2.the cellular concept

8/14/2019 2.the Cellular Concept

1/21

Mobile Communications by Ahmad Hijazi 1

The Cellular Concept1 Frequency reuseConcept of Cell : Divide coverage area into cells . Group a limited number of cells in a cluster .Distribute available bandwidth among cells in same cluster. Repeat clusters along with theirfrequency bands this is called frequency reuse .

Result : increase capacity of system (number of users) without requiring extra bandwidth.

See example in previous chapter.

In the next figure every 7 cells aregrouped in one cluster. Then we repeatclusters to cover more area. Here werepeated the cluster 6 times. If theavailable bandwidth offers 21 channels,

then every cell takes 3 channels (21/7),so every cell can support 3simultaneous calls. But since we have 6clusters, the total number ofsimultaneous calls in the whole regionis 376=216=126 calls at the sametime. If we covered the whole regionwith only one cluster then the totalnumber of simultaneous calls would be only 37=21. By this frequency reuse (repeating theclusters) we increased the capacity of the system by 6 times.

In doing so we must adjust the level of transmitted power from base stations so that it does notcause interference with cells using the same channels (frequencies) ex: cell D from one clusterwith cell D from another cluster. Smaller cells imply less transmitted power.

In the previous example, number of cells in one cluster is N=7. Number of channels (duplexchannels) in one cell is k=3. Number of total available channels is S=21. In general we have

S=kN Number of times a cluster is repeated (number of clusters in whole region) is M=6. Total numberof available channels (simultaneous calls) in the whole region or capacity C=126. In general wehave

C=MS=MkN

Now some important notes.

Q: Why do we use hexagonal shape for a cell.A: Consider that the base station is at the center of a cell, and that the base station antenna radiates

power in all directions uniformly. Consider also that for transmission to be successful the mobile

A

B

E

C

DF

G A

B

E

C

DF

G

A

B

E

C

DF

G

A

B

E

C

DF

G

A

B

E

C

DF

G

A

B

E

C

DF

G


2/21


R

R

R

R

Uncoveredarea

BS1 BS2

BS3 BS4

BS1BS2

BS3 BS4

station must receive a minimum level of power. Then there is a maximum allowable distance R between MS and BS that makes communication possible . This means the coverage area of a cell isa circle of radius R.

Now if we decide to use circles for cells then there would be some uncovered areas between thesecircles, since the distance from these areas to nearest base station is larger than maximumallowable distance R. We want full coverage, what to do?

Solution: Allow circles to overlap until the whole area is covered. Assign coverage area for eachBS based on nearest distance as shown.

We can use several shapes. If one cell is surrounded by 4 other cells we end up with square cells, but if one cell is surrounded by 6 other cells we end up with hexagonal cells as shown in figure.The hexagonal shape has the least overlap between cells and we can cover an area with less basestations using hexagons than any other shape. From now on we will consider cell shape as hexagonfor design purposes.Hexagon shape is ideal. In practice, cell shapes are not exactly hexagons due to terrain andsurrounding environment. However, hexagon shape gives a very good approximation for initialdesign purposes.


3/21


Number of cells in one cluster, N, is not arbitrary but must obey the following rule.

N = i 2 + j2 + ij i,j integers; i,j 0

This rule ensures covering an area by repeating the same cluster exactly and at the same timewithout leaving any gaps.

Ex: i=1, j=1 , then N=1 2+12+11=3i=1, j=2 (same if we choose i=2, j=1) , then N=1 2+2 2+12=7

i=3, j=0 , then N=3 2+02+30=9

N=8 or N= 11 for example are not possible.

How to find cells of same channels (same

letter)? Note that any cell is surrounded by 6 cells, this is because cell shape is hexagon. To find repeatedcells of same channel allocation go i steps in anyof the 6 directions, turn 120 o counterclockwisethen move j steps, you are in a repeated cell. Thefollowing example is for i=3, j=2. Note that allthese are equidistant from the original cell.

