Boolean Functions
The following steps are generally followed while designing the logic circuits using Boolean functions.
 Analysis of the given statement of the problem to find the number of variables
 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)
 Simplify the Boolean expression
 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 minterms.
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. Minterms 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 minterns 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:
 Y(abc) = a’bc + ab’c + abc’ OR
 Y(abc) = ∑(M3, M5, M6) OR
 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 maxterms. So, a POS equation is formed by ANDing the maxterms.
Maxterms:
The input combinations for which the output is ‘0’ are called as maxterms. 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:
 Complement the variables having ‘1’ value
 Keep the variable in uncomplement form having ‘0’ value
The maxterms are:
 (a+b+c), (a +b’ + c), ( a’ + B + c) and (a’ + b’ + c’)
OR
 M_{0}, M_{2}, M_{4 } and M_{7}
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 DeMorgan’s theorem for converting SOP to POS and vice versa

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)

Converting ANDOR 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 ANDOR equation/ logic circuit can be implemented using only the NAND gates

Converting ORAND 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 ORAND 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 uncomplemented form. Based on this definition we can define canonical forms for SOP and POS
 Standard SOP form or Minterm canonical form
 Standard POS form or Maxterm canonical form
Procedure for converting SOP form to standard SOP form
 Check out for the missing literals in each minterm
 Perform AND of each product term with OR of missing literal and its complement
 Expand using distributive law
 Remove the repetitive terms
Example:
F(ABCD) = AC + AB + BC
Step1: B is missing in minterm AC, C is missing in minterm AB, and A is missing in minterm BC
Step2: AC(B+B’) + AB(C+C’) + BC(A+A’)
Step3: Expand – ABC + AB’C + ABC + ABC’ + ABC + A’BC
Step4: remove redundant terms
Therefore the canonical SOP is Y= ABC + AB’C + ABC’ + A’BC
Procedure for converting SOP form to standard SOP form
Step1: Check out for the missing literals in each Maxterm.
Step2: Perform OR of each sum term with ANDING of missing literal and its complement
Step3: Expand using distributive law.
Step4: Remove the repetitive terms
Example:
Suppose Y=(A+C)(B+C)
 Here B is missing in (A+C) and A is missing in (B+C)
 (A+C +B.B’)(B+C + AA’)
 Y = (A+B+C)(A+B’+C)(A+B+C)(A’+B+C)
 Here we find the (A+B+C) is repeating, use it only once.
 Therefore canonical SOP= Y = (A+B+C)(A+B’+C) (A’+B+C)
Quiz
 The SOP expression is written using:
 anding the Maxterms
 ORing the minterms
 ANDing the minterms
 Using any of the above
 A POS equation is writing :
 ANDing the minterms
 ANDing the Maxterms
 ORing the minterms
 ORing the Maxtrms