Electromagnetic Plane Wave Propagation In the transmission line section, we looked at wave propagation in 1 spatial coordinate. Here, we will build on our knowledge of Maxwell's Equation and model a uniform EM wave propagating in the z direction.
Then, we will extend the model to account for the uniform EM wave travelling in any direction.
We will examine uniform plane wave's various properties such as polarization, carried power, and propagation in different materials.
Integral Form
$$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon}$$
$$\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$$
$$\oint \vec{B} \cdot d\vec{A} = 0$$
$$\oint \vec{B} \cdot d\vec{l} = \mu i + \mu \epsilon \frac{d \phi_E}{dt}$$
Differential Form
$$\vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon}$$
$$\vec{\nabla} \times \vec{E} = -\mu \frac{\partial \vec{H}}{\partial t}$$
$$\vec{\nabla} \cdot \vec{H} = 0$$
$$\vec{\nabla} \times \vec{H} = \epsilon \frac{\partial \vec{E}}{\partial t}$$
where $\mu_0 \epsilon_0 \frac{d \phi_E}{dt}$ is the displacement current.
Intuition : If we only look at the dynamic equations, the time-varying H field induces the circulating E field. Time-varying E field induces circulating H field.
2. Uniform Plane Wave Solution Assume $\vec{E} = E_x(z, t) \hat{x}$ and $\vec{H} = H_y(z, t) \hat{y}$ , so they are uniform in planes of constant z with medium $\epsilon$ and $\mu$ .
We will get
$$\vec{\nabla} \times \vec{E} = \frac{\partial{E_x}}{\partial z} \hat{y} $$
$$\vec{\nabla} \times \vec{H} = -\frac{\partial H_y}{\partial z}\hat{x}$$
If we plug these into the source-free vacuum equations, we will eventually get
$$\frac{\partial E_x}{\partial z} = -\mu \frac{\partial H_y}{\partial t}$$
$$\frac{\partial H_y}{\partial z} = -\epsilon \frac{\partial E_x}{\partial t}$$
they are very similar to the transmission line equation.
Furthermore, we could manipulate the equations into two wave equations . For example,
$$
\begin{align}
& \frac{\partial E_x}{\partial z} = -\mu \frac{\partial H_y}{\partial t} \\
& \implies \frac{\partial}{\partial z}\frac{\partial E_x}{\partial z} = -\mu \frac{\partial}{\partial z}\frac{\partial H_y}{\partial t} \\
& \implies \frac{\partial^2 E_x}{\partial z^2} = -\mu \frac{\partial}{\partial t}\frac{\partial H_y}{\partial z} \\
& \implies \frac{\partial^2 E_x}{\partial z^2} = \mu \epsilon \frac{\partial^2 E_x}{\partial t^2}\\
\end{align}
$$
The phase velocity (speed of light ) is $\frac{1}{\sqrt{\mu \epsilon}}$ , which is parallel to $\frac{1}{\sqrt{L' C'}}$ from the transmission line.
The solution to this wave equation is
$$E_x = E_0 cos(\omega t - k z)$$
$$H_y = \frac{E_0}{\eta} cos(\omega t - k z)$$
k is the wave number, $\frac{2 \pi}{\lambda}$
The Intrinsic Impedance relates the magnitude of the E field to H field, just like the Characteristic Impedance relates the voltage and the current.
$$\eta = \sqrt{\frac{\mu}{\epsilon}}$$
where $\mu$ is the magnetic constant and $\epsilon$ is the electric constant,
and
$$|E_x| = \eta |H_y| $$
In vacuum,
$$\eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} = 377 \space \Omega$$
Parallel Between the Transmission Line to Free Space 1-D Voltage and Current waves now turn to Electric and Magnetic field waves.
$$v(z, t) \implies \vec{E}(x, y, z, t)$$
$$i(z, t) \implies \vec{H}(x, y, z, t)$$
In source free ($\rho = 0, \vec{J} = 0$ ) vacuum, we have
$$\vec{\nabla} \times \vec{E} = -\mu_0 \frac{\partial \vec{H}}{\partial t}$$
$$\vec{\nabla} \times \vec{H} = \epsilon_0 \frac{\partial \vec{E}}{\partial t}$$
which parallels to the lossless voltage and current PDE seen in the transmission line section, where
$$\frac{\partial v(z, t)}{\partial z} = - L'\frac{d i(z, t)}{d t}$$
$$\frac{\partial i(z, t)}{\partial z} = - C' \frac{dv(z, t)}{dt}$$
3. Phasors and Maxwell's Equations $\widetilde{E}$ denotes a phasor quantity, and $\widetilde{\mathbf{E}}$ denotes a phasor that is also a vector.
