JEE Physics 2023 Question with Solution

Given: The logical circuit is formed using NAND gates and the input waveforms of $A$ and $B$ are provided.

Find: The correct output waveform $Y$.

Given circuit represent XOR.

For an XOR gate,

$$Y = A \oplus B$$

The output is $1$ when the inputs are different and $0$ when the inputs are same.

Its truth table is

$$\begin{array}{c|c|c} A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}$$

So the waveform must be high exactly in those time intervals where $A \ne B$.

Therefore, the correct option is D.

Given: The circuit uses NAND gates with inputs $A$ and $B$.

Find: Which option gives the output waveform $Y$.

The circuit provided in the image represents an XOR gate. Using NAND-gate construction:

$$\text{NAND1} = \overline{A \cdot B}$$ $$\text{NAND3} = \overline{A \cdot \overline{A \cdot B}}$$ $$\text{NAND4} = \overline{B \cdot \overline{A \cdot B}}$$ $$Y = \overline{\left(\overline{A \cdot \overline{A \cdot B}} \cdot \overline{B \cdot \overline{A \cdot B}}\right)}$$

This simplifies to

$$Y = A \oplus B$$

Now use the XOR truth table:

$$\begin{array}{c|c|c} A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}$$

Hence $Y$ is high whenever the two input waveforms differ and low whenever they are equal.

Therefore, the output waveform matches Option D.