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Tutorial-1The Cellular ConceptSystem Design Fundamentals

Fundamentals ofCellular and Wireless Networks

Lecture ID: ET- IDA-113/114

06.06.2012 , v10

Prof. W. Adibfolieq.drwPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network Engineering1234567Single clusterCluster size N=712345671234567123456712345671345671234567123456712345671234567DCellular Design Fundamentals SummaryFrequency reuse distance D = R 3NNetwork cell layoutFirst interference tierDRR: Cell radiusCluster size N = i2 + ij + j2Where i, j are integersCo-channel interference ratio S/I:S/I = R-n / DinSimplified S/I = Qn / 6 = 1/6 (3N)n/2=> N = 1/3 (6 S/I)2/nCo-Channel reuse ratio Q = D/R = 3N

Approx. W.c. S/IFrequency reuse factor is 1/NRbfolieq.drwPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network EngineeringProblem 1.1:If a total of 33MHz of bandwidth is allocated to a particular FDD cellular telephone system whichuses 25 kHz simplex channels to provide full duplex voice and control channels.1.Compute the number of channels available per cell for a system cluster size N= 4, 7 and 12 cells2.Compute the total channel capacity of the network if only 50 base stations are to be installed.3. Finde the frequency reuse distance D if the if an area of 250 square kilometer is to be covered assuming the use of hexagonal cell shape.Solution 1.1:1.Channel bandwidth = 25 kHz x 2 simplex channels= 50 kHz/duplex channel The total available channels= 33 000 /50 = 660 channels

For N=4 number of channels/cell = 660/4 165 channels For N=7 number of channels/cell = 660/7 95 channelsFor N=12 number of channels/cell = 660/12 55 channels

For N=4, 165 channels/cell x 50 = 8250 channels frequency reuse factor 1/N= =0.25For N=7, 95 channels/cell x 50 = 4750 channels frequency reuse factor 1/N= 1/7=0.14For N=12, 55 channels/cell x 50 = 2750 channels frequency reuse factor 1/N= 1/12=0.08

Every base station covers 250/50 = 5 square kilometers 2.6 R2 = 5 => R = 1.39 kmFrequency resuse distance for N= 4 D = R 3N = 1.39 3x4 = 4.78 kmFrequency resuse distance for N= 7 D = R 3N = 1.39 3x7 = 6.32 kmFrequency resuse distance for N= 12 D = R 3N = 1.39 3x12 = 8.28 km

bfolieq.drwPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network EngineeringProblem 1.2:What is the number of RF channels per cell in the GSM network for the frequencyspectrum allocation as shown belowCompute the cluster size N for GSM system if the required signal to interference ratio S/I is 14 dB and the path loss exponent n= 4.What is the number of channels per cell in GSM under the given S/I condition?What is the number of omnidirectional base stations having 2KM coverage radius for an area of 312 square kilometer. Use the hexagonal cell layout model.What is the maximum number of simultanious users in the networkWhat would be the maximum number of simultanious users in the network if the signal to interference ratio S/I is reduced to 10 dB ?Comput the frequency reuse distance for both S/I cases.

890-915 MHz for uplink (reverse) -> 25 MHz bandwidth935-960 MHz for downlink (forward) -> 25 MHz bandwidth200 KHz per digital channel8 voice channels per digital channelbfolieq.drwPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network EngineeringSolution 1.2:As shown in the above Figure, we have 25 000 KHz/200 KHz=125 channels, one of which is reserved for guard band and 124 channels remain for users. On each channel 8 voice channels are carried by using 8 time-multiplexing. Total number of voice channels=124 x 8 = 992

Cluster size for a given S/I is : N = 1/3 (6 S/I)2/n14 dB S/I=25.11N = 1/3 (6 x 25.11)2/4 4.09 => N=4

Maximum number of simultaneous users per cell is:124 (RF channels) x 8 users/channel = 248 voice channels per cell 4 Cells/cluster

3Cell area is = 2.6 R2 = 2.6 x 22 = 10.4 KM2 Number of required base stations = 312/10.4 = 30 base stations

Maximum number of simultaneous users (channels) = 248 user/cell x 30 cells = 7440 channels

