Question

What does the following combination of gates produce?

• A. NAND gate
• B. NOR gate
• C. XOR gate
• D. AND gate

Hint:

Two NOT gates feed a NOR followed by a NOT. Write the Boolean expression and use De Morgan's theorem.

Answer 1

The Correct Option is A

Step 1: Read the circuit
Input \(A\) goes through gate \(G_1\), a NOT gate. Input \(B\) goes through \(G_2\), also a NOT gate. Their outputs \(\bar A\) and \(\bar B\) enter \(G_3\), a NOR gate. The output of \(G_3\) goes to \(G_4\), a NOT gate, giving \(Y\).

Step 2: Write the expression
\(G_3\) output: \(\overline{\bar A+\bar B}\). \(G_4\) inverts it: \[ Y=\overline{\overline{\bar A+\bar B}}=\bar A+\bar B \]

Step 3: Apply De Morgan's theorem
\(\bar A+\bar B=\overline{A\cdot B}\), so \(Y=\overline{AB}\).

Step 4: Truth table check
A=0,B=0 gives \(Y=1\). A=0,B=1 gives \(Y=1\). A=1,B=0 gives \(Y=1\). A=1,B=1 gives \(Y=0\). This is the NAND truth table.

Step 5: Why the others fail
NOR gives 1 only for 0,0. XOR gives 0 for 0,0. AND gives 1 only for 1,1. None matches the table above.

Final Answer:
The combination acts as a NAND gate, option (A). \[ \boxed{Y=\overline{A\cdot B}\ \text{(NAND)}} \]