*各种波导

常见的有矩形，圆形，同轴，平行双线，平板（微带线）等

*波导中的导波模式

一般取沿着波导的传播方向为z轴

E⃗=E⃗(x,y)eiβz\vec{E} = \vec{E}(x,y) e^{i \beta z}

H⃗=H⃗(x,y)eiβz\vec{H} = \vec{H}(x,y) e^{i \beta z}

有两种解法

  • 考虑

∇×E⃗=iωμH⃗∇×H⃗=−iωεE⃗\begin{aligned} &\nabla \times \vec{E} = i \omega \mu \vec{H}\\ &\nabla \times \vec{H} = -i \omega \varepsilon \vec{E} \end{aligned}

使用纵横分解法（见教材）

  • 考虑

∇2E⃗+k2E⃗=0∇⋅E⃗=0H⃗=1iωμ∇×E⃗\begin{aligned} &\nabla^2 \vec{E} + k^2 \vec{E} = 0\\ &\nabla \cdot \vec{E} = 0\\ & \vec{H} = \frac{1}{i\omega\mu} \nabla \times \vec{E} \end{aligned}

分离变量法求E⃗\vec{E}，再求H⃗\vec{H}

*纵横分解法

E⃗=E⃗(x,y)eiβz=(E⃗t+E⃗z)eiβz\vec{E} = \vec{E}(x,y)e^{i \beta z} =(\vec{E}_t + \vec{E}_z)e^{i \beta z}

H⃗=H⃗(x,y)eiβz=(H⃗t+H⃗z)eiβz\vec{H} = \vec{H}(x,y)e^{i \beta z} =(\vec{H}_t + \vec{H}_z)e^{i \beta z}

这里的E⃗t\vec{E}_t是横向，E⃗z\vec{E}_z是纵向
并且我们考虑

∇=∇t+e^z∂∂z\nabla = \nabla_t + \hat{e}_z \frac{\partial}{\partial z}

那么有

∇t2Ez+kc2Ez=0∇t2Hz+kc2Hz=0E⃗t=ikc2(β∇tEz−ωμe^z×∇tHz)H⃗t=ikc2(β∇tHz+ωεe^z×∇tEz)\begin{aligned} &\nabla^2_t E_z + k^2_c E_z = 0\\ &\nabla^2_t H_z + k^2_c H_z = 0\\ &\vec{E}_t = \frac{i}{k_c^2}(\beta \nabla_t E_z - \omega \mu \hat{e}_z \times \nabla_t H_z)\\ &\vec{H}_t = \frac{i}{k_c^2}(\beta \nabla_t H_z + \omega \varepsilon \hat{e}_z \times \nabla_t E_z)\\ \end{aligned}

前两个方程是纵向场方程，其中kc2=k2−β2k_c^2 = k^2-\beta^2
求解出纵向场后，便可通过后两个方程求出横向场
如果kc≠0k_c \neq 0

  • Ez=0E_z = 0,Hz=0H_z = 0，称为TE模式
  • Ez≠0E_z \neq 0,Hz=0H_z = 0，称为TM模式
  • Ez=0E_z = 0,Hz=0H_z = 0，必然有kc=0k_c = 0，称为TEM模式
    此时有

∇t×E⃗t=0∇t⋅E⃗t=0Ht=kωμe^z×E⃗t=e^z×E⃗tηTEM\begin{aligned} &\nabla_t \times \vec{E}_t = 0\\ &\nabla_t \cdot \vec{E}_t = 0\\ &H_t = \frac{k}{\omega \mu} \hat{e}_z \times \vec{E}_t = \hat{e}_z \times \frac{\vec{E}_t}{\eta_{TEM}} \end{aligned}

其中ηTEM=με\eta_{TEM} = \sqrt{\frac{\mu}{\varepsilon}}
只有当波导横截面支持非零二维静电场时，波导才支持TEM模式
单导体封闭波导一定不能支持TEM模式,多导体波导可以支持TEM模式

*例

考虑矩形波导中的TM模式
纵向电场EzE_z满足：

∇t2Ez+kc2Ez=0\nabla^2_t E_z + k^2_c E_z = 0

设Ez=X(x)Y(y)E_z = X(x)Y(y)

1Xd2Xdx2=−kx21Yd2Ydy2=−ky2kx2+ky2=kc2\begin{aligned} &\frac{1}{X} \frac{d^2 X}{dx^2} = -k_x^2\\ &\frac{1}{Y} \frac{d^2 Y}{dy^2} = -k_y^2\\ &k_x^2 + k_y^2 = k_c^2 \end{aligned}

由边界条件Ez∣x=0=Ez∣x=a=0⇒X(0)=0,X(a)=0E_z |_{x=0} = E_z |_{x=a} = 0 \Rightarrow X(0) = 0,X(a) = 0
所以

