SlideShare a Scribd company logo
Charles Kime & Thomas Kaminski
© 2004 Pearson Education, Inc.
Terms of Use
(Hyperlinks are active in View Show mode)
Chapter 2 – Combinational
Logic Circuits
Part 1 – Gate Circuits and Boolean Equations
Logic and Computer Design Fundamentals
Chapter 2 - Part 1 2
Overview
 Part 1 – Gate Circuits and Boolean Equations
• Binary Logic and Gates
• Boolean Algebra
• Standard Forms
 Part 2 – Circuit Optimization
• Two-Level Optimization
• Map Manipulation
• Multi-Level Circuit Optimization
 Part 3 – Additional Gates and Circuits
• Other Gate Types
• Exclusive-OR Operator and Gates
• High-Impedance Outputs
Chapter 2 - Part 1 3
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 foundation for
designing and analyzing digital systems!
Chapter 2 - Part 1 4
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
Chapter 2 - Part 1 5
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.
Chapter 2 - Part 1 6
 Examples:
• is read “Y is equal to AAND 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”).
 B
A
Y 
y
x
z 

A
X 
Chapter 2 - Part 1 7
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
1
0
0
1
Chapter 2 - Part 1 8
0
1
1
0
X
NOT
X
Z
Truth Tables
 Truth table  a 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:
1
1
1
0
0
1
0
1
0
0
0
0
Z = X·Y
Y
X
AND OR
X Y Z = X+Y
0 0 0
0 1 1
1 0 1
1 1 1
Chapter 2 - Part 1 9
 Using Switches
• For inputs:
 logic 1 is switch closed
 logic 0 is switch open
• For outputs:
 logic 1 is light on
 logic 0 is light off.
• NOT uses a switch such
that:
 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
Chapter 2 - Part 1 10
 Example: Logic Using Switches
 Light is on (L = 1) for
L(A, B, C, D) =
and off (L = 0), otherwise.
 Useful model for relay circuits and for CMOS
gate circuits, the foundation of current digital
logic technology
Logic Function Implementation (Continued)
B
A
D
C
Chapter 2 - Part 1 11
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.
Chapter 2 - Part 1 12
Logic Gates (continued)
 Implementation of logic gates with transistors (See
Reading Supplement  CMOS Circuits)
 Transistor or tube implementations of logic functions are called logic gates or just
gates
 Transistor gate circuits can be modeled by switch circuits
•
F
+V
X
Y
+V
X
+V
X
Y
•
•
•
•
•
• •
•
• •
•
•
(a) NOR
G = X +Y
(b) NAND (c) NOT
X .Y
X
•
•
•
Chapter 2 - Part 1 13
Logic Gate Symbols and Behavior
 Logic gates have special symbols:
 And waveform behavior in time as follows:
(b) Timing diagram
X 0 0 1 1
Y 0 1 0 1
X · Y
(AND) 0 0 0 1
X 1 Y
(OR) 0 1 1 1
(NOT) X 1 1 0 0
(a) Graphic symbols
OR gate
X
Y
Z= X+ Y
X
Y
Z= X · Y
AND gate
X Z= X
NOT gate or
inverter
Chapter 2 - Part 1 14
Logic Diagrams and Expressions
 Boolean equations, truth tables and logic diagrams describe
the same function!
X
Y F
Z
Logic Diagram
Equation
Z
Y
X
F 

Truth Table
1
1 1 1
1
1 1 0
1
1 0 1
1
1 0 0
0
0 1 1
0
0 1 0
1
0 0 1
0
0 0 0
X Y Z Z
Y
X
F 


Chapter 2 - Part 1 15
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
 An algebraic structure defined on a set of at least two elements, B,
together with three binary operators (denoted +, · and ) that
satisfies the following basic identities:
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
Chapter 2 - Part 1 16
 The identities above are organized into pairs. These pairs
have names as follows:
1-4 Existence of 0 and 1 5-6 Idempotence
7-8 Existence of complement 9 Involution
10-11 Commutative Laws 12-13 Associative Laws
14-15 Distributive Laws 16-17 DeMorgan’s Laws
Some Properties of Identities & the 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.
Chapter 2 - Part 1 17
 Unless it happens to be self-dual, the dual of an
expression does not equal the expression itself.
 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 =
 Are any of these functions self-dual?
Some Properties of Identities & the Algebra
(Continued)
Chapter 2 - Part 1 18
 There can be more that 2 elements in B, i. e.,
elements other than 1 and 0. What are some
common useful Boolean algebras with more
than 2 elements?
1.
2.
 If B contains only 1 and 0, then B is called the
switching algebra which is the algebra we use
most often.
Some Properties of Identities & the Algebra
(Continued)
Algebra of Sets
Algebra of n-bit binary vectors
Chapter 2 - Part 1 19
Boolean Operator Precedence
 The order of evaluation in a Boolean
expression is:
1. Parentheses
2. NOT
3. AND
4. OR
 Consequence: Parentheses appear
around OR expressions
 Example: F = A(B + C)(C + D)
Chapter 2 - Part 1 20
Example 1: Boolean Algebraic Proof
 A + A·B = A (Absorption Theorem)
Proof Steps Justification (identity or theorem)
A + A·B
= A · 1 + A · B X = X · 1
= A · ( 1 + B) X · Y + X · Z = X ·(Y + Z)(Distributive Law)
= A · 1 1 + X = 1
= A 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.
Chapter 2 - Part 1 21
 AB + AC + BC = AB + AC (Consensus Theorem)
Proof Steps Justification (identity or
theorem)
AB + AC + BC
= AB + AC + 1 · BC ?
= AB +AC + (A + A) · BC ?
=
Example 2: Boolean Algebraic Proofs
Chapter 2 - Part 1 22
Example 3: Boolean Algebraic Proofs

