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
- 电压反射系数
- 传播常数和相速
- 驻波
- 电压传播的一般形式
- 开路线、短路线、四分之一波长传输线
- - 终端短路传输线
  - 终端开路的传输线
  - 1/4波长传输线
  - 1/2波长的端接传输线
传输线常用公式和近似值★★★
作业

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	Ω/m\Omega/mΩ/m
单位长度上的电导	G	S/mS/mS/m
单位长度上的电感	L	H/mH/mH/m
单位长度上的电容	C	F/mF/mF/m

电压、电流时域方程
v(z,t)=RΔzi(z,t)+LΔz∂i(z,t)∂t+v(z+Δz,t)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)=GΔzv(z+Δz,t)+CΔz∂v(z+Δz,t)∂t+i(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)

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

Δz→0\Delta z \to 0Δz→0
∂v(z,t)∂z+Ri(z,t)+L∂i(z,t)∂t=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
∂i(z,t)∂z+Gv(z,t)+C∂v(z,t)∂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)=RΔzI(z)+jωLΔzI(z)+V(z+Δz)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)=GΔzV(z+Δz)+jωCΔzV(z+Δz)+I(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)

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

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

dV(z)dz+[R+jωL]I(z)=0(1)\frac{{dV(z)}}{{dz}} + \left[ {R + j\omega L} \right]I(z) = 0(1)dzdV(z)+[R+jωL]I(z)=0(1)
dI(z)dz+[G+jωC]V(z)=0(2)\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）
d2V(z)dz2−[R+jωL][G+jωC]V(z)=0(3)\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)
d2I(z)dz2−[R+jωL][G+jωC]I(z)=0(4)\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)
γ=[R+jωL][G+jωC]=α+jβ,α≥0,β∈R\gamma {\rm{ = }}\sqrt {\left[ {R + j\omega L} \right]\left[ {G + j\omega C} \right]} {\rm{ = }}\alpha {\rm{ + }}j\beta,\alpha \ge 0,\beta∈Rγ=[R+jωL][G+jωC]=α+jβ,α≥0,β∈R★

传输方程
d2V(z)d2z−γ2V(z)=0\frac{{{d^2}V(z)}}{{{d^2}z}} - {\gamma ^2}V(z) = 0d2zd2V(z)−γ2V(z)=0★
d2I(z)d2z−γ2I(z)=0\frac{{{d^2}I(z)}}{{{d^2}z}} - {\gamma ^2}I(z) = 0d2zd2I(z)−γ2I(z)=0★

传输方程的解
V(z)=V0+e−γz+V0−eγzV(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma z}}V(z)=V0+e−γz+V0−eγz★
I(z)=I0+e−γz+I0−eγzI(z) = {I_0}^ + {e^{ - \gamma z}} + {I_0}^ - {e^{\gamma z}}I(z)=I0+e−γz+I0−eγz★
其中V0+{V_0}^ +V0+V0−{V_0}^ -V0−I0+{I_0}^ +I0+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
dV(z)dz+[R+jωL]I(z)=0\frac{{dV(z)}}{{dz}} + \left[ {R + j\omega L} \right]I(z) = 0dzdV(z)+[R+jωL]I(z)=0
I(z)=−1R+jωLdV(z)dzI(z){\rm{ = }}\frac{{ - 1}}{{R + j\omega L}}\frac{{dV(z)}}{{dz}}I(z)=R+jωL−1dzdV(z)
I(z)=γR+jωL(V0+e−γz−V0−eγz)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
= 入射波电压 / 入射波电流 = - 反射波电压 / 反射波电流 ★
仅由自身分布参数决定
与信号源和负载无关
一般为复数，与频率有关
Z0=R+jωLG+jωC=r+jx=V0+I0+=−V0−I0−,r≥0,x∈R{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^ - }},r\ge 0,x∈RZ0=G+jωCR+jωL=r+jx=I0+V0+=I0−−V0−,r≥0,x∈R★

