Paper II — Q1
(a) A system is described by the following state equations : dotx₁ = x₁ + x₂ + 3x₃ dotx₂ = 2x₁ + 3x₂ + u₁ dotx₃ = 2x₂ + x₃ +…
A system is described by the following state equations : dotx₁ = x₁ + x₂ + 3x₃ dotx₂ = 2x₁ + 3x₂ + u₁ dotx₃ = 2x₂ + x₃ + u₂ Check the controllability of the system. 10 marks
A single phase, single line diagram of a power system is shown in figure. Find the sending end voltage and the value of load resistance in p.u. referred to sending end if the voltage across load resistance is 9·8 KV. 10 marks
Explain the following related to computer programming : Machine Language (ii) Assembly Language (iii) Compiler
Interpreter (v) ASCII 10 marks
A current of (0·5 + 0·3 sinωt – 0·2 sin 2ωt) amps is passed through the circuit shown in figure. Determine the reading of each instrument if ω = 10⁶ rad/sec. 10 marks
A DPCM system uses a linear predictor with a single tap. The normalized autocorrelation function of the input signal for a lag of one sampling interval is 0·75. The predictor is designed to minimize the prediction error variance. Determine the processing gain attained by the use of this predictor. 10 marks
हिंदी में प्रश्न पढ़ें
एक तंत्र को निम्नलिखित अवस्था समीकरणों द्वारा वर्णित किया गया है : dotx₁ = x₁ + x₂ + 3x₃ dotx₂ = 2x₁ + 3x₂ + u₁ dotx₃ = 2x₂ + x₃ + u₂ तंत्र की नियंत्रणीयता की जाँच करें । (10 अंक)
एक एकल कला विद्युत शक्ति प्रणाली को एकल रेखीय आरेख द्वारा चित्र में दर्शाया गया है । यदि भार प्रतिरोध पर वोल्टता 9·8 KV हो तो प्रेषण सिरे की वोल्टता व प्रेषण सिरे के संदर्भ में भार प्रतिरोध का मान प्रति इकाई में ज्ञात करें । (10 अंक)
अभिकलित्र प्रक्रमण के संदर्भ में निम्नलिखित की व्याख्या करें : मशीन भाषा (ii) समन्वयोजन (असेंबली) भाषा (iii) संकलक (कम्पाइलर)
भाषांतरक (v) ASCII (10 अंक)
चित्र में दर्शाये गये परिपथ में (0·5 + 0·3 sinωt – 0·2 sin 2ωt) Amp विद्युत धारा का प्रवाह हो रहा है । यदि ω = 10⁶ rad/sec हो तो प्रत्येक मापन के पठन (रीडिंग) का निर्धारण करें । (10 अंक)
एक DPCM तंत्र एक रैखिक प्राक्सूचक को एकल टैप के साथ प्रयोग करता है । एक प्रतिचयन अंतराल पश्चता के लिए सामान्यीकृत स्वतः सह संबंध फलन के निवेश का मान 0·75 है । प्राक्सूचक को, पूर्वानुमान त्रुटि के प्रसरण को निम्नतम करने के लिए अभिकल्प किया गया है । इस प्राक्सूचक के प्रयोग करने से प्राप्त संसाधन लब्धि का निर्धारण करें । (10 अंक)
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.
(b) A single-line diagram of a power system. On the left, a source terminal labeled Vs is connected to the primary side of a transformer. This transformer is labeled '11/33 KV', '10 MVA', and 'X = 10%'. The secondary side of this transformer is connected to the primary side of a second transformer. The second transformer is labeled '33/10 KV', '10 MVA', and 'X = 12%'. The secondary side of the second transformer is connected to a node labeled Vr. From node Vr, a load resistor with a value of 8 Ohms is connected to ground.
(d) A series circuit diagram. Starting from the left, a 1 mH inductor is connected in series with a 1000 Ohm resistor. The current 'i' flows through the circuit from left to right. An ammeter labeled 'MC' (Moving Coil) is connected in series with the resistor. Following the MC ammeter, another ammeter labeled 'MI' (Moving Iron) is connected in series. Below the inductor, a voltmeter labeled 'V' is connected in parallel across the inductor. Below the resistor, a second voltmeter labeled 'V' is connected in parallel across the resistor. Both voltmeters have an arrow pointing to the text 'Electro-static'.
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) Controllability of the state model ẋ = Ax + Bu with A = [1 1 3; 2 3 0; 0 2 1] and B = [0 0; 1 0; 0 1] (rows separated by semicolons). AB = [1 3; 3 0; 2 1] and A²B = [10 6; 11 6; 8 1]. The controllability matrix is Q_c = [B AB A²B] (3 × 6). Its first column (0, 1, 0)ᵀ and second column (0, 0, 1)ᵀ are the second and third unit vectors; the third column (1, 3, 2)ᵀ supplies the first. The minor from columns 1, 2 and 3 has determinant 1 ≠ 0, so rank Q_c = 3 = n. By Kalman's criterion the system is completely controllable.
(b) Sending-end voltage and load resistance in p.u. Choose 10 MVA as base and 11 kV at the sending end. The transformer ratios 11/33 and 33/10 give a load-side base of 10 kV, so Z_base(load) = 10²/10 = 10 Ω. The transformers are rated 10 MVA, so X₁ = 0.10 and X₂ = 0.12 pu on this base, giving X = 0.22 pu. R_L = 8/10 = 0.8 pu (equal to 9.68 Ω referred to the sending end, 9.68/12.1 = 0.8 pu). V_R = 9.8/10 = 0.98 pu, so I = 0.98/0.8 = 1.225 pu. V_S = V_R + jIX = 0.98 + j1.225 × 0.22 = 0.98 + j0.2695, |V_S| = 1.016 pu = 1.016 × 11 = 11.18 kV.
