Post on 29-Mar-2018
Chapter 3
Cellular Concept
3.1 Introduction 3.2 Frequency Reuse 3.3 Channel Assignment Strategies 3.4 Handoff Strategies 3.5 Interference and System Capacity 3.6 Trunking and Grade of Service 3.7 Improving Coverage and Capacity
in Cellular Systems
Contents
3.1 Introduction
The cellular concept was a major breakthrough in solving the problem of spectral congestion and user capacity.
It offered very high capacity in a limited spectrum allocation without any major technological changes.
The cellular concept is a system-level idea.
3.2 Frequency Reuse 3.1 Introduction 3.2 Frequency Reuse 3.3 Channel Assignment Strategies 3.4 Handoff Strategies 3.5 Interference and System
Capacity 3.6 Trunking and Grade of Service 3.7 Improving Coverage and
Capacity in Cellular Systems
Cellular Structure: Fig 1
5
1
6 4
2
3 7
Basic Properties of Hexagon
R
Rd 3=
2
233
2*
23*6Area RRR
==
Minimal distance of the same frequency cells
2 2 2 2 2
2 2
Center coordinates:(0,0); Target :( , )
(( , ), (0,0)) 2 cos(120 )
id jd
D id jd i d j d ijd
d i j ij
= + −
= + +
(2d,1d)
(0,0)
(2d,2d) (1d,2d)
120°
Frequency reuse concept Frequency reuse concept S: Total of duplex channels in a cluster k: A group of channels for one cell N: Number of cells to use S channels S=kN (3.1) Cluster size: N (区群大小） N: The number of cells to use the complete
available frequency. Total number of duplex channels: C C = MkN = MS M: reuse times of cluster within the system.
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1416
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Calculational equation of N: N = i2 + ij + j2
Example:
i=2, j=1 N=7
dD 7=
The capacity of a cellular system is directly proportional to the number of times a cluster is replicated in a fixed service area.（The N cells which collectively use the complete set of available frequencies is called a cluster）.
The frequency reuse factor of the cellular system is given by 1/N.
Important Formula
NRD
NRNdDijjiN
ijjidD
33
22
22
=
==++=
++=
Example 3.1 Total bandwidth= 33MHz， Channel BW=25kHz×2 simplex channels
=50kHz/duplex channel Total available channels =33000/50= 660， The total number of cells within service area = 49 1) N = 4, four cells 33MHz bandwidth, total number of
channel available per cell =660/4= 165, duplex channel C=(49/4)×660=8084
2) N = 7,seven cells 33MHz bandwidth, total number of channel available per cell = 660/7=95, duplex channel C=(49/7)×660=4620
3) N = 12, twelve cells 33MHz bandwidth, total number of channel available per cell = 660/12=55, duplex channel C=(49/12)×660=2694
3.3 Channel Assignment Strategies
3.1 Introduction 3.2 Frequency Reuse 3.3 Channel Assignment Strategies 3.4 Handoff Strategies 3.5 Interference and System Capacity 3.6 Trunking and Grade of Service 3.7 Improving Coverage and Capacity in
Cellular Systems
Principle: efficient utilization of the radio spectrum
Frequency reuse scheme is consistent with the objectives of increasing capacity and minimizing interference is required.
Assignment strategies: fixed and dynamic. Fixed assignment: The choice of channel assignment strategy
impacts the performance of the system, particularly as to how calls are managed when a mobile used is handed off from one cell to another.
In a fixed channel assignment strategy (borrowing strategy), each cell is allocated a predetermined set of voice channels.
Dynamic assignment: In a dynamic channel assignment strategy,
instead, each time a call request is made, the serving BS requests a channel from the MSC. The switch then allocates a channel to the requested cell.
following an algorithm that takes into account the likelihood of future blocking within the cell, the frequency of use of the candidate channel, the reuse distance of the channel, and other cost functions.
