All About Boolean Algebra DLD.

© 2009 Pearson Education, Upper Saddle River, NJ 07458. All Rights ReservedFloyd, Digital Fundamentals, 10th ed
Digital
Fundamentals
Tenth Edition
Floyd
Chapter 4
© 2008 Pearson Education

Chapter Objectives
• Apply the basic laws and rules of Boolean algebra.
• Apply DeMorgan’s theorems to Boolean expressions
• Describe gate combinations with Boolean expressions
• Evaluate Boolean expressions
• Simplify expressions by using the laws and rules of
Boolean algebra
• Convert any Boolean expression into a sum-of-product
(SOP) and Product-of-sum (POS) form.
• Relate a Boolean expression to a truth table
• Use a karnaugh map to simplify Boolean expressions.

Boolean Operations and Expressions
• Boolean algebra is the mathematics of
digital systems.
• A basic knowledge of Boolean algebra is
indispensable to the study and analysis of
logic.
• In the last lecture, Boolean operations and
expressions in terms of their relationship to
NOT, AND, OR, NAND, and NOR gates
were introduced.

After completing this section:
• You should be able to define variable
• Literal
• Identify a sum term
• Evaluate a sum term
• Identify a product term
• Evaluate a product term
• Explain Boolean addition
• Explain Boolean multiplication

In Boolean algebra, a variable is a symbol used to represent
an action, a condition, or data. A single variable can only
have a value of 1 or 0.
Summary
Boolean Addition
The complement represents the inverse of a variable and is indicated
with an overbar. Thus, the complement of A is A.
A literal is a variable or its complement.
Addition is equivalent to the OR operation. The sum term is 1 if one or
more if the literals are 1. The sum term is zero only if each literal is 0.
Determine the values of A, B, and C that make the sum term
of the expression A + B + C = 0?
Each literal must = 0; therefore A = 1, B = 0 and C = 1.

Boolean Addition

In Boolean algebra, multiplication is equivalent to the AND
operation. The product of literals forms a product term. The
product term will be 1 only if all of the literals are 1.
Summary
Boolean Multiplication
What are the values of the A, B and C if the
product term of A.B.C = 1?
Each literal must = 1; therefore A = 1, B = 0 and C = 0.
What are the values of the A, B and C if the
product term of A.B.C = 1?

Boolean Multiplication

Summary
Commutative Laws
In terms of the result, the order in which variables
are ORed makes no difference.
The commutative laws are applied to addition and
multiplication. For addition, the commutative law states
A + B = B + A
In terms of the result, the order in which variables
are ANDed makes no difference.
For multiplication, the commutative law states
AB = BA

Continued..

Summary
Distributive Law
The distributive law is the factoring law. A common
variable can be factored from an expression just as in
ordinary algebra. That is
AB + AC = A(B+ C)
The distributive law can be illustrated with equivalent
circuits:
B+ C
C
A
X
B
AB
B
X
A
C
A
AC
AB + ACA(B+ C)

Summary
Rules of Boolean Algebra
1. A + 0 = A
2. A + 1 = 1
3. A . 0 = 0
4. A . 1 = A
5. A + A = A
7. A . A = A
6. A + A = 1
8. A . A = 0
9. A = A
=
10. A + AB = A
12. (A + B)(A + C) = A + BC
11. A + AB = A + B

Summary
Rules of Boolean algebra can be illustrated with Venn
diagrams. The variable A is shown as an area.
The rule A + AB = A can be illustrated easily with a diagram. Add
an overlapping area to represent the variable B.
A B
AB
The overlap region between A and B represents AB.
AA BA B
AB
The diagram visually shows that A + AB = A. Other rules can be
illustrated with the diagrams as well.
=

The proof is shown in the table, which shows the
truth table and the resulting logic circuit
simplification

A
Summary
A + AB = A + B
This time, A is represented by the blue area and B
again by the red circle.
B
The intersection represents
AB. Notice that A + AB = A + B
AABA
Illustrate the rule with a Venn
diagram.

