11. Cellular Planning and Capacity
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| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Cellular Planning and Capacity
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
Cellular Planning and Capacity
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Cellular Planning
We want
• complete area coverage
• no dead spots
For economic reasons, hexagons are normally assumed
Use
polygons
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
For economic reasons, hexagons are normally assumed (result in more efficient design than squares or triangles)
Rx 3=
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Frequency Planning and Re-use
• An operator is allocated a certain bandwidth allocation.
• Bandwidth is sub-divided into sub-bands.
– Exact number of sub-bands is determined by
• the interference tolerate of the radio system and
• the type of basestation antenna arrangement employed.
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• the type of basestation antenna arrangement employed.
• The number of sub-bands is referred to as the cluster size
– Cells are planned in a cluster using entire frequency allocation.
– The previous figure assumed a cluster size of four
• Total frequency allocated is divided into four sub-bands and assigned to the individual cells.
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
Hexagonal Axes and Co-ordinate System
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Normalised Distance to cell centre
• The most convenient co-ordinate system for a hexagonal cellular structure are axes inclined at a 60 degree angle.
• Assuming origin is at (0,0) and restricting u and v to integer values (i,j), the distance to the cell centre is:
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
values (i,j), the distance to the cell centre is:
• The normalised distance between any adjacent cell sites is unity (i=1, j=0 or i=0, j=1).
( ) ijjiDn −+=2
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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To find the nearest co-channel cell:
• Move perpendicular to any of
the six surfaces of the original
cell, pass i=4 cells, rotate 60
degrees counter-clockwise,
move along j=2 cells. (red)
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
move along j=2 cells. (red)
• Move perpendicular to any of
the six surfaces of the original
cell, pass j=2 cells, rotate 60
degrees clockwise, move along
i=4 cells. (blue)
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Cellular Geometry and Area
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Cluster Size• Cells designated A are the six nearest cochannel cells of the
centre cell. They are at the vertices of the larger hexagonal cell of radius D. The radius of the larger cell, D, is given by:
( )ijjiRD ++=2222
3
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• Since the area of a hexagon is proportional to the square of its radius, the area of the larger hexagon is given by:
where k represents a fixed constant
( )ijjiRD ++= 3
( )( )ijjiRkA el ++=222
arg 3
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Cluster Size
• Similarly, the area of the small hexagon is given by:
( )2RkAsmall =
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• Therefore, the ratio of the large hexagonal area to the small hexagonal area is given by:
( )small
( )ijjiAA smallel ++=22
arg 3
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Cluster Size
• The larger hexagon has total of 3N enclosed cells.
– the N shaded centre cells (N = 7)
– one third of the six surrounding N cell clusters.
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• Since the enclosed area is proportional to the number of cells we can write:
( )( ) ( )( ) ( )ijjiNhenceijjiRkNkA
RkkhencekA
el
small
++=++==
==
222222arg
222
33.
1.
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Cluster Size
• Since i and j are restricted to integer values, cluster size N is restricted to limited set of discontinuous values. It is not possible to plan a system with N = 2, 5, 6, 8 … etc.
• Combining the above equation for D and N we can show that:
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• Combining the above equation for D and N we can show that:
• This relates the cell radius and reuse distance to the cluster size. The term D/R is referred to as the cochannel reuse ratio.
NR
D 3=
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Cluster size as a function of i and j
i j N
1 0 1
1 1 3
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
2 0 4
2 1 7
3 0 9
3 1 13
4 0 16
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Cluster size as
a function of
Co-channel
Interference
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Cluster size as a function of Cochannel Interference
• System can be modelled based on the wanted signal and the interference from the surrounding ‘first tier’ cochannel cells.
• Worst case scenario: user on cell boundary (distance R from basestation). The first tier interfering cells lie at distance D-R.
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• The signal power from the centre basestation is inversely proportional to the n-th power of the distance– n = propagation path loss index (n = 2 for LOS communications):
nR
S1
α
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Cluster size as a function of Cochannel Interference
• The interference power from each of the first tier cells is inversely proportional to the n-th power of the distance (D-R):
nD
I1
1α
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• Since there are six interfering cells (assuming omni antenna), actual interference is six times that of 1. Resulting S/I ratio is:
nD
( )( )[ ] n
n
ISRDhenceRD
I
S 16
6==
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Cluster size as a function of Cochannel Interference
( )[ ] nISN2
31 6=
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
Thus, N can be found from the S/I ratio needed by the system.
– GSM and DCS-1800: an S/I of 9 dB is required• hence a reuse cluster size of 4.39 (path loss exponent assumed to be 3).
– TACS: an S/I of 17 dB is required• hence a reuse cluster size of 14.9.
– Based on this alone, GSM appears to offer over three times the capacity of TACS. In practice many other factors can affect system capacity.
