JEE Main 2020 · 6 September · morning · Single correct · +4 / −1
Identify the correct output signal Y in the given combination of gates (as shown) for the given inputs A and B.
- (a)
- (b)
- (c)
- (d)
Answer and solution
Answer (d)
Solution
Physics sir · 3 steps
Two NOT gates feeding a NAND is just an OR gate in disguise, so Y is low only when A and B are both low.
- Reduce the circuit. The NAND gate receives Ā and B̄, so Y = Ā·B̄̅. De Morgan's theorem turns this into a plain OR.
Y = Ā·B̄̅ = Ā̄̄ + B̄̄̄ = A + B
- Read the waveforms. Split the time axis wherever A or B changes. Then mark Y as 0 only in the intervals where both inputs are 0.
Time interval A B Y = A + B 0 to 7.5 0, then 1 after 5 1 1 7.5 to 10 0 0 0 10 to 15 0, then 1 after 12.5 1 1 15 to about 17 0 0 0 about 17 onwards 1, then 0 after about 19 0, then 1 after about 19 1 - So. Y is high everywhere except two short gaps, from 7.5 to 10 and from 15 to about 17. Only option (d) has this shape.
The idea to keep. Bubbles on the inputs of a NAND turn it into an OR, and bubbles on the inputs of a NOR turn it into an AND. Simplify the gate first, then read the waveforms interval by interval.
Answer from the NTA final answer key. Solution by PrepWiser.