Dynamic assignment: Dynamic channel assignment reduce the
likelihood of blocking, which increase the trunking capacity of the system. Dynamic channel assignment strategies
require the MSC to collect real-time data on channel occupancy, traffic distribution, and RSSI (Radio Signal Strength Indications) of all channel on a continuous basis.
3.4 Handoff Strategies 3.1 Introduction 3.2 Frequency Reuse 3.3 Channel Assignment Strategies 3.4 Handoff Strategies 3.5 Interference and System Capacity 3.6 Trunking and Grade of Service 3.7 Improving Coverage and Capacity in
Cellular Systems
Handoff: A mobile moves into a different cell
while a conversation is in progress. The MSC transfers the call to a new channel belonging to new BS.
Important task of any cellular systems Handoff threshold of signal level: -
90dbm~-100dbm (-110dbm~-120dbm)
Handoff:
Dwell time (驻留时间） A call may be maintained within cell,
without handoff
Handoff strategy In 1G, handoff is made by BS, supervised
by MSC. In 2G, handoff division is mobile assisted (MAHO)
Intersystem handoff Prioritizing handoff (优先切换）
Handoff algorithm’s parameters:
signal level handoff threshold handoff time dwell time As Fig 3.3
基站1基站2
BA
时间
接收到的信号强度
切换时的信号强度（通话成功转移到BS2）
B点信号强度时间
接收到的信号强度
B点信号强度（通话中断）
切换门限
维持通话的最小可接收信号
A点信号强度不正确的切换情况
正确的切换情况
In 1G analog cellular systems, signal strength measurements are made by the BSs and supervised by the MSC.
The typical time is to make a handoff = 10s, The handoff threshold is at 6-12dB. In today’s 2G systems, in mobile assisted
handoff (MAHO), Typically time required only = 1-2s, Handoff threshold is at 0-6dB. Intersystem handoff: Handoff occurs among different MSC.
3.4.1 Prioritizing Handoffs
One method for giving priority to handoffs is called the guard channel concept (信道守候), whereby a fraction of the total available channels in a cell is reserved exclusively for handoff requests from ongoing calls which may be handed off into the cell.
Queuing of handoff requests (排队请求) is another method to decrease the probability of forced termination of a call due to lack of available channels.
为高速通信设置的“伞状”宏小区 为低速通信设置的微小区
3.4.2 Practical Handoff Considerations
High speed vehicles and Walking people, fig 3.4 Cell dragging (小区拖尾) Handoff decision bases on a wide range of metrics
other than signal strength.
3.5 Interference and System Capacity
3.1 Introduction 3.2 Frequency Reuse 3.3 Channel Assignment Strategies 3.4 Handoff Strategies 3.5 Interference and System Capacity 3.6 Trunking and Grade of Service 3.7 Improving Coverage and Capacity in
Cellular Systems
Interference is the major limiting factor in the performance of cellular radio systems.
Interference Sources: another mobile in the same cell call in progress in a neighboring cell other BS operating in the same frequency or any noncellular system The two major type of cellular interference: co-channel interference adjacent channel interference.
Co-channel cells: The cells that use the same set of
frequencies in a given coverage area. Co-channel interference: The interference between signals from
co-channel cells. To reduce co-channel interference Co-channel cells must be physically
separated by a minimum distance to provide sufficient isolation due to propagation.
3.5.1 Co-channel interference and System Capacity
The co-channel interference is a function of the radius of the cell (R) and the distance (D) between centers of the nearest co-channel cells.
co-channel reuse ratio Q=D/R=(3N)1/2
A small value of Q provides larger capacity since the cluster size N is small, whereas a large value of Q improves the transmission quality, due to smaller level of co-channel interference.
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Co-channel reuse rate : Q (同道复用比）
Q = D/R = (3N)1/2
D: distance to the center of nearest co-channel cell R: radius of the cell. N: cluster size
Signal to Interference Ratio (S/I) for co-channel interference cell
Ii is the co-channel interference from the ith interfering cell.
S is the desired signal power from the desired base station, as following.
