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Unformatted text preview: vector points outofpage. Then, the possible phasor electric ﬁelds can be shown as in the below
ﬁgures. Note that the ﬁgure on the left shows a righthand circular polarization plane whereas
the ﬁgure on the right shows a lefthand circular polarization plane.
x,z
y y x,z As the wave is righthand 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 copolarized component of the
ˆ
timeaveraged 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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 Spring '11
 Kudeki

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