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HW02 - University of Illinois Spring 2011 ECE 361 Problem...

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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 0 = Z T/ 2 - T/ 2 A 2 dt = A 2 T ; E 1 = Z T/ 2 - T/ 2 2 A 2 cos 2 ( πt/T ) dt = A 2 T . (rms amplitude!) h s 0 , s 1 i = Z T/ 2 - T/ 2 2 A 2 cos( πt/T ) dt = 2 A 2 T π sin( πt/T ) T/ 2 - T/ 2 = 2 2 A 2 T π . Hence, P ( E ) = Q r E 0 + E 1 - 2 h s 0 , s 1 i 2 N 0 = Q s A 2 T (1 - 2 2 ) N 0 . (b) Now obviously E 0 = E 1 = A 2 T as before while h s 0 , s 1 i = - 2 2 A 2 T π giving P ( E ) = Q s A 2 T (1 + 2 2 ) N 0 . 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 0 = E 1 = A 2 T as before while h s 0 , s 1 i = 0 giving P ( E ) = Q s A 2 T N 0 ! . The signals are now orthogonal and the error probability is between the answers to parts (a) and (b). Intuitively, orthogonal signals are better than signals that “resemble” each other but antipodal is even better.

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HW02 - University of Illinois Spring 2011 ECE 361 Problem...

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