FrequencyReuse

Prepared By

Vipin Kumar, SL ,ECE

SJPP, Damla.

All the 7 Cells have Different Frequencies(Group of channels, 200 KHz each)

Limitations• 1-Allocation of Fixed Spectrum by• TRAI• 2-Transmitting ,Switching &• Control Resources• 3-Compatibility amongst different• manufacturers• 4-Requirement of more voice & control• channels in each cell

Solution

• Frequency Reuse• OR• Frequency Planning

Frequency Reuse

• The samefrequency band orchannel used in aCell

• Can be “reused”in another Cell aslong as they arefar apart

Frequency Reuse

• Signal strength donot interfere witheach other

• This, in turn,enhances theavailablebandwidth of eachCell

Cluster of Cells

• A cluster of severalCells with nooverlapping area ishere in the figuresyou are going towatch

Cluster of Cells

Reuse Distance• The distance between

two Cells• Using the same

channelknown as the“reuse distance” andis represented by ‘D’

Reuse DistanceEquation

D = √(3N)RHereN = (Number of

Cells in a cluster)

R = (Radius of Cell)

Reuse Distance

Reuse FactorTherefore,the reuse factor isD/R = √(3N)

7-Cell Cluster

Type of FrequencyReuse Factor in GSM

K = 3 with frequencyhopping

K = 4 with/withoutfreq. hopping

K = 7 with or without freq. hoppingK = 3 with each Cell sectoredK = 4 with each Cell sectored

Better value of Kfor Reuse

• It is better to havelarger value of K

• For larger value ofK there will be many cells in onere-use pattern

Grow network ByCell Splitting

• When business becomesgood ,

• We continue adding morecell inside the old cells, byreusing frequencies correctly

Choice ofHexagonal shapeI Additional area

can be covered withadditional clusters

II Without having anyoverlapped area

III Therefore, hexagonal shape hasbeen found suitable

• So this was regarding frequency re-use or frequency planning usinghexagonal cells in the cellularsystem.

• We have already studied about thechoice of hexagonal cells.

• Let us now study about thegeometry of these hexagonal cells .

Geometry of Hexagonal cells

Distance between point C1 and C2D² = (U2 – U1)²

{(cos30)² +(V2 – V1) +(U2 – U1)( sin30)}²

D² = (U2 – U1)² +(V2 – V1)² +(V2 – V1)x(U2 – U1)

>>>>

Distance between point C1 and C2

the actual center to center distance betweentwo adjacent hexagonal cells is

2R Cos30 or √3Rwhere R is center to vertex distance

•If we assume

(V1,U1) = (0,0) &

(V2,U2) to be integer i &j

•Then D²c1c2 = i² + j² + i j

Relation BetweenD and ND is the function ofN1 & S1 (ReceivedSignal – Noise interference)Ratio at the desired mobile receiverand N1 is the number of co-channelinterfere ring cells in the first tier

>>>>

Therefore,

D² = (2RCos30) i² +j² +i j)

= 3R²( i² +j² +i j )

or D²/R² = 3N

D/R = √3N = q

Relation BetweenD and N

Co channel Reuse Ratio VSFrequency Reuse Pattern

• i j N q= D/R• 1 0 1 1.70• 1 1 3 8.00• 2 0 4 3.46• 2 1 7 4.58• 3 0 9 5.20• 2 2 12 6.00• 3 1 13 6.24• 4 0 16 6.93

Cell Geometry

Cells per ClusterN= i x i+ i x j + j x j

here,

i is nos. of cells to betraversed alongdirection ‘i’,j represents the nos.of cells in direction60 degrees

Multiple Choice• If you repeat thisfor all six sides ofthe Reference Cellby substitutingdifferent values of i& j leads toN = 1,3,4,7,9,12,13,16,19,21,28

Most Popular Clusters• Most popular

values are 7 & 3• Many possible

cluster sizeswith differentvalues of N areshown in thenext figure

Popular Cluster

Popular Clusters

This was about frequency Reuse& Cell Geometry. Now Calculate

Channel InterferencePrepared ByVipin Kumar

(Electronics & CommunicationEngineering)

