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Electrical Engineering 2021 Paper I 50 marks Calculate

Paper I — Q3

(a) (i) Explain what happens when a circuit shown in Figure 3(a)(i) below is constructed using logarithmic amplifier. 10 Figure…

(a)
(i)

Explain what happens when a circuit shown in Figure 3(a)(i) below is constructed using logarithmic amplifier. 10 marks

Figure 3(a)(i)

(ii)

Explain what happens if the topology is modified as shown in Figure 3(a)(ii) below. 10 marks

Figure 3(a)(ii)

(b)

For the circuit shown in Figure 3(b), calculate the voltage V₀(t) as function of time,

Figure 3(b)

where V(t) = 10 sin (6t + 60°) V and I(t) = 5 cos (4t + 30°) A. 20 marks

(c)

A mixer (analog multiplier) is used as a process in some analog communication systems. Two signals X₁(t) and X₂(t) are mixed to produce the output y(t) = X₁(t) X₂(t).

If X₁(t) = 10 sin c (10t) and X₂(t) = 2 cos (1000 πt), then calculate and plot the magnitude of the Fourier transform of output signal. Further, specify and prove the property of Fourier transform used in calculations. 10 marks

हिंदी में प्रश्न पढ़ें
(a)
(i)

समझाइए कि जब चित्र 3(a)(i) में प्रदर्शित परिपथ को लघुगणकीय प्रवर्धक का प्रयोग करते हुए निर्मित किया जाएगा, तो क्या होगा ।

चित्र 3(a)(i)

(ii)

समझाइए कि यदि संस्थितिकी को चित्र 3(a)(ii) में दर्शाए अनुसार परिवर्तित कर दिया जाए, तो क्या होगा ।

चित्र 3(a)(ii)

(b)

चित्र 3(b) में प्रदर्शित परिपथ के लिए वोल्टता V₀(t) की गणना समय के फलन के रूप में कीजिए,

चित्र 3(b)

जहाँ V(t) = 10 sin (6t + 60°) V और I(t) = 5 cos (4t + 30°) A हैं।

(c)

एक मिश्रक (अनुकृप गुणक) को किसी अनुकृप संचार प्रणाली के एक प्रक्रम के रूप में प्रयुक्त किया गया है । निर्गत y(t) = X₁(t) X₂(t) निर्मित करने के लिए दो संकेतों X₁(t) तथा X₂(t) को मिश्रित किया गया है ।

यदि X₁(t) = 10 sin c (10t) तथा X₂(t) = 2 cos (1000 πt) हो, तो निर्गत संकेत के फुरिये रूपांतर के परिमाण की गणना कीजिए तथा उसका आरेखण कीजिए । गणना में प्रयुक्त फुरिये रूपांतर के गुणों का उल्लेख कीजिए तथा उन्हें सिद्ध कीजिए ।

Q3 of the 2021 UPSC Mains Electrical Engineering Paper I, as printed
The question as printed in the 2021 Electrical Engineering paper
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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) Circuit diagram labeled Figure 3(a)(ii). An operational amplifier (op-amp) is shown with its non-inverting input (+) connected to ground. The inverting input (-) is connected to a node. A resistor R1 connects a terminal labeled V_REF to this inverting input node. The output of the op-amp is connected to a node labeled V1. Two NPN transistors, Q1 and Q2, are connected in parallel. The emitters of both Q1 and Q2 are connected to ground. The collectors of both Q1 and Q2 are connected to the node V1. The base of Q1 is connected to the inverting input node of the op-amp. The base of Q2 is connected to the collector of Q2 (node V1). An external current source labeled I_X is connected to the collector of Q2, with the arrow pointing into the collector.

