solutions_homework_1_ece15a_2009 - Solutions to Homework#1...

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Solutions to Homework #1 ECE 15a Winter 2009 1. (a) (ab’+bc)(a’b’+ac+bc) = ab’(a’b’+ac)+bc = aa’b’+ab’c+bc = ab’c+bc = ab’c+(a+1)bc = a(b+b’)c+bc = ac+bc (b) (x+y)(x’+y)(x+y’)(x’+y’) = (xx’+y)(xx’+y’) = yy’ = 0 2. (a) x’y+zw’ = (x’y+z)(x’y+w’) = (x’+z)(y+z)(x’+w’)(y+w’) (b) ax’+ay(x+z) = a(x’+y(x+z)) = a(x’+y)(x’+x+z) = a(x’+y).1 = a(x’+y) (c) a’bc+ad = (a’bc+a)(a’bc+d) = (a’+a)(bc+a)(a’+d)(bc+d) = 1.(b+a)(c+a)(a’+d)(b+d)(c+d) = (a+b)(a+c)(a’+d)(b+d)(c+d) 3. (a) (A’+B)’ = AB’ A’: A’+B: (A’+B)’: B’: AB’: (b) X(Y+Z) = XY + XZ Y+Z: X(Y+Z): XY: , XZ: XY+XZ:
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4. (a) (a’+b’+c)(a’b+ac’)’ = (a’+b’+c)[(a’b)’(ac’)’] = (a’+b’+c)(a+b’)(a’+c) = (a’+b’+c)(aa’+ac+a’b’+b’c) = (a’+b’+c)(ac+a’b’+b’c) = aa’c+a’b’+a’b’c+ab’c+a’b’+b’c+ac+a’b’c+b’c = 0+a’b’+(a+a’+1)b’c+ac = a’b’+a’b’c+ab’c+ac = a’b’(1+c)+a (b’+1)c = a’.1.b’+a.1.c = a’b’+ac (b) (x’y+xy+x’y)(x’y+zw) = y(x’+x+x’)(x’y+zw) = y(x’y+zw) = x’y+zyw (c) abx’+abx+x’abx = ab(x’+x+x’x) = ab
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