00
ndS Pd
−
=
ii
S SI I=∑
P0 is the power received from close-in reference point.
d0 is small distance from the antenna of close-in reference point.
Assume user is at the edge of the cell, Signal to Interference Ratio for co-channel interference cell is
∑−
−
=
i
n
ii
n
dDP
dRP
IS
0
00
Di is the distance between the center of the ith interfering cell and the user.
n is the path loss exponent.
When the transmit power of each base station is equal and the path loss exponent is the same, S/I is
∑∑=
−
−
−
−
=
=0
10
00
)(i
i
ni
n
i
n
ii
n
D
R
dDP
dRP
IS
Considering only the first layer of interfering cells and this distance is equal to the distance D between cell centers. S/I is
00
)3()/(iN
iRD
IS nn
==
Consider only the 1st layer of interfering cells (i0=6, as following Fig (N=7))
6/n
RD
IS
=
In dB:
8.7log10 −RDn
R D
1st layer of cells
Example: For the US AMPS, use FM and 30kHz channels, voice quality is provided when SIR>=18dB.
To meet this requirement, N should be at least 6.49.
Assuming n=4, thus a minimum N=7 is required to meet an SIR requirement of 18dB (Actually =17.8dB). N=7, Q=4.6, the worst case SIR=48.56(17dB).
Other consideration If S/I requirement is 18dB, what is the
recommended cluster size ? (Assume n =4)
42.4645.040/8.25log
8.7log4018
===
−=
RD
RD
RD
ijjiNN
NRD
++=
≥
=
22
52.6
3
72,1or 1,2 ===== NjijiNRNdD
ijjiN
ijjidD
3
22
22
==
++=
++=
8.7log10 −RDn
Using an exact cell geometry layout, it can be shown for a seven-cell cluster using the same frequency, with the mobile unit at the cell boundary.
A
R
D+R
D+R
D-RD-R
D
D
A
A
A
A
A
A4
4 4 4
4 4 4
2( ) 2( ) 21
2( 1) 2( 1) 2
RSIRD R D R D
Q D Q
−
− − −
− − −
=− + + +
=− + + +
Example 3.2 SIR=15dB is required for satisfactory forward
channel performance of a cellular system, N=?,Q=?
Assume that there are six co-channel cells in the first tire, and all of them are at the same distance for the mobile.
(a) n=4, consider a seven-cell reuse pattern, D/R=(3N)1/2=4.583, SIR=(1/6)×(4.583)4=75.3=18.66dB
(b) n=3, consider a seven-cell reuse pattern, SIR=(1/6)×(4.583)3=16.04=12.05dB. Since this is less than the minimum required SIR, we need to use a larger N. The next possible value of N=12(I=2, j=2), D/R=6, SIR=(1/6)×(6)3=36=15.56dB>15dB.
f 1f 2f 3
f 2
f 1
f 3 f 1f 2f 3
f 2
f 1
f 3
f 1f 2f 3
f 2
f 1
f 3 f 1f 2f 3
f 2
f 1
f 3
f 1f 2f 3
f 2
f 1
f 3 f 1f 2f 3
f 2
f 1
f 3N=3
f 1f 2
f 3f 4f 3
f 2f 1 f 1f 2
f 3f 4 f 4f 3
f 2f 1
f 4f 3f 4
f 1f 2 f 2f 1
f 4f 3 f 3f 4
f 1f 2
f 3f 4f 3
f 2f 1 f 1f 2
f 3f 4 f 4f 3
N=4
f 1f 5f 4
f 3
f 6
f 2 f 1f 5f 4
f 3
f 6
f 2
f 7 f 9f 8 f 7 f 9f 8
f 1f 5f 4
f 3
f 6
f 2 f 1f 5f 4
f 3
f 6
f 2
f 7 f 9f 8 f 7 f 9f 8
f 1f 5f 4 f 6
f 2
f 7 f 9f 8
f 1f 5f 4
f 3
f 6
f 2
f 7 f 9f 8 f 7
f 1f 5f 4
f 3
f 6
f 2 f 1f 5f 4
f 3
f 6
f 2
f 7 f 9f 8 f 7 f 9f 8
f 1f 5f 4 f 6
f 2
f 7 f 9f 8N=9
Judiciously assigning the appropriate radio channels to each base station is an important process that is much more difficult in practice than in theory.
