Minterm and Maxterm: (PHC504)

Conversion of SOP to POS Form

In digital electronics, Sum of Products (SOP) and Product of Sums (POS) are two canonical forms used to represent Boolean expressions. These forms are essential in the design and analysis of digital circuits as they simplify complex logical expressions and aid in circuit optimization. The conversion of SOP to POS form is a crucial process in logic simplification, allowing for efficient circuit implementation and reduced complexity. This process involves applying distributive, complement, and Boolean algebraic rules to transform the representation while maintaining its logical equivalence. By converting from SOP to POS form, designers can potentially achieve more efficient and optimized implementations of digital circuits.

Sum of Products (SOP) Form:

The Sum of Products (SOP) form is a way to express a Boolean function as a logical sum (OR operation) of multiple terms, where each term is a logical product (AND operation) of literals (variables or their complements). This form is also known as the "minterm" representation.

For instance, consider a Boolean function F(A, B, C) = Σm(0, 1, 2), where m(0), m(1), and m(2) represent the minterms of the function. The SOP representation would be:

F(A, B, C) = ABC + ABC + ABC

Product of Sums (POS) Form:

The Product of Sums (POS) form, on the other hand, expresses a Boolean function as a logical product (AND operation) of multiple terms, where each term is a logical sum (OR operation) of literals. This form is also known as the "maxterm" representation.

Continuing with the example function F(A, B, C), the POS representation would be:

F(A, B, C) = (A + B + C)(A + B + C)(A + B + C)

Conversion from SOP to POS Form:

The conversion from SOP to POS form involves the following steps:

  1. Distribute Over Sum: Apply the distributive property of Boolean algebra, which states that A(B + C) = AB + AC. This step involves distributing each term in the SOP expression over the sum of the remaining terms.

  2. Apply Complement Rule: Utilize the complement rule, which states that A + A = 1. Apply this rule to eliminate terms that are complemented versions of each other.

  3. Factor Out Common Terms: Identify and factor out common terms in the resulting expression. This step aims to group terms that share common literals.

  4. Simplify and Finalize: Continue simplifying the expression by applying additional Boolean algebraic rules until no further simplification is possible.

Conversion Examples:

Let's illustrate the conversion process with some examples:

Example 1: Convert the SOP expression F(A, B, C) = AB + AC + BC to POS form.

Step 1: Distribute Over Sum
AB + AC + BC = AB + AC(B + B) + BC(A + A) = AB + ABC + ABC


Step 2: Apply Complement Rule
AB + ABC + ABC = AB + ABC + ABC


Step 3: Factor Out Common Terms
AB + ABC + ABC = AB + ABC(1 + C)


Step 4: Simplify and Finalize
AB + ABC(1 + C) = AB + ABC

Therefore, the converted POS expression is F(A, B, C) = AB + ABC.

Example 2: Convert the SOP expression F(A, B, C) = AB + AC + BC to POS form.

Step 1: Distribute Over Sum
AB + AC + BC = AB(C + C) + AC(B + B) + BC(A + A) = ABC + AB + AC + BC


Step 2: Apply Complement Rule
ABC + AB + AC + BC = ABC + AB + AC + BC1


Step 3: Factor Out Common Terms
ABC + AB + AC + BC1 = ABC + AB(1 + C) + AC(1 + B) + BC(1 + B)


Step 4: Simplify and Finalize
ABC + AB(1 + C) + AC(1 + B) + BC(1 + B) = ABC + AB + AC + BC

The converted POS expression is F(A, B, C) = ABC + AB + AC + BC.

Example 3: Convert the SOP expression F(A, B, C) = ABC + ABC + ABC to POS form.

Step 1: Distribute Over Sum
ABC + ABC + ABC = AB(C + C) + ABC(A + A) + ABC(B + B)


Step 2: Apply Complement Rule
AB(C + C) + ABC(A + A) + ABC(B + B) = AB + ABC + ABC


Step 3: Factor Out Common Terms
AB + ABC + ABC = AB(1 + C) + ABC(1 + C)


Step 4: Simplify and Finalize
AB(1 + C) + ABC(1 + C) = AB + ABC

The converted POS expression is F(A, B, C) = AB + ABC.

