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hw02_SLN - ECE 211 HW 2 Solutions p 1 of 6...

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ECE 211 HW 2 Solutions p 1 of 6 //home/vdimitrov/25330/91a9f0118ad1a0bf9591e573625e7aee98ab48ac.doc HW Set 2 – Solutions 1. Use DeMorgan’s Laws to obtain the negation of (A + B ' + C ) ( A' + D ) ( E' + F ) Solution: [ (A + B ' + C ) ( A' + D ) ( E' + F) ] ' = ( A + B' + C ) ' + ( A' + D )' + (E' + F )' = ( A' B C' ) + ( A D' ) + ( E F' )
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ECE 211 HW 2 Solutions p 2 of 6 2. Use DeMorgan’s Laws to obtain the negation of (A' B C' ) + ( A D' ) + [ F + ( G H' ) ] Solution: { (A' B C' ) + ( A D' ) + [ F + ( G H' ) ] }' = (A' B C' )' ( A D' )' [ F + ( G H' ) ] ' = (A + B' + C) (A' + D) [ ( F' ) (G H' )' ] = ( A + B' + C ) (A' + D) ( F' ) (G' + H ) ]
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ECE 211 HW 2 Solutions p 3 of 6 3. Write a logic expression for E = "All 3-bit subsets of the four bits b3, b2, b1, b0 have even parity." (One possible subset consists of the bits b3, b2 and b1. There are a total of four different 3-bit subsets.) Solution: All 3-bit subsets of ( b3 b2 b1 b0 ) have even parity. There are four 3-bit subsets: ( b3 b2 b1 ), ( b3 b2 b0 ), ( b3 b1 b0 ), ( b2 b1 b0 ) Subset ( b3 b2 b1 ) has even parity iff ( b3 b2 b1 ) ' All subsets have even parity iff ( b3 b2 b1 ) ' ( b3 b2 b0 ) ' ( b3 b1 b0 ) '
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