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
z
p
2
=

0
.
0938(ˆ
x
z
)
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
.
5sin
2
✓
2
. Thus, we write
P
r
.
1326
⇥
1
.
414
2
4
⇡
⇥
1
.
2
45
o
.
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
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
W
.
f) Now the short dipole is polarized along the
ˆ
y
direction and our electrical ±eld is in xz plane.
Due to the polarization mismatch, the dipole direction is perpendicular to the incident ±eld
polarization (crosspolarization), so
P
r
W.
2.
a) The wavenumber is
k
=
q
(1
/
p
2)
2
+(1
/
p
2)
2
rad
m
. Then, the wavelength is simply
λ
=
2
⇡
k
=
2
⇡
m.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 Kudeki
 Polarization, Circular polarization

Click to edit the document details