Generally, the available mobile radio spectrum is divided into channels (about 5% control channel), which are part of an air interface standard that is used throughout a country or continent.
One of the key features of CDMA systems is that N=1, and frequency planning is not nearly as difficult as for TDMA or first generation cellular systems.
Breathing cell : a dynamic, time varying coverage region which varies depending on the instantaneous number of users.
3.5.2 Channel Planning for Wireless Systems
Interference resulting from signals which are adjacent in frequency to the desired signal is called adjacent channel interference(ACI).
Adjacent channel interference can be minimized through careful filtering and channel assignments.
If N is small, the separation between adjacent channels at the BS may not be sufficient to keep the ACI level within tolerable limits.
In practice, BS receivers are preceded by a high Q cavity filter in order to reject ACI.
3.5.3 Adjacent Channel Interference
Example: If a close-in mobile is 20 times as close to the
BS as another mobile and has energy spill out of its passband, the signal to interference ratio is
SIR=(20)-n, n=4, SIR=-52dB If the IF filter has a slope of 20dB/octave, then
the ACI must be displaced by at least 6 times the passband bandwidth from the center of the receiver frequency passband to achieve 52dB attenuation.
3.5.4 Power Control for Reducing Interference
In practical cellular systems, the power levels transmitted by every subscriber unit are under constant control by the serving base station, and are done to ensure them to maintain a good quality link.
3.6 Trunking and Grade of Service
3.1 Introduction 3.2 Frequency Reuse 3.3 Channel Assignment Strategies 3.4 Handoff Strategies 3.5 Interference and System Capacity 3.6 Trunking and Grade of Service 3.7 Improving Coverage and Capacity in
Cellular Systems
Cellular radio systems rely on trunking to accommodate a large number of users in a limited radio spectrum.
The concept of trunking allows a large number of users to share the relatively small number of channels in a cell by providing access to each user, from a pool of available channels.
Trunking exploits the statistical behavior of users so that a fixed number of channels or circuits may accommodate a large, random user community.
It should be clear that the allocation of channels in a trunked radio system has a major impact on overall system capacity.
The fundamentals of trunking theory were developed by Erlang.
The Grade of Service (GOS) is a measure of the ability of a user access a truncked system during the busies hour.
GOS gives the likelihood that a call is blocked, or the likelihood of a call experiencing a delay greater than a queuing time.
Table3.3 Definitions of common terms used in trunking theory
Set-up time：给请求的用户分配一个中继无线信道所需的时间。 Blocked call：拥塞无法在请求时间完成的呼叫，叫损失呼叫。 Holding time：通话的平均保持时间，表示为H(以秒为单位)。 Traffic intensity：表征信道时间利用率，为信道的平均占用率 以Erlang为单位。是一个无量纲的值，可用来表征单个 或多个信道的利用率。表示为A。
Load：整个系统的话务量强度，以Erlang为单位。 GOS：表征拥塞的量，定义为呼叫阻塞概率 （表示为B，单位为Erlang），或是延迟时间大于某一 特定时间的概率（表示为C,单位为Erlang）。 Request rate：单位时间内平均的呼叫请求次数。表示为λ/秒
Each user generates a traffic intensity of Au Erlangs given by
H is the average duration of a call is the average number of call requests per unit time for each user The total offered traffic intensity:
U is number of containing user in a system
The traffic intensity per channel:
C is the total number of duplex channels
HAu λ=λ
uUAA =
CUAA uc /=
Simple Capacity Analysis Based on Blocking Probability
Concept of Blocked Traffic • Define carried load as the portion of offered
load that successfully obtains channel resources • Since there is no queueing or buffering space,
the remaining traffic will be blocked
trafficBlocked load Carried load Offered +=
Offered load Carried load
Blocked traffic
Erlang B Formula
A formula to predict blocking probability Assume: C is number of channels (servers), Poisson arrival with rate is λ, mean holding time is 1/µ, and X(t) is a Markov process if holding time is exponen- tially distributed to represent the number of occupied channels at time t.
