session 3 - Boolean Functions Minimization.pptx

Boolean functions
and minimization
Dr.S.Joshua Kumaresan M.E, M.S, Ph.D
Associate professor/ECE
R.M.K. Engineering college

 A Boolean function is an expression formed with:
 Binary variables
 Operators (OR, AND, and NOT)
 Parentheses, and equal sign
 The value of a Boolean function can be either 0 or 1
 A Boolean function may be represented as:
 An algebraic expression, or
 A truth table
 A Logic diagram
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Boolean Functions

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• Boolean Expression
Example: F(x, y) = x + y’ z
• Truth Table
All possible combinations
of input variables
• Logic Circuit
x y z F
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1
x
y
z
F
BOOLEAN FUNCTIONS
DIGITAL PRINCIPLES AND SYSTEM DESIGN 4

Boolean Theorems - Recap
Animated Video

Simplification - EXAMPLE
• (A + B)(A + C) = A + BC
• This rule can be proved as follows:
• (A + B)(A + C) = AA + AC + AB + BC Distributive law
• = A + AC + AB + BC AA = A
• = A( 1 + C) + AB + BC 1 + C = 1
• = A. 1 + AB + BC Factoring (distributive law)
• = A(1 + B) + BC 1 + B = 1
• = A. 1 + BC A . 1 = A
• = A + BC

Simpification - EXAMPLE
Simplify the following expression
Y = (A + B) (A’ + C) (B' + C’)
Y = (A + B) (A’ + C) (B' + C’)
= (AA' + AC +A'B +BC) (B' + C') [A.A' = 0]
= (AC + A'B + BC) (B' + C’)
= AB'C + ACC' + A'BB' + A'BC' + BB'C + BCC‘
= AB'C + A'BC

Simplification - EXAMPLE
• Prove that ABC + ABC' + AB'C + A'BC = AB + AC + BC
• ABC + ABC' + AB'C + A'BC
= AB(C + C') + AB'C + A'BC
= AB + AB'C + A'BC
= A (B + B'C) + A'BC
= A (B + C) (B + B') + A'BC
= AB + AC + A'BC
= B (A + A'C) + AC
= B (A + A') (A + C) + AC
= AB + BC + AC
= AB + AC +BC ...Proved

Simplification – Example
•Solve the boolean expression
•(x+ y) (x+ y’)
= x.x+ xy’+ yx+ yy’
= x+ xy’+ xy+ 0 [ x. x= x]; [ y. y’= 0]
= x (1+ y’+ y)
= x (1) [ 1+y= 1 ]
= x.

Simplification - Example
Solve the expression :A’B’+ A’C’D’+ A’B’D+
A’B’CD’
= A’B’ (1+D) + A’C’D’+ A’B’CD’
= A’B’ (1) + A’C’D’+ A’B’CD’ [1+ x = 1]
= A’B’+ A’C’D’+ A’B’CD’
= A’B’+ A’B’CD’+ A’C’D’
= A’B’ (1+CD’) + A’C’D’
= A’B’ (1) + A’C’D’ [1+ x = 1]
= A’B’+ A’C’D’
= A’ (B’+C’D’)

Consensus Theorem
AB + C + BC = AB + C
Proof Steps
AB + C + BC
= AB + C + 1 · BC
= AB + C + (+ A) · BC
= AB + C + BC + ABC
= AB (1+C) +C (1 + B)
= AB . 1 + C . 1
= AB + C

(a)x+y
Ans: x+y
(b)x.+y)
Ans: xy
© x
Ans: x.(
Try out some Boolean Function Minimization

 The complement of a Boolean function is obtained by
interchanging:
 Operators OR and AND
 Complementing each literal
 This is based on De Morgan’s theorems,
whose general form is:
A 1+A 2 +A 3 +...+An = A 1A 2 A 3 ...An
A 1A 2 A 3 ...An = A 1+A 2 +A 3 +...+An
80
Complement of a Boolean Function

80
Complementing a Boolean Function (Example)
Complement
=
=
=

SUM OF PRODUCT
• The sum-of-products (SOP) form is a method (or form) of simplifying the Boolean
expressions of logic gates.
• Sum and product derived from the symbolic representations of the OR and AND
functions.
• OR (+) , AND ( . ) , addition and multiplication.
• Minterms : It is a product term that consists of all the varaibles in complemented or
true form
f(A,B,C) = ABC + A’BC’
SUM
PRODUCT TERM

