$$ \widetilde{Z}{in}=\frac{Z{in}(z)}{Z_0}\ Z_{in}(z)=Z_0\frac{1+\Gamma}{1-\Gamma}\ \widetilde{Z}{in}=\frac{1+\Gamma}{1-\Gamma}\ \Gamma=\frac{\widetilde{Z}{in}-1}{\widetilde{Z}_{in}+1} $$ 无量纲电长度 $$ \tau=\frac{l}{\lambda} $$ 圆图以$|\Gamma(z)|$为基底 $$ 0\le |\Gamma(z)|\le 1 $$
将阻抗(导纳),$\rho$套在等$|\Gamma|$圆上 $$ \Gamma=\Gamma_u+j\Gamma_v\ \Gamma(z)=\Gamma_L e^{-2j\beta z}=|\Gamma_L|e^{j(\phi_L-2\beta z)}\ \Gamma(z)周期为\frac{\lambda}{2},\tau 周期为\frac{1}{2}\ $$ 套覆阻抗圆 $$ \left{ \begin{array}{l} \widetilde{Z}_{in}(z)=\frac{1+\Gamma(z)}{1-\Gamma(z)}=\widetilde{R}+j\widetilde{X}\ \Gamma=\Gamma_u+j\Gamma_v \end{array}\right.\ $$ 等$\widetilde{R}圆$ $$ (\Gamma_u-\frac{\widetilde{R}}{1+\widetilde{R}})^2+\Gamma_v^2=(\frac{1}{1+\widetilde{R}})^2 $$ 等$\widetilde{X}$圆 $$ (\Gamma_u-1)^2+(\Gamma_v-\frac{1}{\widetilde{X}})^2=(\frac{1}{\widetilde{X}})^2 $$
$$ \rho=\frac{1+|\Gamma|}{1-| \Gamma|} $$ 作圆交实轴
匹配点(0,0) $$ \Gamma=0,\widetilde{R}=1,\widetilde{X}=0 $$ 短路点(-1,0) $$ \Gamma=-1,\widetilde{R}=\widetilde{X}=0\ $$ 开路点(1,0) $$ \Gamma=1,\varphi=0 $$ 单位圆 $$ |\Gamma|=1,\widetilde{Z_{in}}=j\widetilde{X} $$ 正实轴 $$ \widetilde{R}=\rho,1\lt\widetilde{R}\lt \infty $$ 负半轴 $$ \widetilde{R}=k $$
$$ \widetilde{Y}{in}(z)=\frac{1}{\widetilde{Z{in}}}=\frac{1-\Gamma(z)}{1+\Gamma(z)}=\frac{1+(-\Gamma(z))}{1-(-\Gamma(z))}=\frac{1+\Gamma_i(z)}{1-\Gamma_i(z)}\ \widetilde{Y}_{in}(z)=\widetilde{G}+j\widetilde{B}\ \widetilde{R}\Leftrightarrow\widetilde{G}\ \widetilde{X}\Leftrightarrow\widetilde{B}\ \Gamma(z)\Leftrightarrow\Gamma_i(z) $$ 短路开路对换 感性容性对换 波腹波节对换
由阻抗求导纳，只需将阻抗点在圆图上以原点为中心转$\pi$再读数
$$ \widetilde{Z}{in}(z)\widetilde{Z}{in}(z+\frac{\lambda}{4})=1 \ \widetilde{Z}{in}(z)=\frac{1}{\widetilde{Y}{in}(z)}\ \therefore \widetilde{Z}{in}(z+\frac{\lambda}{4})=\widetilde{Y}{in}(z) $$
$$ Z_{in}\rightarrow\widetilde{Z}{in}\rightarrow \widetilde{R}{in} ,\widetilde{X}_{in}\rightarrow \rho=\widetilde{R}\rightarrow |\Gamma|\rightarrow(\varphi,\tau) $$