solutions_homework_4_ece15a_10_new

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Solutions to Homework #4 ECE 15a Winter 2010 1. p q = pq’ + p’q 2. (a) f = (a b)(b c) = (ab’+a’b)(bc’+b’c) = abb’c’ + ab’b’c + a’bbc’ + a’bb’c = 0 + ab’c + a’bc’ + 0 = ab’c + a’bc’ (b) g = (a b) (b c) = (ab’+a’b) (bc’+b’c) = (ab’+a’b)(bc’+b’c)’ + (ab’+a’b)’(bc’+b’c) = (ab’+a’b)(bc’)’(b’c)’ + (ab’)’(a’b)’(bc’+b’c) = (ab’+a’b)(b’+c)(b+c’) + (a’+b)(a+b’)(bc’+b’c) = (ab’+a’b)(bc+b’c’) + (ab+a’b’)(bc’+b’c) = abb’c+ab’c’+a’bc+a’bb’c’+ abc’+abb’c+a’bb’c’+a’b’c = 0 + ab’c’ + a’bc + 0 + abc’ + 0 + 0 + a’b’c = ac’(b’+b) + a’c(b+b’) = ac’ + a’c 3. f(A,B,C) = A’B’+B’C’+A’C (a) NOT: f(a,a,a) = a’a’+a’a’+a’a = a’ + a’ + 0 = a’ AND: f(a,b,b) = a’b’ + b’b’ + a’b = a’(b+b’) + b’ = a’ + b’ = (ab)’ b Need a NOT. OR: f(a,b,a) = a’b’ + b’a’ + a’a = a’b’ + 0 = (a + b)’

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This note was uploaded on 03/17/2010 for the course ECE 15A taught by Professor M during the Spring '08 term at UCSB.

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