Boolean Function

Boolean Functions

The following steps are generally followed while designing the logic circuits using Boolean functions.

  1. Analysis of the given statement of the problem to find the number of variables
  2. Writing the truth table from the given statement
  • Conversion of the truth table into a logic function/ Boolean expression using standard product of sums(SOP) or product of sums (POS)
  1. Simplify the Boolean expression
  2. Create the logic circuit using logic gates

SOP and POS equations

Sum of Product (SOP) equation

The SOP equations are formed by ORing the product or the min-terms.

Formation of Minterm and Maxterm
A B C Minterm(mi) Maxterms(Mi)
0 0 0 A’B’C’ =m0 A+B+C   =M0
0 0 1 A’B’C =m1 A+B+C’  =M1
0 1 0 A’BC’ =m2 A+B’+C  =M2
0 1 1 A’BC  =m3 A+B’+C’ =M3
1 0 0 AB’C’ =m4 A’+B+C  =M4
1 0 1 AB’C  =m5 A’+B+C’  =M5
1 1 0 ABC’  = m6 A’+B’+C  =M6
1 1 1 ABC    = m7 A’+B’+C’ =M7

Min terms

The input combinations for which the output equals ‘1’ is called min terms. Min-terms are also called as product terms.  They are formed by the product of variable having ‘1’ value and complement of variable having ‘0’ value.

Suppose the Y(abc) =’1’ for input combinations 011, 101, 110, then the min-terns are a’bc, ab’c, and abc’.

The minterms are also represented by their equivalent decimal values M3, M5, M6

The different ways of writing the SOP equations is:

  1. Y(abc) = a’bc + ab’c + abc’                           OR
  2. Y(abc) = ∑(M3, M5, M6) OR
  3. Y(abc) =

Product of Sums (POS) equation

Product of sums (POS) equation is formed by ANDing the sum terms. These sum terms are also called max-terms. So, a POS equation is formed by ANDing the max-terms.

Max-terms:

The input combinations for which the output is ‘0’ are called as max-terms. They are formed by ORing the variables having ‘0’ value and complemented of variables having ‘1’ value.

Example:

Suppose the output is ‘0’ for the input combinations 000, 010, 100, 111, the max terms are formed as:

  1. Complement the variables having ‘1’ value
  2. Keep the variable in un-complement form having ‘0’ value

The max-terms are:

  1. (a+b+c), (a +b’ + c), ( a’ + B + c) and (a’ + b’ + c’)

OR

  1. M0, M2,         M   and  M7

The POS equation therefore is formed :

Y(abc) = (a+b+c) . (a +b’ + c) . ( a’ + B + c) . (a’ + b’ + c’)

OR

Y(abc) =  πM ( 0, 2, 4, 7)

 

Conversion from SOP to POS and Vice Versa

 

We can use the De-Morgan’s theorem for converting SOP to POS and vice versa

  1. Converting a SOP equation to POS

Suppose the SOP equation is Y = A’BC + AB’C + ABC’ + ABC

Complement of Y = Y’ =  (A’BC + AB’C + ABC’ + ABC)’

Y’ = (A’BC)’ . (AB’C)’ . (ABC’)’  . (ABC)’

= (A + B’ +C’) . (A’ + B + C’) . (A’ + B’ + C)  . (A’ + B’ + C’)

Double Complement Y’’ = Y = ( (A + B’ + C’) . (A’ + B + C’) . (A’ + B’ + C)  . (A’ + B’ + C’) )’

Therefore, ∑M(3,5,6,7) =πM(0,1,2,4)

  1. Converting AND-OR to NAND implementation

Y          =∑m(1,4,5,6)

=A’B’C + AB’C’ +AB’C + ABC’

Y’’        =( A’B’C + AB’C’ +AB’C + ABC’)’’

=(( A’B’C)’ . (AB’C’)’ . (AB’C)’ . (ABC’)’)’

This show that and AND-OR equation/ logic circuit can be implemented using only the NAND gates

  1. Converting OR-AND equation to all NOR implementation

Consider Y= (A+B)(A+C)(B+C)

Double complement Y” = ((A+B)(A+C)(B+C))”

 =( (A+B)’ + (A+C)’ + (B+C)’)’

This shows that an OR-AND equation can be implemented using all NOR gates

Canonical Form (Standard SOP and POS)

Canonical form are special cases of SOP and POS forms. In canonical form each term consist of the literals either in complemented or un-complemented form. Based on this definition we can define canonical forms for SOP and POS

  1. Standard SOP form or Minterm canonical form
  2. Standard POS form or Maxterm canonical form

Procedure for converting SOP form to standard SOP form

  1. Check out for the missing literals in each minterm
  2. Perform AND of each product term with OR of missing literal and its complement
  3. Expand using distributive law
  4. Remove the repetitive terms

Example:

F(ABCD) = AC + AB + BC

Step-1: B is missing in minterm AC, C is missing in minterm AB, and A is missing in minterm BC

Step-2: AC(B+B’) + AB(C+C’) + BC(A+A’)

Step-3: Expand – ABC + AB’C + ABC + ABC’ + ABC + A’BC

Step-4: remove redundant terms

Therefore the canonical SOP is Y= ABC + AB’C + ABC’ + A’BC

Procedure for converting SOP form to standard SOP form

Step-1: Check out for the missing literals in each Maxterm.

Step-2: Perform OR of each sum term with ANDING of missing literal and its complement

Step-3: Expand using distributive law.

Step-4: Remove the repetitive terms

Example:

Suppose Y=(A+C)(B+C)

  1. Here B is missing in (A+C) and A is missing in (B+C)
  2. (A+C +B.B’)(B+C + AA’)
  3. Y = (A+B+C)(A+B’+C)(A+B+C)(A’+B+C)
  4. Here we find the (A+B+C) is repeating, use it only once.
  5. Therefore canonical SOP= Y = (A+B+C)(A+B’+C) (A’+B+C)

Quiz

  1. The SOP expression is written using:
    1. anding the Maxterms
    2. ORing the minterms
    3. ANDing the minterms
    4. Using any of the above
  2. A POS equation is writing :
    1. ANDing the minterms
    2. ANDing the Maxterms
    3. ORing the minterms
    4. ORing the Maxtrms
Updated: August 8, 2019 — 5:34 pm

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