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

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Question

(D) Let the inputs be $A$ and $B$. The first $NAND$ gate produces output $C = \overline{A.B}$.The next two $NAND$ gates produce outputs $D = \overline

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the inputs be $A$ and $B$. The first $NAND$ gate produces output $C = \overline{A.B}$.The next two $NAND$ gates produce outputs $D = \overline{A.C}$ and $E = \overline{C.B}$.The final $NAND$ gate produces output $Y = \overline{D.E}$.Substituting $C = \overline{A.B}$:$D = \overline{A.(\overline{A.B})} = \overline{A.(\overline{A} + \overline{B})} = \overline{A.\overline{A} + A.\overline{B}} = \overline{0 + A.\overline{B}} = \overline{A.\overline{B}} = \overline{A} + B$.Similarly,$E = \overline{(\overline{A.B}).B} = \overline{\overline{A}.B + B.\overline{B}} = \overline{\overline{A}.B + 0} = \overline{\overline{A}.B} = A + \overline{B}$.Now,$Y = \overline{D.E} = \overline{(\overline{A} + B).(A + \overline{B})} = \overline{\overline{A}.A + \overline{A}.\overline{B} + B.A + B.\overline{B}} = \overline{0 + \overline{A}.\overline{B} + A.B + 0} = \overline{\overline{A}.\overline{B} + A.B}$.This is the expression for the $XNOR$ gate,which outputs $1$ when $A=B$ and $0$ when $A 
eq B$.Thus,the truth table is:$A$$B$$Y$$0$$0$$1$$0$$1$$0$$1$$0$$0$$1$$1$$1$

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

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