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# 5s - EE 478 Multiple User Information Theory Handout#20...

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EE 478 Handout #20 Multiple User Information Theory November 6, 2008 Homework Set #5 Solutions 1. Solution: (a) From the lecture notes, the capacity region C BC ( g 1 , g 2 , P ) of the Gaussian broadcast channel is given by the set of rate pairs ( R 1 , R 2 ) such that R 1 C( αS 1 ) , R 2 C parenleftbigg (1 - α ) S 2 αS 2 + 1 parenrightbigg , for some 0 α 1, where S 1 = g 2 1 P and S 2 = g 2 2 P . For the multiple-access channel assume P 1 = βP and P 2 = (1 - β ) P . The capacity region C MAC ( P ) is given by the set of rate pairs ( R 1 , R 2 ) such that R 1 C( βS 1 ) , R 2 C ((1 - β ) S 2 ) , R 1 + R 2 C( βS 1 + (1 - β ) S 2 ) , for some 0 β 1. (b) We show that the corner points of each region is in the other region. Consider ( R 1 , R 2 ) = (C( αS 1 ) , C parenleftBig (1 α ) S 2 αS 2 +1 parenrightBig ) C BC ( g 1 , g 2 , P ). For β = α ( S 2 + 1) αS 2 + 1 , we have C parenleftbigg (1 - α ) S 2 αS 2 + 1 parenrightbigg = C ((1 - β ) S 2 ) . Note that the defined β satisfies β α , therefore R 1 = C( βS 1 ) < C( αS 1 ). Finally, it suffices to show that C( αS 1 ) + C parenleftbigg (1 - α ) S 2 αS 2 + 1 parenrightbigg C( βS 1 + (1 - β ) S 2 ) . Note that C( αS 1 ) + C parenleftbigg (1 - α ) S 2 αS 2 + 1 parenrightbigg = 1 2 log (1 + S 2 )( αS 1 + 1) αS 2 + 1 , and C( βS 1 + (1 - β ) S 2 ) = 1 2 log( α ( S 2 + 1) S 1 αS 2 + 1 + (1 - α ) S 2 αS 2 + 1 + 1) = C( αS 1 ) + C parenleftbigg (1 - α ) S 2 αS 2 + 1 parenrightbigg . 1

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Therefore, C BC ( g 1 , g 2 , P ) C MAC ( g 1 , g 2 , P ). To prove the other direction, let ( R 1 , R 2 ) = (C( βS 1 ) , C( βS 1 + (1 - β ) S 2 ) - C( βS 1 )) be a corner point of the MAC region. Note that C( βS 1 + (1 - β ) S 2 ) - C( βS 1 ) = 1 2 log parenleftbigg βS 1 + (1 - β ) S 2 + 1 βS 1 + 1 parenrightbigg = 1 2 log(1 + (1 - β ) S 2 βS 1 + 1 ) = C( (1 - β ) S 2 βS 1 + 1 ) . Therefore, ( R 1 , R 2 ) C BC . The other possibility for a corner point of the MAC region is ( R 1 , R 2 ) = (C( βS 1 + (1 - β ) S 2 ) - C ((1 - β ) S 2 ) , C ((1 - β ) S 2 )). Note that C( βS 1 + (1 - β ) S 2 ) - C ((1 - β ) S 2 ) = 1 2 log parenleftbigg βS 1 + (1 - β ) S 2 + 1 (1 - β ) S 2 + 1 parenrightbigg = C( βS 1 (1 - β ) S 2 + 1 ) C( βS 1 ) . Hence, C MAC ( g 1 , g 2 , P ) C BC ( g 1 , g 2 , P ) as well. (c) Consider the point ( R 1 , R 2 ) = (C( βS 1 ) , C( βS 1 + (1 - β ) S 2 ) - C( βS 1 )) = parenleftbigg C( βS 1 ) , C( (1 - β ) S 2 βS 1 + 1 ) parenrightbigg , on the boundary of the multiple-access region. The rate pair ( R 1 , R 2 ) can be achieved using successive cancelation by first decoding the message of user 2 assuming the received sequence of user 1 as noise and then decoding message of user 1 after subtracting off the received sequence of user 2. However, R 1 = C( βS 1 ) , R 2 = C parenleftbigg (1 - β ) S 2 βS 1 + 1 parenrightbigg , is also on the boundary of the the broadcast region. Hence the same codebook (with proper power scaling) can be used for the broadcast channel and achieves the same point on the boundary of the region.
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5s - EE 478 Multiple User Information Theory Handout#20...

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