Find the dual of Boolean expression
B+ A'C+ B'A
Answers
Answer:BOOLEAN THEOREMS:
Boolean algebra helps to analyze a logic circuit and express its operation mathematically. We have several Boolean Theorems that helps us to simplify logic expressions and logic circuits.
Single Variable Theorems:
AND Operation OR Operation
A . 0 = 0 A + 0 = A
A . 1 = A A + 1 = 1
A . A = A A + A = A
A . A’ = 0 A + A’ = 1
Multivariable Theorems:
The multivariable theorems involves more than one variable.
Commutative laws:
A + B = B + A
A . B = B . A
Associative laws:
A + (B + C) = (A + B) + C = A + B + C
A (BC) = (AB) C = ABC
Distributive laws:
A (B + C) = AB + AC
A + BC = (A+B) (A+C)
(A + B) (C + D) = AC + AD + BC + BD
Absorption laws:
A + AB = A
A (A + B) = A
Redundant Literal Rule:
A + A’B = A + B
A’ + AB = A’ + B
A(A’ + B) = AB
Demorgan’s Theorem:
(A+B)’ = A’ . B’
(AB)’ = A’ + B’
This law can be extended to any number of variables or combinations of variables.
How to apply Demorgan’s theorem to an expression?
Complement the entire given expression.
Change all the AND’s to OR’s and all the OR’s to AND’s.
Complement each of the individual variables.
Change all 0s to 1s and all 1s to 0s.
This procedure is called as Demorganization or Complementation of switching expressions. In simple words, we can say it as ‘Break the line, change the sign’.
f(A, B, ………, 0, 1, . , +) = f(A’, B’, ………., 1, 0, +, .)
Consensus Theorem:
AB + A’C + BC = AB + A’C
(A + B)(A’ + C)(B + C) = (A + B)(A’ + C)
If one term is containing A and the other term is containing A’ and the third term containing the left-out literals of the first two terms then the third term is redundant. It means the function remains same with and without the third term.
This theorem can be extended to any number of variables.
AB + A’C + BCD = AB + A’C
(A + B)(A’ + C)(B + C + D) = (A + B)(A’ + C)
Transposition Theorem:
AB + A’C = (A + C)(A’ + B)
DUAL OF A BOOLEAN FUNCTION:
The DUAL of a boolean function is obtained by interchanging OR and AND operations and replacing 1’s by 0’s and 0’s by 1’s.
For example, DUAL of (ABCD…F)’ = A’ + B’ + C’ + D’ + … + F’
COMPLEMENT OF A BOOLEAN FUNCTION:
To compute the complement of a boolean function, we use Demorgan’s theorems.
How to complement a boolean function?
Parenthesize product terms.
Take the dual of a function.
Complement each literal.
For example, F(A,B,C) = AB’ + BC’ + A’C’
F'(A,B,C) = (AB’ + BC’ + A’C’)’
= (AB’)’ . (BC’)’ . (A’C’)’
= A’B . B’C . AC
REDUCING BOOLEAN EXPRESSIONS:
Realization of a digital circuit with the minimal expression results in reduction of cost and complexity and the corresponding increase in reliability.
How to reduce Boolean Expressions?
Multiply all variables necessary to remove parentheses.
Look for identical terms. Only one of those terms be retained and all others are dropped. For example, AB+AB+AB+AB=AB
Look for a variable and its negation in the same term. This term can be dropped. For example, A + BB’C = A
Look for pair of terms that are identical except for one variable which may be missing in one of the terms. The larger term can be dropped. For example, ABCD + ABC = ABC
Look for pair of terms which have the same variables with one or more variables complemented. If a variable in one term of such a pair is complemented while in the second term it is not, then such terms can be combined into a single term with that variable dropped. For example, ABC’D’ + ABC’D = ABC’
SUM OF PRODUCTS FORM (SOP):
The sum of products expressions consists of two or more AND terms (products) that are ORed together. Some examples of this form are:
A’BC+DE’F
AB+C’D+EF+G’H
Note that in sum of products expression, one inversion sign cannot cover more than one variable in a term. For example, we cannot have (AB)’C+(XYZ)’.
PRODUCT OF SUMS FORM (POS):
The product of sums expressions consists of two or more OR terms (sums) that are ANDed together. Some examples of this form are:
(A + B’ + C)(D’ + E’)
(A + B)(D + E’)(X + Y)Z
Minterm and Maxterm for three binary variables:
CANONICAL FORM:
Any Boolean function that is expressed as a sum of minterms or as a product of max terms is said to be in its canonical form or standard form.
For sum (OR) of min terms(1), The representation of the equation will be
F(list of variables) = ∑(list of min-term indices)
For product (AND) of max terms(0), The representation of the equation will be
F(list of variables) = π(list of max-term indices
oolean duals are generated by simply replacing ANDs with ORs and ORs with ANDs. The complements themselves are unaffected, where as the complement of an expression is the negation of the variables WITH the replacement of ANDs with ORs and vice versa. "The Dual of an identity is also an identity.
Answer:
In mathematics and the mathematical logic, Instead of the elementary algebra, where the values of variables are numbers and prime operations are addition and multiplication, the main operations of the Boolean algebra are conjunction (and) denoted as ∧, disjunction (or) denoted as ∨, and the negation (not) denoted as ¬. It is thus the formalism for describing logical operations, in same way that elementary algebra describes numerical operations.
Boolean algebra was introduced by Mr. George Boole in first book Mathematical Analysis of Logic (1847), and then set forth more fully in his An Investigation of Laws of Thought (1854). According to the Huntington, term "Boolean algebra" was first suggested by the Sheffer in 1913, although Charles Sanders Peirce gave title "A Boolean Algebra with One Constant" to first chapter of his "The Simplest Mathematics" in 1880. Boolean algebra has been fundamental in development of digital electronics, and is provided in all modern programming languages. It is also used in theory and statistics.
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