Solutions for Exercises (Wireless Communications)Q1.If a total of 33MHz of bandwidth is allocated to a particular FDD cellular telephone system which uses two 25kHz simplex channels to provide full duplex voice and control channels, computer the number of channels available per cell if a system uses: (a) 4-cell reuses, (b) 7-cell reuses, (c) 12 cell reuse. If 1 MHz of the allocated spectrum is dedicated to control channels, determine an equitable distribution of control channels and voice channels is each cell for each of the three systems.Solution Given:Total bandwidth=33MHzChannel bandwidth=25 kHz2 simplex channels=50 kHz/duplex channelTotal available channels = 33000/50 = 660 channels.A 1 MHz spectrum for control channels implies that there are 1000/50 = 20 control channels out of the 660 channels available. In practice, only the 640 voice channels would be allocated since the control channels are allocated separately as 1 per cell. Here, the 640 voice channels must be equitable distributed to each cell within the cluster.(a) For N = 4 , we can have 5 control channels and 160 voice channels per cell. In practice, however, each cell only needs a single control channel (the controls have a greater reuse distance than the voice channels). Thus, one control channel and 160 voice channels would be assigned to each cell.(b) For N=7, each cell have one control channel, four cells would have 91 voice channels, and three cells would have 92 voice channels. (491+392=640)(c) For N=12, each cell would have one control channel, 8 cells would have 53 voice channels, and, 4 cells would have 54 voice channels. (853+454=640)Q2If a signal to interference ratio of 15 dB is required for satisfactory forward channel performance of a cellular system, what is the frequency reuse factor and cluster size that should be used for maximum capacity if the path loss exponent is (a) n = 4, (b) n = 3 ? Assume that there are 6 co-channels cells in the first tier, and all of them are at the same distance from the mobile. Use suitable approximations.Solution (a) n = 4First, let us consider a 7-cell reuse pattern.the co-channel reuse ratio Q=D/R = 4.583.the signal-to-noise interference ratio is given by S/1 = (1/6) (4.583)4 = 75.3 = 18.66dB.Since this is greater than the minimum required S/1 , N = 7 can be used.b) n = 3First, let us consider a 7-cell reuse pattern.Using equation (2.9), the signal-to-interference ratio is given byS/1 = ( 1/6) ( 4.583)3 = 16.04 = 12.05 dB.Since this is less than the minimum required S/1 , we need to use a larger N.Using equation (N=i2+ij+j2), the next possible value of N is 12, (i = j = 2 )The corresponding co-channel ratio Q is given by Q=D/R = 6.0.S/1 = (1/6) (6.0)3 = 36 = 15.56 dB.Since this is greater than the minimum required S/I, N=12 can be used.Q3.How many users can be supported for 0.5% blocking probability for the following number of trunked channels in a blocked calls cleared system? And how many user can be supported per channel? (a) 1, (b) 5, (c) 10, (d) 20, (e) 100. Assume each user generates 0.1 Erlangs of traffic. SolutionFrom Table 2.4 or F2.6 we can find the total capacity in Erlangs for the 0.5% GOS for different numbers of channels. By using the relation A = UAu , we can obtain the total number of users that can be supported in the system.(a) Given C = 1, Au = 0.1, GOS = 0.005From Erlang B formula, we obtain A = 0.005. ( Because 0.995A=0.005 A=0.005 )Therefore, total number of users, U = A/ Au = 0.005/0.1 = 0.05 users.But, actually one user could be supported on one channel. So, U = 1.And 1 user can be supported by 1 channel.(b) Given C = 5, Au = 0.1, GOS = 0.005From Table 2.4 or Figure 2.6, we obtain A = 1.13.Therefore, total number of users, U = A/Au = 1.13/0.1 = 11 users.The number of user per channel: Uc = U/C = 11/5 = 2.2 user/channel(c) Given C = 10, Au = 0.1, GOS = 0.005From Figure 2.6, we. obtain A = 3.96.Therefore, total number of users, U = A/ Au = 3.96/0.1 = 39 users.The number of user per channel: Uc = U/C = 39/10 = 3.9 user/channel (d) Given C = 20, Au = 0.1, GOS = 0.005From Figure 2.6, we obtain A = 11.10 .Therefore, total number of users, U =A/Au = 11.1/0.1 = 110 users.The number of user per channel: Uc = U/C = 110/20 = 5.5 user/channel (e) Given C = 100, Au = 0.1, GOS = 0.005From Figure 2.6, we obtain A = 80.9.Therefore, total number of users, U =A/Au = 80.9/0.1 = 809 users.The number of user per channel: Uc = U/C = 809/100 = 8.09 user/channelQ4.An urban area has a population of 2 hundred thousand residents. Three competing trunked mobile networks (systems A, B, and C) provide cellular service in this area. System A has 394 cells with 19 channels each, system B has 98 cells with 57 channels each, and system C has 49 cells, each with 100 channels. Find the number of users that can be Supported at 2%