Boolean algebra And Logic Gates

Charles Kime & Thomas Kaminski
© 2008 Pearson Education, Inc.
(Edited by Dr. Muhamed Mudawar for COE 202 & EE 200 at KFUPM)
Boolean Algebra
and Logic Gates
Logic and Computer Design Fundamentals

Binary Logic and Gates
 Binary variables take on one of two values.
 Logical operators operate on binary values and
binary variables.
 Basic logical operators are the logic functions
AND, OR and NOT.
 Logic gates implement logic functions.
 Boolean Algebra: a useful mathematical system
for specifying and transforming logic functions.
 We study Boolean algebra as a foundation for
designing and analyzing digital systems!

Binary Variables
 Recall that the two binary values have
different names:
• True/False
• On/Off
• Yes/No
• 1/0
 We use 1 and 0 to denote the two values.
 Variable identifier examples:
• A, B, y, z, or X1 for now
• RESET, START_IT, or ADD1 later

Logical Operations
 The three basic logical operations are:
• AND
• OR
• NOT
 AND is denoted by a dot (·).
 OR is denoted by a plus (+).
 NOT is denoted by an overbar ( ¯ ), a
single quote mark (') after, or (~) before
the variable.

 Examples:
• is read “Y is equal to A AND B.”
• is read “z is equal to x OR y.”
• is read “X is equal to NOT A.”
Notation Examples
 Note: The statement:
1 + 1 = 2 (read “one plus one equals two”)
is not the same as
1 + 1 = 1 (read “1 or 1 equals 1”).
= BAY ⋅
yxz +=
AX =

Operator Definitions
 Operations are defined on the values
"0" and "1" for each operator:
AND
0 · 0 = 0
0 · 1 = 0
1 · 0 = 0
1 · 1 = 1
OR
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 1
NOT
10 =
01=

01
10
X
NOT
XZ=
Truth Tables
 Tabular listing of the values of a function for all
possible combinations of values on its arguments
 Example: Truth tables for the basic logic operations:
111
001
010
000
Z = X·YYX
AND OR
X Y Z = X+Y
0 0 0
0 1 1
1 0 1
1 1 1

Truth Tables – Cont’d
 Used to evaluate any logic function
 Consider F(X, Y, Z) = X Y + Y Z
X Y Z X Y Y Y Z F = X Y + Y Z
0 0 0 0 1 0 0
0 0 1 0 1 1 1
0 1 0 0 0 0 0
0 1 1 0 0 0 0
1 0 0 0 1 0 0
1 0 1 0 1 1 1
1 1 0 1 0 0 1
1 1 1 1 0 0 1

 Using Switches
• Inputs:
 logic 1 is switch closed
 logic 0 is switch open
• Outputs:
 logic 1 is light on
 logic 0 is light off.
• NOT input:
 logic 1 is switch open
 logic 0 is switch closed
Logic Function Implementation
Switches in series => AND
Switches in parallel => OR
C
Normally-closed switch => NOT

 Example: Logic Using Switches
 Light is on (L = 1) for
L(A, B, C, D) =
and off (L = 0), otherwise.
 Useful model for relay and CMOS gate circuits,
the foundation of current digital logic circuits
Logic Function Implementation – cont’d
B
A
D
C
A (B C + D) = A B C + A D

Logic Gates
 In the earliest computers, switches were opened
and closed by magnetic fields produced by
energizing coils in relays. The switches in turn
opened and closed the current paths.
 Later, vacuum tubes that open and close
current paths electronically replaced relays.
 Today, transistors are used as electronic
switches that open and close current paths.

Logic Gate Symbols and Behavior
 Logic gates have special symbols:
 And waveform behavior in time as follows:
X 0 0 1 1
Y 0 1 0 1
X · Y(AND) 0 0 0 1
X + Y(OR) 0 1 1 1
(NOT) X 1 1 0 0
OR gate
X
Y
Z = X + Y
X
Y
Z = X · Y
AND gate
X Z = X
NOT gateor
inverter

Logic Diagrams and Expressions
 Boolean equations, truth tables and logic diagrams describe
the same function!
 Truth tables are unique, but expressions and logic diagrams
are not. This gives flexibility in implementing functions.
X
Y F
Z
Logic Diagram
Logic Equation
ZYXF +=
Truth Table
11 1 1
11 1 0
11 0 1
11 0 0
00 1 1
00 1 0
10 0 1
00 0 0
X Y Z ZYXF ⋅+=

