Low-pass filter behavior above cutoff (fc): in a standard active or passive low-pass filter used in basic op-amp circuits, what happens to the output voltage gain as frequency increases beyond the critical (cutoff) frequency?

Introduction / Context:
Low-pass filters are fundamental in analog signal processing. They pass low-frequency components while attenuating higher frequencies. The “critical” or cutoff frequency (often noted as fc) is commonly defined at the point where the magnitude response falls to 0.707 of the passband value (−3 dB for first order). This question asks what happens to the output gain beyond that point.

Given Data / Assumptions:

• Idealized low-pass filter behavior (active or passive).
• Standard definition of cutoff at the −3 dB corner for first-order networks.
• No special peaking or resonance unless otherwise noted.

Concept / Approach:
A first-order low-pass has a magnitude |H(jω)| = 1 / sqrt(1 + (ω/ωc)^2). For ω > ωc (i.e., f > fc), the denominator grows, so magnitude decreases with frequency. In Bode terms, the slope is approximately −20 dB/decade per pole. Multi-pole low-pass networks attenuate even more steeply (e.g., −40 dB/decade for two poles). Thus, the gain does not increase or remain constant above cutoff; it decreases progressively.

Step-by-Step Solution:

Verification / Alternative check:
You can confirm by evaluating |H| at f = 10 * fc for a single-pole filter: |H| ≈ 1/√(1 + 10^2) ≈ 1/√101 ≈ 0.099 (about −20 dB), which is clearly lower than at fc and much lower than in the passband.

Why Other Options Are Wrong:
Increases: contradicts the standard roll-off. Remains unchanged: only true well below fc. Doubles per 1 kHz: arbitrary and incorrect. Becomes strictly zero above fc: ideal filters asymptotically approach zero; they do not become exactly zero.

Common Pitfalls:
Confusing low-pass with high-pass behavior; assuming “brick-wall” attenuation where output is exactly zero above fc; overlooking multi-pole slopes.

Final Answer:
Decreases