[Circuit] Transmission line

電路知識：Transmission line
工具：Qucs
線上模擬

簡介：Transmission line 的推導

$\alpha$：衰減常數 (attenuation constant) ， napper/m
$\beta$：相位常數 (phase constant) ， rad/m
$$\begin{array}{rl}\mathbf{V}(z)& ={\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}+{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z}\\ \mathbf{I}(z)& ={\mathbf{I}}_{\mathbf{f}}{e}^{-\gamma z}-{\mathbf{I}}_{\mathbf{r}}{e}^{\gamma z}\\ \gamma =\alpha +j\beta & =\sqrt{(R+jwL)(G+jwC)}\\ v_{phase}& =\frac{z_2-z_1}{t_2-t_1}=\frac{w}{\beta }=c=f\cdot \lambda \\ \lambda & =\frac{2\pi }{\beta }\\ {\mathbf{Z}}_{\mathbf{0}}=\frac{{\mathbf{V}}_{\mathbf{f}}}{{\mathbf{I}}_{\mathbf{f}}}=\frac{{\mathbf{V}}_{\mathbf{r}}}{{\mathbf{I}}_{\mathbf{r}}}& =\frac{R+jwL}{\gamma }=\sqrt{\frac{R+jwL}{G+jwC}}\end{array}$$

