{2})~~~~(1)\\ E_{2} &=& -j\omega L I_{2} + \frac{1}{j\omega C}(I_{1}-I_{2})~~~~(2) \end{eqnarray} (2)式より \begin{eqnarray} j\omega C E_{2} &=& I_{1} - j\omega C \left(j\omega L + \frac{1}{j\omega C}\right)I_{2}\\ I_{1}&=& j\omega C E_{2} + (1-\omega^{2}LC)I_{2} \end{eqnarray} (1)式へ代入する。 \begin{eqnarray} E_{1} &=& \left(j\omega L + \frac{1}{j\omega C}\right)I_{1} - \frac{1}{j\omega C}I_{2}\\ E_{1} &=& \left(j\omega L + \frac{1}{j\omega C}\right)\left(j\omega C E_{2} + (1-\omega^{2}LC)I_{2}\right)\nonumber\\ && - \frac{1}{j\omega C}I_{2}\nonumber\\ &=& j\omega C\left(j\omega L + \frac{1}{j\omega C}\right)E_{2}\nonumber\\ &&+ (1-\omega^{2}LC)\left(j\omega L + \frac{1}{j\omega C}\right)I_{2}- \frac{1}{j\omega C}I_{2}\nonumber\\ &=& (1-\omega^{2}LC)E_{2} + \left( j\omega L(1-\omega^{2}LC) - \frac{\omega L}{j} \right)I_{2}\nonumber\\ &=& (1-\omega^{2}LC)E_{2} + \left( j\omega L(1-\omega^{2}LC) + j\omega L \right)I_{2}\nonumber\\ &=& (1-\omega^{2}LC)E_{2} + \left( j\omega L(2-\omega^{2}LC) \right)I_{2} \end{eqnarray} よって，四端子定数は \begin{eqnarray} \begin{bmatrix} E_{1}\\ I_{1} \end{bmatrix} = \begin{bmatrix} 1-\omega^{2}LC & j\omega L(2-\omega^{2}LC) \\ j\omega C & 1-\omega^{2}LC \end{bmatrix} \begin{bmatrix} E_{2}\\ I_{2} \end{bmatrix} \end{eqnarray} から次のようになる。 \begin{eqnarray} && \underline{A = D = 1-\omega^{2}LC,~B = j\omega L(2-\omega^{2}LC)}\nonumber\\ && \underline{C = j\omega C} \end{eqnarray} \noindent (解法2) 2-2'を開放すると $I_{2}=0$ となる。 2-2'間の電圧 $E_{2}$ は \begin{eqnarray} E_{2} = \frac{\frac{1}{j\omega C}}{j\omega L + \frac{1}{j\omega C}}E_{1} = \frac{1}{1-\omega^{2}LC}E_{1} \end{eqnarray} よって，四端子定数の $A$ は \begin{eqnarray} A = \left[\frac{E_{1}}{E_{2}}\right]_{I_{2}=0} = 1-\omega^{2}LC \end{eqnarray} 電流 $I_{1}$ は \begin{eqnarray} I_{1} &=& \frac{1}{j\omega L + \frac{1}{j\omega C}}E_{1} = \frac{j\omega C}{1-\omega^{2}LC}E_{1}\nonumber\\ &=&\frac{j\omega C}{1-\omega^{2}LC}(1-\omega^{2}LC)E_{2}\nonumber\\ &=& j\omega C E_{2} \end{eqnarray} よって，四端子定数の $C$ は %\begin{eqnarray} %C = \left[ \frac{I_{1}}{E_{2}} \right]_{I_{2}=0} = j\omega C %\end{eqnarray} 2-2'を短絡すると $E_{2}=0$ となる。 電流$I_{1}$ は \begin{eqnarray} I_{1} = \frac{E_{1}}{ j\omega L + \frac{j\omega L\frac{1}{j\omega C}}{j\omega L+\frac{1}{j\omega C}}} \end{eqnarray} となり，分流の法則から$I_{2}$ は \begin{eqnarray} I_{2} &=& \frac{\frac{1}{j\omega C}} {j\omega L + \frac{1}{j\omega C}}I_{1} = \frac{\frac{1}{j\omega C}} {j\omega L + \frac{1}{j\omega C}} \frac{E_{1}}{ j\omega L + \frac{j\omega L\frac{1}{j\omega C}}{j\omega L+\frac{1}{j\omega C}}} \nonumber\\ &=& \frac{\frac{1}{j\omega C}E_{1}} {j\omega L\left(j\omega L+\frac{1}{j\omega C}\right)+\frac{j\omega L}{j\omega C} }\nonumber\\ &=& \frac{E_{1}} {j\omega L(1-\omega^{2}LC)+j\omega L} \nonumber\\ &=& \frac{E_{1}} {j\omega L(2-\omega^{2}LC)} \end{eqnarray} よって，四端子定数の $B$ は \begin{eqnarray} B = \left[\frac{E_{1}}{I_{2}}\right]_{E_{2}=0} = j\omega L(2-\omega^{2}LC) \end{eqnarray} また， \begin{eqnarray} I_{2} &=& \frac{\frac{1}{j\omega C}} {j\omega L + \frac{1}{j\omega C}}I_{1} =\frac{1}{1-\omega^{2}LC} \end{eqnarray} よって，四端子定数の $D$ は \begin{eqnarray} D = \left[\frac{I_{1}}{I_{2}}\right]_{E_{2}=0} =1-\omega^{2}LC \end{eqnarray} \item (2) (1)の結果より， \begin{eqnarray} Z_{01} &=& \sqrt{\frac{AB}{CD}} = \sqrt{\frac{(1-\omega^{2}LC)(j\omega L(2-\omega^{2}LC))}{j\omega C(1-\omega^{2}LC)}}\nonumber\\ &=& \sqrt{\frac{L}{C}(2-\omega^{2}LC)} \end{eqnarray} \begin{eqnarray} Z_{02} &=& \sqrt{\frac{BD}{CA}} = \sqrt{\frac{j\omega L(2-\omega^{2}LC)(1-\omega^{2}LC)}{ j\omega C(1-\omega^{2}LC)}}\nonumber\\ &=& \sqrt{\frac{L}{C}(2-\omega^{2}LC)} \end{eqnarray} \begin{eqnarray} \cosh \theta &=& \sqrt{AD} = \sqrt{(1-\omega^{2}LC)(1-\omega^{2}LC)}\nonumber\\ &=& 1-\omega^{2}LC \end{eqnarray} よって， \begin{eqnarray} && \underline{Z_{01}=Z_{02}=\sqrt{\frac{L}{C}(2-\omega^{2}LC)}}\\ && \underline{\theta = \cosh^{-1}(1-\omega^{2}LC)} \end{eqnarray} \end{enumerate} %=image:/media/2014/11/21/141656311252579200.png:図2 %=image:/media/2014/11/21/141656311357540300.png:図3 2-2’ を開放 %=image:/media/2014/11/21/141656311460898500.png:図4 2-2’ を短絡