X(x)=Asin⁡mπaxX(x) = A \sin \frac{m \pi}{a} x

同理有

Y(y)=Bsin⁡nπbyY(y) = B \sin \frac{n \pi}{b} y

所以

Ez=E0sin⁡mπaxsin⁡nπbyE_z = E_0 \sin \frac{m \pi}{a} x \sin \frac{n \pi}{b} y

kc2=m2π2a2+n2π2b2k_c^2 = \frac{m^2 \pi^2}{a^2} + \frac{n^2 \pi^2}{b^2}

E⃗t=iβkc2mπaE0cos⁡mπaxsin⁡nπbye^x+iβkc2nπbE0sin⁡mπaxcos⁡nπbye^y\vec{E}_t = \frac{i\beta}{k_c^2}\frac{m\pi}{a} E_0 \cos \frac{m \pi}{a} x \sin \frac{n \pi}{b} y \hat{e}_x + \frac{i\beta}{k_c^2}\frac{n\pi}{b} E_0 \sin \frac{m \pi}{a} x \cos \frac{n \pi}{b} y \hat{e}_y

H⃗t=−iωεkc2nπbE0sin⁡mπaxcos⁡nπbye^x+iωεkc2mπaE0cos⁡mπaxsin⁡nπbye^y\vec{H}_t = -\frac{i\omega \varepsilon}{k_c^2} \frac{n\pi}{b} E_0 \sin \frac{m \pi}{a} x \cos \frac{n \pi}{b} y \hat{e}_x + \frac{i\omega \varepsilon}{k_c^2} \frac{m\pi}{a} E_0 \cos \frac{m \pi}{a} x \sin \frac{n \pi}{b} y \hat{e}_y

  • iE记做TMmnTM_{mn}模式
  • 基模为TMnTM_{n}
  • 纵向波矢 β=k2−kc2=1cω2−ωc2\beta = \sqrt{k^2 - k_c^2} = \frac{1}{c} \sqrt{\omega^2 - \omega^2_c}

ωc=ckc=cm2π2a2+n2π2b2\omega_c = ck_c = c\sqrt{\frac{m^2 \pi^2}{a^2} + \frac{n^2 \pi^2}{b^2}}

对于某一给定的模式TMmnTM_{mn}，ωc\omega_c一定，只有当ω>ωc\omega > \omega_c时，β\beta才是实数，此时电磁场才是行波（导波）
ωc\omega_c称为截止频率

  • 相速度

Vp=ωβ=c1−ωc2ω2V_p = \frac{\omega}{\beta} = \frac{c}{\sqrt{1 - \frac{\omega_c^2}{\omega^2}}}

称为波导色散（模式色散）

  • 群速度

Vg=dωdβ=c1−ωc2ω2V_g = \frac{d\omega}{d\beta} = c\sqrt{1 - \frac{\omega_c^2}{\omega^2}}

群速度色散

TE模式

基模：TE01TE_{01}或者TE10TE_{10}

TEM模式

  • kc=0,β=kk_c = 0,\beta = k
  • 截止频率ωc=ckc=0\omega_c = ck_c = 0
  • 无模式色散

谐振腔

kx=mπa,ky=nπb,kz=pπck_x = \frac{m\pi}{a},k_y = \frac{n\pi}{b},k_z = \frac{p\pi}{c}

kx2+ky2+kz2=k2=ωμεk_x^2 + k_y^2 + k_z^2 = k^2 = \omega \mu \varepsilon

所以角频率

ω=1μεkx2+ky2+kz2\omega = \frac{1}{\sqrt{\mu \varepsilon}} \sqrt{k_x^2 + k_y^2 + k_z^2}

真空中就有

ω=ckx2+ky2+kz2\omega = c \sqrt{k_x^2 + k_y^2 + k_z^2}

*电磁场的辐射

时域分析：
已知全空间中随着时间变化的ρ,J⃗\rho,\vec{J}
求E⃗,B⃗\vec{E},\vec{B}

∇×E⃗=−∂B⃗∂t∇×B⃗=μ0J⃗+μ0ε0∂E⃗∂t∇⋅E⃗=ρε0∇⋅B⃗=0\begin{aligned} &\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}\\ &\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t}\\ &\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}\\ &\nabla \cdot \vec{B} = 0 \end{aligned}

（一）

∇⋅B⃗=0⇒B⃗=∇×A⃗\nabla \cdot \vec{B} = 0 \Rightarrow \vec{B} = \nabla \times \vec{A}

（二）
B⃗=∇×A⃗\vec{B} = \nabla \times \vec{A}代入，有

∇×E⃗=−∂∂t(∇×A⃗)=−∇×∂A⃗∂t\nabla \times \vec{E} = -\frac{\partial}{\partial t} (\nabla \times \vec{A}) = -\nabla \times \frac{\partial \vec{A}}{\partial t}

∇×(E⃗+∂A⃗∂t)=0⇒E⃗+∂A⃗∂t=−∇ϕ\nabla \times (\vec{E} + \frac{\partial \vec{A}}{\partial t}) = 0 \Rightarrow \vec{E} + \frac{\partial \vec{A}}{\partial t} = -\nabla \phi

