Electrical Engineering 2024 Paper I 50 marks Solve

Q8

(a) In the figure given below, region 1 is the side of the plane y+z=1 containing the origin and in this region, μ_r₁ = 5. In region 2, μ_r₂ = 7. It is given that B⃗₁ = 3·0a⃗_x + 1·0a⃗_y (T). Determine B⃗₂ and H⃗₂. Given, μ₀ = 4π × 10⁻⁷ H/m : (20 marks) (b) The message signal m(t) has a bandwidth of 20 kHz, a power of 20 W and a maximum amplitude of 8. It is desired to transmit this message through a channel to the destination with 80 dB attenuation and additive white noise with power spectral density $S_n(f) = \frac{N_0}{2} = 0.5 \times 10^{-12}$ W/Hz and achieve an SNR at the modulator output of at least 50 dB. What is the required transmitter power and channel bandwidth, if the modulation scheme employed is as under? (i) DSB-SC AM (ii) SSB AM (iii) Conventional DSB AM with modulation index 0·6 (20 marks) (c) An ideal DC-DC converter as shown in the figure has an input voltage of $V_s = 20$ V, the duty ratio $D = 0.25$ and the switching frequency is 20 kHz. The inductance $L = 150 \mu H$ and capacitance $C = 240 \mu F$. The average diode current is 1.2 A. Determine the following : (i) Peak-peak ripple current of the inductor (ii) Peak current through the switch S (10 marks)

हिंदी में प्रश्न पढ़ें

(a) नीचे दिए गए चित्र में समतल y+z=1 के एक ओर क्षेत्र 1 है, जिसमें मूल है और इस क्षेत्र में μ_r₁ = 5 है। क्षेत्र 2 में, μ_r₂ = 7 है। यह दिया गया है कि B⃗₁ = 3·0a⃗_x + 1·0a⃗_y (T) है। B⃗₂ और H⃗₂ ज्ञात कीजिए। दिया गया है कि, μ₀ = 4π × 10⁻⁷ H/m : (20 अंक) (b) संदेश संकेत m(t) की पटिका चौड़ाई 20 kHz, शक्ति 20 W और अधिकतम आयाम 8 है। इस संदेश को 80 dB क्षीणन और S_n(f) = N₀/2 = 0.5 × 10⁻¹² W/Hz शक्ति वर्णक्रम घनत्व के संयोजक श्वेत रव वाले एक चैनल से होकर गंतव्य तक प्रेषित करना वांछित है और मॉडुलक निर्गत पर कम-से-कम 50 dB SNR उपार्जित करना है। निम्नलिखित मॉडुलन योजना होने पर वांछित प्रेषित शक्ति और चैनल पटिका चौड़ाई क्या होगी? (i) DSB-SC AM (ii) SSB AM (iii) 0·6 मॉडुलन सूचकांक के साथ प्रचलित DSB AM (20 अंक) (c) चित्र में दर्शाए अनुसार एक आदर्श DC-DC परिवर्तक की निवेश वोल्टता $V_s = 20$ V, उपयोगिता अनुपात $D = 0.25$ और स्विचन आवृत्ति 20 kHz है। प्रेरक $L = 150 \mu H$ और संधारिता $C = 240 \mu F$ है। औसत डायोड धारा 1.2 A है। निम्नलिखित ज्ञात कीजिए : (i) प्रेरक में शिखर से शिखर उभयांश धारा (ii) स्विच S से गुजरती हुई शिखर धारा (10 अंक)

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Approach

Solve this multi-part numerical problem by allocating approximately 40% time to part (a) on boundary conditions in magnetostatics, 40% to part (b) on AM modulation systems comparison, and 20% to part (c) on DC-DC converter analysis. Begin each part with stated governing equations, show complete derivations with unit tracking, and conclude with physically verified numerical answers.

Key points expected

  • Part (a): Apply magnetic boundary conditions - normal component of B and tangential component of H are continuous across the interface; correctly identify normal vector to plane y+z=1 and decompose B₁ into normal and tangential components
  • Part (b)(i): For DSB-SC AM, use (SNR)₀ = (P_T/P_R)×(P_m/N₀W) with P_R = P_T×10⁻⁸ (80 dB attenuation) and bandwidth = 2W = 40 kHz
  • Part (b)(ii): For SSB AM, use same SNR formula but with bandwidth = W = 20 kHz and note 3 dB SNR advantage or equivalent power saving
  • Part (b)(iii): For conventional AM with m=0.6, account for power in carrier and sidebands using η = m²/(2+m²), total transmitted power includes carrier power
  • Part (c)(i): Calculate inductor ripple current using Δi_L = V_s(1-D)DT_s/L = V_sD(1-D)/(Lf_s) for buck converter operation
  • Part (c)(ii): Determine peak switch current as I_L,avg + Δi_L/2 using relationship between average diode current and load current

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness25%12.5Correctly applies magnetic boundary conditions for (a) identifying normal vector n̂ = (â_y + â_z)/√2; uses proper SNR formulas for AM systems in (b) distinguishing pre-detection and post-detection SNR; identifies buck converter topology and operating modes correctly in (c)Applies some correct boundary conditions but confuses normal/tangential components or makes sign errors; uses generic SNR expressions without modulation-specific adjustments; identifies converter type but confuses continuous/discontinuous conduction modeFundamental errors in applying boundary conditions (e.g., equating all B components); uses incorrect SNR formulas or ignores attenuation factor; misidentifies converter topology or uses wrong ripple formulas
Numerical accuracy25%12.5All calculations yield correct numerical values with proper significant figures: B₂ ≈ 3.0â_x + 0.5â_y + 0.5â_z T, H₂ components correctly scaled; transmitter powers in (b) match within 5% for all three schemes; ripple current ≈ 0.625 A, peak switch current ≈ 2.3125 A in (c)Correct methodology but arithmetic errors in 1-2 parts; final answers within 10-20% of correct values; unit conversion errors (kHz vs Hz, dB to linear)Major calculation errors exceeding 30% deviation; incorrect unit handling (e.g., mW vs W); missing or inconsistent numerical answers across sub-parts
Diagram quality10%5Clear sketch for (a) showing two regions separated by plane y+z=1 with coordinate system and field vectors; block diagrams for (b) showing transmitter-channel-receiver with noise injection; standard buck converter circuit diagram for (c) with current directions markedBasic sketches present but lacking labels or clarity; missing coordinate system in (a); incomplete block diagrams in (b); converter diagram without switch/diode labelingNo diagrams despite figure references; confusing or incorrect sketches that mislead the solution; diagrams contradict written analysis
Step-by-step derivation25%12.5Complete stepwise derivation: for (a) - unit normal calculation, field decomposition, application of continuity conditions, vector reconstruction; for (b) - explicit SNR derivation from first principles with modulation gain factors; for (c) - volt-second balance, ripple derivation, current relationshipsSome steps shown but skips key intermediate steps or assumes results; uses standard formulas without derivation; logical flow present but gaps in reasoningFinal answers stated without derivation; jumps between equations without justification; incorrect or circular reasoning; no clear problem-solving structure
Practical interpretation15%7.5Interprets results physically: comments on field refraction at magnetic interface; compares AM schemes on power-bandwidth tradeoffs relevant to Indian broadcasting standards (All India Radio); discusses converter efficiency, component sizing, and switching losses for typical SMPS applicationsBrief mention of physical significance without elaboration; generic statements about modulation efficiency; standard comments on converter operation without application contextNo physical interpretation; purely mathematical treatment; incorrect or irrelevant practical comments; ignores real-world constraints like component ratings

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