Wave Propagation and Polarization

Normal Incidence: Reflection and Transmission

Summary of Key Equations

• Reflection Coefficient (\Gamma):

  • Perpendicular Polarization: \Gamma{\perp} = \frac{\eta2 \cos(\thetat) - \eta1 \cos(\thetai)}{\eta2 \cos(\thetat) + \eta1 \cos(\theta_i)}
  • Parallel Polarization: \Gamma{||} = \frac{\eta2 \cos(\thetai) - \eta1 \cos(\thetat)}{\eta2 \cos(\thetai) + \eta1 \cos(\theta_t)}
  • Normal Incidence: \Gamma = \frac{\eta2 - \eta1}{\eta2 + \eta1}

• Transmission Coefficient (\tau):

  • Perpendicular Polarization: \tau{\perp} = \frac{2\eta2 \cos(\thetai)}{\eta2 \cos(\thetat) + \eta1 \cos(\theta_i)}
  • Parallel Polarization: \tau{||} = \frac{2\eta2 \cos(\thetai)}{\eta2 \cos(\thetai) + \eta1 \cos(\theta_t)}
  • Normal Incidence: \tau = \frac{2\eta2}{\eta2 + \eta_1}

• Relation between Reflection and Transmission Coefficients:

  • \tau = 1 + \Gamma
  • \tau{||} = (1 + \Gamma{||}) \frac{\cos(\thetai)}{\cos(\thetat)}

• Reflectivity (R):

  • R = |\Gamma|^2
  • Perpendicular Polarization: R{\perp} = |\Gamma{\perp}|^2
  • Parallel Polarization: R{||} = |\Gamma{||}|^2

• Transmissivity (T):

  • T = |\tau|^2 \frac{\eta1}{\eta2}
  • Perpendicular Polarization: T{\perp} = |\tau{\perp}|^2 \frac{\eta2 \cos(\thetat)}{\eta1 \cos(\thetai)}
  • Parallel Polarization: T{||} = |\tau{||}|^2 \frac{\eta2 \cos(\thetat)}{\eta1 \cos(\thetai)}

• Relation between Reflectivity and Transmissivity:

  • T = 1 - R
  • Perpendicular Polarization: T{\perp} = 1 - R{\perp}
  • Parallel Polarization: T{||} = 1 - R{||}

• Additional Notes:

  • \sin(\thetat) = \sqrt{\frac{\mu1 \epsilon1}{\mu2 \epsilon2}} \sin(\thetai)
  • \eta1 = \sqrt{\frac{\mu1}{\epsilon_1}}
  • \eta2 = \sqrt{\frac{\mu2}{\epsilon_2}}
  • For non-magnetic media: \frac{\eta2}{\eta1} = \frac{n1}{n2}

2014 Test 2 Q1

Problem Setup

• A uniform plane wave in air is incident on a dielectric plane boundary at z = 0.
• The electric field is given by: \vec{E}(x, z, t) = \hat{y} E0 \cos(\omega t + kx x - k_z z) \text{ V/m}
• Dielectric constant of the second medium: \epsilon_r = 2.25
• Incident angle: \theta_i = 60^\circ
• Transmitted angle: \theta_t = 35.26^\circ
• Phase constant in air: k_1 = 5 \text{ rad/m}

Task

• Find the incident phasor magnetic field, \vec{H}_i.
• Find the reflected phasor electric field, \vec{E}_r.
• Find the frequency of the wave, f.