Gate Delay
 In actual physical gates, if an input changes that
causes the output to change, the output change
does not occur instantaneously.
 The delay between an input change and the
output change is the gate delay denoted by tG:
tG
tG
Input
Output
Time (ns)
0
0
1
1
0 0.5 1 1.5
tG = 0.3 ns

1.
3.
5.
7.
9.
11.
13.
15.
17.
Commutative
Associative
Distributive
DeMorgan’s
2.
4.
6.
8.
X .
1 X=
X .
0 0=
X .
X X=
0=X .
X
Boolean Algebra
10.
12.
14.
16.
X + Y Y + X=
(X + Y) Z+ X + (Y Z)+=
X(Y + Z) XY XZ+=
X + Y X .
Y=
XY YX=
(XY)Z X(YZ)=
X + YZ (X + Y)(X + Z)=
X .
Y X + Y=
X + 0 X=
+X 1 1=
X + X X=
1=X + X
X= X
 Invented by George Boole in 1854
 An algebraic structure defined by a set B = {0, 1}, together with two
binary operators (+ and ·) and a unary operator ( )
Idempotence
Complement
Involution
Identity element

Some Properties of Boolean Algebra
 Boolean Algebra is defined in general by a set B that can
have more than two values
 A two-valued Boolean algebra is also know as Switching
Algebra. The Boolean set B is restricted to 0 and 1.
Switching circuits can be represented by this algebra.
 The dual of an algebraic expression is obtained by
interchanging + and · and interchanging 0’s and 1’s.
 The identities appear in dual pairs. When there is only
one identity on a line the identity is self-dual, i. e., the
dual expression = the original expression.
 Sometimes, the dot symbol ‘’ (AND operator) is not
written when the meaning is clear

 Example: F = (A + C) · B + 0
dual F = (A · C + B) · 1 = A · C + B
 Example: G = X · Y + (W + Z)
dual G =
 Example: H = A · B + A · C + B · C
dual H =
 Unless it happens to be self-dual, the dual of an
expression does not equal the expression itself
 Are any of these functions self-dual?
(A+B)(A+C)(B+C)=(A+BC)(B+C)=AB+AC+BC
Dual of a Boolean Expression
(X+Y) · (W · Z) = (X+Y) · (W+Z)
(A+B) · (A+C) · (B+C)
H is self-dual

Boolean Operator Precedence
 The order of evaluation is:
1. Parentheses
2. NOT
3. AND
4. OR
 Consequence: Parentheses appear
around OR expressions
 Example: F = A(B + C)(C + D)

Boolean Algebraic Proof – Example 1
 A + A · B = A (Absorption Theorem)
Proof Steps Justification
A + A · B
= A · 1 + A · B Identity element: A · 1 = A
= A · ( 1 + B) Distributive
= A · 1 1 + B = 1
= A Identity element
 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.

 AB + AC + BC = AB + AC (Consensus Theorem)
Proof Steps Justification
= AB + AC + BC
= AB + AC + 1 · BC Identity element
= AB + AC + (A + A) · BC Complement
= AB + AC + ABC + ABC Distributive
= AB + ABC + AC + ACB Commutative
= AB · 1 + ABC + AC · 1 + ACB Identity element
= AB (1+C) + AC (1 + B) Distributive
= AB . 1 + AC . 1 1+X = 1
= AB + AC Identity element
Boolean Algebraic Proof – Example 2

Useful Theorems
 Minimization
X Y + X Y = Y
 Absorption
X + X Y = X
 Simplification
X + X Y = X + Y
 DeMorgan’s
 X + Y = X · Y
 Minimization (dual)
(X+Y)(X+Y) = Y
 Absorption (dual)
X · (X + Y) = X
 Simplification (dual)
X · (X + Y) = X · Y
 DeMorgan’s (dual)
 X · Y = X + Y

Truth Table to Verify DeMorgan’s
X Y X·Y X+Y X Y X+Y X · Y X·Y X+Y
0 0 0 0 1 1 1 1 1 1
0 1 0 1 1 0 0 0 1 1
1 0 0 1 0 1 0 0 1 1
1 1 1 1 0 0 0 0 0 0
X + Y = X · Y X · Y = X + Y
 Generalized DeMorgan’s Theorem:
X1+ X2+ … + Xn = X1 · X2 · … · Xn
X1 · X2 · … · Xn = X1 + X2 + … + Xn

