20121ee132A_1_EE132A_Solutions_HW2

# 20121ee132A_1_EE132A_Solutions_HW2 - Communication Systems...

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Unformatted text preview: Communication Systems EE 132A UCLA Winter quarter 2011/2012 Prof. Suhas Diggavi Handout # 9, Thursday, January 26, 2012 Solutions: Homework Set # 2 Problem 1 (QPSK decision regions) (a) x 1 x 2 s s 2 s 1 s 3 ˆ H = 1 ˆ H = 0 ˆ H = 3 ˆ H = 2 (b) x 1 x 2 s s 2 s 1 s 3 ˆ H = 1 ˆ H = 0 ˆ H = 3 ˆ H = 2 The decision boundary between two hypotheses ˆ H = i and ˆ H = j is given by || y- s i || 2- || y- s j || 2 = 2 σ 2 ln P H ( i ) P H ( j ) If P H ( i ) > P H ( j ), then the hyperplane (in this case it is the line perpendicular to the line joining s i and s j ) is shifted away from s i . (c) 1 x 1 x 2 s = ˜ s s 1 s 3 ˆ H = 1 ˆ H = 0 ˆ H = 3 ˆ H = 2 ˜ s 1 ˜ s 3 s 2 = ˜ s 2 Define a new observarion ˜ Y = ( Y 1 ,Y 2 / 2). The new observation ˜ Y is a sufficient statistic because we can construct Y exactly if we know ˜ Y . Thus the receiver observes ˜ Y = ˜ s i + ˜ Z where ˜ s i = ( s 1 i ,s 2 i / 2) and ˜ Z = ( Z 1 ,Z 2 / 2). Note that in this new setup we have ˜ s = s , ˜ s 1 = s 1 / 2, ˜ s 2 = s 2 , ˜ s 3 = s 3 / 2 and ˜ Z ∼ N (0 ,σ 2 I 2 ). Problem 2 (QAM with Erasure) P 00 = Pr ( { N 1 ≥ - a } ∩ { N 2 ≥ - a } ) = Pr ( { N 1 ≤ a } ) Pr ( { N 2 ≤ a } ) = h 1- Q a σ i 2 ....
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20121ee132A_1_EE132A_Solutions_HW2 - Communication Systems...

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