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.
Simplification of Boolean Functions:
Three Methods
The algebraic method by
using Identities
The graphical method by
using Karnaugh Map method
Karnaugh Maps
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 cellsExamples of K-Maps:
Examples:
Cell numbers are written in the cells.
2-variable K-map
0 1
2 3
0
1
0 1
3-Variable K-Map:
3-variable K-map
0 1 3 2
4 5 7 6
00 01 11 10
0
1
4-variable K-map
4-variable K-map
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
00 01 11 10
00
01
11
10
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
Minterms and Maxterms
For every K-map, each cell has a minterm associatedwith it .
Thus for cell no. 13 in the 4-variable K-map, theminterm 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 0 for exactly one combination of values
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.Adjacent minterms:
Thus in K-4,Cell 0 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: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.Procedure for placing a
minterm in a K-map
Identify the minterm (product term) term
to be mapped.
Write the corresponding binary numeric
value.
Use binary value as an address to place a
1
in the K-map
Repeat steps for other minterms (P-terms
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:
1 0 1 1
0 1 1 1
0 0 1 1
00 01 11 10
00
01
11
Procedure for writing the SOP
Solution
Form largest groups of
1
s possible
covering all minterms. Groups must be a
power of 2.
Write binary numeric value for groups.
Convert binary value to a product term.
Repeat steps for other groups. Each group
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
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)
Procedure for placing a
maxterm in the K-map
Identify the Sum term to be mapped.
Write corresponding binary numeric value.
Form the complement
Use the complement as an address to
place a
0
in the K-map
Repeat for other maxterms (Sum terms
Procedure for writing the POS
Solution
Form largest groups of
0
s possible, covering
all maxterms. Groups must be a power of 2.
Write binary numeric value for group.
Complement binary numeric value for group.
Convert complement value to a sum-term.
Repeat steps for other groups. Each group
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
Essential Prime Implicant:
Essential prime Implicant:
A Prime Implicant that contains an ‘ANDing ofliterals’, that is not contained in any other prime Implicant.
Or a Prime Implicant, representing a group in theCANONIC form of a Boolean Expression
CANONIC: A SOP or POS expression of n variables iscanonic 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.
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:
SIMPLIFICATION using KARNAUGH MAP
Exam 2:
F=∑ m(0,2,8,9,10,11,14,15)
SIMPLIFICATION using KARNAUGH MAP
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
Exam 3:
Full-adder:
Don’t care conditions
It is not always necessary to fill in the
complete truth table for some
real-world problems,
For example BCD (Binary Coded
Decimal). It uses 4 bit representation
but, goes from 0 to 9 but last 6
BCD
A B C D F 0 0 0 0 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 1
0 1 0 0 1 F= ∑ m(0,1,2,3,4,5,6,7,8,9)+∑ d(10,11,12,13,14,15) 0 1 0 1 1
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.
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)
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)
SIMPLIFICATION using KARNAUGH MAP
Designing for N1
F2=C’
F3=C F1 = ∑m (1,2,5,6)
B