The truth table shown below is for which logi | vedclass

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(C) The truth table represents the output $Y$ for inputs $A$ and $B$ as follows:For $A=0, B=0$,$Y=1$.For $A=0, B=1$,$Y=0$.For $A=1, B=0$,$Y=0$.For $A=

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$XNOR$

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(C) The truth table represents the output $Y$ for inputs $A$ and $B$ as follows:For $A=0, B=0$,$Y=1$.For $A=0, B=1$,$Y=0$.For $A=1, B=0$,$Y=0$.For $A=1, B=1$,$Y=1$.This truth table corresponds to the $XNOR$ gate,which outputs $1$ only when both inputs are the same.The Boolean expression for an $XNOR$ gate is $Y = A \odot B = \overline{A \oplus B} = \bar{A}\bar{B} + AB$.Checking the values:$1$. For $(0, 0): Y = \bar{0}\cdot\bar{0} + 0\cdot 0 = 1\cdot 1 + 0 = 1$.$2$. For $(0, 1): Y = \bar{0}\cdot\bar{1} + 0\cdot 1 = 1\cdot 0 + 0 = 0$.$3$. For $(1, 0): Y = \bar{1}\cdot\bar{0} + 1\cdot 0 = 0\cdot 1 + 0 = 0$.$4$. For $(1, 1): Y = \bar{1}\cdot\bar{1} + 1\cdot 1 = 0\cdot 0 + 1 = 1$.Thus,the correct option is $C$.

The truth table shown below is for which logic gate?
$A: 0, 0, 1, 1$
$B: 0, 1, 0, 1$
$Y: 1, 0, 0, 1$

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The truth table shown below is for which logic gate?$A: 0, 0, 1, 1$$B: 0, 1, 0, 1$$Y: 1, 0, 0, 1$