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Unformatted text preview: EE 478 Handout #14 Multiple User Information Theory October 20, 2008 Homework Set #3 Solutions 1. Solution: Let Y = X 1 + X 2 . Then P( Y = 0) = (1 − p 1 )(1 − p 2 ), P( Y = 1) = (1 − p 1 ) p 2 + p 1 (1 − p 2 ), and P( Y = 2) = p 1 p 2 , and H ( Y ) = − summationdisplay y ∈{ , 1 , 2 } P( Y = y )log P( Y = y ) , = f ( p 1 ,p 2 ) . By differentiating with respect to p 1 and p 2 , and solving ∂f ∂p 1 = ∂f ∂p 2 = 0 , we get p 1 = p 2 = 0 . 5 which corresponds to H ( Y ) = 1 . 5. To verify that the entropy of Y is maximized at this point, we also need to investigate the boundary points. But if p i ∈ { , 1 } for i = 1 or i = 2, then H ( Y ) ≤ 1. 2. Solution: (a) The capacity region of this channel is the convex closure of the set of all ( R 1 ,R 2 ) such that R 1 ≤ H ( Y  X 1 ,Q ) , R 2 ≤ H ( Y  X 2 ,Q ) , R 1 + R 2 ≤ H ( Y  Q ) for some p ( q ) p ( x 1  q ) p ( x 2  q ). But since the RHS of each inequality is bounded by 1, the capacity region is the triangle region specified by R 1 + R 2 ≤ 1. If we let X 1 ∼ Bern( p 1 ) and X 2 ∼ Bern( p 2 ), we see that the points on the boundary are achievable when p 1 p 2 = 1 − p 1 p 2 , i.e., p 1 p 2 = 1 / 2. (b) The capacity region of this channel is again the convex closure of the set of all ( R 1 ,R 2 ) such that R 1 ≤ H ( X 1  Q ) , R 2 ≤ H ( X 2  Q ) , R 1 + R 2 ≤ H ( Y  Q ) for some p ( q ) p ( x 1  q ) p ( x 2  q ). But since the RHS of each inequality is bounded by 1, the capacity region is the triangle region specified by R 1 + R 2 ≤...
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This note was uploaded on 10/24/2011 for the course ELECTRICAL ECE 571 taught by Professor Kelly during the Spring '11 term at University of Illinois, Urbana Champaign.
 Spring '11
 Kelly

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