Maxwell's Equations in Phasor Domain If the time variation of the E and H field is sinusoidal with frequency $\omega$ , then we can write the E and H field as:
$$\vec{E}(x, y, z, t) = \text{Re} \{ \widetilde{\mathbf{E}}(x, y, z) e^{j \omega t}\}$$
$$\vec{H}(x, y, z, t) = \text{Re} \{ \widetilde{\mathbf{H}}(x, y, z) e^{j \omega t}\}$$
where $\widetilde{\mathbf{E}}$ and $\widetilde{\mathbf{H}}$ are the E and H field phasors with vector quantity.
We can put the source-free Maxwell's Equation into the phasor domain by swapping the time derivatives with $j \omega$ and vectors with phasors.
$$\vec{\nabla} \cdot \widetilde{\mathbf{E}} = 0$$
$$\vec{\nabla} \times \widetilde{\mathbf{E}} = -\mu j\omega \widetilde{\mathbf{H}}$$
$$\vec{\nabla} \cdot \widetilde{\mathbf{H}} = 0$$
$$\vec{\nabla} \times \widetilde{\mathbf{H}} = \epsilon j\omega \widetilde{\mathbf{E}}$$
Plain EM Wave in Phasor Domain Recall that the plain wave solutions are
$$\vec{E} = E_0 cos(\omega t - kz) \hat{x}$$
$$\vec{H} = \frac{E_0}{\eta}cos(\omega t - kz)\hat{y}$$
Since
$$\vec{E} = E_0 cos(\omega t - k z)\hat{x} = Re{E_0 e^{j(\omega t - k z)}\hat{x}} = Re{E_0e^{-jkz}\hat{x}\space e^{j \omega t}}$$
$$\vec{H} = H_0 cos(\omega t - k z)\hat{y} = Re{H_0 e^{j(\omega t - k z)}\hat{y}} = Re{H_0e^{-jkz}\hat{y}\space e^{j \omega t}}$$
In the phasor domain, they are
$$\widetilde{\mathbf{E}} = E_0 e^{-jkz} \hat{x}$$
$$\widetilde{\mathbf{H}} = \frac{E_0}{\eta} e^{-jkz} \hat{y}$$
4. Time-Harmonic Uniform Plane Waves in Any Direction E Field
Assume we are in source-free condition,
Using the vector calculus property on the curl of E, we get
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{E}) = \vec{\nabla} (\vec{\nabla} \cdot \vec{E}) - \vec{\nabla}^2 \vec{E}$$
Since the divergence of E is 0, we eventually will end up with
$$ \vec{\nabla}^2 \vec{E} = \mu \epsilon \frac{\partial ^2}{\partial t^2}\vec{E}$$
Notice that this is a wave equation in 3-dimensional space.
In the phasor domain, the PDE takes a form similar to the Helmholtz Equation in the transmission line
$$\vec{\nabla}^2 \widetilde{\mathbf{E}} + \mu \epsilon \omega^2\widetilde{\mathbf{E}} = 0$$
A plane wave solution looks like
$$\widetilde{\mathbf{E}}(x, y, z) = \widetilde{\mathbf{E}}_0 e^{-j (k_x x+ k_y y+ k_z z)}$$
where
$$\widetilde{\mathbf{E}}0 = \widetilde{E} {0x}\hat{x} + \widetilde{E}{0y}\widetilde{y} + \widetilde{E} {0z}\hat{z}$$
Note that each component could be a complex number that carries a phase as well. They could also vary spatially.
