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(B) The given circuit consists of an $AND$ gate,a $NOT$ gate,and an $OR$ gate. The inputs to the $AND$ gate are $A$ and $B$,so its output is $A \cdot

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The given circuit consists of an $AND$ gate,a $NOT$ gate,and an $OR$ gate. The inputs to the $AND$ gate are $A$ and $B$,so its output is $A \cdot B$. The input to the $NOT$ gate is $B$,so its output is $\bar{B}$. These two outputs are then fed into an $OR$ gate,so the final output is $Y = A \cdot B + \bar{B}$. Using Boolean algebra,$Y = A \cdot B + \bar{B} = (A + \bar{B}) \cdot (B + \bar{B}) = (A + \bar{B}) \cdot 1 = A + \bar{B}$. Now,we construct the truth table: $A$$B$$A \cdot B$$\bar{B}$$Y = A \cdot B + \bar{B}$$0$$0$$0$$1$$1$$0$$1$$0$$0$$0$$1$$0$$0$$1$$1$$1$$1$$1$$0$$1$

Find the truth table for the function $Y$ with inputs $A$ and $B$ represented in the following figure.

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

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

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

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