传输线模型（分布参数模型）

这里写目录标题

• 试求电压源电压Vs的时域波形表达式vs(t)
• 关于传输线的计算问题☆☆
• 端接负载的无耗传输线
• 传播常数和相速
• 驻波
• 电压传播的一般形式

设无限长均匀传输线为无穷多个无限小线元级联而成。

其中： z为线长度坐标

物理量	符号	单位
线元长度	∆z	m
单位长度上的电阻	R	$\Omega/m$Ω/m
单位长度上的电导	G	$S/m$S/m
单位长度上的电感	L	$H/m$H/m
单位长度上的电容	C	$F/m$F/m

电压、电流时域方程
$v(z,t) = R\Delta zi(z,t) + L\Delta z\frac{{\partial i(z,t)}}{{\partial t}} + v(z + \Delta z,t)$v(z,t)=RΔzi(z,t)+LΔz∂t∂i(z,t)​+v(z+Δz,t)
$i(z,t) = G\Delta zv(z + \Delta z,t) + C\Delta z\frac{{\partial v(z + \Delta z,t)}}{{\partial t}} + i(z + \Delta z,t)$i(z,t)=GΔzv(z+Δz,t)+CΔz∂t∂v(z+Δz,t)​+i(z+Δz,t)

$\frac{{v(z + \Delta z,t) - v(z,t)}}{{\Delta z}} + Ri(z,t) + L\frac{{\partial i(z,t)}}{{\partial t}} = 0$Δzv(z+Δz,t)−v(z,t)​+Ri(z,t)+L∂t∂i(z,t)​=0
$\frac{{i(z + \Delta z,t) - i(z,t)}}{{\Delta z}} + Gv(z + \Delta z,t) + C\frac{{\partial v(z + \Delta z,t)}}{{\partial t}} = 0$Δzi(z+Δz,t)−i(z,t)​+Gv(z+Δz,t)+C∂t∂v(z+Δz,t)​=0

$\Delta z \to 0$Δz→0
$\frac{{\partial v(z,t)}}{{\partial z}} + Ri(z,t) + L\frac{{\partial i(z,t)}}{{\partial t}} = 0$∂z∂v(z,t)​+Ri(z,t)+L∂t∂i(z,t)​=0
$\frac{{\partial i(z,t)}}{{\partial z}} + Gv(z,t) + C\frac{{\partial v(z,t)}}{{\partial t}} = 0$∂z∂i(z,t)​+Gv(z,t)+C∂t∂v(z,t)​=0

复数法（相量法）表示单频正弦信号
$V(z) = R\Delta zI(z) + j\omega L\Delta zI(z) + V(z + \Delta z)$V(z)=RΔzI(z)+jωLΔzI(z)+V(z+Δz)
$I(z) = G\Delta zV(z + \Delta z) + j\omega C\Delta zV(z + \Delta z) + I(z + \Delta z)$I(z)=GΔzV(z+Δz)+jωCΔzV(z+Δz)+I(z+Δz)

$\frac{{V(z + \Delta z) - V(z)}}{{\Delta z}} + \left[ {R + j\omega L} \right]I(z) = 0$ΔzV(z+Δz)−V(z)​+[R+jωL]I(z)=0
$\frac{{I(z + \Delta z) - I(z)}}{{\Delta z}} + \left[ {G + j\omega C} \right]V(z + \Delta z) = 0$ΔzI(z+Δz)−I(z)​+[G+jωC]V(z+Δz)=0

$\Delta z \to 0$Δz→0
$\frac{{dV(z)}}{{dz}} + \left[ {R + j\omega L} \right]I(z) = 0$dzdV(z)​+[R+jωL]I(z)=0
$\frac{{dI(z)}}{{dz}} + \left[ {G + j\omega C} \right]V(z) = 0$dzdI(z)​+[G+jωC]V(z)=0

