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.
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ACorrect
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BIncorrect
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CValid only if the capacitor is ideal
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DValid only when pulse amplitude is small
Answer
Correct Answer: Incorrect
Explanation
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:
Define τ = R * C and pulse width T_p < 5τ.Compute capacitor rise over T_p: V_out(T_p) = V_in * (1 − exp(−T_p / τ)).For T_p ≪ τ, approximate V_out ≈ V_in * (T_p / τ), clearly less than V_in.After the pulse ends, the capacitor discharges back toward zero with time constant τ; it never “catches up” to the input amplitude of that already-ended pulse.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.