Transcript

1

2

3

4

5

6

7

8

9

10 Problem 7.19 Ignoring reflection at the air soil boundary, if the amplitude of a 3-GHz incident wave is 10 V/m at the surface of a wet soil medium, at what depth will it be down to 1 mv/m? Wet soil is characterized by µ r = 1, ε r = 9, and σ = S/m. Solution: E(z) = E 0 e αz = 10e αz, σ ωε = π 2π = Hence, medium is a low-loss dielectric. α = σ µ 2 ε = σ 2 120π = π εr 2 = (Np/m), = 10e 0.032z, ln10 4 = 0.032z, z = m.

11 Problem 8.4 A 200-MHz, left-hand circularly polarized plane wave with an electric field modulus of 5 V/m is normally incident in air upon a dielectric medium with ε r = 4, and occupies the region defined by z 0. (a) Write an expression for the electric field phasor of the incident wave, given that the field is a positive maximum at z = 0 and t = 0. (b) Calculate the reflection and transmission coefficients. (c) Write expressions for the electric field phasors of the reflected wave, the transmitted wave, and the total field in the region z 0. (d) Determine the percentages of the incident average power reflected by the boundary and transmitted into the second medium. Solution: (a) LHC wave: Hence, (b) k 1 = ω 2π = c = 4π 3 rad/m, k 2 = ω = ω εr2 = 4π 8π 4 = u p2 c 3 3 rad/m. Ẽ i = a 0 (ˆx+ŷe jπ/2 )e jkz = a 0 (ˆx+ jŷ)e jkz, E i (z,t) = ˆxa 0 cos(ωt kz) ŷa 0 sin(ωt kz), E i = [a 2 0 cos 2 (ωt kz)+a 2 0 sin 2 (ωt kz)] 1/2 = a 0 = 5 (V/m). Ẽ i = 5(ˆx+ jŷ)e j4πz/3 (V/m). η 1 = η 0 = 120π (Ω), η 2 = η 0 = η 0 = 60π (Ω). εr 2 Equations (8.8a) and (8.9) give (c) Γ = η 2 η 1 60π 120π = η 2 + η 1 60π + 120π = = 1 3, τ = 1+Γ = 2 3. Ẽ r = 5Γ(ˆx+ jŷ)e jk1z = 5 j4πz/3 (ˆx+ jŷ)e (V/m), 3 Ẽ t = 5τ(ˆx+ jŷ)e jk2z = 10 3 (ˆx+ jŷ)e j8πz/3 (V/m),

12 Ẽ 1 = Ẽ i + Ẽ r = 5(ˆx+ jŷ) [e j4πz/3 13 ] e j4πz/3 (V/m). (d) % of reflected power = 100 Γ 2 = = 11.11%, ( 2 3 % of transmitted power = 100 τ 2 η 1 η 2 = 100 ) 2 120π 60π = 88.89%.

13 Problem 7.16 Dry soil is characterized by ε r = 2.5, µ r = 1, and σ = 10 4 (S/m). At each of the following frequencies, determine if dry soil may be considered a good conductor, a quasi-conductor, or a low-loss dielectric, and then calculate α, β, λ, µ p, and η c : (a) 60 Hz (b) 1 khz (c) 1 MHz (d) 1 GHz Solution: ε r = 2.5, µ r = 1, σ = 10 4 S/m. ε f 60 Hz 1 khz 1 MHz 1 GHz ε = σ ωε σ = 2π f ε r ε Type of medium Good conductor Good conductor Quasi-conductor Low-loss dielectric α (Np/m) β (rad/m) λ (m) u p (m/s) η c (Ω) 1.54(1+ j) 6.28(1+ j) j

14 Problem 7.17 In a medium characterized by ε r = 9, µ r = 1, and σ = 0.1 S/m, determine the phase angle by which the magnetic field leads the electric field at 100 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 = π 2π = 2. η c = µ ε ) 1/2 (1 j ε ε = 120π σ (1 j εr ωε 0 ε r ) 1/2 = (1 j2) 1/2 = j44.18 = Therefore θ η = Since H = (1/η c )ˆk E, H leads E by θ η, or by In other words, H lags E by

