homework6b_solution

homework6b_solution - . 125 = 40 . 0 dB. Using the path...

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ECE 5325/6325 Fall 2009: Homework 6(b) Solution For (i), with 1 MHz, we can achieve bit rate R b = log 2 M B 1+ α . BPSK: R b = 1 1 1 . 25 = 0 . 8 Mbps; QPSK R b = 2 1 1 . 25 = 1 . 6 Mbps; 16-QAM: R b = 4 1 1 . 25 = 3 . 2 Mbps; 64-QAM: R b = 6 1 1 . 25 = 4 . 8 Mbps. For (ii), we need to invert the P [bit error] formulas for 10 - 4 probability of bit error. For BPSK and QPSK, E b N 0 = 1 2 b Q - 1 ( 10 - 4 )B 2 = 6 . 92 For 16-QAM, E b N 0 = 5 4 b Q - 1 ( (4 / 3)10 - 4 )B 2 = 16 . 61 For 64-QAM, E b N 0 = 7 2 b Q - 1 ( (12 / 7)10 - 4 )B 2 = 44 . 87 Using S N = E b N 0 R b B , BPSK: S N = 6 . 92 0 . 8 1 = 5 . 53 = 7 . 43 dB; QPSK: S N = 6 . 92 1 . 6 1 = 11 . 06 = 10 . 44 dB; 16-QAM: S N = 16 . 61 3 . 2 1 = 53 . 16 = 17 . 26 dB; 64-QAM: S N = 44 . 87 4 . 8 1 = 215 . 4 = 23 . 3 dB; For the maximum range, we form the link budget with P t = - 10 dBW Gains: 6 dB total antenna gains Losses: a 20 dB fade margin; Π 0 (dB) = 20log 10 4 πd λ = 20log 10 4 π (1)
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Unformatted text preview: . 125 = 40 . 0 dB. Using the path loss exponent model, we also have 30log 10 ( d/ 1m) losses as a function of path length d . Noise: F = 4 in linear. P N = FkT B = 4(1 . 38 10-23 J/K)(294 K)1 MHz = 1 . 62 10-14 W =-137 . 9 dBW. So: S/N = P t + Gains-Losses-P n S/N =-10(dBW) + 6-20-40 .-30log 10 ( d/ 1m) + 137 . 9(dBW) S/N = 73 . 9-30log 10 ( d/ 1m) d = 1 m 10 [73 . 9( dB )-S/N ( dB )] / 30 (1) So based on the S/N values calculated above, BPSK: d = 164 m; QPSK: d = 130 m; 16-QAM: d = 77 m; 64-QAM: d = 49 m....
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This note was uploaded on 02/25/2011 for the course ECE 5325 taught by Professor Dr.nealpatwari during the Spring '10 term at University of Utah.

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