ENC3250 Solutions pg 46

# ENC3250 Solutions pg 46 - = 20 log(4 4 10 7(2.727 x 102 =...

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-46- = 20 log [(4 π × 4 × 10 7 )/(2.727 x 10 –2 ) = –205.3 dB received_power = 10 + 17 + 52.3 – 205.3 = –126 dBW 9.5 The total bandwidth is 500 MHz. The channel bandwidth is 12 × 36 = 432 MHz. So the overhead is ((500 - 432)/500) × 100% = 13.6% 9.6 a. The data rate R = 2 × QPSK baud rate = 120.272 Mbps The frame duration T = 0.002 s The number of frame bits b F = R × T = 1.20272 × 10 8 × 2 × 10 –3 = 240544 bits Overhead calculation: Overhead bits b o = N R b R + N T b p + (N R + N T )b G where N R = number of participating reference stations = 2 b R = number of bits in reference burst = 576 N T = number of participating traffic stations b p = number of preamble bits =560 b G = number guard bits = 24 b o = (2)(576) + 560 N T + (2)(24) + 24 N T = 1200 + 584 N T b F – b o = 240544 – (1200 + 584 N T ) = 239344 – 584 N T N T = (b F – b o )/16512 = (239344 – 584 N T )/16512 (16512 + 584) N T = 239344 Therefore, NT = 14 b. (b F – b o )/b F = (239344 – 584 ×
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