express the bollean function F= xy+x'z in the product of max terms
Answers
Explanation:
F = x + y z = x + (y z) AND (multiply) has a higher precedence than OR (add)
= x(y+y')(z+z') + (x+x')yz expand 1st term by ANDing it with (y + y’)(z + z’), and 2nd term with (x + x’)
= x y z + x y z' + x y' z + x y' z' + x y z + x' y z
= m7 + m6 + m5 + m4 + m3
= Σ(3, 4, 5, 6, 7) sum of 1-minterms
Example. Express the Boolean function F = x + y z as a product of maxterms.
Solution: First, we need to convert the function into the product-of-OR terms by using the distributive law as
follows:
F = x + y z = x + (y z) AND (multiply) has a higher precedence than OR (add)
= (x + y) (x + z) use distributive law to change to product of OR terms
= (x + y + z z') (x + y y' + z) expand 1st term by ORing it with z z', and 2nd term with y y'
= (x + y + z) (x + y + z') (x + y + z) (x + y' + z)
= M0 • M1 • M2
= Π(0, 1, 2) product of 0-maxterms
Example. Express F ' = (x + y z)' as a sum of minterms.
Solution:
F' = (x + y z)' = (x + (y z))' AND (multiply) has a higher precedence than OR (add)
= x' (y' + z') use dual or De Morgan’s Law
= (x' y') + (x' z') use distributive law to change to sum of AND terms
= x' y' (z + z') + x' (y + y') z' expand 1st term by ANDing it with (z + z'), and 2nd term with (y + y')
= x' y' z + x' y' z' + x' y z' + x' y' z'
= m1 + m0 + m2
= Σ(0, 1, 2) sum of 0-minterms
Example. Express F ' = (x + y z)' as a product of maxterms.
Solution:
F' = (x + y z)' = (x + (y z))' AND (multiply) has a higher precedence than OR (add)
= x' (y' + z') use dual or De Morgan’s Law
= (x' + y y' + z z') (x x' + y' + z') expand 1st term by ORing it with y y' and z z', and 2nd term with x x'
= (x' + y + z) (x' + y + z') (x' + y' + z) (x' + y' + z') (x + y' + z') (x' + y' + z')
= M4 • M6 • M5 • M7 • M3
= Π(3, 4, 5, 6, 7) product of 1-maxterms
Answer:
F = π(0,2,4,5)
Explanation:
F= xy+x'z
=(x+x')(x+z)(y+x')(y+z) use distributive law and expand previous term
=1.(x+z+yy')(y+x'+zz')(y+z+xx') add the variable that is not present in each term
=(x+z+y)(x+z+y')(x'+y+z)(x'+y+z')(x+y+z)(x'+y+z) separate with each added variable
=(x+z+y)(x+z+y')(x'+y+z)(x'+y+z')
=M0 M2 M4 M5 write respective maxterm positions
=π(0,2,4,5)