Proof Steps Justification (identity or
theorem)
=
Y
X
Z
)
Y
X
( 

)
Z
X
(
X
Z
)
Y
X
( 


 Y Y
Chapter 2 - Part 1 23
x y

y





Useful Theorems
   n
inimizatio
M
y
y
y
x
y
y
y
x 





  tion
Simplifica
y
x
y
x
y
x
y
x 







  Absorption
x
y
x
x
x
y
x
x 





Consensus
z
y
x
z
y
z
y
x 








         
z
y
x
z
y
z
y
x 








Laws
s
DeMorgan'
x
x 


x x
x x
x x
x x
y x
 y
Chapter 2 - Part 1 24
Proof of Simplification
   y
y
y
x
y
y
y
x 




 x
 x
Chapter 2 - Part 1 25
Proof of DeMorgan’s Laws
 y
x x
 y
 y
x y
x

Chapter 2 - Part 1 26
Boolean Function Evaluation
x y z F1 F2 F3 F4
0 0 0 0 0
0 0 1 0 1
0 1 0 0 0
0 1 1 0 0
1 0 0 0 1
1 0 1 0 1
1 1 0 1 1
1 1 1 0 1
z
x
y
x
F4
x
z
y
x
z
y
x
F3
x
F2
xy
F1





 z
yz

y

Chapter 2 - Part 1 27
Expression Simplification
 An application of Boolean algebra
 Simplify to contain the smallest number
of literals (complemented and
uncomplemented variables):
Chapter 2 - Part 1 28
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 =
F = (x + y + z)(x + y + z)
 Example: Complement G = (a + bc)d + e
G =
x
 z
y
z
y
x
Chapter 2 - Part 1 29
Overview – Canonical Forms
 What are Canonical Forms?
 Minterms and Maxterms
 Index Representation of Minterms and
Maxterms
 Sum-of-Minterm (SOM) Representations
 Product-of-Maxterm (POM) Representations
 Representation of Complements of Functions
 Conversions between Representations
Chapter 2 - Part 1 30
Canonical Forms
 It is useful to specify Boolean functions in
a form that:
• Allows comparison for equality.
• Has a correspondence to the truth tables
 Canonical Forms in common usage:
• Sum of Minterms (SOM)
• Product of Maxterms (POM)
Chapter 2 - Part 1 31
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.
Y
X
XY
Y
X
Y
X
x
Chapter 2 - Part 1 32
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)
Y
X 
Y
X 
Y
X 
Y
X 
Chapter 2 - Part 1 33
 Examples: Two variable minterms and
maxterms.
 The index above is important for describing
which variables in the terms are true and which
are complemented.
Maxterms and Minterms
Index Minterm Maxterm
0 x y x + y
1 x y x + y
2 x y x + y
3 x y x + y
Chapter 2 - Part 1 34
Standard Order
 Minterms and maxterms are designated with a subscript
 The subscript is a number, corresponding to a binary pattern
 The bits in the pattern represent the complemented or
normal state of each variable listed in a standard order.
 All variables will be present in a minterm or maxterm and
will be listed in the same order (usually alphabetically)
 Example: For variables a, b, c:
• Maxterms: (a + b + c), (a + b + c)
• Terms: (b + a + c), a c b, and (c + b + a) are NOT in
standard order.
• Minterms: a b c, a b c, a b c
• Terms: (a + c), b c, and (a + b) do not contain all
variables
Chapter 2 - Part 1 35
Purpose of the Index
 The index for the minterm or maxterm,
expressed as a binary number, is used to
determine whether the variable is shown in the
true form 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”.
Chapter 2 - Part 1 36
Index Example in Three Variables
 Example: (for three variables)
 Assume the variables are called X, Y, and Z.
 The standard order is X, then Y, then Z.
 The Index 0 (base 10) = 000 (base 2) for three
variables). All three variables are complemented
for minterm 0 ( ) and no variables are
complemented for Maxterm 0 (X,Y,Z).
• Minterm 0, called m0
is .
• Maxterm 0, called M0
is (X + Y + Z).
• Minterm 6 ?
• Maxterm 6 ?
Z
,
Y
,
X
Z
Y
X
Chapter 2 - Part 1 37
Index Examples – Four Variables
Index Binary Minterm Maxterm
i Pattern mi Mi
0 0000
1 0001
3 0011
5 0101
7 0111
10 1010
13 1101
15 1111
d
c
b
a d
c
b
a 


d
c
b
a
d
c
b
a 


d
c
b
a d
c
b
a 


d
c
b
a 


d
c
b
a d
c
b
a 


d
b
a
d
c
b
a d
c
b
a 


?
?
?
?
c
Chapter 2 - Part 1 38
 Review: DeMorgan's Theorem
and
 Two-variable example:
and
Thus M2 is the complement of m2 and vice-versa.
 Since DeMorgan's Theorem holds for n variables,
the above holds for terms of n variables
 giving:
and
Thus Mi is the complement of mi.
Minterm and Maxterm Relationship
y
x
y
·
x 
 y
x
y
x 


y
x
M2

 y
x·
m2

i m
M  i i
i M
m 
Chapter 2 - Part 1 39
Function Tables for Both
 Minterms of Maxterms of
2 variables 2 variables
 Each column in the maxterm function table is the
complement of the column in the minterm function
table since Mi is the complement of mi.
x y m0 m1 m2 m3
0 0 1 0 0 0
0 1 0 1 0 0
1 0 0 0 1 0
1 1 0 0 0 1
x y M0 M1 M2 M3
0 0 0 1 1 1
0 1 1 0 1 1
1 0 1 1 0 1
1 1 1 1 1 0
Chapter 2 - Part 1 40
Observations
 In the function tables:
• Each minterm has one and only one 1 present in the 2n terms (a minimum of
1s). All other entries are 0.
• Each maxterm has one and only one 0 present in the 2n terms All other
entries are 1 (a maximum of 1s).
 We can implement any function by "ORing" the minterms
corresponding to "1" entries in the function table. These are called
the minterms of the function.
 We can implement any function by "ANDing" the maxterms
corresponding to "0" entries in the function table. These are called
the maxterms of the function.
 This gives us two canonical forms:
• Sum of Minterms (SOM)
• Product of Maxterms (POM)
for stating any Boolean function.
Chapter 2 - Part 1 41
x y z index m1 + m4 + m7 = F1
0 0 0 0 0 + 0 + 0 = 0
0 0 1 1 1 + 0 + 0 = 1
0 1 0 2 0 + 0 + 0 = 0
0 1 1 3 0 + 0 + 0 = 0
1 0 0 4 0 + 1 + 0 = 1
1 0 1 5 0 + 0 + 0 = 0
1 1 0 6 0 + 0 + 0 = 0
1 1 1 7 0 + 0 + 1 = 1
Minterm Function Example
 Example: Find F1 = m1 + m4 + m7
 F1 = x y z + x y z + x y z
Chapter 2 - Part 1 42
Minterm Function Example
 F(A, B, C, D, E) = m2 + m9 + m17 + m23
 F(A, B, C, D, E) =
Chapter 2 - Part 1 43
Maxterm Function Example
 Example: Implement F1 in maxterms:
F1 = M0 · M2 · M3 · M5 · M6
)
z
y
z)·(x
y
·(x
z)
y
(x
F1 






z)
y
x
)·(
z
y
x
·( 



x y z i M0 M2 M3 M5 M6 = F1
0 0 0 0 0 1 1 1 = 0
0 0 1 1 1 1 1 1 1 = 1
0 1 0 2 1 0 1 1 1 = 0
0 1 1 3 1 1 0 1 1 = 0
1 0 0 4 1 1 1 1 1 = 1
1 0 1 5 1 1 1 0 1 = 0
1 1 0 6 1 1 1 1 0 = 0
1 1 1 7 1