Note also we assume cells of same size, which is

not always the case in practice.

Ex1: Available bandwidth for a cellular system is21 MHz. System uses FDD to accomplish fullduplex communication, and the bandwidth of asingle simplex channel is 25kHz. If area of one cell is A cell=2km

2 and we want to cover a citywhich has an area of A city=168km

2, find the number of available channels in the city if we use N=4,7,12. Repeat solution if area of one cell is reduced to A cell=1km

2.

Sol: # of available simplex channels for the system 840kHz25MHz21

# of available fullduplex channels S 4202

840

For N=4

# of channels per cell 1054

420Sk

# of cells in the whole city 84km2

km1682

2

cell

city

AA

# of clusters in city 214

84cluster oneincellsof #

cityincellsof #M

i=3

j=2


4/21


# of available channels in city 882042021MkNMSC For N=7


420Sk

# of cells in city is 84 (same as for N=4, since A city and A cell is the same)



cityincellsof #M

# of available channels in city 504042012MkNMSC

For N=12

# of channels per cell 3512420S

k



cluster oneincellsof #cityincellsof #M

# of available channels in city 29404207MkNMSC

We can see that number of available channels increases as N decreases, inverse relation. Thisis so because number of clusters in region (M) increases while number of available channelsfor one cluster (S) which is same as number of available channels for system is fixed. Theterm

N

1 is called frequency reuse factor.

Now lets consider the effect of cell area on number of available channels for whole cityFor N=4


420Sk (same as for A cell=2km

2)

# of cells in the whole city 168km1

km1682

2

cell

city

AA (double for A cell=2km 2)

# of clusters in city424


cityincellsof #M

(double)

# of available channels in city 1764042042MkNMSC (double)

For N=7


420Sk (same as for A cell=2km

2)




cityincellsof #M (double)



5/21


For N=12

# of channels per cell 3512420S

k (same as for A cell=2km2)


# of clusters in city1412


cityincellsof #M

(double)


Again we see that number of available channels increases as cell area decreases, and thereason is that multiplicity M increases.

2 Channel Assignment StrategiesTwo strategies: fixed and dynamic .

The previous example represents fixed channel assignment strategy since the number ofchannels per cell is fixed. So if all channels in a particular cell is occupied by mobile users andyou try to initiate a call, your call will be blocked .Some systems using fixed strategy allow borrowing channels from neighboring cells, in thiscase the MSC is responsible for appropriate borrowing procedure.

The other strategy is called dynamic channel allocation strategy. Here the available channelsare not fixed for each cell, instead all available channels are kept by MSC and if a user wantsto initiate a call it sends a request to MSC which decides the exact channel to be assigned forthat user.

Dynamic assignment reduces probability of blocking and increases system efficiency incomparison to fixed assignment, but it increases computational load on MSC.

3 HandoffsThe process of transferring a mobile user engaged in a call from one base station to another is

called handoff . When a MS is handed off to another base station it is transferred to a differentchannel too.

We need to make number of handoffs as minimum as possible. Also handoff must not benoticed by user. In general, handoff (HO) has a higher priority over new call requests in thenew cell.


6/21


Question is: when to start a handoffprocess?

The mobile station requires a minimumlevel of power, P min , received from a basestation to maintain a call. This level isfixed and depends on the equipment used.Usually P min is between 90dBm and 100dBm. As MS moves away from BS1,received power from BS1, P r1, decreases.We need to start handoff before received

power reaches P min and the call is dropped.

We start handoff when power receivedfrom BS1, P r1, reaches P HO (>P min ) while

power received from BS2, P r2, is greaterthan P min . Handoff usually requires

between 1 to 2 seconds to be completed (in1G it required 10s). We need to completethe handoff before received power fromBS1, P r1, reaches P min .

We define =P HO P min .

