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Sagot :
Let's analyze the given truth table step by step to identify which logic gate it represents:
[tex]\[ \begin{array}{|c|c|c|} \hline A & B & \text{Output} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \hline \end{array} \][/tex]
### Step 1: Evaluate Each Row
1. For [tex]\( A = 0 \)[/tex] and [tex]\( B = 0 \)[/tex], the output is [tex]\( 0 \)[/tex].
2. For [tex]\( A = 0 \)[/tex] and [tex]\( B = 1 \)[/tex], the output is [tex]\( 1 \)[/tex].
3. For [tex]\( A = 1 \)[/tex] and [tex]\( B = 0 \)[/tex], the output is [tex]\( 1 \)[/tex].
4. For [tex]\( A = 1 \)[/tex] and [tex]\( B = 1 \)[/tex], the output is [tex]\( 0 \)[/tex].
### Step 2: Compare With Known Logic Gates
- OR Gate:
[tex]\[ \begin{array}{|c|c|c|} \hline A & B & \text{Output} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \hline \end{array} \][/tex]
The OR gate doesn't match because the output is 1 when both A and B are 1.
- AND Gate:
[tex]\[ \begin{array}{|c|c|c|} \hline A & B & \text{Output} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \hline \end{array} \][/tex]
The AND gate doesn't match because the output is 1 only when both A and B are 1.
- NAND Gate:
[tex]\[ \begin{array}{|c|c|c|} \hline A & B & \text{Output} \\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \hline \end{array} \][/tex]
The NAND gate doesn't match because it produces the opposite outputs where the AND gate is 1.
- NOR Gate:
[tex]\[ \begin{array}{|c|c|c|} \hline A & B & \text{Output} \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \\ \hline \end{array} \][/tex]
The NOR gate doesn't match because it would output a 1 only when both inputs are 0.
- NOT Gate:
[tex]\[ \begin{array}{|c|c|c|} \hline A & & \text{Output} \\ \hline 0 & & 1 \\ 1 & & 0 \\ \hline \end{array} \][/tex]
The NOT gate does not apply here since it only deals with a single input.
- XOR Gate:
[tex]\[ \begin{array}{|c|c|c|} \hline A & B & \text{Output} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \hline \end{array} \][/tex]
The XOR gate matches the given truth table exactly, as it produces a 1 only when the inputs are different.
### Conclusion
The given truth table represents the XOR (exclusive OR) gate. Thus, the correct answer is:
[tex]\[ \boxed{\text{XOR}} \][/tex]
[tex]\[ \begin{array}{|c|c|c|} \hline A & B & \text{Output} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \hline \end{array} \][/tex]
### Step 1: Evaluate Each Row
1. For [tex]\( A = 0 \)[/tex] and [tex]\( B = 0 \)[/tex], the output is [tex]\( 0 \)[/tex].
2. For [tex]\( A = 0 \)[/tex] and [tex]\( B = 1 \)[/tex], the output is [tex]\( 1 \)[/tex].
3. For [tex]\( A = 1 \)[/tex] and [tex]\( B = 0 \)[/tex], the output is [tex]\( 1 \)[/tex].
4. For [tex]\( A = 1 \)[/tex] and [tex]\( B = 1 \)[/tex], the output is [tex]\( 0 \)[/tex].
### Step 2: Compare With Known Logic Gates
- OR Gate:
[tex]\[ \begin{array}{|c|c|c|} \hline A & B & \text{Output} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \hline \end{array} \][/tex]
The OR gate doesn't match because the output is 1 when both A and B are 1.
- AND Gate:
[tex]\[ \begin{array}{|c|c|c|} \hline A & B & \text{Output} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \hline \end{array} \][/tex]
The AND gate doesn't match because the output is 1 only when both A and B are 1.
- NAND Gate:
[tex]\[ \begin{array}{|c|c|c|} \hline A & B & \text{Output} \\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \hline \end{array} \][/tex]
The NAND gate doesn't match because it produces the opposite outputs where the AND gate is 1.
- NOR Gate:
[tex]\[ \begin{array}{|c|c|c|} \hline A & B & \text{Output} \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \\ \hline \end{array} \][/tex]
The NOR gate doesn't match because it would output a 1 only when both inputs are 0.
- NOT Gate:
[tex]\[ \begin{array}{|c|c|c|} \hline A & & \text{Output} \\ \hline 0 & & 1 \\ 1 & & 0 \\ \hline \end{array} \][/tex]
The NOT gate does not apply here since it only deals with a single input.
- XOR Gate:
[tex]\[ \begin{array}{|c|c|c|} \hline A & B & \text{Output} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \hline \end{array} \][/tex]
The XOR gate matches the given truth table exactly, as it produces a 1 only when the inputs are different.
### Conclusion
The given truth table represents the XOR (exclusive OR) gate. Thus, the correct answer is:
[tex]\[ \boxed{\text{XOR}} \][/tex]
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