Standard Forms & Terminology

Standard Forms & Terminology (a) [2 marks] The Boolean function below is given as a product of maxterms. Change it into a sum of minterms: F(x, y, z) = ΠΜ (2,3,4) 9+ (b) [3 marks] The Boolean function below is given as a sum of minterms. Express it as a minimal product of sums: F(x, y, z) = Em (0, 2, 3, 4, 7, 9)

The Correct Answer and Explanation is :

Let’s break down both problems step-by-step.

(a) Convert the Boolean function from the product of maxterms to the sum of minterms:

Given Boolean function:
( F(x, y, z) = \Pi M(2, 3, 4, 9) )

This is in the product-of-maxterms (PoM) form, where each term corresponds to a maxterm for which the function is 0. We are asked to express this in a sum-of-minterms (SoM) form, which is the disjunction (OR) of minterms for which the function is 1.

Steps to convert:

1. First, note that the numbers 2, 3, 4, and 9 represent the indices of the maxterms where the function equals 0. These are the locations where the Boolean function is false.
2. The Boolean function is defined on three variables (x, y, z), so it will have ( 2^3 = 8 ) possible combinations of inputs. Maxterms represent the inputs where the function equals 0. Therefore, the remaining minterms (where the function equals 1) will be for the inputs not corresponding to the indices of the maxterms.
3. The indices where the function equals 1 are the complement of the maxterms. Since maxterms 2, 3, 4, and 9 are used, the remaining minterms are:
   ( M(0), M(1), M(5), M(6), M(7) )
4. The sum-of-minterms expression for the given function is:
   ( F(x, y, z) = \sum m(0, 1, 5, 6, 7) )

(b) Express the Boolean function in minimal product-of-sums (PoS) form:

Given Boolean function:
( F(x, y, z) = \sum m(0, 2, 3, 4, 7, 9) )

This is in the sum-of-minterms (SoM) form, where each term corresponds to a minterm for which the function is 1. We are tasked with converting this to the minimal product-of-sums (PoS) form.

Steps to convert:

1. The minterms provided correspond to the Boolean values where the function equals 1. These are for the indices 0, 2, 3, 4, 7, and 9.
2. For each minterm, we need to create a sum-of-products (SOP) that gives a 0 for that minterm. The product-of-sums form means we need to find combinations of inputs that result in the function being 0 for these minterms.
3. Identifying the terms where the function equals 0:
   The Boolean function will be 0 for the remaining minterms, which are the ones not listed:
   ( m(5), m(6) ).
4. We write the corresponding sums (maxterms) for these indices:

• For ( m(5) ), the binary representation is ( 101 ), which translates to the maxterm ( (x’ + y + z’) ) because the corresponding positions are 0, 1, and 0 (for ( x, y, z )).
• For ( m(6) ), the binary representation is ( 110 ), which translates to the maxterm ( (x’ + y’ + z’) ).

1. The minimal product-of-sums form is the product of the maxterms for the indices 5 and 6, so:
   ( F(x, y, z) = (x’ + y + z’)(x’ + y’ + z’) )

Summary of Answers:

(a) ( F(x, y, z) = \sum m(0, 1, 5, 6, 7) )
(b) ( F(x, y, z) = (x’ + y + z’)(x’ + y’ + z’) )

Explanation:

• In part (a), we took the complement of the maxterms (indices 2, 3, 4, 9) to identify the minterms where the function is 1. These minterms were then expressed as a sum.
• In part (b), we identified the minterms where the function is 0 (indices 5 and 6) and expressed the function as a product of their corresponding maxterms. The minimal form is achieved by using only the necessary maxterms.