Question

The inputs and outputs for different time intervals are given below the NAND gate.

\[ \begin{array}{|c|c|c|c|} \hline \text{Time} & \text{Inputs A} & \text{Inputs B} & \text{Outputs Y} \\ \hline t_1 \text{ to } t_2 & 0 & 1 & P \\ t_2 \text{ to } t_3 & 0 & 0 & Q \\ t_3 \text{ to } t_4 & 1 & 0 & R \\ t_4 \text{ to } t_5 & 1 & 1 & S \\ \hline \end{array} \]

The values taken by \(P\), \(Q\), \(R\), and \(S\) are respectively:

• A. 1, 1, 1, 0
• B. 0, 1, 0, 1
• C. 0, 1, 1, 0
• D. 1, 0, 1, 1

Hint:

NAND gate truth: output is LOW (0) only when ALL inputs are HIGH (1). For any other input combination, output is HIGH (1).

Answer 1

The Correct Option is A

To solve this problem, we need to understand the operation of a NAND gate. A NAND gate is a digital logic gate that performs the opposite (negation) of the AND operation. The truth table for a NAND gate is as follows: 

Input A	Input B	Output Y (A NAND B)
0	0	1
0	1	1
1	0	1
1	1	0

Using this truth table, we can determine the outputs \( P, Q, R, \) and \( S \) for the given time intervals. Let's evaluate each interval:

1. From \( t_1 \) to \( t_2 \), Inputs A = 0 and B = 1. According to the truth table, the output Y is 1. Hence, \( P = 1 \).
2. From \( t_2 \) to \( t_3 \), Inputs A = 0 and B = 0. The truth table gives us an output of 1. Thus, \( Q = 1 \).
3. From \( t_3 \) to \( t_4 \), Inputs A = 1 and B = 0. The output, according to the truth table, is 1. So, \( R = 1 \).
4. From \( t_4 \) to \( t_5 \), Inputs A = 1 and B = 1. Referring to the truth table, the output is 0. Therefore, \( S = 0 \).

Thus, the respective values for \( P, Q, R, \) and \( S \) are 1, 1, 1, and 0. The correct answer is 1, 1, 1, 0.