Question

Which of the following correctly represents the truth table of the configuration of gates shown here?

• A. $\begin{array}{ccc}A& B& Y\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 1\end{array}$
• B. $\begin{array}{ccc}A& B& Y\\ 0& 0& 1\\ 0& 1& 0\\ 1& 0& 1\\ 1& 1& 1\end{array}$
• C. $\begin{array}{ccc}A& B& Y\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}$
• D. $\begin{array}{ccc}A& B& Y\\ 0& 0& 1\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}$

Answer 1

The correct option is C $\begin{array}{ccc}A& B& Y\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}$

Both signals A and B go to a NOT gate. Hence their respective outputs will be $\overline{A}$ and $\overline{B}$.

We have two AND gates after the first two NOT gates. The top AND gate has inputs $\overline{A}$ and $B$ and hence the output will be $\overline{A}.B$
Similarly the bottom AND gate has two inputs $\overline{B}$ and $A$ and hence the output will be $A.\overline{B}$

These two output signals finally go through an OR gate. The schematic till the OR gate is shown over here.


Hence the final output is $\overline{A}.B+A.\overline{B}$
Let us use the Boolean algebra and form a table for this output.


Hence option c is correct.