ENC3250 Solutions pg 52

ENC3250 Solutions pg 52 - b(56 – 40(70 – 40 =(A –...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: b. (56 – 40)/(70 – 40) = (A – 31)/(59.13 – 31); A = 46 Erlangs/cell c. Number of calls/hour/cell = 46/(100/3600) = 1656 Number of calls/hour/km2 = 1656/8 = 207 d. Number of users/hour/cell = 1656/1.2 = 1380 Number of users/hour/channel = 1380/56 = 24.6 e. The total number of cells is 4000/8 = 500 η = (46 Erlangs/cell × 500 cells)/(12.5 MHz × 4000 km2 ) = 0.46 Erlangs/MHz/km2 Source: [GARG96] 10.9 (12.5 × 10 6 – 2(10 x 103 )/(30 × 103 ) = 416 10.10 (25 × 10 6 )/((200 x 103 )/8) = 1000 Source: [RAPP96] 10.11 a. From Figure 10.14, we have 156.25 bits in 0.577 ms. Thus, bit duration is: (0.577 x 10–3)/156.25 = 3.6928 µ s b. The delay is the duration of 1 frame, which is 4.615 ms 10.12 Total bits in one timeslot = 156.25. Data bits = 114. Overhead = (156.25 – 114)/156.25 = 0.27 10.13 Slow FHSS = multiple signal elements per hop. In GSM, the frequency is changed one per frame, which is many bits, so GSM uses slow frequency hopping. 10.14 a. The amount of bandwidth allocated to voice channels (BcNt) must be no greater than the total bandwidth (Bw). Therefore ηa ≤ 1. b. x = (30 × 103 × 395)/(12.5 × 106 ) = 0.948 10.15 a. Number of subscribers = Traffic/0.03 Cell number Subscribers 1 1026.7 2 2223.3 3 1620.0 4 1106.7 5 1273.3 6 1260.0 7 1086.7 b. Number of calls per hour per subscriber = λ = A/h = 0.03/(120/3600) = 0.9 c. Multiply results of part (a) by 0.9 Cell number Calls per hour 1 924 2 2001 3 1458 4 996 5 1146 6 1134 7 978 d. The table in the problem statement gives the value of A. Use P = 0.02. Find N. Cell number Channels 1 40 2 78 3 59 4 43 5 48 6 48 7 42 e. Total number of subscribers = the sum of the values from part (a) = 9597 f. From (d), the total number of channels required = 358 Average number of subscribers per channel = 9597/358 = 26.8 g. Subscriber density = 9597/3100 = 3.1 subscribers per km2 h. Total traffic = the sum of the values from table in the problem statement = 287.9 i. Erlangs per km2 = 287.9/3100 = 0.09 -52- ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online