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350sp13hw14sol

350sp13hw14sol - ECE 350 1 Homework 14 Solutions Due Thu 5...

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ECE 350 Homework 14 — Solutions Due: Thu May 02, 2013, 5 PM 1. a) The propagation vector is given by k = - ˆ x - ˆ z . Thus, the wavenumber is k = p 2 + 2 = p 2 = 4 . 4429 rad m . Then, the wavelength is λ = 2 k = 1 . 414 m and the frequency is f = c λ = 212 . 13 MHz. b) The magnitude of time-average Poynting vector is | < S inc > | = 1 2 10 2 120 = 0 . 1326 Watts m 2 . As ˆ k = - ˆ x z p 2 , the time-average Poynting vector is given by < S inc > = - 0 . 1326 ˆ x + ˆ z p 2 = - 0 . 0938(ˆ x + ˆ z ) Watts m 2 . c) The available power in response to the ˜ E inc is given by P r = S inc A ( 2 , φ 2 ) = S inc λ 2 4 G ( 2 , φ 2 ) where G ( 2 , φ 2 ) = 1 . 5 sin 2 2 . Thus, we write P r = 0 . 1326 1 . 414 2 4 1 . 5 sin 2 45 o = 0 . 0158 W . d) If the short dipole at the origin is polarized along the vector ˆ z - ˆ x p 2 , we will be rotating the coordinate system by 45 . Therefore, we have G ( 0 2 , φ 0 2 ) = G ( 2 + 45 o , φ 2 ) . This implies that P r = S inc λ 2 4 G (90 o , 0 o ) = 0 . 0317 W . e) Now the short dipole rotates to along the vector ˆ z x p 2 , For convenience, we also rotate our coordinate system so that the short dipole is still along our new z axis, therefore, we have G ( 0 2 , φ 0 2 ) = G ( 2 - 45 o , φ 2 ) . This implies that P r = S inc λ 2 4 G (0 o , 0 o ) = 0 W . f) Now the short dipole is polarized along the ˆ y direction and our electrical field is in x-z plane. Due to the polarization mismatch, the dipole direction is perpendicular to the incident field polarization (cross-polarization), so P r = 0 W.

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