假設一段無限短傳輸線的等效電路，如圖
$R=\mathrm{單}\mathrm{位}\mathrm{長}\mathrm{度}\mathrm{之}\mathrm{串}\mathrm{聯}\mathrm{電}\mathrm{阻}，\Omega /m$
$L=\mathrm{單}\mathrm{位}\mathrm{長}\mathrm{度}\mathrm{之}\mathrm{串}\mathrm{聯}\mathrm{電}\mathrm{感}，H/m$
$G=\mathrm{單}\mathrm{位}\mathrm{長}\mathrm{度}\mathrm{之}\mathrm{並}\mathrm{聯}\mathrm{電}\mathrm{導}，S/m$
$C=\mathrm{單}\mathrm{位}\mathrm{長}\mathrm{度}\mathrm{之}\mathrm{並}\mathrm{聯}\mathrm{電}\mathrm{容}，F/m$
$$\begin{array}{rl}v-iR\mathrm{\Delta }z-L\mathrm{\Delta }z\frac{\mathrm{d}i}{\mathrm{d}t}& =v+\mathrm{\Delta }v\\ \mathbf{V}-\mathbf{I}R\mathrm{\Delta }z-jwL\mathrm{\Delta }z\mathbf{I}& =\mathbf{V}+\mathrm{\Delta }\mathbf{V}\\ -\mathbf{I}R\mathrm{\Delta }z-jwL\mathrm{\Delta }z\mathbf{I}& =\mathrm{\Delta }\mathbf{V}\\ -\mathbf{I}R-jwL\mathbf{I}& =\frac{\mathrm{\Delta }\mathbf{V}}{\mathrm{\Delta }z}\\ \frac{\mathrm{\Delta }\mathbf{V}}{\mathrm{\Delta }z}=-(R+& jwL)\mathbf{I}\\ \\ i-G\mathrm{\Delta }z(v+\mathrm{\Delta }v)-C\mathrm{\Delta }z\frac{\mathrm{d}(v+\mathrm{\Delta }v)}{\mathrm{d}t}& =i+\mathrm{\Delta }i\\ \mathbf{I}-G\mathrm{\Delta }z(\mathbf{V}+\mathrm{\Delta }\mathbf{V})-jwC\mathrm{\Delta }z(\mathbf{V}+\mathrm{\Delta }\mathbf{V})& =\mathbf{I}+\mathrm{\Delta }\mathbf{I}\\ -G\mathrm{\Delta }z(\mathbf{V}+\mathrm{\Delta }\mathbf{V})-jwC\mathrm{\Delta }z(\mathbf{V}+\mathrm{\Delta }\mathbf{V})& =\mathrm{\Delta }\mathbf{I}\\ -G(\mathbf{V}+\mathrm{\Delta }\mathbf{V})-jwC(\mathbf{V}+\mathrm{\Delta }\mathbf{V})& =\frac{\mathrm{\Delta }\mathbf{I}}{\mathrm{\Delta }z}\\ \frac{\mathrm{\Delta }\mathbf{I}}{\mathrm{\Delta }z}=-(G+jwC)(\mathbf{V}+& \mathrm{\Delta }\mathbf{V})\end{array}$$ 當 $\mathrm{\Delta }\mathbf{V}\to 0$ ， $\mathrm{\Delta }\mathbf{I}\to 0$ ， $\mathrm{\Delta }z\to 0$
$$\begin{array}{rl}& \{\begin{array}{ccc}\frac{\mathrm{d}\mathbf{V}(z)}{\mathrm{d}z}& =& -(R+jwL)\mathbf{I}(z)\\ \frac{\mathrm{d}\mathbf{I}(z)}{\mathrm{d}z}& =& -(G+jwC)\mathbf{V}(z)\end{array}\\ & \{\begin{array}{ccccc}\frac{{\mathrm{d}}^2\mathbf{V}(z)}{\mathrm{d}{z}^2}& =& -(R+jwL)\frac{\mathrm{d}\mathbf{I}(z)}{\mathrm{d}z}& =& (R+jwL)(G+jwC)\mathbf{V}(z)\\ \frac{{\mathrm{d}}^2\mathbf{I}(z)}{\mathrm{d}{z}^2}& =& -(G+jwC)\frac{\mathrm{d}\mathbf{V}(z)}{\mathrm{d}z}& =& (G+jwC)(R+jwL)\mathbf{I}(z)\end{array}\\ & \{\begin{array}{ccc}\frac{{\mathrm{d}}^2\mathbf{V}(z)}{\mathrm{d}{z}^2}-{\gamma }^2\mathbf{V}(z)& =& 0\\ \frac{{\mathrm{d}}^2\mathbf{I}(z)}{\mathrm{d}{z}^2}-{\gamma }^2\mathbf{I}(z)& =& 0\end{array}\\ \ast & \gamma =\sqrt{(R+jwL)(G+jwC)}\end{array}$$ 利用 Laplace 轉換解二階微方
$$\begin{array}{rl}{f}^{″}(x)-{\gamma }^{2}f(x)& =0\\ s^{2}F(s)-sf(0)-f^{{}^{\prime }}(0)-{\gamma }^{2}F(s)& =0\\ (s^2-{\gamma }^2)F(s)& =sf(0)+f^{\prime }(0)\\ F(s)& =\frac{sf(0)+f^{\prime }(0)}{(s^2-{\gamma }^2)}\\ F(s)& =\frac{A}{s-\gamma }+\frac{B}{s+\gamma }\\ A& =(F(s)\times (s-\gamma ))(\gamma )\\ & =(\frac{sf(0)+f^{\prime }(0)}{s+\gamma })(\gamma )\\ & =\frac{\gamma f(0)+f^{\prime }(0)}{\gamma +\gamma }\\ & =\frac{\gamma f(0)+f^{\prime }(0)}{2\gamma }\\ B& =(F(s)\times (s+\gamma ))(-\gamma )\\ & =(\frac{sf(0)+f^{{}^{\prime }}(0)}{s-\gamma })(-\gamma )\\ & =\frac{-\gamma f(0)+f^{{}^{\prime }}(0)}{-\gamma -\gamma }\\ & =\frac{-\gamma f(0)+f^{{}^{\prime }}(0)}{-2\gamma }\\ & =\frac{\gamma f(0)-f^{\prime }(0)}{2\gamma }\\ F(s)& =\frac{\frac{\gamma f(0)+f^{\prime }(0)}{2\gamma }}{s-\gamma }+\frac{\frac{\gamma f(0)-f^{\prime }(0)}{2\gamma }}{s+\gamma }\\ f(x)& =\frac{\gamma f(0)+f^{\prime }(0)}{2\gamma }{e}^{\gamma x}+\frac{\gamma f(0)-f^{\prime }(0)}{2\gamma }{e}^{-\gamma x}\end{array}$$ 目前只能到這步，因不知初始條件無法解
猜測當初也是觀察而來，且是在負載端，所以 $z=0$ 才會在負載端，利用公式反推初始條件
$$\begin{array}{rl}\mathbf{V}(0)& ={\mathbf{V}}_{\mathbf{f}}+{\mathbf{V}}_{\mathbf{r}}\\ {\mathbf{V}}^{\prime }(0)& =-\gamma {\mathbf{V}}_{\mathbf{f}}+\gamma {\mathbf{V}}_{\mathbf{r}}\\ \mathbf{I}(0)& ={\mathbf{I}}_{\mathbf{f}}-{\mathbf{I}}_{\mathbf{r}}\\ {\mathbf{I}}^{\prime }(0)& =-\gamma {\mathbf{I}}_{\mathbf{f}}-\gamma {\mathbf{I}}_{\mathbf{r}}\end{array}$$ 可得公式，$e^{-\gamma z}$ 項表示波沿 +z 方向傳播；反之，$e^{\gamma z}$ 項為波沿著 -z 方向傳播
因波只會隨著距離加長而減弱
$$\begin{array}{rl}\ast \mathbf{V}(z)& ={\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}+{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z}\\ \ast \mathbf{I}(z)& ={\mathbf{I}}_{\mathbf{f}}{e}^{-\gamma z}-{\mathbf{I}}_{\mathbf{r}}{e}^{\gamma z}\end{array}$$ 將 $e^{-\gamma z}$ 單獨拿出來，轉回時域
$$\begin{array}{rl}v(z)& ={\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}\\ v(z,t)& ={\mathbf{V}}_{\mathbf{f}}{e}^{jwt}{e}^{-\gamma z}\\ & ={\mathbf{V}}_{\mathbf{f}}{e}^{jwt}{e}^{-\alpha z-j\beta z}\\ & ={\mathbf{V}}_{\mathbf{f}}{e}^{-\alpha z+j(wt-\beta )z}\end{array}$$ 可以發現 $wt-\beta$ 便是相位速度，也就是波前進的速度
或從單一點來看，是這點來回振盪的速度。為何會振盪，因為波通過此點
因光速為定值
$$\begin{array}{rl}w{t}_1-\beta z_1& =w{t}_2-\beta z_2\\ \ast v_{phase}& =\frac{z_2-z_1}{t_2-t_1}=\frac{w}{\beta }=c=f\cdot \lambda \\ \ast \lambda & =\frac{2\pi }{\beta }\end{array}$$ 那麼特徵阻抗 (Characteristic Impdance)呢？
$$\begin{array}{rl}\frac{\mathrm{d}\mathbf{V}(z)}{\mathrm{d}z}& =-(R+jwL)\mathbf{I}(z)\\ \mathbf{I}(z)& =-\frac{1}{R+jwL}\frac{\mathrm{d}\mathbf{V}(z)}{\mathrm{d}z}\\ & =-\frac{1}{R+jwL}(-\gamma {\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}+\gamma {\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z})\\ & =\frac{\gamma }{R+jwL}({\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}-{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z})\\ & =\frac{\sqrt{(R+jwL)(G+jwC)}}{R+jwL}({\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}-{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z})\\ & =\sqrt{\frac{G+jwC}{R+jwL}}({\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}-{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z})\\ & ={\mathbf{I}}_{\mathbf{f}}{e}^{-\gamma z}-{\mathbf{I}}_{\mathbf{r}}{e}^{\gamma z}\\ \ast {\mathbf{Z}}_{\mathbf{0}}& =\frac{{\mathbf{V}}_{\mathbf{f}}}{{\mathbf{I}}_{\mathbf{f}}}=\frac{{\mathbf{V}}_{\mathbf{r}}}{{\mathbf{I}}_{\mathbf{r}}}=\frac{R+jwL}{\gamma }=\sqrt{\frac{R+jwL}{G+jwC}}\end{array}$$ 故特徵阻抗 (Characteristic Impdance) 不算是認知上的阻抗，而是運算過程中，得到類似阻抗的解

