Problem 7.9 For a wave characterized by the electric field E(z,t) = ˆxa x cos(ωt kz)+ŷa y cos(ωt kz+δ) identify the polarization state, determine the polarization angles (γ, χ), and sketch the locus of E(0,t) for each of the following cases: (a) a x = 3 V/m, a y = 4 V/m, and δ = 0 (b) a x = 3 V/m, a y = 4 V/m, and δ = 80 (c) a x = 3 V/m, a y = 3 V/m, and δ = 45 (d) a x = 3 V/m, a y = 4 V/m, and δ = 35 Solution: y y 4 3 2-4 -3-2 - 2 3 4 - -2-3 -4 x 4 3 2-4 -3-2 - - 2 3 4-2 -3-4 x (a) (b) y y 4 3 2-4 -2 2 4 - -2-3 -4 (c) x 4 3 2-4 -2-2 4-2 -3-4 (d) x Figure P7.9: Plots of the locus of E(0,t).
ψ 0 = tan (a y /a x ), [Eq. (7.60)], tan2γ = (tan2ψ 0 )cosδ sin2χ = (sin2ψ 0 )sinδ [Eq. (7.59a)], [Eq. (7.59b)]. Case a x a y δ ψ 0 γ χ Polarization State (a) 3 4 0 53.3 53.3 0 Linear (b) 3 4 80 53.3 53.3 0 Linear (c) 3 3 45 45 45 22.5 Left elliptical (d) 3 4 35 53.3 56.2 2.37 Right elliptical (a) E(z,t) = ˆx3cos(ωt kz)+ŷ4cos(ωt kz). (b) E(z,t) = ˆx3cos(ωt kz) ŷ4cos(ωt kz). (c) E(z,t) = ˆx3cos(ωt kz)+ŷ3cos(ωt kz+45 ). (d) E(z,t) = ˆx3cos(ωt kz)+ŷ4cos(ωt kz 35 ).
Problem 7.0 is given by The electric field of a uniform plane wave propagating in free space Ẽ = (ˆx+ jŷ)30e jπz/6 (V/m) Specify the modulus and direction of the electric field intensity at the z = 0 plane at t = 0, 5, and 0 ns. Solution: From E(z,t) = Re[Ẽe jωt ] = Re[(ˆx+ jŷ)20e jπz/6 e jωt ] = Re[(ˆx+ŷe jπ/2 )20e jπz/6 e jωt ] = ˆx20cos(ωt πz/6)+ŷ20cos(ωt πz/6+π/2) = ˆx20cos(ωt πz/6) ŷ20sin(ωt πz/6) (V/m), E = [ Ex 2 + Ey] 2 /2 = 20 (V/m), ( ) ψ = tan Ey = (ωt πz/6). E x f = c λ = kc π/6 3 08 = 2π 2π ω = 2π f = 5π 0 7 rad/s. = 2.5 0 7 Hz, At z = 0, 0 at t = 0, ψ = ωt = 5π 0 7 t = 0.25π = 45 at t = 5 ns, 0.5π = 90 at t = 0 ns. Therefore, the wave is LHC polarized.
Problem 7.2 The electric field of an elliptically polarized plane wave is given by Determine the following: (a) The polarization angles (γ, χ). (b) The direction of rotation. Solution: (a) E(z,t) = [ ˆx0sin(ωt kz 60 ) + ŷ30cos(ωt kz)] (V/m) E(z,t) = [ ˆx0sin(ωt kz 60 )+ŷ30cos(ωt kz)] = [ˆx0cos(ωt kz+30 )+ŷ30cos(ωt kz)] (V/m). Phasor form: Ẽ = (ˆx0e j30 + ŷ30)e jkz. Since δ is defined as the phase of E y relative to that of E x, δ = 30, ( ) 30 ψ 0 = tan = 7.56, 0 tan2γ = (tan2ψ 0 )cosδ = 0.65 or γ = 73.5, sin2χ = (sin2ψ 0 )sinδ = 0.40 or χ = 8.73. (b) Since χ < 0, the wave is right-hand elliptically polarized.
Problem 7.7 In a medium characterized by ε r = 9, µ r =, and σ = 0. S/m, determine the phase angle by which the magnetic field leads the electric field at 00 MHz. Solution: The phase angle by which the magnetic field leads the electric field is θ η where θ η is the phase angle of η c. Hence, quasi-conductor. σ ωε = 0. 36π 2π 0 8 0 9 9 = 2. η c = µ ε ) /2 ( j ε ε = 20π σ ( j εr ωε 0 ε r ) /2 = 25.67( j2) /2 = 7.49+ j44.8 = 84.04 3.72. Therefore θ η = 3.72. Since H = (/η c )ˆk E, H leads E by θ η, or by 3.72. In other words, H lags E by 3.72.
