Question
Download Solution PDFIdentify the gate from the truth table
Input |
Output |
|
A |
B |
|
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
1 |
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFXNOR Gate:
Symbol:
Truth Table:
Input A |
Input B |
Output \(Y={\overline{A\oplus B}}\) |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
1 |
Output Equation: \(Y={\overline{A\oplus B}}\)
Key Points:
1) If B is always Low, the output is the inverted value of the other input A, i.e. A̅.
2) The output is low when both the inputs are different.
3) The output is high when both the inputs are the same.
4) XNOR gate produces an output only when the two inputs are the same.
XOR GATE
Symbol:
Truth Table:
Input A |
Input B |
Output Y = A ⊕ B |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
Output Equation: \(Y = {\bf{A}} \oplus {\bf{B}} = \bar AB + A\bar B\)
Key Points:
1) If B is always High, the output is the inverted value of the other input A, i.e. A̅.
1) The output is low when both the inputs are the same.
2) The output is high when both the inputs are different.
NOT GATE
Symbol:
Truth Table:
Input (A) |
Output (A̅) |
0 (Low) |
1 (High) |
1 (High) |
0 (low) |
Output Equation: Y = A̅
Key Points: The output of NOT gate is an invert of the input
AND GATE
Symbol:
Truth Table:
Input A |
Input B |
Output Y = A.B |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
1 |
Output Equation: Y = A.B
Key Points: The output is high only when both the inputs are high
OR GATE
Symbol:
Truth Table:
Input A |
Input B |
Output Y = A + B |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
Output Equation: Y = A + B
Key Points: The output is low only when both the inputs are low
NAND GATE
Symbol:
Truth Table:
Input A |
Input B |
Output \(Y = \overline {AB}\) |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
Output Equation: \(Y = \overline {A.B} = \bar A + \bar B\)
Key Points:
1) The output is low only when both the inputs are high
2) It is a universal gate
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