RC integrator behavior: A simple series R–C network with the output taken across the capacitor acts as an integrator for a sufficiently high input frequency (f >> 1 / (2π R C)), producing an output proportional to the time integral of the input. Is this statement correct?

Introduction / Context:An “RC integrator” is a foundational analog building block used for pulse shaping, ramp generation, and as part of active filter designs. The classic passive form is a series R followed by a capacitor to ground, with output across the capacitor. Under the right frequency conditions, its output approximates the integral of the input waveform.

Given Data / Assumptions:

• Series resistor R, shunt capacitor C, output across C.
• Linear components, small-signal regime.
• Input frequency sufficiently high relative to the corner f_c = 1/(2π R C).

Concept / Approach:The transfer function of the passive network is H(s) = 1 / (1 + s R C). For frequencies well above f_c, the magnitude of H(s) becomes small and the capacitor voltage approximates the time integral of the input scaled by 1/(R C). Step inputs yield exponential ramps initially; narrow pulses produce proportional area at the output, which is the essence of integration over the pulse width.

Step-by-Step Solution:

Verification / Alternative check:Time-domain check: for a narrow input pulse of area A (volt-seconds), the change in capacitor voltage ΔV ≈ A / (R C), consistent with integration of the input.

Why Other Options Are Wrong:Incorrect: contradicts the standard asymptotic behavior. “Only true if R << Xc at all frequencies”: condition is frequency-dependent; at high f, |Xc| becomes small and the approximation holds. “Only true when the input is DC”: integration of DC produces a ramp until limited; the integrator definition is broader and frequency-domain based.

Common Pitfalls:Assuming perfect integration at any frequency; ignoring output amplitude droop and finite RC leakage; forgetting loading effects that spoil the ideal 1/s response.

Final Answer:Correct