I(z)=V0+Z0e−γz−V0−Z0eγzI(z) = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - \gamma z}} - \frac{{{V_0}^ - }}{{{Z_0}}}{e^{\gamma z}}I(z)=Z0V0+e−γz−Z0V0−eγz
传输方程的解为
V(z)=V0+e−γz+V0−eγzV(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma z}}V(z)=V0+e−γz+V0−eγz★
I(z)=V0+Z0e−γz−V0−Z0eγzI(z) = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - \gamma z}} - \frac{{{V_0}^ - }}{{{Z_0}}}{e^{\gamma z}}I(z)=Z0V0+e−γz−Z0V0−eγz★
注意：复数开方有两个值，根据其物理意义，取值要求α≥0,r≥0,β,x\alpha \ge 0,r \ge 0,\beta,xα≥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)=Re{ejωtV(z)}=ejωtV(z)+e−jωtV∗(z)2v(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V0+=∣V0+∣ejφ+=∣V0+∣(cos⁡φ++jsin⁡φ+){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+)
V0−=∣V0−∣ejφ−{V_0}^ - {\rm{ = }}\left| {{V_0}^ - } \right|{e^{j{\varphi ^ - }}}V0−=∣∣V0−∣∣ejφ−
V(z)=V0+e−γz+V0−eγz=∣V0+∣ejφ+e−αz−jβz+∣V0−∣ejφ−eαz+jβzV(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)=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−)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

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
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
传输线上任意点的电压表达式
v(z,t)=∣V0+∣e−αzcos⁡(ωt−βz+ϕ+)+∣V0−∣eαzcos⁡(ωt+βz+ϕ−)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VsV_sVs，终端接负载ZL=Z0=50ΩZ_L=Z0=50ΩZL=Z0=50Ω
若传输线的长度为l=10ml=10ml=10m，γ=0.1+jπ/20（1/m）γ=0.1+jπ/20（1/m）γ=0.1+jπ/20（1/m）
测得负载电压VLV_LVL的波形为vL(t)=cosωtv_L(t)=cosωtvL(t)=cosωt
试求电压源电压VsV_sVs的时域波形表达式vs(t)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VL=1V_L=1VL=1
{V(z)=V0+e−γz+V0−eγzI(z)=V0+Z0e−γz−V0−Z0eγz⇒{V(0)=VL=V0++V0−I(0)=VLZL=V0+Z0−V0−Z0\{ \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)=Z0V0+e−γz−Z0V0−eγz⇒{V(0)=VL=V0++V0−I(0)=ZLVL=Z0V0+−Z0V0−★
V0+=VL2(1+Z0ZL){V_0}^ + = \frac{{{V_L}}}{2}\left( {1 + \frac{{{Z_0}}}{{{Z_L}}}} \right)V0+=2VL(1+ZLZ0)★
V0−=VL2(1−Z0ZL){V_0}^ - = \frac{{{V_L}}}{2}\left( {1 - \frac{{{Z_0}}}{{{Z_L}}}} \right)V0−=2VL(1−ZLZ0)★
代入VL=1,Z0=ZL=50{V_L} = 1,{Z_0}{\rm{ = }}{Z_L}{\rm{ = }}50VL=1,Z0=ZL=50得V0+=1,V0−=0V_0^ + = 1,V_0^ - = 0V0+=1,V0−=0

