Signal A passes through a NOT gate and then through an AND gate where the second input is permanently set to logic 1. What is the final output expression?
$Y = Ā$
To solve this problem, we need to understand how NOT and AND gates function in digital electronics.
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Understanding the NOT Gate: A NOT gate inverts the input signal. If the input signal is A, the output of the NOT gate would be \overline{A} (the complement of A). This means if A = 1, then \overline{A} = 0, and vice versa.
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Understanding the AND Gate: The AND gate outputs a logic 1 only when all inputs are logic 1. In this scenario, the second input of the AND gate is permanently set to logic 1. The inputs to the AND gate are thus \overline{A} and 1.
Therefore, the output of the AND gate is:
- If \overline{A} = 0, the output is 0 \land 1 = 0.
- If \overline{A} = 1, the output is 1 \land 1 = 1.
This illustrates that the output of the AND gate remains \overline{A}.
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Conclusion: Based on the operation of the NOT and AND gates, the final output expression is \overline{A}.
The correct answer is: Y = \overline{A}.
Similar Questions
A 2-input NOR gate has one input permanently connected to logic LOW (0). How does the output behave with respect to the other input?
Important Questions from Logic Gates
Which logical gate is included in a binary adder circuit to enable binary addition and subtraction?
A Circuit that operates in such a way that its output is high only when all its inputs are high
______ is also known as Inverter.
The minimum number of NAND gates required to reduce the expression ((A + B)C) D is:
In which of the following gates, the output is high if and only if all inputs are high?
