Question

The dual of X.Y + Y.0.Z’ is ?​

Answer 1

Answer:

Explanation:

Dual Of 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

 

Dual of Boolean Expression Examples-

 

Following are examples of dual of Boolean Expressions-

 

Example-01:

 

Consensus  theorem is xy + x’z + yz = xy + x’z

Dual of Consensus theorem is (x + y)(x’ + z)(y + z) = (x + y)(x’ + z)

 

Example-02:

 

Boolean expression is xyz + x’yz’ + y’z = 1

Dual of the above Boolean expression is (x + y + z)(x’ + y + z’)(y’ + z) = 0

 

Self-Dual Functions-

 

When a function is equal to its dual, it is called as a Self dual function.

 

Example-

 

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

 

Clearly, F (A , B , C) = Fd (A , B , C)

∴ F (A , B , C) is a self-dual function.

 

Conditions For Self-Dual Function-

 

The necessary and sufficient conditions for any function to be a self-dual function are-

The function must be a Neutral Function.

The function must not contain any mutually exclusive terms.

 

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.

 

Examples-

 

(ABC , A’B’C’) are mutually exclusive terms.

(AB’C , A’BC’) are mutually exclusive terms.

 

Number of Self-Dual Functions-

 

 

Here n = number of Boolean variables in the function.

 

Explanation-

 

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.

 

So, possible number of self-dual functions

= 2 x 2 x 2 x ……. x 2n-1

= 22^(n-1)

 

Relationship Between Neutral Functions & 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.

 

Example-

 

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-

 

Problem-

 

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

 

Solution-

 

Condition-01:

 

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.

So, 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.

Therefore, it can’t be a self-dual function and it 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).

 

Condition-02:

 

According to condition-02, a self-dual function must not contain mutually exclusive terms.

First, let us find which terms are mutually exclusive-

 

A B C Minterms

0 0 0 0 A’B’C’

1 0 0 1 A’B’C

2 0 1 0 A’BC’

3 0 1 1 A’BC

4 1 0 0 AB’C’

5 1 0 1 AB’C

6 1 1 0 ABC’

7 1 1 1 ABC

 

From here, pairs of mutually exclusive terms are (0,7) , (1,6) , (2,5) , (3,4).

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.

Therefore, functions (iii) and (iv) are self-dual functions.

 

Thus, Option (C) is correct.

 

NOTE-

 

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.

This is because self-dual functions are closed under complementation.

Answer 2

Given:

Boolean expression: $X.Y+Y.0.Z'$

To Find:

The dual of the given Boolean expression $X.Y+Y.0.Z'$.

Solution:

The dual of a Boolean expression is an expression obtained by:

• Changing each OR operator $\[\left( + \right)\]$ to AND operator $\[\left( . \right)\]$ and vice-versa.
• Changing each $1$ to $0$ and each $0$ to $1$.

The dual of a function $F$ is denoted by $\[{F^d}\]$.

Hence, the dual of the Boolean expression $\[F\left( {X,Y,Z} \right) = X.Y + Y.0.Z'\]$ is given by:

$\[{F^d} = \left( {X + Y} \right).\left( {Y + 1 + Z'} \right)\]$

Hence, the dual of the given Boolean expression $X.Y+Y.0.Z'$ is $\[\left( {X + Y} \right).\left( {Y + 1 + Z'} \right)\]$.

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