A | B | Y |
1 | 1 | 1 |
1 | 0 | 0 |
0 | 1 | 1 |
0 | 0 | 1 |
A | B | Y |
1 | 1 | 1 |
1 | 0 | 1 |
0 | 1 | 1 |
0 | 0 | 1 |
A | B | Y |
1 | 1 | 1 |
1 | 0 | 1 |
0 | 1 | 1 |
0 | 0 | 0 |
A | B | Y |
1 | 1 | 1 |
1 | 0 | 1 |
0 | 1 | 0 |
0 | 0 | 1 |
Correct Answer: B
Explanation:
The given truth table represents the output behavior of a logic gate based on the inputs A and B. We need to identify the type of gate based on the given outputs.
The first table shows the following output for inputs A and B:
A | B | Y |
---|---|---|
1 | 1 | 1 |
1 | 0 | 0 |
0 | 1 | 1 |
0 | 0 | 1 |
Upon analyzing the truth table, we can observe that the output is 1 when either A is 1 and B is 0, or when A is 0 and B is 1, which matches the behavior of an Exclusive OR (XOR) gate.
The second table represents the truth table for an OR gate, the third table represents the truth table for an AND gate, and the fourth table represents the truth table for a NAND gate. Therefore, the correct answer is B, which corresponds to the XOR gate.
For the circuit shown above, the equivalent gate is: