[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
$$ \begin{align*} \mathbf{V}(z)&=\mathbf{V_f}e^{-\gamma z}+\mathbf{V_r}e^{\gamma z}\\ \mathbf{I}(z)&=\mathbf{I_f}e^{-\gamma z}-\mathbf{I_r}e^{\gamma z}\\ \gamma = \alpha +j\beta &= \sqrt{(R+jwL)(G+jwC)} \\ v_{phase}&=\frac{z_2-z_1}{t_2-t_1}=\frac{w}{\beta }=c=f\cdot \lambda \\ \lambda &= \frac{2\pi}{\beta }\\ \mathbf{Z_0}=\frac{\mathbf{V_f}}{\mathbf{I_f}}=\frac{\mathbf{V_r}}{\mathbf{I_r}}&=\frac{R+jwL}{\gamma }=\sqrt{\frac{R+jwL}{G+jwC}} \end{align*} $$

假設一段無限短傳輸線的等效電路，如圖
\(R=單位長度之串聯電阻，Ω/m\)
\(L=單位長度之串聯電感，H/m\)
\(G=單位長度之並聯電導，S/m\)
\(C=單位長度之並聯電容，F/m\)
$$ \begin{align*} v-iR\Delta z-L\Delta z\frac{\mathrm{d} i}{\mathrm{d} t} &= v+\Delta v\\ \mathbf{V}-\mathbf{I}R\Delta z-jwL\Delta z\mathbf{I} &= \mathbf{V}+\Delta \mathbf{V}\\ -\mathbf{I}R\Delta z-jwL\Delta z\mathbf{I} &= \Delta \mathbf{V}\\ -\mathbf{I}R-jwL\mathbf{I} &= \frac{\Delta \mathbf{V}}{\Delta z}\\ \frac{\Delta \mathbf{V}}{\Delta z}=-(R+&jwL)\mathbf{I}\\ \\ i-G\Delta z(v+\Delta v)-C\Delta z \frac{\mathrm{d} (v+\Delta v)}{\mathrm{d} t}&=i+\Delta i\\ \mathbf{I}-G\Delta z(\mathbf{V}+\Delta \mathbf{V})-jwC\Delta z (\mathbf{V}+\Delta \mathbf{V})&=\mathbf{I}+\Delta \mathbf{I}\\ -G\Delta z(\mathbf{V}+\Delta \mathbf{V})-jwC\Delta z (\mathbf{V}+\Delta \mathbf{V})&=\Delta \mathbf{I}\\ -G(\mathbf{V}+\Delta \mathbf{V})-jwC (\mathbf{V}+\Delta \mathbf{V})&=\frac{\Delta \mathbf{I}}{\Delta z}\\ \frac{\Delta \mathbf{I}}{\Delta z} = -(G+jwC)(\mathbf{V}+&\Delta \mathbf{V})\\ \end{align*} $$ 當 \(\Delta \mathbf{V}\rightarrow 0\) ， \(\Delta \mathbf{I}\rightarrow 0\) ， \(\Delta z\rightarrow 0\)
$$ \begin{align*} &\left\{\begin{matrix} \frac{\mathrm{d} \mathbf{V}(z)}{\mathrm{d} z} &=& -(R+jwL)\mathbf{I}(z) \\ \frac{\mathrm{d} \mathbf{I}(z)}{\mathrm{d} z} &=& -(G+jwC)\mathbf{V}(z) \\ \end{matrix}\right. \\ &\left\{\begin{matrix} \frac{\mathrm{d}^2 \mathbf{V}(z)}{\mathrm{d} z^2} &=& -(R+jwL)\frac{\mathrm{d} \mathbf{I}(z)}{\mathrm{d} z} &=&(R+jwL)(G+jwC)\mathbf{V}(z)\\ \frac{\mathrm{d}^2 \mathbf{I}(z)}{\mathrm{d} z^2} &=& -(G+jwC)\frac{\mathrm{d} \mathbf{V}(z)}{\mathrm{d} z} &=&(G+jwC)(R+jwL)\mathbf{I}(z)\\ \end{matrix}\right. \\ &\left\{\begin{matrix} \frac{\mathrm{d}^2 \mathbf{V}(z)}{\mathrm{d} z^2} -\gamma ^2\mathbf{V}(z)&=&0\\ \frac{\mathrm{d}^2 \mathbf{I}(z)}{\mathrm{d} z^2} -\gamma ^2\mathbf{I}(z)&=&0\\ \end{matrix}\right.