Since P min is fixed by equipment, we needto choose P HO (or ) properly. If is toosmall (P HO very close to P min), then amoving MS (in a fast car for ex) may reacha point where P r1 is less than P min beforehandoff is completed. On the other hand, if is too large (P HO is very large) thenunnecessary handoffs will start whilesignal received from BS1 is sufficient (P r1 well above P min), this is not good becausewe want to minimize number of handoffs.

MAHO (Mobile Assisted Handoff)In 1G systems (ex: AMPS), BSs used tomeasure power from MSs in order todecide when a handoff is necessary. In 2Gsystems (ex: GSM), MS measures powerreceived from several BSs and sends thesemeasurements to MSC (through BS) whichdecides the handoff. MAHO is much faster

than 1G handoff.

BS1BS2

handoff

Received

powerPHO

Pmin

Pr2

Start HO herePr1= P HO, P r2> P min

P r1 slow car (2sec)

HO completes hereP r1,Pr2> P min

Received power

PHO

Pmin

Pr2

Start HO herePr1= P HO, P r2> P min

Pr1 fast car (2secs)

P r1< Pmin before HO iscomplete, HOrequires 2 secs

Successful handoff

Failed handoff


7/21


Practical considerations regarding handoff

Umbrella cell: As we saw in a previous example, we use small cells to increase capacity of asystem. This is used in dense cities. Suppose that a fast car is driving through small cells, thismeans that a handoff is needed every 10 or 15 seconds, which is not good since we want tominimize number of handoffs. Solution: use large cell, umbrella cell , for fast moving cars andat the same location use small cells for stationary or walking people.

Cell Dragging : Sometimes the signal received by MS from BS is strong even though the MShas moved out of the designed border of the cell. This happens when there is a lineofsight(LOS) between BS and MS (in normal situations, communication between MS and BS is dueto cosequtive reflections and not via LOS). In this case no handoff takes place since P r1>P HO .When this happens, MS may get close to a cell that uses same channel (same frequency)which may lead to interference (cochannel interference).Handoff design in systems must take this phenomenon into consideration and decide handoffwhen MS leaves the border of a cell even if it receives strong signal due to LOS.

4 Interference and System CapacityInterference can be from within the system (ex: from another mobile user or from another basestation), or it can be from outside the system (TV, radio,etc). We will consider interferencefrom within the system. It has two types:1) Cochannel interference2) Adjacent channel interference.

Cochannel interference:Cells with same letter (ex: A) use same channels (same frequencies) and we call them co channel cells. A MS receives strong signal from its own cell BS, and weaker signals from co channel cells on the same frequency. This is called Cochannel interference. Similarly, BSin a cell receives strong signals from its own users, and weaker signals from users in co channel cells on same frequencies.As a result we define SignaltoInterference Ratio (SIR) as the ration of (desired) signal

power to total power in interfering signals.


8/21


RD

D

D

D

D D

B

A

C

B

A

C

B

A

C

B

A

C

B

A

C

B

A

C

B

A

C

R

Di

D1

D2

Cochannel interference N=3

Note: can we increase SIR by reducing power transmitted from cochannel cells and keeping power transmitted from our own cell high? NO! If we do so, then we increase our own SIR and decrease SIR in cochannel cells in whichwe have decreased its transmitted power. We need a solution that works for all cells.Solution: make cochannels cells far away from each other.

Lets do some mathematical analysis:If at distance d o you receive an average power P o, then at a distance d you receive power P r

given byn

oor d

dPP

n is called path loss exponent and is equalto 2 for clear lineofsight. In practice,generally no LOS is available betweentransmitter and receiver, so n can have values

between 2 and 4. (we will explain exactreason for this in coming chapters).Assume all cells transmit same power fromtheir BSs, assume also that MS is located atthe border of its own cell, so distance betweenMS and its own BS is equal to cell radius R.assume distance between MS and other co channel BSs is D i , then SIR is given by

OO

i

1i

ni

n

i

1i

n

o

io

n

oo

D

R

dD

P

d

R

PIS

where i o is number of interfering cells. Mostimportant interfering cochannel cells are thosein the first layer around the considered cell, andin the case of hexagonal cells they are 6. This

number is independent of number of cells percluster N. We will consider only first layer co channel cells and take i o=6.To simplify things we will make anapproximation. For distance between MS andits desired BS we will consider MS is at borderof its cell and take the distance as R. But for allDis we will consider MS is at center of its celland all D is are identical and equal to D. D isdistance between desired cell center and anyfirst layer cochannel cell center.