A sum-of-products (SOP) expression is a product term
or several product terms logically added (ORed)
together. Examples are:
F(x,y,z)=
xy+z
F(x,y,z) =x+ yz
F(x,y)= xy+
xy
80
Sum-of-Products (SOP) Expression

A B C F
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 1
CONVERTING FROM TRUTH TABLE
TO BOOLEAN FUNCTION
Truth Table to Boolean Function
F = A’B’C+ AB'C' + AB'C +ABC
Using Minterms

CONVERTING FROM TRUTH TABLE TO
BOOLEAN FUNCTION
• Sum of Minterms
• F =A’B’C + AB’C’ + AB’C+ ABC
• F= m1 + m4 + m5 + m7
A B C F
0 0 0 0 0
1 0 0 1 1
2 0 1 0 0
3 0 1 1 0
4 1 0 0 1
5 1 0 1 1
6 1 1 0 0
7 1 1 1 1
F = ∑(1,4,5,7)

x y z F1
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 1
80
The following 3 combinations of the variables produce a 1:
001, 100, and 111
Expressing a Function in its
Sum-of-Products Form (Example)
(Continued on next slide)

Sum-of-Products Form (Example)
(Continued from previous slide..)
 Their corresponding minterms are:
x y z, x y z, and x y 
z
 Taking the OR of these minterms, we get
F1 =xy z+ x y z+ x y
z=m1+m4 m7
80
F1 xy z=1, 4,7

MINTERMS
⮚ f(A,B,C) = A’B’C’ + A’BC’ + A’BC + ABC
= m0 + m2 + m3 + m7
= Σm(0,2,3,7)
⮚ f(A,B,C)= A’B’C + A’BC + AB’C + ABC
= m1 + m3 + m5 + m7
= Σm(1,3,5,7)
⮚ f(A,B,C)= A’B’C’ + A’BC’ + A’BC + ABC’
= m0 + m2 + m3 + m6
= Σm(0,2,3,6)

CANONICAL FORM
• In SOP or POS form, all individual terms do not
involve all literals.
• For example AB + A’BC the first product term
do not contain literal C.
• If each term in SOP or POS contain all
literals then the expression is known as
canonical form.

CONVERT SOP TO CANONICAL
SOP FORM
•Step 1: Find the missing literal in each product term if any.
•Step 2: And each product term having missing literals with
terms form by ORing the literal and its complement.
•Step 3: Expends the term by applying, distributive
law and reorder the literals.
•Step 4: Reduce the repeated product terms.
Because A + A = A (Theorem 1a ).

EXAMPLE
f (A,B,C) = AB + BC + AC
Step 1: Find the missing literals in each product
term.
f (A,B,C) = AB + BC + AC
Step 2: AND the product term with missing literal + its
complement.
f (A,B,C) = AB . (C+C’) + BC . (A+A’) + AC . (B+B’)
Literal B is missing
Literal A is missing
Literal C is missing
Missing literals and their complements

Step 3: Expends the term and reorder the literals.
f (A,B,C) = AB . (C+C’) + BC . (A+A’) + AC . (B+B’)
Expand & Reorder:
ABC + ABC’ + ABC + A’BC + ABC + AB’C
Step 4: Omit repeated product terms.
f(A,B,C)=ABC + ABC’ + ABC + A’BC + ABC + AB’C
f(A,B,C)= ABC + ABC’ + A’BC + AB’C
Write once

PRODUCT OF SUM
• When two or more sum terms are multiplied by a Boolean OR operation.
• Sum terms are defined by using OR operation and the product term is
defined by using AND operation.
• Maxterms : It is a sum term that consists of all the varaibles in
complemented or true form
f(A,B,C) = (A’+C) . (A’+B+C’)
PRODUCT
SUM TERM

Product-of Sums (POS) Expression
A product-of-sums (POS) expression is a sum term
(maxterm) or several sum terms (maxterms) logically
multiplied (ANDed) together. Examples are:
F(x,y) = x+ yx+ yx+y
F(x,y,z) = x + yx+ y+z
F(x,y) = x+ yx+y 
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21 / 28
A B C F
0 0 0 1
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 0
BOOLEAN FUNCTION
Truth Table to Boolean Function
F = ( A + B +C') ( A' + B + C)( A '+ B + C')( A' + B' + C')
Using Maxterms

BOOLEAN FUNCTION
• Productof Maxterms
• F=(A+B+C).(A+B’+C).(A+B’+C’).(A’+B’+C)
• F= m0 . m2 . m3 . M6
A BC F
0 0 0 0 0
1 0 0 1 1
2 0 1 0 0
3 0 1 1 0
4 1 0 0 1
5 1 0 1 1
6 1 1 0 0
7 1 1 1 1
F = Π (0,2,3,6)

x y z F1
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 1
 The following 5 combinations of variables produce a 0:
000, 010, 011, 101, and
110
Product-of-Sums Form
(Continued on next slide)

Product-of-Sums Form
(Continued from previous slide..)
 Their corresponding maxterms are:
x+ y+ z, x+ y+ z, x+ y+ z,
x+ y+ z and x+ y+ z
 Taking the AND of these maxterms, we get:
F1 =x+ y+ zx+ y+ z x+ y+ z x+
y+z
x+ y+z=M0 M2 M3  M5 M6
F1 x,y,z = Π0,2,3,5,6

MAXTERMS
⮚f(A,B,C) = (A+B+C).(A+B’+C).(A+B’+C’)+(A’+B’+C’)
= M0 + M2 + M3 + M7
= ΠM(0,2,3,7)
⮚f(A,B,C)= (A+B+C’).(A+B’+C’).(A+B’+C’).(A’+B’+C’)
= M1 + M3 + M5 + M7
= ΠM (1,3,5,7)
⮚f(A,B,C)= (A+B+C).(A+B’+C).(A+B’+C’).(A’+B’+C)
= M0 + M2 + M3 + M6
= ΠM (0,2,3,6)

CONVERT POS TO STANDARD
POS FORM
•Step 1: Find the missing literal in each sum term if any.
•Step 2: OR each sum term having missing literals with terms
form by ANDing the literal and its complement.
•Step 3: Expends the term by applying, distributive
law and reorder the literals.
•Step 4: Reduce the repeated product terms. Because A +
A = A (Theorem 1a ).

EXAMPLE
f (A,B,C) = (A + B) . (B + C) . (A + C)
Step 1: Find the missing literals in each sum
term.
f (A,B,C) = (A + B) . (B + C) . (A + C)
Step 2: OR the sum term with missing
literal . its complement.
f (A,B,C) = (A + B)+(C.C’) + (B + C)+(A.A’) + (A +
C)+(B.B’)
Literal B is missing
Literal A is missing
Literal C is missing
Missing literals and their complements

Step 3: Expends the term and reorder the
literals.
f (A,B,C) = (A + B)+(C.C’) + (B + C)+(A.A’) + (A +C)+(B.B’)
Expand & Reorder:
f(A,B,C)=(A+B+C).(A+B+C’).(A+B+C).(A’+B+C).(A+B+C).
(A+B’+C)
Step 4: Omit repeated sum terms.
f(A,B,C)=(A+B+C).(A+B+C’).(A+B+C).(A’+B+C).(A+B+C).(A+B’+C)
f(A,B,C)=(A+B+C).(A+B+C’).(A’+B+C).(A+B’+C)
Write once

=
Σ
To convert from one canonical form to another,
interchange the symbol and list those numbers missing
from the original form.
Example:
Fx,y,z= Π0,2,4,5 = Σ1,3,6,7
Fx,y,z 1,4,7 = Σ0,2,3,5,6
Conversion Between Canonical Forms (Sum-of-
Products and Product-of-Sums)

Note that each minterm is the complement of its corresponding maxterm and vice-versa
Variables Minterms Maxterms
x y z Term Designation Term Designation
0 0 0
x  y z m 0 x  y  z M 0
0 0 1
x  y z m 1 x  y  z M 1
0 1 0
x  y z m 2 x  y  z M 2
0 1 1
x  y z m 3 x  y  z M 3
1 0 0
x  y z m 4 x  y  z M 4
1 0 1
x  y z m 5 x  y  z M 5
1 1 0
x  y z m 6 x  y  z M 6
1 1 1
x  y z m 7 x  y  z M 7
80
Minterms and Maxterms for three Variables

session 3 - Boolean Functions Minimization.pptx

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