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Question
Reduce the following Boolean expression using k-map.
| A | B | C | D | F2 |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |
To SOP & POS Form
Short Answer
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Solution
For F2(A, B, C, D)
From the truth table, F2 = 1 for:
1000, 1010, 1011, 1110, 1111
Therefore,
F2 = Σm(8, 10, 11, 14, 15)
Minimized SOP form:
F2 = AC + AB′D′
For POS, the minimized form is:
F2 = A(C + B′)(C + D′)
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Chapter 12: Boolean Functions and Reduce Forms - EXERCISE [Page 238]
