The output (Y) of the given logic gate is sim | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the inputs to the first $NAND$ gate be $A$ and $A$. The output is $Y_1 = \overline{A \cdot A} = \bar{A}$.Let the inputs to the $NOR$ gate be $

Vedclass · Accepted Answer

$AND$ gate

Vedclass · Answer

(C) Let the inputs to the first $NAND$ gate be $A$ and $A$. The output is $Y_1 = \overline{A \cdot A} = \bar{A}$.Let the inputs to the $NOR$ gate be $B$ and $B$. The output is $Y_2 = \overline{B+B} = \bar{B}$.The final gate is a $NOR$ gate with inputs $Y_1$ and $Y_2$. The output $Y$ is given by:$Y = \overline{Y_1 + Y_2}$Substituting the values of $Y_1$ and $Y_2$:$Y = \overline{\bar{A} + \bar{B}}$Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.Thus,the output $Y = A \cdot B$,which is the output of an $AND$ gate.

The output $(Y)$ of the given logic gate is similar to the output of an/a

Similar Questions

Which logic gate with inputs $A$ and $B$ performs the same operation as that performed by the following circuit?

For the given combination of logic gates,the inputs $A, B$ and $C$ are as follows. If $A=B=C=0$ and $A=B=1, C=0$,then the logic states of output $D$ are respectively:

In a $\text{NAND}$ gate, $A$ and $B$ are inputs and $Y$ is the output, then the correct option is

From the circuit of the following logic gates,the basic logic gate obtained is

The truth table for the given logic circuit is: (Inputs $A, B$ and Output $E$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module