Programming
October 17, 2023GATE 2008
October 18, 2023GATE 2009
| Question 6 |
What is the minimum number of gates required to implement the Boolean function (AB+C) if we have to use only 2-input NOR gates?
| 2 | |
| 3 | |
| 4 | |
| 5 |
Question 6 Explanation:
NOR is Complement of OR
AB+C
= (A+C)(B+C) ← Distribution of + over
= ((A+C)’+(B+C)’)’
1st NOR- (A+C)’. Let X = (A+C)’
2nd NOR- (B+C)’. Let Y = (B+C)’
3rd NOR- (X+Y)’
AB+C
= (A+C)(B+C) ← Distribution of + over
= ((A+C)’+(B+C)’)’
1st NOR- (A+C)’. Let X = (A+C)’
2nd NOR- (B+C)’. Let Y = (B+C)’
3rd NOR- (X+Y)’
Correct Answer: B
Question 6 Explanation:
NOR is Complement of OR
AB+C
= (A+C)(B+C) ← Distribution of + over
= ((A+C)’+(B+C)’)’
1st NOR- (A+C)’. Let X = (A+C)’
2nd NOR- (B+C)’. Let Y = (B+C)’
3rd NOR- (X+Y)’
AB+C
= (A+C)(B+C) ← Distribution of + over
= ((A+C)’+(B+C)’)’
1st NOR- (A+C)’. Let X = (A+C)’
2nd NOR- (B+C)’. Let Y = (B+C)’
3rd NOR- (X+Y)’