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# Converting SSOP to SPOS Form

A Boolean function can be expressed into two forms namely,

Sum of Products (SOP) Form

Product of Sums (SOP) Form

The **SOP (Sum of Products) form** is one in which the Boolean function is expressed as the sum of product terms, while in the **POS (Product of Sums) form**, the Boolean function is expressed as the product of sum terms of the function. But, in the SOP and POS form, each term of the function may not contain all the variables.

For example, consider a Boolean function in three variables,

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

This is the sum of products (SOP) form of the Boolean function (f). It can be noted that it does not contain all the variables in product terms, i.e. variable C is missing in the first product term and the variable A is missing in the second product term.

Similarly, consider a 3-variable Boolean function in POS form as,

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

In this case, the variable C is missing in the first term and the variable A is missing in the second term.

Sometimes, we need to express a Boolean function in such a form in which each term of the function should contain all the variables of the function. Hence, the form of expressing a Boolean function in which each term of the function contains all the variables of the function is referred to as **Standard Form** or **Expanded Form** or **Canonical Form**.

Thus, there are two standard forms in which a Boolean function can be expressed,

Standard Sum of Products (SSOP) Form

Standard Product of Sums (SPOS) Form

## Standard Sum of Products (SSOP) Form

The standard sum of products (SSOP) form is a form of expressing a Boolean function as a sum of minterms. A minterm is a product term of the Boolean function which contains all variables of the function either in complemented or un-complemented form.

Therefore, the SSOP form of a Boolean function is also known as Expanded Sum of Products form because each of its terms contains all the variables of the function.

For example, consider a Boolean function in SOP form as,

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

Then, the SSOP form of this function will be,

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

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

This is the standard sum of products (SSOP) form the given Boolean function. The Boolean function in SSOP form can also be represented as,

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=m_2+m_4+m_5+m_6=\sum m \lgroup 2,4,5,6\rgroup}$$

## Standard Product of Sums (SPOS) Form

The standard product of sums (SPOS) form is a form of expressing a Boolean function as a product of maxterms. Where, a maxterm is a sum term of the Boolean function which contains all the variables of the function either in complemented or un-complemented form.

Therefore, the SPOS form of a Boolean function is also known as Expanded Product of Sums form because each term of the function contains all the variables.

For example, consider a Boolean function in POS form as,

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

Then, the SPOS form of this function will be,

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

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

This is the standard product of sums (SPOS) form the given Boolean function. The SPOS form of Boolean function can also be represented as,

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=M_1.M_2.M_3.M_5=\prod M\lgroup 1,2,3,5\rgroup}$$

Now, let us discuss the conversion of SSOP (Standard Sum of Products) form to SPOS (Standard Product of Sums) form.

## Converting SSOP to SPOS Form

We can convert the SSOP (Standard Sum of Products) form of a Boolean function to SPOS (Standard Product of Sums) form. To covert a Boolean function from SSOP form to SPOS form, we have to follow these two steps −

Interchange the operational symbols Σ and Π.

List those terms which are missing in the given SSOP form.

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\sum m\lgroup 0,1,2,5,7\rgroup =m_0+m_1+m_2+m_5+m_7}$$

In this function, the minterms 3, 4, and 6 are missing. Therefore, they will appear in the SPOS form as maxterms.

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

In this way, we can convert a Boolean function from SSOP form to SPOS form.

We can also convert a Boolean function from SSOP to SPOS form by using complement and De Morgan’s theorem as follows,

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=\sum m\lgroup 0,1,2,5,7\rgroup =m_0+m_1+m _2+m_5+m_7}$$

The complement of this function is,

$$\mathrm{\overline{\mathit{f}(A,B,C)} =\sum m\lgroup 3,4,6\rgroup=m_3+m_4+m_6}$$

I.e. the complement of a Boolean function expressed as a sum of minterms is equal to the sum of minterms missing from the given Boolean function.

Now, if we apply De Morgan’s theorem ,$\mathrm{\lgroup \overline{A+B}=\overline{A}.\overline{B} \rgroup}$ we obtain the function in the following form,

$$\mathrm{\mathit{f}(A,B,C)=\overline{m_3+m_4+m_6}=\overline{m_3}.\overline{m_4}.\overline{m_6}= M_3.M_4.M_6}$$

$$\mathrm{\Rightarrow \mathit{f}\lgroup A,B,C\rgroup=\prod M\lgroup 3,4,6\rgroup }$$

This is the SPOS form of the given Boolean function.

## Conclusion

This is all about converting SSOP (Standard Sum of Products) form to SPOS (Standard Product of Sums) form.