Question:
Draw the output signal Y in the given combination of gates.
Draw the output signal Y in the given combination of gates.
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The NAND gate with both inputs joined acts as NOT A; then OR with B and a final NOT gives \(Y=\overline{\bar A+B}=A\cdot\bar B\).
Updated On: Nov 4, 2026
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The Correct Option is A
Solution and Explanation
The first gate is a NAND with both inputs tied to A, so its output is \(\bar A\). This is ORed with B and passed through a NOT gate:
\(Y=\overline{\bar A+B}=A\cdot\bar B\) (De Morgan).
From the graphs: A is high for 0-2 s, low for 2-3 s, high for 3-5 s. B is high for 1-2 s and 3-4 s, low otherwise.
0-1 s: A=1, B=0, Y=1.
1-2 s: A=1, B=1, Y=0.
2-3 s: A=0, B=0, Y=0.
3-4 s: A=1, B=1, Y=0.
4-5 s: A=1, B=0, Y=1.
So Y is high for 0-1 s, low for 1-4 s, and high for 4-5 s, which is option (1).
Answer: (1) the waveform that is high from 0 to 1 s and from 4 to 5 s, low otherwise
\(Y=\overline{\bar A+B}=A\cdot\bar B\) (De Morgan).
From the graphs: A is high for 0-2 s, low for 2-3 s, high for 3-5 s. B is high for 1-2 s and 3-4 s, low otherwise.
0-1 s: A=1, B=0, Y=1.
1-2 s: A=1, B=1, Y=0.
2-3 s: A=0, B=0, Y=0.
3-4 s: A=1, B=1, Y=0.
4-5 s: A=1, B=0, Y=1.
So Y is high for 0-1 s, low for 1-4 s, and high for 4-5 s, which is option (1).
Answer: (1) the waveform that is high from 0 to 1 s and from 4 to 5 s, low otherwise
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