EM Wave

EM Wave Equations¶

in free space

by Maxwell's equations, we have

\begin{aligned} (1) & \nabla \times \vec{E} & = - μ_{0} \frac{\partial \vec{H}}{\partial t} \\ \nabla \times \vec{H} & = ε_{0} \frac{\partial \vec{E}}{\partial t} \end{aligned}

^equ-1

\begin{aligned} \nabla^{2} \vec{E} & = \nabla (\nabla \cdot \vec{E}) - \nabla \times \nabla \times \vec{E} \\ = 0 - (- μ_{0} \frac{\partial}{\partial t} (\nabla \times \vec{H})) \\ = μ_{0} ε_{0} \frac{\partial^{2} \vec{E}}{\partial t^{2}} \end{aligned}

and similar for $\nabla^{2} \vec{H}$ .

We can now have the wave equations

{\begin{aligned} \nabla^{2} \vec{E} & = μ_{0} ε_{0} \frac{\partial^{2} \vec{E}}{\partial t^{2}} \\ \nabla^{2} \vec{H} & = μ_{0} ε_{0} \frac{\partial^{2} \vec{H}}{\partial t^{2}} \end{aligned}

compare to standard form of wave equation

\begin{matrix} \frac{\partial^{2} \vec{U}}{\partial t^{2}} = c^{2} \nabla^{2} \vec{U} \end{matrix}

we have

\begin{matrix} c = \frac{1}{\sqrt{μ_{0} ε_{0}}} \end{matrix}

General Solutions (Frequency Domain)¶

\begin{matrix} \nabla^{2} \vec{E_{s}} - γ^{2} \vec{E_{s}} = 0 \\ \nabla^{2} \vec{H_{s}} - γ^{2} \vec{H_{s}} = 0 \end{matrix}

in which $γ$ is propagation constant.

\begin{matrix} γ = \sqrt{j ω μ (σ + j ω ε)} \end{matrix}

Thinking

Faraday's Law

\begin{matrix} \nabla \times \vec{E_{s}} = j ω μ \vec{H_{s}} \end{matrix}

Ampere's Law

\begin{matrix} \nabla \times \vec{H_{s}} = σ \vec{E_{s}} + j ω ε \vec{E_{s}} \end{matrix}

General Solutions (Transmission Line)¶

Take the wave propagates through $\hat{a_{z}}$ direction, then

{\begin{array}{r} \vec{E} = E_{x}^{+} (z - u t) \hat{a_{x}} + E_{x}^{-} (z + u t) \hat{a_{x}} \\ \vec{H} = H_{y}^{+} (z - u t) \hat{a_{y}} + H_{y}^{-} (z + u t) \hat{a_{y}} \end{array}

from $equation (a)$ we have

\begin{aligned} μ_{0} \frac{\partial \vec{H}}{\partial t} & = - \nabla \times \vec{E} = - \frac{\partial \vec{E}}{\partial z} \\ = - ({E_{x}^{+}}^{'} (z - u t) \hat{a_{x}} + {E_{x}^{-}}^{'} (z + u t) \hat{a_{x}}) \\ ⟹ \vec{H} & = - \frac{1}{μ_{0}} \int ({E_{x}^{+}}^{'} (z - u t) \hat{a_{x}} + {E_{x}^{-}}^{'} (z + u t) \hat{a_{x}}) d t \\ = \frac{1}{u μ_{0}} (E_{x}^{+} (z - u t) \hat{a_{x}} - E_{x}^{-} (z + u t) \hat{a_{x}}) \\ = \frac{1}{\sqrt{μ_{0} / ε_{0}}} (E_{x}^{+} (z - u t) \hat{a_{x}} - E_{x}^{-} (z + u t) \hat{a_{x}}) \end{aligned}

Thus we can obtain the wave impedance in free space $η_{0}$

\begin{matrix} η_{0} = \sqrt{\frac{μ_{0}}{ε_{0}}} = \frac{E_{x}^{+}}{H_{x}^{+}} = - \frac{E_{x}^{-}}{H_{x}^{-}} \end{matrix}

Power¶

Poynting Vector

\begin{matrix} \vec{P} = \vec{E} \times \vec{H} {W/m}^{2} \end{matrix}

Instantaneous power across $S$

\begin{matrix} P (t) = \int_{S} \vec{P} \cdot d \vec{S} \end{matrix}

complex Poynting vector
not a phasor, but come up by phasors trickily.
notice that complex Poynting vector is not function of $t$ . Since it is generated from phasors, it represents the concept of average.

\begin{matrix} {\vec{P}}_{c} = \frac{1}{2} {\vec{E}}_{s} \times {\vec{H}}_{s}^{*} = \frac{1}{2} {\vec{E}}_{s}^{*} \times {\vec{H}}_{s} \\ {\vec{P}}_{a v e} = Re [{\vec{P}}_{c}] {W/m}^{2} \end{matrix}

