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Unformatted text preview: ECE 402: Communications Engineering Homework 6 Solutions (Fall 2009) 1. (a) For polar signaling, P b = Q ( p 2 E b /N ). From the reference sheet, we see that Q (3 . 719) = 10 4 and Q (4 . 264) = 10 5 . Hence to obtain P b = 10 4 , we must have s 2 E b N = 3 . 719 = E b N dB = 10log 10 (3 . 719) 2 2 ! = 8 . 4 dB . To obtain P b = 10 5 , we need s 2 E b N = 4 . 264 = E b N dB = 10log 10 (4 . 264) 2 2 ! = 9 . 6 dB . (b) Unipolar signaling ( P b = Q ( p E b /N )) requires twice as much (3 dB more) energyperbit than polar signaling for the same P b , so E b /N = 11 . 4 dB for P b = 10 4 and E b /N = 12 . 6 dB for P b = 10 5 . 2. The error probability of BPSK is P b = Q ( p 2 PT/N ). From the reference sheet, we see that Q (4 . 753) = 10 6 so q 2 PT/N = 4 . 753 = P = 11 . 3 N T = 11 . 3 N R Since N = kT r , the required noise temperature is T r = P 11 . 3 Rk = 5 10 18 11 . 3 4740 1 . 38 10 23 = 6 . 76 o K ....
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This note was uploaded on 01/28/2012 for the course ECE 402 taught by Professor Townsend during the Fall '08 term at N.C. State.
 Fall '08
 Townsend

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