--- title: "K-map Σ(0,1,4,8,10,11,12)+d, standard minterm" subject: "Digital Logic" chapter: 3 key: "semester_3/digital_logic:kmap-standard-minterm-01248" source: https://www.bctengineer.com/q/semester_3/digital_logic/kmap-standard-minterm-01248 site: BCT Engineering --- # K-map Σ(0,1,4,8,10,11,12)+d, standard minterm Digital Logic · Chapter 3: Combinational Logic Circuits - Asked in: 2068 Chaitra, 3+5 marks - Video: [Don't Care (X) Conditions in K-Map - Karnaugh Map ( Solved Examples ) - Design using NAND Gate only](https://www.youtube.com/watch?v=5OX-4g6kqTw) — CS Engineering Gyan - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/kmap-standard-minterm-01248 - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-3 ## Question Simplify the function using K-map F = Σ(0,1,4,8,10,11,12) and D = Σ(2,3,6,9,15). Also convert the result into standard minterm. ## Short answer Simplified expression: F = B' + C' D'. Expanding each term with missing variables yields the standard minterm list Σm(0,1,2,3,4,8,9,10,11,12), which includes the don’t-cares set to 1. ## Answer Given $F(A,B,C,D) = \Sigma m(0,1,4,8,10,11,12) + d(2,3,6,9,15)$. #### K-map simplification ```text CD AB 00 01 11 10 +---+---+---+---+ 00 | 1 | 1 | X | X | +---+---+---+---+ 01 | 1 | 0 | 0 | X | +---+---+---+---+ 11 | 1 | 0 | X | 0 | +---+---+---+---+ 10 | 1 | X | 1 | 1 | +---+---+---+---+ ``` Groups (don't-cares 2, 3, 9 complete the octet; 6 and 15 are not used): - Octet m(0,1,2,3,8,9,10,11) → $\overline{B}$ - Quad m(0,4,8,12) → $\overline{C}\,\overline{D}$ $$ F = \overline{B} + \overline{C}\,\overline{D} $$ Check: the octet covers 0, 1, 8, 10, 11 and the quad covers 0, 4, 8, 12. #### Conversion of the result into standard minterm form A standard (canonical) minterm form has every variable in every term. Multiply each term by $(X+\overline{X})$ for each missing variable X. **Term $\overline{B}$** (A, C, D missing): $$ \begin{aligned} \overline{B} &= \overline{B}(A+\overline{A})(C+\overline{C})(D+\overline{D}) \\ &= \overline{A}\,\overline{B}\,\overline{C}\,\overline{D} + \overline{A}\,\overline{B}\,\overline{C}D + \overline{A}\,\overline{B}C\overline{D} + \overline{A}\,\overline{B}CD \\ &\quad + A\overline{B}\,\overline{C}\,\overline{D} + A\overline{B}\,\overline{C}D + A\overline{B}C\overline{D} + A\overline{B}CD \\ &= m_0+m_1+m_2+m_3+m_8+m_9+m_{10}+m_{11} \end{aligned} $$ **Term $\overline{C}\,\overline{D}$** (A, B missing): $$ \begin{aligned} \overline{C}\,\overline{D} &= (A+\overline{A})(B+\overline{B})\overline{C}\,\overline{D} \\ &= \overline{A}\,\overline{B}\,\overline{C}\,\overline{D} + \overline{A}B\overline{C}\,\overline{D} + A\overline{B}\,\overline{C}\,\overline{D} + AB\overline{C}\,\overline{D} \\ &= m_0+m_4+m_8+m_{12} \end{aligned} $$ Combine and drop repeated terms ($m_0$ and $m_8$ appear twice; $X+X=X$): $$ F = \Sigma m(0,1,2,3,4,8,9,10,11,12) $$ $$ \begin{aligned} F = {} & \overline{A}\,\overline{B}\,\overline{C}\,\overline{D} + \overline{A}\,\overline{B}\,\overline{C}D + \overline{A}\,\overline{B}C\overline{D} + \overline{A}\,\overline{B}CD + \overline{A}B\overline{C}\,\overline{D} \\ & + A\overline{B}\,\overline{C}\,\overline{D} + A\overline{B}\,\overline{C}D + A\overline{B}C\overline{D} + A\overline{B}CD + AB\overline{C}\,\overline{D} \end{aligned} $$ This contains all 7 original minterms plus the don't-cares 2, 3, 9 that the simplification set to 1 (don't-cares 6 and 15 were taken as 0). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the simplified F after using don't-cares 2, 3, 9? A. B' + C'D' B. B + C'D' C. B'C' + D' D. A' + C'D' Answer: A. Octet m(0,1,2,3,8,9,10,11) = B' and quad m(0,4,8,12) = C'D'. (Key point: Simplified function) 2. How is the term C'D' converted to standard minterms? A. Multiply by (A+A')(B+B') to get 4 minterms B. Multiply by (C+C')(D+D') C. Leave it as is D. Add A and B Answer: A. A and B are missing, so C'D' becomes m0 + m4 + m8 + m12. (Key point: Converting to standard minterms) 3. Why are minterms 0 and 8 written only once in the final answer? A. Because X + X = X B. Because they are don't-cares C. Because X · X = 0 D. Because they cancel out Answer: A. Both B' and C'D' contain m0 and m8; in OR, a repeated term is counted once. (Key point: Repeated minterms) 4. What is the standard minterm form of the simplified F? A. Σm(0,1,4,8,12) B. Σm(0,1,2,3,4,8,9,10,11,12) C. Σm(0,1,2,3,6,8,9,15) D. Σm(0,1,2,3,4,8,9,10,11,12,15) Answer: B. It contains the 7 original minterms plus don't-cares 2, 3, 9 set to 1. 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