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 n_1 n 1 Medium 2 has refractive index n 2 n_2 n 2 Oblique Incidence: The incident wave, reflected wave, and transmitted wave are not perpendicular to the boundary.θ i \theta_i θ i 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: μ < e m > 1 , ϵ < / e m > 1 , σ < e m > 1 , η < / e m > 1 , β 1 \mu<em>1, \epsilon</em>1, \sigma<em>1, \eta</em>1, \beta_1 μ < e m > 1 , ϵ < / e m > 1 , σ < e m > 1 , η < / e m > 1 , β 1 Medium 2 properties: μ < e m > 2 , ϵ < / e m > 2 , σ < e m > 2 , η < / e m > 2 , β 2 \mu<em>2, \epsilon</em>2, \sigma<em>2, \eta</em>2, \beta_2 μ < e m > 2 , ϵ < / e m > 2 , σ < e m > 2 , η < / e m > 2 , β 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: μ < e m > 1 , ϵ < / e m > 1 , η < e m > 1 , k < / e m > 1 \mu<em>1, \epsilon</em>1, \eta<em>1, k</em>1 μ < e m > 1 , ϵ < / e m > 1 , η < e m > 1 , k < / e m > 1 Medium 2 properties: μ < e m > 2 , ϵ < / e m > 2 , η < e m > 2 , k < / e m > 2 \mu<em>2, \epsilon</em>2, \eta<em>2, k</em>2 μ < e m > 2 , ϵ < / e m > 2 , η < e m > 2 , k < / e m > 2 Total Fields and Boundary Conditions E ~ < e m > 1 ( z ) = E ~ < / e m > i ( z ) + E ~ < e m > r ( z ) = x ^ ( E < / e m > 0 i e − j k < e m > 1 z + E < / e m > 0 r e j k 1 z ) \tilde{E}<em>1(z) = \tilde{E}</em>i(z) + \tilde{E}<em>r(z) = \hat{x}(E</em>0^i e^{-jk<em>1z} + E</em>0^r e^{jk_1z}) E ~ < e m > 1 ( z ) = E ~ < / e m > i ( z ) + E ~ < e m > r ( z ) = x ^ ( E < / e m > 0 i e − j k < e m > 1 z + E < / e m > 0 r e j k 1 z ) H ~ < e m > 1 ( z ) = H ~ < / e m > i ( z ) + H ~ < e m > r ( z ) = y ^ 1 η < / e m > 1 ( E < e m > 0 i e − j k < / e m > 1 z − E < e m > 0 r e j k < / e m > 1 z ) \tilde{H}<em>1(z) = \tilde{H}</em>i(z) + \tilde{H}<em>r(z) = \hat{y}\frac{1}{\eta</em>1}(E<em>0^i e^{-jk</em>1z} - E<em>0^r e^{jk</em>1z}) H ~ < e m > 1 ( z ) = H ~ < / e m > i ( z ) + H ~ < e m > r ( z ) = y ^ η < / e m > 1 1 ( E < e m > 0 i e − j k < / e m > 1 z − E < e m > 0 r e j k < / e m > 1 z ) E ~ < e m > 2 ( z ) = E ~ < / e m > t ( z ) = x ^ E < e m > 0 t e − j k < / e m > 2 z \tilde{E}<em>2(z) = \tilde{E}</em>t(z) = \hat{x}E<em>0^t e^{-jk</em>2z} E ~ < e m > 2 ( z ) = E ~ < / e m > t ( z ) = x ^ E < e m > 0 t e − j k < / e m > 2 z H ~ < e m > 2 ( z ) = H ~ < / e m > t ( z ) = y ^ E < e m > 0 t η < / e m > 2 e − j k 2 z \tilde{H}<em>2(z) = \tilde{H}</em>t(z) = \hat{y}\frac{E<em>0^t}{\eta</em>2}e^{-jk_2z} H ~ < e m > 2 ( z ) = H ~ < / e m > t ( z ) = y ^ η < / e m > 2 E < e m > 0 t e − j k 2 z Boundary conditions at z = 0:Tangential E-field is continuous: E < e m > 1 ( 0 ) = E < / e m > 2 ( 0 ) E<em>1(0) = E</em>2(0) E < e m > 1 ( 0 ) = E < / e m > 2 ( 0 ) Tangential H-field is continuous: H < e m > 1 ( 0 ) = H < / e m > 2 ( 0 ) H<em>1(0) = H</em>2(0) H < e m > 1 ( 0 ) = H < / e m > 2 ( 0 ) Boundary Conditions and Solutions At the boundary z = 0:E < e m > i 0 + E < / e m > r 0 = E t 0 E<em>{i0} + E</em>{r0} = E_{t0} E < e m > i 0 + E < / e m > r 0 = E t 0 E < e m > i 0 η < / e m > 1 − E < e m > r 0 η < / e m > 1 = E < e m > t 0 η < / e m > 2 \frac{E<em>{i0}}{\eta</em>1} - \frac{E<em>{r0}}{\eta</em>1} = \frac{E<em>{t0}}{\eta</em>2} η < / e m > 1 E < e m > i 0 − η < / e m > 1 E < e m > r 0 = η < / e m > 2 E < e m > t 0 Reflection coefficient: Γ = E < e m > r 0 E < / e m > i 0 = η < e m > 2 − η < / e m > 1 η < e m > 2 + η < / e m > 1 \Gamma = \frac{E<em>{r0}}{E</em>{i0}} = \frac{\eta<em>2 - \eta</em>1}{\eta<em>2 + \eta</em>1} Γ = E < / e m > i 0 E < e m > r 0 = η < e m > 2 + η < / e m > 1 η < e m > 2 − η < / e m > 1 Transmission coefficient: T = E < e m > t 0 E < / e m > i 0 = 2 η < e m > 2 η < / e m > 2 + η 1 T = \frac{E<em>{t0}}{E</em>{i0}} = \frac{2\eta<em>2}{\eta</em>2 + \eta_1} T = E < / e m > i 0 E < e m > t 0 = η < / e m > 2 + η 1 2 η < e m > 2 Reflection and Transmission Coefficients Reflection coefficient: Γ = E < e m > r 0 E < / e m > i 0 = η < e m > 2 − η < / e m > 1 η < e m > 2 + η < / e m > 1 \Gamma = \frac{E<em>{r0}}{E</em>{i0}} = \frac{\eta<em>2 - \eta</em>1}{\eta<em>2 + \eta</em>1} Γ = E < / e m > i 0 E < e m > r 0 = η < e m > 2 + η < / e m > 1 η < e m > 2 − η < / e m > 1 Transmission coefficient: T = E < e m > t 0 E < / e m > i 0 = 2 η < e m > 2 η < / e m > 2 + η 1 T = \frac{E<em>{t0}}{E</em>{i0}} = \frac{2\eta<em>2}{\eta</em>2 + \eta_1} T = E < / e m > i 0 E < e m > t 0 = η < / e m > 2 + η 1 2 η < e m > 2 T = 1 + Γ T = 1 + \Gamma T = 1 + Γ For nonmagnetic media: Γ = ϵ < e m > r 1 − ϵ < / e m > r 2 ϵ < e m > r 1 + ϵ < / e m > r 2 \Gamma = \frac{\sqrt{\epsilon<em>{r1}} - \sqrt{\epsilon</em>{r2}}}{\sqrt{\epsilon<em>{r1}} + \sqrt{\epsilon</em>{r2}}} Γ = ϵ < e m > r 1 + ϵ < / e m > r 2 ϵ < e m > r 1 − ϵ < / e m > r 2 Analogy of Normal Incidence to Transmission Lines Plane Wave equivalent to Transmission Line E ~ < e m > 1 ( z ) = x ^ E < / e m > 0 i ( e − j k < e m > 1 