What will be the outputs Y for the inputs (A | vedclass

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(D) Let the output of the first $NAND$ gate be $P = \overline{A \cdot B}$.The inputs to the top $NAND$ gate are $A$ and $P$,so its output is $Q = \ove

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$1, 0, 0$

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(D) Let the output of the first $NAND$ gate be $P = \overline{A \cdot B}$.The inputs to the top $NAND$ gate are $A$ and $P$,so its output is $Q = \overline{A \cdot P} = \overline{A \cdot \overline{A \cdot B}} = \overline{A} + (A \cdot B) = \overline{A} + B$.The inputs to the bottom $NAND$ gate are $P$ and $B$,so its output is $R = \overline{P \cdot B} = \overline{\overline{A \cdot B} \cdot B} = (A \cdot B) + \overline{B} = A + \overline{B}$.The final output $Y$ is the output of the last $NAND$ gate with inputs $Q$ and $R$:$Y = \overline{Q \cdot R} = \overline{Q} + \overline{R} = \overline{\overline{A} + B} + \overline{A + \overline{B}} = (A \cdot \overline{B}) + (\overline{A} \cdot B)$.This is the expression for an $XOR$ gate.$1$. For $(A = 1, B = 0)$: $Y = (1 \cdot \overline{0}) + (\overline{1} \cdot 0) = (1 \cdot 1) + (0 \cdot 0) = 1 + 0 = 1$.$2$. For $(A = 1, B = 1)$: $Y = (1 \cdot \overline{1}) + (\overline{1} \cdot 1) = (1 \cdot 0) + (0 \cdot 1) = 0 + 0 = 0$.$3$. For $(A = 0, B = 0)$: $Y = (0 \cdot \overline{0}) + (\overline{0} \cdot 0) = (0 \cdot 1) + (1 \cdot 0) = 0 + 0 = 0$.Thus,the outputs are $1, 0, 0$.

What will be the outputs $Y$ for the inputs $(A = 1, B = 0)$,$(A = 1, B = 1)$,and $(A = 0, B = 0)$ respectively?

Similar Questions

The combination of gates shown below yields:

The following logic gate circuit is equivalent to which of the following gates?

The truth table given below corresponds to which logic gate?
$A$ $B$ $Y$
$0$ $0$ $0$
$0$ $1$ $1$
$1$ $0$ $1$
$1$ $1$ $0$

The correct truth table for the given logic gate circuit is:

The truth table given below is for which gate?
$A: 0, 0, 1, 1$
$B: 0, 1, 0, 1$
$C: 1, 1, 1, 0$

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