Canonical forms-Boolean Algebra and Digital Logic Design-Lecture Slides, Slides of Digital Logic Design and Programming

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Overview of Last Lecture!

• Gate Implementation!
• Minimization of function!
  • Algebraic Manipulations!
• Complement of a Boolean function!

Today’s Lecture Outline!

• Canonical Form of Boolean Function!
  • Sum of Minters!
  • Product of Maxterms!
• Standard Form!

Minterms

• Any given binary variable can be represented in two forms: - x, its normal form, and - x’, its complement
• If we consider two binary variables and the AND operation, there are four combinations of the variables: - xy - xy’ - x’y - x’y’
• Each of the above four AND terms is called a minterm or a standard product.
• n variables can be combined to form 2 n minterms.

Minterms Expressed

Maxterms Expressed

Truth Table to Expression (Sum of Minterms)

• Any Boolean function can be expressed as a sum of

minterms or sum of products (i.e. the ORing of

terms).

• You can form the function algebraically by forming a minterm for each combination of the variables that produces a 1 in the function. (Each row with output of 1 becomes a product term)
• Sum (OR) product terms together. x! 0! 0! 0! 0! 1! 1! 1! 1! y! 0! 0! 1! 1! 0! 0! 1! 1! z! 0! 1! 0! 1! 0! 1! 0! 1!

G!

xyz + xyz’ + x’yz

• All three formats are equivalent
• Number of 1’s in truth table output column equals AND terms for Sum-of-Products (SOP) x (^) y z x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 G 0 0 0 1 0 0 1 1 G = xyz + xyz’ + x’yz G x x x x x x x x x Equivalent Representations of Circuits

Truth Table to Expression (Product of Maxterms)

• Any Boolean function can be expressed as a

product of maxterms or product of sums (i.e. the

ANDing of terms).

• You can form the function algebraically by forming a maxterm for each combination of the variables that produces a 0 in the function. (Each row with output of 0 becomes a standard sums)
• AND these maxterms together.

Minterms and Maxterms

• Each variable in a Boolean expression is a literal
• Boolean variables can appear in normal (x) or

complement form (x’)

• Each AND combination of terms is a minterm
• Each OR combination of terms is a maxterm
• Example: Minterms x y z Minterm 0 0 0 x’y’z’ m 0 0 0 1 x’y’z m 1 … 1 0 0 xy’z’ m 4 … 1 1 1 xyz m 7 Maxterms x y z Maxterm 0 0 0 x+y+z M 0 0 0 1 x+y+z’ M 1 … 1 0 0 x’+y+z M 4 … 1 1 1 x’+y’+z’ M 7

Obtaining Sum of Minterms Form F = A’B’C’ + A’B’C + A’BC’ + A’BC + AB’C’ + AB’C+ ABC! "= m 0

• m 1
• m 2
• m 3
• m 4
• m 5
• m 7! F(A, B, C) = ∑(0, 1, 2, 3, 4, 5, 7)!

Canonical Form Conversion

• A function represented as Sum of minterms can be represented as the Product of maxterms of the remaining terms.
• The complement of a function expressed in sum of minterms equals the sum of minterms missing from the original function - F(A, B, C) = ∑(0, 3,4) = m 0 +m 3 +m 4 - F’(A, B, C) = ∑(1,2,5,6,7)= m 1 +m 2 +m 5 +m 6 +m 7
• Now if we take the complement of F’ using DeMorgan’s theorem, we obtain F in the product of maxterms form: - (F’)’ = (m 1 +m 2 +m 5 +m 6 +m 7 )’ - F = m 1 ’. m 2 ’. m 5 ’. m 6 ’. m 7 ’ [Complement of minterms] - = M 1 M 2 M 5 M 6 M 7 [maxterms] - = ∏(1,2, 5, 6, 7)
• This implies the following relation:
  • m’j = Mj
• So sum of minterms: ∑(0,3,4) = product of maxterms: ∏(1,2, 5, 6, 7)

Table A: Conversion of Forms Desired Form! Given Form ! Minterm Expansion of F! ! Maxterm Expansion of F! ! Minterm Expansion of F! ! Maxterm Expansion of F! ! Minterm Expansion of F! !

-! maxterm nos are those nos, not on the minterm list of F! List minterms not present in F! Maxterm nos are same as minterm nos of F! Maxterm Expansion of F! minterm nos are those nos, not on the maxterm list of F! ! -! minterm nos are same as maxterm nos of F! ! List maxterms not present in F!

Possible Functions!

• n variable can give how many possible

functions?!

• Answer = 2

2n!

Conversion to Sum of minterms : By algebraic method!

• Each term is inspected to see if it contains all variables. If it misses one or more can be ANDed with an expression such as x+x’, where x is missing!
• Example!
  • Express F in Sum of Minterms! " "F= A + B’ C! "A = A (B+B’) =AB +AB’! " =AB (C+C’) + AB’ (C+C’)! =ABC + ABC’ + AB’C + AB’C’! B’C = B’C (A+A’)! = AB’C + A’B’C! F = ABC + ABC’ + AB’C + AB’C’ + AB’C + A’B’C! = A’B’C + AB’C’ + AB’C + ABC’ + ABC! m1 + m4 + m5 + m6 + m7! F(A,B,C) = ( 1, 4, 5, 6, 7 )!