Step 1: {Understanding the Logic Circuit} The given circuit consists of: - NOT gates applied to inputs \( A \) and \( B \). - AND gate processing the inverted signals. Step 2: {Deriving the Boolean Expression} The circuit implements: \[ Y = \overline{A} \cdot \overline{B} \] Applying De-Morgan’s theorem: \[ Y = \overline{(A + B)} \] Step 3: {Comparing with Output Waveforms} Analyzing the truth table, we match the waveform and determine that the correct output is given by waveform 4. Thus, the correct answer is (D).
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Approach Solution -2
Step 1: Understanding the Logic Circuit The given circuit contains two inputs, \( A \) and \( B \). Each of these inputs is first passed through a NOT gate, producing the inverted signals \( \overline{A} \) and \( \overline{B} \). These inverted outputs are then fed into an AND gate.
Step 2: Deriving the Boolean Expression The output of the AND gate with inverted inputs is: \[ Y = \overline{A} \cdot \overline{B} \] We can now apply De Morgan’s Theorem to this expression: \[ \overline{A} \cdot \overline{B} = \overline{A + B} \] So, the final expression becomes: \[ Y = \overline{(A + B)} \] This is the standard Boolean expression for a NOR gate.
Step 3: Comparing with Output Waveforms / Logic Behavior Let’s interpret the behavior of the output from the expression \( Y = \overline{(A + B)} \). This logic gate produces a high output (1) only when both inputs are 0. If either \( A \), \( B \), or both are 1, the output becomes 0.
Here is the truth table for verification:
A
B
A + B
Y = &overline;(A + B)
0
0
0
1
0
1
1
0
1
0
1
0
1
1
1
0
Conclusion: The circuit's output logic matches that of a NOR gate. Hence, the logic operation performed by the circuit is: \[ \boxed{Y = \overline{A + B}} \Rightarrow \text{NOR gate} \]
