The truth table for the given logic circuit i | vedclass

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(B) The circuit consists of two $NAND$ gates whose outputs are fed into an $AND$ gate. Let the inputs be $A$ and $B$. The upper $NAND$ gate receives i

Vedclass · Accepted Answer

$A$$B$$C$$0$$0$$1$$0$$1$$0$$1$$0$$1$$1$$1$$0$

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(B) The circuit consists of two $NAND$ gates whose outputs are fed into an $AND$ gate. Let the inputs be $A$ and $B$. The upper $NAND$ gate receives inputs $A$ and $B$,so its output is $\overline{A \cdot B}$. The lower $NAND$ gate receives inputs $\bar{A}$ (from the $NOT$ gate) and $B$,so its output is $\overline{\bar{A} \cdot B}$. The final output $C$ is the $AND$ operation of these two outputs: $C = (\overline{A \cdot B}) \cdot (\overline{\bar{A} \cdot B})$.Using De Morgan's theorem,$\overline{A \cdot B} = \bar{A} + \bar{B}$ and $\overline{\bar{A} \cdot B} = A + \bar{B}$.Thus,$C = (\bar{A} + \bar{B}) \cdot (A + \bar{B})$.Using the distributive law,$C = \bar{A}A + \bar{A}\bar{B} + \bar{B}A + \bar{B}\bar{B}$.Since $\bar{A}A = 0$ and $\bar{B}\bar{B} = \bar{B}$,we get $C = 0 + \bar{B}(\bar{A} + A) + \bar{B} = \bar{B}(1) + \bar{B} = \bar{B} + \bar{B} = \bar{B}$.Therefore,the output $C$ is the $NOT$ of $B$ $(\bar{B})$.| $A$ | $B$ | $C$ ||---|---|---|| $0$ | $0$ | $1$ || $0$ | $1$ | $0$ || $1$ | $0$ | $1$ || $1$ | $1$ | $0$ |

The truth table for the given logic circuit is:

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