hw-sln-04

hw-sln-04 - S07 EEE 350 Random Signal Analysis Homework 4...

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S07 EEE 350 Random Signal Analysis Homework 4 Solutions Feb. 15th, 2007 Problem Solutions : Yates and Goodman, 1.10.1 1.10.2 2.2.1 2.2.3 2.2.6 and 2.2.9 Problem 1.10.1 Solution From the problem statement, we can conclude that the device components are confgured in the ±ollowing way. W 1 W 2 W 5 W 3 W 4 W 6 To fnd the probability that the device works, we replace series devices 1, 2, and 3, and parallel devices 5 and 6 each with a single device labeled with the probability that it works. In particular, P [ W 1 W 2 W 3 ] = (1 - q ) 3 , (1) P [ W 5 W 6 ] = 1 - P [ W c 5 W c 6 ] = 1 - q 2 . (2) This yields a composite device o± the ±orm 1-q 2 1-q ( ) 1-q 3 The probability P [ W 0 ] that the two devices in parallel work is 1 minus the probability that neither works: P ± W 0 ² = 1 - q (1 - (1 - q ) 3 ) . (3) Finally, ±or the device to work, both composite device in series must work. Thus, the probability the device works is P [ W ] = [1 - q (1 - (1 - q ) 3 )][1 - q 2 ] . (4) Problem 1.10.2 Solution Suppose that the transmitted bit was a 1. We can view each repeated transmission as an

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This note was uploaded on 09/26/2008 for the course EEE 350 taught by Professor Duman during the Spring '08 term at ASU.

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hw-sln-04 - S07 EEE 350 Random Signal Analysis Homework 4...

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