Complementing Functions
 Use DeMorgan's Theorem:
1. Interchange AND and OR operators
2. Complement each constant and literal
 Example: Complement F =
F = (x + y + z)(x + y + z)
 Example: Complement G = (a + bc)d + e
G = (a (b + c) + d) e
x+ zyzyx

Expression Simplification
 An application of Boolean algebra
 Simplify to contain the smallest number
of literals (variables that may or may not
be complemented)
= 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 (has only 5 literals)
++++ DCBADCADBADCABA

Next … Canonical Forms
 Minterms and Maxterms
 Sum-of-Minterm (SOM) Canonical Form
 Product-of-Maxterm (POM) Canonical Form
 Representation of Complements of Functions
 Conversions between Representations

Minterms
 Minterms are AND terms with every variable
present in either true or complemented form.
 Given that each binary variable may appear
normal (e.g., x) or complemented (e.g., ), there
are 2n
minterms for n variables.
 Example: Two variables (X and Y) produce
2 x 2 = 4 combinations:
(both normal)
(X normal, Y complemented)
(X complemented, Y normal)
(both complemented)
 Thus there are four minterms of two variables.
YX
XY
YX
YX
x

Maxterms
 Maxterms are OR terms with every variable in
true or complemented form.
 Given that each binary variable may appear
normal (e.g., x) or complemented (e.g., x), there
are 2n
maxterms for n variables.
 Example: Two variables (X and Y) produce
2 x 2 = 4 combinations:
(both normal)
(x normal, y complemented)
(x complemented, y normal)
(both complemented)
YX +
YX +
YX +
YX +

 Two variable minterms and maxterms.
 The minterm mi should evaluate to 1 for each
combination of x and y.
 The maxterm is the complement of the minterm
Minterms & Maxterms for 2 variables
x y Index Minterm Maxterm
0 0 0 m0 = x y M0 = x + y
0 1 1 m1 = x y M1 = x + y
1 0 2 m2 = x y M2 = x + y
1 1 3 m3 = x y M3 = x + y

Minterms & Maxterms for 3 variables
M3 = x + y + zm3 = x y z3110
M4 = x + y + zm4 = x y z4001
M5 = x + y + zm5 = x y z5101
M6 = x + y + zm6 = x y z6011
1
1
0
0
y
1
0
0
0
x
1
0
1
0
z
M7 = x + y + zm7 = x y z7
M2 = x + y + zm2 = x y z2
M1 = x + y + zm1 = x y z1
M0 = x + y + zm0 = x y z0
MaxtermMintermIndex
Maxterm Mi is the complement of minterm mi
Mi = mi and mi = Mi

Purpose of the Index
 Minterms and Maxterms are designated with an index
 The index number corresponds to a binary pattern
 The index for the minterm or maxterm, expressed as a
binary number, is used to determine whether the variable
is shown in the true or complemented form
 For Minterms:
• ‘1’ means the variable is “Not Complemented” and
• ‘0’ means the variable is “Complemented”.
 For Maxterms:
• ‘0’ means the variable is “Not Complemented” and
• ‘1’ means the variable is “Complemented”.

Standard Order
 All variables should be present in a minterm or
maxterm and should be listed in the same order
(usually alphabetically)
 Example: For variables a, b, c:
• Maxterms (a + b + c), (a + b + c) are in standard order
• However, (b + a + c) is NOT in standard order
(a + c) does NOT contain all variables
• Minterms (a b c) and (a b c) are in standard order
• However, (b a c) is not in standard order
(a c) does not contain all variables

Sum-Of-Minterm (SOM)
 Sum-Of-Minterm (SOM) canonical form:
Sum of minterms of entries that evaluate to ‘1’
x y z F Minterm
0 0 0 0
0 0 1 1 m1 = x y z
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 1 m6 = x y z
1 1 1 1 m7 = x y z
F = m1 + m6 + m7 = ∑ (1, 6, 7) = x y z + x y z + x y z
Focus on the
‘1’ entries

