The function F = ((A·C)·(B ⊕ C′))′ is 0 for exactly one combination of the inputs. Click that row of the truth table.
- 1ABC = 000
- 2ABC = 001
- 3ABC = 010
- 4ABC = 011
- 5ABC = 100
- 6ABC = 101
- 7ABC = 110
- 8ABC = 111
The function F = ((A·C)·(B ⊕ C′))′ is 0 for exactly one combination of the inputs. Click that row of the truth table.
For the logic circuit shown, how many of the 8 combinations of the inputs A, B and C give the output F = 1? Find N.
In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.
Simplify F = AB′C′ + B′C + ABC′ + A′B′ to its minimal sum-of-products form. How many literals L does the simplified expression contain?
A positive-edge-triggered D flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.
Find the minimal SOP form of F(A, B, C, D) = Σm(0, 3, 6, 7, 8, 11, 12) + d(4, 10), where d marks the don't-care conditions. How many literals L does it contain?
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