Summary
Rule 12, which states that (A + B)(A + C) = A + BC, can
be proven by applying earlier rules as follows:
(A + B)(A + C) = AA + AC + AB + BC
= A + AC + AB + BC
= A(1 + C + B) + BC
= A . 1 + BC
= A + BC
This rule is a little more complicated, but it can also be
shown with a Venn diagram, as given on the following
slide…

Summary
DeMorgan’s Theorem
The complement of a product of variables is
equal to the sum of the complemented variables.
DeMorgan’s 1st Theorem
AB = A + B
Applying DeMorgan’s first theorem to gates:
OutputInputs
A B AB A + B
0
0
1
1
0
1
0
1
1
1
1
0
1
1
1
0
A + B
A
B
AB
A
B
NAND Negative-OR

Summary
DeMorgan’s 2nd Theorem
The complement of a sum of variables is equal to
the product of the complemented variables.
A + B = A . B
Applying DeMorgan’s second theorem to gates:
A B A + B AB
OutputInputs
0
0
1
1
0
1
0
1
1
0
0
0
1
0
0
0
AB
A
B
A + B
A
B
NOR Negative-AND
Remember:
“Break the bar,
change the sign”

Summary
Apply DeMorgan’s theorem to remove the
overbar covering both terms from the
expression X = C + D.
To apply DeMorgan’s theorem to the expression,
you can break the overbar covering both terms and
change the sign between the terms. This results in
X = C . D. Deleting the double bar gives X = C . D.
=

Apply the demorgans theorem to the following
expressions
• (XYZ)' and (X+Y+Z)'
• (X'+Y'+Z')'
• (WXYZ)' AND (W+X+Y+Z)'
• (W'X'Y'Z')'

Apply demorgans theorems to each of
the following expressions
• [(A+B+C)D]‘
• (ABC+DEF)‘
• [AB‘+C'D+EF]'

A
C
D
B
Summary
Boolean Analysis/expression of Logic Circuits
Combinational logic circuits can be analyzed by writing
the expression for each gate and combining the
expressions according to the rules for Boolean algebra.
Apply Boolean algebra to derive the expression for X.
Write the expression for each gate:
Applying DeMorgan’s theorem and the distribution law:
C (A + B )
= C (A + B )+ D
(A + B )
X = C (A B) + D = A B C + D
X

Standard forms of Boolean expressions
• All Boolean expressions, regardless of their form, can be
converted into either of two standard forms
• Sum-of-products (SOP) form
• Also called Minterms
• Product of Sum (POS) form
• Also called Maxterms
• Standardization makes the evaluation, simplification, and
implementation of Boolean expressions much more
systematic and easier.

Sum of product (SOP) Form
• A product term was defined as a term consisting of
the product (Boolean multiplication) of literals
(variables or their complements).
• When two or more product terms are summed by
Boolean addition. the resulting expression is a
sum-of-products (SOP).
• Some examples are
• AB + ABC
• ABC + CDE + B‘CD‘
• A‘B + A‘BC‘ + AC

SOP (Minterms)
• Also, an SOP expression can contain a single-
variable like,
• A + A’BC’ + BC’D’
Domain of a Boolean Expression
• The domain of a general Boolean expression is the
set of variables contained in the expression in
either complemented or un complemented form.
• For example, the domain of the expression A’B +
AB’C is:-
• The set of variables A, B, C.

SOP
AND/OR Implementation of an SOP Expression
• Implementing an SOP expression simply requires
ORing the outputs of two or more AND gates.
• A product term is produced by an AND operation,
and the sum (addition) of two or more product
terms is produced by an OR operation.
• Therefore, an SOP expression can be implemented
by AND-OR logic in which the outputs of a number
of AND gates connect to the inputs of an OR gate.