\\ \ast &\gamma = \sqrt{(R+jwL)(G+jwC)}\\ \end{align*} $$ 利用 Laplace 轉換解二階微方
$$ \begin{align*} {f}''(x)-\gamma ^2f(x)&=0 \\ s^2F(s)-sf(0)-f^{'}(0)-\gamma ^2F(s)&=0\\ (s^2-\gamma ^2)F(s)&=sf(0)+{f}'(0)\\ F(s)&=\frac{sf(0)+{f}'(0)}{(s^2-\gamma ^2)}\\ F(s)&=\frac{A}{s-\gamma}+\frac{B}{s+\gamma}\\ A&=(F(s)\times (s-\gamma))(\gamma) \\ &=(\frac{sf(0)+{f}'(0)}{s+\gamma})(\gamma)\\ &=\frac{\gamma f(0)+{f}'(0)}{\gamma +\gamma}\\ &=\frac{\gamma f(0)+{f}'(0)}{2\gamma}\\ B&=(F(s)\times (s+\gamma))(-\gamma) \\ &=(\frac{sf(0)+f^{'}(0)}{s-\gamma})(-\gamma)\\ &=\frac{-\gamma f(0)+f^{'}(0)}{-\gamma -\gamma}\\ &=\frac{-\gamma f(0)+f^{'}(0)}{-2\gamma}\\ &=\frac{\gamma f(0)-{f}'(0)}{2\gamma}\\ F(s)&=\frac{\frac{\gamma f(0)+{f}'(0)}{2\gamma}}{s-\gamma}+\frac{\frac{\gamma f(0)-{f}'(0)}{2\gamma}}{s+\gamma}\\ f(x)&=\frac{\gamma f(0)+{f}'(0)}{2\gamma}e^{\gamma x}+\frac{\gamma f(0)-{f}'(0)}{2\gamma}e^{-\gamma x} \end{align*} $$ 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
$$ \begin{align*} \mathbf{V}(0)&=\mathbf{V_f}+\mathbf{V_r}\\ {\mathbf{V}}'(0)&=-\gamma \mathbf{V_f}+\gamma \mathbf{V_r}\\ \mathbf{I}(0)&=\mathbf{I_f}-\mathbf{I_r}\\ {\mathbf{I}}'(0)&=-\gamma\mathbf{I_f}-\gamma\mathbf{I_r}\\ \end{align*} $$ 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
$$ \begin{align*} \ast \mathbf{V}(z)&=\mathbf{V_f}e^{-\gamma z}+\mathbf{V_r}e^{\gamma z}\\ \ast \mathbf{I}(z)&=\mathbf{I_f}e^{-\gamma z}-\mathbf{I_r}e^{\gamma z}\\ \end{align*} $$ 將 \(e^{-\gamma z}\) 單獨拿出來，轉回時域
$$ \begin{align*} v(z)&=\mathbf{V_f}e^{-\gamma z}\\ v(z,t)&=\mathbf{V_f}e^{jwt}e^{-\gamma z}\\ &=\mathbf{V_f}e^{jwt}e^{-\alpha z -j\beta z}\\ &=\mathbf{V_f}e^{-\alpha z +j(wt-\beta) z}\\ \end{align*} $$ 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
