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# Soln07 - ECE 450 Turbo Codes and Iterative Decoding Fall...

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ECE 450 Turbo Codes and Iterative Decoding Fall 2005, Tiffany J Li Solutions to Homework 7 1. Problem 7.1 Solution: The amplitudes A m take the values A m = (2 m - 1 - M ) d 2 , m = 1 , . . . M Hence, the average energy is E av = 1 M M X m =1 s 2 m = d 2 4 M E g M X m =1 (2 m - 1 - M ) 2 = d 2 4 M E g M X m =1 [4 m 2 + ( M + 1) 2 - 4 m ( M + 1)] = d 2 4 M E g 4 M X m =1 m 2 + M ( M + 1) 2 - 4( M + 1) M X m =1 m ! = d 2 4 M E g 4 M ( M + 1)(2 M + 1) 6 + M ( M + 1) 2 - 4( M + 1) M ( M + 1) 2 ! = M 2 - 1 3 d 2 4 E g 2. Problem 7.2 Solution: The correlation coefficient between the m th and the n th signal points is γ mn = s m · s n | s m || s n | where s m = ( s m 1 , s m 2 , . . . , s mN ) and s mj = ± q E s N . Two adjacent signal points differ in only one coordinate, for which s mk and s nk have opposite signs. Hence, s m · s n = N X j =1 s mj s nj = X j 6 = k s mj s nj + s mk s nk = ( N - 1) E s N - E s N = N - 2 N E s Furthermore, | s m | = | s n | = ( E s ) 1 2 so that γ mn = N - 2 N

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The Euclidean distance between the two adjacent signal points is d = q | s m - s n | 2 = s ± 2 q E s /N 2 = s 4 E s N = 2 s E s N 3. Problem 7.5 Solution: a) As an orthonormal set of basis functions we consider the set ψ 1 ( t ) = ( 1 0 t < 1 0 o.w ψ 2 ( t ) = ( 1 1 t < 2 0 o.w ψ 3 ( t ) = ( 1 2 t < 3 0 o.w ψ 4 ( t ) = ( 1 3 t < 4 0 o.w
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Soln07 - ECE 450 Turbo Codes and Iterative Decoding Fall...

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