Problem 7.22 The electric field of a plane wave propagating in a nonmagnetic medium is given by E = ẑ25e 30x cos(2π 0 9 t 40x) (V/m) Obtain the corresponding expression for H. Solution: From the given expression for E, From (7.65a) and (7.65b), ω = 2π 0 9 α = 30 β = 40 (Np/m), (rad/m). (rad/s), α 2 β 2 = ω 2 µε = ω 2 µ 0 ε 0 ε r = ω2 c 2 ε r, 2αβ = ω 2 µε = ω2 c 2 ε r. Using the above values for ω, α, and β, we obtain the following: ε r =.6, ε r = 5.47. ) µ /2 η c = ( ε j ε ε = η ( ) /2 0 j ε r = 377 ( j 5.47 ) /2 = 57.9e j36.85 (Ω). ε r.6.6 ε r Ẽ = ẑ25e 30x e j40x, H = ˆk Ẽ = ˆx ẑ25e 30x e j40x = ŷ0.6e 30x e 40x e j36.85, η c 57.9e j36.85 H = Re{ He jωt } = ŷ0.6e 30x cos(2π 0 9 t 40x 36.85 ) (A/m).
Problem 7.24 In a nonmagnetic, lossy, dielectric medium, a 300-MHz plane wave is characterized by the magnetic field phasor H = (ˆx j4ẑ)e 2y e j9y (A/m) Obtain time-domain expressions for the electric and magnetic field vectors. Solution: Ẽ = η c ˆk H. To find η c, we need ε and ε. From the given expression for H, α = 2 β = 9 (Np/m), (rad/m). Also, we are given than f = 300 MHz = 3 0 8 Hz. From (7.65a), whose solution gives Similarly, from (7.65b), α 2 β 2 = ω 2 µε, 4 8 = (2π 3 0 8 ) 2 4π 0 7 ε r 0 9 36π, 2αβ = ω 2 µε, ε r =.95. which gives 2 2 9 = (2π 3 0 8 ) 2 4π 0 7 ε r 0 9 36π, ε r = 0.9. ) µ /2 η c = ( ε j ε ε = η ( 0 j 0.9 ) /2 = 377 (0.93+ j0.2) = 256.9e j2.6. ε r.95.95 Hence, Ẽ = 256.9e j2.6 ŷ (ˆx j4ẑ)e 2y e j9y = (ˆx j4+ẑ)256.9e 2y e j9y e j2.6 = (ˆx4e jπ/2 + ẑ)256.9e 2y e j9y e j2.6,
E = Re{Ẽe jωt } = ˆx.03 0 3 e 2y cos(ωt 9y+02.6 ) + ẑ256.9e 2y cos(ωt 9y+2.6 ) (V/m), H = Re{ He jωt } = Re{(ˆx+ j4ẑ)e 2y e j9y e jωt } = ˆxe 2y cos(ωt 9y)+ẑ4e 2y sin(ωt 9y) (A/m).
