The truth table for the following logic circuit is:
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The logic circuit shown is a standard implementation of an XOR gate using AND, OR, and NOT gates. The output of an XOR gate is true (1) if and only if the inputs are different.
Let's analyze the logic circuit step-by-step.
The circuit has two inputs, A and B. There are two NOT gates, two AND gates, and one OR gate.
The input to the top AND gate are A and the output of the NOT gate connected to B. So, the inputs are A and $\bar{B}$. The output of this AND gate is $X_1 = A \cdot \bar{B}$.
The input to the bottom AND gate are B and the output of the NOT gate connected to A. So, the inputs are B and $\bar{A}$. The output of this AND gate is $X_2 = B \cdot \bar{A}$.
The outputs of the two AND gates, $X_1$ and $X_2$, are the inputs to the final OR gate.
The final output is $Y = X_1 + X_2 = (A \cdot \bar{B}) + (\bar{A} \cdot B)$.
This is the Boolean expression for the Exclusive OR (XOR) gate.
Let's construct the truth table for this expression:
Case 1: A=0, B=0. $Y = (0 \cdot \bar{0}) + (\bar{0} \cdot 0) = (0 \cdot 1) + (1 \cdot 0) = 0 + 0 = 0$.
Case 2: A=0, B=1. $Y = (0 \cdot \bar{1}) + (\bar{0} \cdot 1) = (0 \cdot 0) + (1 \cdot 1) = 0 + 1 = 1$.
Case 3: A=1, B=0. $Y = (1 \cdot \bar{0}) + (\bar{1} \cdot 0) = (1 \cdot 1) + (0 \cdot 0) = 1 + 0 = 1$.
Case 4: A=1, B=1. $Y = (1 \cdot \bar{1}) + (\bar{1} \cdot 1) = (1 \cdot 0) + (0 \cdot 1) = 0 + 0 = 0$.
The final truth table is:
A | B | Y
--|---|--
0 | 0 | 0
0 | 1 | 1
1 | 0 | 1
1 | 1 | 0
This matches the truth table given in option (A).
