Chat with us
Electrical Engineering 2023 Paper II 50 marks Solve

Paper II — Q4

(a) Consider a systematic linear block code with binary elements whose parity check equations are p₁ = m₁ + m₂ + m₃ p₂ = m₂ + m₃…

(a)

Consider a systematic linear block code with binary elements whose parity check equations are

p₁ = m₁ + m₂ + m₃ p₂ = m₂ + m₃ + m₄ p₃ = m₁ + m₃ + m₄ p₄ = m₁ + m₂ + m₄

where mᵢ are message digits and pᵢ are parity check digits.

(i)

Find the generator matrix and parity check matrix for the code.

(ii)

How many errors can this code detect? How many errors can be corrected?

(iii)

If 10100100 is the received code word, find the corresponding transmitted code word assuming that single-bit error has been made during transmission. 20 marks

(b)

A transmission line has the following parameters :

A = D = 1∠5°, B = 88∠75°

(i)

Determine the sending-end voltage and the voltage regulation if the line supplies a load of 40 MW at 0·8 p.f. lagging with receiving-end voltage 132 kV.

(ii)

Find the power and power factor of the load if the voltages at the two ends are 132 kV and with a phase difference of 30°. 20 marks

(c)

Explain four instructions which are used to control interrupt structure of 8085 microprocessor. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

द्वि-आधारी घटकों के साथ एक क्रमबद्ध रैखिक खण्ड कूट पर विचार कीजिए, जिसके सामंजस्य (पैरिटी) जाँच समीकरण हैं

p₁ = m₁ + m₂ + m₃ p₂ = m₂ + m₃ + m₄ p₃ = m₁ + m₃ + m₄ p₄ = m₁ + m₂ + m₄

जहाँ mᵢ संदेश अंक एवं pᵢ सामंजस्य जाँच अंक हैं।

(i)

इस कूट के लिए जनक मैट्रिक्स एवं सामंजस्य जाँच मैट्रिक्स ज्ञात कीजिए।

(ii)

यह कूट कितनी त्रुटियों का पता लगा सकता है? कितनी त्रुटियों को संशोधित किया जा सकता है?

(iii)

यदि 10100100 प्राप्त हुआ कूट-शब्द है, तो यह मानते हुए कि संचरण में एकल-बिट त्रुटि की गयी है, संगत संचारित कूट-शब्द ज्ञात कीजिए। 20 अंक

(b)

एक संचरण लाइन के प्राचल निम्नलिखित हैं :

A = D = 1∠5°, B = 88∠75°

(i)

प्रेषण सिरे की वोल्टता एवं वोल्टता नियमन ज्ञात कीजिए, यदि लाइन अभिग्राही सिरे की 132 kV वोल्टता के साथ एक 40 MW भार को 0·8 पश्चता शक्ति गुणक पर प्रदाय करती है।

(ii)

भार की शक्ति एवं शक्ति गुणक ज्ञात कीजिए, यदि दोनों सिरों पर 132 kV की वोल्टता एवं 30° का कलांतर हो। 20 अंक

(c)

8085 सूक्ष्म-संसाधित्र (माइक्रोप्रोसेसर) की क्रमभंजक (इंटरप्ट) संरचना को नियंत्रित करने वाले चार अनुदेशों की व्याख्या कीजिए। 10 अंक

Q4 of the 2023 UPSC Mains Electrical Engineering Paper II, as printed
The question as printed in the 2023 Electrical Engineering paper
Evaluate your answer to this question → See all 2023 Electrical Engineering questions

Model answer

Written by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key.

(a)(i) Over GF(2), let c = (m₁, m₂, m₃, m₄, p₁, p₂, p₃, p₄). The parity equations give p = A m, where

A = 1 1 1 0 0 1 1 1 1 0 1 1 1 1 0 1

For a systematic code, G = [I₄ | Aᵀ] and H = [A | I₄]. Hence

G = 1 0 0 0 1 0 1 1 0 1 0 0 1 1 0 1 0 0 1 0 1 1 1 0 0 0 0 1 0 1 1 1

H = 1 1 1 0 1 0 0 0 0 1 1 1 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 0 0 0 1

(a)(ii) The columns of H are h₁=1011ᵀ, h₂=1101ᵀ, h₃=1110ᵀ, h₄=0111ᵀ, h₅=1000ᵀ, h₆=0100ᵀ, h₇=0010ᵀ, h₈=0001ᵀ. No column is zero, no two are equal, and no three sum to zero. But h₁+h₂+h₃+h₅=0, so d_min = 4. Therefore the code detects d_min−1 = 3 errors and corrects t = ⌊(4−1)/2⌋ = 1 error.

(a)(iii) For r = 10100100, with m = 1010 and p = 0100, the syndrome S = H rᵀ is computed as

s₁ = p₁+m₁+m₂+m₃ = 0+1+0+1 = 0 s₂ = p₂+m₂+m₃+m₄ = 1+0+1+0 = 0 s₃ = p₃+m₁+m₃+m₄ = 0+1+1+0 = 0 s₄ = p₄+m₁+m₂+m₄ = 0+1+0+0 = 1

So S = 0001, which matches the 8th column h₈ corresponding to p₄. Hence the error is in p₄; flip it from 0 to 1. The transmitted code word is 10100101.

(b)(i) Using V_S = A V_R + B I_R. Take V_R = 132/√3 ∠0° kV = 76.21∠0° kV per phase.

Load current: I_R = P/(√3 V_L p.f.) = 40×10⁶/(√3×132×10³×0.8) = 218.69 A.

