GATE 2025 EE Q.39: solution and answer

GATE 2025 · EEQ.39 · 2 marks · MCQ

In the circuit, shown below, if the values of R and C are very large, the form of the output voltage for a very high frequency square wave input, is best represented by

GATE 2025 EE question 39: Series RC low-pass network. — Figure from the GATE 2025 EE paper. It is part of the question.

(A)
Figure: $V_{o u t}$ vs t is a triangular wave swinging the full range between $+ 1 V$ and $- 1 V$ . Its peaks and troughs touch the dashed 1 V and -1 V lines.
(B)
Figure: $V_{o u t}$ vs t is a square wave alternating between $+ 1 V$ and $- 1 V$ (the same as the input).
(C)
Figure: $V_{o u t}$ vs t is a small-amplitude triangular wave centred on the time axis (symmetric about 0 V). Its peaks sit only a fraction of a volt above and below the axis, far inside the dashed 1 V and -1 V lines.
(D)
Figure: $V_{o u t}$ vs t is a train of exponential spikes. It starts at $+ 1 V$ at t = 0 and decays to 0, stays at 0, jumps to $- 1 V$ at the next edge and decays back to 0, and so on alternately (a differentiated square wave).

Answer and solution

ANSWER

(C): a small-amplitude triangular wave centred on 0 V. With $R C ≫ T /2$ , the RC low-pass network integrates the $\pm 1 V$ square wave into a triangle of peak $tanh (T /4 R C) \approx T / (4 R C) ≪ 1 V$ .

Official answer: CSource: Official answer key, GATE 2025 EE (IIT Roorkee)

The output is taken across C, so the network is a first-order RC low-pass filter with time constant $τ = R C$ . A very high input frequency together with very large R and C means the half-period is tiny compared with the time constant: $\frac{T}{2} ≪ R C$ .
Average (DC) value: in periodic steady state the capacitor voltage returns to the same value every period, so the capacitor carries no average current. With zero average current there is zero average voltage across R, so $\overline{V_{o u t}} = \overline{V_{in}} = \frac{( + 1 ) + ( - 1 )}{2} = 0 V$ . The output is therefore centred on the time axis.
Because $T /2 ≪ R C$ , the capacitor voltage changes very little in one half-cycle. It stays small compared with $1 V$ (step 5 confirms this), so nearly the whole input appears across R. The charging current is then almost constant during each half-cycle: $i \approx \frac{\pm 1 V}{R}$ .
A constant current into a capacitor gives a linear ramp (the integrator approximation): $\frac{d V _{o u t}}{d t} = \frac{i}{C} \approx \pm \frac{1}{R C}$ . So $V_{o u t}$ ramps up while the input is $+ 1 V$ and ramps down while it is $- 1 V$ , which gives a triangular wave.
Peak-to-peak swing over one half-period: $Δ V_{o u t} \approx \frac{1}{R C} \cdot \frac{T}{2} = \frac{T}{2 R C} ≪ 1 V$ Exact check: in steady state $V_{o u t}$ swings between $\pm V_{p}$ . During a $+ 1 V$ half-cycle it charges from $- V_{p}$ towards $+ 1$ , so $V_{p} = 1 - (1 + V_{p}) e^{- T / (2 R C)}$ , which gives $V_{p} = tanh (\frac{T}{4 R C}) \approx \frac{T}{4 R C}$ when $T ≪ R C$ . That agrees with the peak-to-peak value above.
So the output is a small-amplitude triangular wave centred on $0 V$ , far inside the $\pm 1 V$ lines. Option A has the right shape but a full $\pm 1 V$ swing: since $V_{p} = tanh (T /4 R C)$ , that would need $R C ≪ T$ , and then the capacitor follows the input and the output is not triangular. Option B is what you would get across C when $R C ≪ T$ . Option D is the voltage across R (a differentiator) when $R C ≪ T$ . The answer is (C).

KEY IDEA

When the time constant is much longer than the half-period ( $R C ≫ T /2$ ), an RC network with the output across C acts as an integrator. The square wave in becomes a triangular wave out, with zero average and peak $tanh (T /4 R C) \approx T / (4 R C)$ , so the larger $R C$ (or the higher the frequency), the smaller the triangle.

WATCH OUT FOR

Choosing (A) because 'integrator gives a triangle' and forgetting the amplitude. The integrator's output amplitude is about $T / (4 R C)$ , which is tiny here. A full $\pm 1 V$ swing would need $R C ≪ T$ , and in that case the output across C is square, not triangular. The other trap is picking (D): that is the differentiated waveform, taken across R with $R C ≪ T$ .

From the GATE 2025 EE paper. Every question of this paper · Download the full paper · All 20 Transients questions, with notes

GATE 2025 EE Q.39: solution and answer

Answer and solution

More Transients questions