Problem 8.2 A plane wave traveling in medium with ε r = 2.25 is normally incident upon medium 2 with ε r2 = 4. Both media are made of nonmagnetic, nonconducting materials. If the electric field of the incident wave is given by E i = ŷ8cos(6π 0 9 t 30πx) (V/m). (a) Obtain time-domain expressions for the electric and magnetic fields in each of the two media. (b) Determine the average power densities of the incident, transmitted waves. Solution: (a) E i = ŷ8cos(6π 0 9 t 30πx) (V/m), η = η 0 = η 0 = η 0 εr 2.25.5 = 377 = 25.33 Ω,.5 η 2 = η 0 = η 0 = 377 = 88.5 Ω, εr2 4 2 Γ = η 2 η /2 /.5 = η 2 + η /2+/.5 = 0.43, τ = +Γ = 0.43 = 0.857, E r = ΓE i =.4ŷcos(6π 0 9 t + 30πx) (V/m). reflected and Note that the coefficient of x is positive, denoting the fact that E r belongs to a wave traveling in x-direction. E = E i + E r = ŷ[8cos(6π 0 9 t 30πx).4cos(6π 0 9 t + 30πx)] (A/m), H i = ẑ 8 η cos(6π 0 9 t 30πx) = ẑ3.83cos(6π 0 9 t 30πx) (ma/m), H r = ẑ.4 cos(6π 0 9 t + 30πx) = ẑ4.54cos(6π 0 9 t + 30πx) (ma/m), η H = H i + H r = ẑ[3.83cos(6π 0 9 t 30πx)+4.54cos(6π 0 9 t + 30πx)] (ma/m). Since k = ω µε and k 2 = ω µε 2, ε2 4 k 2 = k = 30π = 40π (rad/m), ε 2.25
E 2 = E t = ŷ8τ cos(6π 0 9 t 40πx) = ŷ6.86cos(6π 0 9 t 40πx) (V/m), H 2 = H t = ẑ 8τ η 2 cos(6π 0 9 t 40πx) = ẑ36.38cos(6π 0 9 t 40πx) (ma/m). (b) S i av = ˆx 82 64 = 2η 2 25.33 = ˆx27.3 (mw/m2 ), S r av = Γ 2 S i av = ˆx(0.43) 2 0.27 = ˆx2.6 (mw/m 2 ), S t av = Et 0 2 2η 2 = ˆxτ 2(8)2 2η 2 = ˆx (0.86)2 64 2 88.5 = ˆx24.7 (mw/m2 ). Within calculation error, S i av + S r av = S t av.
Problem 8.6 A 50-MHz plane wave with electric field amplitude of 50 V/m is normally incident in air onto a semi-infinite, perfect dielectric medium with ε r = 36. Determine the following: (a) Γ (b) The average power densities of the incident and reflected waves. (c) The distance in the air medium from the boundary to the nearest minimum of the electric field intensity, E. Solution: (a) η = η 0 = 20π (Ω), µ2 η 2 = = 20π = 20π = 20π ε 2 εr2 6 (Ω), Γ = η 2 η 20π 20π = η 2 + η 20π + 20π = 0.7. Hence, Γ = 0.7 and θ η = 80. (b) (c) In medium (air), From Eqs. (8.6) and (8.7), S i av = Ei 0 2 2η = (50)2 2 20π = 3.32 (W/m2 ), S r av = Γ 2 S i av = (0.7) 2 3.32 =.67 (W/m 2 ). λ = c f = 3 08 5 0 7 = 6 m. l max = θ rλ 4π = π 6 =.5 m, 4π l min = l max λ 4 =.5.5 = 0 m (at the boundary).
Problem 8.5 A 5-MHz plane wave with electric field amplitude of 0 (V/m) is normally incident in air onto the plane surface of a semi-infinite conducting material with ε r = 4, µ r =, and σ = 00 (S/m). Determine the average power dissipated (lost) per unit cross-sectional area in a 2-mm penetration of the conducting medium. Solution: For convenience, let us choose E i to be along ˆx and the incident direction to be +ẑ. With k = ω 2π 5 06 = c 3 0 8 = π (rad/m), 30 we have From Table 7-, E i = ˆx0cos (π 0 7 t π ) 30 z η = η 0 = 377 Ω. ε ε = σ ωε r ε 0 = (V/m), 00 36π π 0 7 4 0 9 = 9 04, which makes the material a good conductor, for which α 2 = π f µσ = π 5 0 6 4π 0 7 00 = 44.43 (Np/m), β 2 = 44.43 η c2 = (+ j) α 2 σ (rad/m), = (+ j) 44.43 00 = 0.44(+ j) Ω. According to the expression for S av2 given in the answer to Exercise 8.3, ( ) S av2 = ẑ τ 2 Ei 0 2 2 e 2α 2z Re. The power lost is equal to the difference between S av2 at z = 0 and S av2 at z = 2 mm. Thus, where z = 2 mm. τ = +Γ η c 2 P = power lost per unit cross-sectional area = S av2 (0) S av2 (z = 2 mm) ( ) = τ 2 Ei 0 2 2 Re [ e 2α 2z ] η c 2 = + η 2 η η 2 + η = + 0.44(+ j) 377 0.44(+ j)+377 0.0023(+ j) = 3.3 0 3 e j45.
( ) ( Re ηc = Re 2 ( = Re P = (3.3 0 3 ) 2 02 2 ) 0.44(+ j) 0.44( j) ) = Re ( ) + j = 0.44 2 0.88 =.4,.4[ e 2 44.43 2 0 3] =.0 0 4 (W/m 2 ).