Boolean Algebra and Logic Circuits — practice questions

Here is a step-by-step simplification with the law used at each stage:

1. Start with the given expression: $F = \overline{(\overline{A} + B)} + (A \cdot \overline{B} \cdot C)$
2. Apply De Morgan's Law to the first term $\overline{(\overline{A} + B)}$: $F = (\overline{\overline{A}} \cdot \overline{B}) + (A \cdot \overline{B} \cdot C)$
3. Apply the Double Negation Law ($\overline{\overline{X}} = X$): $F = (A \cdot \overline{B}) + (A \cdot \overline{B} \cdot C)$
4. Factor out the common term $(A \cdot \overline{B})$ using the Distributive Law: $F = (A \cdot \overline{B}) \cdot (1 + C)$
5. Apply the Annulment Law ($1 + X = 1$): $F = (A \cdot \overline{B}) \cdot 1$
6. Apply the Identity Law ($X \cdot 1 = X$): $F = A \cdot \overline{B}$

Final simplified expression: $F = A \cdot \overline{B}$ (3 marks: 1 for De Morgan's, 1 for factoring, 1 for final simplification).

1. Draw and populate the K-map: We create a 4x4 grid for variables A, B, C, D and place '1's in the cells for the given minterms.

   K-map populated with 1s at positions 0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14.

2. Identify Essential Prime Implicants: We look for '1's that can only be covered by one largest possible group.

   • The '1' at minterm 14 (1110) can be part of a quad group with cells 4, 6, and 12. This group (4, 6, 12, 14) simplifies to $B \cdot \overline{D}$ (since B=1 and D=0 are constant within the group).
   • The '1' at minterm 2 (0010) can be part of a quad group with cells 0, 4, and 6. This group (0, 2, 4, 6) simplifies to $\overline{A} \cdot \overline{D}$ (since A=0 and D=0 are constant).

3. Cover Remaining '1's: After making these two groups, the '1's at 1, 5, 8, 9, 13 are still uncovered. We can cover all of them with a single large group.

   • The octet (group of 8) covering cells (0, 1, 4, 5, 8, 9, 12, 13) simplifies to $\overline{C}$ (since C=0 is the only constant variable across all cells in this group). Note that we re-use some '1's already covered, which is allowed to form a larger group.

4. Combine the terms: The final simplified expression is the sum (OR) of the terms from each group.

Final simplified expression: $F = \overline{C} + B \cdot \overline{D} + \overline{A} \cdot \overline{D}$ (4 marks: 1 for correctly drawn map, 1 for each correct group term).