Loss Tangent¶

for lossless medium, $σ = 0$ .
propagation constant $γ$

\begin{matrix} γ = \sqrt{j ω μ (σ + j ω ε)} \end{matrix}

Loss Tangent

\begin{matrix} \tan θ = \frac{J_{c s}}{J_{d s}} = \frac{| σ E_{s} |}{| j ω ε E_{s} |} = \frac{σ}{ω ε} \end{matrix}

Loss Angle

\begin{matrix} θ = \tan^{- 1} \frac{σ}{ω ε} \end{matrix}

Skin Depth

\begin{matrix} δ = \frac{1}{α} \end{matrix}

Wave Polarization¶

\begin{matrix} \vec{E} = E_{x} \hat{a_{x}} + E_{y} \hat{a_{y}} \\ = E_{o x} \cos (ω t - β z + ϕ_{x}) \hat{a_{x}} + E_{o y} \cos (ω t - β z + ϕ_{y}) \hat{a_{y}} \\ Δ ϕ = ϕ_{y} - ϕ_{x} \end{matrix}

Linear Polarization¶

\begin{aligned} \vec{E} & = E_{x} \hat{a_{x}} + E_{y} \hat{a_{y}} \\ = E_{1} \hat{a_{1}} + E_{2} \hat{a_{2}} \end{aligned}

$Δ ϕ = n π, n \in Z^{+}$

\begin{matrix} \frac{E_{y}}{E_{x}} = \frac{E_{o y} \cos (ω t - β z + ϕ_{x} + Δ ϕ)}{E_{o x} \cos (ω t - β z + ϕ_{x})} = \frac{E_{o y}}{E_{o x}} \end{matrix}

no necessarily perpendicular

Circular Polarization¶

$E_{o x} = E_{o y}$
$Δ ϕ = (\frac{1}{2} + n) π$

\begin{matrix} \frac{E_{y}}{E_{x}} = \frac{E_{o y} \cos (ω t - β z + ϕ_{x} + Δ ϕ)}{E_{o x} \cos (ω t - β z + ϕ_{x})} = \pm \frac{E_{o y}}{E_{o x}} \tan (ω t - β z + ϕ_{x}) \end{matrix}

thus,

\begin{matrix} ϕ = \tan^{- 1} \frac{E_{y}}{E_{x}} = \pm (ω t - β z + ϕ_{x}) ⟹ circular \end{matrix}

necessarily perpendicular

Elliptical Polarization¶

if $Δ ϕ > 0$ , $E_{y}$ leads $E_{x}$ , (left-handed)
if $Δ ϕ < 0$ , $E_{y}$ lags $E_{x}$ , (right-handed)

Reflection and Transmission¶

Snell's Law¶

lossless¶

\begin{matrix} k_{i} \sin θ_{i} = k_{r} \sin θ_{r} = k_{t} \sin θ_{t} \end{matrix}

lossy¶

\begin{matrix} γ_{i} \sin θ_{i} = γ_{r} \sin θ_{r} = γ_{t} \sin θ_{t} \end{matrix}

True Angle of Refraction¶

Since $γ$ is complex, $\sin θ_{t}$ and $\sin θ_{r}$ is basically complex as well.

Thus

\begin{aligned} {\vec{γ}}_{t} \cdot \vec{r} & = γ_{t} \sin θ_{t} x + γ_{t} \cos θ_{t} z \\ = ℜ [γ_{t} \sin θ_{t}] x + ℜ [γ_{t} \cos θ_{t}] z + j [ℑ [γ_{t} \sin θ_{t}] x + ℑ [γ_{t} \cos θ_{t}] z] \end{aligned}

True angle of refraction $ψ_{β}$

\begin{matrix} ψ_{β} = \tan^{- 1} \frac{ℑ [γ_{t} \sin θ_{t}]}{ℑ [γ_{t} \cos θ_{t}]} \end{matrix}

Polarization¶

Parallel Polarization¶

$E$ is parallel to surface formed by ( $k_{i}, k_{r}, k_{t}$ )

at $z = 0$ ,

solve,

{\begin{aligned} E_{i 0} \cos θ_{i} + E_{r 0} \cos θ_{r} = E_{t 0} \cos θ_{t} \\ H_{i 0} - H_{r 0} = H_{t 0} \end{aligned}

then we have

\begin{matrix} Γ_{∥} = \frac{E_{r 0}}{E_{i 0}} = \frac{η_{2} \cos θ_{t} - η_{1} \cos θ_{i}}{η_{2} \cos θ_{t} + η_{1} \cos θ_{i}} \\ τ_{∥} = \frac{E_{t 0}}{E_{i 0}} = \frac{2 η_{2} \cos θ_{i}}{η_{2} \cos θ_{t} + η_{1} \cos θ_{i}} \end{matrix}

