Low-pass filter behavior — as the input frequency increases beyond the cutoff, the output (across the load) decreases in magnitude. Is this statement correct?

Introduction / Context:Frequency response curves characterize how filters pass or attenuate signals versus frequency. A low-pass filter (LPF) passes low-frequency components while attenuating higher frequencies above its cutoff (fc).

Given Data / Assumptions:

• Simple first-order or higher-order LPF topology.
• Linear, time-invariant behavior around operating point.

Concept / Approach:The magnitude response |H(jω)| of an LPF decreases with ω beyond fc. For a first-order RC LPF, |H(jω)| = 1 / sqrt(1 + (ωRC)^2), clearly decreasing as ω increases.

Step-by-Step Solution:

Verification / Alternative check:Bode plots show −20 dB/decade per pole beyond cutoff for first-order LPFs.

Why Other Options Are Wrong:

Common Pitfalls:Assuming “low-pass” means no attenuation at high frequencies; real filters have finite roll-off and phase shift.

Final Answer:Correct