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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 1023 J/K)(294 K)1 MHz = 1 . 62 1014 W =137 . 9 dBW. So: S/N = P t + GainsLossesP n S/N =10(dBW) + 62040 .30log 10 ( d/ 1m) + 137 . 9(dBW) S/N = 73 . 930log 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; 16QAM: d = 77 m; 64QAM: 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.
 Spring '10
 Dr.NealPatwari

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