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Question
Reduce the following Boolean expression using k-map.
| A | B | C | F1 |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 0 |
To SOP & POS form
Short Answer
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Solution
For F1(A, B, C)
From the truth table, F1 = 1 for:
000, 001, 011, 101
Therefore,
F1 = Σm(0, 1, 3, 5)
Minimized SOP form:
F1 = A′B′ + A′C + B′C
For POS, F1 = 0 for:
010, 100, 110, 111
F1 = ΠM(2, 4, 6, 7)
Minimized POS form:
F1 = (A′ + B′)(A′ + C)(B′ + C)
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Chapter 12: Boolean Functions and Reduce Forms - EXERCISE [Page 238]