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Capacity Evaluation
Capacity = Function [total bandwidth, channel bandwidth, cluster size, cell area]
• Channel bandwidth = average bandwidth needed to support a single communications channel.
– For TDMA systems, divide carrier bandwidth by number of user slots.
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• Cluster size is a function of C/I tolerance and sectorisation strategy.
• Since capacity is often quoted ‘per km squared’, the average cell radius is also used in the formulation.
2// kmMHzErlangs
bandwidthTotalareaCellsizeCluster
channelsofNumberTotalCapacity
××=
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Capacity Evaluation
• Channel bandwidth varies depending on technology.
– TACS , AMPS (FM based) used ≈ 25 – 30 kHz per voice channel.
– GSM support 8 or 16 voice channels in 200 kHz band � average channel bandwidths of ≈ 25 kHz and 12.5 kHz.
– In newer analogue systems, bandwidth reduced to 12.5 kHz.
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
– There is interest in reducing the required bandwidth still further.
• Channel bandwidth is reduced at expense of C/I performance. Total number of calls is directly related to channel bandwidth. However, more bandwidth efficient schemes often need larger cluster sizes and reuse distances due to their greater sensitivity to cochannel interference.
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Capacity Evaluation• Capacity is also a function of cell area.
– higher capacity can be obtained with smaller cell sizes. These
microcells aim to offer high subscriber densities. They offer high
capacity (in terms of subscribers / MHz / km square), but many
basestations are needed for continuous wide coverage.
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
basestations are needed for continuous wide coverage.
• To determine number of subscribers that can be supported,
average traffic density per user can be included in equation.
• Average subscriber creates ≈ 0.02 Erlangs of traffic. Hence:
2// kmMHzssubscriberusersErlangsbandwidthTotalareaCellsizeCluster
channelsofNumberTotalCapacity
×××=
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Example
A digital cellular system has a total bandwidth of 15 MHz and operates with a re-use factor of 4 and a cell radius of 2 km. The network comprises 20 basestations and each 200 kHz carrier is made up of 8 TDMA voice channels. Assume that each user represents a traffic load of 0.02 Erlangs.
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
each user represents a traffic load of 0.02 Erlangs.
– Calculate the total subscriber capacity in each cell.
– Next determine the total capacity in terms of subscribers per MHz per kilometre square.
– Finally, what is the total subscriber capacity of the network?
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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ExamplePart 1:
• Calls per cell = ( 15 / ( ( 0.200 / 8 ) × 4) = 150
• Subscribers per cell = 150 / 0.02 = 7500
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
Part 2:
• Assuming 200kHz call bandwidth, 15MHz supports 600 calls.
• Channels are used over cluster area 4 × 12.6 = 50.4
• � Total capacity = 600 / (15 × 50.4) = 0.794 calls / MHz /
• Subscriber capacity = (0.794/0.02) = 39.68 subscribers/MHz/
2km
2km
2km
2km
| University of Dar es Salaam | Department of Electronics and Telecommunications Engineering |
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Example
Part 3:
• Total subscriber capacity = 7500 × 20 = 150,000
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
Or = 39.68 × 20 × 12.6 × 15 = 150,000.
• Total coverage area = 20 × 12.6 = 252 km2.
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Approaches To Cope With Increasing Capacity
• Adding new channels
• Frequency borrowing – Frequencies are taken from adjacent cells by congested cells
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 26/05/2014 |
• Microcells – Antennas move to buildings, hills, and lamp posts
• Cell Splitting – Allows an orderly growth of the cellular system.
– Increases the number of base stations in order to increase capacity.
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Approaches To Cope With Increasing Capacity
• Sectoring
– Uses directional antennas to further control the interference and
frequency reuse of channels.
– Rely on base station antenna placements to improve capacity
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
– Rely on base station antenna placements to improve capacity
by reducing co-channel interference.
• Coverage zone approaches
– Distributes the coverage of a cell and extends the cell boundary
to hard-to-reach places.
– Rely on base station antenna placements to improve capacity
by reducing co-channel interference.
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Microcellular Systems
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
Microcellular Systems
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Why Microcellular Systems
We want
We have
• Limited radio spectrum allocation
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
We want
• coverage
• capacityCellular
We want
• more capacity Microcellular
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Microcellular Systems Defined
• Not formally defined.
• Generally, small cells < 1 km.
• Urban area.
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• Urban area.
• Base station antennas significantly lower than surrounding roof tops.
• Coverage area largely governed by effects of surrounding buildings (urban shielding).
• Lower transmit power at base station.
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Microcellular Systems
The lower transmit power, combined with theeffects of urban shielding, results in a shorterreuse distance, hence cell frequencies can be
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
reuse distance, hence cell frequencies can bereused more often in a given area. A high densityof reuse cells will naturally increase systemcapacity.
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Microcellular Systems Propagation Models
• Propagation models should take into account the clutter introduced by individual buildings.
