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Unformatted text preview: ECE 6325/5325 Spring 2010: Exam 3 Solutions 1. (20 pts) (a) For d = [1 , , 1], we need to find d = c G . In this case d = [1 , , 1 , 1 , 2 , 1] mod 2 = [1 , , 1 , 1 , , 1]. (b) The coded bits [1 , , 1 , , , 0] are received. We need to find r S = [1 , 2 , 1] mod 2 = [1 , , 1]. Because the product is not all zeros, we know there was an error. This error code [1 , , 1] is the 2nd row of S, so we know the 2nd bit was in error, that is, [1 , 1 , 1 , , , 0] should have been received. This means that [1 , 1 , 1] were the data bits sent. 2. (20 pts) From the starting state: s 1 = 0, s 2 = 1, s 3 = 0, see the above figure to see the subsequent states at each time. We were asked for the first seven outputs, so you only need the first seven. But, the 8th time should be the same as time 1, since the PN code repeats after 2 3 1 = 7 steps. Final solution: 0 1 0 1 1 1 0 3. (10 pts) We have a fourantenna receiver using selection combining. We have set the fade margin to 10 dB. What is the probability that our received signal power will be lower than 10 dB below itsto 10 dB....
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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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