sol.HW4

sol.HW4 - EE 7615 Solutions to Problem Set # 4 10/26/2010...

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Unformatted text preview: EE 7615 Solutions to Problem Set # 4 10/26/2010 1. Let ( t ) = q 2 T cos(2 f t ) for 0 t T . Then we get the following signal vectors. s i ( t ) = d i + 1- q 2 ( t ) , s i = d i + 1- q 2 (a) For q = 4, s s 1 s 3 s 2 h (t) g(r) r(t) t=T r V ^ r where h r ( t ) = q 2 T cos(2 f t ) and g ( r ) = m r - d m 1- d < r m 2 < r d m 3 d < r (b) For the two signals on the outside, we have P ( E | s ) = P ( E | s q- 1 ) = P ( N d 2 ) = Q d 2 N ! For all the other q- 2 signals, ( i 6 = 0 ,q- 1) P ( E | s i ) = P ( N d 2 or N - d 2 ) = 2 P ( N d 2 ) = 2 Q d 2 N ! Thus P ( E ) = 1 q q- 1 X i =0 P ( E | s i ) = 2( q- 1) q Q ( d 2 N ) 2 Q ( d 2 N ) 1 (c) E ave = 1 q q- 1 X i =0 E i = 1 q "- q + 1 2 2 d 2 +- q + 3 2 2 d 2 + + q- 1 2 2 d 2 # = 1 q d 2 2 q/ 2 X i =1 (2 i- 1) 2 = d 2 2 q 4( q/ 2) 3- q/ 2 3 = d 2 12 ( q 2- 1) (d) Let A k denote the amplitude of the signal in the...
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sol.HW4 - EE 7615 Solutions to Problem Set # 4 10/26/2010...

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