Digital Electronics Fundamentals: Boolean Algebra and Logic Simplication

CHAPTER 4

BOOLEAN ALGEBRA & LOGI

DFV10133 SIMPLIFICATION

DIGITAL

ELECTRONICS

FUNDAMENTALS

PRESENTED BY:
SITI ZALEHA ABD AZIZ

LEARNING OUTCOMES

After completing the unit, students should be able to:

1. Determine the Boolean equation for combinational logic

applications
2. Utilize Boolean algebra laws and rule for simplifying
combinational logic circuit.
3. Apply De Morgan’s Theorem to complex Boolean equations
to arrive at simplified equivalent equation.
4. Utilize the Karnaugh mapping procedure to systematically
reduce complex Boolean equations to their simplest form.

4.1 Basic Operations

• Basic operation boolean algebra , AND, OR & NOT
NOT Operation
0�=1 and 1�=0
̅=1 if X=0 and ̅=0 if X=1

4.1 Basic Operations

• Basic operation boolean algebra , AND, OR & NOT

AND Operation

- 0.0 = 0, 0.1 = 0, 1.0 = 0, 1. 1 = 1

- Denotes AND , Boolean expression

C=A.B, the given value AB C=AB
- Note; C=1 if A & B are both 1 00 0
0
- Boolean expression usually write AB 0 1 0
1
10

11

4.1 Basic Operations

• Basic operation boolean algebra , AND, OR & NOT

OR Operation 1+1=1
- 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 0,
- Denotes OR , Boolean expression AB C=A+B
C=A+B, the given value 00 0
- Note; C=1 if and only if A or B 01 1
10 1
(or both) is 1 11 1
- Boolean expression , A+B

4.1 Basic Operations

• Boolean Theorems – help to simplify logic
expression & logic circuits.

4.2 Boolean Algebra and Law

• The basics laws of Boolean algebra are
a) the commutative law
b) the associates law
c) the distribute law

4.2 Boolean Algebra and Law

Commutative law
- Addition of two variables is written algebraically as :

A+B=B+A
- Law states,
i) OR = No difference
ii) Boolean algebra terminology applied to logic
circuits, addition and the OR operation are the same.

4.2 Boolean Algebra and Law

Commutative law
- Commutative Law of Addition ; A+B = B+A

- Commutative Law of Multiplication; AB=BA

4.2 Boolean Algebra and Law

Associative law
- Addition of two variables is written algebraically as :

A+(B+C)=(A+B)+C
- Law states,
i) OR , of more than two variables, the result is the same

4.2 Boolean Algebra and Law

Associative law
- Associative Law of Addition ; A+(B+C) = (A+B)+C

- Associative Law of Multiplication; A(BC)=(BC)A

4.2 Boolean Algebra and Law

Distributive Law
- written for three variables ;

A(B+C)=AB+AC
- Law states,
i) OR (two or more variables) & AND = AND and OR
ii) express the process of factoring common variable A

4.2 Boolean Algebra and Law

Distributive law
- Distributive Law of Addition ; A(B+C) = AB+AC

4.2 Boolean Algebra and Law

Simplification of Logic Function Using Boolean
Algebra

Example 4.1; Prove that X+XY=X

4.2 Boolean Algebra and Law

Simplification of Logic Function Using Boolean
Algebra
Example 4.2; Simplify; y=A � D+A � �

4.2 Boolean Algebra and Law

Simplification of Logic Function Using Boolean
Algebra
Exercise 1; Simplify; y=ACD+ B̅ CD

4.3 De Morgan’s Theorem

- To simplifying expressions , 2 theorems .
Theorem 1

When OR sum of two or more variablesis inverted=invertingeach variable
individually and the AND theseinvertedvariables.

Theorem 2

When AND productof two variables is inverted=invertingeach variable

individuallyand then OR them.

4.3 De Morgan’s Theorem

-Example 4.4 :

Simplify the expression ; to one having only
single variable inverted.

4.3 De Morgan’s Theorem

-Exercise 1:
Simplify the expression ; .

4.4 Truth Table

- For describing how logic circuits output depend on the
input.

Example 4.5
Truth Table for 2 input OR gate
Solution:

X Y Z=X+Y

4.4 Truth Table

Example 4.6
Prove the following expression using truth table
a. A (B+C)=AB+AC

4.4 Truth Table

Exercise 1
Prove the following expression using truth table
a. A●0=0
b. A+AB=A

4.5 Karnaugh Map

- Alternate approach to representing a boolean
function.
- Similar truth table
- Can be used to minimize boolean function graphically
- Each cell position in Karnaugh Map is fixed by a
binary value of the input variables.
- Number of cell = , n=number of input variable.

