boolean algebra [compatibility mode]
DESCRIPTION
DIGITALTRANSCRIPT
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1Boolean Algebra and Logic Gates 1
1.3.5.7.9.
11.13.15.17.
CommutativeAssociativeDistributiveDeMorgans
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 + XX = 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 ( )
IdempotenceComplement
Involution
Identity element
Boolean Algebra and Logic Gates 2
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 0s and 1s.
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
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2Boolean Algebra and Logic Gates 3
Example: F = (A + C) B + 0dual F = (A C + B) 1 = A C + B
Example: G = X Y + (W + Z)dual G =
Example: H = A B + A C + B Cdual 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 Algebra and Logic Gates 4
Boolean Operator Precedence The order of evaluation is:
1. Parentheses2. NOT3. AND4. OR
Consequence: Parentheses appeararound OR expressions
Example: F = A(B + C)(C + D)
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3Boolean Algebra and Logic Gates 5
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.
Boolean Algebra and Logic Gates 6
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
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4Boolean Algebra and Logic Gates 7
Useful Theorems
MinimizationX Y + X Y = Y
AbsorptionX + X Y = X
SimplificationX + X Y = X + Y
DeMorgans X + Y = X Y
Minimization (dual)(X+Y)(X+Y) = Y
Absorption (dual)X (X + Y) = X
Simplification (dual)X (X + Y) = X Y
DeMorgans (dual) X Y = X + Y
Boolean Algebra and Logic Gates 8
Truth Table to Verify DeMorgans
X Y XY X+Y X Y X+Y X Y XY X+Y0 0 0 0 1 1 1 1 1 10 1 0 1 1 0 0 0 1 11 0 0 1 0 1 0 0 1 11 1 1 1 0 0 0 0 0 0
X + Y = X Y X Y = X + Y
Generalized DeMorgans Theorem:X1 + X2 + + Xn = X1 X2 XnX1 X2 Xn = X1 + X2 + + Xn
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5Boolean Algebra and Logic Gates 9
Complementing Functions
Use DeMorgan's Theorem:1. Interchange AND and OR operators2. 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
Boolean Algebra and Logic Gates 10
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
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6Boolean Algebra and Logic Gates 11
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
Boolean Algebra and Logic Gates 12
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) produce2 x 2 = 4 combinations:
(both normal)(X normal, Y complemented)(X complemented, Y normal)(both complemented)
Thus there are four minterms of two variables.
YXXY
YXYX
x
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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) produce2 x 2 = 4 combinations:
(both normal)(x normal, y complemented)(x complemented, y normal)(both complemented)
YX +YX +YX +YX +
Boolean Algebra and Logic Gates 14
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 Maxterm0 0 0 m0 = x y M0 = x + y0 1 1 m1 = x y M1 = x + y1 0 2 m2 = x y M2 = x + y1 1 3 m3 = x y M3 = x + y
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Minterms & Maxterms for 3 variables
M3 = x + y + zm3 = x y z3110M4 = x + y + zm4 = x y z4001M5 = x + y + zm5 = x y z5101M6 = x + y + zm6 = x y z6011
1
100y
1
000x
1
010z
M7 = x + y + zm7 = x y z7
M2 = x + y + zm2 = x y z2M1 = x + y + zm1 = x y z1M0 = x + y + zm0 = x y z0
MaxtermMintermIndex
Maxterm Mi is the complement of minterm miMi = mi and mi = Mi
Boolean Algebra and Logic Gates 16
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.
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9Boolean Algebra and Logic Gates 17
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
Boolean Algebra and Logic Gates 18
Sum-Of-Minterm (SOM) Sum-Of-Minterm (SOM) canonical form:
Sum of minterms of entries that evaluate to 1x y z F Minterm0 0 0 00 0 1 1 m1 = x y z0 1 0 00 1 1 01 0 0 01 0 1 01 1 0 1 m6 = x y z1 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
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Boolean Algebra and Logic Gates 19
+ 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
Boolean Algebra and Logic Gates 20
Product-Of-Maxterm (POM) Product-Of-Maxterm (POM) canonical form:
Product of maxterms of entries that evaluate to 0x y z F Maxterm0 0 0 10 0 1 10 1 0 0 M2 = (x + y + z)0 1 1 11 0 0 0 M4 = (x + y + z)1 0 1 11 1 0 0 M6 = (x + y + z)1 1 1 1
Focus on the 0 entries
F = M2M4M6 = (2, 4, 6) = (x+y+z) (x+y+z) (x+y+z)
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Boolean Algebra and Logic Gates 21
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)
Boolean Algebra and Logic Gates 22
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.
