Large Scale Fading and Network Deployment.pdf

Lecture 5
Large Scale Fading and Network Deployment

+
Large Scale Fading
n “Large” scale variation of signal
strength with distance
n Consider average signal
strength values
n The average is computed
either over short periods of
time or short lengths of
distance
n A straight line is fit to the
average values
n The slope and the intercept give
you the expression for the path
loss
n The variation around the fit is
the shadow fading component
Received
signal
strength
Log distance
Slope &
Intercept
Variation
2

+
Path Loss Models
n Path Loss Models are commonly used to estimate link budgets, cell
sizes and shapes, capacity, handoff criteria etc.
n “Macroscopic” or “large scale” variation of RSS
n Path loss = loss in signal strength as a function of distance
n Terrain dependent (urban, rural, mountainous), ground reflection, diffraction,
etc.
n Site dependent (antenna heights for example)
n Frequency dependent
n Line of sight or not
n Simple characterization: PL = L0 + 10a log10(d)
n L0 is termed the frequency dependent component
n The parameter a is called the “path loss gradient” or exponent
n The value of a determines how quickly the RSS falls with distance
3

+
The Free Space Loss
n Assumption
n Transmitter and receiver are in free space
n No obstructing objects in between
n The earth is at an infinite distance!
n The transmitted power is Pt, and the received power is Pr
n The path loss is Lp = Pt (dB) – Pr (dB)
n Isotropic antennas
n Antennas radiate and receive equally in all directions with unit gain
d
4

+
The Free Space Model
n The relationship between Pt and Pr is given by
Pr = Pt l2/(4pd)2
n The wavelength of the carrier is l = c/f
n In dB
Pr (dBm)= Pt (dBm) - 21.98 + 20 log10(l) – 20 log10(d)
Lp(d) = Pt – Pr = 21.98 – 20 log10(l) + 20 log10(d)
= L0 + 20 log10(d)
n L0 is called the path loss at the first meter (put d = 1)
n We say there is a 20 dB per decade loss in signal strength
5

+
A simple explanation of free space
loss
n Isotropic transmit antenna: Radiates signal
equally in all directions
n Assume a point source
n At a distance d from the transmitter, the
area of the sphere enclosing the Tx is: A =
4pd2
n The “power density” on this sphere is: Pt/
4pd2
n Isotropic receive antenna: Captures power
equal to the density times the area of the
antenna
n Ideal area of antenna is
Aant = l2/4p
n The received power is:
Pr = Pt/ 4pd2 ´ l2/4p = Pt l2/(4pd)2
6

+
Isotropic and Real Antennas
n Isotropic antennas are “ideal” and cannot be achieved in
practice
n Useful as a theoretical benchmark
n Real antennas have gains in different directions
n Suppose the gain of the transmit antenna in the direction of
interest is Gt and that of the receive antenna is Gr
n The free space relation is:
Pr = Pt Gt Gr l2/(4pd)2
n The quantity Pt Gt is called the effective isotropic radiated
power (EIRP)
n This is the transmit power that a transmitter should use
were it having an isotropic antenna
7

+
Summary: Free space loss
n Transmit power Pt and received power Pr
n Wavelength of the RF carrier l = c/f
n Over a distance d the relationship between Pt and Pr is given by:
n where d is in meters
2
2
2
)
4
( d
P
P t
r
p
l
=
In dB, we have:
Pr (dBm)= Pt (dBm) - 21.98 + 20 log10 (l) – 20 log10 (d)
Path Loss = Lp = Pt – Pr = 21.98 - 20log10(l) + 20log10 (d)
8

+
Free Space Propagation
n Notice that factor of 10 increase in distance
n => 20 dB increase in path loss (20 dB/decade)
n Note that higher the frequency the greater the path
loss for a fixed distance
Distance Path Loss at 880 MHz
1 km 91.29 dB
10 km 111.29 dB
Distance 880 MHz 1960 MHz
1 km 91.29 dB 98.25 dB
7 dB greater path loss for PCS band compared
to cellular band in the US
9

