(a) A uniform plane wave travels in vacuum along +y direction. The electric field of the wave at some instant is given as E⃗ =…

(a) For a uniform plane wave in vacuum, the intrinsic impedance is η₀ = √(μ₀/ε₀) = √[(4π × 10⁻⁷)/((1/(36π)) × 10⁻⁹)] = 120π Ω. The wave travels along +y, so âk = ây. The magnetic field is obtained from the plane-wave relation H⃗ = (1/η₀)(ây × E⃗). Given E⃗ = 4x̂ + 3ẑ, ây × E⃗ = ây × (4x̂ + 3ẑ) = 4(ây × x̂) + 3(ây × ẑ). Using ây × x̂ = −ẑ and ây × ẑ = x̂, ây × E⃗ = −4ẑ + 3x̂ = 3x̂ − 4ẑ. Therefore H⃗ = (3x̂ − 4ẑ)/(120π) A/m = x̂/(40π) − ẑ/(30π) A/m.

Final answer: H⃗ = x̂/(40π) − ẑ/(30π) A/m.

(b) Rated transformer: S = 200 kVA. At maximum efficiency, iron loss equals copper loss. Let the common loss be P. At 75% full load and unity power factor, output is Pout = 0.75 × 200 × 1 = 150 kW. ηmax = 0.98 = 150/(150 + 2P). So 150 + 2P = 150/0.98 = 153.0612 kW, giving P = 1.5306 kW. Thus iron loss Pi = 1.5306 kW and copper loss at 75% load is also 1.5306 kW. Full-load copper loss: Pcu,FL = 1.5306/(0.75²) = 2.7211 kW. On the transformer base, per-unit resistance is Req = Pcu,FL/S = 2.7211/200 = 0.013605 pu. Given leakage impedance |Zeq| = 0.10 pu, Xeq = √(0.10² − Req²) = √(0.01 − 0.0001851) = 0.099070 pu. At full load, I = 1 pu, power factor 0.8 lagging, so cosθ = 0.8, sinθ = 0.6. Exact regulation using phasor addition: Vs = 1 + I(Req cosθ + Xeq sinθ) + jI(Xeq cosθ − Req sinθ) = 1 + 0.070326 + j0.071093.

|Vs| = √(1.070326² + 0.071093²) = 1.072685. %VR = (|Vs| − 1) × 100 = 7.2685%.

The usual first-order approximate formula gives %VR ≈ (Req cosθ + Xeq sinθ) × 100 = 7.0326%.

Final answer: Exact voltage regulation at full load, 0.8 pf lagging ≈ 7.27%. By the common approximate formula, it is 7.03%.

(c)(i) In the series L-C circuit, the capacitor initially has VC(0) = V₀ = 120 V. When S is closed, the diode conducts in the direction of current i. The natural angular frequency is ω₀ = 1/√(LC), and the characteristic impedance is Z₀ = √(L/C). With initial current i(0) = 0, the current is i(t) = (V₀/Z₀) sin(ω₀t). Peak current: Ipeak = V₀/Z₀ = V₀√(C/L). Here √(C/L) = √(12 × 10⁻⁶ / 48 × 10⁻⁶) = √(1/4) = 1/2. Hence Ipeak = 120 × 1/2 = 60 A.

Final answer: Peak value of current = 60 A.

(c)(ii) The diode conducts for the first positive half-cycle of the sinusoidal current and stops when i(t) first becomes zero. Thus ω₀t = π, so conduction time is tc = π/ω₀ = π√(LC). Now LC = 48 × 10⁻⁶ × 12 × 10⁻⁶ = 576 × 10⁻¹² s². √(LC) = 24 × 10⁻⁶ s = 24 μs. Therefore tc = 24π μs ≈ 75.40 μs.

At that instant the capacitor voltage has swung to −120 V, and the diode blocks any reverse current, so the capacitor stays at −120 V.

Final answer: Conduction time of the diode = 24π μs ≈ 75.40 μs.

