Deep Dive: E9D01
The correct answer is D: 6 dB. When the operating frequency of an ideal parabolic reflector antenna is doubled, the gain increases by 6 dB. This is because gain is proportional to (diameter/λ)², so doubling frequency (halving wavelength) quadruples the gain ratio, which is 6 dB. For parabolic reflectors, gain = (πD/λ)² × efficiency, where D is diameter and λ is wavelength. If frequency doubles, wavelength halves. So gain becomes (πD/(λ/2))² = (2πD/λ)² = 4 × (πD/λ)². This is 4× the original gain, which is 6 dB (10 log₁₀(4) = 6 dB). This relationship is important for understanding how parabolic antennas scale with frequency. Higher frequencies allow smaller dishes to achieve the same gain, or the same dish to achieve higher gain.
Why Other Answers Are Wrong
Option A: Incorrect. 2 dB would be too small. Doubling frequency quadruples the gain ratio, which is 6 dB. Option B: Incorrect. 3 dB would be half the correct value. The gain increases by 6 dB when frequency doubles. Option C: Incorrect. 4 dB is close but not the exact value. The correct increase is 6 dB (4× gain ratio).
Exam Tip
Parabolic: Double frequency = +6 dB gain. Remember: When operating frequency doubles, ideal parabolic reflector gain increases by 6 dB (gain is proportional to frequency squared, so 2² = 4× = 6 dB).
Memory Aid
**P**arabolic **D**ouble **F**requency = **+6** **d**B (think 'PDF = +6dB')
Real-World Example
You have a 2-meter (144 MHz) parabolic dish with 20 dBi gain. You use the same dish at 70 cm (432 MHz, 3× frequency). The gain increases by about 9.5 dB (3² = 9× = 9.5 dB), so you'd have about 29.5 dBi. If you doubled frequency (288 MHz), gain would increase by 6 dB to 26 dBi.
Source & Coverage
Question Pool: 2024-2028 Question Pool
Subelement: E9D
Reference: FCC Part 97.3
Key Concepts
Verified Content
Question from the official FCC Extra Class pool. Explanation reviewed by licensed amateur radio operators and mapped to the E9D topic.