To obtain the truth-table shown, from the fol | vedclass

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(D) The output of the circuit is given by $Y = A + C$, where $C$ is the output of gate $G$ with inputs $A$ and $B$. Thus, $C = A 	ext{ (gate } G) B$.

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$NOR$

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(D) The output of the circuit is given by $Y = A + C$, where $C$ is the output of gate $G$ with inputs $A$ and $B$. Thus, $C = A 	ext{ (gate } G) B$.Let us test the options for gate $G$:If $G$ is $NOR$, then $C = \overline{A+B}$.The output $Y = A + \overline{A+B}$.Let's verify this for all inputs:$1$. For $A=0, B=0$: $C = \overline{0+0} = 1$. Then $Y = 0 + 1 = 1$. (Matches)$2$. For $A=0, B=1$: $C = \overline{0+1} = 0$. Then $Y = 0 + 0 = 0$. (Matches)$3$. For $A=1, B=0$: $C = \overline{1+0} = 0$. Then $Y = 1 + 0 = 1$. (Matches)$4$. For $A=1, B=1$: $C = \overline{1+1} = 0$. Then $Y = 1 + 0 = 1$. (Matches)Since all values match the given truth table, the gate $G$ is $NOR$.

To obtain the truth-table shown, from the following logic circuit, the gate $G$ should be
$A$ $B$ $Y$
$0$ $0$ $1$
$0$ $1$ $0$
$1$ $0$ $1$
$1$ $1$ $1$

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To obtain the truth-table shown, from the following logic circuit, the gate $G$ should be$A$$B$$Y$$0$$0$$1$$0$$1$$0$$1$$0$$1$$1$$1$$1$