Question

The truth table for the given logic circuit is:

• A. A B C
  0 0 0
  0 1 1
  1 0 1
  1 1 0
• B. A B C
  0 0 1
  0 1 0
  1 0 0
  1 1 1
• C. A B C
  0 0 1
  0 1 0
  1 0 1
  1 1 0
• D. A B C
  0 0 0
  0 1 1
  1 0 0
  1 1 1

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Logic circuits are analyzed by determining the Boolean output of each gate based on its inputs. NAND gate logic is \(\overline{A \cdot B}\).
Step 2: Detailed Explanation:
The given circuit uses four NAND gates.
1. Let the output of the first NAND gate (fed by A and B) be \(Y_1 = \overline{A \cdot B}\).
2. The second NAND gate is fed by A and \(Y_1\). Its output is:
\[ Y_2 = \overline{A \cdot Y_1} = \overline{A \cdot (\overline{A \cdot B})} = \overline{A \cdot (\overline{A} + \overline{B})} = \overline{A\overline{A} + A\overline{B}} = \overline{0 + A\overline{B}} = \overline{A\overline{B}} \]
3. The third NAND gate is fed by B and \(Y_1\). Its output is:
\[ Y_3 = \overline{B \cdot Y_1} = \overline{B \cdot (\overline{A \cdot B})} = \overline{B \cdot (\overline{A} + \overline{B})} = \overline{B\overline{A} + B\overline{B}} = \overline{B\overline{A} + 0} = \overline{\overline{A}B} \]
4. The final NAND gate is fed by \(Y_2\) and \(Y_3\). The final output \(C\) is:
\[ C = \overline{Y_2 \cdot Y_3} = \overline{\overline{A\overline{B}} \cdot \overline{\overline{A}B}} = (A\overline{B}) + (\overline{A}B) \]
The expression \(A\overline{B} + \overline{A}B\) is the Boolean equation for an **XOR gate**.
XOR gate truth table:
- \(A=0, B=0 \implies C = 0 \cdot 1 + 1 \cdot 0 = 0\)
- \(A=0, B=1 \implies C = 0 \cdot 0 + 1 \cdot 1 = 1\)
- \(A=1, B=0 \implies C = 1 \cdot 1 + 0 \cdot 0 = 1\)
- \(A=1, B=1 \implies C = 1 \cdot 0 + 0 \cdot 1 = 0\)
This matches Option (3).
Step 3: Final Answer:
The circuit behaves as an XOR gate. The correct truth table is 00 \(\rightarrow\) 0, 01 \(\rightarrow\) 1, 10 \(\rightarrow\) 1, 11 \(\rightarrow\) 0.