9/21


In this case we write SIR as

o

n

i

1i

n

n

iR

D

D

R IS

o

The ration (D/R) is called cochannel reuse ratio . For hexagonal cells this ration is given by

N3R D

Q

o

n

o

n

o

n

iQ

i N3

iR

D

IS

As we can see, to increase SIR we need to increase N. But at the same time increasing N willdecrease system capacity as we saw in example. We make compromise between SIR and

capacity. Example:

A minimum signaltointerference ratio of 15dB is required in a particular system. Findminimum cluster size (N) if path loss exponent n=3. Repeat for n=4.

Solution: SIR[dB]=10log(SIR) 15dB=10log(SIR) SIR=10 1.5=31.62for n=3

11 N62.31

6

N362.31

i

N3

I

S3

o

n

N=11 is not possible (N can be 3,4,7,9,12,19.etc). So we choose N=12 (i=2,j=2). Now for n=4

59.4 N62.31

6 N3

62.31i N3

IS

4

o

n

So we choose N=7 (i=2,j=1)

Adjacent Channel Interference:Interference between channels next to each other in frequency (neighbor channels) is calledadjacent channel interference. Remember that filters are not ideal and there must be atransitionband before and after the cutoff frequencies. Signals with frequencies in thistransitionband can pass along with the desired signal.

Nearfar effect : interfering MSoperating in an adjacent channel isnearer to BS than desired MS, sosignal received from interfering MS isstronger than that from desired MS.

So what?: Do NOT use adjacent channelsin same cell, keep channels used insame cell separated from each other. freq

Nonideal filter

Interfering signal(strong)

Desired signal(weak)

Nearfar effect


10/21


Example: A cellular system is assigned the frequency band [425440]MHz. Reverse channelsare given the band [425431.3]MHz, while forward channels are given [433.7440]MHz.System uses 50kHz for one voice channel, and a minimum SIR of 17dB. 21 channels of thetotal available channels are used for control purposes and the remaining are for voicecommunications. Control and voice channels use same bandwidth of 50kHz. Path loss

exponent was found to be 4 in that region. Find appropriate channel allocation scheme thatsatisfies cochannel interference requirement and minimizes adjacent channel interference.

Solution:

# of reverse CHs = # of forward CHs 12605.0

7.43344005.0

4253.431

# of voice channels S = 12621 = 105Cluster size N must satisfy cochannel interference condition

7 N72.6 N106

N3 10178.3

# of channels per cell = k = S/N = 105/7= 15 Now we want to allocate channels to cells in a way to minimize adjacent channel interference.First lets give numbers to the available 126 channelsCH1 : reverse [425425.05] MHz , forward [433.7433.75] MHzCH2 : reverse [425.05425.1] MHz , forward [433.75433.8] MHzCH3 : reverse [425.1425.15] MHz , forward [433.8433.85] MHz

.

.

.CH125: reverse [431.2431.25] MHz , forward [439.9439.5] MHzCH126: reverse [431.25431.3] MHz , forward [439.95440] MHzControl channels are CH106 to CH126

To minimize adjacent channel interference we do NOT use adjacent channels in same cell. We followthe following procedure:

CH1 to cell A, CH2 to B, CH3 to C, CH4 to D,CH5 to E, CH6 to F, CH7 to G. Now back to cellACH8 to A, CH9 to B, CH10 to C and so on

until..CH103 to E, CH104 to F, CH105 to GChannel numbers and cells are shown in table.

