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Boolean Algebra
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)
(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 |
Some More Questions From Boolean Algebra Chapter
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)
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)
( 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)
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
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)
F(X,Y,Z,W)= (2,6,7,8,9,10,11,13,14,15)
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