))((Prob)( itXtPi ==The blocking probability
For there is no queueing or buffering space, the remaining traffic will be blocked
Erlang B Formula (cont.)
ttPttP
dttdP ii
ti
∆−∆+
= →∆)()(lim)(
0
))((Prob)( itXtPi ==
For the equilibrium state, that is, dPi /dt=0 for all i:
0
![blocking] 80, 3
!
C
r C k
k
ACP GOS P figure
Ak
=
= = − −
∑.6
Trunking System Capacity Count (1) The Erlang B formula is derived in Appendix A.1
0
![blocking] 80, 3
!
C
r C k
k
ACP GOS P figure
Ak
=
= = − −
∑.6
C is the number of trunked channels A is the total offered traffic
Erlang B chart describes the relation between system capacity, offered load, and blocking probability
( )0
[ ]! 1 !
[ ] [ ]exp( ( ) / ) 81,
The average delay for all calls in a queued system is [ ]
C
r C kC
k
r r
r
AP delayA AA C C k
P delay P delay C A t H p figureHD P
C A
=
=+ −
= − − − −
−
∑>0
>t >0 3.7
=延迟>0
Trunking System Capacity Count (2)
The Erlang C formula is given in Appendix A.2 (a queue is provided to hold calls which are blocked).
For there is buffering space, the remaining traffic will be queued.
Example 3.4 How many users can be supported for 0.5% blocking
probability for the following number of trunked channels in a blocked calls cleared system?(Assume each user generates 0.1Erlangs of traffic)
(a) C=1, Au=0.1, GOS=0.005. From fig. 3.6, A=0.005, U=A/Au=0.005/0.1=0.05users, actually, U=1 user.
(b) C=5, Au=0.1, GOS=0.005. From fig. 3.6, A=1.13, U=A/Au=1.13/0.1=11 users.
(c) C=10, Au=0.1, GOS=0.005. From fig. 3.6, A=3.96, U=A/Au=3.96/0.1=39 users.
(d) C=20, Au=0.1, GOS=0.005. From fig. 3.6, A=11.1, U=A/Au=11.13/0.1=110 users.
(e) C=100, Au=0.1, GOS=0.005. From fig. 3.6, A=80.9, U=A/Au=80.9/0.1=809 users.
Example 3.5 An urban has two million residents. Three mobile
networks provide cellular service. System A has 394 cells, each with 19 channel, system B has 98 cells, each with 57 channels, and system C has 49 cells, each with 100 channels.
Find the number of users that can be supported at 2% blocking if each user average two calls per hour at an average call duration of 3 minutes.
Assuming that all three systems are operated at maximum capacity, compute the percentage market penetration.
Solution: (a) GOS=0.02, C=19, Au=λH=2×(3/60)=0.1 Erlangs. From fig. 3.6, A=12, U=A/Au=12/0.1=120 users. Since there are 394 cells in system A, the total number of users that can be supported =120×394＝47280.
(b) GOS=0.02, C=57, Au=λH=2×(3/60)=0.1 Erlangs. From fig. 3.6, A=45, U=A/Au=45/0.1=450 users. Since there are 98 cells in system B, the total number of users that can be supported =450×98＝44100.
(c) GOS=0.02, C=100, Au=λH=2×(3/60)=0.1 Erlangs. From fig. 3.6, A=88, U=A/Au=88/0.1=880 users. Since there are 49 cells in system C, the total number of users that can be supported =880×49＝43120.