(d) In an FM receiver, the noise power spectral density at the discriminator output increases as f². Hence high audio frequencies suffer a poorer signal-to-noise ratio than low frequencies. A linear pre-emphasis filter at the transmitter boosts the high-frequency components of the modulating signal before modulation. At the receiver, after demodulation, a de-emphasis filter with the inverse frequency response attenuates those high frequencies back to their original relative amplitudes. If Hp(f) is the pre-emphasis response and Hd(f) is the de-emphasis response, then for undistorted signal recovery one requires Hp(f)Hd(f) ≈ 1 over the audio band. The signal is therefore restored to its flat spectrum, but the FM noise, whose power spectral density rises as f², is attenuated at high frequencies by |Hd(f)|². Thus the output SNR improves. The improvement factor is I = ∫₀^fm f² df / ∫₀^fm f² |Hd(f)|² df for a baseband 0 to fm, assuming ideal FM demodulation. This shows that the SNR improvement depends directly on the de-emphasis response. However, since the de-emphasis must be the inverse of the pre-emphasis to avoid signal distortion, the improvement depends on both responses as a matched pair. For a typical RC pre-emphasis with break frequency f₁ and de-emphasis 1/(1 + jf/f₁), the improvement is approximately I ≈ (fm/f₁)²/3 for fm ≫ f₁. The exact result for the RC pair is I = (W/f₁)³ / {3[(W/f₁) − tan⁻¹(W/f₁)]}, with W the audio bandwidth. For broadcast FM (75 µs time constant, f₁ ≈ 2.1 kHz, W = 15 kHz) this gives I ≈ 21, about 13 dB.

Final answer: Pre-emphasis boosts high-frequency signal components at the transmitter; de-emphasis attenuates them at the receiver, reducing high-frequency FM noise while restoring the signal. Yes, the SNR improvement depends on both pre-emphasis and de-emphasis responses, since they must act as inverse filters.

(e)(i) The load reflection coefficient is ΓL = (ZL − Z₀)/(ZL + Z₀). Given ZL = 50 + j30 Ω and Z₀ = 40 Ω, ΓL = (50 + j30 − 40)/(50 + j30 + 40) = (10 + j30)/(90 + j30). Divide numerator and denominator by 10: ΓL = (1 + j3)/(9 + j3). Rationalising: ΓL = (1 + j3)(9 − j3)/(9² + 3²) = (18 + j24)/90 = 0.2 + j0.2667. Magnitude:

|ΓL| = √(0.2² + 0.2667²) = 1/3. Angle: θ = tan⁻¹(4/3) = 53.13°.

Final answer: ΓL = 0.2 + j0.2667 = (1/5) + j(4/15), |ΓL| = 1/3, angle = 53.13°.

(e)(ii) Wave velocity: u = 0.8c = 0.8 × 3 × 10⁸ = 2.4 × 10⁸ m/s. Frequency f = 2 MHz = 2 × 10⁶ Hz. Wavelength: λ = u/f = (2.4 × 10⁸)/(2 × 10⁶) = 120 m. Line length l = 25 m, so βl = 2πl/λ = 2π × 25/120 = 5π/12 = 75°. Thus tan βl = tan 75° = 2 + √3. The input impedance is Zin = Z₀ (ZL + jZ₀ tan βl)/(Z₀ + jZL tan βl). Substituting, Zin = 40[(50 + j30) + j40(2 + √3)]/[40 + j(50 + j30)(2 + √3)]. Numerator: 50 + j[30 + 40(2 + √3)] = 50 + j(110 + 40√3). Denominator: 40 + j(50 + j30)(2 + √3) = 40 − 30(2 + √3) + j50(2 + √3) = (−20 − 30√3) + j(100 + 50√3). Therefore Zin = 40[50 + j(110 + 40√3)]/[(-20 − 30√3) + j(100 + 50√3)] Ω. Numerically, Zin ≈ 29.857 − j22.232 Ω.

Final answer: Zin ≈ (29.86 − j22.23) Ω.