(b) Circuit diagram labeled Figure 3(b). The circuit consists of three parallel branches connected between a top node and a bottom reference node. The voltage across the parallel branches is labeled V0 with the positive terminal at the top. The left branch contains an AC voltage source V(t) in series with a 10 Ohm resistor. The middle branch contains a 0.1 F capacitor in parallel with an AC current source I(t) pointing upwards. The right branch contains a 4 H inductor in series with a DC voltage source of 20 V, with the positive terminal of the DC source at the top.

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)(i) Behaviour of the circuit of Figure 3(a)(i) Q₁ (base grounded, collector at the inverting node, emitter at V₁) sits in the op-amp feedback. The inverting node is a virtual ground, so R₁ carries V_REF/R₁ (V_REF > 0) and all of it flows through Q₁: V_REF/R₁ = I_S exp(−V₁/V_T). This is the logarithmic stage: V₁ = −V_T ln(V_REF/(R₁I_S)). Q₂ has its base at V₁ and its emitter at −V_B (battery as drawn, negative terminal at the emitter), so V_BE2 = V₁ + V_B. For matched devices, I_X = I_S exp((V₁ + V_B)/V_T) = (R₁I_S²/V_REF) e^(V_B/V_T). The log stage followed by an exponential stage gives a reciprocal (divider) function: I_X ∝ 1/V_REF, with the scale fixed by V_B. Accuracy depends on matched I_S and on temperature (V_T and I_S).

(a)(ii) Behaviour of the modified circuit Now Q₂'s base is grounded and its emitter joins V₁, exactly as in Q₁. Both devices see V_BE = −V₁, so I_C2 = I_S2 exp(−V₁/V_T) and I_C1 = V_REF/R₁ give I_X = (V_REF/R₁)(I_S2/I_S1), which equals V_REF/R₁ for matched devices. The circuit is a current mirror (current source): I_X is set by V_REF and R₁ and is independent of the load while Q₂ stays in the active region (collector at or above about 0 V). Because the bases are grounded, base current causes no mirror error; the Early effect (finite output resistance) causes a small dependence on Q₂'s collector voltage, and temperature tracks because both devices share V_T.

(b) V₀(t) by superposition The sources have different frequencies (DC, 6 and 4 rad/s), so each acts alone. DC (20 V): the inductor is a short and the capacitor an open, so V₀ = 20 V. V(t) alone, ω = 6: V = 10∠−30° V (cosine reference); the 20 V source and the open current source drop out, so 4 H goes to ground. jωC = j0.6 S, 1/(j24) = −j0.0417 S. V₀(0.1 + j0.6 − j0.0417) = V/10, so V₀ = 1∠−30°/0.567∠79.8° = 1.763∠−109.8° V. I(t) alone, ω = 4: I = 5∠30° A injected into the node, V(t) shorted. jωC = j0.4 S, 1/(j16) = −j0.0625 S. V₀(0.1 + j0.3375) = I, so V₀ = 5∠30°/0.352∠73.5° = 14.20∠−43.5° V. V₀(t) = 20 + 1.763 cos(6t − 109.8°) + 14.20 cos(4t − 43.5°) V.

(c) Mixer spectrum Take sinc(x) = sin x/x. Then X₁(t) = 10 sinc(10t) ↔ X₁(ω) = π rect(ω/20) (height π, from −10 to +10 rad/s) and X₂(t) ↔ 2π[δ(ω − 1000π) + δ(ω + 1000π)]. Property: multiplication in time is convolution in frequency, F{x₁x₂} = (1/2π) X₁(ω) * X₂(ω). Proof: substitute x₁(t) = (1/2π)∫X₁(λ)e^(jλt)dλ in ∫x₁(t)x₂(t)e^(−jωt)dt and interchange the integrals: Y(ω) = (1/2π)∫X₁(λ)[∫x₂(t)e^(−j(ω−λ)t)dt]dλ = (1/2π)∫X₁(λ)X₂(ω − λ)dλ. Convolving with deltas shifts the pulse (the frequency-shifting property), so Y(ω) = π rect((ω − 1000π)/20) + π rect((ω + 1000π)/20).

|Y(ω)| is two rectangles of height π and width 20 rad/s centred at ±1000π rad/s (±500 Hz), zero elsewhere. (With the normalised sinc, sin πx/πx, each rectangle has height 1 and width 10 Hz.)

