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Boolean Algebra

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Logic and Computer Design Fundamentals
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)
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.
Notation Examples
Examples:
• Y  A  B is read “Y is equal to A AND B.”
• z  x  y is read “z is equal to x OR y.”
• X  A is read “X is equal to NOT A.”
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”).
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
NOT
0+0=0
0+1=1
1+0=1
1+1=1
0 1
1 0
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:
X
0
0
1
1
AND
Y Z = X·Y
0
0
1
0
0
0
1
1
X
0
0
1
1
Y
0
1
0
1
OR
Z = X+Y
0
1
1
1
NOT
X
0
1
Z X
1
0
Truth Tables – Cont’d
Used to evaluate any logic function
Consider F(X, Y, Z) = X Y + Y Z
X
Y
Z
XY
Y
YZ
F = X Y + YZ
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
0
0
0
0
0
1
1
1
1
0
0
1
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
1
1
Logic Function Implementation
Using Switches
Switches in parallel => OR
• Inputs:
logic 1 is switch closed
logic 0 is switch open
• Outputs:
Switches in series => AND
logic 1 is light on
logic 0 is light off.
• NOT input:
Normally-closed switch => NOT
C
logic 1 is switch open
logic 0 is switch closed
Logic Function Implementation – cont’d
Example: Logic Using Switches
B
C
A
D
Light is on (L = 1) for
L(A, B, C, D) = A (B C + D) = A B C + A D
and off (L = 0), otherwise.
Useful model for relay and CMOS gate circuits,
the foundation of current digital logic circuits
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:
X
X
Z= X·Y
Y
Z = X+ Y
Y
AND gate
NOT gateor
inverter
OR gate
And waveform behavior in time as follows:
X
0
0
1
1
Y
0
1
0
1
X ·Y
0
0
0
1
(OR)
X+ Y
0
1
1
1
(NOT)
X
1
1
0
0
(AND)
Z= X
X
Logic Diagrams and Expressions
Truth Table
Logic Equation
XYZ
F  X  YZ
000
0
001
1
010
0
011
0
100
1
101
1
Y
110
1
Z
111
1
F  X Y Z
Logic Diagram
X
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.
F
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:
1
Input 0
Output1
tG
tG
tG = 0.3 ns
0
0
0.5
1
1.5
Time (ns)
Boolean Algebra
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 ( )
1.
3.
5.
7.
9.
10.
12.
14.
16.
X+0= X
Identity element
X .1 = X
X + X =X
2.
4.
6.
X .X =X
Idempotence
X+X=1
8.
X .X = 0
Complement
X+1=1
X= X
X+Y=Y+X
(X + Y) + Z = X + (Y + Z)
X(Y + Z) = XY + XZ
X + Y = X .Y
X .0 = 0
Involution
11. XY = YX
Commutative
Associative
13. (XY) Z = X(YZ)
Distributive
15. X + YZ = (X + Y)(X + Z)
DeMorgan’s
17. X . Y = X + Y
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
Dual of a Boolean Expression
Example: F = (A + C) · B + 0
dual F = (A · C + B) · 1 = A · C + B
Example: G = X · Y + (W + Z)
dual G = (X+Y) · (W · Z) = (X+Y) · (W+Z)
Example: H = A · B + A · C + B · C
dual H = (A+B) · (A+C) · (B+C)
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? H is self-dual
(A+B)(A+C)(B+C)=(A+BC)(B+C)=AB+AC+BC
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
Proof Steps
A + A ·B
= A · 1 + A ·B
= A · ( 1 + B)
=A ·1
=A
(Absorption Theorem)
Justification
Identity element: A · 1 = A
Distributive
1+B=1
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.
Boolean Algebraic Proof – Example 2
AB + AC + BC = AB + AC (Consensus Theorem)
Proof Steps
Justification
AB + AC + BC
Identity element
= AB + AC + 1 · BC
Complement
= AB + AC + (A + A) · BC
Distributive
= AB + AC + ABC + ABC
Commutative
= AB + ABC + AC + ACB
= AB · 1 + ABC + AC · 1 + ACB Identity element
Distributive
= AB (1+C) + AC (1 + B)
1+X = 1
= AB . 1 + AC . 1
Identity element
= AB + AC
Useful Theorems
Minimization
XY+ XY=Y
Minimization (dual)
(X+Y)(X+Y) = Y
Absorption
X+XY=X
Absorption (dual)
X · (X + Y) = X
Simplification
X +XY= X+Y
Simplification (dual)
X · (X + Y) = X ·Y
DeMorgans’
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 X+Y X · Y X·Y X+Y
0
0
1
1
0
1
0
1
0
0
0
1
0
1
1
1
1
1
0
0
1
0
1
0
1
0
0
0
1
0
0
0
1
1
1
0
Generalized DeMorgan’s Theorem:
X1+ X2+ … + Xn= X1 · X2 · … ·Xn
X1· X2 · … · Xn = X1+ X2+ … + Xn
1
1
1
0
Complementing Functions
Use DeMorgan's Theorem:
1. Interchange AND and OR operators
2. Complement each constant and literal
Example: Complement F = xyz
 xyz
F = (x + y + z)(x + y + z)
Example: Complement G = (a + bc)d + e
G = (a (b + c) + d) e
Expression Simplification
An application of Boolean algebra
Simplify to contain the smallest number
of literals (variables that may or may not
be complemented)
AB  ACD  ABD  AC D  ABCD
= 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)
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., x), there
are 2nminterms for n variables.