E⃗=−∇ϕ−∂A⃗∂t\vec{E} = -\nabla \phi - \frac{\partial \vec{A}}{\partial t}

即存在标势ϕ\phi和矢势A⃗\vec{A}，使得

E⃗=−∇ϕ−∂A⃗∂tB⃗=∇×A⃗\begin{aligned} \vec{E} = -\nabla \phi - \frac{\partial \vec{A}}{\partial t}\\ \vec{B} = \nabla \times \vec{A} \end{aligned}

（三）将E⃗,B⃗\vec{E},\vec{B}代入Maxwell方程组

−∇⋅(∇ϕ+∂A⃗∂t)=ρε0∇2ϕ+∂∂t(∇⋅A⃗)=−ρε0\begin{aligned} -\nabla \cdot (\nabla \phi + \frac{\partial \vec{A}}{\partial t}) &= \frac{\rho}{\varepsilon_0}\\ \nabla^2 \phi + \frac{\partial}{\partial t} (\nabla \cdot \vec{A}) &= -\frac{\rho}{\varepsilon_0}\\ \end{aligned}

（四）将E⃗,B⃗\vec{E},\vec{B}代入（三）的结论

∇×(∇×A⃗)=μ0J⃗+μ0ε0∂∂t(−∇ϕ−∂A⃗∂t)∇2A⃗−1c2∂2A⃗∂t2−∇(∇⋅A⃗+1c2∂ϕ∂t)=−μ0J⃗\begin{aligned} &\nabla \times (\nabla \times \vec{A}) = \mu_0 \vec{J} + \mu_0 \varepsilon_0 \frac{\partial}{\partial t} (-\nabla \phi - \frac{\partial \vec{A}}{\partial t})\\ &\nabla^2 \vec{A} - \frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2} -\nabla (\nabla \cdot \vec{A} + \frac{1}{c^2} \frac{\partial \phi}{\partial t}) = -\mu_0 \vec{J} \end{aligned}

ϕ,A⃗\phi,\vec{A}不唯一
假设他们满足

E⃗=−∇ϕ−∂A⃗∂tB⃗=∇×A⃗\begin{aligned} &\vec{E} = -\nabla \phi - \frac{\partial \vec{A}}{\partial t}\\ &\vec{B} = \nabla \times \vec{A} \end{aligned}

取（规范变换）

φ′=φ−∂A⃗∂tA⃗′=A⃗+∇Ψ\begin{aligned} &\varphi' = \varphi - \frac{\partial \vec{A}}{\partial t}\\ &\vec{A}' = \vec{A} + \nabla \Psi \end{aligned}

则：

E⃗′=−∇φ′−∂A⃗′∂t=E⃗B⃗′=∇×A⃗′=B⃗\begin{aligned} &\vec{E}' = -\nabla \varphi' - \frac{\partial \vec{A}'}{\partial t} = \vec{E}\\ &\vec{B}' = \nabla \times \vec{A}' = \vec{B} \end{aligned}

附加约束条件：

  • ∇⋅A⃗=0\nabla \cdot \vec{A} = 0（Coulomb规范）

∇2φ=−ρε0∇2A⃗−1c2∂2A⃗∂t2=−μ0J⃗+1c2∂∂t(∇φ)\begin{aligned} &\nabla^2 \varphi = -\frac{\rho}{\varepsilon_0}\\ &\nabla^2 \vec{A} - \frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2} = -\mu_0 \vec{J} + \frac{1}{c^2} \frac{\partial}{\partial t}(\nabla \varphi) \end{aligned}

φ(x⃗,t)=14πε0∫ρ(x′⃗,t)rdV′\varphi(\vec{x},t) = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho(\vec{x'},t)}{r} dV'

  • ∇⋅A⃗+1c2∂φ∂t=0\nabla \cdot \vec{A} + \frac{1}{c^2} \frac{\partial \varphi}{\partial t} = 0（Lorentz规范）

∇2φ−1c2∂2φ∂t2=−ρε0∇2A⃗−1c2∂2A⃗∂t2=−μ0J⃗\begin{aligned} &\nabla^2 \varphi - \frac{1}{c^2} \frac{\partial^2 \varphi}{\partial t^2} = -\frac{\rho}{\varepsilon_0}\\ &\nabla^2 \vec{A} - \frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2} = -\mu_0 \vec{J} \end{aligned}

即d’Alembert方程
推迟势解：

φ(x⃗,t)=14πε0∫ρ(x′⃗,tr)rdV′A⃗(x⃗,t)=μ04π∫J⃗(x′⃗,tr)rdV′\begin{aligned} &\varphi(\vec{x},t) = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho(\vec{x'},t_r)}{r} dV'\\ &\vec{A}(\vec{x},t) = \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{x'},t_r)}{r} dV' \end{aligned}

其中tr=t−rct_r = t - \frac{r}{c}，r=∣x⃗−x′⃗∣r = |\vec{x} - \vec{x'}|
被称为推迟时间