特別情況

無損耗傳輸線
$$\begin{array}{rl}\gamma =\alpha +j\beta & =jwLC\Rightarrow \{\begin{array}{ccc}\alpha & =& 0\\ \beta & =& w\sqrt{LC}\end{array}\\ v_{phase}& =\frac{1}{\sqrt{LC}}\\ \lambda & =\frac{2\pi }{w\sqrt{LC}}\\ {\mathbf{Z}}_{\mathbf{0}}& =\sqrt{\frac{L}{C}}\end{array}$$

$$\begin{array}{rl}\gamma & =\alpha +j\beta \\ & =\sqrt{(R+jwL)(G+jwC)}\\ & =\sqrt{(0+jwL)(0+jwC)}\\ & =jwLC\Rightarrow \{\begin{array}{ccc}\alpha & =& 0\\ \beta & =& w\sqrt{LC}\end{array}\\ \\ v_{phase}& =\frac{w}{\beta }=\frac{1}{\sqrt{LC}}\\ \lambda & =\frac{2\pi }{\beta }=\frac{2\pi }{w\sqrt{LC}}\\ \\ {\mathbf{Z}}_{\mathbf{0}}& =\sqrt{\frac{R+jwL}{G+jwC}}\\ & =\sqrt{\frac{0+jwL}{0+jwC}}\\ & =\sqrt{\frac{L}{C}}\end{array}$$

有終端負載的傳輸線

反射係數 $\mathrm{\Gamma }$
$$\begin{array}{rl}\mathrm{\Gamma }(z=0)& =\frac{{\mathbf{Z}}_{\mathbf{L}}\mathbf{-}{\mathbf{Z}}_{\mathbf{0}}}{{\mathbf{Z}}_{\mathbf{L}}\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}}\\ \mathrm{\Gamma }(z)& =\frac{{\mathbf{V}}_{\mathbf{r}}}{{\mathbf{V}}_{\mathbf{f}}}{e}^{2\gamma z}=\mathrm{\Gamma }(z=0)e^{2\gamma z}\\ \\ {\mathbf{Z}}_{\mathbf{in}}(z)& =\frac{{\mathbf{Z}}_{\mathbf{L}}(e^{-\gamma z}+e^{\gamma z})+{\mathbf{Z}}_{\mathbf{0}}(e^{-\gamma z}-e^{\gamma z})}{{\mathbf{Z}}_{\mathbf{L}}(e^{-\gamma z}-e^{\gamma z})+{\mathbf{Z}}_{\mathbf{0}}(e^{-\gamma z}+e^{\gamma z})}{\mathbf{Z}}_{\mathbf{0}}=\frac{{\mathbf{Z}}_{\mathbf{L}}cosh(\gamma z)-{\mathbf{Z}}_{\mathbf{0}}sinh(\gamma z)}{-{\mathbf{Z}}_{\mathbf{L}}sinh(\gamma z)+{\mathbf{Z}}_{\mathbf{0}}cosh(\gamma z)}{\mathbf{Z}}_{\mathbf{0}}\end{array}$$

