Unformatted text preview: 3. Losses: There is the 40 dB loss to 1 m, then an additional 10(3 . 0) log 10 d dB. So98(dBm) = 0(dBm) + 6(dB)40(dB)30 log 10 ( d/ 1m) (2) Solving for d , we have . d = (1m)10 64 / 30 = 136m . (3) For part two of this question, now the 98 is reduced to 99 dBm. E±ectively, the LHS of (2) becomes one lower. This increases the 64 on the RHS of (3) to 65, and so d = (1m)10 65 / 30 = 147m....
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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 Utah.
 Spring '10
 Dr.NealPatwari

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