2-Boolean-Algebra.ppt logic and computer design fundamental

Charles Kime & Thomas Kaminski
© 2004 Pearson Education, Inc.
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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

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!

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 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 

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

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

 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

 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

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.

(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 5 X 1 Y
X
Y
Z 5 X · Y
AND gate
X Z 5 X
Logic Gate Symbols and Behavior
 Logic gates have special symbols:
 And waveform behavior in time as follows:

Logic Diagrams and Expressions
 Boolean equations, truth tables and logic diagrams describe the same function!
 Truth tables are unique; expressions and logic diagrams are not. This gives
flexibility in implementing functions.
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 



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

 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
 If the meaning is unambiguous, we leave out the symbol “·”
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.

 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)

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)

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.

 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

Example 3: Boolean Algebraic Proofs

Proof Steps Justification (identity or
theorem)
=
Y
X
Z
)
Y
X
( 

)
Z
X
(
X
Z
)
Y
X
( 


 Y Y

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

Proof of Simplification
   y
y
y
x
y
y
y
x 




 x
 x

Proof of DeMorgan’s Laws
 y
x x
 y
 y
x y
x


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


Expression Simplification
 An application of Boolean algebra
 Simplify to contain the smallest number
of literals (complemented and
uncomplemented variables):
= 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 5 literals



 D
C
B
A
D
C
A
D
B
A
D
C
A
B
A

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

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

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)

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

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 

 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

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

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”.

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

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

 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 

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

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.

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

Minterm Function Example
 F(A, B, C, D, E) = m2 + m9 + m17 + m23
 F(A, B, C, D, E) =

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

























Maxterm Function Example

 F(A, B,C,D) =
14
11
8
3 M
M
M
M
)
D
,
C
,
B
,
A
(
F 




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 

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 


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 


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

 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

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



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



 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 


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.

 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 


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

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.

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