+ a b c d
Sum-Of-Minterm Examples
 F(a, b, c, d) = ∑(2, 3, 6, 10, 11)
 F(a, b, c, d) = m2 + m3 + m6 + m10+ m11
 G(a, b, c, d) = ∑(0, 1, 12, 15)
 G(a, b, c, d) = m0 + m1 + m12 + m15
+ a b c da b c d + a b c d+ a b c d + a b c d
+ a b c da b c d + a b c d

Product-Of-Maxterm (POM)
 Product-Of-Maxterm (POM) canonical form:
Product of maxterms of entries that evaluate to ‘0’
x y z F Maxterm
0 0 0 1
0 0 1 1
0 1 0 0 M2 = (x + y + z)
0 1 1 1
1 0 0 0 M4 = (x + y + z)
1 0 1 1
1 1 0 0 M6 = (x + y + z)
1 1 1 1
Focus on the
‘0’ entries
F = M2·M4·M6 = ∏ (2, 4, 6) = (x+y+z) (x+y+z) (x+y+z)

 F(a, b, c, d) = ∏(1, 3, 6, 11)
 F(a, b, c, d) = M1 · M3 · M6 · M11
 G(a, b, c, d) = ∏(0, 4, 12, 15)
 G(a, b, c, d) = M0 · M4 · M12 · M15
Product-Of-Maxterm Examples
(a+b+c+d) (a+b+c+d) (a+b+c+d) (a+b+c+d)
(a+b+c+d) (a+b+c+d) (a+b+c+d) (a+b+c+d)

Observations
 We can implement any function by "ORing" the minterms
corresponding to the ‘1’ entries in the function table. A
minterm evaluates to ‘1’ for its corresponding entry.
 We can implement any function by "ANDing" the maxterms
corresponding to ‘0’ entries in the function table. A maxterm
evaluates to ‘0’ for its corresponding entry.
 The same Boolean function can be expressed in two
canonical ways: Sum-of-Minterms (SOM) and Product-of-
Maxterms (POM).
 If a Boolean function has fewer ‘1’ entries then the SOM
canonical form will contain fewer literals than POM.
However, if it has fewer ‘0’ entries then the POM form will
have fewer literals than SOM.

Converting to Sum-of-Minterms Form
 A function that is not in the Sum-of-Minterms form
can be converted to that form by means of a truth table
 Consider F = y + x z
x y z F Minterm
0 0 0 1 m0 = x y z
0 0 1 1 m1 = x y z
0 1 0 1 m2 = x y z
0 1 1 0
1 0 0 1 m4 = x y z
1 0 1 1 m5 = x y z
1 1 0 0
1 1 1 0
F = ∑(0, 1, 2, 4, 5) =
m0 + m1 + m2 + m4 + m5 =
x y z + x y z + x y z +
x y z + x y z

Converting to Product-of-Maxterms Form
 A function that is not in the Product-of-Minterms form
can be converted to that form by means of a truth table
 Consider again: F = y + x z
x y z F Minterm
0 0 0 1
0 0 1 1
0 1 0 1
0 1 1 0 M3 = (x+y+z)
1 0 0 1
1 0 1 1
1 1 0 0 M6 = (x+y+z)
1 1 1 0 M7 = (x+y+z)
F = ∏(3, 6, 7) =
M3 · M6 · M7 =
(x+y+z) (x+y+z) (x+y+z)

Conversions Between Canonical Forms
F = m1+m2+m3+m5+m7 = ∑(1, 2, 3, 5, 7) =
x y z + x y z + x y z + x y z + x y z
F = M0· M4· M6 = ∏(0, 4, 6) = (x+y+z)(x+y+z)(x+y+z)
x y z F Minterm Maxterm
0 0 0 0 M0 = (x + y + z)
0 0 1 1 m1 = x y z
0 1 0 1 m2 = x y z
0 1 1 1 m3 = x y z
1 0 0 0 M4 = (x + y + z)
1 0 1 1 m5 = x y z
1 1 0 0 M6 = (x + y + z)
1 1 1 1 m7 = x y z