Co-channel Interferenceand RemedyI All the Cells using the

same channel arephysically locatedapart by at least reusedistance

II Control carefully so that,co-channels do not create a problemfor each other

Co-channelInterference• D is the function of N1and S/I(Received Signal to NoiseInterference)

Therefore co-channel RF signal willmake its effect on nearby cellscalled co-channel interference

Effective InterferingCells For Cell-1

Interference• Interference due tosignal strength of theCells

• With cluster of 7 Cells,there will be 6 Cells usingco-channel at reuse distance

• Unlike thermal noise,which can beovercome byincreasing S/N ratio

• Co-channel cannot be combatedsimply by increasing the power oftransmitter

Interference

Interference• This is becausean increase incarrier transmitpower increases theinterference to neighboringco-channel cells

• To reduce CCI• Co-channel cellsmust be physicallyseparated by aminimum distance

• To provide sufficient isolation dueto propagation

Interference

• The parameter q,called co-channelreuse ratio, isrelated to cluster size

• D/R = √3N = q

Interference

D/R = √3N = qA small value of qprovides largercapacity since thecluster size is smallwhere as large valueof q improves the transmission quality,due to small level of CC interferenceTherefore, compromise is requiredin actual cellular design

Co-channel Interference Ratio (S/I)S/I=Carrier/ Interference

•• M

• S/I = Σ (Ik)• k=1

Here Ik is co-channel interference from BSk andM is the maximum number of interfering

cells

Worst case of co-channelinterference when q=4.6 of 7 cells

• So this was regarding co-channelinterference using hexagonal cellsin the cellular system.

• We have already studied about the• Geometry of hexagonal cells

• Let us now study about the Remedyof co-channel interference in thecellular system .

Remedy for Interference

1. CELL SPLITTING

2. CELL SECTORING

Cell Splitting• The Base Station

of all Cellstransmitinformation at thesame power

• Net coverage area of all theCells is same

Improving Coverage&Capacity Using CS

CS ( Cell Sectoring )

is the process of

subdividing a

congested cell into

smaller cells, each with its own base

station and corresponding reduction in

antenna height and transmit power >>

Improving Coverage &CapacityCS increases the

capacity of a

Cellular System since

it increases the number of times that

channels are reused >>>

• Let the transmitting power of new cellsis Pr with radius half that of theOriginal Cell (As an example)

• Pr (at Old Cell

Boundary)

≈ Pt1R¯ⁿ• Pr (at New Cell

Boundary)

≈ Pt2 (R/2)¯ⁿ

• Where Pt1 and Pt2 are thetransmit powers of larger andsmaller cell base stationsrespectively and n is the pathloss exponent.

• If we take n = 4and set thereceived powerequal to each other,then Pt2 = Pt1/16

• Therefore

• Transmitted powermust be reduced by12dB (16 times) inorder to fill in theoriginal coveragearea with microcells

• While maintainingS/I requirements

Resource Demand• Cost effective• Concentration of

users in the givenarea

Cope WithIncreased Traffic

• One way to copewith increasedtraffic is to split cellinto smaller cells

Additional Base Stations

• Additional BSs(Base Stations)need to beestablished at theCentre of each new Cellthat has been added

Cell Density• So that, the higherdensity of cells canbe handledeffectively

Reducing Co-channelinterference• Coverage area ofnew cell is smaller

• Transmittingpower levels are lower

• This helps in reducing CCI

Cell SectoringFor equal powerin all directionour concentrationhas to be onOmni directional antennas

Antenna

• It is difficult todesign such anantenna, andmost of the time,an antenna covers an area of60 degrees or 120 degrees

Sectored Cells• These are called

directional antennas

end cells served by

these are called

Sectored Cells

• Different sizes of Sectored Cells areshown for your reference

Antenna radiation pattern for60 and 120 Degree

Sectoring for 60 and 120Degree

Antennas Mounting• Sectored antennasare mounted on asingle microwavetower located in theCenter of the Cell

Antennas Placement• An adequate nos.of antennas areplaced to coverall of 360 degreesof a Cell

Omni Directional Antenna• The effect of anODA can beachieved byemployingseveral directional antennasto cover the whole of 360 degrees

Advantage of Sectoring• It requires coverageof smaller area byeach antennae andhence lower power isrequired in transmittingradio signals