+
Example
n Consider Design of a Point-to-Point
link connecting LANs in separate
buildings across a freeway
n Distance .25 mile
n Line of Sight (LOS)
communication
n Unlicensed spectrum – 802.11b
at 2.4GHz
n Maximum transmit power of
802.11 AP is Pt = 24 dBm
n The minimum received signal
strength (RSS) for 11 Mbps
operation is -80 dBm
n Will the signal strength be
adequate for communication?
n Given LOS
n Can approximate propagation
with Free Space Model
10

+
Example (Continued)
n Example
n Distance .25 mile ~ 400m; Receiver Sensitivity Threshold = - 80dBm
n The Received Power Pr is given by: Pr = Pt - Path Loss
Pr = Pt - 21.98 + 20 log10 (l) – 20 log10 (d)
= 24 – 21.98 + 20 log10 (3x108/2.4x109) – 20 log10 (400)
= 24 -21.98 -18.06 -52.04
= 24 – 92.08 = -68.08 dBm
Pr is well above the required -80 dBm for communication at the
maximum data rate – so link should work fine
L0 = 40 dB
at 2.4 GHz
11

+
Cell/Radio Footprint
n The Cell is the area covered by a
single transmitter
n Path loss model roughly
determines the size of cell
n What does “covered” mean?
12

+
Link Budget
n Typical Factors in Link Budget
n Transmit Power (in dBm),
n Antenna Gain, Diversity Gain
n Receiver Sensitivity
n Margins
n Shadow Margin, Interference Margin, Fading Margin
n Losses
n Vehicle Penetration Loss (3-6 dB)
n Body Loss (2-3 dB)
n Building Penetration Loss (5-20 dB depending on building
material
n Electronic Losses: Combiner Loss, Filter Loss, etc.
n Gains are added, Losses are subtracted (e.g.,f = 1900 MHz)
13

+
Example of Link Budget
Link Uplink Downlink
Transmit power 30 dBm 30 dBm
Antenna gain 3 dBi 5 dBi
Diversity gain 5 dB 0 dB
Shadow margin 10 dB 10 dB
Body penetration 2 dB 2 dB
Vehicle penetration 5 dB 5 dB
Receiver sensitivity -105 dBm -90 dBm
Path Loss Budget 126 dB 108 dB
Typical Cellular System is Downlink Limited!
14

+
Calculation of link Budget:
Uplink
Transmit Power 30 dBm
Diversity Gain 5 dBi
33 dBm
38 dBm
Shadow
Margin 10 dB
28 dBm
Body Loss 2 dB
Vehicle Loss 5 dB 21 dBm
Receiver Sensitivity -105 dBm
Path Loss Budget = 126 dB
Antenna Gain 3 dBi
126 dB
15

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Determining Coverage
n Link Budget
n Used to plan useful coverage of cells
n Roundtrip performance of satellites,
etc.
n Simply a balance sheet of all gains
and losses on a transmission
path.
n Gains are added (transmit power,
antenna gains)
n Losses are subtracted (path loss)
n Used to find max allowable path
loss in each link (i.e., uplink and
downlink)
n Ensure adequate RSS at end of each
link
n Simple Example
n The path loss budget is 108 dB
n The path loss model is given by
Lp = 98 + 32 log10d
(d is in km)
n The cell radius should be
98 + 32 log10d = 108 => log10d = 10
d = 10(10/32) = 2.05 km
16

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General Formulation of Path Loss
n Depending on the environment, it is seen that the path
loss (or the RSS) varies as some power of the distance
from the transmitter d
n Here a is called the path-loss exponent or the path-loss
gradient or the distance-power gradient
n The quantity L0 is a constant that is computed at a
reference distance d0
n This reference distance is 1m in indoor areas and 100m or 1
km in outdoor areas
OR
Pr(d) /
✓
Pt
d
◆
Pr(d) =
✓
Pt
L0(d/d0)
◆
17

