K maping of Logic operations

1. Digital System Design
BTEC-302-18
 Topic: K-Map
1
Dr. Sajjan Singh
HOD-ECE

2. 2
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.

3. 3
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.

4. 4
Examples of K-Maps:
 Examples:
Cell numbers are written in the cells.
 2-variable K-map
0 1
2 3
0
1
0 1
A
B

5. 5
3-Variable K-Map:
 3-variable K-map
0 1 3 2
4 5 7 6
00 01 11 10
0
1
A
BC

6. 6
4-variable K-map
 4-variable K-map 1001(ABCD)1101
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
00 01 11 10
00
01
11
10
AB
CD

7. 7
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.

8. 8
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.

9. 9
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.

10. 10
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):

11. 11
Example: Truth Table

12. 12
Example: TruthTable
Dec
number
A B C D F
0 0 0 0 0 1
1 0 0 0 1 0
2 0 0 1 0 1
3 0 0 1 1 1
4 0 1 0 0 0
5 0 1 0 1 1

13. 13
Example: TruthTable (continued)
Dec
number
A B C D F
6 0 1 1 0 1
7 0 1 1 1 1
8 1 0 0 0 1
9 1 0 0 1 0
10 1 0 1 0 1
11 1 0 1 1 1

14. 14
Example: TruthTable (continued)
Dec
Number A B C D F
12 1 1 0 0 0
13 1 1 0 1 0
14 1 1 1 0 1
15 1 1 1 1 1

15. 15
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.

16. 16
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
1 0 1 1
00 01 11 10
00
01
11
10
AB
CD

17. 17
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.

18. 18
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.

19. 19
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.

20. 20
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.

21. 21
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.

22. 22
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

23. 23
SIMPLIFICATION using KARNAUGH MAP
Exam 2:
F=∑ m(0,2,8,9,10,11,14,15)
F= A.B’+A.C+B’.D’

24. 24
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:
Carry= A.C+A.B+B.C
S=A.B’.C’ + A’.B’.C+A.B.C+A’.B.C’

25. 25
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.

26. 26
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)
N1 N2
A
B
C
F
w
x
y
z

27. 27
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.

28. 28

29. 29
SIMPLIFICATION using KARNAUGH MAP
Designing for N1
F1=B’C+BC’
F2=C’
F3=C
F1 = ∑m (1,2,5,6)
F2 =∑ m(0,2,4,6)
F3 =∑m (1,3,5,7)

30. 30
K- Map for F: Designing for N2

31. 31
B
N1 N2