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Unformatted text preview: EE 7615 Solutions to Problem Set # 4 10/26/2010 1. Let φ ( t ) = q 2 T cos(2 πf t ) for 0 ≤ t ≤ T . Then we get the following signal vectors. s i ( t ) = d i + 1 q 2 ¶ φ ( 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 πf 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 ˆ 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 ˆ 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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This note was uploaded on 03/05/2012 for the course EE 7615 taught by Professor Naragipour during the Fall '11 term at LSU.
 Fall '11
 Naragipour

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