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Question
Examine the given truth table and then write both SOP & POS boolean expressions describing the output.
| A | B | C | Output |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
Long Answer
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Solution
For the given truth table:
Output = 1 for:
000, 010, 101, 110
SOP Expression
F = A′B′C′ + A′BC′ + AB′C + ABC′
F = Σm(0, 2, 5, 6)
Simplified SOP:
F = A′C′ + BC′ + AB′C
POS Expression
Output = 0 for:
001, 011, 100, 111
F = (A + B + C′)(A + B′ + C′)(A′ + B + C)(A′ + B′ + C′)
F = ΠM(1, 3, 4, 7)
Simplified POS:
F = (A + C′)(B′ + C′)(A′ + B + C)shaalaa.com
Is there an error in this question or solution?
