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POS and SOP form representation of a Boolean Function

Here, we are going to learn about the POS and SOP form representation of a Boolean Function in digital electronics.
Submitted by Saurabh Gupta, on November 27, 2019

There are four ways in which a Boolean function can be expressed which are as follows,

  1. Product of Sum (POS) Form
  2. Sum of Product (SOP) Form
  3. Canonical Form of POS/SOP Form

1) Product of Sum (POS) Form

As the name suggests, A POS expression contains the sum of various terms ANDed/multiplied together.

Example: Y = (A + B). (C + D). (E + F)

2) Sum of Product (SOP) Form

As the name suggests, A SOP expression is a group of product terms ORed/added together.

Example: Y = (A. B) + (C. D) + (E. F)

3) Canonical Form of POS/SOP Form

If each term of SOP/POS expression contains all the literals in the Boolean function, then they are said to be in canonical form. Suppose, we have a Boolean function (Y) having three literals A, B and C, then canonical Boolean expressions can be written as,

Y = ABC + A. B. C + A.B.C

This is an example of canonical SOP Form, because each term in the SOP form contains all the literals A, B and C.

Similarly, Y = (A+B+C). (A + B + C). (A+B+C)

This is an example of canonical POS form, because each term of the POS form contains all the literals.

Now, let's see a few problems on canonical form.

Example 1: Convert the following expressions in their canonical form.

  1. Y (A, B, C) = AB + BC + CA
  2. Y (X, Y, Z) = X. (X + Y). (X + Y + Z)

Solution (i):

Y (A, B, C) = AB + BC + CA, this expression is a SOP expression, since we notice the Boolean function has three literals A, B and C, so each term of the Boolean expression must contain all the three literals to convert it into canonical SOP form. Therefore,

= Y (A, B, C) = AB + BC + CA
= AB. (C + C) + BC. (A + A) + CA. (B + B)	[Since, C + C = 1]
= ABC + AB. C + ABC + A. BC + ABC + A. B. C
= ABC + AB. C + A. BC + A. B. C
Hence, 
Y = ABC + AB. C + A. BC + A. B.C 
is the required canonical SOP form representation.

Solution (ii):

Y (X, Y, Z) = X. (X + Y). (X + Y + Z), is an example of POS expression, since all the sum terms in the expression doesn't have all the literals X, Y and Z, so we have to express it in such a way that it will have all the three literals in each term.

= Y (X, Y, Z) = X. (X + Y). (X + Y + Z)
= (X + Y. Y + Z. Z) (X + Y + Z. Z). (X + Y + Z)
= (X + Y. Y + Z) (X + Y. Y + Z) (X + Y + Z) (X + Y + Z) (X + Y + Z)
= (X + Y +Z) (X + Y + Z) (X + Y + Z) (X + Y + Z) (X + Y + Z) (X + Y + Z)
= (X + Y +Z) (X + Y + Z) (X + Y + Z) (X + Y + Z)
Hence, 
Y = (X + Y +Z) (X + Y + Z) (X + Y + Z) (X + Y + Z) 
is the required canonical POS form representation.





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