z + Γ e j k < / e m > 1 z ) \tilde{E}<em>1(z) = \hat{x}E</em>0^i(e^{-jk<em>1z} + \Gamma e^{jk</em>1z}) E ~ < e m > 1 ( z ) = x ^ E < / e m > 0 i ( e − j k < e m > 1 z + Γ e j k < / e m > 1 z ) V < e m > 1 ( z ) = V < / e m > 0 + ( e − j β < e m > 1 z + Γ e j β < / e m > 1 z ) V<em>1(z) = V</em>0^+(e^{-j\beta<em>1 z} + \Gamma e^{j\beta</em>1 z}) V < e m > 1 ( z ) = V < / e m > 0 + ( e − j β < e m > 1 z + Γ e j β < / e m > 1 z ) H ~ < e m > 1 ( z ) = y ^ E < / e m > 0 i η < e m > 1 ( e − j k < / e m > 1 z − Γ e j k < e m > 1 z ) \tilde{H}<em>1(z) = \hat{y}\frac{E</em>0^i}{\eta<em>1}(e^{-jk</em>1z} - \Gamma e^{jk<em>1z}) H ~ < e m > 1 ( z ) = y ^ η < e m > 1 E < / e m > 0 i ( e − j k < / e m > 1 z − Γ e j k < e m > 1 z ) I < / e m > 1 ( z ) = V < e m > 0 + Z < / e m > 01 ( e − j β < e m > 1 z − Γ e j β < / e m > 1 z ) I</em>1(z) = \frac{V<em>0^+}{Z</em>{01}}(e^{-j\beta<em>1 z} - \Gamma e^{j\beta</em>1 z}) I < / e m > 1 ( z ) = Z < / e m > 01 V < e m > 0 + ( e − j β < e m > 1 z − Γ e j β < / e m > 1 z ) E ~ < e m > 2 ( z ) = x ^ T E < / e m > 0 i e − j k < e m > 2 z \tilde{E}<em>2(z) = \hat{x}TE</em>0^i e^{-jk<em>2z} E ~ < e m > 2 ( z ) = x ^ T E < / e m > 0 i e − j k < e m > 2 z V < / e m > 2 ( z ) = V < e m > 0 + e − j β < / e m > 2 z V</em>2(z) = V<em>0^+ e^{-j\beta</em>2 z} V < / e m > 2 ( z ) = V < e m > 0 + e − j β < / e m > 2 z H ~ < e m > 2 ( z ) = y ^ T E < / e m > 0 i η < e m > 2 e − j k < / e m > 2 z \tilde{H}<em>2(z) = \hat{y}T\frac{E</em>0^i}{\eta<em>2}e^{-jk</em>2z} H ~ < e m > 2 ( z ) = y ^ T η < e m > 2 E < / e m > 0 i e − j k < / e m > 2 z I < e m > 2 ( z ) = T V < / e m > 0 + Z < e m > 02 e − j β < / e m > 2 z I<em>2(z) = T\frac{V</em>0^+}{Z<em>{02}}e^{-j\beta</em>2 z} I < e m > 2 ( z ) = T Z < e m > 02 V < / e m > 0 + e − j β < / e m > 2 z Γ = η < e m > 2 − η < / e m > 1 η < e m > 2 + η < / e m > 1 \Gamma = \frac{\eta<em>2 - \eta</em>1}{\eta<em>2 + \eta</em>1} Γ = η < e m > 2 + η < / e m > 1 η < e m > 2 − η < / e m > 1 Γ = Z < e m > 02 − Z < / e m > 01 Z < e m > 02 + Z < / e m > 01 \Gamma = \frac{Z<em>{02} - Z</em>{01}}{Z<em>{02} + Z</em>{01}} Γ = Z < e m > 02 + Z < / e m > 01 Z < e m > 02 − Z < / e m > 01 T = 1 + Γ T = 1 + \Gamma T = 1 + Γ T = 1 + Γ T = 1 + \Gamma T = 1 + Γ k < e m > 1 = ω μ < / e m > 1 ϵ < e m > 1 k<em>1 = \omega\sqrt{\mu</em>1\epsilon<em>1} k < e m > 1 = ω μ < / e m > 1 ϵ < e m > 1 k < / e m > 2 = ω μ < e m > 2 ϵ < / e m > 2 k</em>2 = \omega\sqrt{\mu<em>2\epsilon</em>2} k < / e m > 2 = ω μ < e m > 2 ϵ < / e m > 2 β < e m > 1 = ω μ < / e m > 1 ϵ < e m > 1 \beta<em>1 = \omega\sqrt{\mu</em>1\epsilon<em>1} β < e m > 1 = ω μ < / e m > 1 ϵ < e m > 1 β < / e m > 2 = ω μ < e m > 2 ϵ < / e m > 2 \beta</em>2 = \omega\sqrt{\mu<em>2\epsilon</em>2} β < / e m > 2 = ω μ < e m > 2 ϵ < / e m > 2 η < e m > 1 = μ < / e m > 1 ϵ < e m > 1 \eta<em>1 = \sqrt{\frac{\mu</em>1}{\epsilon<em>1}} η < e m > 1 = ϵ < e m > 1 μ < / e m > 1 η < / e m > 2 = μ < e m > 2 ϵ < / e m > 2 \eta</em>2 = \sqrt{\frac{\mu<em>2}{\epsilon</em>2}} η < / e m > 2 = ϵ < / e m > 2 μ < e m > 2 Power Transfer Incident wave, reflected wave, and transmitted wave. Average Poynting vector in medium 1: S < e m > a v 1 ( z ) = 1 2 R e [ E ~ < / e m > 1 ( z ) × H ~ < e m > 1 ∗ ( z ) ] = z ^ ∣ E < / e m > 0 i ∣ 2 2 η 1 ( 1 − ∣ Γ ∣ 2 ) S<em>{av1}(z) = \frac{1}{2}Re[\tilde{E}</em>1(z) \times \tilde{H}<em>1^*(z)] = \hat{z}\frac{|E</em>0^i|^2}{2\eta_1}(1 - |\Gamma|^2) S < e m > a v 1 ( z ) = 2 1 R e [ E ~ < / e m > 1 ( z ) × H ~ < e m > 1 ∗ ( z )] = z ^ 2 η 1 ∣ E < / e m > 0 i ∣ 2 ( 1 − ∣Γ ∣ 2 ) S < e m > a v 1 = S < / e m > i + S < e m > r = ∣ E < / e m > 0 i ∣ 2 2 η < e m > 1 − ∣ E < / e m > 0 i ∣ 2 2 η 1 ∣ Γ ∣ 2 S<em>{av1} = S</em>i + S<em>r = \frac{|E</em>0^i|^2}{2\eta<em>1} - \frac{|E</em>0^i|^2}{2\eta_1}|\Gamma|^2 S < e m > a v 1 = S < / e m > i + S < e m > r = 2 η < e m > 1 ∣ E < / e m > 0 i ∣ 2 − 2 η 1 ∣ E < / e m > 0 i ∣ 2 ∣Γ ∣ 2 S < e m > a v r = − ∣ Γ ∣ 2 S < / e m > a v i S<em>{av}^r = -|\Gamma|^2 S</em>{av}^i S < e m > a v r = − ∣Γ ∣ 2 S < / e m > a v i Power Transfer (Continued) Average Poynting vector in medium 2: S < e m > a v 2 ( z ) = 1 2 R e [ E ~ < / e m > 2 ( z ) × H ~ < e m > 2 ∗ ( z ) ] = z ^ ∣ E < / e m > 0 i ∣ 2 2 η 2 T 2 S<em>{av2}(z) = \frac{1}{2}Re[\tilde{E}</em>2(z) \times \tilde{H}<em>2^*(z)] = \hat{z}\frac{|E</em>0^i|^2}{2\eta_2}T^2 S < e m > a v 2 ( z ) = 2 1 R e [ E ~ < / e m > 2 ( z ) × H ~ < e m > 2 ∗ ( z )] = z ^ 2 η 2 ∣ E < / e m > 0 i ∣ 2 T 2 For lossless media: S < e m > a v 1 = S < / e m > a v 2 S<em>{av1} = S</em>{av2} S < e m > a v 1 = S < / e m > a v 2 ∣ E < e m > 0 i ∣ 2 2 η < / e m > 1 ( 1 − ∣ Γ ∣ 2 ) = ∣ E < e m > 0 i ∣ 2 2 η < / e m > 2 T 2 ⟹ η < e m > 2 η < / e m > 1 = 1 − ∣ Γ ∣ 2 \frac{|E<em>0^i|^2}{2\eta</em>1}(1 - |\Gamma|^2) = \frac{|E<em>0^i|^2}{2\eta</em>2}T^2 \implies \frac{\eta<em>2}{\eta</em>1} = 1 - |\Gamma|^2 