COLUMBIA UNIVERSITY Department of Electrical Engineering EE E4703:
Wireless Communications (Spring 2005)
Solution for Homework Assignment No. 2
Problem 2
n
Find the directivity D = D(n) of an antenna pattern described by dP/d W = cos q,
forward only (i.e., for 0 £ q £ p/2 but for all azimuths 0 £ j < 2p). The answer should
be given as a function of the integer n .
dP
= cos n q
dW
We’ll find the directivity function D(q ,j ) and then maximize it to find the directivity
D.
D(q ,j ) =
4p dP
P dW
p
p
2p
� dP �
n
n
P = �� �
�dW = � � cos q sin qd qdj = 2p � cos q sin q d q
dW
Ł
ł
q =0 j =0
q=0
2
2
u = cosq
du = - sin qd q
1
P = 2p � u n du = 2p
u =0
1
u n+1
2p
=
n + 1 u= 0 n + 1
4p dP
= 2 (n + 1)cos n q
P dW
D(n ) = max D (q , j ) = 2(n + 1)
D(q ,j ) =
The directivity of the antenna is D(n)=2(n+1).
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Problem 3
Suppose we succeeded in generating a sectoral radiation pattern for a wireless cell
that were described as follows. It is uniform (i.e., a constant) in all directions within
both the horizontal sector 0 � < j < 120� and in the range 80 � < q < 90 � of elevation
from the vertical; it is zero for all directions outside these intervals. What is the
directivity (in dB) of this pattern?
Let’s assume we have R w/rad2 radiation in the given sector.
We shall integrate it over the sector to find the emitted power.
First, convert the angles to radians:
0 < j < 2p
3
4p < q < p
9
2
dP
=R
dW , max
4p dP
D(q ,j ) =
P dW
p
2p
3
2p
� dP �
P = �� �
� cosq
�dW = � � R sin q dqdj = R �
3
Ł dW ł
q = 49p j =0
D = D(q ,j ), max =
2
4p
9
p
2
= 0 .3637 R
4p dP
4p
=
R = 34 .55 = 15 .38 dB
P dW , max 0 .3637 R
The directivity of this radiation pattern is 15.38 dB.
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Problem 4
The CTS satellite has an 11.7 GHz transmitter on board, which provides 200 mW to
its 19.3 dB gain antenna. What power (in watts) can be received by a ground station
antenna (a 3.66 m diameter parabolic reflector) with 50.4 dB gain? The satellite is in
synchronous orbit 36,941 km above the ground station.
f = 11.7Ghz
\l = c
f
Pt = 0.2w
= 0.0256 m
Gt = 19.3dB = 85.11
Gr = 50.4dB = 109648
r = 36,941km = 36.941 �10 6 m
For a satellite-to-ground problem we can apply the free space communications model
Prec = Ptr
Gt G r l2
(4pr )
2
= 5 .676 � 10 -15 w
The received power is 5.676·10 -15 watts.
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Problem 5
Rappaport, p. 168, Problem 4.10
hr = 2 m
q D , min = 6 . 261 rad
q i , max = 5 o
f = 900 Mhz
\ l = c = 0 . 333 m
f
qD =
q D , min
2 pD
2p 2 h t h r
4 ph t h r
h
=
=
= 75 . 473 t
l
l
ld
d
d
h t , min
= 75 . 473
= 6 . 261
d min
In order to use the ?i constraint, we must consider the geometry of the model:
ht
hr
?i
dt
?i
dr
hr
= tanq i = tan 5 o
dr
hr
dr =
= 22 .86 m
tan 5 o
h t , min
ht
ht
=
=
= tan 5 o = 0 .0875
d t d - d r d min - 22 .86
75 .473
ht ,min
d min
h t , min
d min - 22 .86
= 6 .261
= 0.0875
Solve the 2 equations for dmin and ht,min
d min = 443 .5m
ht , min = 36 .8 m
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Problem 6
Rappaport, p. 168, Problem 4.11
To completely cancel each other, the signals must have a 2p phase difference (or an
integral multiple of 2p).
2p
2p 2 hr ht
s=
= 2pn
l
l d
n = 1,2,3....
qD =
\d 0 =
2 hr ht
ln
The signal nulls will occur when d=d0, above.