Conversion between Canonical Forms

When a logical expression or Boolean function is expressed as a sum of minterms or as a product of maxterms, then it is called the canonical form of the expression or function.

A canonical form of the Boolean expression is also known as standards form, i.e. Standard Sum of Products (SSOP) Form and Standard Product of Sums (SPOS) Form. The canonical form of a Boolean function involves minterms and maxterms.

• A minterm is a product term which contains all the variables of the Boolean function either in complemented or un-complemented form.

• A maxterm is a sum term which contains all the variables of a Boolean function either in complemented or un-complemented form.

We can convert one canonical form to another one. In this tutorial, we will discuss the conversion between canonical forms.

Conversion between Canonical Forms

The compliment of a Boolean function which is expressed as the sum of minterms is equal to the sum of minterms which are missing from the original Boolean function. This is because, the original Boolean function has only those minterms that make the function equal to 1. While, the complement of the Boolean function is equal to 1 for those minterms that make the function equal to 0.

For example, consider a Boolean function in 3-varaibles,

$$\mathit{f}\mathrm{\lgroup A,B,C\rgroup=\sum m\lgroup 1,2,4,5,7\rgroup }$$

The complement of this function is,

$$\mathrm{\overline{\mathit{f}\lgroup A,B,C\rgroup} =\sum m\lgroup 0,3,6) =m_0+m_3+m_6}$$

Now, if we take the complement of the function (f’) by using De Morgan’s Theorem, $\lgroup\overline{A+B}=\overline{A}.\overline{B}\rgroup$, we obtain the original Boolean function (f) in a different form, i.e.

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\lgroup\overline{ m_0+m_3+m_6}\rgroup=\overline{m_0}.\overline{m_3}.\overline{m_6}}$$

$$\mathrm{\Rightarrow\mathit{f}\lgroup A,B,C\rgroup=M_0.M_3.M_6=\prod \lgroup 0,3,6\rgroup}$$

Actually,

$$\mathrm{\overline{m_i} =M_i}$$

I.e. the complemented minterm with a subscript i is equal to the maxterm with the same subscript i and vice-versa.

Thus, from this discussion, we can write the steps of converting between canonical forms. These steps are as follows −

• Interchange the operational symbols Σ and Π.

• Use De Morgan’s Theorem to write the terms which are not present in the original form of the Boolean function.

Now, let us discuss some solved examples to understand the concept in more detail.

Example 1

Convert the following Boolean function from canonical SOP form to canonical POS form.

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\sum m\lgroup 0,1,2,5,7\rgroup }$$

Solution

The given Boolean function is,

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\sum m\lgroup 0,1,2,5,7\rgroup }$$

$$\mathrm{\Rightarrow \mathit{f}\lgroup A,B,C\rgroup=m_0+m_1+m_2+m_5+m_7}$$

In the canonical SOP form, the minterms 3, 4, and 6 are missing. So in the canonical POS form, the maxterms 3, 4, and 6 we be present. Therefore, the canonical POS form of the given Boolean function is,

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\prod M\lgroup 3,4,6\rgroup=M_3.M_4.M_6}$$

Example 2

Convert the following Boolean function from canonical POS form to canonical SOP form.

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\prod M\lgroup 0,2,4,7\rgroup}$$

Solution

The given Boolean function is,

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\prod M\lgroup 0,2,4,7\rgroup}$$

$$\mathrm{\Rightarrow \mathit{f}\lgroup A,B,C\rgroup=M_0.M_2.M_4.M_7}$$

In the canonical POS form, the maxterms 1, 3, 5, and 6 are missing. Thus, in the canonical SOP form, the minterms 1, 3, 5, and 6 will be present.

Therefore, the canonical SOP form of the given Boolean function is,

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\sum m\lgroup 1,3,5,6\rgroup =m_1+m_3+m_5+m_6}$$

Conclusion

This is all about the conversion between canonical forms. In the above sections of this article, we have seen that the canonical SOP form of a Boolean function can be converted in the canonical POS form and vice-versa, just by interchanging the operational symbols Σ and Π, and by listing the missing terms from the original Boolean function.