Example 3.6 A certain city has an area of 1,300 square miles and is covered
by a cellular system using a seven cell reuse pattern. Each cell has a radius of four miles and the city is allocated 40 MHZ of spectrum with a full duplex channel bandwidth of 60 kHz. Assume a GOS of 2% for an Erlang B system is specified. If the offered traffic per user is 0.03 Erlangs, compute
(a) the number of cells in the service area, (b) the number of channels per cell, (c) traffic intensity of each cell, (d) the maximum carried traffic, (e) the total number of users that can be served for 2% GOS, (f) the number of mobiles per unique channel (where it is
understood that channels are reused), (g) the theoretical maximum number of users that could be
served at one time by the system.
Solution: (a) Given: Total coverage area=1300 miles, and cell radius =4 miles The area of a cell (hexagon)can be shown to be 2.598 R2, thus each cell covers 2.598 × (4)2 =41.57 sq.mi. Hence, the total number of cells are Nc=1300/41.57=31 cells. (b) The total number of channels per cell (C) = allocated spectrum/(channel width ×frequency reuse factor) = 40,000,000/(60,000*7)=95 channels/cell
(c) Give : C=95, and GOS=0.02 From the Erlang B chart, we have traffic intensity per cell A=84 Erlangs/cell
(d) Maximum carried traffic = number of cells × traffic intensity per cell = 31 × 84 =2604 Erlangs.
(e) Given traffic per user = 0.03 Erlangs Total number of users = Total traffic/traffic per user = 2604/0.03 =86,800 users. (f) Number of mobiles per channel =number of users / number of channels = 86,800/666=130 mobiles /channel.
(g) The theoretical maximum number of served mobiles is the number of available channels in the system (all channels occupied)
= C × Nc= 95 × 31= 2945 users, which is =2945/86800=3.4% of the customer base.
Example 3.7
A hexagonal cell within a four-cell system has a radius of 1.389km. A total of 60 channels are used. Compute the following for an Erlang C system that has a 5% probability of a delayed call: (Assume the load per user is 0.029 Erlangs and λ=1 call/hour)
(a) how many users/km2 will this system support? (b) what is the probability that a delayed call will
have to wait for more than 10s? (c) what is the probability that a call will be delayed
for more than 10s?
Solution:
(a) R=1.387km, number of cells per cluster=4, C=60/4=15, area covered per cell=2.598×(1.378)^2 =5 sq km, Au=0.029, GOS=0.05. From fig. 3.7, A=9, U=A/Au=9/0.029= 310 users/5 sq km=62.
(b) λ=1, holding time H= Au / λ=0.029 hour=104.4s, Pr[delay>t|delay]=exp(-(C-A)t/H)
=exp(-(15- 9)10/104.4) =56.29%
(c) Pr[delay>0]=5%=0.05 Pr[delay>t]=0.05×0.5629=2.81%.
3.7 Improving Coverage and Capacity in Cellular Systems
3.1 Introduction 3.2 Frequency Reuse 3.3 Channel Assignment Strategies 3.4 Handoff Strategies 3.5 Interference and System Capacity 3.6 Trunking and Grade of Service 3.7 Improving Coverage and Capacity in
Cellular Systems
Cell Splitting is the process of subdividing a congested cell into smaller cells, each with its own base station and a corresponding reduction in antenna height and transmitter power.
It increases the capacity of a cellular system since it increases the number of times that channels are reused.
Theoretically, if all cells were microcells having half the radius of the original cell, the capacity increase would approach four.
3.7.1 Cell Splitting （ D/R is constant, R is decreased ）
B
C
F
G
E
D
E
D
F
E
G
F
BG
C
D
C
A
E
G
BC
D
If the new cells are smaller in size with radius half that of the original cells, n=4, Pt1 and Pt2 are the transmit powers of the larger and smaller cell base stations, then
The powers at old cell boundary and new cell boundary are
161
2t
tPP =
ntr RPboundarycelloldatP −∝ 1][
ntr RPboundarycellnewatP −∝ )2/(][ 2
Example of Cell Splitting Arrangement II
Example of Cell Splitting Arrangement III
Example 3.8 Consider Figure 3.9. Assume each base station
uses 60 channels, regardless of cell size. If each original cell has a radius of 1 km and each microcell has a radius of 0.5 km, find the number of channels contained in a 3 km by 3 km square centered around A under the following conditions: (a) without the use of microcells; (b) when the lettered microcells as shown in Figure 3.9 are used; and (c) if all the original base stations are replaced by microcells. Assume cells on the edge of the square to be contained within the square.
