Let's analyze the table step by step: | X | Y | | OUTPUT | ---|---|---|---------| | 1 | 1 | 1 | 1 | | | 1 | |1 | | 1 | 1 | | | | | | | | 1 | |1 | 1 | | 1 | 1 | 1 | | | | | 1 | | | | 1 | 1 | --- ### 1. Find a function of this table Identify rows where output is 1: - Row 1: X=1, Y=1, Z=1 - Row 2: X=, Y=1, Z= - Row 5: X=1, Y=, Z=1 - Row 6: X=1, Y=1, Z= - Row 7: X=, Y=, Z= - Row 8: X=, Y=, Z=1 Express these as minterms: | Row | Minterm Expression | |-------|-------------------------------------------| | 1 | X·Y·Z | | 2 | ¬X·Y·¬Z | | 5 | X·¬Y·Z | | 6 | X·Y·¬Z | | 7 | ¬X·¬Y·¬Z | | 8 | ¬X·¬Y·Z | ### 2. Derive the sum of minterms: Output = \[ YZ + \; \neg X Y \neg Z + \; X \neg Y Z + \; X Y \neg Z + \; \neg X \neg Y \neg Z + \; \neg X \neg Y Z \] --- ### 3. Simplify using Boolean algebra terms: - \( YZ \) covers rows 1, 2 (when Y=1, Z=1- \( \neg X Y \neg Z \) covers row 2 - \( X \neg Y Z \ covers row 5 - \( X Y \neg Z \) covers row 6 - \( \neg Xneg Y \neg Z \) covers row 7 - \( \neg X \neg Y Z \) covers row 8 Combine: \[ \text{Output} = YZ + \neg X Y \neg Z + X \neg Y Z + X Y \neg Z + \neg X \neg Y \neg Z + \neg X \neg Y Z \] --- ### 4. Simplify further - Note that \( YZ \) already covers \( Y=1, Z=1 \), so \( \neg X Y Z \) is included. - \( \neg X Y \neg Z \) is unique. - \( X \neg Y Z \) is unique. - \( X Y \neg Z \) is unique. - \( \neg X \neg Y \neg Z \) is unique. - \( \neg X \neg Y Z \) is unique. Final simplified form (by combining expressions): \[ \text{Output} = Y Z + \neg X Y \neg Z + X \neg Y Z + X Y \neg Z + \neg X \neg Y \neg Z \] --- ### 5. Final Boolean function: \[ \boxed{ \text{Output} = Y Z + \neg X Y \neg Z + X \neg Y Z + X Y \neg Z + \neg X \neg Y \neg Z } \] --- ### Summary: - **Function:** The output is high when any of these conditions are true. - **Boolean expression:** As above. Would you like me to continue with the next steps (simplification, Karnaugh map, or schematic)?