GATE EE 2025 Set 1 — Question 39

In the circuit the input is applied to a series resistor $R$ and the output is taken across the capacitor $C$, i.e. it is a first-order low-pass (integrating) network:$$\frac{V_{out}(s)}{V_{in}(s)} = \frac{1/sC}{R + 1/sC} = \frac{1}{1 + sRC}$$Condition given. $R$ and $C$ are very large, so the time constant $\tau = RC$ is very large, while the square-wave input has a very high frequency, so its period $T$ is very small. Hence$$\tau = RC \gg T$$Behaviour during each half cycle. The capacitor voltage is small compared with the input, so the resistor current is almost constant during a half cycle:$$i \approx \frac{V_{in} - V_{out}}{R} \approx \pm\frac{1\ \text{V}}{R}$$A constant charging current makes the capacitor voltage change linearly:$$V_{out}(t) = \frac{1}{C}\int i\,dt \approx \pm\frac{t}{RC}$$So the output ramps up during the positive half cycle and ramps down during the negative half cycle: a triangular waveform.Amplitude of the triangle. The peak-to-peak excursion in half a period is$$\Delta V = \frac{1\ \text{V}}{RC}\cdot\frac{T}{2} \ll 1\ \text{V}$$because $RC \gg T$. The output is therefore a small-amplitude triangular wave about the mean level of the input (zero), not a full $\pm 1$ V waveform.Option (A) shows a triangular wave that swings the full $+1$ V to $-1$ V, which would require $RC$ comparable to $T$. Option (B) reproduces the square wave, which is what a very small $RC$ would give. Option (D) shows the exponential/spiky response of a differentiating (high-pass) network. Option (C) is the small-amplitude triangular wave predicted above.Answer: (C)