For achieving a specified high directivity, how does the required physical size (aperture) of an antenna vary as the operating wavelength decreases?

Difficulty: Easy

Correct Answer: It decreases as wavelength decreases

Explanation:


Introduction / Context:
High-directivity antennas concentrate radiated energy into narrow beams. Dishes, horn antennas, and phased arrays are common examples. Designers often ask how physical dimensions scale when changing frequency (or wavelength) for a target directivity and beamwidth.


Given Data / Assumptions:

  • Comparing antennas of the same type and efficiency.
  • Goal: maintain the same directivity while changing operating wavelength λ.
  • Free-space conditions; mutual coupling and platform constraints are ignored.


Concept / Approach:

For an effective aperture antenna, directivity D is approximately D ≈ 4π * A_e / λ^2, where A_e is effective aperture (proportional to physical area A times efficiency). Holding D constant implies A_e ∝ λ^2. Therefore, as λ decreases (frequency rises), the required area drops with λ^2.


Step-by-Step Solution:

1) Start from D ≈ 4π * A_e / λ^2.2) For fixed D and efficiency, A_e ∝ λ^2.3) Physical aperture area A scales like A_e, so A ∝ λ^2.4) Hence when λ decreases, A (and linear dimensions such as diameter) decrease proportionally (diameter ∝ λ for a given beamwidth).


Verification / Alternative check:

Practical systems demonstrate this scaling: a 1 m dish at X-band provides far greater directivity than at L-band; conversely, the same directivity at X-band needs a much smaller dish than at L-band.


Why Other Options Are Wrong:

  • Increases: Opposite of aperture scaling for fixed D.
  • Unaffected: Ignores λ^2 dependence in aperture theory.
  • Depends only on polarization: Polarization does not set required physical aperture for D.
  • Oscillates with wavelength: No such periodic relation in basic aperture theory.


Common Pitfalls:

Mixing up gain scaling with frequency-dependent efficiency; comparing dissimilar antenna types; overlooking edge-taper and blockage effects which slightly modify constants but not the λ^2 law.


Final Answer:

It decreases as wavelength decreases

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