Difficulty: Easy
Correct Answer: Picowatts
Explanation:
Introduction / Context:
Despite large Effective Isotropic Radiated Power (EIRP) at the satellite and high-gain earth-station antennas, the enormous free-space path loss over ~36,000 km reduces received carrier power to very small levels.
Given Data / Assumptions:
Concept / Approach:
Friis transmission formula indicates Pr = Pt * Gt * Gr * (lambda / (4 * pi * R))^2. Even with large Gt and Gr, R is so large that Pr is in the pico- to nanowatt range. “A few picowatts” is a common order-of-magnitude description for many GEO downlinks.
Step-by-Step Solution:
Estimate FSPL at, say, 12 GHz and R ≈ 36,000 km.Compute lambda = c/f ≈ 0.025 m; (lambda/(4piR))^2 ~ very small.Combine with typical EIRP and G/T to infer Pr ~ 10^−12…10^−9 W.
Verification / Alternative check:
Link budget examples in textbooks yield received carrier powers typically in the picowatt to nanowatt range for GEO links.
Why Other Options Are Wrong:
Milliwatts/Watts/Kilowatts are many orders of magnitude too large for post-FSPL carrier at LNB input under normal operation. Microjoules are energy units, not power.
Common Pitfalls:
Confusing EIRP with received power; ignoring FSPL magnitude at GEO distances.
Final Answer:
Picowatts
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