 1 1 1 1 = 1
1
























Chapter 2 - Part 1 44
Maxterm Function Example

 F(A, B,C,D) =
14
11
8
3 M
M
M
M
)
D
,
C
,
B
,
A
(
F 



Chapter 2 - Part 1 45
Canonical Sum of Minterms
 Any Boolean function can be expressed as a
Sum of Minterms.
• For the function table, the minterms used are the
terms corresponding to the 1's
• For expressions, expand all terms first to explicitly
list all minterms. Do this by “ANDing” any term
missing a variable v with a term ( ).
 Example: Implement as a sum of
minterms.
First expand terms:
Then distribute terms:
Express as sum of minterms: f = m3 + m2 + m0
y
x
x
f 

y
x
)
y
y
(
x
f 


y
x
y
x
xy
f 


v
v 
Chapter 2 - Part 1 46
Another SOM Example
 Example:
 There are three variables, A, B, and C which we
take to be the standard order.
 Expanding the terms with missing variables:
 Collect terms (removing all but one of duplicate
terms):
 Express as SOM:
C
B
A
F 

Chapter 2 - Part 1 47
Shorthand SOM Form
 From the previous example, we started with:
 We ended up with:
F = m1+m4+m5+m6+m7
 This can be denoted in the formal shorthand:
 Note that we explicitly show the standard
variables in order and drop the “m”
designators.
)
7
,
6
,
5
,
4
,
1
(
)
C
,
B
,
A
(
F m


C
B
A
F 

Chapter 2 - Part 1 48
Canonical Product of Maxterms
 Any Boolean Function can be expressed as a Product of
Maxterms (POM).
• For the function table, the maxterms used are the terms
corresponding to the 0's.
• For an expression, expand all terms first to explicitly list all
maxterms. Do this by first applying the second distributive law ,
“ORing” terms missing variable v with a term equal to and then
applying the distributive law again.
 Example: Convert to product of maxterms:
Apply the distributive law:
Add missing variable z:
Express as POM: f = M2 · M3
y
x
x
)
z
,
y
,
x
(
f 

y
x
)
y
(x
1
)
y
)(x
x
(x
y
x
x 








 
z
y
x
)
z
y
x
(
z
z
y
x 







v
v
Chapter 2 - Part 1 49
 Convert to Product of Maxterms:
 Use x + y z = (x+y)·(x+z) with ,
and to get:
 Then use to get:
and a second time to get:
 Rearrange to standard order,
to give f = M5
· M2
Another POM Example
B
A
C
B
C
A
C)
B,
f(A, 


B
z 
)
B
C
B
C
)(A
A
C
B
C
(A
f 




y
x
y
x
x 


)
B
C
C
)(A
A
BC
C
(
f 




)
B
C
)(A
A
B
C
(
f 




C)
B
)(A
C
B
A
(
f 




A
y
C),
B
(A
x 

 C
Chapter 2 - Part 1 50
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 forms.
 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 )
7
,
5
,
3
,
1
(
)
z
,
y
,
x
(
F m


)
6
,
4
,
2
,
0
(
)
z
,
y
,
x
(
F m


)
7
,
5
,
3
,
1
(
)
z
,
y
,
x
(
F M


Chapter 2 - Part 1 51
Conversion Between Forms
 To convert between sum-of-minterms and product-of-
maxterms form (or vice-versa) we follow these steps:
• Find the function complement by swapping terms in the list
with terms not in the list.
• Change from products to sums, or vice versa.
 Example:Given F as before:
 Form the Complement:
 Then use the other form with the same indices – this
forms the complement again, giving the other form of
the original function:
)
7
,
5
,
3
,
1
(
)
z
,
y
,
x
(
F m


)
6
,
4
,
2
,
0
(
)
z
,
y
,
x
(
F m


)
6
,
4
,
2
,
0
(
)
z
,
y
,
x
(
F M


Chapter 2 - Part 1 52
 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
B
C
B
A
C
B
A 

C
·
)
C
B
(A
·
B)
(A 


C)
(A
C)
B
(A 

B)
(A
C
A
C
B
A 

Chapter 2 - Part 1 53
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,
and
• The second level is a single OR gate (with
fewer than 2n
inputs).
 This form often can be simplified so that
the corresponding circuit is simpler.
Chapter 2 - Part 1 54
 A Simplification Example:

 Writing the minterm expression:
F = A B C + A B C + A B C + ABC + ABC
 Simplifying:
F =
 Simplified F contains 3 literals compared to 15 in
minterm F
Standard Sum-of-Products (SOP)
)
7
,
6
,
5
,
4
,
1
(
m
)
C
,
B
,
A
(
F 

Chapter 2 - Part 1 55
AND/OR Two-level Implementation
of SOP Expression
 The two implementations for F are shown
below – it is quite apparent which is simpler!
F
B
C
A
Chapter 2 - Part 1 56
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 two-level
implementations
 Questions:
• How can we attain a “simplest” expression?
• Is there only one minimum cost circuit?
• The next part will deal with these issues.
Chapter 2 - Part 1 57
Terms of Use
 © 2004 by Pearson Education,Inc. All rights reserved.
 The following terms of use apply in addition to the standard
Pearson Education Legal Notice.
 Permission is given to incorporate these materials into classroom
presentations and handouts only to instructors adopting Logic and
Computer Design Fundamentals as the course text.
 Permission is granted to the instructors adopting the book to post
these materials on a protected website or protected ftp site in
original or modified form. All other website or ftp postings,
including those offering the materials for a fee, are prohibited.
 You may not remove or in any way alter this Terms of Use notice
or any trademark, copyright, or other proprietary notice,
including the copyright watermark on each slide.
 Return to Title Page
Chapter 1 58
This Summary is an Online Content from this Book:
Morris Mano, DIGITAL DESIGN, 4th Edition, Prentice Hall, 2007
It is edited for
Logic Analysis and Design Course
6803213-3
by:
T.Mariah Sami Khayat
Teacher Assistant @ Adam University College
For Contacting:
mskhayat@uqu.edu.sa
Kingdom of Saudi Arabia
Ministry of Education
Umm AlQura University
Adam University College
Computer Science Department
‫السعودية‬ ‫العربية‬ ‫المملكة‬
‫التعليم‬ ‫وزارة‬
‫القرى‬ ‫أم‬ ‫جامعة‬
‫أضم‬ ‫الجامعية‬ ‫الكلية‬
‫اآللي‬ ‫الحاسب‬ ‫قسم‬

