Question

Show that the dual of the exclusive -OR is equal to its compliment

Answer 1

HELLO THERE!

Look at the picture to get your answer!

Expression of XOR Gate is :

$A^{\prime}B+AB^{\prime}$ .....(i)

To find the Dual of an expression, change the '.' sign with '+' and '+' sign with '.', you can also apply De-Morgan's Theorem.

So, Dual of the expression of XOR gate is:

$(A^{\prime}+B).(A+B^{\prime})$ ......(ii)

From the Truth Table that I have given, it is clear that (ii) is the complement of (i).

HOPE THIS HELPS!

Thanks...

Answer 2

Explanation:

HELLO THERE!

Look at the picture to get your answer!

Expression of XOR Gate is :

A^{\prime}B+AB^{\prime}A

′

B+AB

′

.....(i)

To find the Dual of an expression, change the '.' sign with '+' and '+' sign with '.', you can also apply De-Morgan's Theorem.

So, Dual of the expression of XOR gate is:

(A^{\prime}+B).(A+B^{\prime})(A

′

+B).(A+B

′

) ......(ii)

From the Truth Table that I have given, it is clear that (ii) is the complement of (i).

HOPE THIS HELPS!

Thanks...