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 timeaverage Poynting vector is

<
S
inc
>

=
1
2
10
2
120
⇡
= 0
.
1326
Watts
m
2
.
As
ˆ
k
=

ˆ
x
+ˆ
z
p
2
, the timeaverage 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
+ˆ
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 xz plane.
Due to the polarization mismatch, the dipole direction is perpendicular to the incident field
polarization (crosspolarization), so
P
r
= 0
W.
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 Spring '11
 Kudeki
 Polarization, Circular polarization

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