Improving Capacity
• Sectoring
• Cell Splitting
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Sectoring
• One way to increase capacity is to use
directional antennas rather than
omnidirectional antennas
• Most common choices are 120 degree and
60 degree sectoring
3 6
5 1
1
2 4 2
3
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120 Degree Sectoring
• For 120 deg sectoring, only two first-tier
9/9/98cochannel interferers instead of six! 3
60 Degree Sectoring
• For 60 deg sectoring, only one first-tier
9/9/98cochannel interferer instead of six! 4
Estimated C/I
• For a simple comparison, use the
approximation:
C
=
n
3N
I i0
• For omnidirectional antennas, i0=6
• For 120 deg sectored antennas, i0=2
• For 60 deg sectored antennas, i0=1
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Estimated C/I (2)
• For omnidirectional antennas, C/I=18 dB
requires N to be at least 7 for n=4
• For 120 deg sectored antennas, the same
performance can be achieved with N=4.
N C/I omni (dB) C/I sectored
(dB)
4 13.8 18.6
7 18.7 23.4
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Effects on System Capacity
• Reducing the cluster size increases the
system capacity by improving frequency
reuse
• However, sectoring reduces the trunking
efficiency
• Sectoring is more expensive since
additional antennas and transmitters are
required
• Bottom line: net win in capacity, but costs
more
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Example
• A particular system has 28 cells and a total
of 56 channels. Assuming a path loss
exponent of n=4, a required GOS of 2%,
and blocked calls cleared, find:
• (a) the maximum system traffic intensity for
N=7 and omnidirectional antennas
• (b) the maximum system traffic intensity for
N=4 and 120 degree sectored antennas
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Cell Splitting
• As the number of subscribers increases, the
capacity of some cells may be exceeded
• Desired GOS can be restored by
subdividing the cells
• Care must be taken to properly assign
channel sets to preserve the frequency reuse
scheme
• Power can be reduced with the same
received signal levels in smaller cells
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Cell Splitting
D E
G
E
F B
F F
D A
C C
G
E B C
B G
F
• Example: Radius of new cell is half that of old
• Assign new channel sets exactly halfway between existing
9/9/98channel sets 10
Reduce Transmitter Power
• Performance will be preserved if the power
at the new cell boundary is the same as that
at the old cell boundary
= Pt 2 ( R / 2)
−n −n
Pt1 R
for n = 4,
Pt1
Pt 2 =
16
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What if all cells are not split?
• If all cells use Pt1, smaller cells will
experience excess cochannel interference
• If all cells use Pt2, all portions of the larger
cells may not be covered
• Solution: Divide into two channel groups:
one for small cells and one for large
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