Question:
The output \(Y\) for the inputs \(A\) and \(B\) of circuit is given by 
The output \(Y\) for the inputs \(A\) and \(B\) of circuit is given by
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Label the NAND outputs: \(N_1 = \overline{AB}\), \(N_2 = \overline{A N_1}\), \(N_3 = \overline{B N_1}\), \(Y = \overline{N_2 N_3}\). This is the standard 4-NAND XOR circuit.
Updated On: Nov 2, 2026
A B Y 0 0 0 0 1 1 1 0 1 1 1 0 A B Y 0 0 1 0 1 0 1 0 0 1 1 1 A B Y 0 0 1 0 1 1 1 0 1 1 1 0 A B Y 0 0 0 0 1 1 1 0 1 1 1 1
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The Correct Option is A
Solution and Explanation
Circuit (from the figure): gate 1 takes A and B: \(N_1 = \overline{A\cdot B}\). Gate 2 takes A and \(N_1\): \(N_2 = \overline{A\cdot N_1}\). Gate 3 takes B and \(N_1\): \(N_3 = \overline{B\cdot N_1}\). The last gate takes \(N_2, N_3\): \(Y = \overline{N_2\cdot N_3}\).
A=0, B=0: \(N_1=1\), \(N_2 = 1\), \(N_3=1\), \(Y=0\).
A=0, B=1: \(N_1=1\), \(N_2=1\), \(N_3=\overline{1\cdot1}=0\), \(Y=\overline{1\cdot0}=1\).
A=1, B=0: \(N_1=1\), \(N_2=0\), \(N_3=1\), \(Y=1\).
A=1, B=1: \(N_1=0\), \(N_2=1\), \(N_3=1\), \(Y=0\).
So Y = 0, 1, 1, 0 for AB = 00, 01, 10, 11, which is XOR. This matches option (A).
Answer: (1) the table with Y = 0, 1, 1, 0 for (A,B) = (0,0), (0,1), (1,0), (1,1)
A=0, B=0: \(N_1=1\), \(N_2 = 1\), \(N_3=1\), \(Y=0\).
A=0, B=1: \(N_1=1\), \(N_2=1\), \(N_3=\overline{1\cdot1}=0\), \(Y=\overline{1\cdot0}=1\).
A=1, B=0: \(N_1=1\), \(N_2=0\), \(N_3=1\), \(Y=1\).
A=1, B=1: \(N_1=0\), \(N_2=1\), \(N_3=1\), \(Y=0\).
So Y = 0, 1, 1, 0 for AB = 00, 01, 10, 11, which is XOR. This matches option (A).
Answer: (1) the table with Y = 0, 1, 1, 0 for (A,B) = (0,0), (0,1), (1,0), (1,1)
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