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1、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
2、: (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=
3、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
4、 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 contro
5、ls 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
6、, 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
7、 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-c
8、ell 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
9、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
10、 = (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
11、, (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 suppor
12、ted 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
13、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
14、.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 p
15、er 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 h
16、undred 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%