+
More Comments
n Path loss is a function of a variety of parameters
n Terrain
n Frequency of operation
n Antenna heights
n Extremely site specific
n Varies depending on environment
n Example: indoor Vs outdoor
n Example: microcell Vs macrocell
n Example: rural Vs dense urban
n Large number of measurement results are available for
different scenarios, frequencies and sites
n Empirical models are popular
18

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Environment Based Path Loss
n Basic characterization: Lp = L0 + 10a log10(d)
n L0 is frequency dependent component (often path loss at 1m)
n The parameter a is called the “path loss gradient” or exponent
n The value of a determines how quickly the RSS falls with d
n a determined by measurements in typical environment
n For example
n a = 2.5 might be used for rural area
n a = 4.8 might be used for dense urban area (downtown Pittsburgh)
n Variations on this approach
n Try and add more terms to the model
n Directly curve fit data
n Two popular measurement based models are Okumura-Hata, and COST 231
n Do some measurements and feed it into simulations (ray tracing)
19

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Okumura-Hata Model
n Okumura collected measurement data and plotted a set of curves for path
loss in urban areas around 900 MHz
n Hata came up with an empirical model for Okumura’s curves
Lp = 69.55 + 26.16 log fc – 13.82 log hte – a(hre) + (44.9 –6.55 log hte)log d
a(hre) = 3.2 (log [11.75 hre])2 – 4.97 dB
n Note: fc is in MHz, d is in km, and antenna heights are in meters
n This is valid only for 400 £ fc £ 1500 MHz for a large city
n 30 £ hte £ 200 m; 1 £ hre £ 10 m;
n Other forms depending on the scenario
20

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Example of Hata’s Model
n Consider the parameters
n hre = 2 m – receiver antenna’s height
n hte = 100 m – transmitter antenna’s height
n fc = 900 MHz – carrier frequency
n Lp = 118.14 + 31.8 log d
n The path loss exponent for this particular case is a = 3.18
n What is the path loss at d = 5 km?
n d = 5 km è Lp = 118.14 + 31.8 log 5 = 140.36 dB
n If the maximum allowed path loss is 120 dB, what distance can the
signal travel?
n Lp = 120 = 118.14 + 31.8 log d => d = 10(1.86/31.8) = 1.14 km
21

+
COST 231 Model
n Models developed by COST
n European Cooperative for Science and Technology
n Collected measurement data
n Plotted a set of curves for path loss in various areas around the 1900
MHz band
n Developed a Hata-like model
Lp = 46.3 + 33.9 log fc – 13.82 log hte - a(hre) + (44.9 –6.55 log hte)log d + C
n C is a correction factor
n C = 0 dB in dense urban; -5 dB in urban; -10 dB in suburban; -17 dB in
rural
n Note: fc is in MHz (between 1500 and 2000 MHz), d is in km, hte is
effective base station antenna height in meters (between 30 and 200m),
hre is mobile antenna height (between 1 and 10m)
22

+
Indoor Path Loss Models
n Indoor applications
n Wireless PBXs
n Wireless Local Area Networks
n Approach is similar to outdoor models
n Distances are smaller
n Site specificity is more important
n Variety of obstructions
n Walls, floors, vending machines, bookcases, human beings etc.
23

+
Motley-Keenan and Rappaport Models
n Assume that the path loss exponent a = 2
n Draw a straight line between the transmitter and receiver
n Assign a loss of some dB to each obstruction that is intersected by this straight line
n Example: Concrete wall 7 dB, Cubicle partition 4 dB
n The path loss is given by:
n mi is the number of partitions of type i and Wi is the loss associated with that partition
n nj is the number of floors of type j and Fj is the loss associated with that floor
n L0 is determined as before (the path loss at one meter)
∑
∑ +
+
+
=
j
j
j
i
i
i
p F
n
W
m
d
L
L log
20
0
24

