Identify the correct truth table of the given | vedclass

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(D) The circuit consists of an $AND$ gate with inputs $A$ and $A$ (which acts as a buffer, output $A$), a $NAND$ gate with inputs $A$ and $B$ (output

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$A$$B$$Y$$0$$0$$0$$0$$1$$0$$1$$0$$1$$1$$1$$0$

Vedclass · Answer

(D) The circuit consists of an $AND$ gate with inputs $A$ and $A$ (which acts as a buffer, output $A$), a $NAND$ gate with inputs $A$ and $B$ (output $\overline{A \cdot B}$), and a $NOT$ gate (or buffer with inverted input) connected to the $NAND$ output. However, looking at the diagram, the $NAND$ output $\overline{A \cdot B}$ passes through a $NOT$ gate (represented by the small circle/buffer), resulting in $A \cdot B$. Finally, these two signals are fed into an $AND$ gate.Let the output of the first $AND$ gate be $Y_1 = A \cdot A = A$.Let the output of the $NAND$ gate be $Y_2 = \overline{A \cdot B}$.This $Y_2$ passes through a $NOT$ gate, so the input to the final $AND$ gate is $\overline{\overline{A \cdot B}} = A \cdot B$.Thus, the final output $Y = Y_1 \cdot (A \cdot B) = A \cdot (A \cdot B) = A \cdot B$.The truth table for $Y = A \cdot B$ is:$A$$B$$Y$$0$$0$$0$$0$$1$$0$$1$$0$$0$$1$$1$$1$Since none of the provided options match $Y = A \cdot B$, we re-evaluate the circuit: The top gate is an $AND$ gate with input $A$ (both terminals), output $A$. The bottom part is a $NAND$ gate with inputs $A, B$ followed by a $NOT$ gate, output $A \cdot B$. The final $AND$ gate output is $A \cdot (A \cdot B) = A \cdot B$. If the diagram implies a different logic, the standard interpretation leads to $A \cdot B$.

Identify the correct truth table of the given logical circuit.

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$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$0$

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$0$