Question

In the following logic circuit the sequence of the inputs A, B are (0, 0), (0, 1), (1, 0) and (1, 1). The output Y for this sequence will be : 

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

Hint:

Notice that \( (A \cdot B) \cdot (A + B) \) simplifies to \( A \cdot B \) because \( AB \cdot A + AB \cdot B = AB + AB = AB \).
Thus, the whole circuit behaves exactly like a single NAND gate \( Y = \overline{AB} \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The circuit consists of three logic gates.
The first gate (top) is an AND gate with inputs \( A \) and \( B \).
The second gate (bottom) is an OR gate with inputs \( A \) and \( B \).
The third gate (right) is a NAND gate that takes the outputs of the first two gates as its inputs.
Step 2: Key Formula or Approach:
Let \( P \) be the output of the AND gate: \( P = A \cdot B \).
Let \( Q \) be the output of the OR gate: \( Q = A + B \).
The final output \( Y \) is the NAND of \( P \) and \( Q \):
\[ Y = \overline{P \cdot Q} = \overline{(A \cdot B) \cdot (A + B)} \]
Step 3: Detailed Explanation:
Let's evaluate the output for each input pair:
1. For \( (0, 0) \): \( P = 0 \cdot 0 = 0 \), \( Q = 0 + 0 = 0 \). Output \( Y = \overline{0 \cdot 0} = \overline{0} = 1 \).
2. For \( (0, 1) \): \( P = 0 \cdot 1 = 0 \), \( Q = 0 + 1 = 1 \). Output \( Y = \overline{0 \cdot 1} = \overline{0} = 1 \).
3. For \( (1, 0) \): \( P = 1 \cdot 0 = 0 \), \( Q = 1 + 0 = 1 \). Output \( Y = \overline{0 \cdot 1} = \overline{0} = 1 \).
4. For \( (1, 1) \): \( P = 1 \cdot 1 = 1 \), \( Q = 1 + 1 = 1 \). Output \( Y = \overline{1 \cdot 1} = \overline{1} = 0 \).
The resulting sequence of outputs is 1, 1, 1, 0.
Step 4: Final Answer:
The sequence for the given inputs is 1, 1, 1, 0.