Question
Draw the Logic Circuit of the following Boolean Expression using only NOR Gates:
( A+B).(C+D)
Solution
Draw the Logic Circuit of the following Boolean Expression using only NOR Gates:
( A+B).(C+D)
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)
Draw the Logic Circuit of the following Boolean Expression using only NAND Gates:
X.Y + Y.Z
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 |
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)
State DeMorgan’s Laws of Boolean Algebra and verify them using truth table.
Draw the Logic Circuit of the following Boolean Expression using only NOR Gates:
( A+B).(C+D)
Derive a Canonical POS expression for a Boolean function G, represented by the following truth table:
| X | Y | X | G(X,Y,Z) |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
Reduce the following Boolean expression to its simplest form using K-Map:
E(U,V,Z,W)= Σ (2,3,6,8,9,10,11,12,13)
Verify the following using Boolean Laws.
A’+ B’.C = A’.B’.C’+ A’.B.C’+ A’.B.C + A’.B’.C+ A.B’.C
Write the Boolean Expression for the result of the Logic Circuit as shown below:
Mock Test Series
