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ClassNotes5 - ECE595 Wireless Communications Lecture 5...

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Unformatted text preview: ECE595 Wireless Communications Lecture 5 Spring 2009 [Review of log-normal fading] dB = 10log | ch | 2 , P dB = 1 2 dB e- ( dB- m dB ) 2 2 2 dB , m dB = dB- 10 n log d [Example] Best path loss fit is- 31 . 54 dB- 37 . 1log d for the data below. Find 2 dB , variance of log-normal fading d P r /P t 10 m-70 dB 20 m-75 dB 50 m-90 dB 100 m-110 dB 300 m-125 dB [Solution] Take the average out and average the rest for variance: 2 dB = 1 5 [(- 70 + 31 . 54 + 37 . 1log 10) 2 + + (- 75 + 31 . 54 + 37 . 1log 20) 2 + = 13 . 29 dB = 3 . 65 2 Lecture 5 [Outage Probability] prob { P r dB P min } P r dB = P T dB + dB (Normal distribution) prob { P r dB P min } = 1- Q P min- dB- 10 n log d z }| { ( P T dB + K dB- 10 log d ) dB Q ( z ) = 1 2 Z z e- x 2 / 2 dx [Example 2.5 of Goldsmiths] Find outage probability at 150 m, P min =- 110 . 5 dBm ,P t = 10mW for previous example....
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ClassNotes5 - ECE595 Wireless Communications Lecture 5...

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