More Related Content

PPT
LCDF4_Chap_02_P1 computer engineering.ppt
PPT
IntroductiontoDigitalElectronics IntroductiontoDigitalElectronics
PPT
LCDF3_Chaaaaaaaaaaaaaaaaaaaaap_02_P1.ppt
PPT
digital logic tutorial combinational logic
PPT
LCDF3_Chap_02_P1.ppt Digital Login Design
PPTX
sop_pos(DE).pptx
PPT
2-Boolean-Algebra.ppt logic and computer design fundamental
PPT
Boolean_Algebra and digital circuits .ppt
LCDF4_Chap_02_P1 computer engineering.ppt
IntroductiontoDigitalElectronics IntroductiontoDigitalElectronics
LCDF3_Chaaaaaaaaaaaaaaaaaaaaap_02_P1.ppt
digital logic tutorial combinational logic
LCDF3_Chap_02_P1.ppt Digital Login Design
sop_pos(DE).pptx
2-Boolean-Algebra.ppt logic and computer design fundamental
Boolean_Algebra and digital circuits .ppt

Similar to combinational logic -lecture chapter 2.... (20)

PPT
102_2_digitalSystem_Chap_2_part_1.ppt
PPT
5. Combinational Circuits_P1 lecturee of dld
PPT
Boolean algebra And Logic Gates
PDF
DLD Chapter-2.pdf
PPT
new logic gates presentation basic gate universal gate.ppt
PPT
Boolean Algebra and Logic gates ( Chapter 2)
PPT
lecture_binary_logic_and_logic_gates.ppt
PDF
Lect 2 boolean algebra (4 5-21)
PPT
booleanalgebra-140914001141-phpapp01 (1).ppt
PPTX
FALLSEM2025-26_BEEE206L_TH_VL2025260101038_2025-07-09_Reference-Material-II.pptx
PPTX
Fundamentals of Digital Design_Class_12th April.pptx
PPT
Boolean Algebra
PDF
18 pc09 1.2_ digital logic gates _ boolean algebra_basic theorems
PPTX
Chapter Data Representation DIGITAL LOGIC CIRCUITS
PDF
Boolean Algebra for Computer Science Student.pdf
PPTX
EEE241DLDlect02 Digital logic and design Boolean function
PDF
Chapter 4 logic design
PDF
Boolean Algebra SOP POS_Computer Architecture.pdf
PDF
digital logic design Chapter 2 boolean_algebra_&_logic_gates
PPTX
Chapter 2.pptx
102_2_digitalSystem_Chap_2_part_1.ppt
5. Combinational Circuits_P1 lecturee of dld
Boolean algebra And Logic Gates
DLD Chapter-2.pdf
new logic gates presentation basic gate universal gate.ppt
Boolean Algebra and Logic gates ( Chapter 2)
lecture_binary_logic_and_logic_gates.ppt
Lect 2 boolean algebra (4 5-21)
booleanalgebra-140914001141-phpapp01 (1).ppt
FALLSEM2025-26_BEEE206L_TH_VL2025260101038_2025-07-09_Reference-Material-II.pptx
Fundamentals of Digital Design_Class_12th April.pptx
Boolean Algebra
18 pc09 1.2_ digital logic gates _ boolean algebra_basic theorems
Chapter Data Representation DIGITAL LOGIC CIRCUITS
Boolean Algebra for Computer Science Student.pdf
EEE241DLDlect02 Digital logic and design Boolean function
Chapter 4 logic design
Boolean Algebra SOP POS_Computer Architecture.pdf
digital logic design Chapter 2 boolean_algebra_&_logic_gates
Chapter 2.pptx
Ad

More from PradeepYata1 (17)

PPTX
EMI-Unit1-L1-YP.pptx----------------------
PPT
ch3a-binary-numbers.ppt-BINARY SYSTEM---
PPTX
EMI UNIT II Lec-6-Oscillators and Singal Generators-yp.pptx
PPTX
EMI UNIT II -PART-2-Oscillators-signaal Genarators-YP.pptx
PPT
cmos technology-micrwind layout tool-demo.ppt
PPTX
EMI-Unit1-Lecture-13 UNIT 1EMI PPT........
PPTX
EMI-Unit1-Lecture-12--std version-UNIT-1-EMI
PPTX
EMI UNIT II -Signal Analyzers-yp 30-9-21.pptx
PPTX
EMI UNIT II -PART-2-Oscillators-Signal Genarators-YP.pptx
PPTX
VLSID unit-2 Scaling of mosfets -yp.pptx
PPT
unit-1 vlsi-CMOS INVERTER ANALYSIS & DESIGN.ppt
PPTX
Zup/Zpd Ratio Determination -vlsi design.
PPTX
NMOS inverter Design and analysis unit 1...
PPT
Bicmos inverter-and latchup problem-u1 ppt
PPT
cmos design and analysis-UNIT1............
PPTX
number system _ppt unit1.................
PPT
DIGITAL ELECTRONICS FOR IOT CSE STUDENTS
EMI-Unit1-L1-YP.pptx----------------------
ch3a-binary-numbers.ppt-BINARY SYSTEM---
EMI UNIT II Lec-6-Oscillators and Singal Generators-yp.pptx
EMI UNIT II -PART-2-Oscillators-signaal Genarators-YP.pptx
cmos technology-micrwind layout tool-demo.ppt
EMI-Unit1-Lecture-13 UNIT 1EMI PPT........
EMI-Unit1-Lecture-12--std version-UNIT-1-EMI
EMI UNIT II -Signal Analyzers-yp 30-9-21.pptx
EMI UNIT II -PART-2-Oscillators-Signal Genarators-YP.pptx
VLSID unit-2 Scaling of mosfets -yp.pptx
unit-1 vlsi-CMOS INVERTER ANALYSIS & DESIGN.ppt
Zup/Zpd Ratio Determination -vlsi design.
NMOS inverter Design and analysis unit 1...
Bicmos inverter-and latchup problem-u1 ppt
cmos design and analysis-UNIT1............
number system _ppt unit1.................
DIGITAL ELECTRONICS FOR IOT CSE STUDENTS
Ad

Recently uploaded (20)