Solution

• The wave vector is given by: \vec{k} = -\hat{x} kx + \hat{z} kz
• Given values:
  • kx = ki \sin(\theta_i) = 5 \sin(60^\circ) = 4.33 \text{ rad/m}
  • kz = ki \cos(\theta_i) = 5 \cos(60^\circ) = 2.5 \text{ rad/m}
• The incident magnetic field is: \vec{H} = \frac{1}{\eta} \hat{k} \times \vec{E}
• In air, \eta = \eta_0 = 377 \Omega
• \vec{H}i = \frac{E0}{\eta0} (-\hat{x} \cos(\thetai) - \hat{z} \sin(\theta_i)) e^{-j(4.3x + 2.5z)}
• \vec{H}i = \frac{E0}{377} (-0.5 \hat{x} - 0.866 \hat{z}) e^{-j(4.3x + 2.5z)}
• The reflection coefficient for perpendicular polarization is:
  • \Gamma{\perp} = \frac{\eta2 \cos(\thetat) - \eta1 \cos(\thetai)}{\eta2 \cos(\thetat) + \eta1 \cos(\theta_i)}
  • Since \eta = \sqrt{\frac{\mu}{\epsilon}}, and for non-magnetic materials, \mu is constant, then \eta \propto \frac{1}{\sqrt{\epsilon}}.
  • \frac{\eta1}{\eta2} = \sqrt{\frac{\epsilon2}{\epsilon1}} = \sqrt{2.25} = 1.5 \Rightarrow \eta2 = \frac{\eta1}{1.5} = \frac{377}{1.5} = 251.33
  • \Gamma_{\perp} = \frac{251.33 \cos(35.26^\circ) - 377 \cos(60^\circ)}{251.33 \cos(35.26^\circ) + 377 \cos(60^\circ)} = \frac{205.4 - 188.5}{205.4 + 188.5} = \frac{16.9}{393.9} = 0.042
• Given that \Gamma = -0.42
• Reflected electric field: \vec{E}r = \Gamma{\perp} E0 \hat{y} e^{-j(4.3x - 2.5z)} = -0.42 E0 \hat{y} e^{-j(4.3x - 2.5z)}
• Frequency: f = \frac{c}{\lambda} = \frac{ck}{2\pi}
• Since k = \frac{\omega}{c}, then \omega = ck and f = \frac{ck}{2\pi}
• f = \frac{3 \times 10^8 \cdot 5}{2\pi} = 238.7 \text{ MHz}

2013 Test 2 Q2

• A plane wave is incident at 30 degrees from medium 1 to medium 2.
• Medium 1 has parameters \mu0 and \epsilon0 \epsilon_r
• Medium 2 has parameters \mu0 and \epsilon0
• Polarization is not explicitly stated but inferred from the problem.
• Brewster angle occurs when \Gamma_{||} = 0
• \tan(\thetaB) = \sqrt{\frac{\epsilon2}{\epsilon_1}}
• Given \thetai = 30^\circ, 0 = \Gamma{||} = \frac{\frac{\eta2}{\cos(\thetat)} - \frac{\eta1}{\cos(\thetai)}}{\frac{\eta2}{\cos(\thetat)} + \frac{\eta1}{\cos(\thetai)}}
• \epsilon_{r1} = 3

2011 Test 1 Q2

• Given that \tan(\thetai) = \frac{4}{3}, so \thetai = 53.13^\circ
• Therefore, \sin(\thetai) = \frac{4}{5}, \cos(\thetai) = \frac{3}{5}
• The wave vector \vec{k} = \hat{x} kx + \hat{z} kz = \hat{x} k \sin(\thetai) + \hat{z} k \cos(\thetai)
• \vec{H} = \frac{1}{\eta} \hat{y} \times \vec{E} = \frac{1}{\eta} (-\hat{z} \cos(\thetai) + \hat{x} \sin(\thetai)) E0 e^{-j(kx x + k_z z)}
• \sin(\thetat) = \sqrt{\frac{\mu1 \epsilon1}{\mu2 \epsilon2}} \sin(\thetai)
• \theta_t = 11.537^\circ
• Reflection coefficient: \Gamma = \frac{\eta2 \cos(\thetai) - \eta1 \cos(\thetat)}{\eta2 \cos(\thetai) + \eta1 \cos(\thetat)} = 0.355 using \eta = 120\pi
• Incident electric field: \vec{E}_i = 12 (0.6 \hat{x} + 0.8 \hat{z}) e^{-j(4x + 3z)} \text{ V/m}
• Reflected electric field: \vec{E}r = \Gamma Ei = 0.42 \cdot 12 (0.6 \hat{x} - 0.8 \hat{z}) e^{-j(4x - 3z)} \text{ V/m}
• Reflected magnetic field: \vec{H}r = \frac{-\Gamma E0}{\eta} (\hat{z} \cos(\thetai) + \hat{x} \sin(\thetai)) e^{-j(4x - 3z)} = 1.33 \text{ mA/m}

