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X y zx y \ z \ 00011110 01 111 x \ y \ z z y x x yz 3.3 Simplify the following Boolean expression using three- variable maps: d) F(x, y, z) = x y z + x.

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Presentation on theme: "X y zx y \ z \ 00011110 01 111 x \ y \ z z y x x yz 3.3 Simplify the following Boolean expression using three- variable maps: d) F(x, y, z) = x y z + x."— Presentation transcript:

1 x y zx y \ z \ x \ y \ z z y x x yz 3.3 Simplify the following Boolean expression using three- variable maps: d) F(x, y, z) = x y z + x \ y \ z + x y \ z \

2 w \ y \ z w x y x z \ z y w x wx yz 3.5 Simplify the following Boolean expression using four- variable maps: a) F(w,x, y, z) =  (1,4,5,6,12,14,15) F =w \ y \ z+w x y+ x z \

3 A B \ D A C A \ C \ D D C A B B \ C \ D AB CD 3.5 Simplify the following Boolean expression using four- variable maps: b) F(A,B,C,D)=  (1,5,9,10,11,14,15) F =A \ C \ D+AC+(AB \ D or B \ C \ D)

4 x \ y \ w \ x z y w x wx yz 3.5 Simplify the following Boolean expression using four- variable maps: c) F(w,x, y, z) =  (0,1,4,5,6,7,8,9) F =x \ y \ +w \ x

5 A \ B B D B \ D \ D C A B A \ D \ AB CD 3.5 Simplify the following Boolean expression using four- variable maps: d) F(A,B,C,D)=  (0,2,4,5,6,7,8,10,13,15) F =B \ D \ +BD+(A \ B or A \ D \ )

6 A B C \ A \ B D B \ D \ D C A B AB CD 3.6 Simplify the following Boolean expression using four- variable maps: a)A \ B \ C \ D \ +AC \ D \ +B \ CD \ +A \ BCD+BC \ D F =B \ D \ +A \ BD+ABC \

7 x \ z x y \ z y w x wx yz 3.6 Simplify the following Boolean expression using four- variable maps: b) x \ z+w \ xy \ +w(x \ y+xy \ ) = x \ z+w \ xy \ +wx \ y+wxy \ F =x \ z+xy \ +wx \ y w x \ y

8 B C D A \ B D B \ D \ D C A B A \ B C AB CD F =B \ D \ +A \ BD+BCD+(A \ BC or A \ CD \ ) A \ C D \ 3.6 Simplify the following Boolean expression using four- variable maps: c)A \ B \ C \ D \ +A \ CD \ +AB \ D \ +ABCD+A \ BD

9 A C D \ B C \ D A \ B \ D \ D C A B A B \ C AB CD F =A \ B \ D \ +BC \ D+ACD \ +AB \ C 3.6 Simplify the following Boolean expression using four- variable maps: d)A \ B \ C \ D \ +AB \ C+B \ CD \ +ABCD \ +BC \ D

10 A \ B \ D essential A C essential B C \ essential D C A B A B non-essential AB CD F =BC \ +AC+A \ B \ D 3.10 Simplify the following Boolean function by first finding the essential prime implicants:c) F(A,B,C,D)=  (1,3,4,5,10,11,12,13,14,15) A \ C \ D non-essential B \ C D non-essential Redundant

11 A B \ A C C \ D D C A B AB CD F = C \ D+AC+AB \ 3.13 Simplify the following Boolean expression to (1) sum-of-products (2) product-of-sums b) ACD \ +C \ D+AB \ +ABCD 1)

12 B C \ D \ A \ C A \ D \ D C A B AB CD 3.13 Simplify the following Boolean expression to (1) sum-of-products (2) product-of-sums b) ACD \ +C \ D+AB \ +ABCD 2) F \ = A \ D \ +A \ C+BC \ D \ F = (A+D)(A+C \ )(B \ +C+D)


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