+
Sample numbers
Signal attenuation of 2.4 GHz through dB
Window in brick wall 2
Metal frame, glass wall into building 6
Office wall 6
Metal door in office wall 6
Cinder wall 4
Metal door in brick wall 12.4
Brick wall next to metal door 3
Source: Harris Semiconductors
25

Example of Partition Dependent Model
n Example:
n The straight line intersects
two brick walls and one
cubicle partition
n Lp = L0 + 20 log d + 2Wbrick +
Wcubicle
n In some models, the path
loss exponent a is different
from 2
TX
RX
d
Brick
Cubicle
Brick
26

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Some Notes
n Empirical models have their disadvantages
n Example: Okumura-Hata model applies to cities that are like Tokyo (what does
that mean? When is a city like Tokyo?)
n Depends on the interpretation of people
n Some people may consider Pittsburgh to be a small city
n Others may think of it as a medium city
n Some models have limited applicability
n Example: COST-231 model cannot be used if hte < hroof where hroof is the average
height of buildings in the area
n There are many other models
n Models for microcellular environments
n Terrain dependent (e.g., Longley-Rice)
27

+
Shadow Fading
n Shadowing occurs when line of sight is blocked - Modeled by a random signal
component Xs
n Measurement studies show that Xs can be modeled with a lognormal distribution è
normal in dB with mean = zero and standard deviation s dB
n Thus at the “designed cell edge” only 50% of the locations have adequate RSS
n Since Xs can be modeled in dB as normally distributed with mean = zero and standard
deviation s dB, s determines the behavior
Pr = Pt – Lp +Xs
28

+
How shadow fading affects system
design
n Typical values for σ are
n Rural 3 dB, suburban 6 dB, urban 8 dB, dense urban 10 dB
n Since X is normal in dB Pr is normal
n Pr = Pt – Lp +Xσ
n Prob {Pr (d) > Threshold } can be found from a normal
distribution table with mean Pr and standard deviation σ
n In order to make at least Y% of the locations have adequate
RSS
n Reduce cell size
n Increase transmit power
n Make the receiver more sensitive
29

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Example of Shadowing
Calculations
n The path loss of a system is given by Lp = 47 + 40 log10 d – 20 log10 hb
where hb = 10m, Pt = 0.5 W, receiver sensitivity = -100 dBm. What is
the cell radius?
n Pt = 10 log10500 = 27 dBm; The permissible path loss is 27-(-100) = 127 dB
n 20 log10hb = 20 log1010 = 20 dB
n 127 = 47 + 40 log10d – 20 => d = 316m
n But the real path loss at any location is
n 127 + X where X is a random variable representing shadowing
n Negative X = better RSS; Positive X = worse RSS
n If the shadow fading component is normally distributed with mean zero and
standard deviation of 6 dB. What should be the shadow margin to have
acceptable RSS in 90% of the locations at the cell edge?
30

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Example again
n Let X be the shadow fading component
n X = N(0,6) and we need to find F such that P{X > F } = 0.1 or we need to
solve Q(F/s) = 0.1
n Use tables or software
n In this example F = 7.69 dB
n Increase transmit power to 27 + 7.69 = 34.69 dBm = 3 W
n Make the receiver sensitivity -107.69 dBm
n Reduce the cell size to 203.1 m
n In practice use .9 or .95 quantile values to determine the Shadow
Fading Margin
Fading Margin is the amount of
extra path loss added to the path
loss budget to account for
shadowing
.9 à SFM = 1.282s
.95 à SFM = 1.654s
31

+
Cell Coverage modeling
n Simple path loss model based on
environment used as first cut for
planning cell locations
n Refine with measurements to
parameterize model
n Alternately use ray tracing:
approximate the radio
propagation by means of
geometrical optics- consider line
of sight path, reflection effects,
diffraction etc.
n CAD deployment tools widely used
to provide prediction of coverage
and plan/tune the network
32

+
Cellular CADTools
n Use GIS terrain data base, along with vehicle
traffic/population density overlays and propagation models
n Output map with cell coverage at various signal levels and
interference values
n To plan out cell coverage area, cell placement, handoff areas,
interference level frequency assignment
33