$$ \begin{align*} wt_1-\beta z_1&=wt_2-\beta z_2\\ \ast v_{phase}&=\frac{z_2-z_1}{t_2-t_1}=\frac{w}{\beta }=c=f\cdot \lambda\\ \ast \lambda &= \frac{2\pi}{\beta } \end{align*} $$ 那麼特徵阻抗 (Characteristic Impdance)呢？
$$ \begin{align*} \frac{\mathrm{d} \mathbf{V}(z)}{\mathrm{d} z} &= -(R+jwL)\mathbf{I}(z)\\ \mathbf{I}(z)&=-\frac{1}{R+jwL}\frac{\mathrm{d} \mathbf{V}(z)}{\mathrm{d} z} \\ &=-\frac{1}{R+jwL}(-\gamma \mathbf{V_f}e^{-\gamma z}+\gamma \mathbf{V_r}e^{\gamma z})\\ &=\frac{\gamma}{R+jwL}(\mathbf{V_f}e^{-\gamma z}-\mathbf{V_r}e^{\gamma z})\\ &=\frac{\sqrt{(R+jwL)(G+jwC)}}{R+jwL}(\mathbf{V_f}e^{-\gamma z}-\mathbf{V_r}e^{\gamma z})\\ &=\sqrt{\frac{G+jwC}{R+jwL}}(\mathbf{V_f}e^{-\gamma z}-\mathbf{V_r}e^{\gamma z})\\ &=\mathbf{I_f}e^{-\gamma z}-\mathbf{I_r}e^{\gamma z}\\ \ast \mathbf{Z_0}&=\frac{\mathbf{V_f}}{\mathbf{I_f}}=\frac{\mathbf{V_r}}{\mathbf{I_r}}=\frac{R+jwL}{\gamma }=\sqrt{\frac{R+jwL}{G+jwC}} \end{align*} $$ 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
$$ \begin{align*} \gamma = \alpha +j\beta &= jwLC \Rightarrow \left\{\begin{matrix} \alpha &=& 0 \\ \beta &=& w\sqrt{LC} \end{matrix}\right.\\ v_{phase}&=\frac{1}{\sqrt{LC}}\\ \lambda &= \frac{2\pi}{w\sqrt{LC}}\\ \mathbf{Z_0}&=\sqrt{\frac{L}{C}}\\ \end{align*} $$

$$ \begin{align*} \gamma & = \alpha +j\beta \\ &= \sqrt{(R+jwL)(G+jwC)} \\ &= \sqrt{(0+jwL)(0+jwC)} \\ &= jwLC \Rightarrow \left\{\begin{matrix} \alpha &=& 0 \\ \beta &=& w\sqrt{LC} \end{matrix}\right.\\ \\ v_{phase}&=\frac{w}{\beta }=\frac{1}{\sqrt{LC}}\\ \lambda &= \frac{2\pi}{\beta }= \frac{2\pi}{w\sqrt{LC}}\\ \\ \mathbf{Z_0}&=\sqrt{\frac{R+jwL}{G+jwC}}\\ &=\sqrt{\frac{0+jwL}{0+jwC}}\\ &=\sqrt{\frac{L}{C}}\\ \end{align*} $$

有終端負載的傳輸線

反射係數 \(\Gamma\)
$$ \begin{align*} \Gamma (z=0)&=\frac{\mathbf{Z_L-Z_0}}{\mathbf{Z_L+Z_0}}\\ \Gamma(z)&=\frac{\mathbf{V_r}}{\mathbf{V_f}}e^{2\gamma z}=\Gamma (z=0)e^{2\gamma z}\\ \\ \mathbf{Z_{in}}(z)&= \frac{\mathbf{Z_L}(e^{-\gamma z}+e^{\gamma z})+\mathbf{Z_0}(e^{-\gamma z}-e^{\gamma z})}{\mathbf{Z_L}(e^{-\gamma z}-e^{\gamma z})+\mathbf{Z_0}(e^{-\gamma z}+e^{\gamma z})}\mathbf{Z_0}=\frac{\mathbf{Z_L}cosh(\gamma z)-\mathbf{Z_0}sinh(\gamma z)}{-\mathbf{Z_L}sinh(\gamma z)+\mathbf{Z_0}cosh(\gamma z)}\mathbf{Z_0}\\ \end{align*} $$