任意點的反射係數，也就是往前的波跟往後的波的比例
$$\begin{array}{rl}\mathrm{\Gamma }(z)& =\frac{{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z}}{{\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}}\\ & =\frac{{\mathbf{V}}_{\mathbf{r}}}{{\mathbf{V}}_{\mathbf{f}}}{e}^{2\gamma z}\end{array}$$ 那麼在 $z=0$ 的反射係數呢？ $$\begin{array}{rl}\mathbf{V}(z)& ={\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}+{\mathbf{V}}_{\mathbf{f}}{e}^{\gamma z}\\ \mathbf{I}(z)& ={\mathbf{I}}_{\mathbf{f}}{e}^{-\gamma z}-{\mathbf{I}}_{\mathbf{r}}{e}^{\gamma z}\\ & =\frac{{\mathbf{V}}_{\mathbf{f}}}{{\mathbf{Z}}_{\mathbf{0}}}{e}^{-\gamma z}-\frac{{\mathbf{V}}_{\mathbf{r}}}{{\mathbf{Z}}_{\mathbf{0}}}{e}^{\gamma z}\end{array}$$ $$\begin{array}{rl}{\mathbf{Z}}_{\mathbf{L}}& =\frac{\mathbf{V}(z=0)}{\mathbf{I}(z=0)}\\ & =\frac{{\mathbf{V}}_{\mathbf{f}}+{\mathbf{V}}_{\mathbf{r}}}{{\mathbf{I}}_{\mathbf{f}}-{\mathbf{I}}_{\mathbf{r}}}\\ & =\frac{{\mathbf{V}}_{\mathbf{f}}+{\mathbf{V}}_r}{{\mathbf{V}}_{\mathbf{f}}-{\mathbf{V}}_{\mathbf{r}}}{\mathbf{Z}}_{\mathbf{0}}\\ \\ ({\mathbf{V}}_{\mathbf{f}}-{\mathbf{V}}_{\mathbf{r}}){\mathbf{Z}}_{\mathbf{L}}& =({\mathbf{V}}_{\mathbf{f}}+{\mathbf{V}}_{\mathbf{r}}){\mathbf{Z}}_{\mathbf{0}}\\ {\mathbf{V}}_{\mathbf{f}}({\mathbf{Z}}_{\mathbf{L}}-{\mathbf{Z}}_{\mathbf{0}})& ={\mathbf{V}}_{\mathbf{r}}({\mathbf{Z}}_{\mathbf{L}}+{\mathbf{Z}}_{\mathbf{0}})\\ {\mathbf{V}}_{\mathbf{r}}& =\frac{{\mathbf{Z}}_{\mathbf{L}}\mathbf{-}{\mathbf{Z}}_{\mathbf{0}}}{{\mathbf{Z}}_{\mathbf{L}}\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}}{\mathbf{V}}_{\mathbf{f}}=\mathrm{\Gamma }(z=0){\mathbf{V}}_{\mathbf{f}}\\ \ast \frac{{\mathbf{V}}_{\mathbf{r}}}{{\mathbf{V}}_{\mathbf{f}}}& =\mathrm{\Gamma }(z=0)=\frac{{\mathbf{Z}}_{\mathbf{L}}\mathbf{-}{\mathbf{Z}}_{\mathbf{0}}}{{\mathbf{Z}}_{\mathbf{L}}\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}}\end{array}$$ 任意點的反射係數可被重寫
$$\begin{array}{rl}\ast \mathrm{\Gamma }(z)& =\frac{{\mathbf{V}}_{\mathbf{r}}}{{\mathbf{V}}_{\mathbf{f}}}{e}^{2\gamma z}=\mathrm{\Gamma }(z=0)e^{2\gamma z}\\ \\ \mathrm{\Gamma }(z=-l)& =\frac{{\mathbf{V}}_{\mathbf{r}}}{{\mathbf{V}}_{\mathbf{f}}}{e}^{2\gamma (-l)}=\mathrm{\Gamma }(z=0)e^{-2\gamma l}\end{array}$$ 那麼 ${\mathbf{Z}}_{\mathbf{in}}(z)$ 呢？
$$\begin{array}{rl}{\mathbf{Z}}_{\mathbf{in}}(z)& =\frac{\mathbf{V}(z)}{\mathbf{I}(z)}\\ & =\frac{{\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}+{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z}}{{\mathbf{I}}_{\mathbf{f}}{e}^{-\gamma z}-{\mathbf{I}}_{\mathbf{r}}{e}^{\gamma z}}\\ & =\frac{{\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}+{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z}}{\frac{{\mathbf{V}}_{\mathbf{f}}}{{\mathbf{Z}}_{\mathbf{0}}}{e}^{-\gamma z}-\frac{{\mathbf{V}}_{\mathbf{r}}}{{\mathbf{Z}}_{\mathbf{0}}}{e}^{\gamma z}}\\ & =\frac{{\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}+{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z}}{{\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}-{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z}}{\mathbf{Z}}_{\mathbf{0}}\\ & =\frac{{\mathbf{V}}_{\mathbf{f}}(e^{-\gamma z}+\mathrm{\Gamma }(z=0)e^{\gamma z})}{{\mathbf{V}}_{\mathbf{f}}(e^{-\gamma z}-\mathrm{\Gamma }(z=0)e^{\gamma z})}{\mathbf{Z}}_{\mathbf{0}}\\ & =\frac{1+\mathrm{\Gamma }(z=0)e^{2\gamma z}}{1-\mathrm{\Gamma }(z=0)e^{2\gamma z}}{\mathbf{Z}}_{\mathbf{0}}\\ & =\frac{1+\frac{{\mathbf{Z}}_{\mathbf{L}}\mathbf{-}{\mathbf{Z}}_{\mathbf{0}}}{{\mathbf{Z}}_{\mathbf{L}}\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}}{e}^{2\gamma z}}{1-\frac{{\mathbf{Z}}_{\mathbf{L}}\mathbf{-}{\mathbf{Z}}_{\mathbf{0}}}{{\mathbf{Z}}_{\mathbf{L}}\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}}{e}^{2\gamma z}}{\mathbf{Z}}_{\mathbf{0}}\\ & =\frac{{\mathbf{Z}}_{\mathbf{L}}\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}+({\mathbf{Z}}_{\mathbf{L}}\mathbf{-}{\mathbf{Z}}_{\mathbf{0}})e^{2\gamma z}}{{\mathbf{Z}}_{\mathbf{L}}\mathbf{+}{\mathbf{Z}}_{\mathbf{0}}-({\mathbf{Z}}_{\mathbf{L}}\mathbf{-}{\mathbf{Z}}_{\mathbf{0}})e^{2\gamma z}}{\mathbf{Z}}_{\mathbf{0}}\\ & =\frac{{\mathbf{Z}}_{\mathbf{L}}(1+e^{2\gamma z})+{\mathbf{Z}}_{\mathbf{0}}(1-e^{2\gamma z})}{{\mathbf{Z}}_{\mathbf{L}}(1-e^{2\gamma z})+{\mathbf{Z}}_{\mathbf{0}}(1+e^{2\gamma z})}{\mathbf{Z}}_{\mathbf{0}}\\ & =\frac{{\mathbf{Z}}_{\mathbf{L}}(e^{-\gamma z}+e^{\gamma z})+{\mathbf{Z}}_{\mathbf{0}}(e^{-\gamma z}-e^{\gamma z})}{{\mathbf{Z}}_{\mathbf{L}}(e^{-\gamma z}-e^{\gamma z})+{\mathbf{Z}}_{\mathbf{0}}(e^{-\gamma z}+e^{\gamma z})}{\mathbf{Z}}_{\mathbf{0}}\\ & =\frac{{\mathbf{Z}}_{\mathbf{L}}(2cosh(\gamma z))+{\mathbf{Z}}_{\mathbf{0}}(-2sinh(\gamma z))}{{\mathbf{Z}}_{\mathbf{L}}(-2sinh(\gamma z))+{\mathbf{Z}}_{\mathbf{0}}(2cosh(\gamma z))}{\mathbf{Z}}_{\mathbf{0}}\\ \ast & =\frac{{\mathbf{Z}}_{\mathbf{L}}cosh(\gamma z)-{\mathbf{Z}}_{\mathbf{0}}sinh(\gamma z)}{-{\mathbf{Z}}_{\mathbf{L}}sinh(\gamma z)+{\mathbf{Z}}_{\mathbf{0}}cosh(\gamma z)}{\mathbf{Z}}_{\mathbf{0}}\end{array}$$

