Sum of Products and Product of Sums

Boolean Algebra

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In this tutorial we will learning about Sum of Products and Product of Sums.

Canonical Expression

A boolean expression consisting entirely either of minterm or maxterm is called canonical expression.

Example
if we have two variables X and Y then,

Following is a canonical expression consisting of minterms XY + X’Y’
and
Following is a canonical expression consisting of maxterm (X+Y) . (X’ + Y’)

Different Forms of Canonical Expression

There are two forms of canonical expression.

  1. Sum of Products (SOP)
  2. Product of Sums (POS)

Sum of Products (SOP)

A boolean expression consisting purely of Minterms (product terms) is said to be in canonical sum of products form.

Example
lets say, we have a boolean function F defined on two variables A and B. So, A and B are the inputs for F and lets say, output of F is true i.e., F = 1 when any one of the input is true or 1. Now we draw the truth table for F.



Now we will create a column for the minterm using the variables A and B. If input is 0 we take the complement of the variable and if input is 1 we take the variable as is.



To get the desired canonical SOP expression we will add the minterms (product terms) for which the output is 1.

F = A’B + AB’ + AB

Converting Sum of Products (SOP) to shorthand notation

From the previous example we have
F = A’B + AB’ + AB
Now, lets say we want to express the SOP using shorthand notation.

we have F = A’B + AB’ + AB

First we need to denote the minterms in shorthand notation.

A’B = (01)2 = m1
AB’ = (10)2 = m2
AB = (11)2 = m3

We saw the conversion of SOP to shorthand notation. Lets check the conversion of shorthand notation to SOP.

Converting shorthand notation to Sum of Products (SOP)

Lets say, we have a boolean function F defined on two variables A and B. So, A and B are the inputs for F and lets say, the minterms are expressed as shorthand notation given below.
F = ∑(1, 2, 3)
our task is to get the SOP.

F has two input variables A and B and output of F = 1 for m1, m2 and m3 i.e., 2nd, 3rd and 4th combination.

we have,
F = ∑(1, 2, 3)
= m1 + m2 + m3
= 01 + 10 + 11

To convert from shorthand notation to SOP we follow the given rules. If the variable is 1 then it is taken "as is" and if the variable is 0 then we take its "complement".

F = ∑(1, 2, 3)
= A’B + AB’ + AB

And we have the required SOP

Product of Sums (POS)

A boolean expression consisting purely of Maxterms (sum terms) is said to be in canonical product of sums form.

Example
Lets say, we have a boolean function F defined on two variables A and B. So, A and B are the inputs for F and lets say, output of F is true i.e., F = 1 when only one of the input is true or 1.

now we draw the truth table for F



Now we will create a column for the maxterm using the variables A and B. If input is 1 we take the complement of the variable and if input is 0 we take the variable as is.



To get the desired canonical POS expression we will multiply the maxterms (sum terms) for which the output is 0.

F = (A+B) . (A’+B’)

Converting Product of Sums (POS) to shorthand notation

From the previous example we have
F = (A+B) . (A’+B’)
Now, lets say we want to express the POS using shorthand notation.

we have F = (A+B) . (A’+B’)

First we need to denote the maxterms in shorthand notation.

A+B = (00)2 = M0
A’+B’ = (11)2 = M3

Now we express F using shorthand notation.

F = M0 . M3

This can also be written as F = ∏(0, 3)

We saw the conversion of POS to shorthand notation. Lets check the conversion of shorthand notation to POS.

Converting shorthand notation to Product of Sums (POS)

Lets say, we have a boolean function F defined on two variables A and B so, A and B are the inputs for F and lets say, the maxterm are expressed as shorthand notation given below.

F = ∏(1, 2, 3)

Our task is to get the POS.

F has two input variables A and B and output of F = 0 for M1, M2 and M3 i.e., 2nd, 3rd and 4th combination.

we have, F = ∏(1, 2, 3)
= M1 . M2 . M3
= 01 . 10 . 11

To convert from shorthand notation to POS we follow the given rules. If the variable is 0 then it is taken as is and if the variable is 1 then we take its complement.

we have, F = ∏(1, 2, 3)
= (A+B’) . (A’+B) . (A’+B’)

And we have the required POS.