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Simplification of Boolean Functions:. An implementation of a Boolean Function requires the use of logic gates. A smaller number of gates, with each gate (other then Inverter) having less number of inputs, may reduce the cost of the implementation.

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simplification of boolean functions
Simplification of Boolean Functions:
  • An implementation of a Boolean Function requires the use of logic gates.
  • A smaller number of gates, with each gate (other then Inverter) having less number of inputs, may reduce the cost of the implementation.
  • There are 2 methods for simplification of Boolean functions.
simplification of boolean functions two methods
Simplification of Boolean Functions: Two Methods
    • The algebraic method by

using Identities

    • The graphical method by

using Karnaugh Map method

  • The K-map method is easy and straightforward.
  • A K-map for a function of n variables
    • consists of 2n cells, and,
    • in every row and column, two adjacent cells should differ in the value of only one of the logic variables.
examples of k maps
Examples of K-Maps:
  • Examples:

Cell numbers are written in the cells.

  • 2-variable K-map

B

0

1

A

0

1

3 variable k map
3-Variable K-Map:
  • 3-variable K-map

BC

A

00 01 11 10

0

1

4 variable k map
4-variable K-map
  • 4-variable K-map

CD

AB

00 01 11 10

00

01

11

10

literal minterm of n variable
Literal, minterm of n variable:

Literal:

  • A variable or its complement is called a literal.

Minterm of n variable:

  • A product of n literals
    • in which each variable appears exactly once, in either its true or its complemented form, but not in both, and,
    • which is equal to 1 for exactly one combination of values of the n variables.
minterms and maxterms
Minterms and Maxterms
  • For every K-map, each cell has a minterm associated with it .
  • Thus for cell no. 13 in the 4-variable K-map, the minterm is A.B.C’.D Or

m13 = A.B.C’.D.Maxterm of n variables:

  • A sum of n literals
    • in which each variable appears exactly once, in either its true or its complemented form, but not in both
    • which has a value of O for exactly one combination of values of the n variables.
maxterms continued
Maxterms (continued):
  • For every K-map, each cell has one Maxterm

associated with it.

  • Thus for cell no.13 in the 4-variable K-map,

M13 = A’ + B’ + C + D’

By De Morgan’s theorem,

mi = Mi’

ADJACENT minterms (Maxterms):

  • Minterm which are identical, except for one variable, are considered to be adjacent to one another.
  • In a K-map, the corresponding cells are said to be adjacent cells.
adjacent minterms
Adjacent minterms:
  • Thus in K-4,

Cell O is adjacent to cells 1, 4, 2 and 8.

  • In a K-map, the corresponding cells in the top and the bottom rows are adjacent to each other.

Similarly the corresponding cells in the leftmost column and the rightmost column are adjacent to each other.

  • An Example:

A function F, of 4 variables, is defined by the truth table given in the next slide. ( and again given in the next 3 slides):

sum of products form
Sum of Products form:
  • The above table can be described by

F =  m(0, 2, 3, 5, 6, 7, 8, 10, 11, 14, 15)

The function can be written as:

F = A’B’C’D’ + A’B’CD’ + A’B’CD + A’BC’D + A’BCD’ + A’BCD + AB’C’D’ + AB’CD’ + AB’CD + ABCD’ + ABCD

…………………………………(1)

  • Each term on the RHS is a minterm.
  • The above function can be simplified by using the Identities.
the graphical method steps
The graphical method steps:
  • The graphical method steps:
  • Insert 1 in those cells where the function F has a value of 1. Put 0 in the other cells.

Examples:

CD

00 01 11 10

AB

00

01

11

10

steps of graphical method continued
Steps of graphical method (continued):
  • Combine adjacent 1’s into group of 2n each such that
    • Each group contains only 1’s.
    • The group is not completely a part of a larger group.
  • Choose the minimum number of the largest sized groups needed to cover all the 1’s.
  • Each group is represented by an expression which is an intersection of the minterm in the group.
  • The simplified solution is a logical OR of the expressions of all the groups chosen in steps 3 above.
product of sums form using maxterms
Product of Sums Form: Using Maxterms

For the same example,

F=  M(1,4,9,12,13)

= (A + B + C+ D’).(A + B’+ C + D).(A’+ B + C +D’).

