The function F = ((B ⊕ C′) + (A′·B)′)′ is 1 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 = ((B ⊕ C′) + (A′·B)′)′ is 1 for exactly one combination of the inputs. Click that row of the truth table.
The inputs of the logic circuit are A = 1, B = 1 and C = 0. Exactly one of its gates has the output 1. Click that gate.
Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(0, 2, 6). How many sum terms (factors) N does it have?
Simplify F = AB′C + A′BC + A′B′C to its minimal sum-of-products form. How many literals L does the simplified expression contain?
Find the minimal SOP form of F(A, B, C, D) = Σm(6, 9, 12, 13) + d(1, 2, 4, 10), where d marks the don't-care conditions. How many literals L does it contain?
Find the minimal SOP form of F(A, B, C, D) = Σm(4, 5, 6, 8, 9, 13, 14, 15) + d(3, 11), where d marks the don't-care conditions. How many product terms N does it have?
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