若傳輸線長度為 $l$ or $\mathrm{\infty }$
$$\begin{array}{rl}\mathrm{\Gamma }(z=-l)& =\mathrm{\Gamma }(z=0)e^{-2\gamma l}\\ \mathrm{\Gamma }(z=-\mathrm{\infty })& =0\\ \\ {\mathbf{Z}}_{\mathbf{in}}(z=0)& =\frac{2{\mathbf{Z}}_{\mathbf{L}}}{2{\mathbf{Z}}_{\mathbf{0}}}{\mathbf{Z}}_{\mathbf{0}}={\mathbf{Z}}_{\mathbf{L}}\\ {\mathbf{Z}}_{\mathbf{in}}(z=-l)& =\frac{{\mathbf{Z}}_{\mathbf{L}}(e^{\gamma l}+e^{-\gamma l})+{\mathbf{Z}}_{\mathbf{0}}(e^{\gamma l}-e^{-\gamma l})}{{\mathbf{Z}}_{\mathbf{L}}(e^{\gamma l}-e^{-\gamma l})+{\mathbf{Z}}_{\mathbf{0}}(e^{\gamma l}+e^{-\gamma l})}{\mathbf{Z}}_{\mathbf{0}}\\ {\mathbf{Z}}_{\mathbf{in}}(z=-\mathrm{\infty })& =\frac{{\mathbf{Z}}_{\mathbf{L}}+{\mathbf{Z}}_{\mathbf{0}}}{{\mathbf{Z}}_{\mathbf{L}}+{\mathbf{Z}}_{\mathbf{0}}}{\mathbf{Z}}_{\mathbf{0}}={\mathbf{Z}}_{\mathbf{0}}\end{array}$$

