The truth table for the system of four NAND g | vedclass

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(A) Let the output of the first $NAND$ gate be $Y_1 = \overline{A \cdot B}$.This $Y_1$ is fed as an input to the next two $NAND$ gates.The outputs of

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the output of the first $NAND$ gate be $Y_1 = \overline{A \cdot B}$.This $Y_1$ is fed as an input to the next two $NAND$ gates.The outputs of these two gates are $Y_2 = \overline{A \cdot Y_1} = \overline{A \cdot \overline{A \cdot B}}$ and $Y_3 = \overline{B \cdot Y_1} = \overline{B \cdot \overline{A \cdot B}}$.These are then fed into the final $NAND$ gate,so $Y = \overline{Y_2 \cdot Y_3} = \overline{\overline{A \cdot \overline{A \cdot B}} \cdot \overline{B \cdot \overline{A \cdot B}}}$.Using De Morgan's theorem,$Y = \overline{\overline{A \cdot \overline{A \cdot B}}} + \overline{\overline{B \cdot \overline{A \cdot B}}} = (A \cdot \overline{A \cdot B}) + (B \cdot \overline{A \cdot B})$.$Y = (A \cdot (\overline{A} + \overline{B})) + (B \cdot (\overline{A} + \overline{B})) = (A \cdot \overline{A} + A \cdot \overline{B}) + (B \cdot \overline{A} + B \cdot \overline{B})$.Since $A \cdot \overline{A} = 0$ and $B \cdot \overline{B} = 0$,we get $Y = A \cdot \overline{B} + \overline{A} \cdot B$,which is the $XOR$ gate operation.The truth table for $XOR$ is:- If $A=0, B=0$,then $Y=0$.- If $A=0, B=1$,then $Y=1$.- If $A=1, B=0$,then $Y=1$.- If $A=1, B=1$,then $Y=0$.This matches option $A$.

The truth table for the system of four $NAND$ gates as shown in the figure is:

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

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

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

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