2 η < / e m > 1 ∣ E < e m > 0 i ∣ 2 ( 1 − ∣Γ ∣ 2 ) = 2 η < / e m > 2 ∣ E < e m > 0 i ∣ 2 T 2 ⟹ η < / e m > 1 η < e m > 2 = 1 − ∣Γ ∣ 2 Lossless - Lossy - Normal Incidence: Lossy Medium 1 properties: μ < e m > 1 , ϵ < / e m > 1 , η < e m > 1 , k < / e m > 1 \mu<em>1, \epsilon</em>1, \eta<em>1, k</em>1 μ < e m > 1 , ϵ < / e m > 1 , η < e m > 1 , k < / e m > 1 Medium 2 properties: μ < e m > 2 , ϵ < / e m > 2 , σ < e m > 2 , η < / e m > 2 , β 2 \mu<em>2, \epsilon</em>2, \sigma<em>2, \eta</em>2, \beta_2 μ < e m > 2 , ϵ < / e m > 2 , σ < e m > 2 , η < / e m > 2 , β 2 Normal Incidence: PEC Perfect Electric Conductor (PEC). n ^ × E ~ = 0 \hat{n} \times \tilde{E} = 0 n ^ × E ~ = 0 n ^ ⋅ H ~ = 0 \hat{n} \cdot \tilde{H} = 0 n ^ ⋅ H ~ = 0 Normal Incidence: Radomes A half-wave section can be used as a dielectric window. η < e m > 2 = η < / e m > 0 \eta<em>2 = \eta</em>0 η < e m > 2 = η < / e m > 0 d = λ 2 d = \frac{\lambda}{2} d = 2 λ Medium 1 properties: μ < e m > 0 , ϵ < / e m > 0 \mu<em>0, \epsilon</em>0 μ < e m > 0 , ϵ < / e m > 0 Medium 2 properties: μ < e m > 0 , ϵ < / e m > 2 \mu<em>0, \epsilon</em>2 μ < e m > 0 , ϵ < / e m > 2 Medium 3 properties: μ < e m > 0 , ϵ < / e m > 0 \mu<em>0, \epsilon</em>0 μ < e m > 0 , ϵ < / e m > 0 Z < e m > i n = Z < / e m > L Z<em>{in} = Z</em>L Z < e m > in = Z < / e m > L Normal Incidence: Lens and Prism Both applications are frequency sensitive and the matching section is only λ 4 \frac{\lambda}{4} 4 λ λ 2 \frac{\lambda}{2} 2 λ η < e m > 2 = η < / e m > 0 η 3 \eta<em>2 = \sqrt{\eta</em>0\eta_3} η < e m > 2 = η < / e m > 0 η 3 Medium 1 properties: μ < e m > 0 , ϵ < / e m > 0 \mu<em>0, \epsilon</em>0 μ < e m > 0 , ϵ < / e m > 0 Coating properties: μ < e m > 2 , ϵ < / e m > 2 \mu<em>2, \epsilon</em>2 μ < e m > 2 , ϵ < / e m > 2 Lens properties: μ < e m > 3 , ϵ < / e m > 3 \mu<em>3, \epsilon</em>3 μ < e m > 3 , ϵ < / e m > 3 λ 4 \frac{\lambda}{4} 4 λ λ 4 \frac{\lambda}{4} 4 λ 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 ϵ < e m > r = 9 \epsilon<em>r = 9 ϵ < e m > r = 9 μ < / e m > r = 1 \mu</em>r = 1 μ < / e m > r = 1 Choose its thickness d d d Structural integrity requires d d d 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 λ 2 2 \frac{\lambda_2}{2} 2 λ 2 d = n λ 2 2 d = n\frac{\lambda_2}{2} d = n 2 λ 2 Example 8-2: Yellow Light Incident upon a Glass Surface A beam of yellow light with wavelength 0.6 μ \mu μ 