15 Problem 7.18 Generate a plot for the skin depth δ s versus frequency for seawater for the range from 1 khz to 10 GHz (use log-log scales). The constitutive parameters of seawater are µ r = 1, ε r = 80, and σ = 4 S/m. Solution: δ s = 1 α = 1 ω ω = 2π f, µε 2 ( ) ε ε 1/2 µε = µ 0 ε 0 ε r = ε r c 2 = 80 c 2 = 80 ( ) 2, ε ε = σ ωε = σ 4 36π = ωε 0 ε r 2π f = f 109. See Fig. P7.18 for plot of δ s versus frequency., 10 1 Skin depth vs. frequency for seawater 10 0 Skin depth (m) Frequency (MHz) Figure P7.18: Skin depth versus frequency for seawater.

16 Problem 8.1 A plane wave in air with an electric field amplitude of 20 V/m is incident normally upon the surface of a lossless, nonmagnetic medium with ε r = 25. Determine the following: (a) The reflection and transmission coefficients. (b) The standing-wave ratio in the air medium. (c) The average power densities of the incident, reflected, and transmitted waves. Solution: (a) η 1 = η 0 = 120π (Ω), η 2 = η 0 = 120π = 24π (Ω). εr 5 From Eqs. (8.8a) and (8.9), (b) Γ = η 2 η 1 24π 120π = η 2 + η 1 24π + 120π = = 0.67, τ = 1+Γ = = (c) According to Eqs. (8.19) and (8.20), S = 1+ Γ 1 Γ = = 5. S i av = Ei 0 2 2η 0 = π = 0.52 W/m2, S r av = Γ 2 S i av = (0.67) = 0.24 W/m 2, S t av = τ 2 Ei 0 2 2η 2 = τ 2 η 1 η 2 S i av = (0.33) 2 120π 24π 0.52 = 0.28 W/m2.

17 Problem 8.2 A plane wave traveling in medium 1 with ε r1 = 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 10 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 10 9 t 30πx) (V/m), η 1 = η 0 = η 0 = η 0 εr = 377 = Ω, 1.5 η 2 = η 0 = η 0 = 377 = Ω, εr2 4 2 Γ = η 2 η 1 1/2 1/1.5 = η 2 + η 1 1/2+1/1.5 = 0.143, τ = 1+Γ = = 0.857, E r = ΓE i = 1.14ŷcos(6π 10 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 1 = E i + E r = ŷ[8cos(6π 10 9 t 30πx) 1.14cos(6π 10 9 t + 30πx)] (A/m), H i = ẑ 8 η 1 cos(6π 10 9 t 30πx) = ẑ31.83cos(6π 10 9 t 30πx) (ma/m), H r = ẑ 1.14 cos(6π 10 9 t + 30πx) = ẑ4.54cos(6π 10 9 t + 30πx) (ma/m), η 1 H 1 = H i + H r = ẑ[31.83cos(6π 10 9 t 30πx)+4.54cos(6π 10 9 t + 30πx)] (ma/m). Since k 1 = ω µε 1 and k 2 = ω µε 2, ε2 4 k 2 = k 1 = 30π = 40π (rad/m), ε

18 E 2 = E t = ŷ8τ cos(6π 10 9 t 40πx) = ŷ6.86cos(6π 10 9 t 40πx) (V/m), H 2 = H t = ẑ 8τ η 2 cos(6π 10 9 t 40πx) = ẑ36.38cos(6π 10 9 t 40πx) (ma/m). (b) S i av = ˆx = 2η = ˆx127.3 (mw/m2 ), S r av = Γ 2 S i av = ˆx(0.143) = ˆx2.6 (mw/m 2 ), S t av = Et 0 2 2η 2 = ˆxτ 2(8)2 2η 2 = ˆx (0.86) = ˆx124.7 (mw/m2 ). Within calculation error, S i av + S r av = S t av.