+
Use GIS maps
nThis shows
possible
location of cell
site and
possible
location of
users where
signal strength
prediction is
desired
34

+
Outdoor Model
CAD Tools
provide a
variety of
propagation
models: free
space,
Okumura-Hata,
etc.
35

+ Typical City pattern
Microcell diamond
Radiation pattern
36

+
Cellular CADTools
nCAD tool – first cut cell site placement,
augmented by extensive measurements to
refine model and tune location and antenna
placement/type
Temporary cell
39

+
Signal strength prediction for
IndoorWLANS
n Motorola LAN Planner
n Lucent: WiSE tool
n Given building/space to be
covered and parameters of
building and AP – predicts signal
coverage
40

+
Site SurveyTools
n Software to measure signal strength and recording in order to construct a
coverage map of structure – must drive/walk around structure to gather
data
n NOKIA site survey tool, Ekahau Site Survey, Motorola LAN survey, etc.
41

+
How about Interference?
nCoverage implies there is enough signal
strength
nBut how about competing signal strength from
a different base station?
nInterference has a significant impact on the
quality of a radio channel
nNext we look at interference and
frequency reuse
42

+
Basic Interference Scenario –
Two links
Pr =
KPt
d↵
= KPtd ↵
Sr =
Pr d
Pr I
=
KPt dR ↵
KPt ID ↵
=
Pt d
Pt I
✓
D
R
◆↵
43
K = const

+
Design and Deployment in Cellular
Networks
n Ad hoc networks
n Usually no architectural design
n Most design is at the protocol level – routing, MAC etc.
n Infrastructure networks
n Deploy a cellular topology based on some requirements
n Frequency reuse
n Start with large cells initially
n As demand increases
n Capacity enhancement techniques
n Reuse partitioning
n Sectored cells
n Migration to digital systems
n Dynamic channel allocation
44

+
Design Challenge
nHow can we reuse frequency bands such
that
nInterference is not so high as to make
communications impossible
nThe available spectrum is reused to make the
best use of capacity
45

+
Cellular Concept
n Proposed by Bell Labs in 1971
n Geographic Service divided into
smaller “cells”
n Neighboring cells do not use same
set of frequencies to prevent
interference
n Often approximate coverage area
of a cell by an idealized hexagon
n Increase system capacity by
frequency reuse
46

+
The Cellular Concept
n Deploy a large number of low-power transmitters
(Base Stations) each having a limited coverage area
n Reuse the spectrum several times in the area to be
covered to increase capacity
n Issues:
n Capacity (traffic load) in a cell
nOne measure = number of communication
channels that are available
n Performance
n Call blocking probability, handoff dropping probability,
throughput etc.
n Interference
47

+
Cellular Concept
n Why not a large radio tower and large service area?
n Number of simultaneous users would be very limited (to
total number of traffic channels T)
n Mobile handset would have greater power requirement
n Cellular concept - small cells with frequency reuse
n Advantages
n Lower power handsets
n Increases system capacity with frequency reuse
n Drawbacks:
n Cost of cells
n Handoffs between cells must be supported
n Need to track user to route incoming call/message
48

+
Recap: Communication Channel
n A frequency band allocated for voice or data
communications
n Simplest example: Frequency division multiple access (FDMA) with
Frequency Division Duplexing (FDD)
n 30 kHz bands are allocated for one conversation
n Separate bands are allocated for uplink (MH to BS) and downlink (
BS to MH)
n A set of time slots allocated for voice or data
communications
n A set of spread-spectrum codes allocated for voice or
data communications
49

+
Types of Interference
nTDMA/FDMA based systems
n Co-channel interference
n Interference from signals transmitted by another
cell using the same radio spectrum
n Adjacent channel interference
n Interference from signals transmitted in the same
cell with overlapping spectral sidelobes
nCDMA systems
n Interference from within the cell
n Interference from outside the cell
50

