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Question
An X-OR gate has following truth table:
| A | B | Y |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
It is represented by following logic relation `Y = barA.B + A.barB`. Build this gate using AND, OR and NOT gates.
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Solution
X-OR gate can be realized by the combination of two NOT gates, two AND gates and one OR gate. According to the problem, the logic relation for the given truth table is
When `Y = barA * B + A * barB = Y_1 + Y_2`
`Y_1 = A * B` and `Y_2 = A * barB`
Y1 can be obtained as output of AND gate 1 for which one input is of A through NOT gate and another input is of B. Y2 can be obtained as the output of AND pate II for which one input is of A and the other input is of B through NOT gate.
Now Y can be obtained as output from OR gate, where Y1 and Y2 are inputs of OR gate.
Thus, the logic circuit of this relation is given below.
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Useful Constants and Relations:
| 1. | Charge of a proton | (e) | =1.6 × 10-19C |
| 2. | Planck's constant | (h) | = 6·6 × 10-34 Js |
| 3. | Mass of an electron | (m) | = 9·1× 10-31 kg |
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| 5. | `(1/(4pi∈_0))` | =9 ×109 mF-1 | |
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| 7. | `((mu_0)/(4pi))` | =1 × 10-7 Hm-1 | |
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