Discussion
A combinational circuit has inputs A, B and C and its K-map is as shown, the output of the circuit will be
| C \ AB | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | 0 | 1 | 0 | 1 |
| 1 | 1 | 0 | 1 | 0 |
Answer: D
The output is the three-variable exclusive OR, \(A \oplus B \oplus C\) (option D). Reading the 1 cells: for C = 0 they are at AB = 01 and 10, and for C = 1 they are at AB = 00 and 11.
\[ f = (\overline{A}B + A\overline{B})\overline{C} \]
\[ \quad + (\overline{A}\,\overline{B} + AB)C \]
\[ f = (A \oplus B)\overline{C} + \overline{(A \oplus B)}\,C \]
\[ f = A \oplus B \oplus C \]
Option A keeps only the C = 1 term with the wrong pairs, option B has only the C = 0 term with the wrong pairs, and option C is the single minterm 000, which is 0 in the map.
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