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

如果ZL=100ΩZ_L=100ΩZL=100Ω，传输线参数不变，则
V0+=VL2(1+Z0ZL)=12(1+50100)=34{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+ZLZ0)=21(1+10050)=43
V0−=VL2(1−Z0ZL)=14{V_0}^ - = \frac{{{V_L}}}{2}\left( {1 - \frac{{{Z_0}}}{{{Z_L}}}} \right) = \frac{1}{4}V0−=2VL(1−ZLZ0)=41
VS=34e1+jπ/2+14e−1−jπ/2=j4(3e−e−1){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=43e1+jπ/2+41e−1−jπ/2=4j(3e−e−1)
vs(t)=14(e−1−3e)sin⁡ωt(V){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Z0{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V0−{V_0}^ -V0−，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)=V0+e−γz+V0−eγzV(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma z}}V(z)=V0+e−γz+V0−eγz★
I(z)=V0+Z0e−γz−V0−Z0eγzI(z) = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - \gamma z}} - \frac{{{V_0}^ - }}{{{Z_0}}}{e^{\gamma z}}I(z)=Z0V0+e−γz−Z0V0−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
Z0=50,γ=0.01+j0.2π{Z_0}{\rm{ = 50 }},\gamma = 0.01 + j0.2\piZ0=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=3l=3l=3
求终端负载电阻ZL{Z_L}ZL
建立z坐标
电源点z=0
终端点z=l=3z=l=3z=l=3
边界条件V(0)=10，I(0)=0.2
解出V0−{V_0}^ -V0−，V0+{V_0}^ +V0+
求出V(3)，I(3)
ZL=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=4l=4l=4，终端端接电阻ZL=60{Z_L}=60ZL=60
求电源输出电流
电源点为z=−l=−4z=-l=-4z=−l=−4
终端点为z=0
边界条件ZL=V(0)/I(0)=60{Z_L} = V(0)/I(0)=60ZL=V(0)/I(0)=60建立V0−{V_0}^ -V0−，V0+{V_0}^ +V0+的关系
再利用边界条件V(-4)=10，求出V0−{V_0}^ -V0−，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
γ=(R+jωL)(G+jωC)=jωLC=jβ\gamma = \sqrt {\left( {R + j\omega L} \right)\left( {G + j\omega C} \right)} = j\omega \sqrt {LC} = j\betaγ=(R+jωL)(G+jωC)=jωLC=jβ虚数★
特征阻抗
Z0=(R+jωL)(G+jωC)=LC{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)=V0+e−γz+V0−eγzV(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma z}}V(z)=V0+e−γz+V0−eγz★
I(z)=V0+Z0e−γz−V0−Z0eγzI(z) = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - \gamma z}} - \frac{{{V_0}^ - }}{{{Z_0}}}{e^{\gamma z}}I(z)=Z0V0+e−γz−Z0V0−eγz★

V0+=VL2(1+Z0ZL){V_0}^ + = \frac{{{V_L}}}{2}\left( {1 + \frac{{{Z_0}}}{{{Z_L}}}} \right)V0+=2VL(1+ZLZ0)★
V0−=VL2(1−Z0ZL){V_0}^ - = \frac{{{V_L}}}{2}\left( {1 - \frac{{{Z_0}}}{{{Z_L}}}} \right)V0−=2VL(1−ZLZ0)★

电压反射系数

Γ(z)=反射电压波入射电压波=V−eγzV+e−γz=V−V+e2γz\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
Γ(z)=Γ0ej2βz\Gamma \left( z \right) = {\Gamma _0}{e^{j2\beta z}}Γ(z)=Γ0ej2βz★
Γ0=Γ(0)=V−V+{\Gamma _0} = \Gamma (0) = \frac{{{V^ - }}}{{{V^ + }}}Γ0=Γ(0)=V+V−★
令d=−zd=-zd=−z，沿反射波传播方向定义的电压反射系数
Γ(d)=Γ0e−j2βd\Gamma \left( d \right) = {\Gamma _0}{e^{ - j2\beta d}}Γ(d)=Γ0e−j2βd★
在端接点上
{V0+=VL2(1+Z0ZL)V0−=VL2(1−Z0ZL)\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.⎩⎨⎧V0+=2VL(1+ZLZ0)V0−=2VL(1−ZLZ0)
Γ0=V0−V0+=ZL−Z0ZL+Z0{\Gamma _0} = \frac{{{V_0}^ - }}{{{V_0}^ + }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}Γ0=V0+V0−=ZL+Z0ZL−Z0★(2.48)(2.52)
Z(0)=ZL=Z01+Γ01−Γ0Z(0)={Z_L} = {Z_0}\frac{{1 + {\Gamma _0}}}{{1 - {\Gamma _0}}}Z(0)=ZL=Z01−Γ01+Γ0★(2.51)
推广到传输线的任意点上Γ(z)=Z(z)−Z0Z(z)+Z0{\rm{ }}\Gamma (z) = \frac{{Z(z) - {Z_0}}}{{Z(z) + {Z_0}}}Γ(z)=Z(z)+Z0Z(z)−Z0★
Z(z)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
V(z)=V0+e−γz+V0−eγz=V0+(e−jβz+Γ0ejβz)=V0+e−jβz[1+Γ(z)]V(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma 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]V(z)=V0+e−γz+V0−eγz=V0+(e−jβz+Γ0ejβz)=V0+e−jβz[1+Γ(z)]★
I(z)=V0+Z0e−γz−V0−Z0eγz=V0+Z0(e−jβz−Γ0ejβz)=V0+Z0e−jβz[1−Γ(z)]I(z) = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - \gamma z}} - \frac{{{V_0}^ - }}{{{Z_0}}}{e^{\gamma 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]I(z)=Z0V0+e−γz−Z0V0−eγz=Z0V0+(e−jβz−Γ0ejβz)=Z0V0+e−jβz[1−Γ(z)]★

