Question

Express the Boolean function F = XY + X'Z in a product of maxterm form?

Answer 1

Answer:

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Answer 2

Answer:Example. Express the Boolean function 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+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

F and F' are shown in the following truth table:

x y z Minterms Maxterms F F '

0 0 0 m0=x' y' z' M0=x + y + z 0 1

0 0 1 m1=x' y' z M1=x + y + z' 0 1

0 1 0 m2=x' y z' M2=x + y' + z 0 1

0 1 1 m3=x' y z M3=x + y' + z' 1 0

1 0 0 m4=x y' z' M4=x' + y + z 1 0

1 0 1 m5=x y' z M5=x' + y + z' 1 0

1 1 0 m6=x y z' M6=x' + y' + z 1 0

1 1 1 m7=x y z M7=x' + y' + z' 1 0

Explanation: