Topic Name : Boolean Algebra

Question

Minimum no of NOR gate required to implement the fxn :

1) y=AB'+ A'B.

2)y=(A'+B)(B+C')(A'+D)(C'+D)

Answer 1

In y=AB'+A'B, we need two extra NOT gate to convert into NOT (to get A' and B').

y=(A'+B)(B+C')(A'+D)(C'+D), we need two NOT gate to convert into A' and C'