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Boolean Algebra and Logic Gates 23
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 Minterm0 0 0 1 m0 = x y z0 0 1 1 m1 = x y z0 1 0 1 m2 = x y z0 1 1 01 0 0 1 m4 = x y z1 0 1 1 m5 = x y z1 1 0 01 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
Boolean Algebra and Logic Gates 24
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 Minterm0 0 0 10 0 1 10 1 0 10 1 1 0 M3 = (x+y+z)1 0 0 11 0 1 11 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)
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Boolean Algebra and Logic Gates 25
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 zF = M0 M4 M6 = (0, 4, 6) =
(x+y+z)(x+y+z)(x+y+z)
x y z F Minterm Maxterm0 0 0 0 M0 = (x + y + z)0 0 1 1 m1 = x y z0 1 0 1 m2 = x y z0 1 1 1 m3 = x y z1 0 0 0 M4 = (x + y + z)1 0 1 1 m5 = x y z1 1 0 0 M6 = (x + y + z)1 1 1 1 m7 = x y z
Boolean Algebra and Logic Gates 26
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 yf = x y + x y + x yf = m3 + m2 + m0 = (0, 2, 3)
Example 2: g = a + b c (3 variables)g = a (b + b)(c + c) + (a + a) b cg = a b c + a b c + a b c + a b c + a b c + a b cg = a b c + a b c + a b c + a b c + a b cg = m1 + m4 + m5 + m6 + m7 = (1, 4, 5, 6, 7)
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Boolean Algebra and Logic Gates 27
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)
Boolean Algebra and Logic Gates 28
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)
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Boolean Algebra and Logic Gates 29
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
Boolean Algebra and Logic Gates 30
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(AB)(A +++
C)(AC)B(A ++B)(ACACBA ++
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Boolean Algebra and Logic Gates 31
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.
Boolean Algebra and Logic Gates 32
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 + AF = 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 S=
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Boolean Algebra and Logic Gates 33
AND/OR Two-Level Implementation
The two implementations for F are shown below
F
ABC
ABC
ABC
ABC
ABC
F
BC
A
It is quite apparent which
is simpler!
Boolean Algebra and Logic Gates 34
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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Boolean Algebra and Logic Gates 35
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Boolean Algebra and Logic Gates 36
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!
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Boolean Algebra and Logic Gates 37
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
Boolean Algebra and Logic Gates 38
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.
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Boolean Algebra and Logic Gates 39
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 as1 + 1 = 1 (read 1 or 1 equals 1).
= BAY yxz +=
AX =
Boolean Algebra and Logic Gates 40
Operator Definitions
Operations are defined on the values "0" and "1" for each operator:
AND0 0 = 00 1 = 01 0 = 01 1 = 1
OR
0 + 0 = 00 + 1 = 11 + 0 = 11 + 1 = 1
NOT
10 =01 =
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Boolean Algebra and Logic Gates 41
0110
XNOT
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:
111001010000
Z = XYYXAND OR
X Y Z = X+Y0 0 00 1 11 0 11 1 1
Boolean Algebra and Logic Gates 42
Truth Tables Contd
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 Z0 0 0 0 1 0 00 0 1 0 1 1 10 1 0 0 0 0 00 1 1 0 0 0 01 0 0 0 1 0 01 0 1 0 1 1 11 1 0 1 0 0 11 1 1 1 0 0 1
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Boolean Algebra and Logic Gates 43
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
XY
Z = X + YX
YZ = X Y
AND gate
X Z = X
NOT gate orinverter
Boolean Algebra and Logic Gates 44
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
ZYX F +=
Truth Table
11 1 111 1 011 0 111 0 000 1 100 1 010 0 100 0 0
X Y Z ZYX F +=