Introduction / Context:
Antenna power gain combines directivity and radiation efficiency. Knowing the radiation and loss resistances allows us to compute efficiency, and from the stated power gain we can determine the directivity. This problem tests core antenna relationships among gain, directivity, and efficiency.
Given Data / Assumptions:
- Radiation resistance Rr = 54 Ω.
- Loss resistance Rl = 6 Ω.
- Stated power gain G = 10 (dimensionless).
- Power gain G = η * D, where η is radiation efficiency and D is directivity.
Concept / Approach:
Radiation efficiency η equals the fraction of input RF power actually radiated: η = Rr / (Rr + Rl). Power gain equals the product of efficiency and directivity: G = η * D. Rearranging yields D = G / η.
Step-by-Step Solution:
Compute efficiency: η = Rr / (Rr + Rl) = 54 / (54 + 6) = 54 / 60 = 0.9.Relate gain and directivity: G = η * D.Solve for D: D = G / η = 10 / 0.9 = 11.111…Round appropriately: D ≈ 11.11.
Verification / Alternative check:
If efficiency were 100% (η = 1), D would equal G (10). With 90% efficiency, D must be slightly higher than 10 to yield the same G, consistent with 11.11.
Why Other Options Are Wrong:
9 or 8.5: These are less than gain; they would imply η > 1 which is impossible physically.10: Only if η = 1; here η = 0.9.Data not sufficient: All required quantities are provided (Rr, Rl, and G).
Common Pitfalls:
Mixing up gain and directivity; forgetting to compute efficiency from resistances; using dBi units accidentally instead of linear values.
Final Answer:
11.11
Discussion & Comments