$\frac{{dV(z)}}{{dz}} + \left[ {R + j\omega L} \right]I(z) = 0(1)$dzdV(z)​+[R+jωL]I(z)=0(1)
$\frac{{dI(z)}}{{dz}} + \left[ {G + j\omega C} \right]V(z) = 0(2)$dzdI(z)​+[G+jωC]V(z)=0(2)
（1）式对z求导再将（2）代入得（3），同理可得（4）
$\frac{{{d^2}V(z)}}{{d{z^2}}} - \left[ {R + j\omega L} \right]\left[ {G + j\omega C} \right]V(z) = 0(3)$dz2d2V(z)​−[R+jωL][G+jωC]V(z)=0(3)
$\frac{{{d^2}I(z)}}{{d{z^2}}} - \left[ {R + j\omega L} \right]\left[ {G + j\omega C} \right]I(z) = 0(4)$dz2d2I(z)​−[R+jωL][G+jωC]I(z)=0(4)
$\gamma {\rm{ = }}\sqrt {\left[ {R + j\omega L} \right]\left[ {G + j\omega C} \right]} {\rm{ = }}\alpha {\rm{ + }}j\beta$γ=[R+jωL][G+jωC]​=α+jβ

传输方程
$\frac{{{d^2}V(z)}}{{{d^2}z}} - {\gamma ^2}V(z) = 0$d2zd2V(z)​−γ2V(z)=0
$\frac{{{d^2}I(z)}}{{{d^2}z}} - {\gamma ^2}I(z) = 0$d2zd2I(z)​−γ2I(z)=0

传输方程的解为
$V(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma z}}$V(z)=V0​+e−γz+V0​−eγz☆☆
$I(z) = {I_0}^ + {e^{ - \gamma z}} + {I_0}^ - {e^{\gamma z}}$I(z)=I0​+e−γz+I0​−eγz☆☆
其中${V_0}^ +$V0​+${V_0}^ -$V0​−${I_0}^ +$I0​+${I_0}^ -$I0​−为复常数，与传输距离z无关
由传输线边界条件确定
四个参数中只有两个是独立的
方程中只有z是实数，其他均为复数

根据前面的方程（1）得
$\frac{{dV(z)}}{{dz}} + \left[ {R + j\omega L} \right]I(z) = 0$dzdV(z)​+[R+jωL]I(z)=0
$I(z){\rm{ = }}\frac{{ - 1}}{{R + j\omega L}}\frac{{dV(z)}}{{dz}}$I(z)=R+jωL−1​dzdV(z)​
$I(z) = \frac{\gamma }{{R + j\omega L}}\left( {{V_0}^ + {e^{ - \gamma z}} - {V_0}^ - {e^{\gamma z}}} \right)$I(z)=R+jωLγ​(V0​+e−γz−V0​−eγz)
特征阻抗${Z_0}{\rm{ = }}\sqrt {\frac{{R + j\omega L}}{{G + j\omega C}}} = r + jx = \frac{{V_0^ + }}{{I_0^ + }} = \frac{{ - V_0^ - }}{{I_0^ - }}$Z0​=G+jωCR+jωL​​=r+jx=I0+​V0+​​=I0−​−V0−​​
$I(z) = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - \gamma z}} - \frac{{{V_0}^ - }}{{{Z_0}}}{e^{\gamma z}}$I(z)=Z0​V0​+​e−γz−Z0​V0​−​eγz
传输方程的解为
$V(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma z}}$V(z)=V0​+e−γz+V0​−eγz☆☆
$I(z) = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - \gamma z}} - \frac{{{V_0}^ - }}{{{Z_0}}}{e^{\gamma z}}$I(z)=Z0​V0​+​e−γz−Z0​V0​−​eγz☆☆
注意：复数开方有两个值，根据其物理意义，取值要求$\alpha \ge 0{\rm{ ; }}r \ge 0{\rm{ ; }}\beta,x$α≥0;r≥0;β,x为任意实数（z的方向从源到负载）

