solutions_hw7_09

solutions_hw7_09 - Homework #7 ECE 15a Solutions Problem-1...

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Homework #7 ECE 15a Winter 2009 Solutions Problem-1 (15p) A B C D E Y 1 Y 2 Y 3 Y 4 Y 5 Y 6 Y 7 0 0 0 0 0 X X X X X X X 0 0 0 0 1 X X X X X X X 0 0 0 1 0 X X X X X X X 0 0 0 1 1 1 0 1 1 1 1 1 0 0 1 0 0 X X X X X X X 0 0 1 0 1 0 0 0 0 1 1 0 0 0 1 1 0 1 1 1 0 1 0 1 0 0 1 1 1 X X X X X X X 0 1 0 0 0 X X X X X X X 0 1 0 0 1 1 1 1 0 1 1 0 0 1 0 1 0 0 1 0 1 1 1 0 0 1 0 1 1 X X X X X X X 0 1 1 0 0 1 1 1 1 0 1 0 0 1 1 0 1 X X X X X X X 0 1 1 1 0 X X X X X X X 0 1 1 1 1 X X X X X X X 1 0 0 0 0 X X X X X X X 1 0 0 0 1 1 1 1 1 0 1 1 1 0 0 1 0 1 0 0 0 1 1 0 1 0 0 1 1 X X X X X X X 1 0 1 0 0 1 1 1 1 1 1 1 1 0 1 0 1 X X X X X X X 1 0 1 1 0 X X X X X X X 1 0 1 1 1 X X X X X X X 1 1 0 0 0 1 1 1 1 1 1 0 1 1 0 0 1 X X X X X X X 1 1 0 1 0 X X X X X X X 1 1 0 1 1 X X X X X X X 1 1 1 0 0 X X X X X X X 1 1 1 0 1 X X X X X X X 1 1 1 1 0 X X X X X X X 1 1 1 1 1 X X X X X X X Y 1 = A’ (BD’ + B’D) + A Y 2 = A’ (B + E’) + A D’ Y 3 = A’ (BD’ + B’D) + AD’ Y 4 = A’ (D’E’ + C’D) + AD’ Y 5 = A’ (D + E) + AE’ Y 6 = A’ (B + E) + A Y 7 = A’B’D + AB’D’
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A B C D E X Y W Z 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 1 0 1 1 0 1 1 0 0 0 0 1 1 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 1 1 0 0 1 0 0 0 0 1 1 0 0 1 1 0 0 0 1 1 0 1 0 0 0 0 0 1 1 0 1 0 1 0 0 0 1 1 0 1 1 0 0 0 0 1 1 0 1 1 1 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 0 1 0 0 0 1 1 1 0 1 0 0 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 0 1 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 1 X = A’BCD’E Y = A’BCDE’ W = A’BC’DE’ Z = A + A’ (BD’E’ + BDE + BC’D’) or Z = A + A’ (BD’E’ + BDE + BC’E)
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This note was uploaded on 02/02/2011 for the course ECE 15A taught by Professor M during the Spring '08 term at UCSB.

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solutions_hw7_09 - Homework #7 ECE 15a Solutions Problem-1...

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