Electrical Engineering 2022 Paper II 50 marks Solve

Paper II — Q7

(a) The figure below shows the encoder for a rate r = 1/2, constraint length k = 4 convolutional code : [Figure: input feeding…

(a)
(i)

The figure below shows the encoder for a rate r = 1/2, constraint length k = 4 convolutional code : [Figure: input feeding three flip-flops in cascade, an upper modulo-2 adder fed from the input and the outputs of all three flip-flops giving g⁽¹⁾(x), a lower modulo-2 adder fed from the input, the first flip-flop output and the third flip-flop output giving g⁽²⁾(x), and a commutator switching the output between them] Determine the encoder output produced by the message sequence 11011 ....

(ii)

Also, construct the code tree for the encoder. Trace the path through the tree that corresponds to the message 11011 .... Compare the resulting output with that determined in (i). 20 marks

(b)
(i)

Explain in brief about the necessity of automatic load frequency control in the power generating system. Draw the schematic diagram of load frequency and voltage control regulators. 10 marks

(ii)

Two generators rated 200 MW and 400 MW are operating in parallel. The droop characteristics of their governors are 4% and 6% respectively from no load to full load. Assuming that the governors are operating at 50 Hz at no load, how would a load of 600 MW be shared between them? What would be the corresponding system frequency at this load? Assume free governor operation. 10 marks

(c)

Explain in brief about different types of HVDC links, their advantages and disadvantages. 10 marks

हिंदी में प्रश्न पढ़ें
(a)
(i)

नीचे चित्र में दर r = 1/2 और प्रतिबंध लंबाई k = 4 के संवलक कूट का कूटज दर्शाया गया है : [figure] संदेश अनुक्रम 11011 ... द्वारा रचित कूटज निर्गम निर्धारित कीजिए।

(ii)

कूटज के लिए कूट वृक्ष की रचना भी कीजिए। कूट वृक्ष में संदेश 11011 ... से संगत मार्ग अनुसरित कीजिए। परिणामी निर्गम की तुलना (i) में प्राप्त निर्गम से कीजिए। (20 अंक)

(b)
(i)

शक्ति जनित्र निकाय में भार आवृत्ति के स्वतः नियंत्रण की आवश्यकता के बारे में संक्षेप में समझाइए। भार आवृत्ति एवं वोल्टता नियंत्रण प्रणाली का योजनाबद्ध आरेख भी बनाइए। (10 अंक)

(ii)

200 MW एवं 400 MW के दो जनित्र समानांतर क्रम में प्रचालित हैं। शून्य भार से पूर्ण भार तक उनके अधिनियंत्रकों के निपाती अभिलक्षण क्रमशः 4% एवं 6% हैं। यह मानते हुए कि अधिनियंत्रक शून्य भार पर 50 Hz पर प्रचालित होते हैं, 600 MW का भार इनके मध्य किस प्रकार वितरित होगा? इस भार पर संगत निकाय आवृत्ति क्या होगी? मान लीजिए कि अधिनियंत्रक मुक्त रूप से प्रचालित है। (10 अंक)

(c)

एचवी डी सी सी सौं शृंखलाओं के विभिन्न प्रकारों, उनके फायदे एवं नुकसान के बारे में संक्षेप में समझाइए। (10 अंक)

Q7 of the 2022 UPSC Mains Electrical Engineering Paper II, as printed
The question as printed in the 2022 Electrical Engineering paper

The figure this question refers to, in words

The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.

(a) Rate 1/2, K = 4 convolutional encoder: three F/F stages, upper adder taps input, F/F1, F/F2, F/F3 (g1 = 1111), lower adder taps input, F/F1, F/F3 (g2 = 1101), output commutator.

Model answer

Drafted with AI 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. How we build rubrics and answers.

(a) The rate-1/2, constraint-length-4 convolutional encoder

The encoder has three flip-flops in cascade. Let m₀ be the current message bit at the input, and m₁, m₂, m₃ the contents of the first, second and third flip-flop (the previous three bits). Reading the taps off the figure, the upper modulo-2 adder is fed from the input, from the first and second flip-flop outputs and from the third; the lower adder is fed from the input, from the first flip-flop output and from the third. The generator sequences are therefore

g⁽¹⁾ = (1, 1, 1, 1), g⁽¹⁾(x) = 1 + x + x² + x³, output c₁ = m₀ ⊕ m₁ ⊕ m₂ ⊕ m₃, g⁽²⁾ = (1, 1, 0, 1), g⁽²⁾(x) = 1 + x + x³, output c₂ = m₀ ⊕ m₁ ⊕ m₃.

The commutator sends c₁ then c₂ for every input bit, giving rate 1/2 with constraint length K = 4.