4.5 Karnaugh Map

Example 4.5.1
1) Draw the K-Map from thetrue tablegiven

Exercise 4.5.1
2)

or
or etc.

4.5 Karnaugh Map

4.5.1 Adjacency in K-Map
- Cell in Karnaugh Map are arrange – only single
variable change between adjacent cells.

Adjacent Cell
– single variable change
- immediately next to it on any of its four sides

Not Adjacent Cell
– Cell diagonally touch any of its corners (known as
wrap-around adjacency)

4.5 Karnaugh Map

a) Cell that differ by one variable are adjacent
- Example ; 3 variable K-Map

010 cell is adjacent to 000 cell, 011 cell & 110 cell

4.5 Karnaugh Map

b) Cell that differ by more than one variable are not
adjacent
- Example ; 3 variable K-Map

010 cell is not adjacent to 001 cell, 111 cell , 100 cell
or 101 cell

4.5 Karnaugh Map

4.5.2 K-Map Simplification of (Product of Sum) POS
Boolean Expression
- Standard to mapping POS expression as follows:

a) determine the binary value of each sum term in the
standard POS expression.

b) Place a ‘0’ on the K-map in the corresponding cell

4.5 Karnaugh Map

4.5.2 K-Map Simplification of (Product of Sum) POS
Boolean Expression

Example 4.7 :
Draw K Map for boolean expression;

F=(A+B+C)(A+B+ ̅)( +̅ B+ ̅)( +̅ � +C)

Solution;

4.5 Karnaugh Map

4.5.2 K-Map Simplification of (Product of Sum) POS
Boolean Expression

Exercise 1 :
Draw K Map for boolean expression;

F=( +̅ � + ̅)( +̅ � +C)( +̅ B+C)

4.5 Karnaugh Map

4.5.3 K-Map Representation of Sum of Product (SOP)
expression
- The steps of mapping – non-standard SOP expression
as follow :

Step 1: expand each product term numerically
Step 2 : for each binary value obtained in Step 1,

place a 1 in the cell that corresponds to the
binary value.

4.5 Karnaugh Map

4.5.3 K-Map Representation of Sum of Product (SOP)
expression

Example 4.8 : Draw K Map for boolean expression
given using SOP Standard.
F= ̅ + � + ̅

Answer :

4.5 Karnaugh Map

4.5.3 K-Map Representation of Sum of Product (SOP)
expression

Exercise 1 : Draw K Map for boolean expression given
using SOP Standard.
1. F= A � + ̅
2. F= A � + + ̅ ̅

4.5 Karnaugh Map

4.5.3 K-Map Representation of Sum of Product (SOP)
expression
Example 2 : Given 3 Input variable K-Map, find boolean
expression – SOP Standard

Answer .
F=A � + + ̅ ̅

4.5 Karnaugh Map

4.5.3 K-Map Representation of Sum of Product (SOP)
expression
Exercise : Given 3 Input variable K-Map, find boolean
expression – SOP Standard

Answer .
F= + �

4.5 Karnaugh Map

4.5.4 K-Map Simplification of SOP Expressions

Step 1: Map the SOP expression
Step 2 : Do groupingthe 1s by enclosingthose

adjacentcells containing1s
i. Rule 1 : A group must contain 2n cells which contain 1s where

n=0,1,2…
ii. Rule2 : The size of each groupsis maximum
iii. Rule3 : The number of groups is minimum
iv. Rule4 : Each 1 on the map is includedin at least one group
v. Rule5: The overlappinggroups include non-common1s
Step 3 : Determinethe minimum SOP

4.5 Karnaugh Map

4.5.4 K-Map Simplification of SOP Expressions

Exercise 1 :
Simplify the boolean expression given using K-Map -
Standard POS.
Y=(A+ + ̅)(A+B+C)(A+ � +C)( ̅+ � +C)(A+ � + ̅)

Answer :
Y=A( � + )

4.5 Karnaugh Map

4.5.4 K-Map Simplification of POS Expressions

Exercise : Given 3 input expression,
F=AB ̅ + ABC + A � , simplify it by using K-Map.

Answer :
F=AB+AC

4.5 Karnaugh Map

4.5.4 K-Map Simplification of POS Expressions

Exercise : Given 2 input expression,
F= ̅ + A � + AB, simplify it by using K-Map.

Answer :
F=A+B

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