Cellular and Wireless Networks System Design Fundamentals Fundamentals of Cellular and Wireless Networks Lecture ID: ET- IDA-113/114 06.06.2016 , v13 Prof. W. Adi Tutorial-1 The Cellular Concept System Design Fundamentals

Cellular Design Fundamentals Summary 1 2 3 4 5 6 7 R Hexagon area ≈ 2,6 R2 D R 7 6 2 7 Cluster size N = i2 + ij + j2 Where i, j are integers 1 6 2 7 1 5 3 6 2 1 Single cluster Cluster size N=7 4 5 3 7 4 5 3 R: Cell radius 4 6 2 7 D Frequency reuse factor is 1/N 1 6 7 5 3 Frequency reuse distance D = R  3N 1 6 2 7 4 5 3 1 Co-Channel reuse ratio Q = D/R = 3N 4 5 3 6 2 7 7 4 1 6 2 Co-channel interference ratio S/I: S/I = R-n /  Di–n Simplified S/I = Qn / 6 = 1/6 (3N)n/2 => N = 1/3 (6 S/I)2/n 5 3 1 6 2 4 5 3 1 4 5 3 First interference tier 4 Network cell layout Approx. W.c. S/I

Problem 1.1: If a total of 33MHz of bandwidth is allocated to a particular FDD cellular telephone system which uses 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 cells 2. 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 channels For 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= ¼=0.25 For N=7, 95 channels/cell x 50 = 4750 channels frequency reuse factor 1/N= 1/7=0.14 For 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 km Frequency resuse distance for N= 4  D = R  3N = 1.39  3x4 = 4.78 km Frequency resuse distance for N= 7  D = R  3N = 1.39  3x7 = 6.32 km Frequency resuse distance for N= 12  D = R  3N = 1.39  3x12 = 8.28 km

Problem 1.2: What is the number of RF channels per cell in the GSM network for the frequencyspectrum allocation as shown below Compute 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 network What 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 bandwidth 935-960 MHz for downlink (forward) -> 25 MHz bandwidth 200 KHz per digital channel 8 voice channels per digital channel

Solution 1.2: Co-channel interference ratio S/I: S/I = R-n /  Di–n 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/n 14 dB  S/I=25.11 N = 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 3 Cell 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 channels Co-channel distance D = R  3N = 2 KM  3x4 = 6.93 KM for N=4 Co-channel distance D = R  3N = 2 KM  3x3 = 6.0 KM for N=3 Co-channel interference ratio S/I: S/I = R-n /  Di–n Simplified S/I = Qn / 6 = 1/6 (3N)n/2 => N = 1/3 (6 S/I)2/n

Problem 1.3: What is the number of channels per cell in the AMPS network for the frequencyspectrum allocation as shown below Compute 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 network What 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. AMPS Advanced Mobile Phone System Frequency specrum allocation

Solution 1.3: Co-channel interference ratio S/I: S/I = R-n /  Di–n 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/n 18 dB  S/I=63 N = 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 3 Cell 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 channels Co-channel distance D = R  3N = 2 KM  3x7 = 9.16 KM for N=7 Co-channel distance D = R  3N = 2 KM  3x4 = 6.93 KM for N=4 Co-channel interference ratio S/I: S/I = R-n /  Di–n Simplified S/I = Qn / 6 = 1/6 (3N)n/2 => N = (1/3) (6 S/I)2/n

Problem 1.4: find the S/I ratio for the cellular model shown below taking into account first and second tier interfering cells. Use i=2, j=1 and N=7 cluster size. D2 D1

Solution 1.4: 30° D2 D1 cos 30°+ D1 D1 cos 30°+ 30°

D1 = R  3N => D2 = 2 D1 cos 30° = 2 D1 ½  3 = D1  3 =3R  N D2 =3R  N

Problem 1.5: Solution 1.5: Pt R=1390 m P0 = Pt - 45 d0 = 10 m Pr Thus: 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: Pt R=1390 m P0 = Pt - 45 Thus: Pt = P0 + 45 Pt = 5.72+ 45 Pt = 50.72 dBm Pt (dBm)= 10 log10 Pt (mW) 50.72(dBm) = 10 log10 Pt Pt= 105.07= 118048 mw Pt= 118.048 watt d0 = 10 m P0 = ? dBm Pr 45 dB loss -80dBm Pr = P0 ( dr/d0) –n for n=4 10 log Pr = 10 log P0 - 40 log dr + 40 log d0 Pr(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 - 40 P0(dBm) = - 80 dBm + 125.72 – 40 = 5.72 dBm