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Boolean Algebra
Question
State Distributive Laws of Boolean Algebra and verify them using truth table.
(i) X. (Y+Z)= X.Y + X.Z
(ii) X + Y.Z= (X + Y). (X+Z)
Solution
(i)
| X | Y | Z | Y+Z | X.(Y+Z) | X.Y | X.Z | X.Y+X.Z |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
(ii)
| X | Y | Z | Y.Z | X+Y.Z | (X+Y) | (X+Z) | (X+Y).(X+Z) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Question
Draw the Logic Circuit of the following Boolean Expression using only NAND Gates:
X.Y + Y.Z
Solution
Question
Derive a Canonical SOP expression for a Boolean function F, represented by the following truth table:
| U | V | W | F(U,V,W) |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
Solution
F(U,V,W)= U’V’W’ + U’VW’ + U’VW + UVW’
Question
Reduce the following Boolean Expression to its simplest form using K-Map:
F(X,Y,Z,W)= Σ (0,1,2,3,4,5,10,11,14)
Solution
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Mock Test Series
Mock Test Series