Name the logic gate which generates high output when at least one input is high.
Answer
647.1k+ views
Hint: We fetch the output of all types of logic gates. Consider the logic gate has two inputs and check which logic gate output suits the given question. As per given in the question, the output is high when at least one input is high. That means if both the inputs are high and output is also high, the condition is still valid.
Complete step by step answer:
The types of logic gates are: AND gate, NAND gate, OR gate, NOR gate, NOT gate, Ex-OR gate and Ex-NOR gate.
Suppose A and B are the two inputs of the logic gate.
The output of AND gate is,
\[Y = A \cdot B\]
Therefore, if any one of the inputs is zero, the output of the AND gate will become zero.
The output of NAND gate is,
\[Y = \overline {A \cdot B} \]
If one of the inputs is high, the term \[A \cdot B\] will become zero and the term \[\overline {A \cdot B} \] will become high. But also, if both the inputs are zero, the output will be high. Therefore, it is not the NAND gate we are asked.
The output of the OR gate is,
\[Y = A + B\]
The truth table of the OR gate is,
Therefore, the OR gate generates high output when at least one input is high and even generates high output when both the inputs are high.
Note: We are asked “at least” one input is high, that means if both inputs are high, the condition is valid. We denote “1” for high and “0” for low to draw the truth table of the logic gates. The NAND gate is valid until both the inputs are low. It gives high output for both low inputs which does not follow the condition given in the question.
Complete step by step answer:
The types of logic gates are: AND gate, NAND gate, OR gate, NOR gate, NOT gate, Ex-OR gate and Ex-NOR gate.
Suppose A and B are the two inputs of the logic gate.
The output of AND gate is,
\[Y = A \cdot B\]
Therefore, if any one of the inputs is zero, the output of the AND gate will become zero.
The output of NAND gate is,
\[Y = \overline {A \cdot B} \]
If one of the inputs is high, the term \[A \cdot B\] will become zero and the term \[\overline {A \cdot B} \] will become high. But also, if both the inputs are zero, the output will be high. Therefore, it is not the NAND gate we are asked.
The output of the OR gate is,
\[Y = A + B\]
The truth table of the OR gate is,
| A | B | \[Y = A + B\] |
| low | low | low |
| low | high | high |
| high | low | high |
| high | high | high |
Therefore, the OR gate generates high output when at least one input is high and even generates high output when both the inputs are high.
Note: We are asked “at least” one input is high, that means if both inputs are high, the condition is valid. We denote “1” for high and “0” for low to draw the truth table of the logic gates. The NAND gate is valid until both the inputs are low. It gives high output for both low inputs which does not follow the condition given in the question.
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