Filter definition — do filters selectively pass certain frequency components while attenuating others?

Difficulty: Easy

Correct Answer: True

Explanation:


Introduction / Context:
Filters are used in audio, radio frequency, power supplies, and data conversion to shape spectra. The fundamental idea is frequency selectivity, not uniform transmission of all frequencies. This item asks whether a filter necessarily passes some frequencies and rejects others by design.


Given Data / Assumptions:

  • Linear time-invariant filter context.
  • Frequency response H(jω) characterizes gain versus frequency.
  • Idealized categories include low-pass, high-pass, band-pass, and band-stop.


Concept / Approach:
A filter’s magnitude response has regions of relatively high transmission (passbands) and regions of significant attenuation (stopbands). Transition bands connect these regions. Even an all-pass network is a special type that preserves magnitude but still alters phase; however, the general term “filter” in practice implies amplitude selectivity at least in one band.


Step-by-Step Solution:

Low-pass: passes frequencies below cutoff fc; attenuates above.High-pass: passes above fc; attenuates below.Band-pass: passes between f1 and f2; attenuates outside that interval.Band-stop: attenuates a region near f0; passes frequencies outside that region.


Verification / Alternative check:
Plot |H(jω)| for a simple RC filter to observe passband and stopband behavior. Measurements with a signal generator and spectrum analyzer will confirm gain differences across frequencies that match filter design equations.


Why Other Options Are Wrong:

  • “False” suggests no frequency selectivity, which contradicts the purpose and function of filters in communication and control systems.


Common Pitfalls:
Confusing ideal with real filters. Real filters have finite roll-off and ripple, but they still implement frequency discrimination sufficient for the application.


Final Answer:
True

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion