Algebraic Manipulation

Minterms and Maxterms

Any boolean expression may be expressed in terms of either minterms or maxterms. To do this we must first define the concept of a literal. A literal is a single variable within a term which may or may not be complemented. For an expression with N variables, minterms and maxterms are defined as follows :

• A minterm is the product of N distinct literals where each literal occurs exactly once.
• A maxterm is the sum of N distinct literals where each literal occurs exactly once.

For a two-variable expression, the minterms and maxterms are as follows

 X Y Minterm Maxterm 0 0 X'.Y' X+Y 0 1 X'.Y X+Y' 1 0 X.Y' X'+Y 1 1 X.Y X'+Y'

For a three-variable expression, the minterms and maxterms are as follows

 X Y Z Minterm Maxterm 0 0 0 X'.Y'.Z' X+Y+Z 0 0 1 X'.Y'.Z X+Y+Z' 0 1 0 X'.Y.Z' X+Y'+Z 0 1 1 X'.Y.Z X+Y'+Z' 1 0 0 X.Y'.Z' X'+Y+Z 1 0 1 X.Y'.Z X'+Y+Z' 1 1 0 X.Y.Z' X'+Y'+Z 1 1 1 X.Y.Z X'+Y'+Z'

This allows us to represent expressions in either Sum of Products or Product of Sums forms

Sum Of Products (SOP)

The Sum of Products form represents an expression as a sum of minterms.

F(X, Y, ...) = Sum (ak.mk)

where ak is 0 or 1 and mk is a minterm.

To derive the Sum of Products form from a truth table, OR together all of the minterms which give a value of 1.

Example - SOP

Consider the truth table

 X Y F Minterm 0 0 0 X'.Y' 0 1 0 X'Y 1 0 1 X.Y' 1 1 1 X.Y

Here SOP is f(X.Y) = X.Y' + X.Y

Product Of Sum (POS)

The Product of Sums form represents an expression as a product of maxterms.

F(X, Y, .......) = Product (bk + Mk), where bk is 0 or 1 and Mk is a maxterm.

To derive the Product of Sums form from a truth table, AND together all of the maxterms which give a value of 0.

Example - POS

Consider the truth table from the previous example.

 X Y F Maxterm 0 0 1 X+Y 0 1 0 X+Y' 1 0 1 X'+Y 1 1 1 X'+Y'

Here POS is F(X,Y) = (X+Y')

Exercise

Give the expression represented by the following truth table in both Sum of Products and Product of Sums forms.

 X Y Z F(X,Y,X) 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0

Conversion between POS and SOP

Conversion between the two forms is done by application of DeMorgans Laws.

Simplification

As with any other form of algebra you have encountered, simplification of expressions can be performed with Boolean algebra.

Example

Show that X.Y.Z' + X'.Y.Z' + Y.Z = Y

X.Y.Z' + X'.Y.Z' + Y.Z = Y.Z' + Y.Z = Y

Example

Show that (X.Y' + Z).(X + Y).Z = X.Z + Y.Z

(X.Y' + Z).(X + Y).Z

= (X.Y' + Z.X + Y'.Z).Z

= X.Y'Z + Z.X + Y'.Z

= Z.(X.Y' + X + Y')

= Z.(X+Y')

 Copyright © 1998-2014 Deepak Kumar Tala - All rights reserved Do you have any Comment? mail me at:deepak@asic-world.com