5.10 dB S/I=10, N = 1/3 (6 x 10)2/4 2.58 cells/cluster => N=3 124 (RF channels) x 8 users/channel = 330 channels per cell 3 Cells/cluster Maximum number of system channels = 330 user/cell x 30 cells = 9900 channelsCo-channel distance D = R 3N = 2 KM 3x4 = 6.93 KM for N=4Co-channel distance D = R 3N = 2 KM 3x3 = 6.0 KM for N=3

Co-channel interference ratio S/I:S/I = R-n / DinSimplified S/I = Qn / 6 = 1/6 (3N)n/2=> N = 1/3 (6 S/I)2/nbfolieq.drwPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network Engineering

AMPS Advanced Mobile Phone System Frequency specrum allocationProblem 1.3:What is the number of channels per cell in the AMPS network for the frequencyspectrum allocation as shown belowCompute the cluster size N for AMPS system if the required signal to interference ratio S/I is 18 dB and the path loss exponent n= 4.What is the number of channels per cell in AMPS under the given S/I condition?What is the number of omnidirectional base stations having 2KM coverage radius for an area of 312 square kilometer. Use the hexagonal cell layout model.What is the maximum number of simultanious users in the networkWhat would be the maximum number of simultanious users in the network if the signal to interference ratio S/I is reduced to 14 dB ?Comput the frequency reuse distance for both S/I cases.

bfolieq.drwPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network EngineeringSolution 1.3:As shown in the Figure, we have 416 channels: 21 of which are control channels and 395 channels remain for users

Cluster size for a given S/I is : N = 1/3 (6 S/I)2/n18 dB S/I=63N = 1/3 (6 x 63)2/4 6.48 => N=7

Maximum number of simultaneous users per cell (channels per cell) is:395 (voice channels) 56.42 = 56 channels per cell 7 Cells/cluster

3Cell area is = 2.6 R2 = 2.6 x 22 = 10.4 KM2 Number of required base stations = 312/10.4 = 30 base stations

Maximum number of simultaneous users = 56 user/cell x 30 cells = 1,680 channels

5.14 dB S/I=25.11, N = 1/3 (6 x 25.11)2/4 4.09 cells/cluster => N=4 395 (voice channels) 98.75 = 99 channels per cell 4 Cells/cluster Maximum number of system channels = 99 user/cell x 30 cells = 2,970 channelsCo-channel distance D = R 3N = 2 KM 3x7 = 9.16 KM for N=7Co-channel distance D = R 3N = 2 KM 3x4 = 6.93 KM for N=4

Co-channel interference ratio S/I:S/I = R-n / DinSimplified S/I = Qn / 6 = 1/6 (3N)n/2=> N = (1/3) (6 S/I)2/nbfolieq.drwPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network EngineeringProblem 1.4:find the S/I ratio for the cellular model shown below taking into account first and second tierinterfering cells. Use i=2, j=1 and N=7 cluster size.

D1D2bfolieq.drwPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network EngineeringSolution 1.4:

3030D1D2D1 cos 30+D1 cos 30+bfolieq.drwPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network Engineering

D1 = R 3N => D2 = 2 D1 cos 30 = 2 D1 3 = D1 3 =3R N D2 =3R N bfolieq.drwPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network Engineeringd0 = 10 mR=1390 mPtP0 =Pr-80dBm? dBmPr = P0 ( dr/d0) n for n=410 log Pr = 10 log P0 - 40 log dr + 40 log d0Pr(dBm) = P0(dBm) - 40 log 1390 + 40 log 10-80 dBm = P0(dBm) - 40 log 1390 + 40

P0(dBm) = - 80 dBm + 40 x 3.14 - 40P0(dBm) = - 80 dBm + 125.72 40 = 5.72 dBm

45 dB lossP0 = Pt - 45Thus:Pt = P0 + 45

Pt = 5.72+ 45Pt = 50.72 dBm

Pt (dBm)= 10 log10 Pt (mW)50.72(dBm) = 10 log10 PtPt= 105.07= 118048 mw

Pt= 118.048 wattProblem 1.5:Referring to problem 1.1. If the cell radius R in a cellular network is 1390 meter and the mobile device sensitivity is 80 dBm (Pt (dBm)= 10 log10 Pt ) . Find the required transmitter power of a base station assuming that the received power at the distance of 10 meter from the base station antenna is 45 dB less than the transmitted power. (use the simplified propagation model with n=4).Solution 1.5:bfolieq.drwPage : N Cellular & Wireless NetworksTechnical University of BraunschweigIDA: Institute of Computer and Network Engineering