19 Problem 8.3 A plane wave traveling in a medium with ε r1 = 9 is normally incident upon a second medium with ε r2 = 4. Both media are made of nonmagnetic, nonconducting materials. If the magnetic field of the incident plane wave is given by H i = ẑ2cos(2π 10 9 t ky) (A/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, reflected, and transmitted waves. Solution: (a) In medium 1, u p = c εr1 = = (m/s), k 1 = ω u p = 2π 109 = 20π (rad/m), H i = ẑ2cos(2π 10 9 t 20πy) (A/m), η 1 = η 0 = 377 = Ω, εr1 3 η 2 = η 0 = 377 = Ω, εr2 2 E i = ˆx2η 1 cos(2π 10 9 t 20πy) = ˆx251.34cos(2π 10 9 t 20πy) (V/m), Γ = η 2 η = η 2 + η = 0.2, τ = 1+Γ = 1.2, E r = ˆx cos(2π 10 9 t + 20πy) = ˆx50.27cos(2π 10 9 t + 20πy) (V/m), H r = ẑ cos(2π 10 9 t + 20πy) η 1 = ẑ0.4cos(2π 10 9 t + 20πy) (A/m), E 1 = E i + E r = ˆx[25.134cos(2π 10 9 t 20πy)+50.27cos(2π 10 9 t + 20πy)] (V/m),

20 H 1 = H i + H r = ẑ[2cos(2π 10 9 t 20πy) 0.4cos(2π 10 9 t + 20πy)] (A/m). In medium 2, ε2 4 40π k 2 = k 1 = 20π = (rad/m), ε ( E 2 = E t = ˆx251.34τ cos 2π 10 9 t 40πy ) 3 ( = ˆx301.61cos 2π 10 9 t 40πy ) (V/m), 3 H 2 = H t = ẑ ( cos 2π 10 9 t 40πy ) η 2 3 ( = ẑ1.6cos 2π 10 9 t 40πy ) (A/m). 3 (b) S i av = ŷ E 0 2 2η 1 = ŷ (251.34) = ŷ (W/m2 ), S r av = ŷ Γ 2 (251.34) = ŷ10.05 (W/m 2 ), S t av = ŷ( ) = ŷ (W/m 2 ).

21 Problem 8.4 A 200-MHz, left-hand circularly polarized plane wave with an electric field modulus of 5 V/m is normally incident in air upon a dielectric medium with ε r = 4, and occupies the region defined by z 0. (a) Write an expression for the electric field phasor of the incident wave, given that the field is a positive maximum at z = 0 and t = 0. (b) Calculate the reflection and transmission coefficients. (c) Write expressions for the electric field phasors of the reflected wave, the transmitted wave, and the total field in the region z 0. (d) Determine the percentages of the incident average power reflected by the boundary and transmitted into the second medium. Solution: (a) LHC wave: Hence, (b) k 1 = ω 2π = c = 4π 3 rad/m, k 2 = ω = ω εr2 = 4π 8π 4 = u p2 c 3 3 rad/m. Ẽ i = a 0 (ˆx+ŷe jπ/2 )e jkz = a 0 (ˆx+ jŷ)e jkz, E i (z,t) = ˆxa 0 cos(ωt kz) ŷa 0 sin(ωt kz), E i = [a 2 0 cos 2 (ωt kz)+a 2 0 sin 2 (ωt kz)] 1/2 = a 0 = 5 (V/m). Ẽ i = 5(ˆx+ jŷ)e j4πz/3 (V/m). η 1 = η 0 = 120π (Ω), η 2 = η 0 = η 0 = 60π (Ω). εr 2 Equations (8.8a) and (8.9) give (c) Γ = η 2 η 1 60π 120π = η 2 + η 1 60π + 120π = = 1 3, τ = 1+Γ = 2 3. Ẽ r = 5Γ(ˆx+ jŷ)e jk1z = 5 j4πz/3 (ˆx+ jŷ)e (V/m), 3 Ẽ t = 5τ(ˆx+ jŷ)e jk2z = 10 3 (ˆx+ jŷ)e j8πz/3 (V/m),

22 Ẽ 1 = Ẽ i + Ẽ r = 5(ˆx+ jŷ) [e j4πz/3 13 ] e j4πz/3 (V/m). (d) % of reflected power = 100 Γ 2 = = 11.11%, ( 2 3 % of transmitted power = 100 τ 2 η 1 η 2 = 100 ) 2 120π 60π = 88.89%.