Simplification of Boolean Functions:

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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.
• There are 2 methods for simplification of Boolean
  functions.

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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.

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Examples of K-Maps

• Examples
• Cell numbers are written in the cells.
• 2-variable K-map

B
0
1
A
0 1
2 3
0
1
4
3-Variable K-Map

• 3-variable K-map

BC
00 01 11 10
A
0 1 3 2
4 5 7 6
0 1
5
4-variable K-map

• 4-variable K-map

CD
AB
00 01 11 10
00
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
01
11
10
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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.

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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.

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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 Morgans 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.

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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)

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Example Truth Table
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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
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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
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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
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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 ABCD ABCD ABCD ABCD ABCD
  ABCD ABCD ABCD ABCD ABCD ABCD
  
• 
  (1)
• Each term on the RHS is a minterm.
• The above function can be simplified by using the
  Identities.

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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
1 0 1 1
0 1 1 1
0 0 1 1
1 0 1 1
00
01
11
10
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Steps of graphical method (continued)

• Combine adjacent 1s into group of 2n each such
  that
• Each group contains only 1s.
• 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 1s.
• 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.

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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.

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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.

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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.

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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.

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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

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SIMPLIFICATION using KARNAUGH MAP

• Exam 2
• F? m(0,2,8,9,10,11,14,15)

F A.BA.CB.D
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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.CA.BB.C
SA.B.C A.B.CA.B.CA.B.C
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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 dont care inputs,
  since these inputs do not have any effect on F.

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Dont 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 ? M(2,4,5) ? D(1,6)

A
w
F
N1
N2
xy
B
C
z
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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.
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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)
F2C
F3C
F1BCBC
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K- Map for F Designing for N2
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B
N2
N1