Conversion of POS to SOP form

Converting a Boolean expression from POS (Product of Sums) form to SOP (Sum of Products) form is a fundamental operation in digital electronics. Converting from POS to SOP form involves identifying complements, applying De Morgan's theorem to distribute complements, and finally ORing the distributed terms. This conversion is an essential skill for optimizing digital circuits and reducing complexity. POS and SOP are two common ways to represent Boolean functions, and understanding how to convert between them is essential for optimizing and simplifying digital circuits.

Conversion Process from POS to SOP:

  1. Given POS Expression: Start with the given Boolean expression in POS form. A POS expression consists of multiple product terms (ANDed terms) that are ORed together.
  2. Identify Complements: Identify the variables and their complements within each product term.
  3. Apply De Morgan's Theorem: Apply De Morgan's theorem to distribute the complements inside each product term and rewrite them in SOP form.
  4. OR the Distributed Terms: Finally, OR all the distributed terms obtained from each product term to obtain the equivalent SOP expression.

Conversion examples:

Example 1: Given POS Expression: F = (A + B)(C + D)

1. Identify Complements: A, B, C, D
2. Apply De Morgan's Theorem: F = (ABC) + (ABD) + (BCD)
3. OR the Distributed Terms: F = (ABC) + (ABD) + (BCD)

Example 2: Given POS Expression: F = (A + B)(C + D)(E + F)

1. Identify Complements: C, D, F
2. Apply De Morgan's Theorem: F = (A + B + C)(A + B + D)(A + B + E) + (A + B + C)(A + B + D)(A + B + F)
3. OR the Distributed Terms: F = (A + B + C)(A + B + D)(A + B + E) + (A + B + C)(A + B + D)(A + B + F)

Example 3: Given POS Expression: F = (A + B + C)(D + E + F)

1. Identify Complements: None
2. Apply De Morgan's Theorem: No complements, so no changes needed.
3. OR the Distributed Terms: F = (A + B + C)(D + E + F)

Example 4: Given POS Expression: F = (A + B)(C + D)(E + F)

1. Identify Complements: None
2. Apply De Morgan's Theorem: No complements, so no changes needed.
3. OR the Distributed Terms: F = (A + B)(C + D)(E + F)

Example 5: Given POS Expression: F = (A + B + C)(D + E + F)

1. Identify Complements: A, B, C, D, E, F
2. Apply De Morgan's Theorem: F = (A + B + C)(D + E + F)
3. OR the Distributed Terms: F = (A + B + C)(D + E + F)

Example 6: Given POS Expression: F = (A + B + C)(D + E)(F + G)

1. Identify Complements: C, F
2. Apply De Morgan's Theorem: F = (A + B + C)(D + E)(F + G)
3. OR the Distributed Terms: F = (A + B + C)(D + E)(F + G)

Example 7: Given POS Expression: F = (A + B)(C + D)(E + F)

1. Identify Complements: B, F
2. Apply De Morgan's Theorem: F = (A + B + C)(A + B + D)(A + B + F)
3. OR the Distributed Terms: F = (A + B + C)(A + B + D)(A + B + F)

Example 8: Given POS Expression: F = (A + B)(C + D)(E + F)

1. Identify Complements: B, C, D, E
2. Apply De Morgan's Theorem: F = (A + B + C)(A + B + D)(A + B + E)(A + B + F)
3. OR the Distributed Terms: F = (A + B + C)(A + B + D)(A + B + E)(A + B + F)

Example 9: Given POS Expression: F = (A + B)(C + D)(E + F)

1. Identify Complements: A, B
2. Apply De Morgan's Theorem: F = (A + B + C)(A + B + D)(A + B + E)(A + B + F)
3. OR the Distributed Terms: F = (A + B + C)(A + B + D)(A + B + E)(A + B + F)

Example 10: Given POS Expression: F = (A + B)(C + D)(E + F)

1. Identify Complements: B, D, E, F
2. Apply De Morgan's Theorem: F = (A + B + C)(A + B + D)(A + B + E)(A + B + F)
3. OR the Distributed Terms: F = (A + B + C)(A + B + D)(A + B + E)(A + B + F)