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Unformatted text preview: University of Illinois Spring 2011 ECE 361: Problem Set 2: Solutions Matched and Unmatched Filtering, and Signal Design Throughout this Problem Set, assume that the bits being transmitted are equally likely to be 0 or 1. 1. [Matched Filtering] (a) E = Z T/ 2 T/ 2 A 2 dt = A 2 T ; E 1 = Z T/ 2 T/ 2 2 A 2 cos 2 ( t/T ) dt = A 2 T . (rms amplitude!) h s ,s 1 i = Z T/ 2 T/ 2 2 A 2 cos( t/T ) dt = 2 A 2 T sin( t/T ) T/ 2 T/ 2 = 2 2 A 2 T . Hence, P ( E ) = Q r E + E 1 2 h s ,s 1 i 2 N = Q s A 2 T (1 2 2 / ) N . (b) Now obviously E = E 1 = A 2 T as before while h s ,s 1 i = 2 2 A 2 T giving P ( E ) = Q s A 2 T (1 + 2 2 / ) N . This is smaller than the answer to part (a) because the argument of Q ( ) is larger. Simplistically, the signals in part (a) resemble each other while the signals in part (b) look more like antipodal signals which are known to give the smallest error probability. (c) Now E = E 1 = A 2 T as before while h s ,s 1 i = 0 giving P ( E ) = Q s A 2 T N !...
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 Spring '09

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