Step 1: Understanding the Concept:
The circuit diagram consists exclusively of NAND gates (D-shaped symbols with a small inversion bubble at the output).
We need to trace the boolean logic through each stage of the circuit from inputs \(X\) and \(Y\) to the final output. Step 2: Key Formula or Approach:
The boolean expression for a NAND gate with inputs \(A\) and \(B\) is \(\overline{A \cdot B}\).
If both inputs of a NAND gate are tied together (i.e., \(A = B\)), it acts as a NOT gate: \(\overline{A \cdot A} = \overline{A}\).
De Morgan's Theorem: \(\overline{A \cdot B} = \overline{A} + \overline{B}\) and \(\overline{A + B} = \overline{A} \cdot \overline{B}\). Step 3: Detailed Explanation:
Stage 1: The input \(X\) is fed into both terminals of the first NAND gate.
Output of the top gate = \(\overline{X \cdot X} = \overline{X}\).
Similarly, the input \(Y\) is fed into both terminals of the second NAND gate.
Output of the bottom gate = \(\overline{Y \cdot Y} = \overline{Y}\).
Stage 2: These two signals, \(\overline{X}\) and \(\overline{Y}\), are used as inputs for the third NAND gate.
Output of the third gate = \(\overline{\overline{X} \cdot \overline{Y}}\).
Using De Morgan's theorem, we simplify this:
\(\overline{\overline{X} \cdot \overline{Y}} = \overline{\overline{X}} + \overline{\overline{Y}} = X + Y\).
(Notice that this intermediate stage creates an OR gate).
Stage 3: The output \(X + Y\) is fed into both terminals of the final fourth NAND gate.
Output of the final gate = \(\overline{(X + Y) \cdot (X + Y)} = \overline{X + Y}\).
The boolean expression \(\overline{X + Y}\) represents the logic of a NOR gate. Step 4: Final Answer:
The given circuit works as a NOR gate.
