RC integrator — single narrow pulse behavior: “The output voltage of an RC integrator will eventually equal the amplitude of a single input pulse if the pulse width is less than 5 time constants.” Decide whether this statement is valid.

Introduction / Context:
An RC integrator is designed so that the capacitor voltage represents the time integral (area) of the input waveform, particularly when the input changes much faster than the circuit’s time constant τ = R * C. The prompt asks whether a single narrow pulse (width < 5τ) will make the output eventually equal the pulse amplitude. This clarifies a common misconception between an integrator and a follower.

Given Data / Assumptions:

• Series RC integrator with output taken across the capacitor.
• Single rectangular input pulse of width T_p where T_p < 5τ.
• Initial capacitor voltage is zero; ideal components, linear operation.

Concept / Approach:
For a finite pulse width T_p, the capacitor charges only for time T_p, reaching V_out(T_p) = V_in * (1 − exp(−T_p / τ)). If T_p is small compared to τ, the exponential term exp(−T_p / τ) ≈ 1 − (T_p / τ), giving V_out ≈ V_in * (T_p / τ) which is much less than V_in. Thus, the output does not “equal the pulse amplitude.” Instead, it is a smaller value proportional to the pulse area V_in * T_p divided by τ.

Step-by-Step Solution:

Verification / Alternative check:
Simulate with τ = 10 ms and T_p = 2 ms: V_out(T_p) = V_in * (1 − e^(−0.2)) ≈ 0.181 * V_in, far below V_in. As T_p approaches several τ, V_out approaches V_in, but the statement restricts T_p to be less than 5τ and claims “equal,” which is not guaranteed nor typical.

Why Other Options Are Wrong:

• Correct: Not correct; integrators respond to pulse area, not necessarily to amplitude.
• Valid only if the capacitor is ideal / pulse amplitude small: Ideal parts or small amplitude do not make V_out equal to V_in for short pulses.

Common Pitfalls:
Confusing RC charging to source (a follower with long pulses) with integrating behavior for short pulses; assuming the 5τ “near steady-state” rule for steps applies to narrow pulses.

Final Answer:
Incorrect.