The combination of the gates shown in the following figure yields
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A Bubbled NAND gate (NAND with inverted inputs) acts as an OR gate. Similarly, a Bubbled AND gate acts as a NOR gate. Remembering these "bubbled" equivalencies saves time in logic circuit analysis!
Step 1: Understanding the Concept:
The given logic circuit consists of three NAND gates. A NAND gate with both inputs tied together functions as a NOT gate. We need to find the final boolean expression for the output of the circuit by evaluating the output at each stage. Step 2: Key Formula or Approach:
For a NAND gate with inputs $X$ and $Y$, the output is $\overline{X \cdot Y}$.
If the inputs are tied together ($X = Y$), the output becomes $\overline{X \cdot X} = \overline{X}$, which is the operation of a NOT gate.
De Morgan's Theorem: $\overline{X \cdot Y} = \overline{X} + \overline{Y}$. Step 3: Detailed Explanation:
Let's trace the signals through the circuit:
1. The top-left gate is a NAND gate with both inputs tied to $A$. Its output is $\overline{A \cdot A} = \overline{A}$.
2. The bottom-left gate is a NAND gate with both inputs tied to $B$. Its output is $\overline{B \cdot B} = \overline{B}$.
3. The final gate is a NAND gate taking the outputs of the first two gates as its inputs.
The final output $X$ is given by:
\[ X = \overline{\overline{A} \cdot \overline{B}} \]
Using De Morgan's Theorem, we can simplify this expression:
\[ X = \overline{\overline{A}} + \overline{\overline{B}} \]
\[ X = A + B \]
This boolean expression corresponds to an OR gate. Step 4: Final Answer:
The combination of gates acts as an OR gate. Therefore, the correct option is (B).