PPTX
Infosys Presentation by1.Riyan Bagwan 2.Samadhan Naiknavare 3.Gaurav Shinde 4...
PDF
Embodied AI: Ushering in the Next Era of Intelligent Systems
PPTX
Sustainable Sites - Green Building Construction
PPTX
UNIT 4 Total Quality Management .pptx
PDF
Enhancing Cyber Defense Against Zero-Day Attacks using Ensemble Neural Networks
PPTX
bas. eng. economics group 4 presentation 1.pptx
PDF
Mohammad Mahdi Farshadian CV - Prospective PhD Student 2026
PPTX
KTU 2019 -S7-MCN 401 MODULE 2-VINAY.pptx
PDF
PPT on Performance Review to get promotions
PPTX
MCN 401 KTU-2019-PPE KITS-MODULE 2.pptx
DOCX
ASol_English-Language-Literature-Set-1-27-02-2023-converted.docx
PDF
keyrequirementskkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
PPTX
Construction Project Organization Group 2.pptx
PPT
CRASH COURSE IN ALTERNATIVE PLUMBING CLASS
PPTX
MET 305 2019 SCHEME MODULE 2 COMPLETE.pptx
PDF
The CXO Playbook 2025 – Future-Ready Strategies for C-Suite Leaders Cerebrai...
PPT
Project quality management in manufacturing
PPTX
web development for engineering and engineering
PDF
TFEC-4-2020-Design-Guide-for-Timber-Roof-Trusses.pdf
PPTX
UNIT-1 - COAL BASED THERMAL POWER PLANTS
Infosys Presentation by1.Riyan Bagwan 2.Samadhan Naiknavare 3.Gaurav Shinde 4...
Embodied AI: Ushering in the Next Era of Intelligent Systems
Sustainable Sites - Green Building Construction
UNIT 4 Total Quality Management .pptx
Enhancing Cyber Defense Against Zero-Day Attacks using Ensemble Neural Networks
bas. eng. economics group 4 presentation 1.pptx
Mohammad Mahdi Farshadian CV - Prospective PhD Student 2026
KTU 2019 -S7-MCN 401 MODULE 2-VINAY.pptx
PPT on Performance Review to get promotions
MCN 401 KTU-2019-PPE KITS-MODULE 2.pptx
ASol_English-Language-Literature-Set-1-27-02-2023-converted.docx
keyrequirementskkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
Construction Project Organization Group 2.pptx
CRASH COURSE IN ALTERNATIVE PLUMBING CLASS
MET 305 2019 SCHEME MODULE 2 COMPLETE.pptx
The CXO Playbook 2025 – Future-Ready Strategies for C-Suite Leaders Cerebrai...
Project quality management in manufacturing
web development for engineering and engineering
TFEC-4-2020-Design-Guide-for-Timber-Roof-Trusses.pdf
UNIT-1 - COAL BASED THERMAL POWER PLANTS

combinational logic -lecture chapter 2....