Γ0=V0−V0+=ZL−Z0ZL+Z0{\Gamma _0} = \frac{{{V_0}^ - }}{{{V_0}^ + }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}Γ0=V0+V0−=ZL+Z0ZL−Z0
反射系数的几个常用值★ [-1,1] ？
终端匹配ZL=Z0⇒Γ0=0{Z_L} = {Z_0} \Rightarrow {\Gamma _0} = 0ZL=Z0⇒Γ0=0
终端短路ZL=0⇒Γ0=−1{Z_L} = 0 \Rightarrow {\Gamma _0} = - 1ZL=0⇒Γ0=−1
终端开路ZL=∞⇒Γ0=1{Z_L} = \infty \Rightarrow {\Gamma _0} = 1ZL=∞⇒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)=∣V0+∣cos⁡(ωt−βz+ϕ+){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ωt1−βz1+ψ1=0\omega {t_1} - \beta \,{z_1} + {\psi _1} = 0ωt1−βz1+ψ1=0
当t=t1t = t_1t=t1时，A点在z1z_1z1处
ωt1−βz1+ψ1\omega {t_1} - \beta \,{z_1} + {\psi _1}ωt1−βz1+ψ1
当t=t1+△tt = t_1+ △tt=t1+△t时，A点在z=z1+△zz= z_1+△zz=z1+△z处
ω(t1+Δt)−β(z1+Δz)+ψ1\omega ({t_1} + \Delta t) - \beta ({z_1} + \Delta z) + \psi {\,_1}ω(t1+Δt)−β(z1+Δz)+ψ1

相同相位点
ωt1−βz1+ψ1=ω(t1+Δt)−β(z1+Δz)+ψ1\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
ωΔt=βΔz\omega \Delta t = \beta \Delta zωΔt=βΔz
相同相位点的移动速度为相位速度即相速
vp=lim⁡Δt→0ΔzΔt=ωβ=1εμ=Cεrμr=λf=λω/2π=ω/k{v_p} = \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta z}}{{\Delta t}} = \frac{\omega }{\beta }= \frac{1}{{\sqrt {\varepsilon \mu } }} = \frac{C}{{\sqrt {{\varepsilon _r}{\mu _r}} }} = \lambda f = \lambda \omega /2\pi = \omega /kvp=Δt→0limΔtΔz=βω=εμ1=εrμrC=λf=λω/2π=ω/k★
群速度vg=dωdk{v_g} = \frac{{d\omega }}{{dk}}vg=dkdω★
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)=∣V0+∣cos⁡(ωt−βz+ϕ+){v^ + }(z,t) = \left| {V_0^ + } \right|\cos (\omega t - \beta z + {\phi ^ + })v+(z,t)=∣∣V0+∣∣cos(ωt−βz+ϕ+)
λβ=2π=ωT\lambda \beta = 2\pi = \omega Tλβ=2π=ωT★
T=1/fT=1/fT=1/f★
λ=2πβ=ωβT=vT=vf\lambda = \frac{{2\pi }}{\beta } = \frac{\omega }{\beta }T = vT{\rm{ = }}\frac{v}{f}λ=β2π=βωT=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

Γ0=V0−V0+=ZL−Z0ZL+Z0{\Gamma _0} = \frac{{{V_0}^ - }}{{{V_0}^ + }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}Γ0=V0+V0−=ZL+Z0ZL−Z0
反射系数的几个常用值
终端匹配ZL=Z0⇒Γ0=0⇒SWR=1+∣Γ0∣1−∣Γ0∣=1{Z_L} = {Z_0} \Rightarrow {\Gamma _0} = 0 \Rightarrow SWR = \frac{{1 + \left| {{\Gamma _0}} \right|}}{{1 - \left| {{\Gamma _0}} \right|}}=1ZL=Z0⇒Γ0=0⇒SWR=1−∣Γ0∣1+∣Γ0∣=1
终端短路ZL=0⇒Γ0=−1⇒SWR=1+∣Γ0∣1−∣Γ0∣=∞{Z_L} = 0 \Rightarrow {\Gamma _0} = - 1 \Rightarrow SWR = \frac{{1 + \left| {{\Gamma _0}} \right|}}{{1 - \left| {{\Gamma _0}} \right|}}=\inftyZL=0⇒Γ0=−1⇒SWR=1−∣Γ0∣1+∣Γ0∣=∞全反射
终端开路ZL=∞⇒Γ0=1⇒SWR=1+∣Γ0∣1−∣Γ0∣=∞{Z_L} = \infty \Rightarrow {\Gamma _0} = 1 \Rightarrow SWR = \frac{{1 + \left| {{\Gamma _0}} \right|}}{{1 - \left| {{\Gamma _0}} \right|}}=\inftyZL=∞⇒Γ0=1⇒SWR=1−∣Γ0∣1+∣Γ0∣=∞全反射