(i) Encoder output for 11011. Start with the register cleared, and follow the message with three zeros to flush it (the state after the fifth bit is shown so the tail can be checked). State = (m₁ m₂ m₃).

bit 1, state 000: c₁ = 1, c₂ = 1 → 11; next state 100 bit 1, state 100: c₁ = 1⊕1 = 0, c₂ = 1⊕1 = 0 → 00; next state 110 bit 0, state 110: c₁ = 0⊕1⊕1 = 0, c₂ = 0⊕1 = 1 → 01; next state 011 bit 1, state 011: c₁ = 1⊕0⊕1⊕1 = 1, c₂ = 1⊕0⊕1 = 0 → 10; next state 101 bit 1, state 101: c₁ = 1⊕1⊕0⊕1 = 1, c₂ = 1⊕1⊕1 = 1 → 11; next state 110 flush 0, state 110: → 01; next state 011 flush 0, state 011: → 01; next state 001 flush 0, state 001: → 11; next state 000

The encoded sequence is 11 00 01 10 11 for the five message bits, followed by 01 01 11 for the tail: 1100011011 010111.

The same result follows by polynomial multiplication. The message polynomial is m(x) = 1 + x + x³ + x⁴, and

c₁(x) = g⁽¹⁾(x) m(x) = (1 + x + x² + x³)(1 + x + x³ + x⁴) = 1 + x³ + x⁴ + x⁷ (all other terms cancel in pairs modulo 2), i.e. c₁ = 1 0 0 1 1 0 0 1; c₂(x) = g⁽²⁾(x) m(x) = (1 + x + x³)(1 + x + x³ + x⁴) = 1 + x² + x⁴ + x⁵ + x⁶ + x⁷, i.e. c₂ = 1 0 1 0 1 1 1 1.

Interleaving c₁ and c₂ gives 11 00 01 10 11 01 01 11, as before.

(ii) Code tree. The tree starts at the cleared state. From every node two branches leave: the upper branch for an input 0 and the lower branch for an input 1, each labelled with the two output bits (c₁c₂) that the input produces from that node's state. From the root, input 0 gives 00 and input 1 gives 11. From the node reached by 1 (state 100), input 0 gives 11 and input 1 gives 00. From state 110, input 0 gives 01 and input 1 gives 10; from state 010, 0 gives 10 and 1 gives 01; and so on for each of the eight states. Because the encoder has only 2³ = 8 states, from the fourth level on the tree repeats itself: the pattern of branch labels leaving any node depends only on the state, so the tree can be folded into the eight-state trellis.

Diagram: Code tree for the rate 1/2, K = 4 convolutional encoder

Tracing the message 11011 through the tree: lower branch (11), lower branch (00), upper branch (01), lower branch (10), lower branch (11). The labels read off the path are 11 00 01 10 11, exactly the output found in (i); the tree is the encoder's state machine drawn out in time, so the two must agree.

(b)(i) Automatic load frequency control

The frequency of a power system is set by the balance between mechanical power input and electrical load: when load rises the generators slow down and frequency falls, and the reverse when load falls. Frequency must be held within tight limits (49.9–50.05 Hz under the Indian Grid Code) because turbine blades, motors, clocks and the whole synchronous operation of the grid depend on it; a sustained deviation trips under-frequency relays and can cascade into a blackout. Load changes continuously and unpredictably, and the primary governor action alone leaves a steady-state error proportional to the droop, so automatic load frequency control is needed to (1) restore frequency to nominal after every load change, (2) hold the scheduled tie-line interchange between control areas, and (3) share the load change among generators economically. The frequency and voltage loops are separate because real power affects frequency while reactive power affects voltage.

The schematic has two regulators on one machine. The load-frequency regulator: a speed sensor on the shaft compares the frequency with the reference, the error, together with the tie-line power error (weighted to form the area control error, ACE = ΔPtie + B Δf), is passed through an integrator and a speed changer that adjusts the governor set point; the governor operates the hydraulic amplifier and steam valve, and the turbine output follows. The automatic voltage regulator: a potential transformer at the terminals feeds a comparator with the voltage reference, the amplified error drives the exciter, which sets the field current and hence the terminal voltage. The two loops are drawn side by side on the generator, the AVR acting on the field and the ALFC on the prime mover.

Diagram: Schematic of load frequency control and automatic voltage regulator

(b)(ii) Sharing a 600 MW load between two generators

Droop is the fall in frequency from no load to full load as a fraction of rated frequency. Generator 1 (200 MW, 4%) drops 2 Hz over 200 MW, so its characteristic is P₁ = (200/2)(50 − f) = 100(50 − f) MW; generator 2 (400 MW, 6%) drops 3 Hz over 400 MW, so P₂ = (400/3)(50 − f) = 133.3(50 − f) MW. Both run at the common system frequency f, so

P₁ + P₂ = 233.3(50 − f) = 600 → 50 − f = 2.571 Hz, f = 47.43 Hz.

Then P₁ = 100 × 2.571 = 257.1 MW and P₂ = 133.3 × 2.571 = 342.9 MW.

