Boolean Algebra
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Transcript of Boolean Algebra
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Boolean Algebra
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We will find a lot of similarities between Boolean algebra and
"normal" algebra, the kind of algebra involving so-called real numbers.
Just bear in mind that the system of numbers defining Boolean algebra
is severely limited in terms of scope, and that there can only be one of
two possible values for any Boolean variable: 1 or 0. Consequently, the
"Laws" of Boolean algebra often differ from the "Laws" of real-number
algebra, making possible such statements as 1 + 1 = 1, which would
normally be considered absurd. Once you comprehend the premise of
all quantities in Boolean algebra being limited to the two possibilities of
1 and 0, and the general philosophical principle of Laws depending on
quantitative definitions, the "nonsense" of Boolean algebra disappears.
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Boolean Algebra
• Invented by George Boole in 1854
• Considered as the backbone of computer circuit
analysis and design
• It is a branch of mathematics that is directly
applicable to digital design.
• It is a set of elements, a set of operators that act
on these elements, and a set of axioms or
postulates that govern the actions of these
operators on these elements.
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Theorems and Postulates
• The postulates are basic axioms of the algebraic
structure and need no proof.
Basic Identity (a) x+ 0 = x (b) x. 1 = x
Commutative
Property
(a) x + y = y + x (b) xy = yx
Distributive
Property
(a) x(y+z) = xy+xz (b) x+yz = (x+y )
(x+z)
Basic Identity (a) x+ x’ = 1 (b) xx’ = 0
Postulates of Boolean Algebra
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• The theorems must be proven from the
postulates
Theorem 1. Identity x + x = x xx = x
Theorem 2. Identity x+ 1= 1 x.0 = 0
Theorem 3. Involution (x’)’ = x
Theorem 4.
Associative
x+(y+z) = (x+y) + z x(yz) = (xy)z
Theorem 5. De
Morgan’s
(x+y)’ = x’.y’ (xy)’ = x’ + y’
Theorem 6.
Absorption
x + xy = x x(x+y) = x
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Operations with Boolean Algebra
• Complementation (‘)
– To complement a variable is to reverse its value
– Represented by placing a bar over the variable or by the (‘) symbol after the variable.
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• Addition (+)
– Boolean Addition is equivalent to logical OR.
– The (+) plus symbol is used to indicate the addition or ORing
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• Multiplication (.)
– Boolean multiplication is equivalent to a logical AND operation
Note:
Prohibited Operations: The operations of subtraction and division do not exist in Boolean Algebra
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Truth Table
• It lists every possible combination of inputs and the
output corresponding to each combination of inputs.
• Given a function N inputs, the possible combinations will
then be equal to 2N.
• Example: using a truth table prove that x (x+y) = x.
x y x+y x(x+y)
0 0 0 0
0 1 1 0
1 0 1 1
1 1 1 1
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Boolean Functions
• It is defined using an equal sign and an
expression comprise of binary variables,
two binary operators OR and AND, the
unary operator NOT and the parentheses.
• It may also be represented by a truth
table, showing all the possible 1’s and 0’s
combination of the N binary variable
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Example: 1. F1=x(y+z)
• F1=1 only when : x=1, y=1, z =1 or
x=1, y =1, z=0 and
x=1, y=0, z=1
X y z y+z F1
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 1 0
1 0 0 0 0
1 0 1 1 1
1 1 0 1 1
1 1 1 1 1
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Algebraic Manipulation
• A Boolean function may be transformed
from an algebraic expression into a logic
diagram composed of AND, OR and NOT.
• There is an AND gate for each term in the
expression, and an OR gate is used to
combined two or more terms.
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Examples
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Complement of a Function
• The function F and F’ are complementary if:
– they depend on the same set of input variables
– for every combination of values of the input variables, the values of F and F’ are inverses to each other.
The complement of a function may be derived algebraically using De Morgan’s theorem.
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Canonical Forms
• A binary variable may appear either in its normal
form (x) or in its complemented form (x’).
• Combining x and y with the AND operator, the
following 4 combinations will result: x’y’, x’y, xy’
and xy.
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Canonical Form
X Y
X’Y’
XY
XY’ X’Y
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Possible Combination for Three
Variables x y Z Minterm Designation Maxterm Designation
0 0 0 x’y’z’ m0 x+y+z M0
0 0 1 x’y’z m1 x+y+z’ M1
0 1 0 x’yz’ m2 x+y’+z M2
0 1 1 x’yz m3 x+y’+z’ M3
1 0 0 xy’z’ m4 x’+y+z M4
1 0 1 xy’z m5 x’+y+z’ M5
1 1 0 xyz’ m6 x’+y’+z M6
1 1 1 xyz m7 x’+y’+z’ M7
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• Minterm or Standard Products
– Obtained from an AND terms of the N
variables
– It is represented by the symbol mj where j
denotes the decimal equivalent of the binary
value
• Maxterm or Standard Sums
– Obtained from an OR term of the N variables
– It is represented by the symbol Mj where j
denotes the decimal equivalent of the binary
value
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Sum of Minterms
• The ORing of terms which result in
a 1-value of the function. Ex.
• Boolean Function:
F1=x’y’z + xy’z’ + xy’z
=m1 + m4 +m5
or F1=∑(1, 4, 5)
• Complement of the Function
F1’=x’y’z’ + x’yz’ + x’yz + xyz’ + xyz
= m0, m2, m3, m6, m7
or F1’ =∑(0, 2, 3, 6, 7)
x y z F1
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 0
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Product of Maxterms
• The ANDing of terms which
each result in a 0-value of the
function
• Ex. Express the Boolean
function F= AB + AC’ as a
product of maxterms
• F= π(M0,M1, M2,M3,M5)
• F=(A + B +C) (A+B+C’)
(A+B’+C) (A+B’+C’) (A’+B+C’)
A B C F
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 1
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Standard Forms:
• Two types of Standard Form
– Sum of Products
• A Boolean expression containing AND terms of one or more
literals each.
• The sum denotes the ORing of the AND terms.
• Ex. F1= y + x’y’z + xz
– Product of Sums
• A Boolean expression containing OR terms of one or more
literals each.
• The sum denotes the ANDing of the OR terms.
• Ex. F2=y (x’+z) (x+y+z) (x’+y+z+w’)
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Exercises
1. Simplify the following Boolean
expressions to a minimum using Boolean
Algebra
a. (x’y’ + z)’ + z + xy + wz
b. x’ + xy + xz’ + xy’z’
c. AB(C’D +CD’) + A’B’(C’+D)(C+D’)
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2. For the Boolean function F given in the
truth table, find the following:
a) List the minterms of the function.
b) List the minterms of F’
c) Express F in sum of minterms in
algebraic form
d) Simply the function to an expression
with the minimum number of literals.
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x y z F
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 1
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3. Given the following Boolean function:
F=xy’z +x’y’z + w’xy + wx’y + wxy
a) Obtain the truth table of the function.
b) Draw the logic diagram using the original
Boolean expression.
c) Simplify the function to a minimum number of
literals using Boolean algebra.
d) Obtain the truth table of the simplifed
expression and compare with the original table.