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Γ0=±1{\Gamma _0} = \pm 1Γ0=±1，零点设在端接负载上
V(z)=V0+e−γz+V0−eγz=V0+(e−jβz+Γ0ejβz)=V0+e−jβz[1+Γ(z)]V(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma 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]V(z)=V0+e−γz+V0−eγz=V0+(e−jβz+Γ0ejβz)=V0+e−jβz[1+Γ(z)]★
V(d)=V0+(ejβd±e−jβd)={2V0+cos⁡βd开路j2V0+sin⁡βd短路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.V(d)=V0+(ejβd±e−jβd)={2V0+cosβd开路j2V0+sinβd短路★

时域式
v(t,d)=Re{ejωtV(d)}={2∣V0+∣cos⁡βdcos⁡(ωt+φ+)−2∣V0+∣sin⁡βdsin⁡(ωt+φ+)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)=Acos⁡βdcos⁡(ωt+φ)=∣Acos⁡βd∣cos⁡(ωt+φ(z))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⁡∣Acos⁡βd∣=∣A∣\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⁡∣Acos⁡βd∣=0\min \left| {A\cos \beta d} \right|{\rm{ = }}0min∣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

驻波比SWR∈[1,∞)SWR∈[1,∞)SWR∈[1,∞) ？
SWR=∣VMAX∣∣VMIN∣=1+∣Γ0∣1−∣Γ0∣⇒∣Γ0∣=SWR−1SWR+1SWR = \frac{{\left| {{V_{MAX}}} \right|}}{{\left| {{V_{MIN}}} \right|}} = \frac{{1 + \left| {{\Gamma _0}} \right|}}{{1 - \left| {{\Gamma _0}} \right|}} \Rightarrow \left| {{\Gamma _0}} \right| = \frac{{SWR - 1}}{{SWR + 1}}SWR=∣VMIN∣∣VMAX∣=1−∣Γ0∣1+∣Γ0∣⇒∣Γ0∣=SWR+1SWR−1★

开路线、短路线、四分之一波长传输线

V(z)=V0+e−γz+V0−eγz=V0+e−γz{1+Γ0e2γz}V(z) = {V_0}^ + {e^{ - \gamma z}} + {V_0}^ - {e^{\gamma z}} = {V_0}^ + {e^{ - \gamma z}}\left\{ {1 + {\Gamma _0}^{}{e^{2\gamma z}}} \right\}V(z)=V0+e−γz+V0−eγz=V0+e−γz{1+Γ0e2γz}★
I(z)=V0+Z0e−γz−V0−Z0eγz=V0+Z0e−γz{1−Γ0e2γz}I(z) = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - \gamma z}} - \frac{{{V_0}^ - }}{{{Z_0}}}{e^{\gamma z}} = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - \gamma z}}\left\{ {1 - {\Gamma _0}^{}{e^{2\gamma z}}} \right\}I(z)=Z0V0+e−γz−Z0V0−eγz=Z0V0+e−γz{1−Γ0e2γz}★

Zd=V(−d)I(−d)=Z01+Γ0e−2γd1−Γ0e−2γd=Z01+Γ(−d)1−Γ(−d){Z_d} = \frac{{V( - d)}}{{I( - d)}} = {Z_0}\frac{{1 + {\Gamma _0}^{}{e^{ - 2\gamma d}}}}{{1 - {\Gamma _0}^{}{e^{ - 2\gamma d}}}} = {Z_0}\frac{{1 + \Gamma ( - d)}}{{1 - \Gamma ( - d)}}Zd=I(−d)V(−d)=Z01−Γ0e−2γd1+Γ0e−2γd=Z01−Γ(−d)1+Γ(−d)★

