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au10_312ps8_sol

# au10_312ps8_sol - ECE312 Autumn 2010 Problem Set 8...

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ECE312 Autumn 2010: Problem Set 8 Solutions Problem 1 : Z A = 75 j 10 Ω, R loss = 1 Ω. (a) We know that Z A = R A + jX A , and that R A = R rad + R loss . Therefore R rad = 75 1 = 74 Ω, and the efficiency is R rad /R A = 74 / 75 = 98 . 67 % (b) Power radiated = | ˜ I | 2 R rad / 2 = (0 . 5) 2 (74) / 2 = 9 . 25 W (c) To obtain a 74 Ω radiation resistance for a Hertzian dipole, we need R rad = 80 π 2 parenleftbigg l λ parenrightbigg 2 l λ = 1 π radicalbigg 74 80 = 0 . 3061 Given λ = c/ (3 × 10 9 ) = 10 cm, we find l = (0 . 3061)(10) = 3.06 cm . Since this antenna would not have l/λ 1, it is not a true Hertzian dipole. Hertzian dipoles have small R rad values. Problem 2 : f = 820 kHz, ˜ I = 8 A. ˆ x is pointing North, ˆ y West. (a) Given f = 820 kHz, λ = (3 × 10 8 ) / (820 × 10 3 ) = 365 . 85 m. 2.5 m long antenna is short compared to λ ( l/λ = 0 . 0068) so modeling as a Hertzian dipole appears reasonable. (In reality the presence of the ground would cause changes from our basic dipole fields.) (b) The electric field of a Hertzian dipole in the far field is ˜ E = j ˜ Iklη 4 π e jkR R ˆ θ sin θ Plugging in our information ˜ I

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