SOP form
NAND/NAND Implementation of an SOP Expression
• NAND gates can be used to implement an SOP expression.
• Using only NAND gates, an AND/OR function can be
accomplished.
• The first level of NAND gates feed into a NAND gate that
acts as a negative-OR gate.
• The NAND and negative-OR inversions cancel and the
result is effectively an AND/OR circuit.

Conversion of a General Expression to SOP Form
• Any logic expression can be changed into
SOP form by applying Boolean algebra
techniques.
• For example. the expression A(B + CD) can
be converted to SOP form by applying the
distributive law.
• A(B + CD) = AB + ACD

Convert each of the following Boolean expressions to SOP form
• AB + B(CD + EF)
= AB + BCD + BEF
• (A + B)(B + C + D)
= AB + AC + AD + BB + BC + BD
• [( A + B )’ + C ]’
= (( A + B )’)’C’ = (A + B)C’
= AC’ + BC’.

Standard SOP Form
• SOP expressions in which some of the product
terms do not contain some of the variables in the
domain of the expression.
• For example
• A’ BC’ + AB’ D + A’ B C’ D
• Domain made up of the variables A, B, C. and D.
• Complete set of variables in the domain is not
represented in the first two terms of the expression.
• D or D’ is missing from the first term
• C and C’ is missing from the second term.

SOP form
• A standard SOP expression is one in which
all the variables in the domain appear in
each product term in the expression.
• For example.
• AB’CD + A’ B’ CD’ + ABC’ D’

STANDARD SOP
• Converting Product Terms to Standard SOP
• Step 1.Multiply each nonstandard product term by a term
made up of the sum of a missing variable and its
complement. This results in two product terms. As you know
you can multiply anything by 1 with out changing its value.
• Step 2.Repeat Step 1 until all resulting product terms
contain all variables in the domain in either complemented
or un complemented form.
• In converting a product term to standard form, the number of
product terms is doubled for each missing variable

Convert the following Boolean expression into standard SOP form:
• AB’ C + A’ B’+ ABC’ D
• Solution
• AB’ C = AB’ C(D + D’ ) = AB’ CD + AB’ CD’
• In this case, two standard product terms are
the result.
• The second term, A’B’, is missing variables C
or C’ and D or D’, so first multiply the second
term by C + C’ as follows.

Example
• A‘ B‘ = A‘B‘(C + C‘) = A‘ B‘ C + A‘ B‘ C ‘
• The two resulting terms are missing variable D or D’, so
multiply both terms by D + D’ as follows:
• A’B’ = A’B’C + A’B’C’ = A’B’C(D + D’) +
• A’B’C’(D + D’)
• = A’ B’ CD + A’ B’ CD’ + A’B’C’D + A’B’C’D’
• In this case, four standard product terms are the result.
The third term, ABCD, is already in standard form. The
complete standard SOP form of the original expression
is as follows:

Example contd
• Accumulated terms are,
• AB’ C + A’ B’+ ABC’ D = AB’CD + AB’CD’ +
A’B’CD + A’ B’CD’ + A’B’C’D + A’B’C’D’ +
ABC’D.
• Convert the expression WX’ Y + X’ YZ’ +
WX Y’to standard SOP form.

The Product-of-Sums (POS) Form
• When two or more sum terms are multiplied,
the resulting expression is a product-of-
sums (POS).
• (A’ + B)(A + B’ + C)
• (A + B’ + C)( C + D’ + E’)(B + C’ + D)
• (A’ + B)(A’ + B’ + C’)(A + C’)
• A POS expression can contain a single-
variable term, as in the following i.e. A’,
• A’ (A + B’ + C)(B’+ C’ + D).

Converting Standard SOP to Standard POS form.

The equivalent POS expression is
• (A+B+C’)(A’+B+C)(A’+B’+C)

Converting SOP expressions to the truth table format

Converting POS Expressions to Truth Table Format

All About Boolean Algebra DLD.

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