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Computer Organization I
Lecture 5: Boolean Algebra
Overview
Why Studying Boolean Algebra in Circuit?
Basic Operation Rules in Boolean Algebra
How to Simplify Logic Expression?
Objectives
Understand Boolean algebra is the theoretic foundation
to analyze the operation of logic circuits
Familiar with the rules and theorems of Boolean algebra
Know how to simplify the logic expression
Why Studying Boolean Algebra?
• Boolean algebra used to simplify or optimize logic circuit
given logic expression Y = A + B C, truth table is unique,
however, the expression and logic circuit diagram are
not…for example:
Truth Table
Y =A
+ B· C
is equivalent to
Y = AB +AB + B· C
ABC
Y = A + B
000
0
001
1
010
0
011
0
100
1
101
1
110
1
111
1
·C
Why Studying Boolean Algebra?
• Boolean algebra used to simplify or optimize logic circuit
A
B
Logic Diagram Y = A
+ B· C
Y
C
A
Logic Diagram Y = AB +AB + B· C
Y
B
C
Comparing these two
diagrams, it is obvious that
the circuit of the second
expression is more complex
than the circuit in the first
expression even though both
expressions implement the
same function.
Using boolean algebra can
get the simplest expression
and thus get the optimized
circuit
Basic Operations in Boolean Algebra
- Boolean Addition?
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.
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 of 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.
Basic Operations in Boolean Algebra
- Boolean Multiplication?
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.
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.
Operation Rules in Boolean Algebra
- Basic Rules?
• Existence of 0-1
1. A + 0 = A
A=1
0
Y=1
A=0
0
Y=0
Y=1
A=0
1
Y=1
2. A + 1 = 1
A=1
1
3. A . 0 = 0
A=1
0
Y=0
A=0
0
Y=0
Y=1
A=0
1
Y=0
4. A . 1 = A
A=1
1
Operation Rules in Boolean Algebra
- Basic Rules?
• Idempotence
5. A + A = A
A=1
A=1
Y=1
A=0
A=0
Y=0
A=0
A=0
Y=0
6. A . A = A
A=1
A=1
Y=1
Operation Rules in Boolean Algebra
- Basic Rules?
• Existence of Complement
7. A + A = 1
A=1
A=0
Y=1
A=0
A=1
Y=1
8. A . A = 0
A=1
A=0
Y=0
A=0
A=1
Y=0
Operation Rules in Boolean Algebra
- Basic Rules?
• Involution
=
9. A = A
A=0
A=1
A=0
A=1
A=0
A=1
Operation Rules in Boolean Algebra
- Commutative Laws?
The commutative laws are applied to addition and
multiplication. For addition, the commutative law states
In terms of the result, the order in which variables are
ORed makes no difference.
A+B=B+A
A
B
is equivalent to
A+B
B
A
B+A
Operation Rules in Boolean Algebra
- Commutative Laws?
For multiplication, the commutative law states
In terms of the result, the order in which variables are
ANDed makes no difference.
AB = BA
A
B
is equivalent to
AB
B
A
BA
Operation Rules in Boolean Algebra
- Associative Laws?
The associative laws are also applied to addition and
multiplication. For addition, the associative law states
When ORing more than two variables, the result is
the same regardless of the grouping of the variables.
A + (B +C) = (A + B) + C
A+(B+C)
A
B
C
B+C
A
B
C
is equivalent to
A+B
(A+B)+C
Operation Rules in Boolean Algebra
- Associative Laws?
For multiplication, the associative law states
When ANDing more than two variables, the result is
the same regardless of the grouping of the variables.
A(BC) = (AB)C
A(BC)
A
B
C
BC
A
B
C
is equivalent to
AB
(AB)C
Operation Rules in Boolean Algebra
- Distributive Laws?
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
A
B
B+ C
X
X
A
A(B+ C)
AB
A
C
AC
AB + AC
Operation Rules in Boolean Algebra
- Boolean Operator Precedence?