(A’ + B’ + C + D).( A’ + B’ + C +D’) …………..(2)

The simplification process is a dual of the

process for the SOP form.

some definitions
Some definitions:

The definitions: Given a function F of n variables.

Implicant:

  • A minterm P is an implicant of F if and only if, for the combination of values of the n variables, for which P = 1, F is also equal to 1.

Prime Implicant : An implicant is a Prime Implicant if after deleting any literal from it , the remaining product term is no longer an implicant.

Or an implicant whose group in the K-map is not completely covered by another implicant, represented by a larger group.

essential prime implicant
Essential Prime Implicant:

Essential prime Implicant:

  • A Prime Implicant that contains an ‘ANDing of literals’, that is not contained in any other prime Implicant.
  • Or a Prime Implicant, representing a group in the K-map, such that at least one cell of the group is not covered by any other Prime Implicant.
canonic form of a boolean expression
CANONIC form of a Boolean Expression

CANONIC: A SOP or POS expression of n variables is canonic if each product or sum has exactly n literals.

SOP format: F = ‘ORing’ of minterms -----(3)

POS format: F = ‘ORing’ of minterms -----(4)

The sum of the number of terms on the RHS of equations (3) and (4) is always equal to 2n.

A minterm that is covered by only one PI is called a distinguished minterm.

A Maxterm that is covered by only one PI is called a distinguished Maxterm.

Equations (1) and (2) show the canonic form of the Boolean expression for the example given on slide 10.

use of karnaugh map for simplification of logic functions
Use of KARNAUGH MAP for Simplification of Logic Functions

SOL: On reading the three sets of adjacent boxes of 8, 4 and 2 cells respectively, we get:

F = C + B’.D’ + A’.B.D

simplification using karnaugh map
SIMPLIFICATION using KARNAUGH MAP

Exam 2:

F=∑ m(0,2,8,9,10,11,14,15)

F= A.B’+A.C+B’.D’

simplification using karnaugh map23
SIMPLIFICATION using KARNAUGH MAP

Exam 3:

Full-adder:

A B C S Carry

0 0 0 0 0

0 0 1 1 0

0 1 0 1 0

0 1 1 0 1

1 0 0 1 0

1 0 1 0 1

1 1 0 0 1

1 1 1 1 1

Carry= A.C+A.B+B.C

S=A.B’.C’ + A’.B’.C+A.B.C+A’.B.C’

multistage logic circuit
Multistage Logic Circuit
  • Multistage Logic Circuit:

N1 and N2 ; Two logic circuits.

W, X, Y, Z: independent logic variables

For each of the 16 possible combination of values for W, X, Y and Z, some specific value of A, B and C would be the outputs.

Three variables normally have 8 possible sets of values .However, in the above circuit N1 may constrain the values to a smaller set. The remaining set of values for A, B and C would not affect the output of N2.Thus for N2, the non available inputs are called ‘don’t care’ inputs, since these inputs do not have any effect on F.

don t care condition
Don’t care condition

Example 1:

Let A,B and C never have 001 or 110 values. Then for F, values of 001 and 110 for A, B and C are not of any importance.

Exam 2: All possible input combinations are present. But the output is used in such a way that we do not care whether it is 0 or 1 for certain input combinations.

F= ∑ m(0,3,7)+ ∑ d(1,6)

Or F = Π M(2,4,5) Π D(1,6)

w

A

F

N1

N2

xy

B

z

C

simplification using karnaugh map26
SIMPLIFICATION using KARNAUGH MAP

Example: Given the Characteristic Table for a 2-stage network. (Please see the Figure in the next slide.)

Solution:

F1 = ∑m (1,2,5,6)

F2 =∑ m(0,2,4,6)

F3 =∑m (1,3,5,7)

F= ∑m (1, 2, 6 ), d(0, 3, 4, 7)

Solution is continued in the next 3 slides.

simplification using karnaugh map designing for n1
SIMPLIFICATION using KARNAUGH MAP Designing for N1

F1 = ∑m (1,2,5,6)

F2 =∑ m(0,2,4,6)

F3 =∑m (1,3,5,7)

F2=C’

F3=C

F1=B’C+BC’

slide30

B

N2

N1