# FinalSoln - University of Illinois Spring 2011 ECE 361...

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Unformatted text preview: University of Illinois Spring 2011 ECE 361: Solutions to the Final Examination Friday May 6, 2011, 1:30 p.m. – 4:30 p.m. This is a closed-book closed-notes examination except that three 8.5” × 11” sheets of notes are permitted: both sides may be used. 1. [45 points] Consider a binary digital communication system operating over an AWGN chan- nel with two-sided power spectral density N / 2 per Hz. Four possible transmitted signal sets { s ( t ) ,s 1 ( t ) } of duration T are shown below where all eight signals have value for t < and t > T . The constants A , B , C , and D are to be chosen by you to satisfy various constraints as specified in parts (a)-(c). The question to be answered in each case is: Subject to the stated constraint, which signal set has the smallest error probability (when the appropriate optimum receiver is used) and what is this smallest error probability? E and E 1 denote the signal energies, and P = E /T and P 1 = E 1 /T denote the average signal powers in the two signals. Set I s ( t ) = A, s 1 ( t ) =- √ 3 · A · ( t/T ) , Set II s ( t ) = √ 2 · B · cos( πt/T ) , s 1 ( t ) = √ 2 · B · sin( πt/T ) Set III s ( t ) = √ 2 · C · cos( πt/T ) , s 1 ( t ) =- √ 2 · C · cos( πt/T ) Set IV s ( t ) = D, s 1 ( t ) =- D · ( √ 2)- 1 and all signals have value 0 for t < 0 and t > T . Solution: The data bits are equally likely, and the minimum error probability achieved by a signal set is P ( E ) = Q r E + E 1- 2 h s ,s 1 i 2 N . These numbers are shown below E E 1 ( P + P 1 ) / 2 h s ,s 1 i E + E 1- 2 h s ,s 1 i Set I A 2 T A 2 T A 2- ( √ 3 / 2) A 2 T (2 + √ 3) A 2 T Set II B 2 T B 2 T B 2 2 B 2 T Set III C 2 T C 2 T C 2- C 2 T 4 C 2 T Set IV D 2 T D 2 T/ 2 3 D 2 / 4- D 2 T/ √ 2 (1 . 5 + √ 2) D 2 T (a) The total average power ( P + P 1 ) / 2 is no more than a given positive number P , or equivalently, E + E 1 ≤ 2 P T . Solution: Here, A 2 = B 2 = C 2 = 3 D 2 / 4 = P and so we can express E + E 1- 2 h s ,s 1 i as (2+ √ 3) P T , 2 P T , 4 P T , and (4 / 3)(1 . 5+ √ 2) P T for the 4 sets. Obviously, 2 < 2+ √ 3 < 4, and since (1 . 5+ √ 2) < 3, we see that (4 / 3)(1 . 5+ √ 2) < 4 also, and so Set III has smallest error probability Q q 2 P T/N . (b) Both signal powers P and P 1 are no more than a given positive number P . Solution: The only difference from part (a) is that now D 2 = P which increases the error probability for Set IV. Thus, Set III has smallest error probability Q q 2 P T/N . (c) Both signals must satisfy a peak instantaneous power constraint or equivalently, an am- plitude constraint | s ( t ) | ≤ α and | s 1 ( t ) | ≤ α for all t . Solution: Now √ 3 A = √ 2 B = √ 2 C = D = α and so we can express E + E 1- 2 h s ,s 1 i as [(2+ √ 3) / 3] α 2 T , α 2 T , 2 α 2 T , and (1 . 5+ √ 2) α 2 T for the 4 sets. Since (1 . 5+ √ 2) > 2, Set IV has smallest error probability Q q (0 . 75 + 0 . 5 √ 2) α 2 T/N ....
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FinalSoln - University of Illinois Spring 2011 ECE 361...

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