Which of the following represents the correct output V_oof circuit shown in Figure? Assume RC = 1.

Question

Question

This question was previously asked in

ISRO Scientist Electronics 2024 Official Paper

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Answer 1

Concept:

The given circuit is an op-amp differentiator. For an ideal differentiator with input \( v_i(t) \), the output is given by:

\( v_o(t) = -RC \cdot \frac{d}{dt}v_i(t) \)

Given that \( RC = 1 \), the equation simplifies to:

\( v_o(t) = -\frac{d}{dt}v_i(t) \)

Given:

The input signal \( v_i(t) \) is a triangular waveform.

During the rising edge (slope = +1): \( \frac{d}{dt}v_i(t) = 1 \Rightarrow v_o(t) = -1 \)
During the falling edge (slope = –1): \( \frac{d}{dt}v_i(t) = -1 \Rightarrow v_o(t) = +1 \)

Thus, the output waveform will be a square wave alternating between +1 and –1, switching at every point where the slope of \( v_i(t) \) changes.

Answer:

Option 2: Square wave output alternating between +1 and –1 (in response to triangular wave input)