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 | 0 | 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 (a_{k}.m_{k})
where a_{k} is 0 or 1 and m_{k} 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 (b_{k} + M_{k}), where b_{k} is 0 or 1 and M_{k} 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')