13 January 2011

Canonical Forms and Standard Forms

Standard Forms

  • each term in the function can have any number of literals.

Example, F1 = a +b’c + cde

  • There are 5 variables in F1 (a,b,c,d,e)
  • Canonical form terms should be written as m0, m1… (sum of Products) or M0, M1, M2… (Product of sums).
  • Each term in the canonical form should have all the literals.

Example: F1 = ab’c’d’e’ + abcd’e’ + abcde

More about Canonical Forms

Minterms

  • It is Sum of Products
  • Canonical form is Sum of Minterms
  • three variable minterms are shown below
x a b c minterms
0 0 0 0 m0=a’.b’.c’
1 0 0 1 m1=a’.b’.c
2 0 1 0 m2=a’.b.c’
3 0 1 1 m3=a’.b.c
4 1 0 0 m4=a.b’.c’
5 1 0 1 m5=a.b’.c
6 1 1 0 m6=a.b.c’
7 1 1 1 m7=a.b.c

Maxterms

  • Product of Sum (PoS)
  • Canonical form is Product of MaxTerms
  • three variable maxterms are shown below
x a b c minterms
0 0 0 0 M0=(a+b+c)
1 0 0 1 M1=(a+b+c’)
2 0 1 0 M2=(a+b’+c)
3 0 1 1 M3=(a+b’+c’)
4 1 0 0 M4=(a’+b+c)
5 1 0 1 M5=(a’+b+c’)
6 1 1 0 M6=(a’+b’+c)
7 1 1 1 M7=(a’+b’+c’)

usually

Mi = (mj)’

Express the boolean function F = A + BC in a sum of minterms.

The function has three variables,

so F = A + BC will be

F = A(B + B’) + (A +A’) BC  [since, x + x’ =1]

F = AB + AB’ + ABC + A’BC

F = AB(C+C’) + AB’(C+C’) + ABC + A’BC

F = ABC + ABC’ + AB’C + AB’C’ + ABC + A’BC

F = ABC + ABC’ + AB’C + AB’C’ + A’BC [since x + x = x]

F= m0 + m6 + m5 + m4 + m3


1 comment:

  1. wonderful explanation. plz do visit this page once. its really helpful.

    ReplyDelete

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