2019 Exam Q5

Problem Setup

• A uniform plane wave with \vec{H} = \hat{y} H0 e^{-j(10x + 10\sqrt{3}z)} \text{ A/m} is incident from a dielectric medium with \mu1 = \mu0, \epsilon1 = \epsilon0 and \sigma1 = 0 onto a semi-infinite dielectric medium with \mu2 = \mu0, \epsilon2 = 9\epsilon0 and \sigma_2 = 0.

Task

• Determine the angle of incidence, \theta_i.
• Determine the phase constant, k_1, in the incident wave in medium 1.
• Determine the intrinsic impedance, \eta_2, in medium 2.
• Determine the average power density in the incident wave in medium 1.

Solution

• Angle of incidence: \tan(\thetai) = \frac{kx}{kz} = \frac{10}{10\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow \thetai = 30^\circ
• Phase constant: k1 = \sqrt{kx^2 + k_z^2} = \sqrt{10^2 + (10\sqrt{3})^2} = \sqrt{100 + 300} = \sqrt{400} = 20 \text{ rad/m}
• Intrinsic impedance: \eta2 = \sqrt{\frac{\mu2}{\epsilon2}} = \sqrt{\frac{\mu0}{9\epsilon0}} = \frac{1}{3} \sqrt{\frac{\mu0}{\epsilon0}} = \frac{1}{3} \eta0 = \frac{1}{3} \cdot 120\pi = 40\pi \approx 125.67 \Omega
• Average power density: \vec{S}{av} = \frac{1}{2} \text{Re} [\vec{E} \times \vec{H}^* ] = \frac{1}{2} \text{Re} [(\eta \vec{H}) \times \vec{H}^* ] = \frac{1}{2} \eta |H0|^2 \hat{k}
• \vec{S}_{av} = \frac{1}{2} (120\pi) (1)^2 (0.5 \hat{x} + 0.5\sqrt{3} \hat{z}) = 60\pi (0.5 \hat{x} + 0.5\sqrt{3} \hat{z}) = (30\pi \hat{x} + 30\sqrt{3}\pi \hat{z}) \text{ W/m}^2

2013 Exam Q2

• A plane wave is incident at 50 degrees.
• Medium 1: \mu0, \epsilon0 \epsilon_r
• Medium 2: \mu0, \epsilon0
• Good conductor: \frac{\sigma}{\omega \epsilon} >> 1
• If \frac{\sigma}{\omega \epsilon} = 200, then it qualifies as a good conductor.
• Calculate \alpha and \eta:
  • \alpha = \sqrt{\pi f \mu \sigma}
  • \eta = \sqrt{\frac{j \omega \mu}{\sigma + j \omega \epsilon}} \approx \sqrt{\frac{j \omega \mu}{\sigma}} = (1+j) \sqrt{\frac{\pi f \mu}{\sigma}}

Problem Setup - Electric Field in Medium 2:

• The general form for the electric field in medium 2 is:
  \mathbf{E}{t} (x, z) = E{0} e^{-\alpha z} e^{-j(\beta{x} x + \beta{z} z)} \hat{y}

• For a good condutor \beta \approx \alpha = \sqrt{ \pi f \mu \sigma}

• Given \sigma = 2.15 \cdot \omega \cdot \epsilon_0 \cdot (1 + j)

• It uses generic values and results to illustrate that \alpha \approx \beta for a good conductor.