Diagram: Load sharing of two generators by governor droop

The 600 MW load equals the combined rating, but under free governor action the machine with the smaller droop takes the larger share in proportion to its rating: generator 1 is overloaded by 57 MW (29%) while generator 2 is 57 MW under its rating, and the frequency has fallen to 47.43 Hz. For rating-proportional sharing the two droops would have to be equal, and the frequency would have to be restored by supplementary (secondary) control raising the set points.

(c) HVDC links

Monopolar link: a single conductor, usually at negative polarity to reduce corona and radio interference, with the return through the earth or sea; two converter stations, one at each end. Cheapest, used for cable crossings and as the first stage of a bipolar scheme; the disadvantages are the continuous ground current (corrosion of buried structures, interference with telecommunications) and no redundancy.

Bipolar link: two conductors, one positive and one negative with respect to earth, each with its own converter, the mid-points grounded at one or both ends. In normal operation the earth carries only the small unbalance current; if one pole fails the other continues at half power with ground return. This is the standard for long overhead lines (Rihand–Dadri ±500 kV, Talcher–Kolar ±500 kV, the ±800 kV Champa–Kurukshetra and Raigarh–Pugalur links). It costs more than monopolar but gives redundancy and low earth current.

Homopolar link: two or more conductors of the same polarity, usually negative, with ground return. It has the corona advantage of negative polarity and allows both conductors to be used when one converter is out, but the permanent ground current is its weakness and it is now rare.

Back-to-back link: both converters in the same station with no DC line, used to tie asynchronous systems or systems of different frequency (the Vindhyachal, Chandrapur, Sasaram and Gazuwaka stations that joined India's regional grids). Multi-terminal links connect three or more converter stations (the North-East–Agra ±800 kV scheme).

Advantages common to HVDC: no stability limit on line length, lower line cost and losses beyond the break-even distance (about 600–800 km overhead, 40–50 km cable), no charging current, control of power flow, asynchronous interconnection, no contribution to short-circuit level, and fewer conductors and narrower right of way. By converter technology, line-commutated (LCC, thyristor) links suit bulk long-distance transfer, while voltage-source (VSC, IGBT) links control active and reactive power independently, need no strong AC grid and suit offshore wind and city in-feeds (the Pugalur–Thrissur ±320 kV VSC link). Disadvantages: expensive converter stations, harmonic generation needing filters, large reactive-power demand at the converters (about 50–60% of the DC power), difficulty of breaking DC current and hence of tapping or building meshed networks, and the need for voltage transformation to be done on the AC side.

How this answer will be evaluated

What earns full marks: Complete working with correct diagrams, clear assumptions, and accurate calculations.

Evaluation rubric

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

  1. (a) Determine encoder output for 11011 and construct/trace the code tree. 20 marks

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

    Must cover

    • Identify generator polynomials from the encoder diagram
    • Output for 11011: 11 00 01 10 11 (tail 01 01 11), from g1 = 1111, g2 = 1101
    • Construct the code tree for the encoder
    • Trace the path for 11011 and compare with (i)

    Loses marks

    • Incorrect generator polynomial identification
    • Missing comparison between tree and direct calculation

    Earns more

    • Correctly identifies F/F connections and modulo-2 adders
    • Shows intermediate states in the code tree

    Extra mark

    • Draws the state diagram alongside the tree
  2. (b)(i) Explain necessity of automatic load frequency control and draw regulators. 10 marks

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

    Must cover

    • Explains why frequency control is necessary in power systems
    • Draws schematic of load frequency control regulator
    • Draws schematic of voltage control regulator

    Loses marks

    • Missing schematic diagrams
    • Vague explanation without specific control mechanisms

    Earns more

    • Mentions governor droop characteristics
    • Labels components in the schematic diagrams

    Extra mark

    • Includes a block diagram of the control loop
  3. (b)(ii) Calculate load sharing and system frequency for two parallel generators. 10 marks

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

    Must cover

    • Uses droop characteristics to calculate load sharing
    • Calculates system frequency at 600 MW load
    • P1 ≈ 257.1 MW, P2 ≈ 342.9 MW at f ≈ 47.43 Hz

    Loses marks

    • Formula application without equivalent circuit/model
    • Sign errors in frequency deviation calculation

    Earns more

    • States assumptions clearly (free governor operation)
    • Uses per-unit system where appropriate

    Extra mark

    • Draws the governor droop curves
  4. (c) Explain different types of HVDC links, their advantages and disadvantages. 10 marks

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

    Must cover

    • Identifies different types of HVDC links
    • Lists advantages of each type
    • Lists disadvantages of each type

    Loses marks

    • Missing advantages or disadvantages
    • Vague description without specific HVDC types

    Earns more

    • Mentions specific HVDC technologies (LCC, VSC)
    • Provides examples of HVDC applications

    Extra mark

    • Compares monopolar, bipolar, homopolar and back-to-back links point by point
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