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# 6s - EE 478 Handout#22 Multiple User Information Theory...

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Unformatted text preview: EE 478 Handout #22 Multiple User Information Theory November 13, 2008 Homework Set #6 Solutions 1. Solution: (a) Codebook generation: For each m ∈ [1 : 2 nR ], m 01 ∈ [1 : 2 nR 01 ], and m 02 ∈ [1 : 2 nR 02 ] generate u n ( m ,m 01 ,m 02 ) ∼ producttext n i =1 p U ( u i ). Choose ˜ R 11 > R 11 and ˜ R 22 > R 22 , and for each ( m ,m 01 ,m 02 ) generate a Marton table as described in the Lecture note, i.e., for each ℓ 11 ∈ [1 : 2 n ˜ R 11 ] generate u n 1 ( ℓ 11 | m ,m 01 ,m 02 ) ∼ producttext n i =1 p U 1 | U ( u 1 i | u i ) and for each ℓ 22 ∈ [1 : 2 n ˜ R 22 ] generate u n 2 ( ℓ 22 | m ,m 01 ,m 02 ) ∼ producttext n i =1 p U 2 | U ( u 2 i | u i ). Divide 2 n ˜ R 11 u n 1 sequences into 2 nR 11 equal size bins and divide 2 n ˜ R 22 u n 2 sequences into 2 nR 22 equal size bins. For each message ( m ,m 1 ,m 2 ), define B ( m ,m 1 ,m 2 ) = { ( u n ( m ,m 01 ,m 02 ) ,u n 1 ( ℓ 11 | m ,m 01 ,m 02 ) ,u n 2 ( ℓ 22 | m ,m 01 ,m 02 )) ∈ T ( n ) ǫ ( U ,U 1 ,U 2 ) : ( m 11 − 1)2 n ( ˜ R 11 − R 11 ) + 1 ≤ ℓ 11 ≤ m 11 2 n ( ˜ R 11 − R 11 ) , ( m 22 − 1)2 n ( ˜ R 22 − R 22 ) + 1 ≤ ℓ 22 ≤ m 22 2 n ( ˜ R 22 − R 22 ) bracerightBig . For each message ( m ,m 1 ,m 2 ) look into B ( m ,m 1 ,m 2 ) and if it is non empty then choose an arbitrary pair ( u n ,u n 1 ,u n 2 ) ∈ B ( m ,m 1 ,m 2 ), and generate x n ∼ producttext n i =1 p X | U ,U 1 ,U 2 ( x i | u ,u 1 i ,u 2 i ). If it is empty choose an arbitrary sequence x n . Decoding: The decoder i looks for some ( ˆ m , ˆ m 01 , ˆ m 02 , ˆ ℓ ii ) such that ( u n ( ˆ m , ˆ m 01 , ˆ m 02 ) ,u n i ( ˆ ℓ ii | ˆ m , ˆ m 01 , ˆ m 02 ) ,y n i ) ∈ T ( n ) ǫ . Finally, it declares the bin number of ˆ ℓ ii as message ˆ m 11 . Analysis of probability of error: Assume without loss of generality that m = m 01 = m 02 = m 11 = m 22 = 1. The decoding probability of error of decoder i can be upper bounded as 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 2 , 1) ,Y n 2 ) ∈ T ( n ) ǫ for some ˜ m 22 negationslash = 1 } ....
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6s - EE 478 Handout#22 Multiple User Information Theory...

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