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

Difficulty: Easy

Correct Answer: Correct

Explanation:


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:

1) Identify the filter as low-pass.2) Recognize cutoff frequency fc = 1/(2πRC) for RC case.3) For f ≫ fc, output magnitude falls approximately as 1/(ωRC) per pole.4) Therefore, as input frequency increases, output decreases.


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


Why Other Options Are Wrong:

Statements tying behavior to resonance, DC, or ideal inductors confuse LPFs with band-pass or RLC resonant circuits.


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


Final Answer:
Correct

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