JNTU B.Tech II PULSE AND DIGITAL CIRCUITS Previous Question Paper (March - 2022 - R13)

A low pass RC circuit acts as a frequency-dependent voltage divider. When a sinusoidal input voltage v_i(t) = V_m sin(ωt) is applied, the output voltage v_o(t) is determined by the complex impedance of the capacitor X_c = 1 / (jωC). The transfer function is H(jω) = V_o / V_i = (1 / jωC) / (R + 1 / jωC) = 1 / (1 + jωRC). This can be written in polar form as H(jω) = (1 / sqrt(1 + (ωRC)²)) ∠ -tan^-1(ωRC).

The output voltage is v_o(t) = (V_m / sqrt(1 + (ωRC)²)) sin(ωt - φ), where φ = tan^-1(ωRC). The circuit attenuates higher frequencies more than lower frequencies, with the cutoff frequency defined as f_c = 1 / (2πRC). At frequencies well below f_c, the output follows the input with minimal attenuation and phase shift. At frequencies well above f_c, the output amplitude decreases significantly, and the phase shift approaches -90 degrees.

Given a square wave with peak-to-peak amplitude V_pp = 2V, the peak values are +1V and -1V. The period is T, and the time constant τ = RC = T/2. For an integrating circuit, the output voltage v_o(t) during the interval 0 ≤ t ≤ T/2 is given by v_o(t) = V₁ + (V₂ - V₁)(1 - e^-t/τ), where V₁ is the initial voltage at t=0 and V₂ is the input voltage level.

In steady state, the output is symmetric. Let the peak values of the output be V_max and V_min. Since the input is ±1V, the output at t = T/2 (which is t = τ) is V_min = V_max + (V_in - V_max)(1 - e^-τ/τ) = V_max + (-1 - V_max)(1 - e^-1). Given symmetry, V_max = -V_min. Substituting V_min = -V_max: -V_max = V_max - (1 + V_max)(1 - 0.368). Solving this, 2V_max = (1 + V_max)(0.632), which yields 2V_max = 0.632 + 0.632V_max, so 1.368V_max = 0.632, resulting in V_max ≈ 0.462V. The peak-to-peak output voltage is 2 * V_max ≈ 0.924V.

The steady-state waveform is a triangular-like wave consisting of exponential segments rising from -0.462V to +0.462V and falling from +0.462V to -0.462V, with the curvature defined by the time constant τ = T/2.