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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.
y y -x,-z As the wave is right-hand circularly polarized, the imaginary part of the electric ﬁeld phasor
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
E = j x + 2ˆ + j z .
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
2 Considering that the antenna is z -polarized, we can write the co-polarized component of the
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
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- Spring '11