B
C
F
G
E
D
E
D
F
E
G
F
BG
C
D
C
A
E
G
BC
D
Solution (a) without the use of microcells: A cell radius of 1 km implies that the sides of the larger hexagons are also 1 km in length. To cover the 3 km by 3 km square centered around base station A, we need to cover 1.5 km (1.5 times the hexagon radius toward the right, left, top,and bottom of base station A. this is shown in Figure 3.9. From Figure 3.9, we see that this area contains five base stations. Since each base station has 60 channels, the total number of channels without cell splitting is equal to 5 ×60=300 channels.
Solution (b) with the use of the microcells as shown in Figure3.9 In Figure 3.9 the base station A is surrounded by six microcells. Therefore, the total number of base stations in the squarde area under study is equal to 5+6=11. Since each base station has 60 channels, the total number of channels will be equal to 11×60=660 channels. This is a 2.2 times increase in capacity when compared to case (a).
Solution (c) if all the base stations are replaced by microcelles: From Figure 3.9, we see there are a total of 5+12=17 base stations in the square region under study. Since each base station has 60 channels, the total number of channels will be equal to 17 ×60 =1020channels. This is a 3.4 times increase in capacity compared to case (a). Theoretically, if all cells were microcells having half the radius of the original cell, the capacity increase would approach four.
It increases capacity by keeping the cell radius nuchanged and seek methods to decrease the D/R ratio.
It increases SIR so that the cluster size may be reduced and thus increases the frequency reuse.
The SIR is improved using directional antenna. By using 1200 sectors, the number of interferences in
the first tier is reduced from 6 to 2. In practical systems, further improvement in SIR is
achieved by down tilting the sector antennas such that the radiation pattern in the vertical (elevation) plane has a notch at the neatest co-channel cell distance.
It will increases the number of antennas and it breaks up the available trunked channel pool into several smaller pools, and decreases trunking efficiency.
3.7.2 Sectoring （ R is constant, D/R is decreased ）
7
62
3
4
1
5
5
5
5
5
5
5
It needs to provide dedicated coverage for hard-to-reach
areas, such as within buildings, or in valleys or tunnels. Repeaters are often used to provide such range extension
capabilities. Repeaters are bi-direction, they amplify and reradiate the
BS signals. Repeaters do not add capacity to the system. In practice, directional antennas or distributed antenna
systems(DAS) are connected to the inputs or outputs of repeaters for localized spot coverage, particularly in tunnels or buildings.
Determining the proper location for repeaters and distributed antenna systems within buildings requires careful planning.
3.7.3 Repeaters for Range Extension
More than one zone sites (TX/RX) are connected to a single BS and share the same radio equipment.
Multiple zones and a single BS make up a cell. Unlike in sectoring, a handoff is not required at MSC
when the mobile travels between zones. The BS simply switches the channel to a different zone
site. BS radiation is localized and interference is reduced. The advantage is that while the cell maintains a
particular coverage radius, CCI is reduced since a large central BS is replaced by several lower powered transmitters.
Decreased CCI improves the signal quality and also leads to an increase in capacity without the degradation in trunking efficiency caused by sectoring.
3.7.4 A Microcell Zone Concept
微小区选择器
微波和光缆电路
基站
Tx/Rx
Tx/Rx
Tx/Rx
R
R1
D
D1
D
If N=7，SIR=18dB，then it needs D1/R1=4.6. While by using zone microcell concept, D/R=3, N=3. From N=7 to N=3, amounts to a 2.33 times increase in capacity.