JEE Main 2021 · 26 February · evening · Single correct · +4 / −1
Draw the output signal Y in the given combination of gates.
- (a)
- (b)
- (c)
- (d)
Answer and solution
Answer (a)
Solution
Physics sir · 3 steps
Reduce the gates to one Boolean expression first, then read the waveforms interval by interval.
- NAND as NOT. Both inputs of the NAND gate are tied to A, so its output is A · A̅ = A̅. A NAND with its inputs joined is just a NOT gate.
- Build Y. The OR gate gives A̅ + B, and the final NOT inverts it. By De Morgan's theorem, A̅ + B̅ = A · B̅.
Y = A̅ + B̅ = A · B̅
Interval A B Y = A · B̅ 0 to 1 s 1 0 1 1 to 2 s 1 1 0 2 to 3 s 0 0 0 3 to 4 s 1 1 0 4 to 5 s 1 0 1 - So. Y is high only from 0 to 1 s and from 4 to 5 s, and low from 1 s to 4 s. That is the waveform in option (a).
The idea to keep. With a gate combination, never trace waveforms gate by gate blindly. Simplify to one expression like Y = AB̅, then check only where it is 1.
Answer from the NTA final answer key. Solution by PrepWiser.