ECE594I, Prof. Brown, Fall Quarter 2009
1
HW#3 Solutions
1) To evaluate the feedhorn, start with F(
θ,φ
) = exp(10
θ
2
) ,
(a) Halfpower beam width defined by
]
)
2
/
(
10
exp[
5
.
0
2
β
−
=
where
β
is the full width at the
halfpower points.
Solving f or
β
, we get
2
/
1
]
10
/
)
5
.
0
ln(
[
*
2
−
= 0.53 rad = 30.2
o
.
(b) The pattern solid angle is
( )
,s
i
n
p
Fd
d
θ
φθ
φ
Ω=
∫∫
But this is not integrable in
θ
(try mathematically to verify this using a symbolic integration
code, such as Mathmatica or Maple).
So we take advantage of fact that the pattern given has but
a single main lobe, and approximate by spherical trigonometry
Y
X
P
≈
Ω
.
From part (a)
β
=
β
X
=
β
Y
= 0.527, so
Ω
P
= 0.277 steradians.
But a better way to estimate
Ω
P
(as discussed in
lecture) is possible since the beam is symmetric
∫∫
⋅
=
π
ρπ
2
0
2
/
0
2
00
sin
sin
)
,
(
d
d
d
d
F
where F(
θ,φ
) is approximated as unity over the cone full angle
β
.
This so called “icecream
cone” approximation leads to
Ω
P
≈
2
/
0
cos
2
−
= 2
π
(1cos
β
/2) = 0.217 steradian.
(c) The antenna directivity is:
D = 4
π
/
Ω
P
≈
4
π
/
0.277 = 45.4
(according to pencil beam approximation)
D = 4
π
/
0.217 = 57.9
(according to “icecreamcone” approximation)
(d) S
max
at range of 10 cm is just P
rad
/(4
π
r
2
)
⋅
D
= 4.6x10
5
W/cm
2
in “icecreamcone”
approximation.
2) Dish with circular beamwidth of 1.5
o
.
(a) For pencilbeam case, D
≈
4
π
/(
β
x
β
y).
For the dish,
β
= 1.5*
π
/180
o
= 0.026 rad.
So, D
≈
4
π
/(0.026)
2
= 1.84x10
4
or D = 10log10[1.84x10
4
] = 42.6 dB
(b)
If dish area is doubled, we expect the directivity to also double consistent with
D
max
= 4
π
A/
λ
2
,
and the beamwidth should decrease consistent with 4
π
/(
β
X
β
Y
) = 4
π
A/
λ
2
.
So new directivity
4
3.66 10
45.6
Dd
B
=×
⇒
.
And the new beamwidth
β
= 1.5
o
/(2)
1/2
= 1.06
o
= 0.018 rad.
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 Spring '09
 O
 Dipole antenna, radiation pattern, Beamwidth

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