Dual of a Boolean Expression-
To get a dual of any Boolean Expression, replace-
- OR with AND i.e. + with .
- AND with OR i.e. . with +
- 1 with 0
- 0 with 1
We know, Consensus theorem is-
xy + x’z + yz = xy + x’z
The dual of Consensus theorem is-
(x + y)(x’ + z)(y + z) = (x + y)(x’ + z)
Consider any Boolean Expression-
xyz + x’yz’ + y’z = 1
The dual of this Boolean expression is-
(x + y + z)(x’ + y + z’)(y’ + z) = 0
|When a function is equal to it dual, it is called as a self dual function.|
Consider the function-
F (A , B , C) = AB + BC + CA
The dual of this function is-
Fd (A , B , C)
= (A + B)(B + C)(C + A)
= AB + BC + CA
Since, F (A , B , C) = Fd (A , B , C)
∴ F (A , B , C) is a self-dual function.
Conditions for a self-dual function-
The necessary and sufficient conditions for a function to be a self-dual function are-
- The function must be a Neutral function.
- The function must not contain mutually exclusive terms.
What are mutually exclusive terms?
Consider we have any term X consisting of some variables, then a term obtained by complementing each variable of term X is called as its mutually exclusive term.
Number of Self-dual Functions-
where n = number of Boolean variables in the function
- We know, for a function to be a self-dual function, the function must be a neutral function.
- For a function to be a neutral function, number of minterms must be equal to number of maxterms. So, we choose half of the terms i.e. 2n / 2 = 2n-1 terms.
- Now, for each of these terms, we have two choices whether to include it or not in the self-dual function.
Possible number of self-dual functions
= 2 x 2 x 2 x ……. x 2n-1
Relationship between Neutral Functions and Self-dual Functions-
Every self-dual function is surely a neutral function but every neutral function need not be a self-dual function.
Important Property of Self-dual Functions-
Self-duality is closed under complementation.
If the function F (A , B , C) = ∑ (0 , 1 , 2 , 4) is a self-dual function, then-
Its complement function F’ (A , B , C) = ∑ (3 , 5 , 6 , 7) will also be a self-dual function.
PRACTICE PROBLEM BASED ON SELF-DUAL FUNCTIONS-
Consider the following functions-
- F (A , B , C) = ∑ (0 , 2 , 3)
- F (A , B , C) = ∑ (0 , 1 , 6 , 7)
- F (A , B , C) = ∑ (0 , 1 , 2 , 4)
- F (A , B , C) = ∑ (3 , 5 , 6 , 7)
Which of the above functions are self-dual functions?
- Only (iii)
- Only (ii)
- Only (iii) and (iv)
- All are self-dual functions
According to condition-01, for a function to be a self-dual function, the function must be a neutral function.
- In all the given options, we have functions of 3 variables- A, B and C.
- For a function with 3 Boolean variables, neutral function must contain exactly 2n-1 = 23-1 = 4 minterms and 4 maxterms.
- But, Function-(i) contains only 3 minterms. So, it is not a neutral function and therefore it can’t be a self-dual function. So, Function-(i) gets eliminated.
- We are now left with three other functions which satisfies condition-01 and are all neutral functions. We will now use 2nd condition to eliminate the incorrect option(s).
According to condition-02, a self-dual function must not contain mutually exclusive terms.
First, let us find which terms are mutually exclusive-
From here, pairs of mutually exclusive terms are-
(0,7) , (1,6) , (2,5) , (3,4)
- Because mutually exclusive terms are not allowed in self-dual functions, therefore, terms inside the pairs can not appear together.
- But terms 0 and 7 appear together in the function-(ii). So, it can not be a self-dual function.
- But functions (iii) and (iv) do not contain any mutually exclusive terms.
∴ Functions (iii) and (iv) are self-dual functions.
Option (C) is correct.
Functions (iii) and (iv) are complementary functions. So, if one function is a self-dual function, the other function will also be a self-dual function because self-dual functions are closed under complementation.
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