A B C D E F G1 2 3 4 5 6 78 9 10 11 12 13 1415 16 17 18 19 20 2122 23 24 25 26 27 2829 30 31 32 33 34 3536 37 38 39 40 41 4243 44 45 46 47 48 4950 51 52 53 54 55 5657 58 59 60 61 62 6364 65 66 67 68 69 7071 72 73 74 75 76 7778 79 80 81 82 83 8485 86 87 88 89 90 9192 93 94 95 96 97 9899 100 101 102 103 104 105

1 2 125 126 1 2 125 126

Reverse channels Forward channels

425

425.05

425.1 431.2 431.3

431.25

433.7433.8 439.9 440

439.95

433.75

.................... ....................


11/21


12/21


If average number of calls initiated per hour by one user is increased (two calls per hourinstead of one in the example), we also expect blocking probability to increase.

On the other hand if number of available channels increases, GOS decreases.

Lets study this in a mathematical way.

Consider in the previous example that each user initiates 2 calls per hour (on average ) insteadof one, and that each call lasts 5 minutes (also on average ). We call the average call durationthe holding time (H). In this example H=5 mins=(1/12) hour. We call the average number ofcall requests per unit time the request rate (). In this example =2 per hour. So the averagechannel occupancy (or traffic intensity) by each user A u is 35=10 mins per hour, or (1/6)hour per hour, or (1/6) Erlang.

Au=H (Erlang)

The Erlang is a dimensionless quantity and it represents the ratio of time a user is utilizing thesystem. is measured in (hr) 1 or (sec) 1 . H is measured in (hr) or (sec). Note that you need touse the same units for both to compute A u correctly. A u is always less than or equal to 1.

If you have U users (U=40 in example) using C channels (C=10 in ex.), then the total trafficintensity offered to system is

A=UA u (Erlang)

A can be greater than 1, in the example A=40(1/6)=6.67 Erlangs.

We define traffic intensity per channel as

C

UA

C

AA uc

In general A c can be more than 1 also. In the ex. A c=6.67/10=0.667 Erlangs.

We have two types of trunking systems Blocked Calls Cleared (Erlang B): in this type, when no channel is available, the call is

blocked and it is up to user to repeat the call request. Blocked Calls Delayed (Erlang C): in this type, when no channel is available, the call is

held in a queue and waits until a channel becomes available. If waiting time exceeds aspecific limit then call is blocked and then it is up to user to repeat the call request.

Both types are of the fixed assignment strategy

Blocked Calls Cleared Now a formula to compute GOS (probability of blocking) or Pr[blocking] for the case ofBlocked Calls Cleared. This formula is called the Erlang B formula

C

0k

k

C

!k A!C

A

blockingPr GOS


13/21


In our example

10

0k

k

10

!k

67.6!10

67.6

blockingPr GOS

The summation in the denominator is difficult to compute and may need a computer program.Instead, Erlang B charts are available to evaluate GOS given A and C. Using the exact formula GOS=0.0659, while using the chart GOS=0.06 which is a goodapproximation

Lets take some examples.

Example:3.4How many users can be supported if the required blocking probability is 0.5% and each userhas a traffic intensity of 0.1 Erlang. Solve if number of available channels is: 1, 5, 10, 20, 100.

Erlang B chart

Number of channels C=10

Approximate position of 6.67A


14/21


Solution :For the case of C=1 it is easy to use the exact formula

005.0A005.0A1

A

!k A!1

A

blockingPr 1

0k

k

1

user 05.0U1.0U005.0UAA u Number of users can not be less than one so U=1.

For C=5 we use chart. We go along the curve that corresponds to C=5 and find its intersectionwith the horizontal line corresponding to Pr[blocking]=0.005. On the xaxis we find A1.1

users11U1.0U1.1UAA u

For C=10 we go along the curve labeled with C=10 and find the point that corresponds toPr[blocking]=0.005. Here we find A 4.

users40U1.0U4UAA u


users110U1.0U11UAA u


users800U1.0U80UAA u Clearly, an increase in number of channels (C) leads to an increase in capacity or number ofusers (U).