We will introduce a vector, the wave vector , which is
$$\vec{k} = k_x \hat{x} + k_y \hat{y} + k_z \hat{z}$$
and
$$\vec{k} \cdot \vec{r}(x, y, z) = k_x x + k_y y + k_z z$$
$$|\vec{k}| = k_x^2 + k_y^2 + k_z^2$$
If we put the general solution back to the phasor PDE, we get
$$|\vec{k}| = \sqrt{\mu \epsilon} \omega = \frac{2 \pi}{\lambda}$$
H field
By working with the E field phasor, we will get the magnetic field phasor
$$\widetilde{\mathbf{H}} (\vec{r}) = \frac{1}{\eta} \hat{k} \times \widetilde{\mathbf{E}} (\vec{r})$$
where $\hat{k}$ points in the direction of $\vec{k}$
We can also find for the E field phasor that
$$\widetilde{\mathbf{E}}(\vec{r}) = -\eta \hat{k} \times \widetilde{\mathbf{H}}(\vec{r})$$
Geometric Restriction
If we recall the Maxwell Equation in phasor form,
$$\vec{\nabla} \cdot \widetilde{\mathbf{E}} = 0$$
$$\implies (\frac{\partial}{\partial x}\hat{x} + \frac{\partial}{\partial y}\hat{y} + \frac{\partial}{\partial z}\hat{z}) \cdot \widetilde{\mathbf{E}}_0 e^{-j (k_x x+ k_y y+ k_z z)} = 0$$
$$\implies -(jk_x)\widetilde{E}{0x}e^{-j(k_x x+ k_y y+ k_z z)} -(jk_y)\widetilde{E} {0y}e^{-j(k_x x+ k_y y+ k_z z)} -(jk_z)E_{0z}e^{-j(k_x x+ k_y y+ k_z z)} = 0$$
$$\implies (-j)(\vec{k} \cdot \widetilde{\mathbf{E}}) = 0$$
Similarly,
$$(-j)(\vec{k} \cdot \mathbf{\widetilde{H}}) = 0$$
We can conclude that
$$\vec{k} \perp \mathbf{\widetilde{E}} $$
$$\vec{k} \perp \mathbf{\widetilde{H}} $$
From the cross-product definition of the E and H fields, we further know that
$$\mathbf{\widetilde{E}} \perp \mathbf{\widetilde{H}} $$
Intuition
When $\vec{k} \cdot \vec{r}$ is constant, or $\vec{k} \cdot \vec{r}_1 = \vec{k} \cdot \vec{r}_2$ , there's a plane perpendicular to $\vec{k}$ where the field is constant (Uniform ). However, the field can change from plane to plane.
5. Wave Polarization of a Uniform Plane Wave The polarization of a uniform plane wave describes the locus (collection of point) traced by the tip of the $\vec{E}$ vector (in the plane orthogonal to propagation) at a given point in space as a function of time.
Assumptions:
The wave propagates in the z-direction ($\vec{k}$ points to the z-direction)
The plane wave is uniform, meaning the E field is constant in magnitude and direction on a plane at a given time.
Since the wave vector points to the z direction, the E field must lie in the x-y plane, where polarization occurs.
$$\mathbf{\widetilde{E}}(x, y, z) = \mathbf{\widetilde{E}}0 e^{-j (k_x x+ k_y y+ k_z z)} = (\widetilde{E} {0x}\hat{x} + \widetilde{E}_{0y}\hat{y})e^{-j (k_z z)}$$
$$\mathbf{\widetilde{E}}(z) = \hat{x} \widetilde{E}{0x} e^{-jk_z z} + \hat{y} \widetilde{E} {0y} e^{-jk_z z}$$
Let $\widetilde{E}{0x} = a_x$ and $\widetilde{E} {0y} = a_ye^{j\delta}$,
$$\mathbf{\widetilde{E}}(z) = (\hat{x} a_x + \hat{y} a_y e^{j \delta})e^{-jk_z z}$$
In the time domain:
$$\vec{E}(z, t) = \text{Re} \{\mathbf{\widetilde{E}}(z)e^{j\omega t}\}$$
$$= \hat{x} a_x cos(\omega t - k_z z) + \hat{y} a_y cos(\omega t - k_z + \delta)$$
$$= \hat{x}E_x(z, t) + \hat{y}E_{y}(z, t)$$
Two quantities describe the wave polarization:
$$\vec{E}(z = z_0, t) = \sqrt{E_{x}^2(z, t) + E_{y}^2(z, t)}$$
where $z_0$ is a constant
$$\psi (z = z_0, t) = \arctan\Bigg(\frac{E_y(z, t)}{E_x(z, t)}\Bigg)$$
$\vec{E}(z, t)$ traces out a line
Consider $\delta = 0$ or $\delta = \pi$ ,
$$\mathbf{\widetilde{E}}(z) = (\hat{x} a_x + \hat{y} a_y e^{j \delta})e^{-jk_z z}$$
$$= (\hat{x} a_x \pm \hat{y} a_y)e^{-jk_z z}$$
In the time domain at a fixed position,
$$\vec{E}(z = z_0, t) = (\hat{x} a_x \pm \hat{y} a_y) \cos(\omega t)$$
where $z_0$ is a constant.
