The output y of the given logic gate network | vedclass

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

Question

(D) Let the inputs to the two $NAND$ gates be $P$ and $Q$.From the circuit diagram,the upper $NAND$ gate has inputs $\bar A$ and $B$. Its output is $P

Vedclass · Accepted Answer

None

Vedclass · Answer

(D) Let the inputs to the two $NAND$ gates be $P$ and $Q$.From the circuit diagram,the upper $NAND$ gate has inputs $\bar A$ and $B$. Its output is $P = \overline{\bar A \cdot B} = A + \bar B$.The lower $NAND$ gate has inputs $A$ and $\bar B$. Its output is $Q = \overline{A \cdot \bar B} = \bar A + B$.The final gate is a $NOR$ gate,so the output $y = \overline{P + Q}$.Substituting the values of $P$ and $Q$:$y = \overline{(A + \bar B) + (\bar A + B)}$$y = \overline{(A + \bar A) + (B + \bar B)}$Since $A + \bar A = 1$ and $B + \bar B = 1$:$y = \overline{1 + 1} = \overline{1} = 0$.Thus,the output $y = 0$.

The output $y$ of the given logic gate network is:

Similar Questions

In the case of a $NAND$ gate,if $A$ and $B$ are the inputs and $Y$ is the output,then:

Which logic gate is represented by the following symbol?

In the following combination of logic gates,the output $Y$ can be written in terms of inputs $A$ and $B$ as:

The Boolean equation for the circuit given in the figure is

Write the truth table of $NAND$ gate.

Vedclass Test Series

Exam Paper Generator

Online Exam Module