+
Clustering inTDMA/FDMA
n Adjacent cells CANNOT use the same channels
n Co-channel interference will be too severe
n The available spectrum is divided into chunks (sub-
bands) that are distributed among the cells
n Cells are grouped into clusters
n Each cluster of cells employ the entire available radio
spectrum
n The spatial allocation of sub-bands has to be done to
minimize interference
51

+
Cellular Concept (cont)
nLet T = total number of duplex channels
Nc cells = size of cell cluster (typically 4, 7, 9, 12, 21)
L = T/Nc = number of channels per cell
nFor a specific geographic area, if clusters are
replicated M times, then total number of
channels
n System capacity = M×T
n Choice of Nc determines distance between cells using the
same frequencies – termed co-channel cells
n Nc depends on how much interference can be tolerated by
mobile stations and path loss
52

+
Cell Design - Reuse Pattern
n Example: cell cluster size Nc = 7, frequency reuse factor =
1/7;
n Assume T = 490 total channels, L = T/Nc = 70 channels per
cell
B
A
E
C
D
G
F
B
A
E
C
D
G
F
B
A
E
C
D
G
F
Assume T = 490 total channels,
Nc = 7, N = 70 channels/cell
Clusters are replicated M=3
times
System capacity = 3 x 490 = 1470
total channels
53

+
Cellular Geometry
n Propagation models represent cell as a circular area
n Approximate cell coverage with a hexagon - allows easier analysis
n Frequency assignment of F MHz for the system
n The multiple access techniques translates F to T traffic channels
n Cluster of cells Nc = group of adjacent cells which use all of the systems frequency assignment
54

+
Cellular Geometry
nCells do not have a “nice” shape in reality
nA model is required for
n Planning the architecture
n Evaluating performance
n Predict future requirements
nSimple Model:
n All cells are identical
n There are no ambiguous areas
n There are no areas that are NOT covered by any cell
55

+
Possibilities for cell geometry
model
nEquilateral triangle, square or regular
hexagon
56

+
Why hexagon?
nAmong the three choices, the hexagon is
the closest approximation to a circle
nFor a given radius (largest possible distance
from center of a polygon to its edge) a
hexagon has the largest area
nA circle is sometimes used when
continuous distributions are being
considered
57

+
Determining co-channel cells and the
reuse factor
-1
-1
0,
1
2
3
1
2
3
4
u
v n Co-channel cells must be
placed as far apart as
possible for a given cluster
size
n Hexagonal geometry has
some properties that can be
employed to determine the
co-channel cell
n Co-ordinate system: u and v
co-ordinates
Cells are placed so that their
centers have integer co-ordinates
58

+
Finding (placing) Co-channel cells
(continued)
n Move a distance i along the u
direction and a distance j along
the v direction
n The cluster size Nc = i2 + ij + j2
,
A
u
A
A
A
A
A
A
59

+
C
D
B
A
E
G
F
C
D
B
A
E
G
F
C
D
B
A
E
G
F
C
D
B
A
E
G
F
C
D
B
A
E
G
F
C
D
B
A
E
G
F
C
D
B
A
E
G
F
Example:
i = 2, j = 1
Cluster size
Nc= 7
Used in
Advanced Mobile
Phone Service (AMPS)
60

+
More Examples
1
3
2
4
3 1
4
2
1
2
3
4
1
3
1
4
2
6 7
5
1
1
1
1
1
1
Nc = 4 (i =2, j=0)
Nc = 7 (i =2, j =1)
2
9
8
6
7
1
3
10
11
12
4
5
6
5
8
6
7
9
8
12
4
5
3
10
11
12
4
9
10
11
Nc = 12 (i=2, j=2)
61

+
Some results
n Nc = number of cells in a cluster
n R = radius of a cell
n D = distance between co-channel cells
n Nc can only take values that are of the form i2 + ij + j2 ;
i, j are integers
n There are exactly six co-channel cells for a hexagonal
geometry
D
R
=
p
3Nc
62