模擬結果
簡單範例

之前推導的皆是穩態 (因有轉換為 Phasor domain，穩態原因)

1. 以暫態來看，Power 一開始送出，不會知道導線多長，所以會認為導線無限長
   此時 ${\mathbf{Z}}_{\mathbf{in}}(z=-\mathrm{\infty })={\mathbf{Z}}_{\mathbf{0}}$，電壓會是分壓下的結果 $1\times \frac{150}{200}=0.75V$
2. 訊號到達負載端，因 $\mathbf{V}(0)={\mathbf{V}}_{\mathbf{f}}+{\mathbf{V}}_{\mathbf{r}}={\mathbf{V}}_{\mathbf{f}}+\mathrm{\Gamma }(z=0){\mathbf{V}}_{\mathbf{f}}$
   故負載電壓為 $0.75+0.75\times \frac{50-150}{200}=0.375V$，反射電壓為 $0.75\times \frac{50-150}{200}=-3.75V$
3. 訊號又回到起點，因可看作從負載端送來 ${\mathbf{V}}_{\mathbf{f}}$，仍可使用 $\mathbf{V}(0)={\mathbf{V}}_{\mathbf{f}}+{\mathbf{V}}_{\mathbf{r}}={\mathbf{V}}_{\mathbf{f}}+\mathrm{\Gamma }(z=0){\mathbf{V}}_{\mathbf{f}}$
   故起點為 $-3.75+(-3.75)\times \frac{50-150}{200}=-0.1875V$，反射電壓為 $-3.75\times \frac{50-150}{200}=0.1875V$

從範例可知特徵阻抗 (Characteristic Impdance) 不完全算是阻值，比較像是能量傳遞的介質

並聯範例

1. 以暫態來看，Power 一開始送出，不會知道導線多長，又無電阻，所以全力送出 $1V$
2. 到達 $\mathrm{vin}2$ 時，因後兩端導線長度未知，所以會認為導線無限長
   此時因 ${\mathbf{Z}}_{\mathbf{in}}(z=-\mathrm{\infty })={\mathbf{Z}}_{\mathbf{0}}$，${\mathbf{Z}}_{\mathbf{in}}=50+50//50=75\mathrm{\Omega }$
   故 $\mathrm{vin}2={\mathbf{V}}_{\mathbf{f}}+{\mathbf{V}}_{\mathbf{r}}={\mathbf{V}}_{\mathbf{f}}+\mathrm{\Gamma }(z=0){\mathbf{V}}_{\mathbf{f}}=1.2V$
   反射電壓為 $\mathrm{\Gamma }(z=0){\mathbf{V}}_{\mathbf{f}}=\frac{75-50}{75+50}\times 1=0.2V$
   因電阻分壓 $\mathrm{vmid}=1.2\times \frac{50//50}{{\mathbf{Z}}_{\mathbf{in}}}=0.4V$
3. 到達 $\mathrm{vout}1$ & $\mathrm{vout}2$ 時，因阻抗匹配，故完全接受為 $0.4V$

