(a) The plane wave E = 30 cos (ωt – z) aₓ V/m in air normally hits a lossless medium (μ = μ₀, ε = 4ε₀) at z = 0. (i) Find…

(a)(i) Reflection and Transmission Coefficients, SWR For normal incidence, the wave impedance of medium 1 (air) is η₁ = √(μ₀/ε₀) = η₀ = 120π Ω ≈ 377 Ω. For medium 2, μ₂ = μ₀, ε₂ = 4ε₀: η₂ = √(μ₀/(4ε₀)) = η₀/2 = 60π Ω ≈ 188.5 Ω. Using the boundary condition at z = 0, Γ = (η₂ − η₁)/(η₂ + η₁) = (60π − 120π)/(60π + 120π) = −60π/180π = −1/3. The electric-field transmission coefficient is τ = 1 + Γ = 1 − 1/3 = 2/3. The standing wave ratio in air is S = (1 + |Γ|)/(1 − |Γ|) = (1 + 1/3)/(1 − 1/3) = (4/3)/(2/3) = 2. Power check: |Γ|² + (η₁/η₂)|τ|² = 1/9 + 2 × 4/9 = 1.

(a)(ii) Reflected Electric and Magnetic Fields The incident field is E_i = 30 cos(ωt − z) aₓ V/m. The reflected electric field has amplitude E_r0 = Γ E_i0 = (−1/3)(30) = −10 V/m. Since the reflected wave travels in the −z direction in air (z ≤ 0), E_r = −10 cos(ωt + z) aₓ V/m. For the magnetic field, use H = (1/η₁)(a_k × E), with a_k = −a_z for the reflected wave. H_r = (1/η₀)(−a_z) × [−10 cos(ωt + z) aₓ] = (10/η₀) cos(ωt + z) a_y = (10/(120π)) cos(ωt + z) a_y = (1/(12π)) cos(ωt + z) a_y A/m ≈ 0.0265 cos(ωt + z) a_y A/m. This holds in air, z ≤ 0, for the lossless dielectric interface.

(b) Buck Regulator Design The printed Figure 6(b) shows a boost arrangement (series inductor, shunt switch, series diode) labelled Va = 5 V, which cannot step 16 V down. The text asks for a buck regulator with 16 V in and 4 V out, so the design below follows the text. For an ideal buck regulator in continuous conduction mode, the voltage conversion ratio is V₀ = D Vₛ, where D is the switch duty cycle. Given Vₛ = 16 V and V₀ = 4 V (as stated in the text), D = V₀/Vₛ = 4/16 = 0.25 = 25%. Switching period: T = 1/f = 1/(20 × 10³) = 50 μs.

For the inductor, the peak-to-peak ripple current is ΔI = V₀(1 − D)/(f L). Solve for L: L = V₀(1 − D)/(f ΔI) = 4(1 − 0.25)/(20 × 10³ × 0.75) = 4 × 0.75/(15 × 10³) = 3/(15 × 10³) = 2 × 10⁻⁴ H = 200 μH.

For the capacitor, the output ripple voltage is ΔV₀ = ΔI/(8 f C). Thus C = ΔI/(8 f ΔV₀) = 0.75/(8 × 20 × 10³ × 30 × 10⁻³) = 0.75/4800 = 1/6400 F = 156.25 μF.

Hence L = 200 μH, C = 156.25 μF, duty cycle D = 0.25. These results assume ideal switch, diode, inductor and capacitor, and continuous inductor current.

(c) Superposition: Voltage across 5 Ω Use the superposition theorem. Let the top-left node voltage across the 5 Ω resistor be V, and the top-right node be B. Deactivate current sources by open-circuiting them.

Case 1: Only 2 A source active. At the left node, the 5 Ω resistor is in parallel with the series path 10 Ω + 2 Ω = 12 Ω. R_eq1 = (5 × 12)/(5 + 12) = 60/17 Ω. V₁ = 2 × 60/17 = 120/17 V.

Case 2: Only 4 A source active. At node B, the 2 Ω resistor is in parallel with 10 Ω + 5 Ω = 15 Ω. R_eq2 = (2 × 15)/(2 + 15) = 30/17 Ω. V_B = 4 × 30/17 = 120/17 V. This voltage divides across 10 Ω and 5 Ω, so V₂ = V_B × 5/(10 + 5) = (120/17) × (1/3) = 40/17 V.

By superposition, V = V₁ + V₂ = 120/17 + 40/17 = 160/17 = 9.41 V approximately, positive at the top terminal of the 5 Ω resistor.