+
Revisiting Signal to interference
ratio calculation
nGeneral:
nOne desired signal and one interfering
signal at distances d1 and d2
Sr =
Pdesired
PInterference,i
i
∑
a
a
a
÷
÷
ø
ö
ç
ç
è
æ
=
= -
-
1
2
2
1
d
d
d
KP
d
KP
S
t
t
r
d1
d2
63

+
Sr in a hexagonal architecture
n With Js interfering base stations
n Js = 6 for a hexagonal architecture
n a = 4 for urban areas
n Maximum distance of the MS from a desired BS is R
n Approximate distance of the MS from each of the co-
channel interferers is D
n The expression for Sr is:
å
=
= s
J
n
n
r
d
d
S
1
0
a
a
Sr ≈
R−4
JsD−4
=
R−4
6D−4
=
1
6
D
R
#
$
%
&
'
(
4
=
3
2
Nc
2
Solve for D/R
64

+
Sr as a function of the cluster size
4 6 8 10 12 14 16 18
Frequency Reuse Factor
12
14
16
18
20
22
24
26
28
SIR
in
dB 65

+
Issues Revisited
n Cluster size Nc determines
n The co-channel interference
n The number of channels
allocated to a cell
n Larger Nc is, smaller is the
co-channel interference
n Larger Nc is, smaller is the
number of channels
available for a given cell
n Capacity reduces
n What Nc should we use
based on SIR or C/I?
66

+
Example: AMPS
n Voice channels occupy 30 kHz and use frequency modulation (FM)
n 25 MHz is allocated to the uplink and 25 MHz for the downlink
n 12.5 MHz is allocated to non-traditional telephone service providers (Block A)
n 12.5 MHz / 30 kHz = 416 channels
n 395 are dedicated for voice and 21 for control
824
MHz
849
MHz
869
MHz
894
MHz
45 MHz
Uplink Band Downlink Band
Block A Block B
21 Control & 395
Voice Channels
30 kHz
67

+
Reuse in AMPS
nSubjective voice quality tests indicate that
Sr = 18 dB is needed for good voice quality
nThis implies Nc = 7
nSee next slide also
nCells do not actually conform to a
hexagonal shape and usually a reuse factor
of Nc = 12 is needed
68

+
Frequency Reuse
Example: Consider cellular system with
• Sr requirement of 18 dB
• Suburban propagation environment with a = 4.
Determine the minimum cluster size.
18 dB è 18 = 10 log10 (x) è
1.8 = log10 (x) è x = 101.8 è
x = 63.0957.
Nc = 1/3 × (6 × 63.0957)0.5 = 6.4857
Since Nc must be an integer, you round up to
nearest feasible cluster size
=> Nc = 7
Solving for D/R results in
Remember ,
which results in
D
R
= 6Sr
( )
1/α
Nc =
1
3
6Sr
( )
2/α
D / R = 3Nc
69

+
AMPS: Adjacent channel
interference
n Cluster size is Nc = 7
n Consider the 395 voice channels
n 1: 869.00 – 869.03 MHz
n 2: 869.03 – 869.06 MHz …
n Cell A is allocated channels 1,8,15…
n Cell B is allocated channels 2,9,16…
n Channels within the cell have sufficient separation so
that adjacent channel interference is minimized
70

+
Frequency Assignment
n Typical C/I values used in practice
are 13-18 dB.
n Once the frequency reuse cluster
size Nc is determined, frequencies
must be assigned to cells
n Must maintain C/I pattern
between clusters
n Within a cluster – seek to
minimize adjacent channel
interference
n Adjacent channel interference is
interference from frequency
adjacent in the spectrum
n Example: You are operating a cellular
network with 25KHz NMT traffic
channels 1 through 12.
n Label the traffic channels as {f1, f2,
f3, f4, f5, f6, f7, f8, f9, f10, f11, f12}
n Place the traffic channels in the
cells above such that a frequency
reuse cluster size of 4 is used and
adjacent channel interference is
minimized
71

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