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Unformatted text preview: EE 478 Handout #18 Multiple User Information Theory Thursday, October 30, 2008 Homework Set #6 Due: Tuesday, November 11, 2008. 1. Marton inner bound with common messages. Consider the Marton inner bound with common messages described on page 537 of the lecture notes. Decoder j = 1 , 2 looks for the unique message ( m ,m 01 ,m 02 ,ℓ jj ) such that ( U n ( ˆ m , ˆ m 01 , ˆ m 02 ) ,U n 1 ( ˆ m , ˆ m 01 , ˆ m 02 , ˆ l jj ) ,Y n j ) ∈ T ( n ) ǫ . Then ˆ m jj is estimated as the bin index of ˆ l jj in the Marton table corresponding to U n ( ˆ m , ˆ m 01 , ˆ m 02 ). (a) Prove that if the following set of inequalities are satisfied then the probability of error goes to zero as n → ∞ R + R 01 + R 02 + ˜ R 11 < I ( U ,U 1 ; Y 1 ) , R + R 01 + R 02 + ˜ R 22 < I ( U ,U 2 ; Y 2 ) , ˜ R 11 < I ( U 1 ; Y 1  U ) , ˜ R 22 < I ( U 2 ; Y 2  U ) , R 11 + R 22 − ˜ R 11 − ˜ R 22 < I ( U 1 ; U 2  U ) . Hint: The last inequality ensures that encoding is successful. Assume without loss of generality that m = m 01 = m 02 = m 11 = m 22 = 1, and upper bound the probability of error as follows: P ( n ) e < P( E 1 ) + P( E 2 ∩E c 1 ) + P( E 01 ∩ E c 1 ) + P( E 11 ∩ E c 1 ) + P( E 02 ∩ E c 1 ) + P( E 22 ∩ E c 1 ) , where E 1 = { ( U n (1 , 1 , 1) ,U n 1 (1 , 1 , 1) ,U n 2 (1 , 1 , 1) ,X n (1 , 1 , 1) ,Y n 1 ,Y n 2 ) T ( n ) ǫ } , E 01 = { ( U n ( ˜ m , ˜ m 01 , ˜ m 02 ) ,U n 1 ( ˜ m , ˜ m 1 , ˜ m 02 ) ,Y n 1 ) ∈ T ( n ) ǫ for some ( ˜ m , ˜ m 01 , ˜ m 02 ) negationslash = (1 , 1 , 1) and some ˜ m 11 } , E 11 = { ( U n (1 , 1 , 1) ,U n 1 (1 , ˜ m 1 , 1) ,Y n 1 ) ∈ T ( n ) ǫ for some ˜ m 11 negationslash = 1 } , E 02 = { ( U n ( ˜ m , ˜ m 01 , ˜ m 02 ) ,U n 2 ( ˜ m , ˜ m 2 , ˜ m 01 ) ,Y n 2 ) ∈ T ( n ) ǫ for some ( ˜ m , ˜ m 01 , ˜ m 02 ) negationslash = (1 , 1 , 1) and some ˜ m 22 } , E 22 = { ( U n (1 , 1 , 1) ,U n 2 (1 , ˜ m...
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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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