A system of four gates is set up as shown. Th | vedclass

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(A) In the given system, all four gates are $NOR$ gates. Let the output of the first gate be $Y' = \overline{A+B}$.This $Y'$ is fed into the next two

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

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(A) In the given system, all four gates are $NOR$ gates. Let the output of the first gate be $Y' = \overline{A+B}$.This $Y'$ is fed into the next two $NOR$ gates along with inputs $A$ and $B$ respectively.The outputs of these two gates are $Y'' = \overline{A+Y'}$ and $Y''' = \overline{B+Y'}$.Finally, these are fed into the last $NOR$ gate, so $Y = \overline{Y''+Y'''}$.Truth Table Calculation:$A$$B$$Y' = \overline{A+B}$$Y'' = \overline{A+Y'}$$Y''' = \overline{B+Y'}$$Y = \overline{Y''+Y'''}$$0$$0$$1$$0$$0$$1$$0$$1$$0$$1$$0$$0$$1$$0$$0$$0$$1$$0$$1$$1$$0$$0$$0$$1$This corresponds to the $XNOR$ gate logic.

$A$ system of four gates is set up as shown. The truth table corresponding to this system is

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

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

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

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