Wave Propagation and Reflection

Signal Refraction at Boundaries

This note covers signal refraction at boundaries, radiation by antennas, wave propagation in lossless and lossy media, antenna reception, and wave refraction across boundaries.

Normal and Oblique Incidence

• Normal Incidence: The incident wave, reflected wave, and transmitted wave are all perpendicular to the boundary between medium 1 and medium 2.
  • Medium 1 has refractive index n_1.
  • Medium 2 has refractive index n_2.
• Oblique Incidence: The incident wave, reflected wave, and transmitted wave are not perpendicular to the boundary.
  • \theta_i: Angle of incidence.

Normal Incidence Analysis

• Transmission-line concepts from Chapter 2 can be used to analyze plane wave reflection and transmission at interfaces between dissimilar media.

Modeling Normal Incidence

• Traveling wave and standing wave patterns are formed due to the interference of incident and reflected waves.
• Medium 1 properties: \mu1, \epsilon1, \sigma1, \eta1, \beta_1
• Medium 2 properties: \mu2, \epsilon2, \sigma2, \eta2, \beta_2

Individual Waves

• Lossless Medium
• Notation conventions:
  • - sign: indicates wave propagation in the +z direction.
  • + sign: indicates wave propagation in the -z direction.
• Medium 1 properties: \mu1, \epsilon1, \eta1, k1
• Medium 2 properties: \mu2, \epsilon2, \eta2, k2

Total Fields and Boundary Conditions

• \tilde{E}1(z) = \tilde{E}i(z) + \tilde{E}r(z) = \hat{x}(E0^i e^{-jk1z} + E0^r e^{jk_1z})
• \tilde{H}1(z) = \tilde{H}i(z) + \tilde{H}r(z) = \hat{y}\frac{1}{\eta1}(E0^i e^{-jk1z} - E0^r e^{jk1z})
• \tilde{E}2(z) = \tilde{E}t(z) = \hat{x}E0^t e^{-jk2z}
• \tilde{H}2(z) = \tilde{H}t(z) = \hat{y}\frac{E0^t}{\eta2}e^{-jk_2z}
• Boundary conditions at z = 0:
  • Tangential E-field is continuous: E1(0) = E2(0)
  • Tangential H-field is continuous: H1(0) = H2(0)

Boundary Conditions and Solutions

• At the boundary z = 0:
  • E{i0} + E{r0} = E_{t0}
  • \frac{E{i0}}{\eta1} - \frac{E{r0}}{\eta1} = \frac{E{t0}}{\eta2}
• Reflection coefficient: \Gamma = \frac{E{r0}}{E{i0}} = \frac{\eta2 - \eta1}{\eta2 + \eta1}
• Transmission coefficient: T = \frac{E{t0}}{E{i0}} = \frac{2\eta2}{\eta2 + \eta_1}

Reflection and Transmission Coefficients

• Reflection coefficient: \Gamma = \frac{E{r0}}{E{i0}} = \frac{\eta2 - \eta1}{\eta2 + \eta1} (normal incidence)
• Transmission coefficient: T = \frac{E{t0}}{E{i0}} = \frac{2\eta2}{\eta2 + \eta_1} (normal incidence)
• T = 1 + \Gamma (normal incidence)
• For nonmagnetic media: \Gamma = \frac{\sqrt{\epsilon{r1}} - \sqrt{\epsilon{r2}}}{\sqrt{\epsilon{r1}} + \sqrt{\epsilon{r2}}}