对于无耗传输线 注意公式中 d 的方向。
Zd=Z01+Γ0e−j2βd1−Γ0e−j2βd=Z01+ZL−Z0ZL+Z0e−j2βd1−ZL−Z0ZL+Z0e−j2βd=Z0(ZL+Z0)+(ZL−Z0)e−j2βd(ZL+Z0)−(ZL−Z0)e−j2βd=Z0ZLcos⁡βd+jZ0sin⁡βdZ0cos⁡βd+jZLsin⁡βd=Z0ZL+jZ0tgβdZ0+jZLtgβd{Z_d} = {Z_0}\frac{{1 + {\Gamma _0}^{}{e^{ - j2\beta d}}}}{{1 - {\Gamma _0}^{}{e^{ - j2\beta d}}}} = {Z_0}\frac{{1 + \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}{e^{ - j2\beta d}}}}{{1 - \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}{e^{ - j2\beta d}}}}\\ = {Z_0}\frac{{\left( {{Z_L} + {Z_0}} \right) + \left( {{Z_L} - {Z_0}} \right){e^{ - j2\beta d}}}}{{\left( {{Z_L} + {Z_0}} \right) - \left( {{Z_L} - {Z_0}} \right){e^{ - j2\beta d}}}} \\ = {Z_0}\frac{{{Z_L}\cos \beta d + j{Z_0}\sin \beta d}}{{{Z_0}\cos \beta d + j{Z_L}\sin \beta d}} \\ = {Z_0}\frac{{{Z_L} + j{Z_0}tg\beta d}}{{{Z_0} + j{Z_L}tg\beta d}}Zd=Z01−Γ0e−j2βd1+Γ0e−j2βd=Z01−ZL+Z0ZL−Z0e−j2βd1+ZL+Z0ZL−Z0e−j2βd=Z0(ZL+Z0)−(ZL−Z0)e−j2βd(ZL+Z0)+(ZL−Z0)e−j2βd=Z0Z0cosβd+jZLsinβdZLcosβd+jZ0sinβd=Z0Z0+jZLtgβdZL+jZ0tgβd
Zin≡Zd=Z0ZL+jZ0tgβdZ0+jZLtgβd{Z_{in}} \equiv {Z_d} = {Z_0}\frac{{{Z_L} + j{Z_0}tg\beta d}}{{{Z_0} + j{Z_L}tg\beta d}}Zin≡Zd=Z0Z0+jZLtgβdZL+jZ0tgβd★
阻抗匹配时Zin=Z0{Z_{in}} = {Z_0}Zin=Z0★
d=∣l∣{\rm{ }}d = \left| l \right|d=∣l∣
βd=2πλd\beta d = \frac{{2\pi }}{\lambda }dβd=λ2πd

终端短路传输线

ZL=0{Z_L} = 0ZL=0★
Zin0=jZ0tgβd{Z_{in0}} = j{Z_0}tg\beta dZin0=jZ0tgβd
d=λ4⇒βd=π2⇒Zin→∞d = \frac{\lambda }{4} \Rightarrow \beta d = \frac{\pi }{2} \Rightarrow {Z_{in}} \to \inftyd=4λ⇒βd=2π⇒Zin→∞
d=λ2⇒βd=π⇒Zin=0d = \frac{\lambda }{2} \Rightarrow \beta d = \pi \Rightarrow {Z_{in}} = 0d=2λ⇒βd=π⇒Zin=0
Zin(d){Z_{in}}(d)Zin(d)的周期=λ2\frac{\lambda }{2}2λ

终端开路的传输线

ZL→∞{Z_L} \to \inftyZL→∞★
Zin∞=Z0ZL+jZ0tgβdZ0+jZLtgβd=−jZ0ctgβd{Z_{in\infty}} = {Z_0}\frac{{{Z_L} + j{Z_0}tg\beta d}}{{{Z_0} + j{Z_L}tg\beta d}} = {\rm{ - }}j{Z_0}ctg\beta dZin∞=Z0Z0+jZLtgβdZL+jZ0tgβd=−jZ0ctgβd
1/j=−j1/j=-j1/j=−j

