GATE 2022 EC Q.42: solution and answer

GATE 2022 · ECQ.42 · 2 marks · MCQ

A circuit with an ideal OPAMP is shown. The Bode plot for the magnitude (in dB) of the gain transfer function $(A_{V} (j ω) = V_{o u t} (j ω) / V_{in} (j ω))$ of the circuit is also provided (here, $ω$ is the angular frequency in rad/s). The values of $R$ and $C$ are ________.

GATE 2022 EC question 42: Left: the circuit. — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
$R = 3 k Ω, C = 1 μ F$
(B)
$R = 1 k Ω, C = 3 μ F$
(C)
$R = 4 k Ω, C = 1 μ F$
(D)
$R = 3 k Ω, C = 2 μ F$

Answer and solution

ANSWER

(A) $R = 3 k Ω, C = 1 μ F$ . The 12 dB flat level is a gain of 4, which gives $1 + R /1 k Ω = 4$ . The 9 dB point at $ω = 1 0^{3} rad/s$ is the corner, which gives $C = 1/ (1 0^{3} \times 1 0^{3}) = 1 μ F$ .

Official answer: ASource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Assumptions. The op-amp is ideal: infinite input impedance, zero input current, infinite open-loop gain. It works in its linear region, so the ±15 V rails play no part.
Input RC network (voltage divider). The op-amp draws no input current, so the series 1 kΩ resistor and $C$ form an unloaded low-pass divider: $V_{+} = V_{in} \frac{1/ ( j ω C )}{1 k Ω + 1/ ( j ω C )} = \frac{V _{in}}{1 + j ω ( 1 k Ω ) C}$
Non-inverting amplifier. Negative feedback holds a virtual short at the inputs, so $V_{-} = V_{+}$ . $R$ and the 1 kΩ resistor to ground divide the output: $V_{-} = V_{o u t} \frac{1 k Ω}{1 k Ω + R}$ . Therefore $V_{o u t} = (1 + \frac{R}{1 k Ω}) V_{+}$ .
Overall transfer function. $A_{V} (j ω) = (1 + \frac{R}{1 k Ω}) \frac{1}{1 + j ω / ω _{c}}, ω_{c} = \frac{1}{( 1 k Ω ) C}$ This is a first-order low-pass response. It is flat at the DC gain, 3 dB below that level at $ω_{c}$ , and falls toward a 20 dB/decade asymptote beyond. $R$ sets only the DC gain. $C$ , with the series 1 kΩ, sets only the corner.
DC gain from the plot. The flat level is 12 dB, so $∣ A_{V} (0) ∣ = 1 0^{12/20} \approx 3.98 \approx 4$ . Then $1 + R /1 k Ω = 4$ , which gives $R = 3 k Ω$ . This rules out (B), whose gain of 2 would sit at about 6 dB, and (C), whose gain of 5 would sit at about 14 dB.
Corner frequency from the plot. The curve reaches 3 dB below the flat level ( $12 - 3 = 9 dB$ ) at $lo g_{10} ω = 3$ , so $ω_{c} = 1 0^{3} rad/s$ .
Solve for C. $C = \frac{1}{( 1 k Ω ) ω _{c}} = \frac{1}{1 0 ^{3} \times 1 0 ^{3}} = 1 0^{- 6} F = 1 μ F$ .
Check (D). With $C = 2 μ F$ the corner would be $ω_{c} = 500 rad/s$ ( $lo g_{10} ω \approx 2.7$ ). At $ω = 1 0^{3} rad/s$ the magnitude would be $12 - 10 lo g_{10} (1 + 2^{2}) \approx 5 dB$ , not the 9 dB on the plot. So the answer is $R = 3 k Ω, C = 1 μ F$ , option (A).

KEY IDEA

The ideal op-amp draws no current, so the input RC is an unloaded first-order low-pass with $ω_{c} = 1/ ((1 k Ω) C)$ . The non-inverting stage multiplies it by $1 + R /1 k Ω$ and does not move the pole. Read the flat level for $R$ (12 dB is a gain of 4, so $R = 3 k Ω$ ). Read the point 3 dB below that level for $C$ (9 dB at $ω = 1 0^{3} rad/s$ , so $C = 1 μ F$ ).

WATCH OUT FOR

One slip is using the inverting-gain formula $R /1 k Ω = 4$ , which gives $R = 4 k Ω$ and lands on option (C). Another is putting the feedback $R$ into the time constant; only the series 1 kΩ and $C$ set $ω_{c}$ . A third is taking the corner from the wrong point: where the curve first visibly bends (about $lo g_{10} ω = 2.3$ ), or where it reaches 3 dB absolute (about $lo g_{10} ω = 3.4$ ). The corner is the point 3 dB below the flat level, which is 9 dB here.

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 13 Filters & waveforms questions, with notes

GATE 2022 EC Q.42: solution and answer

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