Analogy of Normal Incidence to Transmission Lines

• Plane Wave equivalent to Transmission Line
• \tilde{E}1(z) = \hat{x}E0^i(e^{-jk1z} + \Gamma e^{jk1z}) equivalent to V1(z) = V0^+(e^{-j\beta1 z} + \Gamma e^{j\beta1 z})
• \tilde{H}1(z) = \hat{y}\frac{E0^i}{\eta1}(e^{-jk1z} - \Gamma e^{jk1z}) equivalent to I1(z) = \frac{V0^+}{Z{01}}(e^{-j\beta1 z} - \Gamma e^{j\beta1 z})
• \tilde{E}2(z) = \hat{x}TE0^i e^{-jk2z} equivalent to V2(z) = V0^+ e^{-j\beta2 z}
• \tilde{H}2(z) = \hat{y}T\frac{E0^i}{\eta2}e^{-jk2z} equivalent to I2(z) = T\frac{V0^+}{Z{02}}e^{-j\beta2 z}
• \Gamma = \frac{\eta2 - \eta1}{\eta2 + \eta1} equivalent to \Gamma = \frac{Z{02} - Z{01}}{Z{02} + Z{01}}
• T = 1 + \Gamma equivalent to T = 1 + \Gamma
• k1 = \omega\sqrt{\mu1\epsilon1}, k2 = \omega\sqrt{\mu2\epsilon2}
• \beta1 = \omega\sqrt{\mu1\epsilon1}, \beta2 = \omega\sqrt{\mu2\epsilon2}
• \eta1 = \sqrt{\frac{\mu1}{\epsilon1}}, \eta2 = \sqrt{\frac{\mu2}{\epsilon2}}

Power Transfer

• Incident wave, reflected wave, and transmitted wave.
• Average Poynting vector in medium 1: S{av1}(z) = \frac{1}{2}Re[\tilde{E}1(z) \times \tilde{H}1^*(z)] = \hat{z}\frac{|E0^i|^2}{2\eta_1}(1 - |\Gamma|^2)
• S{av1} = Si + Sr = \frac{|E0^i|^2}{2\eta1} - \frac{|E0^i|^2}{2\eta_1}|\Gamma|^2
• S{av}^r = -|\Gamma|^2 S{av}^i

Power Transfer (Continued)

• Average Poynting vector in medium 2: S{av2}(z) = \frac{1}{2}Re[\tilde{E}2(z) \times \tilde{H}2^*(z)] = \hat{z}\frac{|E0^i|^2}{2\eta_2}T^2
• For lossless media: S{av1} = S{av2}
• \frac{|E0^i|^2}{2\eta1}(1 - |\Gamma|^2) = \frac{|E0^i|^2}{2\eta2}T^2 \implies \frac{\eta2}{\eta1} = 1 - |\Gamma|^2

Lossless - Lossy - Normal Incidence: Lossy

• Medium 1 properties: \mu1, \epsilon1, \eta1, k1
• Medium 2 properties: \mu2, \epsilon2, \sigma2, \eta2, \beta_2

Normal Incidence: PEC

• Perfect Electric Conductor (PEC).
• \hat{n} \times \tilde{E} = 0
• \hat{n} \cdot \tilde{H} = 0

Normal Incidence: Radomes

• A half-wave section can be used as a dielectric window.
• \eta2 = \eta0 and the matching section d = \frac{\lambda}{2}.
• Medium 1 properties: \mu0, \epsilon0
• Medium 2 properties: \mu0, \epsilon2
• Medium 3 properties: \mu0, \epsilon0
• Z{in} = ZL

Normal Incidence: Lens and Prism

• Both applications are frequency sensitive and the matching section is only \frac{\lambda}{4} or \frac{\lambda}{2} at one frequency.
• \eta2 = \sqrt{\eta0\eta_3}
• Medium 1 properties: \mu0, \epsilon0
• Coating properties: \mu2, \epsilon2
• Lens properties: \mu3, \epsilon3
• \frac{\lambda}{4} coating:
• \frac{\lambda}{4} coating:

Example 8-1: Radar Radome Design

• A 10-GHz aircraft radar uses a narrow-beam scanning antenna mounted on a gimbal behind a dielectric radome.
• Radome material is a lossless dielectric with \epsilonr = 9 and \mur = 1.
• Choose its thickness d such that the radome appears transparent to the radar beam.
• Structural integrity requires d to be greater than 2.3 cm.

Example 8-1: Radar Radome Solution

• From transmission lines, since Media 1 and 3 are the same (air), no net reflection will occur at z = -d if the radome thickness is an integer multiple of \frac{\lambda_2}{2}.
• d = n\frac{\lambda_2}{2}

Example 8-2: Yellow Light Incident upon a Glass Surface

• A beam of yellow light with wavelength 0.6 \mum is normally incident in air upon a glass surface.
• The surface is situated in the plane z = 0 and the relative permittivity of glass is 2.25.
• Determine:
  • the locations of the electric field maxima in medium 1 (air),
  • the standing-wave ratio, and
  • the fraction of the incident power transmitted into the glass medium.