Diagram: Spectrum of the mixer output y(t) = X1(t) X2(t)

How this answer will be evaluated

What earns full marks: All parts solved with correct method, clear steps, and proper interpretation.

Evaluation rubric

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

  1. (a)(i) Behaviour of the circuit of Figure 3(a)(i): log stage Q₁ in the op-amp feedback followed by Q₂ with a battery at its emitter. 10 marks

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

    Must cover

    • Show that Q₁ in the feedback with a virtual ground gives V₁ = −V_T ln(V_REF/(R₁I_S)).
    • Find V_BE2 = V₁ + V_B and I_X = (R₁I_S²/V_REF) e^(V_B/V_T).
    • Conclude that the circuit realises a reciprocal function, I_X ∝ 1/V_REF.

    Loses marks

    • Treating the circuit as a constant-current source or ignoring the virtual ground.
    • Wrong sign in the log expression or in V_BE2.

    Earns more

    • Current in R₁ = V_REF/R₁ with a virtual ground at the inverting node.
    • Dependence on matching and temperature (V_T, I_S) noted.

    Extra mark

    • Translinear statement I_X·(V_REF/R₁) = I_S² e^(V_B/V_T).
  2. (a)(ii) Behaviour of the modified topology of Figure 3(a)(ii), where both transistors have grounded bases and emitters joined at V₁. 10 marks

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

    Must cover

    • Read the topology correctly: bases grounded, emitters at V₁, so V_BE1 = V_BE2 = −V₁.
    • Derive I_X = (V_REF/R₁)(I_S2/I_S1), equal to V_REF/R₁ for matched devices.
    • Identify a current mirror whose output is independent of the load while Q₂ is active.

    Loses marks

    • Assuming grounded emitters instead of the drawn connection.
    • Treating Q₁ and Q₂ as independent devices.

    Earns more

    • Early effect and the compliance limit of Q₂'s collector noted.
    • Statement that area ratio scales the output.

    Extra mark

    • Mention of temperature tracking since both devices share V_T.
  3. (b) V₀(t) in the circuit of Figure 3(b) with DC, 6 rad/s and 4 rad/s sources. 20 marks

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

    Must cover

    • Use superposition, one source at a time, with the other sources suppressed.
    • Write the nodal equations with jωC and 1/(jωL) at ω = 6 and ω = 4.
    • Combine as V₀(t) = 20 + 1.763 cos(6t − 109.8°) + 14.20 cos(4t − 43.5°) V.

    Loses marks

    • Using one frequency for all impedances.
    • Adding phasors of different frequencies.

    Earns more

    • Correct phasor conversion of sin to cosine reference (10∠−30° V).
    • Correct DC value of 20 V.

    Extra mark

    • Phasor diagram for the node currents.
  4. (c) Magnitude spectrum of y(t) = X₁(t)X₂(t), with the property used stated and proved. 10 marks

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

    Must cover

    • Find X₁(ω) = π rect(ω/20) and X₂(ω) = 2π[δ(ω − 1000π) + δ(ω + 1000π)].
    • State and prove F{x₁x₂} = (1/2π) X₁ * X₂ and obtain Y(ω) = π rect((ω − 1000π)/20) + π rect((ω + 1000π)/20).
    • Plot two rectangles of height π and width 20 rad/s at ±1000π rad/s.

    Loses marks

    • Missing 1/2π factor in the convolution.
    • Proof of the property absent.

    Earns more

    • Convention for sinc stated.
    • Frequency shift identified as the delta-convolution result.

    Extra mark

    • Result also given for the normalised sinc.
What "Calculate" is asking you to do

Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.

Structure that answers it

Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units

Where marks are lost

Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.

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