• The capacity can be deterministically evaluated by
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• The capacity can be deterministically evaluated by studying the effective coverage area of each individual cell.
• The propagation coverage for each cell site can be obtained either statistically (by fitting curves to measured results) or deterministically by methods such as ray-tracing.
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Microcell Capacity Evaluation
• First: estimate average cell coverage area.
– Assume system is noise limited, use measurements or
models to determine how the average power varies with
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
models to determine how the average power varies with
distance.
– Average cell radius = average distance from the basestation
where the required signal power and outage probabilities can
be maintained.
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Microcell Capacity Evaluation
• For a cochannel limited system, it is important to design the network to have the minimum practical distance between frequencies.
• This minimum distance depends on many factors, e.g.
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• This minimum distance depends on many factors, e.g.
• the number of cochannel cells in the vicinity of the centre cell,
• the type of geographic terrain,
• the antenna height,
• the antenna pattern,
• the sectorisation and
• the transmit power at each base site.
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Determination of Average Reuse Distance
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Determination of Average Reuse Distance
• A wanted cell receives interference from a cochannel cell.
• With reference to the required signal to interference ratio,six cochannel cells can be assumed and then movedcloser towards the centre cell as shown in the previousfigure.
• The diagram shows the received signal level relative to
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• The diagram shows the received signal level relative tothe received cochannel interference level for a givencochannel separation distance.
• As the cochannel cells are moved closer, there will comea point where the interference ratio and outageprobabilities are no longer met. This point represents theminimum practical reuse distance.
• Based on knowledge of the cell size, R, and the reusedistance, D, the cluster size can be determined.
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Case Study – Microcellular study implemented
Microcellular study
based on the
building database
for a typical UK
urban environment
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
On the left is a map
of the microcellular
area - 500 × 500 m.
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Case Study – Microcellular study implemented
Antenna height = 10 m, vertical dipole is assumed.
To study coverage, transmitter placed at centre (x=225m, y=305m).
Using suitable propagation
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
Using suitable propagation model, coverage for surrounding area is found as shown to the left.
o = base stationwhite = good signal levelblack = poor signal level
Signal level better in one direction.
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Case Study – Microcellular study implemented
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
From the previous figure, the average signal propagation
pathloss is derived as a function of distance.
Average Street Propagation Pathloss (f = 1800MHz)
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Case Study – Microcellular study implemented
When minimum received signal level = -70dBm (10% outage),
evaluated average cell radius = 160m
average reuse distance = 600m (for certain C/I threshold)
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
average reuse distance = 600m (for certain C/I threshold)
mobile receiver height = 1.5m
mobile antenna pattern = omni
cluster size = 4.68 (7)
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| University of Dar es Salaam | Department of Telecommunications Engineering |
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Example
A digital microcellular system has a total bandwidth of 15 MHz and operates with a re-use factor of 7 and a cell radius of 160m. The network comprises 20 basestations and each 200 kHz carrier is made up of 8 TDMA voice channels. Assume that each user represents a traffic load of 0.02 Erlangs and
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
that each user represents a traffic load of 0.02 Erlangs and that the average reuse distance is 600m.
Calculate the total call and subscriber capacity in each cell and for the entire network.
| University of Dar es Salaam | Department of Telecommunications Engineering |
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Soln:
• Calls per cell = ( 15 / ( 7 × 0.2 ) ) × 8 = 85.71
• Subscribers per cell = 4,285.7
• Assuming 200kHz call bandwidth, 15MHz supports 600 calls.
• Channels are used over cluster area of 7× 0.0804 km2 = 0.563 km2
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
• Channels are used over cluster area of 7× 0.0804 km2 = 0.563 km2
• Thus, total capacity = 600 / (15 × 0.563) = 71.04 calls / MHz / km2
• Subscriber capacity = (71.04 / 0.02) = 3,552 subscribers/MHz/km2
• Total subscriber capacity = 4,285.7 × 20 = 85,714
• Or = 3,552 × 20 × 0.0804 × 15 = 85,714
• Total coverage area = 20 × 0.0256 = 1.608 km2
| University of Dar es Salaam | Department of Telecommunications Engineering |
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• Example illustrates how future microcellular systems will achieve
very high traffic densities over very small coverage areas.
– System in example supports > 85,000 subscribers in 1.6 km2 area (system has capacity > 3,500 subscribers/MHz/km2).
– In contrast, the traditional cellular systems example achieved a capacity of 39.68 subscribers/Mhz/km2. However, the system offered a larger
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| TE 412 Introduction to Wireless Communications | Christine Mwase | 10/03/2014 |
of 39.68 subscribers/Mhz/km . However, the system offered a larger coverage area (252km2).
– In practice, the choice of cell size and capacity must be accurately matched to the number of users in a given area.
– Future mobile communication systems are likely to employ a mixture of small cells in the cities and large cells for more rural areas. Large cells are also more appropriate for vehicular use to reduce the number of handovers expected.