Question

Four NOR gates are connected as shown in figure. The truth table for the given figure is : 

• A. A
• B. B
• C. C
• D. D

Hint:

When analyzing complex logic circuits, it's often easier to use Boolean algebra to simplify the expression rather than testing each input combination through every gate. Remembering De Morgan's theorems (\(\overline{A+B} = \bar{A}\cdot\bar{B}\) and \(\overline{A\cdot B} = \bar{A}+\bar{B}\)) is crucial for simplification. Recognizing standard gate combinations (like this XNOR gate) saves time.

Answer 1

The Correct Option is A

Step 1: Understanding the Question: 
We are given a logic circuit made of four NOR gates and two inputs, A and B. We need to find the truth table for the final output Y. 
Step 2: Key Formula or Approach: 
We will trace the logic signals through the circuit for each possible input combination (00, 01, 10, 11). The Boolean expression for a NOR gate with inputs X and Z is \( \overline{X+Z} \). 
Step 3: Detailed Explanation: 
Let's label the intermediate outputs. 
- The output of the top-left NOR gate (inputs A, B) be \( Y_1 \). \[ Y_1 = \overline{A+B} \] - The output of the middle NOR gate (inputs A, \(Y_1\)) be \( Y_2 \). \[ Y_2 = \overline{A+Y_1} = \overline{A + \overline{A+B}} \] - The output of the bottom NOR gate (inputs B, \(Y_1\)) be \( Y_3 \). \[ Y_3 = \overline{B+Y_1} = \overline{B + \overline{A+B}} \] - The output of the final NOR gate (inputs \(Y_2\), \(Y_3\)) is Y. \[ Y = \overline{Y_2+Y_3} = \overline{(\overline{A + \overline{A+B}}) + (\overline{B + \overline{A+B}})} \] Let's simplify this using Boolean algebra: 
First, simplify \( Y_2 \) and \( Y_3 \). Using De Morgan's theorem (\(\overline{X+Z} = \bar{X}\cdot\bar{Z}\)): 
\[ Y_2 = \overline{A + \overline{A+B}} = \bar{A} \cdot \overline{(\overline{A+B})} = \bar{A} \cdot (A+B) = \bar{A}A + \bar{A}B = 0 + \bar{A}B = \bar{A}B \] \[ Y_3 = \overline{B + \overline{A+B}} = \bar{B} \cdot \overline{(\overline{A+B})} = \bar{B} \cdot (A+B) = A\bar{B} + \bar{B}B = A\bar{B} + 0 = A\bar{B} \] Now, substitute these into the expression for Y: 
\[ Y = \overline{Y_2 + Y_3} = \overline{\bar{A}B + A\bar{B}} \] The expression \( \bar{A}B + A\bar{B} \) is the definition of the XOR operation (\(A \oplus B\)). 
Therefore, \( Y = \overline{A \oplus B} \), which is the XNOR operation. 
Let's construct the truth table for Y = XNOR(A, B): 
- If A=0, B=0: \( Y = \overline{0 \oplus 0} = \overline{0} = 1 \) 
- If A=0, B=1: \( Y = \overline{0 \oplus 1} = \overline{1} = 0 \) 
- If A=1, B=0: \( Y = \overline{1 \oplus 0} = \overline{1} = 0 \) 
- If A=1, B=1: \( Y = \overline{1 \oplus 1} = \overline{0} = 1 \) 
Step 4: Final Answer: 
The resulting truth table is: 
This matches the truth table in option (A).