任意點的反射係數，也就是往前的波跟往後的波的比例
$$ \begin{align*} \Gamma(z)&=\frac{\mathbf{V_r}e^{\gamma z}}{\mathbf{V_f}e^{-\gamma z}}\\ &=\frac{\mathbf{V_r}}{\mathbf{V_f}}e^{2\gamma z}\\ \end{align*} $$ 那麼在 \(z=0\) 的反射係數呢？ $$ \begin{align*} \mathbf{V}(z)&=\mathbf{V_f}e^{-\gamma z}+\mathbf{V_f}e^{\gamma z}\\ \mathbf{I}(z)&=\mathbf{I_f}e^{-\gamma z}-\mathbf{I_r}e^{\gamma z}\\ &=\frac{\mathbf{V_f}}{\mathbf{Z_0}}e^{-\gamma z}-\frac{\mathbf{V_r}}{\mathbf{Z_0}}e^{\gamma z}\\ \end{align*} $$ $$ \begin{align*} \mathbf{Z_L}&=\frac{\mathbf{V}(z=0)}{\mathbf{I}(z=0)} \\ &=\frac{\mathbf{V_f}+\mathbf{V_r}}{\mathbf{I_f}-\mathbf{I_r}}\\ &=\frac{\mathbf{V_f}+\mathbf{V}_r}{\mathbf{V_f}-\mathbf{V_r}}\mathbf{Z_0} \\ \\ (\mathbf{V_f}-\mathbf{V_r})\mathbf{Z_L}&=(\mathbf{V_f}+\mathbf{V_r})\mathbf{Z_0} \\ \mathbf{V_f}(\mathbf{Z_L}-\mathbf{Z_0})&=\mathbf{V_r}(\mathbf{Z_L}+\mathbf{Z_0}) \\ \mathbf{V_r}&=\frac{\mathbf{Z_L-Z_0}}{\mathbf{Z_L+Z_0}}\mathbf{V_f}=\Gamma (z=0) \mathbf{V_f}\\ \ast \frac{\mathbf{V_r}}{\mathbf{V_f}}&=\Gamma (z=0)=\frac{\mathbf{Z_L-Z_0}}{\mathbf{Z_L+Z_0}} \end{align*} $$ 任意點的反射係數可被重寫
$$ \begin{align*} \ast \Gamma(z)&=\frac{\mathbf{V_r}}{\mathbf{V_f}}e^{2\gamma z}=\Gamma (z=0)e^{2\gamma z}\\ \\ \Gamma (z=-l)&=\frac{\mathbf{V_r}}{\mathbf{V_f}}e^{2\gamma (-l)}=\Gamma (z=0)e^{-2\gamma l} \end{align*} $$ 那麼 \(\mathbf{Z_{in}}(z)\) 呢？
$$ \begin{align*} \mathbf{Z_{in}}(z)&= \frac{\mathbf{V}(z)}{\mathbf{I}(z)}\\ &= \frac{\mathbf{V_f}e^{-\gamma z}+\mathbf{V_r}e^{\gamma z}}{\mathbf{I_f}e^{-\gamma z}-\mathbf{I_r}e^{\gamma z}}\\ &= \frac{\mathbf{V_f}e^{-\gamma z}+\mathbf{V_r}e^{\gamma z}}{\frac{\mathbf{V_f}}{\mathbf{Z_0}}e^{-\gamma z}-\frac{\mathbf{V_r}}{\mathbf{Z_0}}e^{\gamma z}}\\ &= \frac{\mathbf{V_f}e^{-\gamma z}+\mathbf{V_r}e^{\gamma z}}{\mathbf{V_f}e^{-\gamma z}-\mathbf{V_r}e^{\gamma z}}\mathbf{Z_0}\\ &= \frac{\mathbf{V_f}(e^{-\gamma z}+\Gamma (z=0) e^{\gamma z})}{\mathbf{V_f}(e^{-\gamma z}-\Gamma (z=0)e^{\gamma z})}\mathbf{Z_0}\\ &= \frac{1+\Gamma (z=0) e^{2\gamma z}}{1-\Gamma (z=0) e^{2\gamma z}}\mathbf{Z_0}\\ &= \frac{1+\frac{\mathbf{Z_L-Z_0}}{\mathbf{Z_L+Z_0}} e^{2\gamma z}}{1-\frac{\mathbf{Z_L-Z_0}}{\mathbf{Z_L+Z_0}} e^{2\gamma z}}\mathbf{Z_0}\\ &= \frac{\mathbf{Z_L+Z_0}+(\mathbf{Z_L-Z_0})e^{2\gamma z}}{\mathbf{Z_L+Z_0}-(\mathbf{Z_L-Z_0}) e^{2\gamma z}}\mathbf{Z_0}\\ &= \frac{\mathbf{Z_L}(1+e^{2\gamma z})+\mathbf{Z_0}(1-e^{2\gamma z})}{\mathbf{Z_L}(1-e^{2\gamma z})+\mathbf{Z_0}(1+e^{2\gamma z})}\mathbf{Z_0}\\ &= \frac{\mathbf{Z_L}(e^{-\gamma z}+e^{\gamma z})+\mathbf{Z_0}(e^{-\gamma z}-e^{\gamma z})}{\mathbf{Z_L}(e^{-\gamma z}-e^{\gamma z})+\mathbf{Z_0}(e^{-\gamma z}+e^{\gamma z})}\mathbf{Z_0}\\ &= \frac{\mathbf{Z_L}(2cosh(\gamma z))+\mathbf{Z_0}(-2sinh(\gamma z))}{\mathbf{Z_L}(-2sinh(\gamma z))+\mathbf{Z_0}(2cosh(\gamma z))}\mathbf{Z_0}\\ \ast &= \frac{\mathbf{Z_L}cosh(\gamma z)-\mathbf{Z_0}sinh(\gamma z)}{-\mathbf{Z_L}sinh(\gamma z)+\mathbf{Z_0}cosh(\gamma z)}\mathbf{Z_0}\\ \end{align*} $$

若傳輸線長度為 \(l\) or \(\infty\)
$$ \begin{align*} \Gamma (z=-l)&=\Gamma (z=0)e^{-2\gamma l}\\ \Gamma (z=-\infty )&=0\\ \\ \mathbf{Z_{in}}(z=0)&=\frac{2\mathbf{Z_{L}}}{2\mathbf{Z_{0}}}\mathbf{Z_0}=\mathbf{Z_L}\\ \mathbf{Z_{in}}(z=-l)&=\frac{\mathbf{Z_L}(e^{\gamma l}+e^{-\gamma l})+\mathbf{Z_0}(e^{\gamma l}-e^{-\gamma l})}{\mathbf{Z_L}(e^{\gamma l}-e^{-\gamma l})+\mathbf{Z_0}(e^{\gamma l}+e^{-\gamma l})}\mathbf{Z_0}\\ \mathbf{Z_{in}}(z=-\infty )&=\frac{\mathbf{Z_L}+\mathbf{Z_0}}{\mathbf{Z_L}+\mathbf{Z_0}}\mathbf{Z_0}=\mathbf{Z_0}\\ \end{align*} $$

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 \(\mathbf{Z_{in}}(z=-\infty )=\mathbf{Z_0}\)，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 \(\mathbf{V}(0)=\mathbf{V_f}+\mathbf{V_r}=\mathbf{V_f}+\Gamma (z=0)\mathbf{V_f}\)
   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 \(\mathbf{V_f}\)，仍可使用 \(\mathbf{V}(0)=\mathbf{V_f}+\mathbf{V_r}=\mathbf{V_f}+\Gamma (z=0)\mathbf{V_f}\)
   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 \(\mathrm{vin2}\) 時，因後兩端導線長度未知，所以會認為導線無限長
   此時因 \(\mathbf{Z_{in}}(z=-\infty )=\mathbf{Z_0}\)，\(\mathbf{Z_{in}} = 50+50//50=75 \Omega\)
