Neutral Functions | Minterms and Maxterms

Minterms and Maxterms-

 

For any function consisting of n Boolean variables,

Number of minterms possible = Number of maxterms possible = 2n

 

Minterms and Maxterms Examples-

 

The examples of minterms and maxterms are-

 

Example-01:

 

For any function consisting of 2 Boolean variables A and B, we have-

  • Number of minterms possible = 22 =  4.
  • Number of maxterms possible = 22 =  4.

 

The following table shows the minterms and maxterms-

 

ABMintermsMaxterms
00A’B’A + B
01A’BA + B’
10AB’A’ + B
11ABA’ + B’

 

Example-02:

 

For any function consisting of 3 Boolean variables A, B and C, we have-

  • Number of minterms possible = 23 =  8.
  • Number of maxterms possible = 23 =  8.

 

The following table shows the minterms and maxterms-

 

ABCMintermsMaxterms
000A’B’C’A + B + C
001A’B’CA + B + C’
010A’BC’A + B’ + C
011A’BCA + B’ + C’
100AB’C’A’ + B + C
101AB’CA’ + B + C’
110ABC’A’ + B’ + C
111ABCA’ + B’ + C’

 

Neutral Functions-

 

A neutral function is a function for which number of minterms and number of maxterms are same.

 

Example-

 

The following function is an example of a neutral function-

f( A, B) = A ⊕ B

 

This is because this function has equal number of minterms and maxterms as shown by the following table-

 

ABA ⊕ B
000
011
101
110

 

Number Of Neutral Functions Possible-

 

 

Here n = Number of Boolean variables in the function.

 

Explanation-

 

  • For n variables, total number of terms possible = number of combinations of n variables = 2n.
  • Since maximum number of terms possible = 2n, so we choose half of the terms i.e 2n / 2 = 2n-1.
  • We assign them the output logic ‘1’.
  • We assign ‘0’ to rest half of the terms.
  • Thus, number of neutral functions possible with n Boolean variables = C ( 2n , 2n-1 ).

 

PRACTICE PROBLEM BASED ON NEUTRAL FUNCTIONS-

 

Problem-

 

Consider any function consisting of 2 Boolean variables A and B-

  1. Calculate the total number of neutral functions that are possible.
  2. Write all the neutral functions.

 

Solution-

 

We know, number of neutral functions possible with n Boolean variables = C ( 2n , 2n-1 ).

 

Substituting n = 2, we get-

Number of neutral functions possible with 2 Boolean variables

= C ( 22 , 22-1 )

= C ( 4 , 2 )

= 6

 

Thus, total 6 Boolean functions are possible with 2 Boolean variables.

The 6 Boolean Functions are-

  • f ( A , B ) = A
  • f ( A , B ) = A’
  • f ( A , B ) = B
  • f ( A , B ) = B’
  • f ( A , B ) = A ⊙ B
  • f ( A , B ) = A ⊕ B

 

VariablesBoolean Functions
ABAA’BB’A⊙BA⊕B
00010110
01011001
10100101
11101010

 

The above table clearly shows that for each function, number of minterms = number of maxterms.

So, they all are neutral functions.

 

To gain better understanding about Neutral Functions,

Watch this Video Lecture

 

Next Article- Self-Dual Functions

 

Get more notes and other study material of Digital Design.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Summary
Neutral Functions | Minterms and Maxterms
Article Name
Neutral Functions | Minterms and Maxterms
Description
A neutral function is a function for which number of minterms and number of maxterms are same. Number of neutral functions possible with n Boolean variables = C ( 2n , 2n-1 ).
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Gate Vidyalay
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