复函数表达式中解出对应的时域函数
$v(z,t) = {\mathop{\rm Re}\nolimits} \left\{ {{e^{j\omega t}}V(z)} \right\} = \frac{{{e^{j\omega t}}V(z) + {e^{ - j\omega t}}{V^ * }(z)}}{2}$v(z,t)=Re{ejωtV(z)}=2ejωtV(z)+e−jωtV∗(z)​
假设${V_0}^ + {\rm{ = }}\left| {{V_0}^ + } \right|{e^{j{\varphi ^ + }}}{\rm{ = }}\left| {{V_0}^ + } \right|\left( {\cos {\varphi ^ + } + j\sin {\varphi ^ + }} \right)$V0​+=∣∣​V0​+∣∣​ejφ+=∣∣​V0​+∣∣​(cosφ++jsinφ+)
${V_0}^ - {\rm{ = }}\left| {{V_0}^ - } \right|{e^{j{\varphi ^ - }}}$V0​−=∣∣​V0​−∣∣​ejφ−
$V(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma z}} = \left| {{V_0}^ + } \right|{e^{j{\varphi ^ + }}}{e^{ - \alpha z - j\beta z}} + \left| {{V_0}^ - } \right|{e^{j{\varphi ^ - }}}{e^{\alpha z + j\beta z}}$V(z)=V0​+e−γz+V0​−eγz=∣∣​V0​+∣∣​ejφ+e−αz−jβz+∣∣​V0​−∣∣​ejφ−eαz+jβz
$v(z,t) = {\mathop{\rm Re}\nolimits} \left\{ {{e^{j\omega t}}V(z)} \right\} = \left| {{V_0}^ + } \right|{e^{ - \alpha z}}{e^{j\left( {\omega t - \beta z + {\varphi ^ + }} \right)}} + \left| {{V_0}^ - } \right|{e^{\alpha z}}{e^{j\left( {\omega t + \beta z + {\varphi ^ - }} \right)}} = \left| {V_0^ + } \right|{e^{ - \alpha z}}\cos (\omega t - \beta z + {\phi ^ + }) + \left| {V_0^ - } \right|{e^{\alpha z}}\cos (\omega t + \beta z + {\phi ^ - })$v(z,t)=Re{ejωtV(z)}=∣∣​V0​+∣∣​e−αzej(ωt−βz+φ+)+∣∣​V0​−∣∣​eαzej(ωt+βz+φ−)=∣∣​V0+​∣∣​e−αzcos(ωt−βz+ϕ+)+∣∣​V0−​∣∣​eαzcos(ωt+βz+ϕ−)
同理可得电流波形的表达式

方程的物理意义

1. 方程以复数形式描述了在单频正弦波激励下传输线各点上的电压和电流的幅度和相位
2. 方程中z为实数。z的方向及电压电流的方向见参考图，从源向终端负载
3. 传输线上任意点的电压表达式
   $v(z,t) = \left| {V_0^ + } \right|{e^{ - \alpha z}}\cos (\omega t - \beta z + {\phi ^ + }) + \left| {V_0^ - } \right|{e^{\alpha z}}\cos (\omega t + \beta z + {\phi ^ - })$v(z,t)=∣∣​V0+​∣∣​e−αzcos(ωt−βz+ϕ+)+∣∣​V0−​∣∣​eαzcos(ωt+βz+ϕ−)
   称方程中第一项为电压入射波，传播方向与z相同
   称方程中第二项为电压反射波，传播方向与z相反
   各点均由入射波与反射波叠加而成。

试求电压源电压Vs的时域波形表达式vs(t)

例：设有限长传输线的源端接理想电压源$V_s$Vs​，终端接负载$Z_L=Z0=50Ω$ZL​=Z0=50Ω
若传输线的长度为$l=10m$l=10m，$γ=0.1+jπ/20（1/m）$γ=0.1+jπ/20（1/m）
测得负载电压$V_L$VL​的波形为$v_L(t)=cosωt$vL​(t)=cosωt
试求电压源电压$V_s$Vs​的时域波形表达式$v_s(t)$vs​(t)
首先定义z的坐标