  • 1. Charles Kime & Thomas Kaminski © 2004 Pearson Education, Inc. Terms of Use (Hyperlinks are active in View Show mode) Chapter 2 – Combinational Logic Circuits Part 1 – Gate Circuits and Boolean Equations Logic and Computer Design Fundamentals
  • 2. Chapter 2 - Part 1 2 Overview  Part 1 – Gate Circuits and Boolean Equations • Binary Logic and Gates • Boolean Algebra • Standard Forms  Part 2 – Circuit Optimization • Two-Level Optimization • Map Manipulation • Multi-Level Circuit Optimization  Part 3 – Additional Gates and Circuits • Other Gate Types • Exclusive-OR Operator and Gates • High-Impedance Outputs
  • 3. Chapter 2 - Part 1 3 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 foundation for designing and analyzing digital systems!
  • 4. Chapter 2 - Part 1 4 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
  • 5. Chapter 2 - Part 1 5 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.
  • 6. Chapter 2 - Part 1 6  Examples: • is read “Y is equal to AAND 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”).  B A Y  y x z   A X 
  • 7. Chapter 2 - Part 1 7 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 1 0 0 1
  • 8. Chapter 2 - Part 1 8 0 1 1 0 X NOT X Z Truth Tables  Truth table  a 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: 1 1 1 0 0 1 0 1 0 0 0 0 Z = X·Y Y X AND OR X Y Z = X+Y 0 0 0 0 1 1 1 0 1 1 1 1
  • 9. Chapter 2 - Part 1 9  Using Switches • For inputs:  logic 1 is switch closed  logic 0 is switch open • For outputs:  logic 1 is light on  logic 0 is light off. • NOT uses a switch such that:  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
  • 10. Chapter 2 - Part 1 10  Example: Logic Using Switches  Light is on (L = 1) for L(A, B, C, D) = and off (L = 0), otherwise.  Useful model for relay circuits and for CMOS gate circuits, the foundation of current digital logic technology Logic Function Implementation (Continued) B A D C
  • 11. Chapter 2 - Part 1 11 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.
  • 12. Chapter 2 - Part 1 12 Logic Gates (continued)  Implementation of logic gates with transistors (See Reading Supplement  CMOS Circuits)  Transistor or tube implementations of logic functions are called logic gates or just gates  Transistor gate circuits can be modeled by switch circuits • F +V X Y +V X +V X Y • • • • • • • • • • • • (a) NOR G = X +Y (b) NAND (c) NOT X .Y X • • •
  • 13. Chapter 2 - Part 1 13 Logic Gate Symbols and Behavior  Logic gates have special symbols:  And waveform behavior in time as follows: (b) Timing diagram X 0 0 1 1 Y 0 1 0 1 X · Y (AND) 0 0 0 1 X 1 Y (OR) 0 1 1 1 (NOT) X 1 1 0 0 (a) Graphic symbols OR gate X Y Z= X+ Y X Y Z= X · Y AND gate X Z= X NOT gate or inverter
  • 14. Chapter 2 - Part 1 14 Logic Diagrams and Expressions  Boolean equations, truth tables and logic diagrams describe the same function! X Y F Z Logic Diagram Equation Z Y X F   Truth Table 1 1 1 1 1 1 1 0 1 1 0 1 1 1 0 0 0 0 1 1 0 0 1 0 1 0 0 1 0 0 0 0 X Y Z Z Y X F   
  • 15. Chapter 2 - Part 1 15 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  An algebraic structure defined on a set of at least two elements, B, together with three binary operators (denoted +, · and ) that satisfies the following basic identities: 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
  • 16. Chapter 2 - Part 1 16  The identities above are organized into pairs. These pairs have names as follows: 1-4 Existence of 0 and 1 5-6 Idempotence 7-8 Existence of complement 9 Involution 10-11 Commutative Laws 12-13 Associative Laws 14-15 Distributive Laws 16-17 DeMorgan’s Laws Some Properties of Identities & the 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.
  • 17. Chapter 2 - Part 1 17  Unless it happens to be self-dual, the dual of an expression does not equal the expression itself.  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 =  Are any of these functions self-dual? Some Properties of Identities & the Algebra (Continued)
  • 18. Chapter 2 - Part 1 18  There can be more that 2 elements in B, i. e., elements other than 1 and 0. What are some common useful Boolean algebras with more than 2 elements? 1. 2.  If B contains only 1 and 0, then B is called the switching algebra which is the algebra we use most often. Some Properties of Identities & the Algebra (Continued) Algebra of Sets Algebra of n-bit binary vectors
  • 19. Chapter 2 - Part 1 19 Boolean Operator Precedence  The order of evaluation in a Boolean expression is: 1. Parentheses 2. NOT 3. AND 4. OR  Consequence: Parentheses appear around OR expressions  Example: F = A(B + C)(C + D)
  • 20. Chapter 2 - Part 1 20 Example 1: Boolean Algebraic Proof  A + A·B = A (Absorption Theorem) Proof Steps Justification (identity or theorem) A + A·B = A · 1 + A · B X = X · 1 = A · ( 1 + B) X · Y + X · Z = X ·(Y + Z)(Distributive Law) = A · 1 1 + X = 1 = A 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.
  • 21. Chapter 2 - Part 1 21  AB + AC + BC = AB + AC (Consensus Theorem) Proof Steps Justification (identity or theorem) AB + AC + BC = AB + AC + 1 · BC ? = AB +AC + (A + A) · BC ? = Example 2: Boolean Algebraic Proofs
  • 22. Chapter 2 - Part 1 22 Example 3: Boolean Algebraic Proofs  Proof Steps Justification (identity or theorem) = Y X Z ) Y X (   ) Z X ( X Z ) Y X (     Y Y
  • 23. Chapter 2 - Part 1 23 x y  y      Useful Theorems    n inimizatio M y y y x y y y x         tion Simplifica y x y x y x y x           Absorption x y x x x y x x       Consensus z y x z y z y x                    z y x z y z y x          Laws s DeMorgan' x x    x x x x x x x x y x  y
  • 24. Chapter 2 - Part 1 24 Proof of Simplification    y y y x y y y x       x  x
  • 25. Chapter 2 - Part 1 25 Proof of DeMorgan’s Laws  y x x  y  y x y x 
  • 26. Chapter 2 - Part 1 26 Boolean Function Evaluation x y z F1 F2 F3 F4 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 0 1 1 0 1 0 1 1 1 0 1 1 1 1 1 0 1 z x y x F4 x z y x z y x F3 x F2 xy F1       z yz  y 
  • 27. Chapter 2 - Part 1 27 Expression Simplification  An application of Boolean algebra  Simplify to contain the smallest number of literals (complemented and uncomplemented variables):
  • 28. Chapter 2 - Part 1 28 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 = F = (x + y + z)(x + y + z)  Example: Complement G = (a + bc)d + e G = x  z y z y x
  • 29. Chapter 2 - Part 1 29 Overview – Canonical Forms  What are Canonical Forms?  Minterms and Maxterms  Index Representation of Minterms and Maxterms  Sum-of-Minterm (SOM) Representations  Product-of-Maxterm (POM) Representations  Representation of Complements of Functions  Conversions between Representations
  • 30. Chapter 2 - Part 1 30 Canonical Forms  It is useful to specify Boolean functions in a form that: • Allows comparison for equality. • Has a correspondence to the truth tables  Canonical Forms in common usage: • Sum of Minterms (SOM) • Product of Maxterms (POM)
  • 31. Chapter 2 - Part 1 31 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. Y X XY Y X Y X x
  • 32. Chapter 2 - Part 1 32 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) Y X  Y X  Y X  Y X 
  • 33. Chapter 2 - Part 1 33  Examples: Two variable minterms and maxterms.  The index above is important for describing which variables in the terms are true and which are complemented. Maxterms and Minterms Index Minterm Maxterm 0 x y x + y 1 x y x + y 2 x y x + y 3 x y x + y
  • 34. Chapter 2 - Part 1 34 Standard Order  Minterms and maxterms are designated with a subscript  The subscript is a number, corresponding to a binary pattern  The bits in the pattern represent the complemented or normal state of each variable listed in a standard order.  All variables will be present in a minterm or maxterm and will be listed in the same order (usually alphabetically)  Example: For variables a, b, c: • Maxterms: (a + b + c), (a + b + c) • Terms: (b + a + c), a c b, and (c + b + a) are NOT in standard order. • Minterms: a b c, a b c, a b c • Terms: (a + c), b c, and (a + b) do not contain all variables
  • 35. Chapter 2 - Part 1 35 Purpose of the Index  The index for the minterm or maxterm, expressed as a binary number, is used to determine whether the variable is shown in the true form 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”.
