RC low-pass intuition: For a first-order RC low-pass filter where the output is taken across the capacitor, does the output amplitude decrease as the input frequency increases (i.e., higher f ⇒ lower |Vout|/|Vin|)?

Introduction / Context:
First-order RC low-pass filters are fundamental in smoothing, anti-aliasing, and bandwidth limiting. The hallmark of a low-pass is that it passes low frequencies with little attenuation and attenuates high frequencies. This question verifies that intuition for the common configuration with output across the capacitor.

Given Data / Assumptions:

• Series R, shunt C, output measured across C.
• Ideal linear components, sinusoidal steady state.
• Cutoff frequency f_c = 1 / (2π R C).

Concept / Approach:
The transfer function is H(jω) = 1 / (1 + jωRC). Its magnitude |H(jω)| = 1 / sqrt(1 + (ωRC)^2). As frequency increases, the denominator grows, so the magnitude decreases. Therefore, the output amplitude drops with higher input frequency, which is exactly what a low-pass filter is designed to do.

Step-by-Step Solution:

Verification / Alternative check:
At ω = 0, |H| = 1 (no attenuation). At ω = ω_c, |H| = 1/√2 (−3 dB). As ω → ∞, |H| → 0. Scope measurements on a simple RC confirm the roll-off behavior.

Why Other Options Are Wrong:
Incorrect: contradicts the low-pass definition.
Inductive loads / frequency constraint options do not alter the basic monotonic trend for the ideal RC low-pass with output across C.

Common Pitfalls:
Confusing the low-pass with the high-pass (output taken across R); forgetting that loading by the next stage can shift the effective corner frequency.

Final Answer:
Correct