HW1_Sol

HW1_Sol - ECE15a HW#1-Solutions Q-1(a(ab bc(ab ac bc =...

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ECE15a HW#1-Solutions Q-1 (a) (ab’+bc)(a’b’+ac+bc) = ab’(a’b’+ac+bc)+bc(a’b’+ac+bc) (Using distributive law-3a) = ab’c+abc+bc (Using law of complementation-6a) = ab’c+bc (Using law of absorbtion-6b) (b) (x+y)(x’+y)(x+y’)(x’+y’) = [(x+y)(x’+y)] [(x+y’)(x’+y’)] = [x(x’+y)+y(x’+y)] [x(x’+y’)+y’(x’+y’)] [Using distributive law-3a] = [xy+x’y+y] [xy’+x’y’+y’] [Using law of complementation-6a] = y y’ [Using law of absorbtion-6b] = 0 [Using law of complementation-6a] Q-2 (a) x’y+zw’ = (x’y+z)(x’y+w’) Using distributive law (3b) [x+yz=(x+y)(x+z)] = (x’+z)(y+z)(x’+w’)(y+w’) Using distributive law (3b) [x+yz=(x+y)(x+z)] (b) ax’+ay(x+z) = a [x’+y(x+z)] = a [(x’+y)(x’+x+z)] Using distributive law (3b) [x+yz=(x+y)(x+z)] = a [(x’+y)(1+z)] Using Law of complementation (6b) = a [ ( x + y ) 1 ] U s i n g l a w ( 9 b ) = a ( x + y ) U s i n g l a w ( 1 0 a ) (c) a’bc+ad = (a’bc+a)(a’bc+d) Using distributive law (3b) [x+yz=(x+y)(x+z)] = [(a’+a)(b+a)(c+a)] [(a’+d)(b+d)(c+d)] Using distributive law (3b) [x+yz=(x+y)(x+z)] = 1 (b+a)(c+a)(a’+d)(b+d)(c+d) Using law of complementation (6b)

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= (a+b)(a+c)(a’+d)(b+d)(c+d) Using law (10a) and commutative law (1b) Q-3 (a) (A’+B)’ = AB’ (b) X(Y+Z)=XY+XZ A B B A A B (A’ + B) (A’ + B)’ B’ A AB’ B A A B A B A B A’ B X (Y+Z) X(Y+Z)
Q-4 (a) (a’+b’+c)(a’b+ac’)’ = (a’+b’+c)[(a’b)’(ac’)’] Using De Morgan’s law (8b) = (a’+b’+c)(a+b’)(a’+c) Using De Morgan’s law (8a) = [a’(a+b’)+b’(a+b’)+c(a+b’)] (a’+c) Using Distributive law (3a) = [0+a’b’+ab’+b’+ac+b’c](a’+c) Using Distributive law (3a) and law of

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HW1_Sol - ECE15a HW#1-Solutions Q-1(a(ab bc(ab ac bc =...

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