定义余弦函数为相位参考，峰值为幅度参考，则$V_L=1$VL​=1
$\{ \begin{array}{l} V(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma z}}\\ I(z) = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - \gamma z}} - \frac{{{V_0}^ - }}{{{Z_0}}}{e^{\gamma z}} \end{array} \Rightarrow \{ \begin{array}{l} V(0) = {V_L} = {V_0}^ + + {V_0}^ - \\ I(0) = \frac{{{V_L}}}{{{Z_L}}} = \frac{{{V_0}^ + }}{{{Z_0}}} - \frac{{{V_0}^ - }}{{{Z_0}}} \end{array}${V(z)=V0​+e−γz+V0​−eγzI(z)=Z0​V0​+​e−γz−Z0​V0​−​eγz​⇒{V(0)=VL​=V0​++V0​−I(0)=ZL​VL​​=Z0​V0​+​−Z0​V0​−​​
${V_0}^ + = \frac{{{V_L}}}{2}\left( {1 + \frac{{{Z_0}}}{{{Z_L}}}} \right)$V0​+=2VL​​(1+ZL​Z0​​)☆☆
${V_0}^ - = \frac{{{V_L}}}{2}\left( {1 - \frac{{{Z_0}}}{{{Z_L}}}} \right)$V0​−=2VL​​(1−ZL​Z0​​)☆☆
代入${V_L} = 1,{Z_0}{\rm{ = }}{Z_L}{\rm{ = }}50$VL​=1,Z0​=ZL​=50得$V_0^ + = 1,V_0^ - = 0$V0+​=1,V0−​=0

$V(z) = {e^{ - \gamma z}} \Rightarrow {V_S} = V( - l) = {e^{\gamma l}} = {e^{1 + j\pi /2}} = je$V(z)=e−γz⇒VS​=V(−l)=eγl=e1+jπ/2=je
${v_s}(t) = {\mathop{\rm Re}\nolimits} \left\{ {je\left( {\cos \omega t + j\sin \omega t} \right)} \right\} \approx - 2.7\sin \omega t{\rm{ }}(V)$vs​(t)=Re{je(cosωt+jsinωt)}≈−2.7sinωt(V)

如果$Z_L=100Ω$ZL​=100Ω，传输线参数不变，则
${V_0}^ + = \frac{{{V_L}}}{2}\left( {1 + \frac{{{Z_0}}}{{{Z_L}}}} \right){\rm{ = }}\frac{1}{2}\left( {1 + \frac{{50}}{{100}}} \right) = \frac{3}{4}$V0​+=2VL​​(1+ZL​Z0​​)=21​(1+10050​)=43​
${V_0}^ - = \frac{{{V_L}}}{2}\left( {1 - \frac{{{Z_0}}}{{{Z_L}}}} \right) = \frac{1}{4}$V0​−=2VL​​(1−ZL​Z0​​)=41​
${V_S} = \frac{3}{4}{e^{1 + j\pi /2}} + \frac{1}{4}{e^{ - 1 - j\pi /2}} = \frac{j}{4}\left( {3e - {e^{ - 1}}} \right)$VS​=43​e1+jπ/2+41​e−1−jπ/2=4j​(3e−e−1)
${v_s}(t) = \frac{1}{4}\left( {{e^{ - 1}} - 3e} \right)\sin \omega t{\rm{ }}(V)$vs​(t)=41​(e−1−3e)sinωt(V)

关于传输线的计算问题☆☆

一般说来在传输系数，特征阻抗${Z_0}$Z0​已知的条件下，传输线的计算问题主要是根据边界条件（z=0，z=l等）求出${V_0}^ -$V0​−，${V_0}^ +$V0​+，从而完善电压电流方程
解题时注意z坐标的建立，原点和终点的坐标值的确定
尽管方程描述的是分布参数模型
$V(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma z}}$V(z)=V0​+e−γz+V0​−eγz
$I(z) = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - \gamma z}} - \frac{{{V_0}^ - }}{{{Z_0}}}{e^{\gamma z}}$I(z)=Z0​V0​+​e−γz−Z0​V0​−​eγz
但任何一个z固定的点都可等效为集总参数模型

