Sum of Product and Product of Sum Form

Sum of Product and Product of Sum Form are the two forms of boolean expressions. Generally, Boolean expressions are built with constants and variables. These expressions describe the Boolean function. For example, in an expression f = A + (B . C), f is a function of the variables A, B and C.

In a Boolean function, the variables may appear either in complemented or in normal form. Each occurrence of a variable either in complemented or in normal form is called a literal.

For example, in the Boolean expression f = (A . D) + (B’ . C), the variables A, C and D are in normal form but the variable B is in complemented form. In this expression, A, B’, C and D are called the literals.

How these literals appear in the expression decides the term. Here, the literals appear in product term. A product term is defined as either a literal or a product(conjunction) of literals. For clear understanding, look at the below expression.

In the above expression, F in the Boolean function, A, B, C’ and D are the five literals. A, (B . C’) and (A . D) are the product terms. Similarly, a sum term is defined as either a literal or a sum(disjunction) of literals. Look at the below example,

In this expression, F is the Boolean function, A, A’, B, C’ and D are the five literals. A, (B + C’) and (A’ + D) are the sum terms.

From the literals and the terms that we have discussed above, the Sum of Product(SOP) form and Product of Sum(POS) form can be derived.

Sum of Product(SOP) form

Sum of Product form is a group of product(AND) terms that are summed(ORed) together. Here, the product and sum are not mathematical operations but are binary operations. The SOP expression may consists of two or more product(AND) terms, all of which are ORed together. The SOP form is also called as disjunctive normal form.

Standard or Canonical SOP Form

If each term in the SOP form contains all the literals in either complemented or uncomplemented form, then the SOP is referred to as Canonical or standard SOP form, where each individual term is called minterm.

In the above example, ABC, A’B’C, ABC’, A’BC and A’BC’ are the five minterms, where each term has all the literals in the Boolean function either in complemented or uncomplemented form. So, it is said to be a standard or canonical form.

Minterm representation

Minterms are the short hand notations used to express the product terms in SOP form. That is, each individual term in SOP form is called minterm. For a Boolean function having n variables, there will be 2ⁿ minterms.

For example, a Boolean function with 3 variables will have 2³ = 8 minterms and a Boolean unction with 4 variables will have 2⁴ = 16 minterms.

The following table shows the minterms for 3 variable Boolean function.

The canonical SOP expression F = ABC + A’B’C + ABC’ + A’BC +A’BC’ can be written as follows in terms of minterm notation.

F = m₇ +m₁ + m₆ + m₃ + m₂

F = ∑(m₇, m₁, m₆, m₃, m₂), where ∑ denotes sum of product.

This implies that the given Boolean function is logically true for the minterms (m₇, m₁, m₆, m₃, m₂).

Non-canonical SOP form

The non-canonical or normal SOP form is a simplified version of standard or canonical SOP form. The normal form is obtained by doing simplification to the standard form. Consider the above equation and see how it is simplified to get the non-canonical SOP form.

The obtained expression F + AB + A’B + A’B’C is in SOP form but literal C is missing in 1^st and 2^nd terms. Hence, it is a non-canonical SOP form.

Minimal SOP Form

It is the most simplified version of the canonical SOP form. It can be derived by applying the Postulates and theorems of Boolean algebra but it can be made easy by applying Karnaugh map(K map).

Let us consider the above SOP expression F = ABC + A’B’C + ABC’ + A’BC +A’BC’. Using K map, the given expression can be simplified as below.

The obtained expression from K map is F = A’C + B. Thus we can observe that, an expression consists of 5 product terms has been simplified to an expression consisting of only 2 product terms, which need only 1 AND gate and 1 OR gate.

Among all the three forms(Canonical SOP, Non-canonical SOP and Minimal SOP form), Minimal SOP Form is preferred because it reduces the complexity of the circuit, thereby reducing the circuit cost, size and time to fabricate the circuit.

Conversion of non-canonical SOP to canonical SOP form

Below are the steps to convert non canonical SOP to Canonical or standard SOP.

Find the missing literal in each product term.
AND each product term to the term having missing literal by ORing the missing literal and its complement.
Expand the terms and rearrange the literals in the product terms.
Reduce the expression by omitting the repeated terms if any(i.e. A+A=A)

Let us see an example here.

Convert the given expression F(A, B, C) = A + B’C into canonical SOP form.

In the given expression, literal B and C are missing in the 1^st product term. So (B+B’) and (C+C’) are multiplied with the term A. Similarly, literal A is missing in the 2^nd product term. So (A+A’) is multiplied with the product term B’C.

Problem on Conversion of non-canonical SOP to canonical SOP form

Thus we have obtained the canonical SOP or standard SOP form.