The order of evaluation in a Boolean expression is:
1.
2.
3.
4.
Parentheses
NOT
AND
OR
Consequence: Parentheses appear around OR expressions
Example: F = A(B + C)(C + D)
Operation Rules in Boolean Algebra
- Absorption Theorem?
A + AB = A
Proof Steps
A + A·B
= A·1+A·B
= A · ( 1 + B)
= A·1
= A
Justification (identity or theorem)
X=X·1
X·Y+X·Z=
X ·(Y + Z) (Distributive Law)
1+X=1
X·1=X
• Our primary reason for doing proofs is to learn:
– Careful and efficient use of the identities and theorems of Boolean
algebra, and
– How to choose the appropriate identity or theorem to apply to
make forward progress, irrespective of the application.
Operation Rules in Boolean Algebra
- Absorption Theorem?
According to absorption theorem A + A·B = A, we can derive:
Rule 1: A · (A + B) = A
Rule 2: A · (A + B) = A · B
Rule 3: (A + B) · (A + C) = A + B · C
(A + B)(A + C) = AA + AC + AB + BC
= A + AC + AB + BC
= A(1 + C + B) + BC
= A . 1 + BC
= A + BC
Rule 4: A + A · B = A + B
Operation Rules in Boolean Algebra
- DeMorgan’s Theorem?
DeMorgan’s 1st Theorem
The complement of a product of variables is equal to the
sum of the complemented variables.
AB = A + B
Inputs
A
0
0
1
1
B
0
1
0
1
Output
AB A + B
1
1
1
1
1
1
0
0
Operation Rules in Boolean Algebra
- DeMorgan’s Theorem?
Proof of DeMorgan’s 1st Theorem
In order to prove AB = A + B we need to prove:
(1) A + B + AB = 1 and (2) (A + B ) · AB = 0
A + B + AB = (A + B + A) (A + B +B)
= (1 + B) (1 + A)
=1
(A + B ) · AB = A AB + B AB = 0
Operation Rules in Boolean Algebra
- DeMorgan’s Theorem?
DeMorgan’s 2nd Theorem
The complement of a sum of variables is equal to the product
of the complemented variables.
A+B=A.B
Inputs
A
0
0
1
1
B
0
1
0
1
Output
A + B AB
1
1
0
0
0
0
0
0
Operation Rules in Boolean Algebra
- DeMorgan’s Theorem?
Proof of DeMorgan’s 2nd Theorem
In order to prove A + B = A . B we need to prove:
(1) A · B + A + B = 1 and (2) (A · B ) · (A + B) = 0
A · B + A + B = (A + B + A) (B + A +B)
= (1 + B) (1 + A)
=1
(A · B ) · (A + B) = A (A · B) + B (A · B) = 0
Operation Rules in Boolean Algebra
- DeMorgan’s Theorem?
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.
Boolean Analysis 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:
A
B
(A + B )
C (A + B )
X = C (A + B )+ D
C
D
Applying DeMorgan’s theorem and the distribution law:
X = C (A B) + D = A B C + D
Logic Expression Simplification
• An application of Boolean algebra
• Simplify to contain the smallest number of
literals (complemented and uncomplemented
variables):
A B + A C D + A BD + A C D + A BC D
= AB + ABCD + A C D + A C D + A B D
= AB + AB(CD) + A C (D + D) + A B D
= AB + A C + A B D = B(A + AD) +AC
= B (A + D) + A C, there are 5 literals
How to Calculate Complementing Functions
• Use DeMorgan's Theorem to complement
a function:
1. Interchange AND and OR operators
2. Complement each constant value and
literal
• Example: Complement F =
xy z + xyz
F = (x + y + z)(x + y + z)
• Example: Complement G = (a + bc)d + e
G=
Chapter 2 - Part 1
27
Summary
• Basic Operation Rules and Laws in Boolean Algebra
• How to Simplify Logic Expression
Thank you
Q&A