  • 36. Chapter 2 - Part 1 36 Index Example in Three Variables  Example: (for three variables)  Assume the variables are called X, Y, and Z.  The standard order is X, then Y, then Z.  The Index 0 (base 10) = 000 (base 2) for three variables). All three variables are complemented for minterm 0 ( ) and no variables are complemented for Maxterm 0 (X,Y,Z). • Minterm 0, called m0 is . • Maxterm 0, called M0 is (X + Y + Z). • Minterm 6 ? • Maxterm 6 ? Z , Y , X Z Y X
  • 37. Chapter 2 - Part 1 37 Index Examples – Four Variables Index Binary Minterm Maxterm i Pattern mi Mi 0 0000 1 0001 3 0011 5 0101 7 0111 10 1010 13 1101 15 1111 d c b a d c b a    d c b a d c b a    d c b a d c b a    d c b a    d c b a d c b a    d b a d c b a d c b a    ? ? ? ? c
  • 38. Chapter 2 - Part 1 38  Review: DeMorgan's Theorem and  Two-variable example: and Thus M2 is the complement of m2 and vice-versa.  Since DeMorgan's Theorem holds for n variables, the above holds for terms of n variables  giving: and Thus Mi is the complement of mi. Minterm and Maxterm Relationship y x y · x   y x y x    y x M2   y x· m2  i m M  i i i M m 
  • 39. Chapter 2 - Part 1 39 Function Tables for Both  Minterms of Maxterms of 2 variables 2 variables  Each column in the maxterm function table is the complement of the column in the minterm function table since Mi is the complement of mi. x y m0 m1 m2 m3 0 0 1 0 0 0 0 1 0 1 0 0 1 0 0 0 1 0 1 1 0 0 0 1 x y M0 M1 M2 M3 0 0 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 1 0
  • 40. Chapter 2 - Part 1 40 Observations  In the function tables: • Each minterm has one and only one 1 present in the 2n terms (a minimum of 1s). All other entries are 0. • Each maxterm has one and only one 0 present in the 2n terms All other entries are 1 (a maximum of 1s).  We can implement any function by "ORing" the minterms corresponding to "1" entries in the function table. These are called the minterms of the function.  We can implement any function by "ANDing" the maxterms corresponding to "0" entries in the function table. These are called the maxterms of the function.  This gives us two canonical forms: • Sum of Minterms (SOM) • Product of Maxterms (POM) for stating any Boolean function.
  • 41. Chapter 2 - Part 1 41 x y z index m1 + m4 + m7 = F1 0 0 0 0 0 + 0 + 0 = 0 0 0 1 1 1 + 0 + 0 = 1 0 1 0 2 0 + 0 + 0 = 0 0 1 1 3 0 + 0 + 0 = 0 1 0 0 4 0 + 1 + 0 = 1 1 0 1 5 0 + 0 + 0 = 0 1 1 0 6 0 + 0 + 0 = 0 1 1 1 7 0 + 0 + 1 = 1 Minterm Function Example  Example: Find F1 = m1 + m4 + m7  F1 = x y z + x y z + x y z
  • 42. Chapter 2 - Part 1 42 Minterm Function Example  F(A, B, C, D, E) = m2 + m9 + m17 + m23  F(A, B, C, D, E) =
  • 43. Chapter 2 - Part 1 43 Maxterm Function Example  Example: Implement F1 in maxterms: F1 = M0 · M2 · M3 · M5 · M6 ) z y z)·(x y ·(x z) y (x F1        z) y x )·( z y x ·(     x y z i M0 M2 M3 M5 M6 = F1 0 0 0 0 0 1 1 1 = 0 0 0 1 1 1 1 1 1 1 = 1 0 1 0 2 1 0 1 1 1 = 0 0 1 1 3 1 1 0 1 1 = 0 1 0 0 4 1 1 1 1 1 = 1 1 0 1 5 1 1 1 0 1 = 0 1 1 0 6 1 1 1 1 0 = 0 1 1 1 7 1         1 1 1 1 = 1 1                        
  • 44. Chapter 2 - Part 1 44 Maxterm Function Example   F(A, B,C,D) = 14 11 8 3 M M M M ) D , C , B , A ( F    
  • 45. Chapter 2 - Part 1 45 Canonical Sum of Minterms  Any Boolean function can be expressed as a Sum of Minterms. • For the function table, the minterms used are the terms corresponding to the 1's • For expressions, expand all terms first to explicitly list all minterms. Do this by “ANDing” any term missing a variable v with a term ( ).  Example: Implement as a sum of minterms. First expand terms: Then distribute terms: Express as sum of minterms: f = m3 + m2 + m0 y x x f   y x ) y y ( x f    y x y x xy f    v v 
  • 46. Chapter 2 - Part 1 46 Another SOM Example  Example:  There are three variables, A, B, and C which we take to be the standard order.  Expanding the terms with missing variables:  Collect terms (removing all but one of duplicate terms):  Express as SOM: C B A F  
  • 47. Chapter 2 - Part 1 47 Shorthand SOM Form  From the previous example, we started with:  We ended up with: F = m1+m4+m5+m6+m7  This can be denoted in the formal shorthand:  Note that we explicitly show the standard variables in order and drop the “m” designators. ) 7 , 6 , 5 , 4 , 1 ( ) C , B , A ( F m   C B A F  
  • 48. Chapter 2 - Part 1 48 Canonical Product of Maxterms  Any Boolean Function can be expressed as a Product of Maxterms (POM). • For the function table, the maxterms used are the terms corresponding to the 0's. • For an expression, expand all terms first to explicitly list all maxterms. Do this by first applying the second distributive law , “ORing” terms missing variable v with a term equal to and then applying the distributive law again.  Example: Convert to product of maxterms: Apply the distributive law: Add missing variable z: Express as POM: f = M2 · M3 y x x ) z , y , x ( f   y x ) y (x 1 ) y )(x x (x y x x            z y x ) z y x ( z z y x         v v
  • 49. Chapter 2 - Part 1 49  Convert to Product of Maxterms:  Use x + y z = (x+y)·(x+z) with , and to get:  Then use to get: and a second time to get:  Rearrange to standard order, to give f = M5 · M2 Another POM Example B A C B C A C) B, f(A,    B z  ) B C B C )(A A C B C (A f      y x y x x    ) B C C )(A A BC C ( f      ) B C )(A A B C ( f      C) B )(A C B A ( f      A y C), B (A x    C
  • 50. Chapter 2 - Part 1 50 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 forms.  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 ) 7 , 5 , 3 , 1 ( ) z , y , x ( F m   ) 6 , 4 , 2 , 0 ( ) z , y , x ( F m   ) 7 , 5 , 3 , 1 ( ) z , y , x ( F M  
  • 51. Chapter 2 - Part 1 51 Conversion Between Forms  To convert between sum-of-minterms and product-of- maxterms form (or vice-versa) we follow these steps: • Find the function complement by swapping terms in the list with terms not in the list. • Change from products to sums, or vice versa.  Example:Given F as before:  Form the Complement:  Then use the other form with the same indices – this forms the complement again, giving the other form of the original function: ) 7 , 5 , 3 , 1 ( ) z , y , x ( F m   ) 6 , 4 , 2 , 0 ( ) z , y , x ( F m   ) 6 , 4 , 2 , 0 ( ) z , y , x ( F M  
  • 52. Chapter 2 - Part 1 52  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 B C B A C B A   C · ) C B (A · B) (A    C) (A C) B (A   B) (A C A C B A  
  • 53. Chapter 2 - Part 1 53 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, and • The second level is a single OR gate (with fewer than 2n inputs).  This form often can be simplified so that the corresponding circuit is simpler.
  • 54. Chapter 2 - Part 1 54  A Simplification Example:   Writing the minterm expression: F = A B C + A B C + A B C + ABC + ABC  Simplifying: F =  Simplified F contains 3 literals compared to 15 in minterm F Standard Sum-of-Products (SOP) ) 7 , 6 , 5 , 4 , 1 ( m ) C , B , A ( F  
  • 55. Chapter 2 - Part 1 55 AND/OR Two-level Implementation of SOP Expression  The two implementations for F are shown below – it is quite apparent which is simpler! F B C A
  • 56. Chapter 2 - Part 1 56 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 two-level implementations  Questions: • How can we attain a “simplest” expression? • Is there only one minimum cost circuit? • The next part will deal with these issues.
  • 57. Chapter 2 - Part 1 57 Terms of Use  © 2004 by Pearson Education,Inc. All rights reserved.  The following terms of use apply in addition to the standard Pearson Education Legal Notice.  Permission is given to incorporate these materials into classroom presentations and handouts only to instructors adopting Logic and Computer Design Fundamentals as the course text.  Permission is granted to the instructors adopting the book to post these materials on a protected website or protected ftp site in original or modified form. All other website or ftp postings, including those offering the materials for a fee, are prohibited.  You may not remove or in any way alter this Terms of Use notice or any trademark, copyright, or other proprietary notice, including the copyright watermark on each slide.  Return to Title Page
  • 58. Chapter 1 58 This Summary is an Online Content from this Book: Morris Mano, DIGITAL DESIGN, 4th Edition, Prentice Hall, 2007 It is edited for Logic Analysis and Design Course 6803213-3 by: T.Mariah Sami Khayat Teacher Assistant @ Adam University College For Contacting: mskhayat@uqu.edu.sa Kingdom of Saudi Arabia Ministry of Education Umm AlQura University Adam University College Computer Science Department ‫السعودية‬ ‫العربية‬ ‫المملكة‬ ‫التعليم‬ ‫وزارة‬ ‫القرى‬ ‫أم‬ ‫جامعة‬ ‫أضم‬ ‫الجامعية‬ ‫الكلية‬ ‫اآللي‬ ‫الحاسب‬ ‫قسم‬