Since p.f. = 0.8 lagging, φ = cos⁻¹0.8 = 36.87°, so I_R = 218.69∠−36.87° A.

A V_R = 1∠5° × 76.21∠0° = 76.21∠5° kV. B I_R = 88∠75° × 218.69∠−36.87° = 19.245∠38.13° kV.

Thus V_S = 76.21∠5° + 19.245∠38.13° kV = 91.06 + j18.53 kV = 92.92∠11.5° kV per phase.

So line-to-line sending voltage: V_S,LL = √3 × 92.92 = 160.95 kV.

Voltage regulation: %VR = (|V_S| − |V_R|)/|V_R| × 100 = (160.95 − 132)/132 × 100 = 21.93%.

(b)(ii) Given |V_S| = |V_R| = 132 kV and phase difference 30°, take V_R = 76.21∠0° kV, V_S = 76.21∠30° kV per phase.

I_R = (V_S − A V_R)/B = [76.21(1∠30° − 1∠5°)]/(88∠75°) kV/Ω.

Now 1∠30° − 1∠5° = 0.432879∠107.5°. Hence numerator = 32.99∠107.5° kV.

So I_R = (32.99/88)∠(107.5° − 75°) kA = 0.3749∠32.5° kA = 374.9∠32.5° A.

Apparent power: S = √3 V_L I_L = √3 × 132×10³ × 374.9 = 85.71 MVA.

The load power factor angle is −32.5°, so p.f. = cos32.5° = 0.843 leading. P = 85.71 × 0.8434 = 72.29 MW, Q = −85.71 × 0.5373 = −46.05 MVAr (leading).

(c) Four 8085 interrupt-control instructions are:

  • EI: Enable Interrupts. It sets the interrupt-enable flip-flop, allowing maskable interrupts such as INTR, RST 5.5, RST 6.5 and RST 7.5 to be recognised unless individually masked. TRAP remains non-maskable.
  • DI: Disable Interrupts. It resets the interrupt-enable flip-flop, inhibiting all maskable interrupts. TRAP is not affected.
  • SIM: Set Interrupt Mask. It sets or resets the mask bits for RST 5.5, RST 6.5 and RST 7.5, can reset the RST 7.5 latch, and controls serial output data (SOD). The mask bits become effective only when the mask-set-enable condition is satisfied.
  • RIM: Read Interrupt Mask. It reads the interrupt mask, pending interrupt status, interrupt-enable status, and serial input data (SID) into the accumulator, allowing the program to monitor and control the interrupt structure.

What "Solve" is asking you to do

Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.

Structure that answers it

Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity

Where marks are lost

Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.

All UPSC directive words, compared →

How this answer will be evaluated

Approach

(a) calculate: given > formula > substitution > result with units > interpretation | (b) calculate: given > formula > substitution > result with units > interpretation | (c) explain: definition/context > points in order > small example > short close Full marks: Correct matrices and phasor calculations with clear steps; precise 8085 instruction details.

Key points expected

  • Construct 4x8 generator matrix G from parity equations
  • Derive 4x8 parity check matrix H
  • Calculate minimum distance d_min for error detection/correction
  • Compute syndrome to identify and correct the single-bit error
  • Apply V_s = AV_r + BV_rI_r equation with correct phasors
  • Calculate voltage regulation percentage using standard formula
  • Determine load current magnitude and phase angle from power/pf
  • Calculate power and pf for the 30-degree phase difference case

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Derive G and H matrices, determine error capability, and correct a specific received codeword. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Construct 4x8 generator matrix G from parity equations
    • Derive 4x8 parity check matrix H
    • Calculate minimum distance d_min for error detection/correction
    • Compute syndrome to identify and correct the single-bit error

    Loses marks

    • Matrix G or H has incorrect dimensions or entries
    • Confuses error detection limit with correction limit
    • Fails to show the syndrome calculation for the received word

    Earns more

    • Explicitly states code parameters (n, k, d_min)
    • Shows binary addition (XOR) steps for parity bits
    • Verifies the corrected codeword satisfies Hc^T = 0

    Extra mark

    • Lists all possible single-bit error syndromes
  2. (b) Compute sending-end voltage, regulation, and load power/pf using ABCD parameters. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Apply V_s = AV_r + BV_rI_r equation with correct phasors
    • Calculate voltage regulation percentage using standard formula
    • Determine load current magnitude and phase angle from power/pf
    • Calculate power and pf for the 30-degree phase difference case

    Loses marks

    • Sign error in the ABCD parameter application
    • Confuses line-to-line voltage with phase voltage
    • Incorrect calculation of load current angle

    Earns more

    • Draws a phasor diagram for the transmission line
    • Clearly defines reference phasor (e.g., V_r at 0 degrees)
    • States units (kV, A, MW) for all intermediate results

    Extra mark

    • Calculates line losses or efficiency
  3. (c) Describe four specific instructions for controlling the 8085 interrupt structure. 10 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Identifies four distinct interrupt control instructions (e.g., SIM, RIM, EI, DI)
    • Explains the function of each instruction (e.g., enabling/disabling, status read)
    • Mentions the specific flags or registers affected (e.g., I flag, status bits)
    • Distinguishes between global and specific interrupt control

    Loses marks

    • Confuses 8085 instructions with 8086 or other microprocessors
    • Lists instructions without explaining their specific interrupt function
    • Fails to mention the I flag or status bits

    Earns more

    • Provides a short code example or sequence
    • Mentions the specific bit positions in the accumulator for SIM/RIM

    Extra mark

    • Draws the interrupt status register layout

Practice this exact question

Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.

Evaluate my answer →

More from Electrical Engineering 2023 Paper II