(a) A uniform plane wave travels in vacuum along +y direction. The electric field of the wave at some instant is given as E⃗ = 4x̂ + 3ẑ. Find the vector magnetic field H⃗. (Given, μ₀ = 4π × 10⁻⁷ H/m, ε₀ = 1/(36π) × 10⁻⁹ F/m) (10 marks)

(b) The maxi

Question

(a) A uniform plane wave travels in vacuum along +y direction. The electric field of the wave at some instant is given as E⃗ = 4x̂ + 3ẑ. Find the vector magnetic field H⃗. (Given, μ₀ = 4π × 10⁻⁷ H/m, ε₀ = 1/(36π) × 10⁻⁹ F/m) (10 marks)

(b) The maximum efficiency of a 200 kVA, 3300/600 V, 50 Hz, single-phase transformer is 98% and occurs at 75% full load and unity power factor. If the leakage impedance is 10%, find the voltage regulation at full load and power factor 0.8 lagging. (10 marks)

(c) A diode circuit with an L-C load is shown in the figure, with the capacitor having an initial voltage V_C(t=0) = 120 V, capacitance C = 12 μF and inductance L = 48 μH. If switch S is closed at t = 0 s, then find the following: (i) Peak value of current i (ii) Conduction time of the diode (10 marks)

(d) How can linear pre-emphasis and de-emphasis filters be employed to improve the performance of an FM system? Is the improvement in output SNR dependent on both the frequency responses of the pre-emphasis filter and the de-emphasis filter? (10 marks)

(e) A transmission line is 25 m long. It has characteristic impedance Z₀ = 40 Ω and operates at 2 MHz. The line is terminated with a load of Z_L = (50 + j30) Ω. If the wave velocity is u = 0.8c (with c = 3×10⁸ m/s) on the line, determine (i) the reflection coefficient and (ii) the input impedance. (10 marks)

UPSC Answer Check · Accepted Answer

(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.

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.

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.

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

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