Example:3.6

Area of city = 1300 mi 2, N=7, radius of cell R=4 mi, available bandwidth=40MHzFullduplex channel BW=60kHz, required GOS=2%, A u=0.03 Erlang

a) Number of cells in cityareacell

areacity

Area of hexagon 222 mi57.4142

33R

233

Number of cells in city = 1300/41.57=31.2731 cells.

b) Number of channels per cell (k) )S(channelsavailableof #

# of available channels 66667.666kHz60

MHz40

BWchannel

BWavailable

# of channels per cell=666/7=95.1495 channels per cell


15/21


c) Traffic intensity (A) for each cell. We use chart with C=95 and Pr[blocking]=0.02. C=95 isnot available so we take midpoint between C=90 and C=100. we find that A 85 Erlangs.

d) Maximum carried traffic in whole system= (# of cells)(traffic intensity per cell)=3185=2635 Erlangs

e) Total number of users (U) that can be served users8783303.0

2635AA

u

f) Number of users per channel channe per users131666

87833)S(channelsavailableof #

usersof #

g) Number of simultaneous calls in whole system = (# of cells)( # of channels per cell)= 3195= 2945 calls at one time

Example:3.5

Area of population of 2 million people, 3 competing companies: A divides area into 394 cellswith 19 channels per cell, B 98 cells with 54 channels per cell, C 49 cells with 100 channels

per cell. Required GOS from all companies is 2%. In average each user has 2 calls per hour,each call of average 3 mins. Find percentage market penetration (ratio of total users to total

population) of each company if operating at full capacity.

Solution :Channels of any cell are independent of channels of other cells, fixed channel allocationstrategy.Traffic intensity for each user is A

u=(23mins)/(1hr)=6mins/60mins=0.1 Erlang

Company A:Each cell has C=19, Pr[blocking]=0.02, using chart A14 Erlangs. U=A/A u=14/0.1=140 user

per cell.Total number of users in company A=140394=55160 users. Penetration=55160/2million=2.76%

Company B:Each cell has C=54, Pr[blocking]=0.02, using chart A45 Erlangs. U=A/A u=45/0.1=450 user

per cell.

Total number of users in company A=45098=44100 users. Penetration=44100/2million=2.21%

Company C:Each cell has C=100, Pr[blocking]=0.02, using chart A90 Erlangs. U=A/A u=90/0.1=900 user

per cell.Total number of users in company A=90049=44100 users. Penetration=44100/2million=2.21%

Problem 3.13

S=300, total # of cells in region=84, traffic intensity per user A u=0.04 Erlang, requiredGOS=1%. Find maximum total number of users served by system if N=4,7,12?


16/21


Solution : N=4 k=300/4=75 chs/cellUsing chart C=75,GOS=0.01 A60 Erlangs U=A/A u=60/0.04=1500 users per cellTotal # of users in system = 150084=126000 users

N=7 k=300/7=42.942 chs/cellUsing chart C=42,GOS=0.01 A30 Erlangs U=A/A u=30/0.04=750 users per cellTotal # of users in system = 75084=63000 users

N=12 k=300/12=25 chs/cellUsing chart C=25,GOS=0.01 A15 Erlangs U=A/A u=15/0.04=375 users per cellTotal # of users in system = 37584=31500 users

Note: Which case can serve more users: 100 channels trunked together in one cell or in 25 cells

with 4 channels each? Consider each user offers A u=0.1 Erlang and required GOS=5% for

both cases.Case1: from chart using C=100 and GOS=0.05 we get A95 Erlangs. U=A/A u=95/0.1=950users total

Case2: from chart using C=4 and GOS=0.05 we get A1.5 Erlangs. U=A/A u=1.5/0.1=15 users per cell total users=1525=375

Result: from the point of view of trunking efficiency, it is always better to put channels in onepool than to divide them into several pools.The way in which channels are assigned to cells is very important in determining the capacity

of the system (in terms of number of users that can be served at a predetermined GOS, or theGOS for a predetermined number of users), its clear now that dynamic assignment strategygives better capacity than fixed assignment.

