INTRODUCTION





Two Valued Boolean Algebra





Postulates of Boolean Algebra





Operator Precedence in Boolean Algebra





BOOLEAN FUNCTIONS





Simplification of Boolean Expressions




Simplify x (x' + y)



Simplify x + (x'y)



Simplify (x+y)(x+y')



Simplify xy + x'z + yz




Complement of a Function




Find complement of a Function F1 = x'yz' + x'y'z


  1. So we have F'1 = (x'yz' + x'y'z)'

  2. = (x'yz')'(x'y'z)'

  3. = (x + y' + z)(x + y + z')


Find complement of a Function F2 = x(y'z' + yz)


  1. So we have F'2 = [x(y'z' + yz)]'

  2. = x' + (y'z' + yz)'

  3. = x' + (y'z')'(yz)'

  4. = x' + (y+z)(y'+z')

  5. = x' + yz' + y'z



Complement of a Function - Dual Method




F1 = x'yz' + x'y'z


  1. The dual of F1 = (x'+y+z')(x'+y'+z)

  2. Complement of each literal: (x+y'+z)(x+y+z') = F'1


F2 = x(y'z' + yz)


  1. The dual of F2 = x + (y'+z')(y+z)

  2. Complement of each literal: x' + (y+z)(y'+z') = F'2



CANONICAL AND STANDARD FORMS




F1 = x'y'z + xy'z' + xyz


  1. From the table above we obtain F1 = m1 + m4 + m7


F2 = x'yz + xy'z + xyz' + xyz


  1. From the table above we obtain F1 = m3 + m5 + m6 + m7


F1 = (x+y+z)(x+y'+z)(x'+y+z')(x'+y'+z)


  1. From the table above we obtain F1 = M0 . M2 . M3 . M5 . m6



Sum of Minterms



Q: F = A + B'C


  1. The term has three variables: A,B,C and he first part of the expression 'A' has two variables missing viz. B and C

  2. So, A = A(B + B') = AB + AB'

  3. It still has one variable missing

  4. AB + AB' = AB(C + C') + AB'(C + C')

  5. = ABC + ABC' + AB'C + AB'C'

  6. Now the second term of the expression is BC'

  7. B'C = B'C(A + A') = B'CA + B'CA'

  8. So we have F = ABC + ABC' + AB'C + AB'C' + B'CA + B'CA'

  9. But AB'C appears twice so finally we get

  10. F = ABC + ABC' + AB'C' + B'CA + B'CA'

  11. F = m1 + m4 + m5 + m6 + m7

  12. A more convenient form isF = ∑(1, 4, 5, 6, 7)



Product of Maxterms



Solve F = xy + x'z


  1. First, convert the func­tion into OR terms by using the distributive law: x + yz = (x + y)(x + z)

  2. So we have, F = xy + x'z = (xy + x')(xy + z)

  3. = (x + x')(y + x')(x + z)(y + z)

  4. = (x' + y)(x + z)(y + z)

  5. The function has three variables: x, y, and z. Each OR term is missing one variable; therefore,

  6. x' + y = x' + y +zz' = (x' + y + z)(x' + y + z')

  7. x + z = x + z + yy' = (x + y + z)(x + y' + z)

  8. y + z = y + z + xx' = (x + y + z)(x' + y + z)

  9. Combining all the terms and removing those which appear more than once, we finally obtain

  10. F = (x + y + z)(x + y' + z)(x' + y + z)(x' + y + z')

  11. Which is same as F = M0M2M4M5

  12. F = ∏(0, 2, 4, 5)

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