特征阻抗Z0=Zin0Zin∞{Z_0} = \sqrt {{Z_{in0}}{Z_{in\infty }}}Z0=Zin0Zin∞
无论对分布参数网络还是集总参数网络都可以这样定义
当终端匹配时，输入阻抗等于特征阻抗

1/4波长传输线

tgβd→∞tg\beta d \to \inftytgβd→∞★
Zin=Z0ZL+jZ0tgβdZ0+jZLtgβd=Z02ZL{Z_{in}} = {Z_0}\frac{{{Z_L} + j{Z_0}tg\beta d}}{{{Z_0} + j{Z_L}tg\beta d}} = \frac{{Z_0^2}}{{{Z_L}}}Zin=Z0Z0+jZLtgβdZL+jZ0tgβd=ZLZ02★
一般说来Z0为纯电阻，实数
因此四分之一波长传输线可看成阻抗变换器，终端端接电容，输入端呈电感特性。一端开路，另一端短路。

★★★		阻抗圆图	导纳圆图
圆心在上半平面1/x>0	电感性	电抗x>0	电纳b<0
圆心在下半平面1/x<0	电容性	电抗x<0	电纳b>0

1/2波长的端接传输线

tgβd=0tg\beta d = 0tgβd=0★
Zin=Z0ZL+jZ0tgβdZ0+jZLtgβd=ZL{Z_{in}} = {Z_0}\frac{{{Z_L} + j{Z_0}tg\beta d}}{{{Z_0} + j{Z_L}tg\beta d}} = {Z_L}Zin=Z0Z0+jZLtgβdZL+jZ0tgβd=ZL

传输线常用公式和近似值★★★

传输线参数一般工程近似值（标称值）：
γ=[R+jωL][G+jωC]=α+jβ,α≈0,β≈ωLC\gamma {\rm{ = }}\sqrt {\left[ {R + j\omega L} \right]\left[ {G + j\omega C} \right]} = \alpha + j\beta,\alpha \approx {\rm{0 }},\beta \approx \omega \sqrt {LC}γ=[R+jωL][G+jωC]=α+jβ,α≈0,β≈ωLC
Z0=[R+jωL][G+jωC]≈LC={50Ω75Ω{Z_0}{\rm{ = }}\sqrt {\frac{{\left[ {R + j\omega L} \right]}}{{\left[ {G + j\omega C} \right]}}} \approx \sqrt {\frac{L}{C}} = \left\{ \begin{array}{l} 50\Omega \\ 75\Omega \end{array} \right.Z0=[G+jωC][R+jωL]≈CL={50Ω75Ω
vp=ωβ≈1LC≥0.77vC{v_p}{\rm{ = }}\frac{\omega }{\beta } \approx \frac{1}{{\sqrt {LC} }} \ge 0.77{v_C}vp=βω≈LC1≥0.77vC
λ=vpf=1fLC=2πββ=2πλ\lambda = \frac{{{v_p}}}{f} = \frac{1}{{f\sqrt {LC} }} = \frac{{2\pi }}{\beta }\beta = \frac{{2\pi }}{\lambda }λ=fvp=fLC1=β2πβ=λ2π

无耗传输线传输线方程及参数关系
V(z)=V0+e−jβz+V0−ejβz=V+(z)+V−(z)V(z) = {V_0}^ + {e^{ - j\beta z}} + {V_0}^ - {e^{j\beta z}}= {V^ + }(z) + {V^ - }(z)V(z)=V0+e−jβz+V0−ejβz=V+(z)+V−(z)
V+(z)=V0+e−jβz{V^ + }(z) = {V_0}^ + {e^{ - j\beta z}}V+(z)=V0+e−jβz
V−(z)=V0−ejβz{V^ - }(z) = {V_0}^ - {e^{j\beta z}}V−(z)=V0−ejβz
I(z)=I0+e−jβz+I0−ejβz=V0+Z0e−jβz−V0−Z0ejβzI(z) = {I_0}^ + {e^{ - j\beta z}} + {I_0}^ - {e^{j\beta z}} = \frac{{{V_0}^ + }}{{{Z_0}}}{e^{ - j\beta z}} - \frac{{{V_0}^ - }}{{{Z_0}}}{e^{j\beta z}}I(z)=I0+e−jβz+I0−ejβz=Z0V0+e−jβz−Z0V0−ejβz
I0+=V0+/V0+Z0Z0{I_0}^ + = {{{V_0}^ + } \mathord{\left/ {\vphantom {{{V_0}^ + } {{Z_0}}}} \right.} {{Z_0}}}I0+=V0+/V0+Z0Z0
I0−=−V0−/V0−Z0Z0{I_0}^ - = - {{{V_0}^ - } \mathord{\left/ {\vphantom {{{V_0}^ - } {{Z_0}}}} \right.} {{Z_0}}}I0−=−V0−/V0−Z0Z0