特別情況
波長 $\lambda \gg$ 線長 $l$

$$\begin{array}{rl}\gamma & =\alpha +j\beta =\alpha +j\frac{2\pi }{\lambda }\\ {\mathbf{Z}}_{\mathbf{0}}& =\frac{R+jwL}{\gamma }=\frac{R+jwL}{\alpha +j\frac{2\pi }{\lambda }}\\ {\mathbf{Z}}_{\mathbf{in}}(z)& =\frac{{\mathbf{Z}}_{\mathbf{L}}(e^{-\gamma z}+e^{\gamma z})+{\mathbf{Z}}_{\mathbf{0}}(e^{-\gamma z}-e^{\gamma z})}{{\mathbf{Z}}_{\mathbf{L}}(e^{-\gamma z}-e^{\gamma z})+{\mathbf{Z}}_{\mathbf{0}}(e^{-\gamma z}+e^{\gamma z})}{\mathbf{Z}}_{\mathbf{0}}\\ & =\frac{{\mathbf{Z}}_{\mathbf{L}}(e^{-(\alpha +j\frac{2\pi }{\lambda })z}+e^{(\alpha +j\frac{2\pi }{\lambda })z})+{\mathbf{Z}}_{\mathbf{0}}(e^{-(\alpha +j\frac{2\pi }{\lambda })z}-e^{(\alpha +j\frac{2\pi }{\lambda })z})}{{\mathbf{Z}}_{\mathbf{L}}(e^{-(\alpha +j\frac{2\pi }{\lambda })z}-e^{(\alpha +j\frac{2\pi }{\lambda })z})+{\mathbf{Z}}_{\mathbf{0}}(e^{-(\alpha +j\frac{2\pi }{\lambda })z}+e^{(\alpha +j\frac{2\pi }{\lambda })z})}{\mathbf{Z}}_{\mathbf{0}}\\ & \because \lambda \gg z\\ & =\frac{{\mathbf{Z}}_{\mathbf{L}}(e^{-\alpha z}+e^{\alpha z})+{\mathbf{Z}}_{\mathbf{0}}(e^{-\alpha z}-e^{\alpha z})}{{\mathbf{Z}}_{\mathbf{L}}(e^{-\alpha z}-e^{\alpha z})+{\mathbf{Z}}_{\mathbf{0}}(e^{-\alpha z}+e^{\alpha z})}{\mathbf{Z}}_{\mathbf{0}}\\ \\ \mathbf{V}(z)& ={\mathbf{V}}_{\mathbf{f}}{e}^{-\gamma z}+{\mathbf{V}}_{\mathbf{r}}{e}^{\gamma z}\\ & ={\mathbf{V}}_{\mathbf{f}}{e}^{-(\alpha +j\frac{2\pi }{\lambda })z}+{\mathbf{V}}_{\mathbf{r}}{e}^{(\alpha +j\frac{2\pi }{\lambda })z}\\ & \because \lambda \gg z\\ & ={\mathbf{V}}_{\mathbf{f}}{e}^{-\alpha z}+{\mathbf{V}}_{\mathbf{r}}{e}^{\alpha z}\\ \\ \mathrm{\Gamma }(z)& =\mathrm{\Gamma }(z=0)e^{2\gamma z}\\ & =\mathrm{\Gamma }(z=0)e^{2(\alpha +j\frac{2\pi }{\lambda })z}\\ & \because \lambda \gg z\\ & =\mathrm{\Gamma }(z=0)e^{2\alpha z}\end{array}$$ 若為無損耗傳輸線 $\alpha =0$
$$\begin{array}{rl}{\mathbf{Z}}_{\mathbf{in}}(z)& =\frac{2{\mathbf{Z}}_{\mathbf{L}}}{2{\mathbf{Z}}_{\mathbf{0}}}{\mathbf{Z}}_{\mathbf{0}}={\mathbf{Z}}_{\mathbf{L}}\\ \mathrm{\Gamma }(z)& =\mathrm{\Gamma }(z=0)\\ \mathbf{V}(z)& ={\mathbf{V}}_{\mathbf{f}}+{\mathbf{V}}_{\mathbf{r}}={\mathbf{V}}_{\mathbf{f}}(1+\mathrm{\Gamma }(z=0))\end{array}$$ 即然在源端看到的負載 ${\mathbf{Z}}_{\mathbf{in}}(-l)={\mathbf{Z}}_{\mathbf{L}}$ 跟負載端一致，所以分壓會是正確的
而 $\mathbf{V}(z)$ 與 $z$ 無關，所以整條線上都是同樣的電壓，也就是正確的分壓

常見的經驗方法認為如果電纜或者電線的長度大於波長的1/10，則需被作為傳輸線處理

參考

傳輸線理論與阻抗匹配
傳輸線模型
Transmission Line vs Lumped Element