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(D) Let the output of the $NOR$ gate be $Y$. The $NOR$ gate performs the operation $Y = \overline{A+B}$. The $NAND$ gate then takes $Y$ and $C$ as inp

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$1$ and $1$

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(D) Let the output of the $NOR$ gate be $Y$. The $NOR$ gate performs the operation $Y = \overline{A+B}$. The $NAND$ gate then takes $Y$ and $C$ as inputs to produce output $D = \overline{Y \cdot C}$.Case $1$: $A=0, B=0, C=0$$Y = \overline{0+0} = \overline{0} = 1$$D = \overline{Y \cdot C} = \overline{1 \cdot 0} = \overline{0} = 1$Case $2$: $A=1, B=0, C=1$$Y = \overline{1+0} = \overline{1} = 0$$D = \overline{Y \cdot C} = \overline{0 \cdot 1} = \overline{0} = 1$Thus, the outputs are $1$ and $1$ respectively.

Input $A$	Input $B$	Output $Y$
$0$	$0$	$1$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

Similar Questions

Draw the output signal $Y$ for the given combination of logic gates shown in the figure.

The truth table of a logic gate is given below. Identify the gate.
Input $A$ Input $B$ Output $Y$
$0$ $0$ $1$
$0$ $1$ $1$
$1$ $0$ $1$
$1$ $1$ $0$

The output $(Y)$ of the given logic implementation is similar to the output of a . . . . . . gate.

The resultant logic gate from the combination of the following gates is:

The truth table for the given circuit is:

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