Deep Dive: E9E06
The correct answer is C: 75 ohms. To construct a quarter-wave Q-section for matching a 100-ohm feed point impedance to a 50-ohm transmission line, you need a transmission line with impedance equal to the geometric mean of the two impedances: √(Z₁ × Z₂) = √(100 × 50) = √5000 ≈ 70.7 ohms. 75 ohms is the closest standard value. A quarter-wave matching section (Q-section) transforms impedance. The Q-section's characteristic impedance must be the geometric mean of the source and load impedances: Z_Q = √(Z_source × Z_load). For 100 ohms to 50 ohms: Z_Q = √(100 × 50) = √5000 ≈ 70.7 ohms. Standard 75-ohm cable is close enough and commonly used. The Q-section must be exactly 1/4 wavelength at the operating frequency to work properly.
Why Other Answers Are Wrong
Option A: Incorrect. 50 ohms is the load impedance, not the Q-section impedance. The Q-section needs the geometric mean, not one of the end impedances. Option B: Incorrect. 62 ohms is close but not the geometric mean. The calculation gives about 70.7 ohms, so 75 ohms is closer. Option D: Incorrect. 90 ohms is too high. The geometric mean of 100 and 50 is about 70.7 ohms.
Exam Tip
Q-section impedance = Geometric mean. Remember: Q-section impedance = √(Z₁ × Z₂). For 100Ω to 50Ω: √(100×50) = √5000 ≈ 71Ω, so 75Ω is the closest standard value.
Memory Aid
**Q**-**S**ection **I**mpedance = **G**eometric **M**ean (think 'QSI = GM', √(100×50) ≈ 75Ω)
Real-World Example
You need to match a 100-ohm antenna to 50-ohm coax. You calculate the Q-section impedance: √(100 × 50) = √5000 ≈ 70.7 ohms. You use 75-ohm cable (closest standard value) cut to exactly 1/4 wavelength. This Q-section transforms the 100-ohm antenna impedance to 50 ohms, matching your coax.
Source & Coverage
Question Pool: 2024-2028 Question Pool
Subelement: E9E
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 E9E topic.