Example 8-2: Solution

• \eta1 = \sqrt{\frac{\mu0}{\epsilon_0}} \approx 120\pi \Omega
• \eta2 = \frac{\eta1}{\sqrt{\epsilon_r}} = \frac{120\pi}{\sqrt{2.25}} \approx 80\pi \Omega
• \Gamma = \frac{\eta2 - \eta1}{\eta2 + \eta1} = \frac{80\pi - 120\pi}{80\pi + 120\pi} = -0.2

Example 8-2: Solution (Continued)

• |\Gamma| = 0.2 and \phi = \pi.
• The electric-field magnitude is maximum at
  z{max} = \frac{\thetar \lambda1}{4\pi} + n\frac{\lambda1}{2} = \frac{\lambda1}{4} + n\frac{\lambda1}{2} \quad (n = 0, 1, 2, …)
  with \lambda_1 = 0.6 \mu m.
• S = \frac{1 + |\Gamma|}{1 - |\Gamma|} = \frac{1 + 0.2}{1 - 0.2} = 1.5
• \frac{S{av2}}{S{av1}} = \frac{\eta1}{\eta2}|T|^2 = \frac{4\eta1\eta2}{(\eta2 + \eta1)^2}
• \frac{S{av2}}{S{av1}} = 1 - |\Gamma|^2 = 1 - (0.2)^2 = 0.96 = 96\%.

Example 8-3: Normal Incidence on a Metal Surface

• A 1-GHz x-polarized plane wave traveling in the +z-direction is incident from air upon a copper surface.
• The air-to-copper interface is at z = 0 and copper has \epsilonr = 1, \mur = 1, and \sigma = 5.8 \times 10^7 S/m.
• The amplitude of the electric field of the incident wave is 12 (mV/m).
• Obtain expressions for the instantaneous electric and magnetic fields in the air medium.

Example 8-3: Solution

• In medium 1 (air), \alpha = 0,
• \beta = k_1 = \frac{\omega}{c} = \frac{2\pi \times 10^9}{3 \times 10^8} = \frac{20\pi}{3} \text{ (rad/m)},
• \eta1 = \eta0 = 377 \Omega,
• \lambda = \frac{2\pi}{k_1} = 0.3 \text{ m}.
• Since \frac{\sigma}{\omega \epsilon} = 1 \times 10^9 > 1, copper is an excellent conductor at f = 1 GHz.

Example 8-3: Solution (Continued)

• \eta_{c2} = (1 + j)\sqrt{\frac{\pi f \mu}{\sigma}} = (1 + j)\sqrt{\frac{\pi \times 10^9 \times 4\pi \times 10^{-7}}{5.8 \times 10^7}} \approx 8.25(1 + j) \text{ ($\Omega$)}.
• Since \eta{c2} is so small compared to \eta0 = 377 \Omega for air, the copper surface acts like a short circuit.
• \Gamma = \frac{\eta{c2} - \eta0}{\eta{c2} + \eta0} \approx -1

Example 8-3: Solution (Continued)

• Upon setting \Gamma = -1
• \tilde{E}1(z) = \hat{x}E0^i(e^{-jk1z} - e^{jk1z}) = -\hat{x}j2E0^i \sin k1z
• \tilde{H}1(z) = \hat{y}\frac{E0^i}{\eta1}(e^{-jk1z} + e^{jk1z}) = \hat{y}2\frac{E0^i}{\eta1}\cos k1z
• E1(z, t) = Re[\tilde{E}1(z)e^{j\omega t}] = \hat{x}2E0^i\sin k1z \sin \omega t = \hat{x}24\sin(\frac{20\pi z}{3})\sin(2\pi \times 10^9 t) \text{ (mV/m)}
• H1(z, t) = Re[\tilde{H}1(z)e^{j\omega t}] = \hat{y}2\frac{E0^i}{\eta1}\cos k_1z \cos \omega t = \hat{y}64\cos(\frac{20\pi z}{3})\cos(2\pi \times 10^9 t) \text{ ($\mu$A/m)}