midterm_solutions

# X n x n p n n p x x 2 p x 0 4 summationdisplay n 1 p

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X | N ( x | n ) p N ( n ) p X ( x ) . 2

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p X (0) = 4 summationdisplay n =1 p ( X,N ) (0 , n ) = 4 summationdisplay n =1 p ( X | N ) (0 | n ) p N ( n ) = 4 summationdisplay n =1 n 1 2 n · n 10 = 4 summationdisplay n =1 n 1 20 = 3 10 . p X (1) = 1 p X (0) = 7 10 . Thus, p N | X ( n | 0) = n - 1 2 n · n 10 3 10 = n 1 6 p N | X ( n | 1) = n +1 2 n · n 10 7 10 = n + 1 14 (c) ˆ N that minimizes the probability of error P e, 1 = P { ˆ N negationslash = N | X = 1 } is ˆ N ( X = 1) = arg max n p N | X ( n | 1) = arg max n n + 1 14 = 4 (d) The resulting P e, 1 is P e, 1 = P { ˆ N negationslash = N | X = 1 } = P { N negationslash = 4 | X = 1 } = 3 summationdisplay n =1 p N | X ( n | 1) = 3 summationdisplay n =1 n + 1 14 = 9 14 . 3. Joshua’s Story on LLR-Based Detection (40 points) Joshua, a fresh graduate from UCSD, joined a cell phone chip company as a Modem Systems Engineer. As Joshua’s first task, his technical lead asks to design a detector of a binary signal X that goes through an AWGN channel. Y = X + Z, where X = braceleftBigg 1 , with probability 1 / 2 1 , with probability 1 / 2 and Z ∼ N (0 , σ 2 ). Unfortunately, the detector that Joshua needs to design cannot directly observe the channel output Y . Instead, it can observe only a processed version of the channel output, L = g ( Y ), where g ( y ) = ln parenleftbigg f Y | X ( y | 1) f Y | X ( y | − 1) parenrightbigg , which is called as the log likelihood ratio (LLR). As a student of ECE 153, can you help Joshua design the optimum detector based on the LLR observation and successfully complete his first task?
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