INTELSAT-IV antenna comparison (1974): The narrow-angle 4.5° beam produced a signal stronger than the 17.34° earth-coverage beam by what factor?

Introduction / Context:
For antennas with similar efficiency and illumination, power gain is inversely proportional to the square of the beamwidth. Comparing narrow and wide beams thus involves squaring the beamwidth ratio.

Given Data / Assumptions:

• Narrow-beam half-power beamwidth: 4.5°.
• Earth-coverage beamwidth: 17.34°.
• Similar aperture and efficiency assumptions.

Concept / Approach:

Gain ratio G_narrow/G_wide ≈ (θ_wide/θ_narrow)^2. Substitute the given beamwidths to obtain the power (signal) ratio directly.

Step-by-Step Solution:

Verification / Alternative check:

HPBW ∝ λ/D ⇒ G ∝ (D/λ)^2 ⇒ G ∝ 1/(HPBW)^2 for comparable patterns; hence the squared relation.

Why Other Options Are Wrong:

• 17.34/4.5: linear ratio, not a power (gain) ratio.
• Product or fourth power: not consistent with first-order beamwidth–gain relationships.
• (4.5/17.34)^2: reciprocal (would be a loss, not a gain increase).

Common Pitfalls:

• Mistaking amplitude (field) ratios for power ratios; power scales with the square.

Final Answer:

(17.34/4.5)^2