Magnitude
$$|\vec{E}(z = z_0, t)| = \sqrt{a_x^2 + a_y^2} \space \cos(\omega t)$$
Inclination Angle
$$\psi (z = z_0, t) = \arctan\Big(\frac{a_y}{a_x}\Big)$$
Left-Hand Circular Polarization (LHC)
$\vec{E}(z, t)$ traces out a circle in the clockwise direction.
Consider $\delta = \frac{\pi}{2}$ and $a_x = a_y = a$ ,
$$\mathbf{\widetilde{E}}(z) = (\hat{x} a + \hat{y} a e^{j \frac{\pi}{2}})e^{-jk_z z} = (\hat{x} a + \hat{y}aj)e^{-jkz}$$
In the time domain,
$$\vec{E}(z, t) = \text{Re}\{\mathbf{\widetilde{E}}(z)e^{j\omega t}\} = \hat{x}a \cos(\omega t - k_z z) + \hat{y} a \cos(\omega t - k_z z + \frac{\pi}{2})$$
$$= \hat{x}a \cos(\omega t - k_z z) - \hat{y} a \sin(\omega t - k_z z)$$
Magnitude
$$|\vec{E}(z = 0, t)| = \sqrt{a^2 \cos^2(\omega t) + a^2 \sin^2(\omega t)} = a$$
Inclination Angle
$$\psi (z = 0, t) = arctan\Big(\frac{-a\space sin(\omega t)}{a\space cos(\omega t)}\Big) = -\omega t$$
Notice that the magnitude of the vector is the same, but the angle decreases with time, so it rotates in the clockwise direction.
Right-Hand Circular Polarization (RHC) $\vec{E}(z, t)$ traces out a circle in the counter-clockwise direction.
If we let $\delta = -\frac{\pi}{2}$ and follow the calculation in LHC, we will get that the
Magnitude
$$|\vec{E}(z = 0, t)| = a$$
Inclination Angle
$$\psi (z = 0, t) = \omega t$$
The electric field vector traces out an ellipse in the plane perpendicular to propagation.
$a_{\xi}$ is the length of the major axis and $a_{\eta}$ is the length of the minor axis.
Ellipticity Angle
The shape and handedness of the ellipse are characterized by the ellipticity angle, $\chi$
$$\tan(\chi) = \pm \frac{a_{\eta}}{a_{\xi}} = \pm \frac{1}{R}$$
$$-\frac{\pi}{4} \leq \chi \leq \frac{\pi}{4}$$
The plus sign corresponds to clockwise rotation and the minus sign corresponds to counter-clockwise rotation.
The following relationship between parameters also holds:
$$\sin2\chi = (\sin2\psi_0)\sin\delta$$
$$-\frac{\pi}{4}\leq \chi \leq \frac{\pi}{4}$$
Rotation Angle
The angle between the major axis of the ellipse to the reference axis, which is the x-axis here.
$$\tan(2 \gamma) = \tan(2 \psi_0) \space \cos(\delta)$$
$$-\frac{\pi}{2} \leq \gamma \leq \frac{\pi}{2}$$
Rule:
$\gamma > 0$ for $\cos(\delta) > 0$
$\gamma < 0$ for $\cos(\delta) < 0$
Auxiliary Angle
$$\tan(\psi_0) = \frac{a_y}{a_x}$$
$$0 \leq \psi_0 \leq \frac{\pi}{2}$$
Nonmagnetic material means $\mu_r = 1$ and $\mu = \mu_0$
Vacuum means $\epsilon = \epsilon_0$ and $\mu = \mu_0$
The frequency of an EM wave is constant regardless of the material it's in.
The wavelength of an EM wave is NOT constant in different materials. It is $$\lambda = \frac{\lambda_0}{\sqrt{\epsilon_r \mu_r}}$$
The following assumptions are made in our analysis:
Time Invariance :
$\epsilon_r$ and $\mu_r$ are constant with time.