(c) Programming terms (i) Machine language: the binary instruction set a CPU executes directly; fast, but hardware-dependent and hard for humans to write. (ii) Assembly language: mnemonics such as MOV and ADD standing for machine instructions, converted to machine code by an assembler; still machine-specific. (iii) Compiler: translates the whole high-level source into an object (machine-code) file before execution, reporting syntax errors and optimising; execution is then fast. (iv) Interpreter: translates and executes the source statement by statement without a separate object file; easier to debug but slower to run. (v) ASCII: American Standard Code for Information Interchange, a 7-bit code for 128 characters (letters, digits, symbols, control codes) used to store and exchange text.
(d) Instrument readings The coil is ideal, X_L = ωL = 10⁶ × 10⁻³ = 1000 Ω at ω and 2000 Ω at 2ω. MC ammeter (average value): 0.5 A. MI ammeter (rms): I_rms = √(0.5² + 0.3²/2 + 0.2²/2) = √0.315 = 0.561 A. Electrostatic voltmeter across 1000 Ω (rms): 0.561 × 1000 = 561 V. Electrostatic voltmeter across 1 mH (the dc component drops no voltage): V = √[(0.3 × 1000/√2)² + (0.2 × 2000/√2)²] = √(212.1² + 282.8²) = 353.6 V.
(e) DPCM processing gain For one tap, the prediction error variance is minimised by a₁ = ρ(1) = 0.75, giving σₑ² = σₓ²(1 − ρ²(1)). G_p = σₓ²/σₑ² = 1/(1 − 0.75²) = 1/0.4375 = 2.29 (≈ 3.6 dB).
How this answer will be evaluated
What earns full marks: Complete method with correct calculations and clear interpretation for all parts.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Controllability of the given third-order state model with two inputs. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Write A = [1 1 3; 2 3 0; 0 2 1] and B = [0 0; 1 0; 0 1].
- Form Q_c = [B AB A²B] with AB = [1 3; 3 0; 2 1] and A²B = [10 6; 11 6; 8 1].
- Show rank Q_c = 3 (a non-zero 3 × 3 minor) and conclude that the system is completely controllable.
Loses marks
- Wrong A or B entries (for example, misplacing the zero in A).
- No rank or determinant evidence for the conclusion.
Earns more
- Correct dimensions (3 × 6) of the controllability matrix.
- Explicit evaluation of AB and A²B.
Extra mark
- Statement of Kalman's rank criterion.
- (b) Sending-end voltage and the load resistance in p.u. for a two-transformer single-phase system. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Choose bases (10 MVA; 11 kV at the sending end, 10 kV at the load) and convert X to 0.22 pu.
- Find R_L = 8/10 = 0.8 pu (9.68 Ω at the sending end).
- Compute V_S = 0.98 + j0.2695 pu, i.e. 1.016 pu = 11.18 kV.
Loses marks
- Using the wrong base voltage for the load zone.
- Not referring the load to the sending end.
Earns more
- Base voltage correct in each zone through the transformer ratios.
- Clear per-unit conversion steps.
Extra mark
- Per-unit equivalent circuit drawn.
- (c) Meaning of machine language, assembly language, compiler, interpreter and ASCII. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Define machine language (binary, executed directly) and assembly language (mnemonics translated by an assembler).
- Distinguish a compiler (whole program to object code before execution) from an interpreter (statement by statement).
- Define ASCII as a 7-bit character code for 128 characters.
Loses marks
- Confusing compiler and interpreter.
- Vague definitions without context.
Earns more
- Translation process mentioned for each item.
- An example of an assembly instruction.
Extra mark
- Mention of the 8-bit extended ASCII.
- (d) Readings of the MC and MI ammeters and two electrostatic voltmeters for a non-sinusoidal current. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Identify the dc term, the fundamental and the second harmonic, with X_L = 1000 Ω and 2000 Ω.
- Give MC = 0.5 A (average) and MI = 0.561 A (rms of all components).
- Give the electrostatic voltmeter readings: 561 V across 1000 Ω and 353.6 V across the coil (no dc drop).
Loses marks
- Ignoring the dc component in the rms calculation.
- Taking the same reactance for the second harmonic.
Earns more
- Correct inductive reactance at each frequency.
- The dc component handled properly in each instrument.
Extra mark
- Remark that the electrostatic voltmeter reads the rms value.
- (e) Processing gain of a DPCM system with a single-tap predictor for ρ(1) = 0.75. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- State the optimum coefficient a₁ = ρ(1) = 0.75.
- Write the error variance σₑ² = σₓ²(1 − ρ²(1)).
- Compute G_p = 1/(1 − 0.5625) = 2.29 (3.6 dB).
Loses marks
- Using the autocorrelation value incorrectly (for example, 1/(1 − ρ)).
- Omitting the error variance step.
Earns more
- Wiener–Hopf (normal) equation reasoning for a₁.
- Clear derivation of the error variance.
Extra mark
- Link to the improvement in signal-to-noise ratio.
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.
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 2021 Paper II
- Q1 (a) A system is described by the following state equations : dotx₁ = x₁ + x₂ + 3x₃ dotx₂…
- Q2 (a) Draw the sequence networks and calculate the load sequence impedances of a load circu…
- Q3 3.(a) A solidly earthed 400 KV, 3 phase busbar system is connected with two incoming and…
- Q4 4.(a)(i) The configuration of a 400 KV 3 phase line is shown in figure. The radius of eac…