Convert POS to Standard POS Form

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When a Boolean expression is represented as a product of sum terms, it is called POS (Product of Sums) form. In POS form, each sum term of the expression may not contain all the variables.

On the other hand, when the Boolean expression is represented as a product of sum terms, where each sum term contains all the variables of the function, it is called Standard Product of Sums (SPOS) form. In the Standard POS form, each sum term of the Boolean expression is called a maxterm.

Now, let us discuss the expansion of a Boolean expression in POS form to Standard POS form.

Conversion of a Boolean Expression in POS Form to Standard POS Form

A Boolean expression in POS form can be converted into the standard POS form as follows

• Write down all the sum terms of the given Boolean expression.
• If one or more variables are missing in any sum term, then add the products of each of the missing variables and its complement to that term.
• Expand the terms according to the rules of Boolean algebra.
• Finally, drop out the redundant terms from the expression.

Let us understand the conversion of a Boolean expression in POS form to Standard POS form with the help of examples.

Example 1

Convert the following 3-variable Boolean expression in POS form to its Standard POS form.

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\lgroup A+\overline{B}\rgroup.\lgroup \overline{B}+C\rgroup.\lgroup A+\overline{C}\rgroup}$$

Solution

The given Boolean function is,

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\lgroup A+\overline{B}\rgroup.\lgroup \overline{B}+C\rgroup.\lgroup A+\overline{C}\rgroup}$$

The given Boolean expression is in its POS form. In this, the variable C is missing from the first term, variable A is missing from the second term, and variable B is missing from the third term.

Therefore, to convert this given function into its SPOS form, we will add the product of missing variable and its compliment to each term of the function, i.e.

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\lgroup A+\overline{B}+C\overline{C} \rgroup.\lgroup \overline{B}+C+A\overline{A}\rgroup.\lgroup A+\overline{C}+B\overline{B} \rgroup}$$

$$\mathrm{\Rightarrow \mathit{f}\lgroup A,B,C\rgroup=\lgroup A+\overline{B}+C\rgroup\lgroup A+\overline{B}+\overline{C} \rgroup\lgroup A+\overline{B}+C\rgroup\lgroup \overline{A}+\overline{B}+C\rgroup\lgroup A+B+\overline{C} \rgroup\lgroup A+\overline{B}+\overline{C} \rgroup}$$

On dropping out the redundant terms, we get,

$$\mathrm{\Rightarrow \mathit{f}\lgroup A,B,C\rgroup=\lgroup A+\overline{B}+C\rgroup\lgroup \overline{A}+\overline{B}+C\rgroup\lgroup A+B+\overline{C}\rgroup\lgroup A+\overline{B}+\overline{C}\rgroup}$$

This is the standard POS form of the given Boolean function.

Example 2

Convert the following 4 variable Boolean function into its standard POS form.

$$\mathrm{\mathit{f}\lgroup A,B,C,D\rgroup=\lgroup A+C+D\rgroup.\lgroup A+\overline{B}+\overline{D} \rgroup.\lgroup A+\overline{C}+D\rgroup}$$

Solution

The given function is in the POS form,

$$\mathrm{\mathit{f}\lgroup A,B,C,D\rgroup=\lgroup A+C+D\rgroup.\lgroup A+\overline{B}+\overline{D} \rgroup.\lgroup A+\overline{C}+D\rgroup}$$

Here, the variable B is missing from first and third terms, and the variable C is missing from the second term of the function. Therefore, in order to get standard POS form of the function, we will add the product of the missing variable and its complement as follows,

$$\mathrm{\mathit{f}\lgroup A,B,C,D\rgroup=\lgroup A+C+D+B\overline{B} \rgroup.\lgroup A+\overline{B}+\overline{D}+C\overline{C} \rgroup.\lgroup A+\overline{C}+D+B\overline{B} \rgroup}$$

$$\mathrm{\Rightarrow \mathit{f}\lgroup A,B,C,D\rgroup=\lgroup A+B+C+D\rgroup.\lgroup A+\overline{B}+C+D\rgroup.\lgroup A+\overline{B}+C+\overline{D} \rgroup.\lgroup A+\overline{B}+\overline{C}+\overline{D}\rgroup.\lgroup A+B+\overline{C}+D\rgroup.\lgroup A+\overline{B}+\overline{C}+D\rgroup}$$

Since, there is no redundant term in the expression. Thus, this is the standard POS form of the given Boolean function.

Example 3

Convert the following 2 variable Boolean function into Standard POS form.

$$\mathrm{\mathit{f}\lgroup A,B\rgroup=\overline{A}.\lgroup A+\overline{B}\rgroup.B}$$

Solution

The given Boolean function is,

$$\mathrm{\mathit{f}\lgroup A,B\rgroup=\overline{A}.\lgroup A+\overline{B}\rgroup.B}$$

This function is in its POS form having three sum terms, where the variable B is missing from the first term and variable A is missing from the third term. So, to convert it into standard POS form, we will add the product of variable and its complement to the term as follows,

$$\mathrm{\mathit{f}\lgroup A,B\rgroup=\lgroup \overline{A}+B\overline{B}\rgroup.\lgroup A+\overline{B} \rgroup.\lgroup A\overline{A}+B\rgroup}$$

$$\mathrm{\Rightarrow \mathit{f}\lgroup A,B\rgroup=\lgroup \overline{A}+B\rgroup\lgroup\overline{A}+\overline{B} \rgroup\lgroup A+\overline{B}\rgroup\lgroup A+B\rgroup\lgroup \overline{A}+B\rgroup}$$

Removing the redundant terms from the expression, we get,

$$\mathrm{\Rightarrow \mathit{f}\lgroup A,B\rgroup=\lgroup \overline{A}+B\rgroup\lgroup\overline{A}+\overline{B} \rgroup\lgroup A+\overline{B}\rgroup\lgroup A+B\rgroup}$$

This is the standard POS form of the given Boolean expression.

Numerical Problems

Q 1. − Convert the following 3 variable Boolean expression in POS form to its Standard POS form.

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\lgroup A+B\rgroup\lgroup A+\overline{B}\rgroup\lgroup \overline{B}+C\rgroup\lgroup A+C\rgroup}$$

Q 2. − Convert the following 4 variable Boolean expression in POS from to its standard POS form.

$$\mathrm{\mathit{f}\lgroup A,B,C,D \rgroup=\lgroup A+B+\overline{C} \rgroup\lgroup A+\overline{B}+\overline{D} \rgroup\lgroup \overline{A}+\overline{B}+D \rgroup\lgroup A+C+\overline{D} \rgroup}$$

Q 3. − Convert the following 2 variable Boolean function into standard POS form.

$$\mathrm{\mathit{f}\lgroup A,B \rgroup=\lgroup A+B \rgroup.A}$$

Conclusion

This is all about the conversion of a Boolean expression in POS (Product of Sums) form to SPOS (Standard Product of Sums) form.