Homogenuous :
$\epsilon_r$ and $\mu_r$ are constant with space.
Isotropy : Material always behaves the same regardless of the field's direction.
Linearity :
$\vec{D} = \epsilon \vec{E}$ and $\vec{B} = \mu \vec{H}$
Source-Free
$\rho, \vec{J} = 0$
In a conductor, ohm's law applies
$$\vec{J} = \sigma \vec{E}$$
Therefore, we cannot assume the wave is traveling in the source-free medium. So the curl of H field becomes:
$$\vec{\nabla} \times \mathbf{\widetilde{H}} = \sigma \mathbf{\widetilde{E}} + j\omega \epsilon \mathbf{\widetilde{E}}$$
$$\implies \vec{\nabla} \times \mathbf{\widetilde{H}} = (\sigma + j\omega \epsilon )\mathbf{\widetilde{E}}$$
$$\implies \vec{\nabla} \times \mathbf{\widetilde{H}} = j \omega (\frac{\sigma}{j \omega} + \epsilon )\mathbf{\widetilde{E}}$$
In a conductor, let $\epsilon_c = \epsilon + \frac{\sigma}{j \omega} = \epsilon' - j \epsilon''$ , we have
$$\vec{\nabla} \times \mathbf{\widetilde{H}} = j\omega \epsilon_c \mathbf{\widetilde{E}}$$
How would this affect our calculation?
$$k_c = \sqrt{\mu \epsilon_c} \omega = k - j \alpha$$
$$\eta_c = \sqrt{\frac{\mu}{\epsilon_c}}$$
Looking back to our plane wave solution,
$$e^{-jk_c z} = e^{-j (k - j\alpha)z} = e^{-jkz} e^{-\alpha z}$$
It means that the wave attenuates as it travels because of the $e^{-\alpha z}$ term, and notice the complex intrinsic impedance introduces a phase shift between the E and H fields.
For both the complex wave number and the intrinsic impedance, both are related a quantity called the "loss tangent" ($\delta_c$ )
$$\tan (\delta_c) = \frac{\epsilon''}{\epsilon'}$$
The more conductive a material is, the greater the $\sigma$ , and the greater the $\epsilon''$ .
Low Loss Dielectric
For low-loss dielectric, $\frac{\epsilon''}{\epsilon'} << 1$
$$\alpha \approx \frac{\sigma}{2} \sqrt{\frac{\mu}{\epsilon'}}$$
$$k = \omega \sqrt{\mu \epsilon'}$$
$$\eta_c = \sqrt{\frac{\mu}{\epsilon'}}$$
Good Conductor
For good conductors, $\frac{\epsilon''}{\epsilon'} >> 1$
$$\alpha = k \approx \sqrt{\frac{\mu \sigma \omega}{2}}$$
$$\eta_c = (e^{j \frac{\pi}{4}}) \frac{\alpha}{\sigma} = (1 + j)\frac{\alpha}{\sigma}$$
The E and H fields are 45 degrees out of phase.
Skin Depth
Tells us how deep the EM wave can penetrate a material without attenuation.
$$\delta_s = \frac{1}{\alpha}$$
7. Power Carried by the EM Wave Time Average Poynting Vector $$\vec{S}_{avg} = \frac{1}{2} \text{Re}\{\mathbf{\widetilde{E} \times \mathbf{\widetilde{H}^*}}\}$$
$\vec{S}_{avg}$ is the power density and has the unit of watts per area squared.
It points in the direction of the wave propagation.
Specifically, for uniform plane wave propagating in the z direction in lossless material ,
$$\vec{S}{avg} = \frac{1}{2\eta}(|\widetilde{E} {0x}|^2 + |\widetilde{E}_{0y}|^2)\hat{z}$$
More generally in lossy material ,
$$\vec{S}{avg} = e^{-2\alpha z }cos(\theta_c) \frac{1}{2|\eta_c|}(|\widetilde{E} {0x}|^2 + |\widetilde{E}_{0y}|^2)\hat{z}$$
$$-\frac{d U_{EM}}{dt} = \frac{d W}{dt} + \oint (\vec{E} \times \vec{H}) d\vec{A}$$
Total Energy of EM Field = Work Done on Charges + Energy Radiating Away
