exam3-retake-s10-5325-solutions

# exam3-retake-s10-5325-solutions - ECE 5325/6325 Spring 2010...

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Unformatted text preview: ECE 5325/6325 Spring 2010: Exam 3 Retake Solutions 1. (25 pts) (a) The coded bits c = d G , so c = [1 , 1 , 1 , 1 , , , 0] 1 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 1 0 1 1 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 1 1 1 1 1 = [1 , 1 , 1 , 1 , , , , 3 , 1 , 3 , 3] mod 2 = [1 , 1 , 1 , 1 , , , , 1 , 1 , 1 , 1] (1) (b) err = [0 , , 1 , 1 , 1 , , , 1 , , , 1] S = [3 , 2 , 2 , 3] mod 2 = [1 , , , 1] (2) which is the 2nd row of S. Thus the 2nd bit was in error, and the coded bits should have been [0 , 1 , 1 , 1 , 1 , , , 1 , , , 1]. The data bits were the first seven, [0 , 1 , 1 , 1 , 1 , , 0]. 2. (20 pts) Answer. Consider M = 1. You need a 30 dB fade margin, not 5 dB. So you increase the transmit power by 25 dB and pay \$500. Consider M = 2, which costs \$100 for the extra antenna....
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exam3-retake-s10-5325-solutions - ECE 5325/6325 Spring 2010...

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