The logic circuit shown has the input wavefor | vedclass

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(D) The circuit consists of two $NOT$ gates,one $AND$ gate,and one $NOR$ gate.Let the inputs be $A$ and $B$.The upper branch has a $NOT$ gate,so the i

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(D) The circuit consists of two $NOT$ gates,one $AND$ gate,and one $NOR$ gate.Let the inputs be $A$ and $B$.The upper branch has a $NOT$ gate,so the input to the $NOR$ gate is $\overline{A}$.The lower branch has a $NOT$ gate for $B$ and an $AND$ gate,so the input to the $NOR$ gate is $A \cdot \overline{B}$.The $NOR$ gate takes these two inputs and performs the $NOR$ operation:$Y = \overline{\overline{A} + (A \cdot \overline{B})}$Using De Morgan's Law: $\overline{X + Y} = \overline{X} \cdot \overline{Y}$$Y = \overline{\overline{A}} \cdot \overline{(A \cdot \overline{B})}$$Y = A \cdot (\overline{A} + \overline{\overline{B}})$$Y = A \cdot (\overline{A} + B)$$Y = (A \cdot \overline{A}) + (A \cdot B)$Since $A \cdot \overline{A} = 0$,we get:$Y = 0 + AB = AB$Thus,the circuit acts as an $AND$ gate. The output waveform $Y$ will be high only when both $A$ and $B$ are high.

$A$	$B$	Output
$1$	$0$	$1$
$1$	$1$	$0$
$0$	$1$	$1$
$0$	$0$	$0$

The logic circuit shown has the input waveforms $A$ and $B$ as shown. Pick out the correct output waveform.

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The correct truth table for the following logic circuit is:

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