GATE 2024 IN Q.60: solution and answer

GATE 2024 · INQ.60 · 2 marks · NAT

In the figure shown, $S_{W}$ is a switch whose position changes from 1 to 0 when $V_{C}$ changes from logic HIGH to LOW and vice versa. The bandwidth of the permanent magnet moving coil (PMMC) type voltmeter is $1 Hz$ . If $V_{sense} = 2 sin (4000 π t) V$ and $V_{ref} = 4 sin (2000 π t) V$ , the voltmeter reading is ______ $V$ (rounded off to nearest integer).

Note: Assume all components are ideal.

GATE 2024 IN question 60: The input node V_sense connects by one wire directly to terminal 1 of the switch S_W, and by a branch to the input of an… — Figure from the GATE 2024 IN paper. It is part of the question.

Answer and solution

ANSWER

The PMMC voltmeter reads 0 V. The switch-and-inverter pair multiplies V_sense by a ±1 square wave at 1 kHz, and a 2 kHz signal multiplied by that square wave has no DC component.

Official answer: 0Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

Comparator with its $-$ input grounded: $V_{C}$ is HIGH when $V_{ref} > 0$ and LOW when $V_{ref} < 0$ . $V_{ref} = 4 sin (2000 π t)$ has frequency $1 kHz$ , so its period is $T = 1 ms$ . For $0 < t < 0.5 ms$ , $V_{C}$ is HIGH and $S_{W}$ is at 1. For $0.5 < t < 1 ms$ , $V_{C}$ is LOW and $S_{W}$ is at 0.
Switch output: position 1 passes $V_{sense}$ and position 0 passes $- V_{sense}$ (through the gain $- 1$ amplifier). So $V_{out} = V_{sense} \cdot s (t)$ , where $s (t) = sgn [sin (2000 π t)]$ is a $\pm 1$ square wave at $1 kHz$ . The circuit is a phase-sensitive (synchronous) detector.
PMMC with $1 Hz$ bandwidth: the meter responds only to the DC (average) value of $V_{out}$ . Every AC component of $V_{out}$ is at $1 kHz$ or above, far outside its bandwidth.
Average over one reference period:
$V_{avg} = \frac{1}{T} [\int_{0}^{T /2} 2 sin (4000 π t) d t - \int_{T /2}^{T} 2 sin (4000 π t) d t]$
Across each half-period ( $0.5 ms$ ), the argument $4000 π t$ advances by $4000 π \times 0.5 \times 1 0^{- 3} = 2 π$ . $V_{sense}$ at $2 kHz$ therefore completes exactly one full cycle in each half, so each integral is $0$ and $V_{avg} = 0 V$ .
Fourier check: the square wave $s (t)$ has no DC term and only odd harmonics ( $1, 3, 5, \dots kHz$ ). Multiplying by a $2 kHz$ sine gives components at $∣2 \pm (2 k + 1) ∣ kHz$ . These are all odd multiples of $1 kHz$ and are never $0 Hz$ , so $V_{out}$ has no DC term.
The voltmeter reading is $0 V$ .

KEY IDEA

The comparator, the gain $- 1$ amplifier and the switch together multiply $V_{sense}$ by a $\pm 1$ square wave locked to $V_{ref}$ . This is a phase-sensitive detector. A PMMC meter with a $1 Hz$ bandwidth reads only the DC part of the product. That DC part is non-zero only when $V_{sense}$ has a component at the reference frequency or at one of its odd harmonics. Here $2 kHz$ is an even harmonic of $1 kHz$ , so the reading is $0$ .

WATCH OUT FOR

One error is to treat the circuit as a full-wave rectifier, taking $V_{out} = ∣ V_{sense} ∣$ and reading $\frac{2 V _{m}}{π} = \frac{4}{π} \approx 1.27 \to 1 V$ . The switch follows $V_{ref}$ , not $V_{sense}$ , so the two signals are at different frequencies and the product is not rectified. Another error is to read an rms value, $\frac{2}{2} \approx 1.41 \to 1 V$ . A PMMC meter responds to the average value, not the rms value.

From the GATE 2024 IN paper. Every question of this paper · Download the full paper · All 8 V, I, power & energy questions, with notes

GATE 2024 IN Q.60: solution and answer

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