EXOR FUNCTION

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• EX-OR FUNCTION
• A EX-OR gate will have two input and a
  output. The EX-OR gate will be logic 1 if
  and only if both inputs are not at same
  logic state.
• The Boolean output expression of EX-OR
  gate is
• f1 x y x y
• f2 ( x y )( x y )
• f3 x y x y
• f4 (x y ) (x y)

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• The truth table of Ex-OR gate

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• EX-NOR FUNCTION
• A EX-NOR gate will have two input and a
  output. The EX-NOR gate will be logic 1 if
  and only if both inputs are at same logic
  state.
• The Boolean output expression of EX-NOR
  gate is
• f1 x y x y
• f2 ( x y )( x y )
• f3 x y x y
• f4 ( x y ) (x y)

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• The truth table of Ex-NOR gate

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• Example 1Using Boolean identities prove the
  following
• x y x y x y x y
• Example 2Using perfect induction method
  prove the following identities.
• f(x, y,z) x z x y ( x y)(x z)
• Example 3 Express the following in min
  term canonical formulas and construct the
  truth table.
• f(x,y,z) ?m(0,2,4,5,7)

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• Example 4 Express the following in maxterm
  canonical formulas and construct the truth
  table.
• f(x,y,z) ? M(2,4,7)
• Example 5Express the following by a min
  -term canonical formulas without constructing
  truth table
• f(x,y,z) x y x z z
• Ans f ?m(0,1,2,3,4,6)

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• Example 6 Minimize the following Boolean
  output expression using Boolean algebra
• f(A, B, C) ( A (A B))(B B C ) ( C
  ABC )

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• CHAPTER 2 SIMPLICATION OF BOOLEAN
  EXPRESSIONS
• SIMPLIFICATION OF BOOLEAN ALGEBRA
• The given Boolean output expression has to
  be simplified in terms its equation nor
  the capacity. The following are the method
  used
• Graphical method- Karnaugh method (K-map
  method)
• Tabular method Quine McCluskey method.

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• FORMULATION OF THE SIMPLIFICATION PROBLEM
  The merit of the network can be evaluated
  referring to the following
• The cost of network.
• The reliability of the network.
• Propagation delay.
• 4. Fabrication density of the network.

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• CRITERIA OF MINIMALITY Minimal overall
  response time- minimum levels, minimum terms
  to represent the Boolean function either
  in SOP or POS, Single output network is to
  be considered.
• SIMPLIFICATION OF PROBLEM Using Boolean
  algebra XX1, X.X 0, X XX and XX X one can
  simplify the given Boolean output
  expression without altering basic function .
  Only the no. Literal sum or product terms
  are minimized.

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• PRIME IMPLICANTS AND IRRENDUNDANT DISJUCTIVE
  EXPRESSION
• IMPLIES Consider two complete function of n
  -variables
• f1 and f2 . The function f1 implies the
  function f2 if there is no assignment of
  values to the n-variables that makes f1
  equal to 1 and f2 equals to 0.
• Ex. f1(x,y,z) xy yz and f2(x,y,z) xy yz
  x z

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• Similarly in the conjunctive normal
  formula if f3 implies f4 when if there is
  no assignment of values to the n-variable
  makes f4 equal to 0 and f3 equals to 1.
• Ex.f3(x,y,z) (xy)(yz)(xz) and
• f4(x,y,z) (xy)(yz)

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• Truth table

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• SUBSUMES It is a comparison between two
  product terms or two sum terms is also
  possible. A term t1 is said to be subsume
  of term t2 if all the literal of the term
  t2 are also literal of the term t1.
• Ex. x y z is subsume of the product x z
• similarly ( x y z ) is subsume of term
  (x z)

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• IMPLICANTS A product term said to be
  implicant of a complete function if the
  product term implies the function, which
  describes the functional values of 1.
• The term x is equals to 0 for the four
  3-tuples of (x,y,z) (0,0,0), (0,0,1), (0,1,0)
  and (0,1,1).
• Hence the x is said to be implicant and
  other implicant that represent value 1 is
  y z , where it represent value 1 for two
  3-tuples(x,y,z) ( 0,0,1) and ( 1,0,1).

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• Truth table

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• PRIME IMPLICANT If the implicant does not
  subsumes any other implicant with fewer
  literals of the same function. In other
  words if we remove prime implicant term
  from the expression the remaining product
  terms no longer implies the function
• Ex. x and y z are prime implicants

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• Theorem 1 When he cost assigned by some
  criterion, for a minimal Boolean formula
  is such that decreasing the number of
  literals in the disjunctive normal formula
  does not increase the cost of the formula,
  there is at lest one minimal disjunctive
  normal formula that corresponds to a sum
  of the prime implicants.

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• IRRENDUNDANT DISJUNCTIVE NORMAL FORMULAS An
  irredundant disjunctive normal formula
  describing a complete function is defined
  as an expression in SOP form such that (1)
  every product term in the expression is a
  prime implicant and (2) no product term
  may be eliminated from the expression
  without changing the function described
  by the expression.

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• IMPLICATE A sum term is said to be
  implicate of a complete Boolean function
  if the function implies the sum term. Which
  describes the function value of 0.
• Ex. A sum term xy has the value 0 for
  two- 3 tuples (x,y,z) (1,0,0) and (1,1,0)
  The sum term x y is implied by the
  function hence it is referred as implicate
  of the function.

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• PRIME IMPLICATE If the implicate does not
  subsumes any other implicate with fewer
  literals of the same function. In other
  words if we remove prime implicate term
  from the expression the remaining sum terms
  no longer implies the function
• Ex. x and (y z) are prime implicates

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• IRRENDUNDANT CONJUNCTIVE NORMAL FORMULAS An
  irredundant conjunctive normal formula
  describing a complete function is defined
  as an expression in POS form such that (1)
  every sum term in the expression is a
  prime implicate and (2) no sum term may
  be eliminated from the expression
  without changing the function described
  by the expression.

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• Note It should be noted that prime
  implicates are dual concept of prime
  implicants and irredundant conjunctive
  normal formulas are dual concept of
  irredundant disjunctive normal formulas. Prime
  implicant f is exactly complement f will
  be prime implicate and vice versa.