Perpendicular Polarization¶

at $z = 0$ ,

solve,

{\begin{aligned} - H_{i 0} \cos θ_{i} + H_{r 0} \cos θ_{r} = - H_{t 0} \cos θ_{t} \\ E_{i 0} + E_{r 0} = E_{t 0} \end{aligned}

then we have

\begin{matrix} Γ_{⊥} = \frac{E_{r 0}}{E_{i 0}} = \frac{η_{2} \cos θ_{i} - η_{1} \cos θ_{t}}{η_{2} \cos θ_{i} + η_{1} \cos θ_{t}} \\ τ_{⊥} = \frac{E_{t 0}}{E_{i 0}} = \frac{2 η_{2} \cos θ_{i}}{η_{2} \cos θ_{i} + η_{1} \cos θ_{t}} \end{matrix}

Brewster Angle¶

Denote as $θ_{B ∥}$
a special $θ_{i}$ that make no reflection.

For nonmagnetic medium ( $μ_{r} = 1$ ), only exists at parallel polarization.

\begin{matrix} Γ_{∥} = \frac{η_{2} \cos θ_{t} - η_{1} \cos θ_{i}}{η_{2} \cos θ_{t} + η_{1} \cos θ_{i}} = 0 \\ ⟹ η_{2} \cos θ_{t} = η_{1} \cos θ_{B ∥} \end{matrix}

\begin{matrix} η_{2}^{2} \cos^{2} θ_{t} = η_{1}^{2} \cos^{2} θ_{B ∥} \\ \frac{μ_{2}}{ε_{2}} (1 - \frac{μ_{1} ε_{1}}{μ_{2} ε_{2}} \sin^{2} θ_{B ∥}) = \frac{μ_{1}}{ε_{1}} (1 - \sin^{2} θ_{B ∥}) \end{matrix}

for non-magnetic medium $μ_{1} = μ_{2} = μ_{0}$

\begin{aligned} ⟹ \sin^{2} θ_{B ∥} & = \frac{\frac{μ_{2}}{ε_{2}} - \frac{μ_{1}}{ε_{1}}}{\frac{μ_{1} ε_{1}}{μ_{2} ε_{2}} \frac{μ_{2}}{ε_{2}} - \frac{μ_{1}}{ε_{1}}} \\ = \frac{\frac{μ_{0}}{ε_{2}} - \frac{μ_{0}}{ε_{1}}}{\frac{ε_{1}}{ε_{2}} \frac{μ_{0}}{ε_{2}} - \frac{μ_{0}}{ε_{1}}} \\ = \frac{ε_{1} - ε_{2}}{\frac{ε_{1}^{2}}{ε_{2}} - ε_{2}} \\ = \frac{ε_{1} - ε_{2}}{\frac{1}{ε_{2}} (ε_{1}^{2} - ε_{2}^{2})} \\ = \frac{ε_{2}}{ε_{1} + ε_{2}} \end{aligned}

then,

\begin{matrix} \tan θ_{B ∥} = \sqrt{\frac{ε_{2}}{ε_{1}}} \end{matrix}

Normal Incidence¶

Power¶

incident power

\begin{matrix} P_{i} = \frac{1}{2} ℜ [E_{i 0} H_{i 0}^{*}] = \frac{1}{2} \frac{| E_{i 0} |^{2}}{ℜ [η_{1}^{*}]} \end{matrix}

reflected power

\begin{aligned} P_{r} & = - \frac{1}{2} ℜ [E_{r 0} H_{r 0}^{*}] \\ = - \frac{1}{2} \frac{| Γ E_{i 0} |^{2}}{ℜ [η_{1}^{*}]} \\ = - | Γ |^{2} P_{i} \end{aligned}

\begin{matrix} E_{t s} = E_{t 0} e^{- γ_{2} z} = E_{i 0} (1 + Γ) e^{- γ_{2} z} \\ H_{t s} = H_{t 0} e^{- γ_{2} z} = \frac{E_{t 0}}{η_{2}} e^{- γ_{2} z} = \frac{E_{i 0} (1 + Γ)}{η_{2}} e^{- γ_{2} z} \\ = \frac{H_{i 0} η_{1} (1 + Γ)}{η_{2}} e^{- γ_{2} z} = H_{i 0} (1 - Γ) e^{- γ_{2} z} \end{matrix}

then,

\begin{aligned} P_{t} & = \frac{1}{2} ℜ [E_{t 0} H_{t 0}^{*}] \\ = P_{i} (1 + Γ) (1 - Γ^{*}) \\ \overset{α = 0}{=} P_{i} (1 - | Γ |^{2}) \end{aligned}

Note

compare to $equation (6)$

\begin{aligned} P_{i} & \leftrightarrow P^{+} \\ P_{r} & \leftrightarrow P^{-} \\ P_{t} & \leftrightarrow P \\ P_{i} + P_{r} = P_{t} & \leftrightarrow P^{+} + P^{-} = P \end{aligned}