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  ../images/main/bullet_green_ball.gif Algebraic Manipulation
   

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  ../images/main/bulllet_4dots_orange.gif Minterms and Maxterms
   

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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

   

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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'

   

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

   

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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'

   

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

   

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  ../images/main/bullet_star_pink.gif Sum Of Products (SOP)
   

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

   

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

   

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

   

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

   

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  ../images/main/4blue_dots_bullets.gif Example - SOP
   

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Consider the truth table

   

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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

   

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  ../images/main/bullet_star_pink.gif Product Of Sum (POS)
   

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

   

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

   

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

   

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  ../images/main/4blue_dots_bullets.gif Example - POS
   

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Consider the truth table from the previous example.

   

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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')

   

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  ../images/main/bullet_star_pink.gif Exercise
   

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

   

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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

   

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  ../images/main/bulllet_4dots_orange.gif Conversion between POS and SOP
   

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

   

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  ../images/main/bullet_star_pink.gif Simplification

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

   

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  ../images/main/4blue_dots_bullets.gif Example
   

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

   

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

   

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  ../images/main/4blue_dots_bullets.gif Example
   

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

   

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(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')

   

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Copyright 1998-2014

Deepak Kumar Tala - All rights reserved

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