Question

Logic symbol

Realization of AND gate using NOR gate.

Answer 1

Hint :In order to solve this question, we are going to apply the NOR gate logic formula for the output at each point for the inputs $ A $ , $ B $ , $ P $ and $ Q $ . For the first NOR gate, two inputs $ A $ and $ A $ enter the OR gate and the negation of the inputs take place, same for the other two gates to get a final output AB.
The logic operation for NOR gate is,
 $ P=\overline{x+y} $ , where $ x $ and $ y $ are the two inputs
According to De Morgan’s laws
 $ \overline{x+y}=\overline{x}\centerdot \overline{y} $
 $ \overline{\overline{x}}=x $ .

Complete Step By Step Answer:
All the three gates that are present here are NOR gates, i.e. the negation of OR gate, which means the negation of the sum of the inputs.
For the first gate, the inputs are $ A $ and $ A $ such that they add and then are negated to give the output $ P $
 $ \begin{align}
  & \overline{A+A}=\overline{A}\centerdot \overline{A} \\
 & \Rightarrow P=\overline{A} \\
\end{align} $
For the second gate, the input is $ B $ which is shorted to get the two inputs $ B $ which are added and then negated from the NOR gate to give the output $ Q $
 $ \begin{align}
  & \overline{B+B}=\overline{B}\centerdot \overline{B} \\
 & \Rightarrow Q=\overline{B} \\
\end{align} $
For the third NOR gate the inputs are $ P $ and $ Q $ , which are equal to $ \overline{A} $ and $ \overline{B} $ respectively, applying logic operation of NOR gate, we get
 $ X=\overline{P+Q}=\overline{\overline{A}+\overline{B}}=\overline{\overline{A}}\centerdot \overline{\overline{B}}=A\centerdot B $
Thus, we can say that the output $ X=AB $ .

Note :
This NOR gate can be used to build the AND gate by using three gates in the order that $ A $ and $ B $ Inputs are put into separate NOR gates and are shorted and then the outputs are put into the third NOR gate which gives the output $ X=AB $ . AND gate can also be obtained from the NAND gate which is a universal gate.