Note that the gure on the left shows a right hand

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Unformatted text preview: vector points out-of-page. Then, the possible phasor electric fields can be shown as in the below figures. Note that the figure on the left shows a right-hand circular polarization plane whereas the figure on the right shows a left-hand circular polarization plane. -x,z y y -x,-z As the wave is right-hand circularly polarized, the imaginary part of the electric field phasor 2 2 ˜ must be along x and z directions. Since Ey + Exz = 1 where Exz is the imaginary part of E, X = j and Z = j . Finally, we can express the electric field phasor at t = 0 at the origin as p ˜ E = j x + 2ˆ + j z . ˆ y ˆ c) The direction of the Poynting vector will be parallel with the direction of propagation vector. Hence, the incident power can be written as < Sinc >= ˜ |Einc |2 x + z ˆˆ p. 2⌘ o 2 Considering that the antenna is z -polarized, we can write the co-polarized component of the ˆ time-averaged Poynting vector associated with E as ˜ Einc = j z e ˆ 2 p j x +z 2 . d) Since the load is matched, q = 1. Then, Pr = PL . Recalling that the power delivered by the antenna to...
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