Algebraic Conversion to Sum-of-Minterms
 Expand all terms first to explicitly list all minterms
 AND any term missing a variable v with (v + v)
 Example 1: f = x + x y (2 variables)
f = x (y + y) + x y
f = x y + x y + x y
f = m3 + m2 + m0 = ∑(0, 2, 3)
 Example 2: g = a + b c (3 variables)
g = a (b + b)(c + c) + (a + a) b c
g = a b c + a b c + a b c + a b c + a b c + a b c
g = a b c + a b c + a b c + a b c + a b c
g = m1 + m4 + m5 + m6 + m7 = ∑ (1, 4, 5, 6, 7)

Algebraic Conversion to Product-of-Maxterms
 Expand all terms first to explicitly list all maxterms
 OR any term missing a variable v with v · v
 Example 1: f = x + x y (2 variables)
Apply 2nd
distributive law:
f = (x + x) (x + y) = 1 · (x + y) = (x + y) = M1
 Example 2: g = a c + b c + a b (3 variables)
g = (a c + b c + a) (a c + b c + b) (distributive)
g = (c + b c + a) (a c + c + b) (x + x y = x + y)
g = (c + b + a) (a + c + b) (x + x y = x + y)
g = (a + b + c) (a + b + c) = M5 . M2 = ∏ (2, 5)

Function Complements
 The complement of a function expressed as a
sum of minterms is constructed by selecting the
minterms missing in the sum-of-minterms
canonical form
 Alternatively, the complement of a function
expressed by a Sum of Minterms form is simply
the Product of Maxterms with the same indices
 Example: Given F(x, y, z) = ∑ (1, 3, 5, 7)
F(x, y, z) = ∑ (0, 2, 4, 6)
F(x, y, z) = ∏ (1, 3, 5, 7)

Summary of Minterms and Maxterms
 There are 2n
minterms and maxterms for Boolean
functions with n variables.
 Minterms and maxterms are indexed from 0 to 2n
– 1
 Any Boolean function can be expressed as a logical
sum of minterms and as a logical product of maxterms
 The complement of a function contains those minterms
not included in the original function
 The complement of a sum-of-minterms is a product-of-
maxterms with the same indices

 Standard Sum-of-Products (SOP) form:
equations are written as an OR of AND terms
 Standard Product-of-Sums (POS) form:
equations are written as an AND of OR terms
 Examples:
• SOP:
• POS:
 These “mixed” forms are neither SOP nor POS
•
•
Standard Forms
BCBACBA ++
C·)CB(A·B)(A +++
C)(AC)B(A ++
B)(ACACBA ++

Standard Sum-of-Products (SOP)
 A sum of minterms form for n variables can
be written down directly from a truth table.
• Implementation of this form is a two-level
network of gates such that:
• The first level consists of n-input AND gates
• The second level is a single OR gate
 This form often can be simplified so that the
corresponding circuit is simpler.

 A Simplification Example:
 Writing the minterm expression:
F = A B C + A B C + A B C + ABC + ABC
 Simplifying:
F = A B C + A (B C + B C + B C + B C)
F = A B C + A (B (C + C) + B (C + C))
F = A B C + A (B + B)
F = A B C + A
F = B C + A
 Simplified F contains 3 literals compared to 15
Standard Sum-of-Products (SOP)
)7,6,5,4,1()C,B,A(F Σ=

AND/OR Two-Level Implementation
 The two implementations for F are shown below
F
B
C
A
It is quite
apparent which
is simpler!

SOP and POS Observations
 The previous examples show that:
• Canonical Forms (Sum-of-minterms, Product-of-Maxterms),
or other standard forms (SOP, POS) differ in complexity
• Boolean algebra can be used to manipulate equations into
simpler forms
• Simpler equations lead to simpler implementations
 Questions:
• How can we attain a “simplest” expression?
• Is there only one minimum cost circuit?
• The next part will deal with these issues

Terms of Use
 All (or portions) of this material © 2008 by Pearson
Education, Inc.
 Permission is given to incorporate this material or
adaptations thereof into classroom presentations and
handouts to instructors in courses adopting the latest
edition of Logic and Computer Design Fundamentals
as the course textbook.
 These materials or adaptations thereof are not to be
sold or otherwise offered for consideration.
 This Terms of Use slide or page is to be included within
the original materials or any adaptations thereof.

Boolean algebra And Logic Gates

More Related Content

What's hot (20)

Similar to Boolean algebra And Logic Gates (20)

More from Kumar (20)

Boolean algebra And Logic Gates

Editor's Notes