Γ0=Γ(0)=V0−V0+=ZL−Z0ZL+Z0{\Gamma _0} = \Gamma (0) = \frac{{{V_0}^ - }}{{{V_0}^ + }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}Γ0=Γ(0)=V0+V0−=ZL+Z0ZL−Z0
ZL=Z01+Γ01−Γ0{Z_L} = {Z_0}\frac{{1 + {\Gamma _0}}}{{1 - {\Gamma _0}}}ZL=Z01−Γ01+Γ0

Γ(z)=Γ0ej2βz=V−(z)V+(z)=Zin(z)−Z0Zin(z)+Z0\Gamma (z){\rm{ = }}{\Gamma _0}{e^{j2\beta z}} = \frac{{{V^ - }(z)}}{{{V^ + }(z)}} = \frac{{{Z_{in}}(z) - {Z_0}}}{{{Z_{in}}(z) + {Z_0}}}Γ(z)=Γ0ej2βz=V+(z)V−(z)=Zin(z)+Z0Zin(z)−Z0
Zin(z)=Z01+Γ(z)1−Γ(z)=Z01+Γ0ej2βz1−Γ0ej2βz=Z0ZL−jZ0tgβzZ0−jZLtgβz=Z0ZL+jZ0tgβdZ0+jZLtgβd,d=−z{Z_{in}}(z) = {Z_0}\frac{{1 + \Gamma (z)}}{{1 - \Gamma (z)}} = {Z_0}\frac{{1 + {\Gamma _0}{e^{j2\beta z}}}}{{1 - {\Gamma _0}{e^{j2\beta z}}}} = {Z_0}\frac{{{Z_L} - j{Z_0}tg\beta z}}{{{Z_0} - j{Z_L}tg\beta z}} = {Z_0}\frac{{{Z_L} + j{Z_0}tg\beta d}}{{{Z_0} + j{Z_L}tg\beta d}},d = - zZin(z)=Z01−Γ(z)1+Γ(z)=Z01−Γ0ej2βz1+Γ0ej2βz=Z0Z0−jZLtgβzZL−jZ0tgβz=Z0Z0+jZLtgβdZL+jZ0tgβd,d=−z

Zin(z)=jZ0tgβd={jZ0tg2πλd,ZL=0短路−jZ0ctg2πλd,ZL→∞开路Z02ZL,d=λ/4,tgβd→∞ZL,d=λ/2,tgβd=0{Z_{in}}(z) = j{Z_0}tg\beta d = \left\{ \begin{array}{l} j{Z_0}tg\frac{{2\pi }}{\lambda }d,{Z_L} = 0短路\\ -j{Z_0}ctg\frac{{2\pi }}{\lambda }d,{Z_L} \to \infty开路\\ \frac{{Z_0^2}}{{{Z_L}}},d = \lambda/{4},tg\beta d \to \infty\\ {Z_L},d =\lambda/{2},tg\beta d = 0 \end{array} \right.Zin(z)=jZ0tgβd=⎩⎪⎪⎨⎪⎪⎧jZ0tgλ2πd,ZL=0短路−jZ0ctgλ2πd,ZL→∞开路ZLZ02,d=λ/4,tgβd→∞ZL,d=λ/2,tgβd=0

SWR=1+∣Γ∣1−∣Γ∣⇒∣Γ∣=SWR−1SWR+1SWR = \frac{{1 + \left| \Gamma \right|}}{{1 - \left| \Gamma \right|}}{\rm{ }} \Rightarrow {\rm{ }}\left| \Gamma \right| = \frac{{SWR - 1}}{{SWR + 1}}SWR=1−∣Γ∣1+∣Γ∣⇒∣Γ∣=SWR+1SWR−1

作业

《射频电路设计——理论与应用》
第二章中习题2.16, 2.19,2.25,2.28 2.31,2.33

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