假设所有参数的单位为标准单位，已知传输线参数
${Z_0}{\rm{ = 50 }},\gamma = 0.01 + j0.2\pi$Z0​=50,γ=0.01+j0.2π

1、若信号源电压幅度V(0)=10，输出电流 I(0)=0.2，传输线长度$l=3$l=3
求终端负载电阻${Z_L}$ZL​
建立z坐标
电源点z=0
终端点$z=l=3$z=l=3
边界条件V(0)=10，I(0)=0.2
解出${V_0}^ -$V0​−，${V_0}^ +$V0​+
求出V(3)，I(3)
${Z_L} = V(3)/I(3)$ZL​=V(3)/I(3)

2、已知信号源电压幅度V(-4)=10，传输线长度$l=4$l=4，终端端接电阻${Z_L}=60$ZL​=60
求电源输出电流
电源点为$z=-l=-4$z=−l=−4
终端点为z=0
边界条件${Z_L} = V(0)/I(0)=60$ZL​=V(0)/I(0)=60建立${V_0}^ -$V0​−，${V_0}^ +$V0​+的关系
再利用边界条件V(-4)=10，求出${V_0}^ -$V0​−，${V_0}^ +$V0​+
最后代入电流方程求出I(-4)

端接负载的无耗传输线

仅由储能器件(C,L)而无耗能器件(R,G)构成的传输线叫做无耗传输线，传输参数中R=G=0
$\gamma = j\beta = \sqrt {\left( {R + j\omega L} \right)\left( {G + j\omega C} \right)} = j\omega \sqrt {LC}$γ=jβ=(R+jωL)(G+jωC)​=jωLC​虚数
${Z_0} = \sqrt {\frac{{\left( {R + j\omega L} \right)}}{{\left( {G + j\omega C} \right)}}} = \sqrt {\frac{L}{C}}$Z0​=(G+jωC)(R+jωL)​​=CL​​实数
$V(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma z}}$V(z)=V0​+e−γz+V0​−eγz☆☆
$I(z) = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - \gamma z}} - \frac{{{V_0}^ - }}{{{Z_0}}}{e^{\gamma z}}$I(z)=Z0​V0​+​e−γz−Z0​V0​−​eγz☆☆
${V_0}^ + = \frac{{{V_L}}}{2}\left( {1 + \frac{{{Z_0}}}{{{Z_L}}}} \right)$V0​+=2VL​​(1+ZL​Z0​​)☆☆
${V_0}^ - = \frac{{{V_L}}}{2}\left( {1 - \frac{{{Z_0}}}{{{Z_L}}}} \right)$V0​−=2VL​​(1−ZL​Z0​​)☆☆