Blocked Calls DelayedIn this type, when no channel is available, the call is held in a queue and waits until a channel

becomes available. If waiting time exceeds a specific limit then call is blocked and then it isup to user to repeat the call request.

Probability that a call will be delayed is given by the Erlang C formula:

Pr[ call is delayed] =1C

0k

k C

C

!k A

)CA

1(!CA

A

Now given a call has not found an available channel and is delayed, what is the probabilitythat it has to wait for more than t seconds Pr[delay > t | call is delayed]?

Pr[delay > t | call is delayed] = et

HAC

This is called exponential distribution. The average delay (given that a call has been delayed)is D:


17/21


ACH

D

Now, whats the probability that a call is delayed and the delay is more than t seconds?

Pr[call is delayed & delay > t]= Pr[ call is delayed] Pr[delay > t | call is delayed]

et

HAC

1C

0k

k C

C

!k A

)CA

1(!CA

A

Again the term for Pr[ call is delayed] is long to evaluate and a chart called Erlang C chart isavailable. The term of Pr[delay > t | call is delayed] is a simple exponent and can be evaluatedeasily.

Example:3.7

Erlang C system, N=4, R=1.387 km, S=60, A u=0.029 Erlang, =1 call/hour,Pr[call is delayed]=5%a) # of users/km 2 b) Pr[delay>10s|call is delayed] c) Pr[call is delayed & delay>10s]

Solution :a) # of chs/cell=60/4=15 ch/cell.

P

r [ c a

l l i s d e

l a y e

d ]

Erlang C Chart

A


18/21


From Erlang C chart using C=15 and Pr[call is delayed]=0.05, we find that A 9 Erlangs,U=A/A u=9/0.029=310.3 310 users/cell

Area of cell A cell= 222 km5387.12

33R

233

# of users/km 2=310/5=62 user/km 2

b) Au=H H=0.029/1=0.029 hour=0.0296060=104.4 seconds

Pr[delay>10s|call is delayed]= et

HAC

= e10

4.104915

=0.5629

c) Pr[call is delayed & delay > 10s]= Pr[ call is delayed] Pr[delay > 10s | call is delayed]

=0.050.5629=0.0281

6 Increasing Capacity of System

Directional AntennaSo far, we have considered Omnidirectional antenna which radiates uniformly in alldirections. We will consider now directional antenna. Every antenna has a pattern that definesthe (relative) power transmitted (or received) in a given direction. We define the beam-width of an antenna as the angle separating the two halfpower points. For simplicity, we willconsider antenna is able to transmit (or receive) only from within beam-width, and zerooutside beam-width (idealization of directional antenna).

Important: radiation pattern is valid for both transmission and reception. That is, if antenna

beamwidth is 30o

, it can receive only within this 30o

, and it can transmit only within this 30o

.

Cell SectoringIn cell sectoring we divide the cell into sectors , sectors maybe 60 o (6 sections/cell) or 120 o (3sections/cell). To do so we use directional antenna of 60 o or 120 o beam-width. Also we divide

channels available to that cells into 6 (for 60o

sectors) or 3 (for 120o

sectors) groups and giveeach section one of these groups (ex: a cell with 30 chs become 3 sectors with 10 chs each).