Example: Two variables (X and Y) produce
2 x 2 = 4 combinations:
XY(both normal)
XY (X normal, Y complemented)
XY (X complemented, Y normal)
XY (both complemented)
Thus there are four minterms of two variables.
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 2nmaxterms for n variables.
Example: Two variables (X and Y) produce
2 x 2 = 4 combinations:
X  Y (both normal)
X  Y (x normal, y complemented)
X  Y (x complemented, y normal)
X  Y (both complemented)
Minterms & Maxterms for 2 variables
Two variable minterms and maxterms.
x y
0 0
Index
0
Minterm
m0 = x y
Maxterm
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
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 3 variables
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
Index
0
1
2
3
4
5
6
7
Minterm
Maxterm
m0 = x y z
m1 = x y z
m2 = x y z
m3 = x y z
m4 = x y z
m5 = x y z
m6 = x y z
m7 = x y z
M0 = x + y + z
M1 = x + y + z
M2 = x + y + z
M3 = x + y + z
M4 = x + y + z
M5 = x + y + z
M6 = x + y + z
M7 = x + y + z
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
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
0
0
0
0
1
1
Minterm
m1 = x y z
Focus on the
‘1’ entries
m6 = x y z
m7 = x y z
F = m1 + m6 + m7 = ∑ (1, 6, 7) = x y z + x y z + x y z
Sum-Of-Minterm Examples
F(a, b, c, d) = ∑(2, 3, 6, 10, 11)
F(a, b, c, d) = m2 + m3 + m6 + m10+ m11
abcd+abcd+abcd+abcd+abcd
G(a, b, c, d) = ∑(0, 1, 12, 15)
G(a, b, c, d) = m0 + m1 + m12+ m15
abcd+abcd+abcd+abcd
Product-Of-Maxterm (POM)
Product-Of-Maxterm (POM) canonical form:
Product of maxterms of entries that evaluate to ‘0’
x
y
z
F
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
1
0
1
0
1
0
1
Maxterm
M2 = (x + y + z)
Focus on the
‘0’ entries
M4 = (x + y + z)
M6 = (x + y + z)
F = M2·M4·M6 = ∏ (2, 4, 6) = (x+y+z) (x+y+z) (x+y+z)
Product-Of-Maxterm Examples
F(a, b, c, d) = ∏(1, 3, 6, 11)
F(a, b, c, d) = M1 · M3 · M6 ·M11
(a+b+c+d) (a+b+c+d) (a+b+c+d)(a+b+c+d)
G(a, b, c, d) = ∏(0, 4, 12, 15)
G(a, b, c, d) = M0 · M4· M12·M15
(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-ofMaxterms (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
0
1
1
1
1
1
1
1
0
1
1
0
0
m0 = x y z
m1 = x y z
m2 = x y z
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
m4 = x y z
m5 = x y z
F = ∑(0, 1, 2, 4, 5) =
m0 + m1 + m2 + m4 + m5 =
x y z + x y z + x y z+
xyz+xyz
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
0
0
0
0
1
1
1
1
1
1
1
0
1
1
0
0
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
Minterm
F = ∏(3, 6, 7) =
M3 · M6 · M7 =
M3 = (x+y+z)
M6 = (x+y+z)
M7 = (x+y+z)
(x+y+z) (x+y+z) (x+y+z)
Conversions Between Canonical Forms
x
y
z
F
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
1
1
0
1
0
1
Minterm
Maxterm
M0 = (x + y + z)
m1 = x y z
m2 = x y z
m3 = x y z
M4 = (x + y + z)
m5 = x y z
M6 = (x + y + z)
m7 = x y z
F = m1+m2+m3+m5+m7 = ∑(1, 2, 3, 5, 7) =
x y z + x y z + x y z + x y z + x yz
F = M0· M4· M6= ∏(0, 4, 6) = (x+y+z)(x+y+z)(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=xy+xy+xy
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 bc
g = a b c + a b c + a b c + a b c + a bc
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 2nddistributive law:
f = (x + x) (x + y) = 1 · (x + y) = (x + y) =M1
Example 2: g = a c + b c + a b g (3 variables)
= (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)
(x + x y = x + y)
g = (c + b + a) (a + c + b)
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 2nminterms 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-ofmaxterms with the same indices
Standard Forms
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: A B C A B C  B
• POS: (A B)· (AB  C )·C
These “mixed” forms are neither SOP nor POS
• (A B  C) (A  C)
• AB C AC(A B)
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.
Standard Sum-of-Products (SOP)
A Simplification Example:
F(A, B, C) 
 (1,4,5,6,7)
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
AND/OR Two-Level Implementation
The two implementations for F are shown below
A
F
B
C
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
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