Editor's Notes

  • #10: L (A, B, C, D) = A ((B C') + D) = A B C' + A D Complementary metal–oxide–semiconductor = CMOS =  is a technology for constructing integrated circuits. أكسيد معدني متمم أشباه الموصلات
  • #11: energizing = making sth active. Coils = wires. vacuum tubes = أنابيب مفرغة
  • #12: The transistor without the “bubble” on its input is an N-type field effect transistor. It acts like a closed switch between its top and bottom terminals with an H (1) applied to its input on its left. It acts like an open switch with an L (0) applied to its input. The transistor with the “bubble” on its input is a P-type field effect transistor. The +V at the top provides an H (1) and the Ground symbol at the bottom provides an L (0). By modeling the two types of field effect transistors as switches, one can see how the series and parallel interconnections can produce 1’s and 0’s on the outputs on the right in response to applied 1’s and 0’s on the inputs on the left. NOR and NAND are OR and AND, each followed by a NOT respectively.
  • #17: Dual G = ((X+Y) · (W · Z)') = ((X+Y) ·(W' + Z') Dual H = (A + B)(A + C)(B + C). Using the Boolean identities, = (A +BC) (B+C) = AB + AC + BC. So H is self-dual. -Distributive Identity then AB + AC + BBC + BCC = //X.X = X AB + AC + BC + BC = //X+X = X AB + AC + BC
  • #18: Algebra of Sets correspondence to switching algebra: Set - variable, Union - OR, Intersect AND ., Universe - 1, Empty set - 0, Complement of Set - NOT, Subset – AND of variables and variable complements. Algebra of n-bit binary vectors correspondence to switching algebra: {n-bit binary vectors} – {0,1}, bitwise OR - OR, bitwise AND - AND, bitwise NOT - NOT, Vector of all 1s – 1, Vector of all 0s – 0.
  • #19: Consequence= result = outcome
  • #20: Absorption = نظرية الامتصاص Justification مبرر أو مسوغ = Identities = المسلمات
  • #21: Justification 1: 1 . X = X Justification 2: X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X إجماع = Consensus
  • #22: = X’ Y’ Z + X Y’ (A + B)’ = A’ . B’ (DeMorgan’s Law) = Y’ X’ Z + Y’ X A . B = B . A (Commutative Law) = Y’ (X’ Z + X) A(B + C) = AB + AC (Distributive Law) = Y’ (X’ + X)(Z + X) A + BC = (A + B)(A + C) (Distributive Law) = Y’ . 1 . (Z + X) A + A’ = 1 = Y’ (X + Z) 1 . A = A, A + B = B + A (Commutative Law)
  • #24: x . y + x’ . y = (x + x’) . y X(Y + Z) = XY + XZ (Distributive Law) = 1 . y X + X’ = 1 = y X . 1 = X x + y . x’ + y X+(YZ) = XY + XZ (Distributive Law) = (x . x’) + y X . X’ = 0 = 0 + y X + 0 = X = y
  • #25: It is important that we do not USE DeMorgan’s Laws in doing this proof. This requires a different proof method. We will show that, x’ . y’, satisfies the definition of the complement of (x + y), defined as (x + y)’ by DeMorgan’s Law. To show this we need to show that A + A’ = 1 and A.A’ = 0 with A = x + y and A’ = x’. y’. This proves that x’. y’ = (x + y)’. (if x+y is a complement of x’.y’, then (x+y)’ is equal to x’.y’ ) That’s the idea ;) Part 1: Show x + y + x’. y’ = 1. x + y + x’. y’ = (x + y + x’) (x + y + y’) X + YZ = (X + Y)(X + Z) (Distributive Law) = (x + x’ + y) (x + y + y’) X + Y = Y + X (Commutative Law) = (1 + y)(x + 1) X + X’ = 1 = 1 . 1 1 + X = 1 = 1 1 . X = 1 Part 2: Show (x + y) . x’. y’ = 0. (x + y) . x’. y’ = (x . x’. y’ + y . x’. y’) X (Y + Z) = XY + XZ (Distributive Law) = (x . x’. y’ + y . y’ . x’) XY = YX (Commutative Law) = (0 . y’ + 0 . x’) X . X’ = 0 = (0 + 0) 0 . X = 0 = 0 X + 0 = X (With X = 0) Based on the above two parts, x’y’ = (x + y)’ The second DeMorgans’ law is proved by duality. Note that DeMorgan’s Law, given as an identity is not an axiom in the sense that it can be proved using the other identities.
  • #26: F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0
  • #28: G’ = (a.b’+c’) + d.e’
  • #36: m6 = X Y Z’ M6 = (X’ + Y’ + Z)
  • #37: M1 = a + b + c + d’ m3 = a’ b’ c d m7 = a’ b c d M 13 = a’ + b’ + c + d’
  • #42: F(A,B,C,D,E) = A’B’C’DE’ + A’BC’D’E + AB’C’D’E + AB’CDE
  • #44: F = (A + B + C’ + D’) (A’ + B + C + D) (A’ + B + C’ + D’) (A’ + B’ + C’ + D)
  • #46: F = A(B + B’)(C + C’) + (A + A’) B’ C = ABC + ABC’ + AB’C + AB’C’ + AB’C + A’B’C //AB’C is repeated and X+X = X = ABC + ABC’ + AB’C + AB’C’ + A’B’C = m7 + m6 + m5 + m4 + m1 = m1 + m4 + m5 + m6 + m7
  • #49: x + x’y = (x+x’)(x+y) = 1 (x+y) = x+y A’+AC’ = A’+C’ and the like.
  • #50: Π=𝑃𝑖 (here Product) Σ=𝑆𝑖𝑔𝑚𝑎
  • #51: Indices = the plural of index.
  • #54: F = A’ B’ C + A (B’ C’ + B C’ + B’ C + B C) = A’ B’ C + A (B’ + B) (C’ + C) = A’ B’ C + A.1.1 = A’ B’ C + A //x’y + y = x + y = B’C + A