max. pow er

half power

half power

Beam-width

Directional Antenna OmniDirectional AntennaIdealized DirectionalAntenna


19/21


B

A0C

B

A2

C

B

A1

C

B

A6

C

B

A5C

B

A4

C

B

A3

C

Sectoring with 120 o Radiation pattern for sector 1 of several

cochannel cells

Sect 1

Sect 1

Sect 1

Sect 1

Sect 1

Sect 1

Sect 1

Sect 3

Sect 2

Sect 3

Sect 2

Why sectoring?Before sectoring each MS (or BS) in acell could receive cochannelinterference from 6 BSs (or MSs)located in cochannel cells because BSs

used Omnidirectional antenna.For example a MS within cell A 0 couldreceive interference from all A 1 to A 6 BSs because BSs in A 1 to A 6 used totransmit through Omnidirectionalantenna. Also, BS of cell A 0 couldreceive interference from all MSslocated in A 0 to A 6 because BS of cell A 0 used to receive through omnidirectionalantenna.After sectoring (3 sectors of 120 oin thisexample), however, a MS within sector 1of cell A 0 now receives interference fromsector 1 of A 4 and sector 1 of A 5 BSsonly, since BSs in sector 1 of A 1, A2,A3, and A 6 do not transmit in thedirection of cell A 0 because they usedirectional antenna of 120 o beamwidth.At the same time, the BS of sector 1 of

A0 now receives interference from MSs located in sector 1 of A 1 and A 2 only, since BS ofsector 1 of A 0 does not receive from the direction of A 3, A 4, A 5, and A 6 again because it usesdirectional antenna of 120 o beamwidth and can receive only within this beamwidth.

Note that channels in sector 1 are different from those in sectors 2 and 3, and hence everysector becomes as if it were an independent cell. When you move from one sector to anotherwithin the same cell you need to be given a new channel in the new sector, a new channel on anew frequency. This is handoff.

Remember that cochannel SIR is given by

o

n

i N3

IS

, and we used i o=6. Now after

sectoring i o becomes equal to 2. This increases cochannel SIR, and enables us to use smaller N to satisfy the minimum SIR required by equipment. Using smaller N leads to increase incapacity and this is the goal of cell sectoring.

If we use 60 o sectors (6 sectors/cell) then i o=1 and further increase in cochannel SIR is possible.

What is the drawback (or disadvantage) of sectoring?

From trunking efficiency point of view, sectoring will lead to higher probability of blocking(or higher probability of delay for Erlang C systems), since putting all channels in one basketis more efficient than dividing them in groups. Trunking efficiency with one complete cell is


20/21


better than with 3 sectors/cell, which is in turn better than with 6 sectors/cell. (See example intrunking section).

Handoff becomes necessary between sectors since they use different channels now. Thisincreases complexity on MSC. Anyway modern equipment allow handoff between sectors ofsame cell without the need of assistance from MSC. Still number of handoffs is increased.

What about channel allocation? How toreduce adjacent channel interference withinsame sector?Again we use same strategy, that is, makechannels used by the same sector as far fromeach other as possible. In a previous examplefor adjacent channel interference, cell B wasassigned 15 channels (2, 9, 16, 23, 30, 37, 44,51, 58, 65, 72, 79, 86, 93, 100). If we use 3sectors/cell then sector 1 of cell B is given (2,23, 44, 65, 86), sector 2 is given (9, 30, 51,72, 93), and sector 3 is given (16, 37, 58, 79,100). This gives maximum separation

between channels of same sector.

Cell SplittingAs we have seen in a previous example, reducing cellarea leads to increasing in capacity. If number of usersin a region is increased and Pr[blocking] is increasedtoo, then we need to assign more channels in that samearea. We do so by dividing the original cells intosmaller cells. This is the idea of cell splitting.

Until now, we have considered base stations to be at centers of cells. Indeed this is not always

the case, and in practical systems it is rarely used. A more common way of distributing BSs isto locate them at corners of cells instead of centers of cells. Each base station has threedirectional antennas with 120 o beamwidth. Each cell is covered by three such antennas atthree base stations as shown, and every antenna covers one third of a cell. Also, each basestation is shared by three neighboring cells, this means that number of BSs (in a city for ex.)is kept the same as before.

Sector 1(2, 23, 44,

65, 86)

Sector 2(9, 30, 51,

72, 93)

Sector 3(16, 37, 58,

79, 100)

B

C

G

F

E

D

A

B

C

DE

F

G


21/21

2.the cellular concept

Documents