   故 \(\mathrm{vin2}=\mathbf{V_f}+\mathbf{V_r}=\mathbf{V_f}+\Gamma (z=0)\mathbf{V_f}=1.2V\)
   反射電壓為 \(\Gamma (z=0)\mathbf{V_f}=\frac{75-50}{75+50} \times 1 = 0.2V\)
   因電阻分壓 \(\mathrm{vmid}=1.2 \times \frac{50//50}{\mathbf{Z_{in}}}=0.4V\)
3. 到達 \(\mathrm{vout1}\) & \(\mathrm{vout2}\) 時，因阻抗匹配，故完全接受為 \(0.4V\)

特別情況
波長 \(\lambda \gg\) 線長 \(l\)

$$ \begin{align*} \gamma &= \alpha +j\beta = \alpha +j\frac{2\pi}{\lambda}\\ \mathbf{Z_0}&=\frac{R+jwL}{\gamma }=\frac{R+jwL}{\alpha +j\frac{2\pi}{\lambda} }\\ \mathbf{Z_{in}}(z)&=\frac{\mathbf{Z_L}(e^{-\gamma z}+e^{\gamma z})+\mathbf{Z_0}(e^{-\gamma z}-e^{\gamma z})}{\mathbf{Z_L}(e^{-\gamma z}-e^{\gamma z})+\mathbf{Z_0}(e^{-\gamma z}+e^{\gamma z})}\mathbf{Z_0}\\ &=\frac{\mathbf{Z_L}(e^{-(\alpha +j\frac{2\pi}{\lambda}) z}+e^{(\alpha +j\frac{2\pi}{\lambda}) z})+\mathbf{Z_0}(e^{-(\alpha +j\frac{2\pi}{\lambda}) z}-e^{(\alpha +j\frac{2\pi}{\lambda}) z})}{\mathbf{Z_L}(e^{-(\alpha +j\frac{2\pi}{\lambda}) z}-e^{(\alpha +j\frac{2\pi}{\lambda}) z})+\mathbf{Z_0}(e^{-(\alpha +j\frac{2\pi}{\lambda}) z}+e^{(\alpha +j\frac{2\pi}{\lambda}) z})}\mathbf{Z_0}\\ &\because \lambda \gg z\\ &=\frac{\mathbf{Z_L}(e^{-\alpha z}+e^{\alpha z})+\mathbf{Z_0}(e^{-\alpha z}-e^{\alpha z})}{\mathbf{Z_L}(e^{-\alpha z}-e^{\alpha z})+\mathbf{Z_0}(e^{-\alpha z}+e^{\alpha z})}\mathbf{Z_0}\\ \\ \mathbf{V}(z)&=\mathbf{V_f}e^{-\gamma z}+\mathbf{V_r}e^{\gamma z}\\ &=\mathbf{V_f}e^{-(\alpha +j\frac{2\pi}{\lambda}) z}+\mathbf{V_r}e^{(\alpha +j\frac{2\pi}{\lambda}) z}\\ &\because \lambda \gg z\\ &=\mathbf{V_f}e^{-\alpha z}+\mathbf{V_r}e^{\alpha z}\\ \\ \Gamma(z)&=\Gamma (z=0)e^{2\gamma z}\\ &=\Gamma (z=0)e^{2(\alpha +j\frac{2\pi}{\lambda}) z}\\ &\because \lambda \gg z\\ &=\Gamma (z=0)e^{2\alpha z}\\ \end{align*} $$ 若為無損耗傳輸線 \(\alpha =0\)
$$ \begin{align*} \mathbf{Z_{in}}(z)&=\frac{2\mathbf{Z_L}}{2\mathbf{Z_0}}\mathbf{Z_0}=\mathbf{Z_L}\\ \Gamma(z)&=\Gamma (z=0)\\ \mathbf{V}(z)&=\mathbf{V_f}+\mathbf{V_r}=\mathbf{V_f}(1+\Gamma (z=0))\\ \end{align*} $$ 即然在源端看到的負載 \(\mathbf{Z_{in}}(-l)=\mathbf{Z_L}\) 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