Advertisements
Advertisements
Question
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 | 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 |
Solution
F(P,Q,R)= (P+Q+R')(P+Q'+R)(P'+Q+R')(P'+Q'+R)
shaalaa.com
Obtaining Sum of Product (SOP) and Product of Sum (POS) Form the Truth Table
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Derive a Canonical POS expression for a Boolean function FN, represented by the following truth truth table:
| X | y | z | FN (X, Y, Z) |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
Derive a Canonical POS expression for a Boolean function G, represented by the following truth table:
| X | Y | Z | 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 |
