What is the minimum number of NAND gates required to implement a 2-input EXCLUSIVE-OR function without using any other logic gate?a)2b)3c)5d)6Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Minimum number of NAND gates required to implement a 2-input EXCLUSIVE-OR function without using any other logic gate is 3.

Explanation:

1. Basic Understanding:
A NAND gate is a universal gate, which means any logic function can be implemented using only NAND gates. The EXCLUSIVE-OR function returns true if the inputs are different, and false if they are the same.

2. Truth Table:
Let's start by creating a truth table for the 2-input EXCLUSIVE-OR function:

| A | B | Output |
|---|---|--------|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |

3. Implementing EXCLUSIVE-OR using NAND gates:
Now, we can use NAND gates to implement the EXCLUSIVE-OR function. We can break this down into multiple steps:

Step 1: Implement the NAND function using only NAND gates. A NAND gate can be represented as (A NAND B) = (A.B)'.

Step 2: Implement the NOT function using only NAND gates. A NOT gate can be represented as (A)' = (A NAND A).

Step 3: Combine the NAND and NOT gates to create the EXCLUSIVE-OR function using the following expression:
Output = ((A NAND (A NAND B)) NAND (B NAND (A NAND B)))

4. Implementation using 3 NAND gates:
Let's now implement the EXCLUSIVE-OR function using 3 NAND gates:

Gate 1: (A NAND B)
Gate 2: (A NAND (A NAND B))
Gate 3: (B NAND (A NAND B))
Output: ((A NAND (A NAND B)) NAND (B NAND (A NAND B)))

5. Conclusion:
Hence, it is evident that the minimum number of NAND gates required to implement a 2-input EXCLUSIVE-OR function without using any other logic gate is 3.

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