This preview shows page 1. Sign up to view the full content.
Unformatted text preview: s still along our new z axis, therefore, we have
0
G(✓2 , 0 ) = G(✓2 45o , 2 ). This implies that
2
2 Pr = Sinc 4⇡ G(0o , 0o ) = 0 W. f) Now the short dipole is polarized along the y direction and our electrical ﬁeld is in xz plane.
ˆ
Due to the polarization mismatch, the dipole direction is perpendicular to the incident ﬁeld
polarization (crosspolarization), so Pr = 0 W.
2.
a) The wavenumber is k =
2⇡ m. q p
p
(1/ 2)2 + (1/ 2)2 = 1 rad . Then, the wavelength is simply
m = 2⇡
k = b) Considering that the wave is a righthand circularly polarized plane TEM wave, we are given
p
the real part of the electric ﬁeld phasor as Ey = y 2 and asked to ﬁnd the imaginary part.
ˆ
Since this is a TEM wave, we can draw the imaginary components and the propagation vector
as shown in the below ﬁgure where the possible unit vectors of the imaginary part of the electric
ﬁeld phasor are displayed in green and red. as 1 z k ˜
Exz x y ˜
Exz Imagine that we rotate the coordinate system in the above ﬁgure in a way that the propagation...
View
Full
Document
 Spring '11
 Kudeki

Click to edit the document details