电压反射系数
$\Gamma (z) = \frac{{反射电压波}}{{入射电压波}}{\rm{ = }}\frac{{{V^ - }{e^{\gamma z}}}}{{{V^ + }{e^{ - \gamma z}}}} = \frac{{{V^ - }}}{{{V^ + }}}{e^{2\gamma z}}$Γ(z)=入射电压波反射电压波​=V+e−γzV−eγz​=V+V−​e2γz
无耗传输线的电压反射系数
$\Gamma \left( z \right) = {\Gamma _0}{e^{j2\beta z}}$Γ(z)=Γ0​ej2βz☆☆
${\Gamma _0} = \Gamma (0) = \frac{{{V^ - }}}{{{V^ + }}}$Γ0​=Γ(0)=V+V−​☆☆
令$d=-z$d=−z，沿反射波传播方向定义的电压反射系数
$\Gamma \left( d \right) = {\Gamma _0}{e^{ - j2\beta d}}$Γ(d)=Γ0​e−j2βd☆☆
在端接点上
$\left\{ \begin{array}{l} {V_0}^ + = \frac{{{V_L}}}{2}\left( {1 + \frac{{{Z_0}}}{{{Z_L}}}} \right)\\ {V_0}^ - = \frac{{{V_L}}}{2}\left( {1 - \frac{{{Z_0}}}{{{Z_L}}}} \right) \end{array} \right. \Rightarrow {\Gamma _0} = \frac{{{V_0}^ - }}{{{V_0}^ + }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}\Rightarrow {Z_L} = {Z_0}\frac{{1 + {\Gamma _0}}}{{1 - {\Gamma _0}}}$⎩⎨⎧​V0​+=2VL​​(1+ZL​Z0​​)V0​−=2VL​​(1−ZL​Z0​​)​⇒Γ0​=V0​+V0​−​=ZL​+Z0​ZL​−Z0​​⇒ZL​=Z0​1−Γ0​1+Γ0​​☆☆
推广到传输线的任意点上${\rm{ }}\Gamma (z) = \frac{{Z(z) - {Z_0}}}{{Z(z) + {Z_0}}}$Γ(z)=Z(z)+Z0​Z(z)−Z0​​☆☆
$Z(z)$Z(z)为传输线z点向负载端看的等效阻抗
无耗传输线上电压或电流波可用反射系数表示为：
$\left\{ \begin{array}{l} V(z) = {V_0}^ + \left( {{e^{ - j\beta z}} + {\Gamma _0}^{}{e^{j\beta z}}} \right) = {V_0}^ + {e^{ - j\beta z}}\left[ {1 + \Gamma (z)} \right]\\ I(z) = \frac{{{V_0}^ + }}{{{Z_0}}}\left( {{e^{ - j\beta z}} - {\Gamma _0}^{}{e^{j\beta z}}} \right) = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - j\beta z}}\left[ {1 - \Gamma (z)} \right] \end{array} \right.${V(z)=V0​+(e−jβz+Γ0​ejβz)=V0​+e−jβz[1+Γ(z)]I(z)=Z0​V0​+​(e−jβz−Γ0​ejβz)=Z0​V0​+​e−jβz[1−Γ(z)]​

${\Gamma _0} = \frac{{{V_0}^ - }}{{{V_0}^ + }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}$Γ0​=V0​+V0​−​=ZL​+Z0​ZL​−Z0​​
反射系数的几个常用值
终端匹配${Z_L} = {Z_0} \Rightarrow {\Gamma _0} = 0$ZL​=Z0​⇒Γ0​=0
终端短路${Z_L} = 0 \Rightarrow {\Gamma _0} = - 1$ZL​=0⇒Γ0​=−1
终端开路${Z_L} = \infty \Rightarrow {\Gamma _0} = 1$ZL​=∞⇒Γ0​=1

传播常数和相速

考察入射波在无耗传输线上的传输(无反射）
${v^ + }(z,t) = \left| {V_0^ + } \right|\cos (\omega t - \beta z + {\phi ^ + })$v+(z,t)=∣∣​V0+​∣∣​cos(ωt−βz+ϕ+)
选两个同相位的点观察，假设$\omega {t_1} - \beta \,{z_1} + {\psi _1} = 0$ωt1​−βz1​+ψ1​=0
当$t = t_1$t=t1​时，A点在$z_1$z1​处
$\omega {t_1} - \beta \,{z_1} + {\psi _1}$ωt1​−βz1​+ψ1​
当$t = t_1+ △t$t=t1​+△t时，A点在$z= z_1+△z$z=z1​+△z处
$\omega ({t_1} + \Delta t) - \beta ({z_1} + \Delta z) + \psi {\,_1}$ω(t1​+Δt)−β(z1​+Δz)+ψ1​

相同相位点
$\omega {t_1} - \beta \,{z_1} + {\psi _1} = \omega ({t_1} + \Delta t) - \beta ({z_1} + \Delta z) + {\psi _1}$ωt1​−βz1​+ψ1​=ω(t1​+Δt)−β(z1​+Δz)+ψ1​
$\omega \Delta t = \beta \Delta z$ωΔt=βΔz
相同相位点的移动速度为相位速度即相速
$v = \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta z}}{{\Delta t}} = \frac{\omega }{\beta }$v=Δt→0lim​ΔtΔz​=βω​

