Boolean Algebra
Sponsor Area
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:
Derive a Canonical POS expression for a Boolean function F, represented by the following truth table:
P | Q | R | F(P, Q, R) |
0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 |
0 | 1 | 0 | 1 |
0 | 1 | 1 | 0 |
1 | 0 | 0 | 0 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 |
Reduce the following Boolean Expression to its simplest form using K-Map:
F(X,Y,Z,W)= (2,6,7,8,9,10,11,13,14,15)
Verify the following using Boolean Laws.
X + Y' = X.Y+ X.Y'+ X'.Y'
Derive a Canonical SOP expression for a Boolean function F, represented by the following truth table :
A | B | C | F(A,B,C) |
0 | 0 | 0 | 1 |
0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 |
0 | 1 | 1 | 1 |
1 | 0 | 0 | 1 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 |
Draw the Logic Circuit for the following Boolean Expression :
(U + V').W' + Z
Sponsor Area
Sponsor Area