Given the function F (X, Y , Z)=Σm(0,1, 2 , 4 , 6)

answer the following questions:

1. Obtain the expression in the Canonical Disjunctive Normal Form

2. Obtain the expression in the Canonical Conjunctive Normal Form

3. Derive the truth table for both the Minterms and Maxterms

4. Obtain the minimized SOP and POS

5. Draw the resultant circuit diagram for the minimized SOP

(1)Minterms complement = XYZ (2) Compliment of Minterms = Σm(0,1, 2 , 4 , 6) (3) (X+Y+Z) (4) Minimized SOP = Z + XY

Manterms = πM

Explanation:

Solution

Recall that:

Given the function F (X, Y , Z)=Σm(0,1, 2 , 4 , 6)

(1) Canonical Disjunctive Normal Form: In boolean algebra, the boolean function can be expressed as Canonical Disjunctive form known as minterms

In Minterm we assign 'I' to each uncomplimented variable and '0' to each complemented/complementary variable

For the given question stated we ave the following:

Minterms = XYZ, XYZ, XYZ, XYZ, XYZ.

(2) Canonical Conjunctive Normal Form: In boolean algebra, the boolean function can be expressed as Canonical Disjunctive form known as maxterms.

In Maxterms we assign '0' to each uncomplimented variable and '1' to each complemented/complementary variable

Compliment of Minterms = Σm(0,1, 2 , 4 , 6)

Maxterms = πM

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