这种向前传播的波称之为行波
幅度不变，相位随z变化
${v^ + }(z,t) = \left| {V_0^ + } \right|\cos (\omega t - \beta z + {\phi ^ + })$v+(z,t)=∣∣​V0+​∣∣​cos(ωt−βz+ϕ+)
$\lambda \, = \frac{{2\pi }}{\beta } = \frac{\omega }{\beta }T = vT{\rm{ = }}\frac{v}{f}$λ=β2π​=βω​T=vT=fv​
任何一个时间t上，z轴上的电压分布为一个正弦函数，该函数一个周期的长度=波长=正弦波在一个周期时间上沿z轴的移动距离

驻波

只在原地振荡，不向前传播
驻波是由入射波和反射波叠加形成的，考察终端开/短路有限长传输线
$V(z) = {V_0}^ + \left( {{e^{ - j\beta z}} + {\Gamma _0}^{}{e^{j\beta z}}} \right)$V(z)=V0​+(e−jβz+Γ0​ejβz)

${\Gamma _0} = \frac{{{V_0}^ - }}{{{V_0}^ + }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}$Γ0​=V0​+V0​−​=ZL​+Z0​ZL​−Z0​​
反射系数的几个常用值
终端匹配${Z_L} = {Z_0} \Rightarrow {\Gamma _0} = 0$ZL​=Z0​⇒Γ0​=0
终端短路${Z_L} = 0 \Rightarrow {\Gamma _0} = - 1$ZL​=0⇒Γ0​=−1
终端开路${Z_L} = \infty \Rightarrow {\Gamma _0} = 1$ZL​=∞⇒Γ0​=1

零点设在端接负载上
${\Gamma _0} = \pm 1{\rm{ }} \Rightarrow {\rm{ }}V(d) = {V_0}^ + \left( {{e^{j\beta d}} \pm {e^{ - j\beta d}}} \right) = \left\{ \begin{array}{l} 2{V_0}^ + \cos \beta d开路\\ j2{V_0}^ + \sin \beta d短路 \end{array} \right.$Γ0​=±1⇒V(d)=V0​+(ejβd±e−jβd)={2V0​+cosβd开路j2V0​+sinβd短路​
时域式
$v(t,d) = {\mathop{\rm Re}\nolimits} \left\{ {{e^{j\omega t}}V(d)} \right\} = \left\{ \begin{array}{l} 2\left| {{V_0}^ + } \right|\cos \beta d\cos \left( {\omega t + {\varphi ^ + }} \right)\\ -2\left| {{V_0}^ + } \right|\sin \beta d\sin \left( {\omega t + {\varphi ^ + }} \right) \end{array} \right.$v(t,d)=Re{ejωtV(d)}={2∣∣​V0​+∣∣​cosβdcos(ωt+φ+)−2∣∣​V0​+∣∣​sinβdsin(ωt+φ+)​

$v(t,d) = A\cos \beta d\cos \left( {\omega t + \varphi } \right){\rm{ = }}\left| {A\cos \beta d} \right|\cos \left( {\omega t + \varphi (z)} \right)$v(t,d)=Acosβdcos(ωt+φ)=∣Acosβd∣cos(ωt+φ(z))
其幅度值在z轴上的分布如图所示：

在传输线的任意点上为幅度不同的正弦波
振幅最大处称为波腹$\max \left| {A\cos \beta d} \right|{\rm{ = }}\left| {\rm{A}} \right|$max∣Acosβd∣=∣A∣
振幅最小处称为波节$\min \left| {A\cos \beta d} \right|{\rm{ = }}0$min∣Acosβd∣=0
波节处没有正弦波的存在
这意味着波仅仅在原地振动，而不向前传播。

电压传播的一般形式

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