The truth table for the following logic circu | vedclass

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Question

(A) The given circuit consists of two $NOT$ gates, two $AND$ gates, and one $OR$ gate. The inputs are $A$ and $B$. The output of the top $AND$ gate is

Vedclass · Accepted Answer

$A = 0$, $B = 0$, $Y = 0$; $A = 0$, $B = 1$, $Y = 1$; $A = 1$, $B = 0$, $Y = 1$; $A = 1$, $B = 1$, $Y = 0$

Vedclass · Answer

(A) The given circuit consists of two $NOT$ gates, two $AND$ gates, and one $OR$ gate. The inputs are $A$ and $B$. The output of the top $AND$ gate is $\bar{A} \cdot B$ and the output of the bottom $AND$ gate is $A \cdot \bar{B}$. These are fed into an $OR$ gate, so the final output is $Y = \bar{A} \cdot B + A \cdot \bar{B}$. This is the Boolean expression for an $XOR$ gate. The truth table is as follows:$A$$B$$Y$$0$$0$$0$$0$$1$$1$$1$$0$$1$$1$$1$$0$

The truth table for the following logic circuit is

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The truth table for the given circuit is

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