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 η < e m > 1 = μ < / e m > 0 ϵ 0 ≈ 120 π Ω \eta<em>1 = \sqrt{\frac{\mu</em>0}{\epsilon_0}} \approx 120\pi \Omega η < e m > 1 = ϵ 0 μ < / e m > 0 ≈ 120 π Ω η < e m > 2 = η < / e m > 1 ϵ r = 120 π 2.25 ≈ 80 π Ω \eta<em>2 = \frac{\eta</em>1}{\sqrt{\epsilon_r}} = \frac{120\pi}{\sqrt{2.25}} \approx 80\pi \Omega η < e m > 2 = ϵ r η < / e m > 1 = 2.25 120 π ≈ 80 π Ω Γ = η < e m > 2 − η < / e m > 1 η < e m > 2 + η < / e m > 1 = 80 π − 120 π 80 π + 120 π = − 0.2 \Gamma = \frac{\eta<em>2 - \eta</em>1}{\eta<em>2 + \eta</em>1} = \frac{80\pi - 120\pi}{80\pi + 120\pi} = -0.2 Γ = η < e m > 2 + η < / e m > 1 η < e m > 2 − η < / e m > 1 = 80 π + 120 π 80 π − 120 π = − 0.2 Example 8-2: Solution (Continued) ∣ Γ ∣ = 0.2 |\Gamma| = 0.2 ∣Γ∣ = 0.2 ϕ = π \phi = \pi ϕ = π The electric-field magnitude is maximum atz < e m > m a x = θ < / e m > r λ < e m > 1 4 π + n λ < / e m > 1 2 = λ < e m > 1 4 + n λ < / e m > 1 2 ( n = 0 , 1 , 2 , … ) z<em>{max} = \frac{\theta</em>r \lambda<em>1}{4\pi} + n\frac{\lambda</em>1}{2} = \frac{\lambda<em>1}{4} + n\frac{\lambda</em>1}{2} \quad (n = 0, 1, 2, …) z < e m > ma x = 4 π θ < / e m > r λ < e m > 1 + n 2 λ < / e m > 1 = 4 λ < e m > 1 + n 2 λ < / e m > 1 ( n = 0 , 1 , 2 , … ) λ 1 = 0.6 μ m \lambda_1 = 0.6 \mu m λ 1 = 0.6 μ m S = 1 + ∣ Γ ∣ 1 − ∣ Γ ∣ = 1 + 0.2 1 − 0.2 = 1.5 S = \frac{1 + |\Gamma|}{1 - |\Gamma|} = \frac{1 + 0.2}{1 - 0.2} = 1.5 S = 1 − ∣Γ∣ 1 + ∣Γ∣ = 1 − 0.2 1 + 0.2 = 1.5 S < e m > a v 2 S < / e m > a v 1 = η < e m > 1 η < / e m > 2 ∣ T ∣ 2 = 4 η < e m > 1 η < / e m > 2 ( η < e m > 2 + η < / e m > 1 ) 2 \frac{S<em>{av2}}{S</em>{av1}} = \frac{\eta<em>1}{\eta</em>2}|T|^2 = \frac{4\eta<em>1\eta</em>2}{(\eta<em>2 + \eta</em>1)^2} S < / e m > a v 1 S < e m > a v 2 = η < / e m > 2 η < e m > 1 ∣ T ∣ 2 = ( η < e m > 2 + η < / e m > 1 ) 2 4 η < e m > 1 η < / e m > 2 S < e m > a v 2 S < / e m > a v 1 = 1 − ∣ Γ ∣ 2 = 1 − ( 0.2 ) 2 = 0.96 = 96 % \frac{S<em>{av2}}{S</em>{av1}} = 1 - |\Gamma|^2 = 1 - (0.2)^2 = 0.96 = 96\% S < / e m > a v 1 S < e m > a v 2 = 1 − ∣Γ ∣ 2 = 1 − ( 0.2 ) 2 = 0.96 = 96% 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 ϵ < e m > r = 1 \epsilon<em>r = 1 ϵ < e m > r = 1 μ < / e m > r = 1 \mu</em>r = 1 μ < / e m > r = 1 σ = 5.8 × 10 7 \sigma = 5.8 \times 10^7 σ = 5.8 × 1 0 7 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), α = 0 \alpha = 0 α = 0 β = k 1 = ω c = 2 π × 10 9 3 × 10 8 = 20 π 3 (rad/m) \beta = k_1 = \frac{\omega}{c} = \frac{2\pi \times 10^9}{3 \times 10^8} = \frac{20\pi}{3} \text{ (rad/m)} β = k 1 = c ω = 3 × 1 0 8 2 π × 1 0 9 = 3 20 π (rad/m) η < e m > 1 = η < / e m > 0 = 377 Ω \eta<em>1 = \eta</em>0 = 377 \Omega η < e m > 1 = η < / e m > 0 = 377Ω λ = 2 π k 1 = 0.3 m \lambda = \frac{2\pi}{k_1} = 0.3 \text{ m} λ = k 1 2 π = 0.3 m Since σ ω ϵ = 1 × 10 9 > 1 \frac{\sigma}{\omega \epsilon} = 1 \times 10^9 > 1 ω ϵ σ = 1 × 1 0 9 > 1 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 η < e m > c 2 \eta<em>{c2} η < e m > c 2 η < / e m > 0 = 377 Ω \eta</em>0 = 377 \Omega η < / e m > 0 = 377Ω Γ = η < e m > c 2 − η < / e m > 0 η < e m > c 2 + η < / e m > 0 ≈ − 1 \Gamma = \frac{\eta<em>{c2} - \eta</em>0}{\eta<em>{c2} + \eta</em>0} \approx -1 Γ = η < e m > c 2 + η < / e m > 0 η < e m > c 2 − η < / e m > 0 ≈ − 1 Example 8-3: Solution (Continued) Upon setting Γ = − 1 \Gamma = -1 Γ = − 1 E ~ < e m > 1 ( z ) = x ^ E < / e m > 0 i ( e − j k < e m > 1 z − e j k < / e m > 1 z ) = − x ^ j 2 E < e m > 0 i sin k < / e m > 1 z \tilde{E}<em>1(z) = \hat{x}E</em>0^i(e^{-jk<em>1z} - e^{jk</em>1z}) = -\hat{x}j2E<em>0^i \sin k</em>1z E ~ < e m > 1 ( z ) = x ^ E < / e m > 0 i ( e − j k < e m > 1 z − e j k < / e m > 1 z ) = − x ^ j 2 E < e m > 0 i sin k < / e m > 1 z H ~ < e m > 1 ( z ) = y ^ E < / e m > 0 i η < e m > 1 ( e − j k < / e m > 1 z + e j k < e m > 1 z ) = y ^ 2 E < / e m > 0 i η < e m > 1 cos k < / e m > 1 z \tilde{H}<em>1(z) = \hat{y}\frac{E</em>0^i}{\eta<em>1}(e^{-jk</em>1z} + e^{jk<em>1z}) = \hat{y}2\frac{E</em>0^i}{\eta<em>1}\cos k</em>1z H ~ < e m > 1 ( z ) = y ^ η < e m > 1 E < / e m > 0 i ( e − j k < / e m > 1 z + e j k < e m > 1 z ) = y ^ 2 η < e m > 1 E < / e m > 0 i cos k < / e m > 1 z E < e m > 1 ( z , t ) = R e [ E ~ < / e m > 1 ( z ) e j ω t ] = x ^ 2 E < e m > 0 i sin k < / e m > 1 z sin ω t = x ^ 24 sin ( 20 π z 3 ) sin ( 2 π × 10 9 t ) (mV/m) E<em>1(z, t) = Re[\tilde{E}</em>1(z)e^{j\omega t}] = \hat{x}2E<em>0^i\sin k</em>1z \sin \omega t = \hat{x}24\sin(\frac{20\pi z}{3})\sin(2\pi \times 10^9 t) \text{ (mV/m)} E < e m > 1 ( z , t ) = R e [ E ~ < / e m > 1 ( z ) e j ω t ] = x ^ 2 E < e m > 0 i sin k < / e m > 1 z sin ω t = x ^ 24 sin ( 3 20 π z ) sin ( 2 π × 1 0 9 t ) (mV/m) H1(z, t) = Re[\tilde{H} 1(z)e^{j\omega t}] = \hat{y}2\frac{E0^i}{\eta 1}\cos k_1z \cos \omega t = \hat{y}64\cos(\frac{20\pi z}{3})\cos(2\pi \times 10^9 t) \text{ (\muA/m)}