
The truth table given below is for (A and B are inputs, y is the outputs)
A B Y 0 0 1 0 1 1 1 0 1 1 1 0
(A) NOR
(B) AND
(C) XOR
(D) NAND
| A | B | Y |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Answer
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Hint: In this question, we need to determine the type of the gate whose inputs and the output have been given. For this, we will check for all the given options and satisfy all the inputs for the gates.
Complete step by step answer:
In binary operation, 0 can also be represented by a bar on the constant
\[
0 \to \bar P \to False \\
\Rightarrow 1 \to P \to True \\
\]
For the given truth table, write the table in a logical operation,
Case 1:
When both the inputs are 0 and output is 1, we can say both the inputs are false, and output is true
\[\bar A + \bar B = Y\]
This can be written as
\[\overline {A \cdot B} = Y\]
Since both inputs are in the form of the product with a bar, we can say it is a NAND gate.
Case 2:
When one the inputs is 0 and other is 1 also the output is 1, we can say one input is false and other is true also the output is true
\[\bar A + B = Y\]
This can be written as
\[\overline {A \cdot \bar B} = Y\]
Since both inputs are in the form of the product with a bar, we can say it is a NAND gate.
Case 3:
When one the inputs is 1 and other is 0 also the output is 1, we can say one input is true and other is false also the output is true
\[A + \bar B = Y\]
This can be written as
\[\overline {\bar A \cdot B} = Y\]
Since both inputs are in the form of the product with a bar, we can say it is a NAND gate.
Case 4:
When both the inputs are 1 and output is 0, we can say both the inputs are true, and output is false
\[A + B = \bar Y\]
This can be written as
\[\overline {\bar A \cdot \bar B} = Y\]
Since both inputs are in the form of the product with a bar, we can say it is a NAND gate.
Hence we can say the given gate is NAND gate and option D is the correct.
Note: In the given truth table inputs and outputs are given, we know binary numbers are 0 and 1 in binary operation where 0 is to denote if the input is low and 1 if the input is high.A gate is an electronic device which is used for logical operation for two or more inputs, where the output is dependent on the input and output changes when input is changed.
Complete step by step answer:
In binary operation, 0 can also be represented by a bar on the constant
\[
0 \to \bar P \to False \\
\Rightarrow 1 \to P \to True \\
\]
For the given truth table, write the table in a logical operation,
Case 1:
When both the inputs are 0 and output is 1, we can say both the inputs are false, and output is true
\[\bar A + \bar B = Y\]
This can be written as
\[\overline {A \cdot B} = Y\]
Since both inputs are in the form of the product with a bar, we can say it is a NAND gate.
Case 2:
When one the inputs is 0 and other is 1 also the output is 1, we can say one input is false and other is true also the output is true
\[\bar A + B = Y\]
This can be written as
\[\overline {A \cdot \bar B} = Y\]
Since both inputs are in the form of the product with a bar, we can say it is a NAND gate.
Case 3:
When one the inputs is 1 and other is 0 also the output is 1, we can say one input is true and other is false also the output is true
\[A + \bar B = Y\]
This can be written as
\[\overline {\bar A \cdot B} = Y\]
Since both inputs are in the form of the product with a bar, we can say it is a NAND gate.
Case 4:
When both the inputs are 1 and output is 0, we can say both the inputs are true, and output is false
\[A + B = \bar Y\]
This can be written as
\[\overline {\bar A \cdot \bar B} = Y\]
Since both inputs are in the form of the product with a bar, we can say it is a NAND gate.
Hence we can say the given gate is NAND gate and option D is the correct.
Note: In the given truth table inputs and outputs are given, we know binary numbers are 0 and 1 in binary operation where 0 is to denote if the input is low and 1 if the input is high.A gate is an electronic device which is used for logical operation for two or more inputs, where the output is dependent on the input and output changes when input is changed.
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