Draw the output signal Y for the given combin | vedclass

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(D) The given circuit consists of a $NAND$ gate,an $OR$ gate,and a $NOT$ gate. Let the inputs be $A$ and $B$. The output of the $NAND$ gate is $\overl

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(D) The given circuit consists of a $NAND$ gate,an $OR$ gate,and a $NOT$ gate. Let the inputs be $A$ and $B$. The output of the $NAND$ gate is $\overline{A \cdot A} = \overline{A}$.This output $\overline{A}$ and input $B$ are fed into an $OR$ gate,giving an intermediate output $Z = \overline{A} + B$.Finally,this is passed through a $NOT$ gate to give the final output $Y = \overline{Z} = \overline{\overline{A} + B}$.Using De Morgan's law,$\overline{\overline{A} + B} = \overline{\overline{A}} \cdot \overline{B} = A \cdot \overline{B}$.Now,we analyze the truth table for $Y = A \cdot \overline{B}$:- For $t = 0$ to $1$ s: $A=1, B=0 \implies Y = 1 \cdot \overline{0} = 1 \cdot 1 = 1$.- For $t = 1$ to $2$ s: $A=1, B=1 \implies Y = 1 \cdot \overline{1} = 1 \cdot 0 = 0$.- For $t = 2$ to $3$ s: $A=0, B=0 \implies Y = 0 \cdot \overline{0} = 0 \cdot 1 = 0$.- For $t = 3$ to $4$ s: $A=1, B=1 \implies Y = 1 \cdot \overline{1} = 1 \cdot 0 = 0$.- For $t = 4$ to $5$ s: $A=1, B=0 \implies Y = 1 \cdot \overline{0} = 1 \cdot 1 = 1$.Thus,the output $Y$ is high $(1)$ from $t = 0$ to $1$ s and from $t = 4$ to $5$ s,and low $(0)$ otherwise. This matches the signal shown in option $D$.

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

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