Product of Sum(POS) Form

Product of Sum form is a group of Sum(OR) terms that are Multiplied(ANDed) together. Here also, the product and sum are not mathematical operations but are binary operations. The POS expression may consists of two or more sum(OR) terms, all of which are ANDed together. The POS form is also called as Conjunctive normal form.

Standard or Canonical POS Form

If each term in the POS form contains all the literals in either complemented or uncomplemented form, then the POS is referred to as Canonical or standard POS form, where each individual term is called maxterm.

In the above example, (A+B+C), (A’+B’+C), (A+B+C’), (A’+B+C) and (A’+B+C’) are the five maxterms, where each term has all the literals in the Boolean function either in complemented or uncomplemented form. So, it is said to be a standard or canonical POS form.

Maxterm representation

Maxterm representation is similar to the minterm representation but here it is used to express the sum terms in POS form. That is, each individual term in POS form is called maxterm. For a Boolean function having n variables, there will be 2ⁿ maxterms.

For example, a Boolean function with 3 variables will have 2³ = 8 maxterms and a Boolean unction with 4 variables will have 2⁴ = 16 maxterms.

The following table shows the maxterms for 3 variable Boolean function.

The canonical POS expression F = (A+B+C) . (A’+B’+C) . (A+B+C’) . (A’+B+C) . (A’+B+C’) can be written as follows in terms of maxterm notation.

F = M₀ +M₆ + M₁ + M₄ + M₅

F = Π(M₀, M₆, M₁, M₄, M₅), where Π denotes product of sum.

This implies that the given Boolean function is logically true for the maxterms (M₀, M₆, M₁, M₄, M₅).

Non-canonical POS form

The non-canonical or normal POS form is a simplified version of standard or canonical POS form. The normal form is obtained by doing simplification to the standard form. Consider the above equation and see how it is simplified to get the non-canonical POS form.

The obtained expression is in POS form but literal C is missing in 1^st and 2^nd terms. Hence, it is a non-canonical POS form.

Minimal POS Form

It is the most simplified version of the canonical POS form. It can be derived by applying the Postulates and theorems of Boolean algebra but it can be made easy by applying Karnaugh map(K map).

Let us consider the above POS expression F = (A+B+C) . (A’+B’+C) . (A+B+C’) . (A’+B+C) . (A’+B+C’). Using K map, the given expression can be simplified as below.

The obtained expression from K map is F = (A’+C) . B. Thus we can observe that, an expression consists of 5 sum terms has been simplified to an expression consisting of only 2 sum terms, which need only 1 AND gate and 1 OR gate.

Among all the three forms(Canonical POS, Non-canonical POS and Minimal POS form), Minimal POS form is preferred because it reduces the complexity of the circuit, thereby reducing the circuit cost, size and time to fabricate the circuit.

Conversion of non-canonical POS to canonical POS form

Below are the steps to convert non canonical POS to Canonical or standard POS.

Find the missing literal in each sum term.
OR each sum term to the term having missing literal by ANDing the missing literal and its complement.
Expand the terms and rearrange the literals in the sum terms.
Reduce the expression by omitting the repeated terms if any(i.e. A . A = A)

Let us see an example here.

Convert the given expression F(A, B, C) = (A+B)(B+C) into canonical POS form.

In the given expression, literal C is missing in the 1^st sum term. So (C.C’) is added with the term (A+B). Similarly, literal A is missing in the 2^nd sum term. So (A.A’) is added with the term (B+C).

Problem on Conversion of non-canonical POS to canonical POS form

Thus we have obtained the canonical POS or standard POS form.

Conversion between Canonical SOP and Canonical POS

Canonical SOP can be converted into canonical POS form and the vice versa can be done in a simple way. To better understand, let us consider the SOP expression F = ABC + A’B’C + ABC’ + A’BC +A’BC’.

Now to convert it into canonical POS expression, follow the steps below.

The above expression in terms of minterm is given by, F = m₇ +m₁ + m₆ + m₃ + m₂.

The minterms and maxterms have a complementary relationship for a given Boolean expression. So, write the expression in terms of maxterm for the remaining terms in the above minterm expression.

We get, F = M₀ + M₄ + M₅

Thus, the canonical POS expression is, F = (A+B+C) . (A’+B+C) . (A’+B+C’)

Let us consider another example and do it very simply.

Consider the POS expression in terms of maxterm. F = Π M(1, 3, 5, 7, 9, 11, 12, 13).

The SOP expression is given by, F = ∑ m(0, 2, 4, 6, 8, 10, 14, 15)

Sum of Product and Product of Sum Form